From 0e59757765c0e3e09e35843ece7755206fa1dd07 Mon Sep 17 00:00:00 2001 From: mhjensen Date: Mon, 3 Sep 2018 16:27:52 +0200 Subject: [PATCH 01/12] Added exercise set 2 --- doc/Projects/2018/hw2/html/hw2-bs.html | 218 ++++++++++++++++ doc/Projects/2018/hw2/html/hw2.html | 167 +++++++++++++ .../2018/hw2/ipynb/ipynb-hw2-src.tar.gz | Bin 0 -> 206 bytes doc/Projects/2018/hw2/pdf/hw2.p.tex | 235 ++++++++++++++++++ doc/Projects/2018/hw2/pdf/hw2.pdf | Bin 0 -> 198044 bytes doc/Projects/2018/hw2/pdf/hw2.tex | 203 +++++++++++++++ doc/src/Projects/2018/Exercises/hw2.do.txt | 63 ++++- doc/web/course.do.txt | 8 + doc/web/course.html | 68 +++-- 9 files changed, 933 insertions(+), 29 deletions(-) create mode 100644 doc/Projects/2018/hw2/html/hw2-bs.html create mode 100644 doc/Projects/2018/hw2/html/hw2.html create mode 100644 doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz create mode 100644 doc/Projects/2018/hw2/pdf/hw2.p.tex create mode 100644 doc/Projects/2018/hw2/pdf/hw2.pdf create mode 100644 doc/Projects/2018/hw2/pdf/hw2.tex diff --git a/doc/Projects/2018/hw2/html/hw2-bs.html b/doc/Projects/2018/hw2/html/hw2-bs.html new file mode 100644 index 000000000..f97cff804 --- /dev/null +++ b/doc/Projects/2018/hw2/html/hw2-bs.html @@ -0,0 +1,218 @@ + + + + + + + +Homework 2 + + + + + + + + + + + + + + + + + + + + + + + +
+
+ + Homework 2 +
+ +
+ +
+
+ + +
+ +

 

 

 

+ + + + + + +
+

Homework 2

+ +

+ + +

+Data Analysis and Machine Learning FYS-STK3155/FYS4155 +
+ +

+ + +

Department of Physics, University of Oslo, Norway
+
+

+

Sep 3, 2018

+
+

+

+ +

Exercise 4

+ +

+This exercise is a continuation of exercise 2 from homework 1. We will +use the same function to generate our data set, still staying with a +simple function \( y(x) \) which we want to fit using linear regression, +but now extending the analysis to include the Ridge and the Lasso +regression methods. You can use the code under the Regression as an example on how to use the Ridge and the Lasso methods, see the regression slides). + +

+We will thus again generate our own dataset for a function \( y(x) \) where +\( x \in [0,1] \) and defined by random numbers computed with the uniform +distribution. The function \( y \) is a quadratic polynomial in \( x \) with +added stochastic noise according to the normal distribution \( \cal{N}(0,1) \). + +

+The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points). +

+ + +

x = np.random.rand(100,1)
+y = 5*x*x+0.1*np.random.randn(100,1)
+
+
    +
  1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \).
    2. +
+ +Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \). + +
    +
  1. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
    2. +
  2. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression.
    3. +
+ +You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11. + +
    +
  1. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    2. +
  2. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    3. +
+ +$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +$$ + +and the \( R^2 \) score function. +If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as +$$ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +$$ + +where we have defined the mean value of \( \hat{y} \) as +$$ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +$$ + +Discuss these quantities as functions of the variable \( \lambda \) in the Ridge and Lasso regression methods. + +

Exercise 5

+ +

+Using the singular value decomposition, show that the variance of the direction vector +\( \hat{z}_i=\hat{X}\hat{v}_i \) is equal to (equation (3.49) of Hastie et al.) +$$ +\mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, +$$ + +where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. + +

+ +

+ + +
+ + + + + + + +
+ © 1999-2018, "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html". Released under CC Attribution-NonCommercial 4.0 license +
+ + + + + + diff --git a/doc/Projects/2018/hw2/html/hw2.html b/doc/Projects/2018/hw2/html/hw2.html new file mode 100644 index 000000000..3e2d44270 --- /dev/null +++ b/doc/Projects/2018/hw2/html/hw2.html @@ -0,0 +1,167 @@ + + + + + + + +Homework 2 + + + + + + + + + + + + + + + + + + + + + + + +

Homework 2

+ +

+ + +

+Data Analysis and Machine Learning FYS-STK3155/FYS4155 +
+ +

+ + +

Department of Physics, University of Oslo, Norway
+
+

+

Sep 3, 2018

+
+ +

Exercise 4

+ +

+This exercise is a continuation of exercise 2 from homework 1. We will +use the same function to generate our data set, still staying with a +simple function \( y(x) \) which we want to fit using linear regression, +but now extending the analysis to include the Ridge and the Lasso +regression methods. You can use the code under the Regression as an example on how to use the Ridge and the Lasso methods, see the regression slides). + +

+We will thus again generate our own dataset for a function \( y(x) \) where +\( x \in [0,1] \) and defined by random numbers computed with the uniform +distribution. The function \( y \) is a quadratic polynomial in \( x \) with +added stochastic noise according to the normal distribution \( \cal{N}(0,1) \). + +

+The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points). +

+ + +

x = np.random.rand(100,1)
+y = 5*x*x+0.1*np.random.randn(100,1)
+
+
    +
  1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \).
    2. +
+ +Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \). + +
    +
  1. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
    2. +
  2. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression.
    3. +
+ +You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11. + +
    +
  1. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    2. +
  2. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    3. +
+ +$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +$$ + +and the \( R^2 \) score function. +If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as +$$ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +$$ + +where we have defined the mean value of \( \hat{y} \) as +$$ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +$$ + +Discuss these quantities as functions of the variable \( \lambda \) in the Ridge and Lasso regression methods. + +

Exercise 5

+ +

+Using the singular value decomposition, show that the variance of the direction vector +\( \hat{z}_i=\hat{X}\hat{v}_i \) is equal to (equation (3.49) of Hastie et al.) +$$ +\mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, +$$ + +where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. + + + + +

+ © 1999-2018, "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html". Released under CC Attribution-NonCommercial 4.0 license +
+ + + + + + diff --git a/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz b/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz new file mode 100644 index 0000000000000000000000000000000000000000..4b1da116e789a1063214110a9cbf73f4e1523659 GIT binary patch literal 206 zcmV;<05Sg`iwFRKM2%Yj1MSbjY63A3#&OR+#XP~1&8B)T=%I&#h%XS?*p01;CK27Y z@2b$A>`hAP|9APAVHjqX^XJ*?yej81dKjhD#2C5cbnaFt*wJu;=!QeB~dp>U@@$MPIrYB*fy@F}_Q8(v{!$ zVJ&TR`O=r<4o9Dy*wBk@TVH$(_G7$H@h#b1uExLp2><{90000000000008_ipKT2s ITmUEl08uJxEdT%j literal 0 HcmV?d00001 diff --git a/doc/Projects/2018/hw2/pdf/hw2.p.tex b/doc/Projects/2018/hw2/pdf/hw2.p.tex new file mode 100644 index 000000000..7824027f3 --- /dev/null +++ b/doc/Projects/2018/hw2/pdf/hw2.p.tex @@ -0,0 +1,235 @@ +%% +%% Automatically generated file from DocOnce source +%% (https://github.com/hplgit/doconce/) +%% +%% +% #ifdef PTEX2TEX_EXPLANATION +%% +%% The file follows the ptex2tex extended LaTeX format, see +%% ptex2tex: http://code.google.com/p/ptex2tex/ +%% +%% Run +%% ptex2tex myfile +%% or +%% doconce ptex2tex myfile +%% +%% to turn myfile.p.tex into an ordinary LaTeX file myfile.tex. +%% (The ptex2tex program: http://code.google.com/p/ptex2tex) +%% Many preprocess options can be added to ptex2tex or doconce ptex2tex +%% +%% ptex2tex -DMINTED myfile +%% doconce ptex2tex myfile envir=minted +%% +%% ptex2tex will typeset code environments according to a global or local +%% .ptex2tex.cfg configure file. doconce ptex2tex will typeset code +%% according to options on the command line (just type doconce ptex2tex to +%% see examples). If doconce ptex2tex has envir=minted, it enables the +%% minted style without needing -DMINTED. +% #endif + +% #define PREAMBLE + +% #ifdef PREAMBLE +%-------------------- begin preamble ---------------------- + +\documentclass[% +oneside, % oneside: electronic viewing, twoside: printing +final, % draft: marks overfull hboxes, figures with paths +10pt]{article} + +\listfiles % print all files needed to compile this document + +\usepackage{relsize,makeidx,color,setspace,amsmath,amsfonts,amssymb} +\usepackage[table]{xcolor} +\usepackage{bm,ltablex,microtype} + +\usepackage[pdftex]{graphicx} + +\usepackage{ptex2tex} +% #ifdef MINTED +\usepackage{minted} +\usemintedstyle{default} +% #endif + +\usepackage[T1]{fontenc} +%\usepackage[latin1]{inputenc} +\usepackage{ucs} +\usepackage[utf8x]{inputenc} + +\usepackage{lmodern} % Latin Modern fonts derived from Computer Modern + +% Hyperlinks in PDF: +\definecolor{linkcolor}{rgb}{0,0,0.4} +\usepackage{hyperref} +\hypersetup{ + breaklinks=true, + colorlinks=true, + linkcolor=linkcolor, + urlcolor=linkcolor, + citecolor=black, + filecolor=black, + %filecolor=blue, + pdfmenubar=true, + pdftoolbar=true, + bookmarksdepth=3 % Uncomment (and tweak) for PDF bookmarks with more levels than the TOC + } +%\hyperbaseurl{} % hyperlinks are relative to this root + +\setcounter{tocdepth}{2} % levels in table of contents + +% --- fancyhdr package for fancy headers --- +\usepackage{fancyhdr} +\fancyhf{} % sets both header and footer to nothing +\renewcommand{\headrulewidth}{0pt} +\fancyfoot[LE,RO]{\thepage} +% Ensure copyright on titlepage (article style) and chapter pages (book style) +\fancypagestyle{plain}{ + \fancyhf{} + \fancyfoot[C]{{\footnotesize \copyright\ 1999-2018, "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html". Released under CC Attribution-NonCommercial 4.0 license}} +% \renewcommand{\footrulewidth}{0mm} + \renewcommand{\headrulewidth}{0mm} +} +% Ensure copyright on titlepages with \thispagestyle{empty} +\fancypagestyle{empty}{ + \fancyhf{} + \fancyfoot[C]{{\footnotesize \copyright\ 1999-2018, "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html". Released under CC Attribution-NonCommercial 4.0 license}} + \renewcommand{\footrulewidth}{0mm} + \renewcommand{\headrulewidth}{0mm} +} + +\pagestyle{fancy} + + +% prevent orhpans and widows +\clubpenalty = 10000 +\widowpenalty = 10000 + +% --- end of standard preamble for documents --- + + +% insert custom LaTeX commands... + +\raggedbottom +\makeindex +\usepackage[totoc]{idxlayout} % for index in the toc +\usepackage[nottoc]{tocbibind} % for references/bibliography in the toc + +%-------------------- end preamble ---------------------- + +\begin{document} + +% matching end for #ifdef PREAMBLE +% #endif + +\newcommand{\exercisesection}[1]{\subsection*{#1}} + + +% ------------------- main content ---------------------- + + + +% ----------------- title ------------------------- + +\thispagestyle{empty} + +\begin{center} +{\LARGE\bf +\begin{spacing}{1.25} +Homework 2 +\end{spacing} +} +\end{center} + +% ----------------- author(s) ------------------------- + +\begin{center} +{\bf \href{{http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html}}{Data Analysis and Machine Learning FYS-STK3155/FYS4155}} +\end{center} + + \begin{center} +% List of all institutions: +\centerline{{\small Department of Physics, University of Oslo, Norway}} +\end{center} + +% ----------------- end author(s) ------------------------- + +% --- begin date --- +\begin{center} +Sep 3, 2018 +\end{center} +% --- end date --- + +\vspace{1cm} + + +\subsection{Exercise 4} + +This exercise is a continuation of exercise 2 from homework 1. We will +use the same function to generate our data set, still staying with a +simple function $y(x)$ which we want to fit using linear regression, +but now extending the analysis to include the Ridge and the Lasso +regression methods. You can use the code under the Regression as an example on how to use the Ridge and the Lasso methods, see the \href{{https://compphysics.github.io/MachineLearning/doc/pub/Regression/html/Regression-bs.html}}{regression slides}). + +We will thus again generate our own dataset for a function $y(x)$ where +$x \in [0,1]$ and defined by random numbers computed with the uniform +distribution. The function $y$ is a quadratic polynomial in $x$ with +added stochastic noise according to the normal distribution $\cal{N}(0,1)$. + +The following simple Python instructions define our $x$ and $y$ values (with 100 data points). +\bpycod +x = np.random.rand(100,1) +y = 5*x*x+0.1*np.random.randn(100,1) +\epycod + +\begin{enumerate} +\item Write your own code for the Ridge method (see chapter 3.4 of Hastie \emph{et al.}, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. +\end{enumerate} + +\noindent +Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. + +\begin{enumerate} +\item Repeat the above but using the functionality of \textbf{scikit-learn}. Compare your code with the results from \textbf{scikit-learn}. Remember to run with the same random numbers for generating $x$ and $y$. + +\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. +\end{enumerate} + +\noindent +You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. + +\begin{enumerate} +\item Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie \emph{et al.}. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of \textbf{scikit-learn}. + +\item Finally, using \textbf{scikit-learn} or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as +\end{enumerate} + +\noindent +\[ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +\] +and the $R^2$ score function. +If $\tilde{\hat{y}}_i$ is the predicted value of the $i-th$ sample and $y_i$ is the corresponding true value, then the score $R^2$ is defined as +\[ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +\] +where we have defined the mean value of $\hat{y}$ as +\[ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +\] +Discuss these quantities as functions of the variable $\lambda$ in the Ridge and Lasso regression methods. + +\subsection{Exercise 5} + +Using the singular value decomposition, show that the variance of the direction vector +$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie \emph{et al.}) +\[ +\mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, +\] +where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. + +% ------------------- end of main content --------------- + +% #ifdef PREAMBLE +\end{document} +% #endif + diff --git a/doc/Projects/2018/hw2/pdf/hw2.pdf b/doc/Projects/2018/hw2/pdf/hw2.pdf new file mode 100644 index 0000000000000000000000000000000000000000..d5d7c1f38df71f3ee7d4e5342a627c79b3246525 GIT binary patch literal 198044 zcma&NL$oLgtYy1x+qP}nwr%5V+qP}nwr$(CRk!~5uk|{$I!U_eWF~8bR6#_HmXVGX zigbQ?WF3l~fPui?$O?*w2Z~26mrgYL48&dC=dbxLDH!;k^A#nw-Ne)NFPIXeFIdzJYQ$Fc5 z3^fufA!S^$exH7kAY8$Mgtj&pLLC@Z%xxXMl(}1z%j-|o^WYmkdtL2bg$tQ;K{0Dn z)>cNNRJqIylXZyV$h8^kSMqoLPWxC#>O||0aqoPG?DF?pmS65VI=`VAt!u}}fftw1 ztdyb|RS_~8(+Y>&99k)gmKNXPR%KniYeABIrj^h6WvnMzMJ6izUzgTwi`4Da=tF;8 ztyNZM$I)(=K0nXy9@ zv%2Y`v2{_>4XvMFZ<%0y>vgUBF~96j$f(ZNN>5%irSIjccApCAUE9r(4yxqHl$X|`mx&GEE}+A-E#>i``4+i!tOW_wxBJhmA1FHIYQMeeh! 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Released under CC Attribution-NonCommercial 4.0 license}} +% \renewcommand{\footrulewidth}{0mm} + \renewcommand{\headrulewidth}{0mm} +} +% Ensure copyright on titlepages with \thispagestyle{empty} +\fancypagestyle{empty}{ + \fancyhf{} + \fancyfoot[C]{{\footnotesize \copyright\ 1999-2018, "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html". Released under CC Attribution-NonCommercial 4.0 license}} + \renewcommand{\footrulewidth}{0mm} + \renewcommand{\headrulewidth}{0mm} +} + +\pagestyle{fancy} + + +% prevent orhpans and widows +\clubpenalty = 10000 +\widowpenalty = 10000 + +% --- end of standard preamble for documents --- + + +% insert custom LaTeX commands... + +\raggedbottom +\makeindex +\usepackage[totoc]{idxlayout} % for index in the toc +\usepackage[nottoc]{tocbibind} % for references/bibliography in the toc + +%-------------------- end preamble ---------------------- + +\begin{document} + +% matching end for #ifdef PREAMBLE + +\newcommand{\exercisesection}[1]{\subsection*{#1}} + + +% ------------------- main content ---------------------- + + + +% ----------------- title ------------------------- + +\thispagestyle{empty} + +\begin{center} +{\LARGE\bf +\begin{spacing}{1.25} +Homework 2 +\end{spacing} +} +\end{center} + +% ----------------- author(s) ------------------------- + +\begin{center} +{\bf \href{{http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html}}{Data Analysis and Machine Learning FYS-STK3155/FYS4155}} +\end{center} + + \begin{center} +% List of all institutions: +\centerline{{\small Department of Physics, University of Oslo, Norway}} +\end{center} + +% ----------------- end author(s) ------------------------- + +% --- begin date --- +\begin{center} +Sep 3, 2018 +\end{center} +% --- end date --- + +\vspace{1cm} + + +\subsection*{Exercise 4} + +This exercise is a continuation of exercise 2 from homework 1. We will +use the same function to generate our data set, still staying with a +simple function $y(x)$ which we want to fit using linear regression, +but now extending the analysis to include the Ridge and the Lasso +regression methods. You can use the code under the Regression as an example on how to use the Ridge and the Lasso methods, see the \href{{https://compphysics.github.io/MachineLearning/doc/pub/Regression/html/Regression-bs.html}}{regression slides}). + +We will thus again generate our own dataset for a function $y(x)$ where +$x \in [0,1]$ and defined by random numbers computed with the uniform +distribution. The function $y$ is a quadratic polynomial in $x$ with +added stochastic noise according to the normal distribution $\cal{N}(0,1)$. + +The following simple Python instructions define our $x$ and $y$ values (with 100 data points). +\begin{print} +x = np.random.rand(100,1) +y = 5*x*x+0.1*np.random.randn(100,1) +\end{print} + +\begin{enumerate} +\item Write your own code for the Ridge method (see chapter 3.4 of Hastie \emph{et al.}, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. +\end{enumerate} + +\noindent +Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. + +\begin{enumerate} +\item Repeat the above but using the functionality of \textbf{scikit-learn}. Compare your code with the results from \textbf{scikit-learn}. Remember to run with the same random numbers for generating $x$ and $y$. + +\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. +\end{enumerate} + +\noindent +You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. + +\begin{enumerate} +\item Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie \emph{et al.}. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of \textbf{scikit-learn}. + +\item Finally, using \textbf{scikit-learn} or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as +\end{enumerate} + +\noindent +\[ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +\] +and the $R^2$ score function. +If $\tilde{\hat{y}}_i$ is the predicted value of the $i-th$ sample and $y_i$ is the corresponding true value, then the score $R^2$ is defined as +\[ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +\] +where we have defined the mean value of $\hat{y}$ as +\[ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +\] +Discuss these quantities as functions of the variable $\lambda$ in the Ridge and Lasso regression methods. + +\subsection*{Exercise 5} + +Using the singular value decomposition, show that the variance of the direction vector +$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie \emph{et al.}) +\[ +\mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, +\] +where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. + +% ------------------- end of main content --------------- + +\end{document} + diff --git a/doc/src/Projects/2018/Exercises/hw2.do.txt b/doc/src/Projects/2018/Exercises/hw2.do.txt index bd2355fab..48412eeb1 100644 --- a/doc/src/Projects/2018/Exercises/hw2.do.txt +++ b/doc/src/Projects/2018/Exercises/hw2.do.txt @@ -1,12 +1,65 @@ -TITLE: Homework 1 +TITLE: Homework 2 AUTHOR: "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio.no/studier/emner/matnat/fys/FYS3155/index-eng.html" {copyright, 1999-present|CC BY-NC} at Department of Physics, University of Oslo, Norway DATE:Today -===== Exercise 1 ===== +===== Exercise 4 ===== -Make sure you have installed all necessary t +This exercise is a continuation of exercise 2 from homework 1. We will +use the same function to generate our data set, still staying with a +simple function $y(x)$ which we want to fit using linear regression, +but now extending the analysis to include the Ridge and the Lasso +regression methods. You can use the code under the Regression as an example on how to use the Ridge and the Lasso methods, see the "regression slides":"https://compphysics.github.io/MachineLearning/doc/pub/Regression/html/Regression-bs.html"). -===== Exercise 2 ===== +We will thus again generate our own dataset for a function $y(x)$ where +$x \in [0,1]$ and defined by random numbers computed with the uniform +distribution. The function $y$ is a quadratic polynomial in $x$ with +added stochastic noise according to the normal distribution $\cal{N}(0,1)$. -===== Exercise 3 ===== +The following simple Python instructions define our $x$ and $y$ values (with 100 data points). +!bc pycod +x = np.random.rand(100,1) +y = 5*x*x+0.1*np.random.randn(100,1) +!ec + +o Write your own code for the Ridge method (see chapter 3.4 of Hastie *et al.*, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. +Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. + +o Repeat the above but using the functionality of _scikit-learn_. Compare your code with the results from _scikit-learn_. Remember to run with the same random numbers for generating $x$ and $y$. + +o Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. +You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of _scikit-learn_ and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie *et al.* and their figure 3.11. + +o Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie *et al.*. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of _scikit-learn_. + +o Finally, using _scikit-learn_ or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as +!bt +\[ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +\] +!et +and the $R^2$ score function. +If $\tilde{\hat{y}}_i$ is the predicted value of the $i-th$ sample and $y_i$ is the corresponding true value, then the score $R^2$ is defined as +!bt +\[ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +\] +!et +where we have defined the mean value of $\hat{y}$ as +!bt +\[ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +\] +!et +Discuss these quantities as functions of the variable $\lambda$ in the Ridge and Lasso regression methods. + +===== Exercise 5 ===== + +Using the singular value decomposition, show that the variance of the direction vector +$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie *et al.*) +!bt +\[ +\mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, +\] +!et +where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. diff --git a/doc/web/course.do.txt b/doc/web/course.do.txt index d98c3a3ac..68763e1ff 100644 --- a/doc/web/course.do.txt +++ b/doc/web/course.do.txt @@ -96,6 +96,14 @@ ${text_types(ch)} * "Plain html":"http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw1/html/hw1.html" * "Bootstrap slide style, easy for reading on mobile devices": "http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw1/html/hw1-bs.html" +=== Second homework set, week 36 === + * LaTeX and PDF: + * "LaTex file":"http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw2/pdf/hw2.tex" + * "PDF file":"http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw2/pdf/hw2.pdf" + * HTML: + * "Plain html":"http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw2/html/hw2.html" + * "Bootstrap slide style, easy for reading on mobile devices": "http://compphysics.github.io/MachineLearning/doc/Projects/2018/hw2/html/hw2-bs.html" + === Project 1, Deadline October 1 === diff --git a/doc/web/course.html b/doc/web/course.html index f17cafa09..b207da6d6 100644 --- a/doc/web/course.html +++ b/doc/web/course.html @@ -99,25 +99,26 @@ div { text-align: justify; text-justify: inter-word; } ('Teach yourself C++', 2, None, '___sec13'), ('Projects and Exercises Fall 2018', 2, None, '___sec14'), ('First homework set, week 35', 3, None, '___sec15'), - ('Project 1, Deadline October 1', 3, None, '___sec16'), - ('Project 2, Deadline November 5', 3, None, '___sec17'), - ('Project 3, Deadline November 30', 3, None, '___sec18'), - ('Course content', 3, None, '___sec19'), - ('Learning outcomes', 2, None, '___sec20'), - ('Prerequisites', 2, None, '___sec21'), - ('The course has two central parts', 2, None, '___sec22'), + ('Second homework set, week 36', 3, None, '___sec16'), + ('Project 1, Deadline October 1', 3, None, '___sec17'), + ('Project 2, Deadline November 5', 3, None, '___sec18'), + ('Project 3, Deadline November 30', 3, None, '___sec19'), + ('Course content', 3, None, '___sec20'), + ('Learning outcomes', 2, None, '___sec21'), + ('Prerequisites', 2, None, '___sec22'), + ('The course has two central parts', 2, None, '___sec23'), ('Statistical analysis and optimization of data', 3, None, - '___sec23'), - ('Machine learning', 3, None, '___sec24'), - ('Recommended textbooks', 2, None, '___sec25'), + '___sec24'), + ('Machine learning', 3, None, '___sec25'), + ('Recommended textbooks', 2, None, '___sec26'), ('"Other ' 'textbooks":"https://github.com/CompPhysics/MachineLearning/tree/master/doc/Textbooks"', 2, None, - '___sec26'), - ('Teaching schedule Fall 2018', 2, None, '___sec27')]} + '___sec27'), + ('Teaching schedule Fall 2018', 2, None, '___sec28')]} end of tocinfo --> @@ -577,7 +578,26 @@ formulas in HTML or ipython notebook files. -

Project 1, Deadline October 1

+

Second homework set, week 36

+ + + +

Project 1, Deadline October 1

  • LaTeX and PDF:
    • @@ -596,7 +616,7 @@ formulas in HTML or ipython notebook files.
-

Project 2, Deadline November 5

+

Project 2, Deadline November 5

  • LaTeX and PDF:
    • @@ -615,7 +635,7 @@ formulas in HTML or ipython notebook files.
-

Project 3, Deadline November 30

+

Project 3, Deadline November 30

  • LaTeX and PDF:
    • @@ -634,7 +654,7 @@ formulas in HTML or ipython notebook files.
-

Course content

+

Course content

Probability theory and statistical methods play a central role in science. Nowadays we are @@ -653,7 +673,7 @@ tools of probability theory, the aim of this course is to expose you to central This course covers thus topics like Monte Carlo methods and Markov chains, Bayesian statistics, error estimates, various linear methods, optimization of data and error analysis and central algorithms in machine learning. The course has several numerical projects and numerical exercises that are meant to illustrate the theory. -

Learning outcomes

+

Learning outcomes

The course introduces a variety of central algorithms and methods @@ -670,19 +690,19 @@ essential for studies of data analysis and machine learning. The course is proje

  • Work on numerical projects to illustrate the theory. The projects play a central role and students are expected to know modern programming languages like Python or C++.
    • -

    Prerequisites

    +

    Prerequisites

    Basic knowledge in programming and numerics. Required courses are the equivalents to the University of Oslo mathematics courses MAT1100, MAT1110, MAT1120 and at least one of the corresponding computing and programming courses INF1000/INF1110 or MAT-INF1100/MAT-INF1100L/BIOS1100/KJM-INF1xxx. -

    The course has two central parts

    +

    The course has two central parts

    1. Statistical analysis and optimization of data
    2. Machine learning
    -

    Statistical analysis and optimization of data

    +

    Statistical analysis and optimization of data

    The following topics will be covered @@ -699,7 +719,7 @@ The following topics will be covered

  • Practical optimization using Singular-value decomposition and least squares for parameterizing data.
    • -

    Machine learning

    +

    Machine learning

    The following topics will be covered @@ -715,14 +735,14 @@ The following topics will be covered All the above topics will be supported by examples, hands-on exercises and project work. -

    Recommended textbooks

    +

    Recommended textbooks

    -

    Other textbooks

    +

    Other textbooks

    General learning book on statistical analysis: @@ -744,7 +764,7 @@ All the above topics will be supported by examples, hands-on exercises and proje -

    Teaching schedule Fall 2018

    +

    Teaching schedule Fall 2018

    Acronyms for textbooks and references to chapter From 6470c2d5f4e85713d530934c9a032e7f16bb5e75 Mon Sep 17 00:00:00 2001 From: mhjensen Date: Mon, 3 Sep 2018 16:35:05 +0200 Subject: [PATCH 02/12] fixing typos as usual --- doc/Projects/2018/hw2/html/hw2-bs.html | 9 ++------- doc/Projects/2018/hw2/html/hw2.html | 9 ++------- .../2018/hw2/ipynb/ipynb-hw2-src.tar.gz | Bin 206 -> 207 bytes doc/Projects/2018/hw2/pdf/hw2.p.tex | 9 ++------- doc/Projects/2018/hw2/pdf/hw2.pdf | Bin 198044 -> 198064 bytes doc/Projects/2018/hw2/pdf/hw2.tex | 9 ++------- doc/src/Projects/2018/Exercises/hw2.do.txt | 5 ++--- 7 files changed, 10 insertions(+), 31 deletions(-) diff --git a/doc/Projects/2018/hw2/html/hw2-bs.html b/doc/Projects/2018/hw2/html/hw2-bs.html index f97cff804..367aaa066 100644 --- a/doc/Projects/2018/hw2/html/hw2-bs.html +++ b/doc/Projects/2018/hw2/html/hw2-bs.html @@ -150,12 +150,7 @@ Compare and analyze your results with those from exercise 2. Study the dependenc

    1. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      2. -
    2. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression.
      3. -
    - -You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11. - -
      +
    1. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    2. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    3. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    @@ -186,7 +181,7 @@ $$ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, $$ -where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise.

    diff --git a/doc/Projects/2018/hw2/html/hw2.html b/doc/Projects/2018/hw2/html/hw2.html index 3e2d44270..81b326f5d 100644 --- a/doc/Projects/2018/hw2/html/hw2.html +++ b/doc/Projects/2018/hw2/html/hw2.html @@ -115,12 +115,7 @@ Compare and analyze your results with those from exercise 2. Study the dependenc

    1. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      2. -
    2. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression.
      3. -
    - -You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11. - -
      +
    1. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    2. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    3. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    @@ -151,7 +146,7 @@ $$ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, $$ -where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. diff --git a/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz b/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz index 4b1da116e789a1063214110a9cbf73f4e1523659..b45bada7796cbc362a3cd933ea09c659efdf149f 100644 GIT binary patch literal 207 zcmb2|=3r=d>y2h$etX_G>#%`9+ru@rU6zd#>tC(mKB?60q`9}DXKK{Xi9wzVzrM}3 zzOd#}tj5y!%@2~>9M9}oBOmVn+NO5d(YqQ;XYHD%IsNS6Iji55gmw2m*I zwrJU+xQ<=-y~3Z*D3J(%S$@p*ao)M!Q}Bm;NBQJ=f4`|RAcGJ4YZ$p!PG8NS H!N33jOh#n7 literal 206 zcmV;<05Sg`iwFRKM2%Yj1MSbjY63A3#&OR+#XP~1&8B)T=%I&#h%XS?*p01;CK27Y z@2b$A>`hAP|9APAVHjqX^XJ*?yej81dKjhD#2C5cbnaFt*wJu;=!QeB~dp>U@@$MPIrYB*fy@F}_Q8(v{!$ zVJ&TR`O=r<4o9Dy*wBk@TVH$(_G7$H@h#b1uExLp2><{90000000000008_ipKT2s ITmUEl08uJxEdT%j diff --git a/doc/Projects/2018/hw2/pdf/hw2.p.tex b/doc/Projects/2018/hw2/pdf/hw2.p.tex index 7824027f3..dd6529e49 100644 --- a/doc/Projects/2018/hw2/pdf/hw2.p.tex +++ b/doc/Projects/2018/hw2/pdf/hw2.p.tex @@ -191,13 +191,8 @@ Compare and analyze your results with those from exercise 2. Study the dependenc \begin{enumerate} \item Repeat the above but using the functionality of \textbf{scikit-learn}. Compare your code with the results from \textbf{scikit-learn}. Remember to run with the same random numbers for generating $x$ and $y$. -\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. -\end{enumerate} +\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. -\noindent -You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. - -\begin{enumerate} \item Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie \emph{et al.}. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of \textbf{scikit-learn}. \item Finally, using \textbf{scikit-learn} or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as @@ -225,7 +220,7 @@ $\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie \emph{et a \[ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, \] -where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interprepation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where $d_i$ are the singular values of the matrix $\hat{X}$. 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ze~=t_vpc1*N+EQawo-)UKusYVo|!{9$7qi3C@WL0L+TbgNNk)CUN0&bK)B2e2w+7} zi)sdEbMX;pfAOh^KIaQQ>Tyn}>CJ^)ex%^Gs3tnq!`YM23gzlCp$OsLLE>pENyVji zp=7SO3mX>bF``H~;x$4M$?ZwzrM$H1evHf)uzV@Yg~nOt|<6)xGGz&%yV$H`3SVin7ap; z@+*v~mNEGZjBT!VSNZbgI$zpsP@{Pcz&4kMYkaAD=Cz40tDb?c%>~;UU*;ZPuFIEn zEnfaP_&Q)uXVKnP3zfAl(k+HKF@V-J1Sn7gM{sWy2*xsG1BJ$H{he_3Ph zo@&F2$;mc$82KE`Jyt%iS>C+}($%YP^=iVpUc>gxJyzka@fN#FJI8FZrW@0f_{F} zUh5b1JDm2~yrAFSwAc0p{RW3O6bJRVZr1i%P~GF_1|rbzd2_y`%`1DR%bfcQ7P>Lryde^HX83_14E+~M@1 z5+CN(ueZ$zClqM!*PCjRelSripMv}spwnDdm%sx9Y(p?GHa0;)G&VCZH#Ij#L_tGE zK{Yo;LoqZ%Mnp6(b$ignTglA^7J}tPX2TgH820jMlHy{?xGguKaHp* lID>PzfJ;c_zqh6&d9i;W^&f Date: Mon, 3 Sep 2018 16:38:00 +0200 Subject: [PATCH 03/12] more typos grrr --- doc/Projects/2018/hw2/html/hw2-bs.html | 9 ++------- doc/Projects/2018/hw2/html/hw2.html | 9 ++------- .../2018/hw2/ipynb/ipynb-hw2-src.tar.gz | Bin 207 -> 207 bytes doc/Projects/2018/hw2/pdf/hw2.p.tex | 9 ++------- doc/Projects/2018/hw2/pdf/hw2.pdf | Bin 198064 -> 198096 bytes doc/Projects/2018/hw2/pdf/hw2.tex | 9 ++------- doc/src/Projects/2018/Exercises/hw2.do.txt | 5 ++--- 7 files changed, 10 insertions(+), 31 deletions(-) diff --git a/doc/Projects/2018/hw2/html/hw2-bs.html b/doc/Projects/2018/hw2/html/hw2-bs.html index 367aaa066..7e1d9a41e 100644 --- a/doc/Projects/2018/hw2/html/hw2-bs.html +++ b/doc/Projects/2018/hw2/html/hw2-bs.html @@ -143,14 +143,9 @@ The following simple Python instructions define our \( x \) and \( y \) values ( y = 5*x*x+0.1*np.random.randn(100,1)
      -
    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \).
      2. -
    - -Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \). - -
      +
    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
    2. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      3. -
    3. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
      4. +
    4. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    5. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    6. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    diff --git a/doc/Projects/2018/hw2/html/hw2.html b/doc/Projects/2018/hw2/html/hw2.html index 81b326f5d..e68f82db2 100644 --- a/doc/Projects/2018/hw2/html/hw2.html +++ b/doc/Projects/2018/hw2/html/hw2.html @@ -108,14 +108,9 @@ The following simple Python instructions define our \( x \) and \( y \) values ( y = 5*x*x+0.1*np.random.randn(100,1)
      -
    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \).
      2. -
    - -Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \). - -
      +
    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
    2. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      3. -
    3. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and computetheir variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
      4. +
    4. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    5. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    6. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    diff --git a/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz b/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz index b45bada7796cbc362a3cd933ea09c659efdf149f..6bb2b7ebf214a27972f81ff43417b60f0a74f477 100644 GIT binary patch literal 207 zcmb2|=3p>)?~P_)etZ647L%dCvBL9qM{N#Hy4(ADX8^w(i(tIMri;GO!oEto^6!T( zw7oie%BBDBKN%i2II|{Ae|>l4ME%MyCSEJuvUfh)GV@MQ==WdCbhI^JM_r$~uGCo6 zR7BWT=JFMsZ(JgCF>s5Ry2h$etX_G>#%`9+ru@rU6zd#>tC(mKB?60q`9}DXKK{Xi9wzVzrM}3 zzOd#}tj5y!%@2~>9M9}oBOmVn+NO5d(YqQ;XYHD%IsNS6Iji55gmw2m*I zwrJU+xQ<=-y~3Z*D3J(%S$@p*ao)M!Q}Bm;NBQJ=f4`|RAcGJ4YZ$p!PG8NS H!N33jOh#n7 diff --git a/doc/Projects/2018/hw2/pdf/hw2.p.tex b/doc/Projects/2018/hw2/pdf/hw2.p.tex index dd6529e49..03165d7b8 100644 --- a/doc/Projects/2018/hw2/pdf/hw2.p.tex +++ b/doc/Projects/2018/hw2/pdf/hw2.p.tex @@ -182,16 +182,11 @@ y = 5*x*x+0.1*np.random.randn(100,1) \epycod \begin{enumerate} -\item Write your own code for the Ridge method (see chapter 3.4 of Hastie \emph{et al.}, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. -\end{enumerate} +\item Write your own code for the Ridge method (see chapter 3.4 of Hastie \emph{et al.}, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. -\noindent -Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. - -\begin{enumerate} \item Repeat the above but using the functionality of \textbf{scikit-learn}. Compare your code with the results from \textbf{scikit-learn}. Remember to run with the same random numbers for generating $x$ and $y$. -\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. +\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and compute their variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. \item Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie \emph{et al.}. Discuss your results and compare with standard regression and the Ridge regression results. 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Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. -\noindent -Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. - -\begin{enumerate} \item Repeat the above but using the functionality of \textbf{scikit-learn}. Compare your code with the results from \textbf{scikit-learn}. Remember to run with the same random numbers for generating $x$ and $y$. -\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. +\item Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of \textbf{scikit-learn} and compute their variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie \emph{et al.} and their figure 3.11. \item Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie \emph{et al.}. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of \textbf{scikit-learn}. diff --git a/doc/src/Projects/2018/Exercises/hw2.do.txt b/doc/src/Projects/2018/Exercises/hw2.do.txt index add3555f6..593182104 100644 --- a/doc/src/Projects/2018/Exercises/hw2.do.txt +++ b/doc/src/Projects/2018/Exercises/hw2.do.txt @@ -22,12 +22,11 @@ x = np.random.rand(100,1) y = 5*x*x+0.1*np.random.randn(100,1) !ec -o Write your own code for the Ridge method (see chapter 3.4 of Hastie *et al.*, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. -Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. +o Write your own code for the Ridge method (see chapter 3.4 of Hastie *et al.*, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. Compare and analyze your results with those from exercise 2. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. o Repeat the above but using the functionality of _scikit-learn_. Compare your code with the results from _scikit-learn_. Remember to run with the same random numbers for generating $x$ and $y$. -o Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of _scikit-learn_ and computetheir variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie *et al.* and their figure 3.11. +o Our next step is to study the variance of the parameters $\beta_1$ and $\beta_2$ (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix $(\hat{X}^T\hat{X})^{-1}+\lambda\hat{I}$) or use the functionality of _scikit-learn_ and compute their variances. Discuss the results of these variances as functions of $\lambda$. In particular, try to link your discussion with the discussion in Hastie *et al.* and their figure 3.11. o Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie *et al.*. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of _scikit-learn_. From acb8aa711d3917e9b3b0ba4d5ceceadc91b9a813 Mon Sep 17 00:00:00 2001 From: mhjensen Date: Tue, 4 Sep 2018 22:04:50 +0200 Subject: [PATCH 04/12] preparing ridge material for slides --- doc/src/Regression/out.txt | 2144 ++++++++++++++++++++++++++++++++++++ 1 file changed, 2144 insertions(+) create mode 100644 doc/src/Regression/out.txt diff --git a/doc/src/Regression/out.txt b/doc/src/Regression/out.txt new file mode 100644 index 000000000..e7253cd8d --- /dev/null +++ b/doc/src/Regression/out.txt @@ -0,0 +1,2144 @@ +# #ifdef LATEX2DOCONCE +This is the result of the doconce latex2doconce program. +The translation from LaTeX is just a helper. The text must +be carefully examined! (Be prepared that some text might also +be lost in the translation - in seldom cases.) + + +## search for CHECK to see if auto editing was correct + +# #endif + +========= Ridge regression ========= label{chap:ridgeRegression} +\pagenumbering{arabic} +# \setcounter{page} +High-throughput techniques measure many characteristics of a single sample simultaneously. The number of characteristics $p$ measured may easily exceed ten thousand. In most medical studies the number of samples $n$ involved often falls behind the number of characteristics measured, i.e: $p > n$. The resulting $(n \times p)$-dimensional data matrix $\mathbf{X}$: +\begin{align*} +\mathbf{X} & = +\left( +X_{\ast,1} \, | \, \ldots \, | \, X_{\ast,p} +\right) +\, \, \, = \, \, \, +\left( +\begin{array}{c} +X_{1,\ast} \\ \vdots \\ X_{n,\ast} +\end{array} \right) +\, \, \, = \, \, \, +\left( +\begin{array}{ccc} +X_{1,1} & \ldots & X_{1,p} +\\ +\vdots & \ddots & \vdots +\\ +X_{n,1} & \ldots & X_{n,p} +\end{array} \right) +\end{align*} +from such a study contains a larger number of covariates than samples. When $p > n$ the data matrix $\mathbf{X}$ is said to be *high-dimensional*. + +In this chapter we adopt the traditional statistical notation of the data matrix. An alternative notation would be $\mathbf{X}^{\top}$ (rather than $\mathbf{X}$), which is employed in the field of (statistical) bioinformatics. In $\mathbf{X}^{\top}$ the rows comprise the samples rather than the covariates. The case for the bioinformatics notation stems from practical arguments. A spreadsheet is designed to have more rows than columns. In case $p > n$ the traditional notation yields a spreadsheet with more columns than rows. When $p > 10000$ the conventional display is impractical. In this chapter we stick to the conventional statistical notation of the data matrix as all mathematical expressions involving $\mathbf{X}$ are then in line with those of standard textbooks on regression. + +The information contained in $\mathbf{X}$ is often used to explain a particular property of the samples involved. In applications in molecular biology $\mathbf{X}$ may contain microRNA expression data from which the expression levels of a gene are to be described. When the gene's expression levels are denoted by $\mathbf{Y} = (Y_{1}, \ldots, Y_n)^{\top}$, the aim is to find the linear relation $Y_i = \mathbf{X}_{i, \ast} \bbeta$ from the data at hand by means of regression analysis. Regression is however frustrated by the high-dimensionality of $\mathbf{X}$ (illustrated in Section ref{sect.ridgeRegression} and at the end of Section ref{sect.constrainedEstimation}). These notes discuss how regression may be modified to accommodate the high-dimensionality of $\mathbf{X}$. First, however, `standard' linear regression is recaputilated. + + +======= Linear regression ======= + +Consider an experiment in which $p$ characteristics of $n$ samples are measured. The data from this experiment are denoted $\mathbf{X}$, with $\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design matrix*. Additional information of the samples is available in the form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is generally referred to as the *response variable*. The aim of regression analysis is to explain $\mathbf{Y}$ in terms of $\mathbf{X}$ through a functional relationship like $Y_i = f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of $f(\cdot)$ is available, it is common to assume a linear relationship between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to the *linear regression model*: + +!bt + +!bt +\begin{align} +Y_{i} & = \mathbf{X}_{i,\ast} \, \bbeta + \varepsilon_i label{form.linRegressionModel} +\\ +\nonumber +& = \beta_1 \, X_{i,1} + \ldots + \beta_{p} \, X_{i, p} + \varepsilon_i. +\end{align} +!et +!et +In model (ref{form.linRegressionModel}) $\bbeta = (\beta_1, \ldots, \beta_p)^{\top}$ is the *regression parameter*. The parameter $\beta_j$, $j=1, \ldots, p$, represents the +effect size of covariate $j$ on the response. That is, for each unit change in covariate $j$ (while keeping the other covariates fixed) the observed change in the response is equal to $\beta_j$. +The second summand on the right-hand side of the model, $\varepsilon_i$, is referred to as the error. It represents the part of the response not explained by the functional part $\mathbf{X}_{i,\ast} \, \bbeta$ of the model (ref{form.linRegressionModel}). In contrast to the functional part, which is considered to be systematic (i.e. non-random), the error is assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal $Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= \mathbf{X}_{i_2,\ast}$. To complete the formulation of model (ref{form.linRegressionModel}) we need to specify the probability distribution of $\varepsilon_i$. It is assumed that $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$ and the $\varepsilon_{i}$ are independent, i.e.: +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, \varepsilon_{i_2}) & = +\left\{ +\begin{array}{lcc} +\sigma^2 & \mbox{if} & i_1 = i_2, +\\ +0 & \mbox{if} & i_1 \not= i_2. +\end{array} +\right. +\end{align*} +The randomness of $\varepsilon_i$ implies that $\mathbf{Y}_i$ is also a random variable. In particular, $\mathbf{Y}_i$ is normally distributed, because $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$ and $\mathbf{X}_{i,\ast} \, \bbeta$ is a non-random scalar. To specify the parameters of the distribution of $\mathbf{Y}_i$ we need to calculate its first two moments. Its expectation equals: +\begin{align*} +\mathbb{E}(Y_i) & = \mathbb{E}(\mathbf{X}_{i, \ast} \, \bbeta) + \mathbb{E}(\varepsilon_i) \, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \bbeta, +\end{align*} +while its variance is: +\begin{align*} +\mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i - \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - [\mathbb{E}(Y_i)]^2 +# \\ +# & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \bbeta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 +\\ +& = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \bbeta)^2 + 2 \varepsilon_i \mathbf{X}_{i, \ast} \, \bbeta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 +\\ +# & = ( \mathbf{X}_{i, \ast} \, \bbeta)^2 + 2 \mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \bbeta + \mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 +# \\ +& = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, \mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +Hence, $Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \bbeta, \sigma^2)$. This formulation (in terms of the normal distribution) is equivalent to the formulation of model +(ref{form.linRegressionModel}), as both capture the assumptions involved: the linearity of the functional part and the normality of the error. + +Model (ref{form.linRegressionModel}) is often written in a more condensed matrix form: + +!bt + +!bt +\begin{align} +\mathbf{Y} & = \mathbf{X} \, \bbeta + \vvarepsilon, label{form.linRegressionModelinMatrix} +\end{align} +!et +!et +where $\vvarepsilon = (\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)^{\top}$ and distributed as $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{p}, \sigma^2 \mathbf{I}_{nn})$. As above model (ref{form.linRegressionModelinMatrix}) can be expressed as a multivariate normal distribution: $\mathbf{Y} \sim \mathcal{N}(\mathbf{X} \, \bbeta, \sigma^2 \mathbf{I}_{nn})$. + +Model (ref{form.linRegressionModelinMatrix}) is a so-called hierarchical model. This terminology emphasizes that $\mathbf{X}$ and $\mathbf{Y}$ are not on a par, they play different roles in the model. The former is used to explain the latter. In model (ref{form.linRegressionModel}) $\mathbf{X}$ is referred as the *explanatory* or *independent* variable, while the variable $\mathbf{Y}$ is generally referred to as the *response* or *dependent* variable. + +The covariates, the columns of $\mathbf{X}$, may themselves be random. To apply the linear model they are temporarily assumed fixed. The linear regression model is then to be interpreted as $\mathbf{Y} \, | \, \mathbf{X} \sim \mathcal{N}(\mathbf{X} \, \bbeta, \sigma^2 \mathbf{I}_{nn})$ + +\begin{example} *Methylation of a tumor-suppressor gene* +\\ +Consider a study which measures the gene expression levels of a tumor-suppressor genes (TSG) and two methylation markers (MM1 and MM2) on 67 samples. A methylation marker is a gene that promotes methylation. Methylation refers to attachment of a methyl group to a nucleotide of the DNA. In case this attachment takes place in or close by the promotor region of a gene, this complicates the transcription of the gene. Methylation may down-regulate a gene. This mechanism also works in the reverse direction: removal of methyl groups may up-regulate a gene. A tumor-suppressor gene is a gene that halts the progression of the cell towards a cancerous state. + +The medical question associated with these data: do the expression levels methylation markers affect the expression levels of the tumor-suppressor gene? To answer this question we may formulate the following linear regression model: +\begin{align*} +Y_{i, `{\footnotesize tsg`}} & = \beta_0 + \beta_{`{\footnotesize mm1`}} X_{i, `{\footnotesize mm1`}} ++ \beta_{`{\footnotesize mm2`}} X_{i, `{\footnotesize mm2`}} + \varepsilon_i, +\end{align*} +with $i = 1, \ldots, 67$ and $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. The interest focusses on $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$. A non-zero value of at least one of these two regression parameters indicates that there is a linear association between the expression levels of the tumor-suppressor gene and that of the methylation markers. + +Prior knowledge from biology suggests that the $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$ are both non-positive. High expression levels of the methylation markers lead to hyper-methylation, in turn inhibiting the transcription of the tumor-suppressor gene. Vice versa, low expression levels of MM1 and MM2 are (via hypo-methylation) associated with high expression levels of TSG. Hence, a negative concordant effect between MM1 and MM2 (on one side) and TSG (on the other) is expected. Of course, the methylation markers may affect expression levels of other genes that in turn regulate the tumor-suppressor gene. The regression parameters $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$ then reflect the indirect effect of the methylation markers on the expression levels of the tumor suppressor gene. +\end{example} + + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +The linear regression model (ref{form.linRegressionModel}) involves the unknown parameters: $\bbeta$ and $\sigma^2$, which need to be learned from the data. The parameters of the regression model, $\bbeta$ and $\sigma^2$ are estimated by means of likelihood maximization. Recall that $Y_i \sim \mathcal{N}( \mathbf{X}_{i,\ast} \, \bbeta, \sigma^2)$ with corresponding density: $ f_{Y_i}(y_i) = (2 \, \pi \, \sigma^2)^{-1/2} \, \exp[ - (y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 / 2 \sigma^2 ]$. The likelihood thus is: +\begin{align*} +L(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = +# f_{\mathbf{Y}}(\mathbf{y}) \, \, \, = \, \, \, \prod_{i=1}^n f_{Y_i}(y_i) \, \, \, = \, \, \, +\prod_{i=1}^n \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, \exp[ - (Y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2 / 2 \sigma^2 ], +\end{align*} +in which the independence of the observations has been used. Because of the concavity of the logarithm, the maximization of the likelihood coincides with the maximum of the logarithm of the likelihood (called the log-likelihood). Hence, to obtain maximum likelihood (ML) estimates of the parameter it is equivalent to find the maximum of the log-likelihood. The log-likelihood is: +\begin{align*} +\mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = +\log[ L(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) ] +\, \, \, = \, \, \, % \log \Big[ \prod_{i=1}^n f_{Y_i}(y_i) \Big] +# \, \, \, = \, \, \, \sum_{i=1}^n \log[ f_{Y_i}(y_i) ] +# \\ +# & = \sum_{i=1}^n [ -\log(\sqrt{2 \, \pi} \, \sigma) - (y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 / 2 \sigma^2 ] +# \\ +# & = +-n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \sum_{i=1}^n (y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2. +\end{align*} +After noting that $\sum_{i=1}^n (Y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2 = \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 \, \, \, = \, \, \, (\mathbf{Y} - \mathbf{X} \, \bbeta)^{\top} \, (\mathbf{Y} - \mathbf{X} \, \bbeta)$, the log-likelihood can be written as: +\begin{align*} +\mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = -n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \, \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. +\end{align*} +In order to find the maximum of the log-likelihood, take its derivate with respect to $\bbeta$: +\begin{align*} +\frac{\partial }{\partial \, \beta} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = - \frac{1}{ 2 \sigma^2} \, \frac{\partial }{\partial \, \beta} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 \, \, \, = \, \, \, \frac{1}{\sigma^2} \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \, \bbeta). +\end{align*} +Equate this derivative to zero gives the estimating equation for $\bbeta$: + +!bt + +!bt +\begin{align} label{form.normalEquation} +\mathbf{X}^{\top} \mathbf{X} \, \bbeta & = \mathbf{X}^{\top} \mathbf{Y}. +\end{align} +!et +!et +Equation (ref{form.normalEquation}) is called to the *normal equation*. Pre-multiplication of both sides of the normal equation by $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ now yields the ML estimator of the regression parameter: $\hat{\bbeta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}$, in which it is assumed that $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ is well-defined. + +Along the same lines one obtains the ML estimator of the residual variance. Take the partial derivative of the log-likelihood with respect to $\sigma^2$: +\begin{align*} +\frac{\partial }{\partial \, \sigma} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = - \frac{n}{\sigma} + \frac{1}{\sigma^3} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. +\end{align*} +Equate the right-hand side to zero and solve for $\sigma^2$ to find $\hat{\sigma}^2 = \tfrac{1}{n} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2$. In this expression $\bbeta$ is unknown and the ML estimate of $\bbeta$ is plugged-in. +\\ +\\ +With explicit expressions of the ML estimators at hand, we can study their properties. The expectation of the ML estimator of the regression parameter $\bbeta$ is: +\begin{align*} +\mathbb{E}(\hat{\bbeta}) & = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] +\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbb{E}[ \mathbf{Y}] +# \\ +\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{X} \, \bbeta +\, \, \, \, \, = \, \, \, \bbeta. +\end{align*} +Hence, the ML estimator of the regression coefficients is unbiased. + +The variance of the ML estimator of $\bbeta$ is: +\begin{align*} +\mbox{Var}(\hat{\bbeta}) & = \mathbb{E} \{ [\hat{\bbeta} - \mathbb{E}(\hat{\bbeta})] [\hat{\bbeta} - \mathbb{E}(\hat{\bbeta})]^{\top} \} +\\ +& = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \bbeta] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \bbeta]^{\top} \} +\\ +# & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}]^{\top} \} - \bbeta \, \bbeta^{\top} +# \\ +# & = \mathbb{E} \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \, \mathbf{Y}^{\top} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \} - \bbeta \, \bbeta^{\top} +# \\ +& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{\top} \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +\\ +& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \{ \mathbf{X} \, \bbeta \, \bbeta^{\top} \, \mathbf{X}^{\top} + \SSigma \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +# \\ +# & = (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \bbeta \, \bbeta^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} +# \\ +# & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \bbeta \bbeta^T +\\ +& = \bbeta \, \bbeta^{\top} + \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1}, +\end{align*} +in which we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{\top}) = \mathbf{X} \, \bbeta \, \bbeta^{\top} \, \mathbf{X}^{\top} + \sigma^2 \, \mathbf{I}_{nn}$. From $\mbox{Var}(\hat{\bbeta}) = \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1}$, one obtains an estimate of the variance of the estimate of the $j$-th regression coefficient: $\hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ [(\mathbf{X}^{\top} \mathbf{X})^{-1}]_{jj} }$. This may be used to construct a confidence interval for the estimates or test the hypothesis $H_0: \beta_j = 0$. In the latter $\hat{\sigma}^2$ should not be the maximum likelihood estimator, as it is biased. It is then to be replaced by the residual sum-of-squares divided by $n-p$ rather than $n$. +\\ +\\ +The prediction of $Y_i$, denoted $\widehat{Y}_i$, is the expected value of $Y_i$ according the linear regression model (with its parameters replaced by their estimates). The prediction of $Y_i$ thus equals $\mathbb{E}(Y_i; \hat{\bbeta}, \hat{\sigma}^2) = \mathbf{X}_{i, \ast} \hat{\bbeta}$. In matrix notation the prediction is: +\begin{align*} +\widehat{\mathbf{Y}} & = \mathbf{X} \, \hat{\bbeta} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, := \, \, \, \mathbf{H} \mathbf{Y}, +\end{align*} +where $\mathbf{H}$ is the *hat matrix*, as it `puts the hat' on $\mathbf{Y}$. Note that the hat matrix is a projection matrix, i.e. $\mathbf{H}^2 = \mathbf{H}$ for +\begin{align*} +\mathbf{H}^2 & = \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}. +\end{align*} +Thus, the prediction $\widehat{\mathbf{Y}}$ is an orthogonal projection of $\mathbf{Y}$ onto the space spanned by the columns of $\mathbf{X}$. + +With $\widehat{\bbeta}$ available, an estimate of the errors $\hat{\varepsilon}_i$, dubbed the *residuals* are obtained via: +\begin{align*} +\hat{\vvarepsilon} & = \mathbf{Y} - \widehat{\mathbf{Y}} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, \hat{\bbeta} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, [ \mathbf{I} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} ] \, \mathbf{Y}. +\end{align*} +Thus, the residuals are a projection of $\mathbf{Y}$ onto the orthogonal complement of the space spanned by the columns of $\mathbf{X}$. The residuals are to be used in diagnostics, e.g. checking of the normality assumption by means of a normal probability plot. +\\ +\\ +For more on the linear regression model confer the monograph of cite{Drap1998}. + + + +======= Ridge regression} label{sect.ridgeRegression ======= +When the design matrix is high-dimensional, the covariates (the columns of $\mathbf{X}$) are super-collinear. Recall *collinearity* in regression analysis refers to the event of two (or multiple) covariates being highly linearly related. Consequently, the subspace spanned by collinear covariates may not be (or close to not being) of full rank. When the subspace (onto which $\mathbf{Y}$ is projected) is (close to) rank deficient, it is (almost) impossible to separate the contribution of the individual covariates. The uncertainty with respect to the covariate responsible for the variation explained in $\mathbf{Y}$ is often reflected in the fit of the linear regression model to data by a large error of the estimates of the regression parameters corresponding to the collinear covariates. + +\begin{example} \mbox{ } \\ +The flotillins (the FLOT-1 and FLOT-2 genes) have been observed to regulate the proto-oncogene ERBB2 *in vitro* \citep{Pust2013}. One may wish to corroborate this *in vivo*. To this end we use gene expression data of a breast cancer study, available as a Bioconductor package: {\tt breastCancerVDX}. From this study the expression levels of probes interrogating the FLOT-1 and ERBB2 genes are retrieved. For clarity of the illustration the FLOT-2 gene is ignored. After centering, the expression levels of the first ERBB2 probe are regressed on those of the four FLOT-1 probes. The R-code below carries out the data retrieval and analysis. +\lstinputlisting{introExampleForNotes.r} +Prior to the regression analysis, we first assess whether there is collinearity among the FLOT-1 probes through evaluation of the correlation matrix. This reveals a strong correlation ($\hat{\rho} = 0.91$) between the second and third probe. All other cross-correlations do not exceed the 0.20 (in an absolute sense). Hence, there is collinearity among the columns of the design matrix in the to-be-performed regression analysis. +\verbatiminput{introExampleRegressionOutput.txt} +The output of the regression analysis above shows the first probe to be significantly associated to the expression levels of ERBB2. The collinearity of the second and third probe reveals itself in the standard errors of the effect size: for these probes the standard error is much larger than those of the other two probes. This reflects the uncertainty in the estimates. Regression analysis has difficulty to decide to which covariate the explained proportion of variation in the response should be attributed. The large standard error of these effect sizes propagates to the testing as the Wald test statistic is the ratio of the estimated effect size and its standard error. Collinear covariates are thus less likely to pass the significance threshold. +\end{example} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +The case of two (or multiple) covariates being perfectly linearly dependent is referred as *super-collinearity*. The rank of a high-dimensional design matrix is maximally equal to $n$: $\mbox{rank}(\mathbf{X}) \leq n$. Consequently, the dimension of subspace spanned by the columns of $\mathbf{X}$ is smaller than or equal to $n$. As $p > n$, this implies that columns of $\mathbf{X}$ are linearly dependent. Put differently, a high-dimensional $\mathbf{X}$ suffers from super-collinearity. + + + +\begin{example} *Super-collinearity* label{example.supercollinearity} +\\ +Consider the design matrix: +\begin{align*} +\mathbf{X} & = \left( +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right) +\end{align*} +The columns of $\mathbf{X}$ are linearly dependent: the first column is the row-wise sum of the other two columns. The rank (more correct, the column rank) of a matrix is the dimension of space spanned by the column vectors. Hence, the rank of $\mathbf{X}$ is equal to the number of linearly independent columns: $\mbox{rank}(\mathbf{X}) = 2$. +\end{example} + + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +Super-collinearity of an $(n \times p)$-dimensional design matrix $\mathbf{X}$ implies[^footnote1][^footnote1]: If the (column) rank of $\mathbf{X$ is smaller than $p$, there exists a non-trivial $\mathbf{v} \in \mathbb{R}^p$ such that $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. Multiplication of this inequality by $\mathbf{X}^{\top}$ yields $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. As $\mathbf{v} \not= \mathbf{0}_{p}$, this implies that $\mathbf{X}^{\top} \mathbf{X}$ is not invertible.} that the rank of the $(p \times p)$-dimensional matrix $\mathbf{X}^{\top} \mathbf{X}$ is smaller than $p$, and, consequently, it is singular. A square matrix that does not have an inverse is called *singular*. A matrix $\mathbf{A}$ is singular if and only if its determinant is zero: $\mbox{det}(\mathbf{A}) = 0$. + +\begin{example} *Singularity* label{example.singular} +\\ +Consider the matrix $\mathbf{A}$ given by: +\begin{align*} +\mathbf{A} & = \left( +\begin{array}{rr} +1 & 2 +\\ +2 & 4 +\end{array} \right) +\end{align*} +Clearly, $\mbox{det}(\mathbf{A}) = a_{11} a_{22} - a_{12} a_{21} = 1 \times 4 - 2 \times 2 = 0$. Hence, $\mathbf{A}$ is singular and its inverse is undefined. +\end{example} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +As $\mbox{det}(\mathbf{A})$ is equal to the product of the eigenvalues $\nu_j$ of $\mathbf{A}$, the matrix $\mathbf{A}$ is singular if one (or more) of the eigenvalues of $\mathbf{A}$ is zero. To see this, consider the spectral decomposition of $\mathbf{A}$: +\begin{align*} +\mathbf{A} & = \sum_{j=1}^p \nu_j \, \mathbf{v}_j \, \mathbf{v}_j^{\top}, +\end{align*} +where $\mathbf{v}_j$ is the eigenvector corresponding to $\nu_j$. The inverse of $\mathbf{A}$ is then: +\begin{align*} +\mathbf{A}^{-1} & = \sum_{j=1}^p \nu_j^{-1} \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. +\end{align*} +The right-hand side is undefined if $\nu_j =0$ for any $j$. + + +\begin{contexample}_ref{example.singular_} \hspace{3pt}*Singularity (continued)* +\\ +Revisit Example ref{example.singular}. Matrix $\mathbf{A}$ has eigenvalues $\nu_1 =5$ and $\nu_2=0$. According to the spectral decomposition, the inverse of $\mathbf{A}$ is: +\begin{align*} +\mathbf{A}^{-1} & = \frac{1}{5} \, \mathbf{v}_1 \, \mathbf{v}_1^{\top} + \frac{1}{0} \, \mathbf{v}_2 \, \mathbf{v}_2^{\top}. +\end{align*} +This expression is undefined as we divide by zero in the second summand on the right-hand side. +\end{contexample} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +In summary, the columns of a high-dimensional design matrix $\mathbf{X}$ are linearly dependent and this super-collinearity causes $\mathbf{X}^{\top} \mathbf{X}$ to be singular. Now recall the ML estimator of the parameter of the linear regression model: + +!bt + +!bt +\begin{align} +\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\end{align} +!et +!et +This estimator is only well-defined if $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits. Hence, when $\mathbf{X}$ is high-dimensional the regression parameter $\bbeta$ cannot be estimated. +\\ +\\ +Above only the practical consequence of high-dimensionality is presented: the expression $( \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ cannot be evaluated numerically. But the problem arising from the high-dimensionality of the data is more fundamental. To appreciate this, consider the normal equations: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} \bbeta & = \mathbf{X}^{\top} \mathbf{Y}. +\end{align*} +The matrix $\mathbf{X}^{\top} \mathbf{X}$ is of rank $n$, while $\bbeta$ is a vector of length $p$. Hence, while there are $p$ unknowns, the system of linear equations from which these are to be solved effectively comprises $n$ degrees of freedom. If $p > n$, the vector $\bbeta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = +\mathbf{X}^{\top} \mathbf{0}_{p} = \mathbf{0}_{n}$, the solution of the normal equations is: +\begin{align*} +\hat{\bbeta} & = ( \mathbf{X}^{\top} \mathbf{X})^{-} \mathbf{X}^{\top} \mathbf{Y} + \mathbf{v} \qquad \mbox{for all } \mathbf{v} \in V, +\end{align*} +where $\mathbf{A}^{-}$ denotes the Moore-Penrose inverse of the matrix $\mathbf{A}$, which is defined as: +\begin{align*} +\mathbf{A}^{-} & = \sum_{j=1}^p \nu_j^{-1} \, I_{\{ \nu_j \not= 0 \} } \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. +\end{align*} +The solution of the normal equations is thus only determined up to an element from a non-trivial space $V$, and there is no unique estimator of the regression parameter. +\\ +\\ +To obtain an estimate of the regression parameter $\bbeta$ when $\mathbf{X}$ is (close to) super-collinearity, cite{Hoer1970} proposed an ad-hoc fix to resolve the (almost) singularity of $\mathbf{X}^{\top} \mathbf{X}$. Simply replace $\mathbf{X}^{\top} \mathbf{X}$ by $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ with $\lambda \in [0, \infty)$. The scalar $\lambda$ is a tuning parameter, henceforth called the *penalty parameter*. + + +\begin{contexample}_ref{example.supercollinearity_} \hspace{3pt}*Super-collinearity (continued)* +\\ +Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Then, for (say) $\lambda = 1$: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} & = \left( +\begin{array}{rrr} +5 & 2 & 2 +\\ +2 & 7 & -4 +\\ +2 & -4 & 7 +\end{array} \right). +\end{align*} +The eigenvalues of this matrix are 11, 7, and 1. Hence, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ has no zero eigenvalue and its inverse is well-defined. +\end{contexample} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +With the ad-hoc fix for the singularity of $\mathbf{X}^{\top} \mathbf{X}$, cite{Hoer1970} proceed to define the *ridge regression estimator*: + +!bt + +!bt +\begin{align} label{form.ridgeRegressionEstimator} +\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}, +\end{align} +!et +!et +for $\lambda \in [0, \infty)$. Clearly, this is - for $\lambda$ strictly positive - a well-defined estimator, even if $\mathbf{X}$ is high-dimensional. However, each choice of $\lambda$ leads to a different ridge regression estimate. The set of all ridge regression estimates $\{ \hat{\bbeta}(\lambda) \, : \, \lambda \in [0, \infty) \}$ is called the *solution* or *regularization path* of the ridge estimator. + + + +\begin{contexample}_ref{example.supercollinearity_} \hspace{3pt}*Super-collinearity (continued)* +\\ +Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Suppose that the corresponding response vector is $\mathbf{Y} = (1.3, -0.5, 2.6, 0.9)^{\top}$. +The ridge regression estimates for, e.g. $\lambda = 1, 2$, and $10$ are then: +\begin{align*} +\hat{\bbeta}(1) & = (0.614, 0.548, 0.066)^{\top}, +\\ +\hat{\bbeta}(2) & = (0.537, 0.490, 0.048)^{\top}, +\\ +\hat{\bbeta}(10) & = (0.269, 0.267, 0.002)^{\top}. +\end{align*} +The full solution path of the ridge estimator is plotted in Figure ref{fig.ridgeSolPathPlusVar}. + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [ridgeSolutionPath_logLambda.eps, width=400 frac=1.0] Solution path of the ridge estimator and its variance. The left panel shows the solution path of the ridge estimator for the data of Example ref{example.supercollinearity}. In the right panel the corresponding variance of the ridge estimator is plotted against the (logarithm of the) penalty parameter.} label{fig.ridgeSolPathPlusVar +# \afterpage{} +\end{contexample} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +Having obtained an estimate of the regression parameter $\bbeta$, one can define the fit $\widehat{\mathbf{Y}}$. It is defined analogous to the standard case: +\begin{align*} +\widehat{\mathbf{Y}}(\lambda) & = \mathbf{X} \hat{\bbeta}(\lambda) +\, \, \, = \, \, \, \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\, \, \, := \, \, \, \mathbf{H}(\lambda) \mathbf{Y}. +\end{align*} +Previously, when using the ML estimator, the fit could be understood as a projection of $\mathbf{Y}$ onto the subspace spanned by the columns of $\mathbf{X}$. The fit $\widehat{\mathbf{Y}}(\lambda)$ corresponding to the ridge estimator is not a projection of $\mathbf{Y}$ onto $\mathbf{X}$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$a). Consequently, the `ridge residuals' $\mathbf{Y} - \widehat{\mathbf{Y}}(\lambda)$ are not orthogonal to the fit $\widehat{\mathbf{Y}}(\lambda)$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$b). + + +======= Eigenvalue shrinkage ======= + +The effect of the ridge penalty may also studied from the perspective of singular values. Let the singular value decomposition of the $(n \times p)$-dimensional design matrix $\mathbf{X}$ be: +\begin{align*} +\mathbf{X} & = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}, +\end{align*} +where $\mathbf{D}_x$ an $(n \times n)$-dimensional diagonal matrix with the singular values, +$\mathbf{U}_x$ an $(n \times n)$-dimensional matrix with columns containing the left singular vectors (denoted $\mathbf{u}_i$), and $\mathbf{V}_x$ a $(p \times n)$-dimensional matrix with columns containing the right singular vectors (denoted $\mathbf{v}_i$). The columns of $\mathbf{U}_x$ and $\mathbf{V}_x$ are orthogonal: $\mathbf{U}_x^{\top} \mathbf{U}_x = \mathbf{I}_{nn} = \mathbf{V}_x^{\top} \mathbf{V}_x$. + +The OLS estimator can then be rewritten in terms of the SVD-matrices as: +\begin{align*} +\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}, +\end{align*} +where $\mathbf{D}_x^{-2} \mathbf{D}_x$ is not simplified further to emphasize the effect of the ridge penalty. Similarly, the ridge estimator can be rewritten in terms of the SVD-matrices as: +\begin{align*} +\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. +\end{align*} +Combining the two results and writing $(\mathbf{D}_x)_{jj} = d_{x,jj}$ we have: +\begin{align*} +d_{x,jj}^{-1} & \geq & \frac{d_{x,jj}}{d_{x,jj}^2 + \lambda} \qquad \mbox{ for all } \lambda > 0. +\end{align*} +Thus, the ridge penalty shrinks the singular values. +\\ +\\ +Return to the problem of the super-collinearity of $\mathbf{X}$ in the high-dimensional setting ($p > n$). The super-collinearity implies the singularity of $\mathbf{X}^{\top} \mathbf{X}$ and prevents the calculation of the OLS estimator of the regression coefficients. However, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ is non-singular, with inverse: +\begin{align*} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} & = \sum_{j=1}^p (d_{x,jj}^2 + \lambda)^{-1} \mathbf{v}_j \mathbf{v}_j^{\top}. +\end{align*} +The right-hand side is well-defined for $\lambda > 0$. + + +===== Principal components regression ===== + +Principal component regression is a close relative to ridge regression that can also be applied in a high-dimensional context. Principal components regression explains the response not by the covariates themselves but by linear combinations of the covariates as defined by the principal components of $\mathbf{X}$. Let $\mathbf{U} \mathbf{D} \mathbf{V}^{\top}$ be the singular value decomposition of $\mathbf{X}$. The $i$-th principal component of $\mathbf{X}$ is then $\mathbf{X} \mathbf{v}_i$, henceforth denoted $\mathbf{z}_i$. Let $\mathbf{Z}_k$ be the matrix of the first $k$ principal components, i.e. $\mathbf{Z}_k = \mathbf{X} \mathbf{V}_k$ where $\mathbf{V}_k$ contains the first $k$ right singular vectors as columns. Principal components regression then amounts to regressing the response $\mathbf{Y}$ onto $\mathbf{Z}_{k}$, that is, it fits the model $\mathbf{Y} = \mathbf{Z}_k \ggamma + \vvarepsilon$. The least squares estimator of $\ggamma$ then is (with some abuse of notation): +\begin{align*} +\hat{\ggamma} & = (\mathbf{Z}_k^{\top} \mathbf{Z}_k)^{-1} \mathbf{Z}_k^{\top} \mathbf{Y} \, \, \, = \, \, \, (\mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{X} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{U} \mathbf{D} \mathbf{V}^{\top} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} +\\ +& = (\mathbf{I}_{kn} \mathbf{D}^2 \mathbf{I}_{nk})^{-1} \mathbf{I}_{kn} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} +\\ +& = \mathbf{D}_k^{-2} \widetilde{\mathbf{D}}_k \mathbf{U}^{\top} \mathbf{Y} \, \, \, = \, \, \ \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}, +\end{align*} +where $\mathbf{D}_k$ and $\widetilde{\mathbf{D}}_k$ are submatrices of $\mathbf{D}$. The matrix $\mathbf{D}_k$ is obtained from $\mathbf{D}$ by removal of the last $n-p$ rows and columsn, while for $\widetilde{\mathbf{D}}_k$ only the last $n-k$ rows are dropped. Similarly, $\mathbf{I}_{kn}$ and $\mathbf{I}_{nk}$ are obtained from $\mathbf{I}_{nn}$ by removal of the last $n-k$ rows and columns, respectively. The principal component regression estimator of $\bbeta$ then is $\hat{\bbeta}_{\mbox{{\tiny pcr}}} = \mathbf{V}_k \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}$. When $k$ is set equal to the column rank of $\mathbf{X}$, and thus to the rank of $\mathbf{X}^{\top} \mathbf{X}$, the +principal component regression estimator $\hat{\bbeta}_{\mbox{{\tiny pcr}}} = (\mathbf{X}^{\top} \mathbf{X})^- \mathbf{X}^{\top} \mathbf{Y}$, where $\mathbf{A}^-$ denotes the Moore-Penrose inverse of matrix $\mathbf{A}$. + +The relation between ridge and principal component regression becomes clear when their corresponding estimators are written in terms of the singular value decomposition of $\mathbf{X}$: +\begin{align*} +\hat{\bbeta}_{\mbox{{\tiny pcr}}} & = \mathbf{V}_x (\mathbf{I}_{nk} \mathbf{D}_x \mathbf{I}_{kn})^{-1} \mathbf{U}_x^{\top} \mathbf{Y}, +\\ +\hat{\bbeta} (\lambda) & = \mathbf{V}_x (\mathbf{D}_x^2 + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. +\end{align*} +Both operate on the singular values of the design matrix. But where principal component regression thresholds the singular values of $\mathbf{X}$, ridge regression shrinks them (depending on their size). Hence, one applies a discrete map on the singular values while the other a continuous one. + + + + + +======= Moments ======= + +The first two moments of the ridge regression estimator are derived. Next the performance of the ridge regression estimator is studied in terms of the mean squared error, which combines the first two moments. + + +===== Expectation ===== + +The left panel of Figure ref{fig.ridgeSolPathPlusVar} shows ridge estimates of the regression parameters converging to zero as the penalty parameter tends to infinity. This behaviour of the ridge estimator does not depend on the specifics of the data set. To see this study the expectation of the ridge estimator: +\begin{align*} +\mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \big] +\\ +& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \big] +\\ +& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta} \big] +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \mathbb{E} ( \hat{\bbeta} ) +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \bbeta. +\end{align*} +Clearly, $\mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] \not= \bbeta$ for any $\lambda > 0$. Hence, the ridge estimator is biased. + +From the expression above it is clear that the expectation of the ridge estimator vanishes as $\lambda$ tends to infinity: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \bbeta \, \, \, = \, \, \, \mathbf{0}_{p}. +\end{align*} +Hence, all regression coefficients are shrunken towards zero as the penalty parameter increases. This also holds for $\mathbf{X}$ with $p > n$. Furthermore, this behaviour is not strictly monotone in $\lambda$: $\lambda_{a} > \lambda_b$ does not necessarily imply $|\hat{\beta}_j (\lambda_a) | < |\hat{\beta}_j (\lambda_b) |$. Upon close inspection this can be witnessed from the ridge solution path of $\beta_3$ in Figure ref{fig.ridgeSolPathPlusVar}. + + +\begin{example} *Orthonormal design matrix* label{example.orthoronormalDesign} +\\ +Consider an orthonormal design matrix $\mathbf{X}$, i.e.: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} & = \mathbf{I}_{pp} \, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1}. +\end{align*} +An example of an orthonormal design matrix would be: +\begin{align*} +\mathbf{X} & = \frac{1}{2} \left( +\begin{array}{rr} +-1 & -1 +\\ +-1 & 1 +\\ +1 & -1 +\\ +1 & 1 +\end{array} \right). +\end{align*} +This design matrix is orthonormal as $\mathbf{X}^{\top} \mathbf{X} = \mathbf{I}_{22}$, which is easily verified: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} & = \frac{1}{4} +\left( +\begin{array}{rrrr} +-1 & -1 & 1 & 1 +\\ +-1 & 1 & -1 & 1 +\end{array} \right) +\left( +\begin{array}{rr} +-1 & -1 +\\ +-1 & 1 +\\ +1 & -1 +\\ +1 & 1 +\end{array} \right) \, \, \, = \, \, \, +\frac{1}{4} +\left( +\begin{array}{rr} +4 & 0 +\\ +0 & 4 +\end{array} \right) \, \, \, = \, \, \, \mathbf{I}_{22}. +\end{align*} +In case of an orthonormal design matrix the relation between the OLS and ridge estimator is: +\begin{align*} +\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\, \, \, = \, \, \, (\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (1 + \lambda)^{-1} \mathbf{I}_{pp} \mathbf{X}^{\top} \mathbf{Y} +\qquad \, \, = \, \, \, (1 + \lambda)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (1 + \lambda)^{-1} \hat{\bbeta}. +\end{align*} +Hence, the ridge estimator scales the OLS estimator by a factor. When taking the expectation on both sides, it is evident that the ridge estimator converges to zero as $\lambda \rightarrow \infty$. +\end{example} + + + +===== Variance ===== + +As for the ML estimate of the regression parameter $\bbeta$ of model (ref{form.linRegressionModelinMatrix}), we derive the second moment of the ridge estimator. Hereto define: +\begin{align*} +\mathbf{W}_{\lambda} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X}. +\end{align*} +Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\bbeta}(\lambda)$ can be expressed as $\mathbf{W}_{\lambda} \hat{\bbeta}$ for: +\begin{align*} +\mathbf{W}_{\lambda} \hat{\bbeta} & = \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ] \}^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \hat{\bbeta}(\lambda). +\end{align*} +The linear operator $\mathbf{W}_{\lambda}$ thus transforms the ML estimator of the regression parameter into the ridge estimator. + +It is now easily seen that: +\begin{align*} +\mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \mbox{Var}[ \mathbf{W}_{\lambda} \hat{\bbeta} ] \qquad \qquad \, \, \, \, \, \, = \, \, \, \mathbf{W}_{\lambda} \mbox{Var}[\hat{\bbeta} ] \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}, +\end{align*} +in which we have used $\mbox{Var}(\mathbf{A} \mathbf{Y}) = \mathbf{A} \mbox{Var}( \mathbf{Y}) \mathbf{A}^{\top}$ for a non-random matrix $\mathbf{A}$, the fact that $\mathbf{W}_{\lambda}$ is non-random, and $ \mbox{Var}[\hat{\bbeta} ] = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}$. + + +Like the expectation the variance of the ridge estimator vanishes as $\lambda$ tends to infinity: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mbox{Var} \big[ \hat{\bbeta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \mathbf{0}_{pp}. +\end{align*} +Hence, the variance of the ridge regression coefficient estimates decreases towards zero as the penalty parameter becomes large. This is illustrated in the right panel of Figure ref{fig.ridgeSolPathPlusVar} for the data of Example ref{example.supercollinearity}. + + +With an explicit expression of the variance of the ridge estimator at hand, we can compare it to that of the OLS estimator: +\begin{align*} +\mbox{Var}[ \hat{\bbeta} ] - \mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \sigma^2 [(\mathbf{X}^{\top} \mathbf{X})^{-1} - \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] +\\ +& = \sigma^2 \mathbf{W}_{\lambda} \{ [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ] (\mathbf{X}^{\top} \mathbf{X})^{-1} [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ]^{\top} - (\mathbf{X}^{\top} \mathbf{X})^{-1} \} \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}. +\end{align*} +The difference is non-negative definite as each component in the matrix product is non-negative definite. Hence, the variance of the ML estimator exceeds (in the positive definite ordering) that of the ridge estimator: + +!bt + +!bt +\begin{align} label{form.VarInequalityMLandRidge} +\mbox{Var}[ \hat{\bbeta} ] & \succeq & \mbox{Var}[ \hat{\bbeta}(\lambda) ], +\end{align} +!et +!et +with the inequality being strict if $\lambda > 0$. In other words, the variance of the ML estimator is larger than that of the ridge estimator (in the sense that their difference is non-negative definite). The variance inequality (ref{form.VarInequalityMLandRidge}) can be interpreted in terms of the stochastic behaviour of the estimator. This is illustrated by the next example. + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [varMLellipses.eps, width=400 frac=1.0] Level sets of the distribution of the ML (left panel) and ridge (right panel) regression estimators.} label{fig.varOLSandRidge +# \afterpage{} + + +\begin{example} *Variance comparison* +\\ +Consider the design matrix: +\begin{align*} +\mathbf{X} & = \left( +\begin{array}{rr} +-1 & 2 +\\ +0 & 1 +\\ +2 & -1 +\\ +1 & 0 +\end{array} \right). +\end{align*} +The variances of the ML and ridge (with $\lambda=1$) estimates of the regression coefficients then are: +\begin{align*} +\mbox{Var}(\hat{\bbeta}) & = \sigma^2 \left( +\begin{array}{rr} +0.3 & 0.2 +\\ +0.2 & 0.3 +\end{array} \right) +\qquad \mbox{and} \qquad +\mbox{Var}[\hat{\bbeta}(\lambda)] \, \, \, = \, \, \, \sigma^2 \left( +\begin{array}{rr} +0.1524 & 0.0698 +\\ +0.0698 & 0.1524 +\end{array} \right). +\end{align*} +These variances can be used to construct levels sets of the distribution of the estimates. The level sets that contain 50\%, 75\% and 95\% of the distribution of the ML and ridge estimates are plotted in Figure ref{fig.varOLSandRidge}. In line with inequality (ref{form.VarInequalityMLandRidge}) the level sets of the ridge estimate are smaller than that of the ML estimate: it thus varies less. +\end{example} + + + + +\begin{contexample}_ref{example.orthoronormalDesign_} *Orthonormal design matrix (continued)* +\\ +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{Var}[ \hat{\bbeta} ] = \sigma^2 \mathbf{I}_{pp}$ and +\begin{align*} +\mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} +\, \, \, = \, \, \, \sigma^2 [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{I}_{pp} \{ [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \}^{\top} +\, \, \, = \, \, \, \sigma^2 (1 + \lambda )^{-2} \mathbf{I}_{pp} . +\end{align*} +As the penalty parameter $\lambda$ is non-negative the former exceeds the latter. In particular, this expression vanishes as $\lambda \rightarrow \infty$. +\end{contexample} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +The full distribution of the ridge regression estimator is now known. The estimator, $\hat{\bbeta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ is a linear estimator, linear in $\mathbf{Y}$. As $\mathbf{Y}$ is normally distributed, so is $\hat{\bbeta}(\lambda)$. Moreover, the normal distribution is fully characterized by its first two moments, which are available. Hence: +\begin{align*} +\hat{\bbeta}(\lambda) & \sim & \mathcal{N} \big( (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X} \, \bbeta, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top} \big). +\end{align*} +Given $\lambda$ and $\bbeta$, the random behavior of the estimator is thus known. + + + +===== Mean squared error} label{sect:ridgeMSE ===== +Previously, we motivated the ridge estimator as an ad hoc solution to collinearity. An alternative motivation comes from studying the Mean Squared Error (MSE) of the ridge regression estimator: for a suitable choice of $\lambda$ the ridge regression estimator may outperform the ML regression estimator in terms of the MSE. Before we prove this, we first derive the MSE of the ridge estimator and quote some auxiliary results. + +Recall that (in general) for any estimator of a parameter $\theta$: +\begin{align*} +\mbox{MSE}( \hat{\theta} ) & = \mathbb{E} [ ( \hat{\theta} - \theta)^2 ] +\, \, \, = \, \, \, \mbox{Var}( \hat{ \theta} ) + [\mbox{Bias} ( \hat{\theta} )]^2. +\end{align*} +Hence, the MSE is a measure of the quality of the estimator. + +The MSE of the ridge estimator is: + +!bt + +!bt +\begin{align} +\mbox{MSE}[\hat{\bbeta}(\lambda)] & = \mathbb{E} [ (\mathbf{W}_{\lambda} \, \hat{\bbeta} - \bbeta)^{\top} \, (\mathbf{W}_{\lambda} \, \hat{\bbeta} - \bbeta) ] \nonumber +\\ +& = \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta) + \mathbb{E} ( \bbeta^{\top} \bbeta) \nonumber +\\ +& = \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta) + \mathbb{E} ( \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta ) \nonumber +\\ +& & - \mathbb{E} ( \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta ) + \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\bbeta}) + \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta) \nonumber +\\ +& & - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta) + \mathbb{E} ( \bbeta^{\top} \bbeta) \nonumber +\\ +& = \mathbb{E} [ ( \hat{\bbeta} - \bbeta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\bbeta} - \bbeta) ] \nonumber +\\ +& & - \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta + \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \bbeta + \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta \nonumber +\\ +& & - \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \bbeta - \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta + \bbeta^{\top} \bbeta \nonumber +\\ +& = \mathbb{E} \big\{ ( \hat{\bbeta} - \bbeta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\bbeta} - \bbeta) \big\} + \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta \nonumber +\\ +& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta. label{form.ridgeMSE} +\end{align} +!et +!et +In the last step we have used $\hat{\bbeta} \sim \mathcal{N}( \bbeta, \sigma^2 \, [\mathbf{X}^{\top} \mathbf{X}]^{-1} )$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ for some nonrandom symmetric positive definite matrix $\LLambda$ is (cf. \citealt{Math1992}): +\begin{align*} +\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) & = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}, +\end{align*} +of course replacing $\vvarepsilon$ by $\hat{\bbeta}$ in this expectation. The first summand in the final derived expression for $\mbox{MSE}[\hat{\bbeta}(\lambda)]$ is the sum of the variances of the ridge estimator, while the second summand can be thought of the ``squared bias'' of the ridge estimator. In particular, $\lim_{\lambda \rightarrow \infty} \mbox{MSE}[\hat{\bbeta}(\lambda)] = \bbeta^{\top} \bbeta$, which is the squared biased for an estimator that equals zero (as does the ridge estimator in the limit). + + + +\begin{example} *Orthonormal design matrix* +\\ +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{MSE}[ \hat{\bbeta} ] = p \, \sigma^2$ and +\begin{align*} +\mbox{MSE}[ \hat{\bbeta}(\lambda) ] & = \frac{p \, \sigma^2}{ (1+ \lambda)^{2}} + \frac{\lambda^2}{ (1+ \lambda)^{2}} \bbeta^{\top} \bbeta. +\end{align*} +The latter achieves its minimum at: $\lambda = p \sigma^2 / \bbeta^{\top} \bbeta$. +\end{example} + + + +The following theorem and proposition are required for the proof of the main result. + +\begin{theorem} *(Theorem 1 of \citealp{Theo1974*}) label{theo.Theobald1} +\\ +Let $\hat{\ttheta}_1$ and $\hat{\ttheta}_2$ be (different) estimators of $\ttheta$ with second order moments: +\begin{align*} +\mathbf{M}_k & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta) (\hat{\ttheta}_k - \ttheta)^{\top} ] \qquad \mbox{for } k=1,2, +\end{align*} +and +\begin{align*} +\mbox{MSE}(\hat{\ttheta}_k) & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta)^{\top} \mathbf{A} (\hat{\ttheta}_k - \ttheta) ] \qquad \mbox{for } k=1,2, +\end{align*} +where $\mathbf{A} \succeq 0$. Then, $\mathbf{M}_1 - \mathbf{M}_2 \succeq 0$ if and only if $\mbox{MSE}(\hat{\ttheta}_1) - \mbox{MSE}(\hat{\ttheta}_2) \geq 0$ for all $\mathbf{A} \succeq 0$. +\end{theorem} + + +\begin{proposition} *(\citealp{Fare1976*}) label{prop.Farebrother} +\\ +Let $\mathbf{A}$ be a $(p \times p)$-dimensional, positive definite matrix, $\mathbf{b}$ be a nonzero $p$ dimensional vector, and $c \in \mathbb{R}_+$. Then, $c \mathbf{A} - \mathbf{b} \mathbf{b}^{\top} \succ 0$ if and only if $\mathbf{b}^{\top} \mathbf{A}^{-1} \mathbf{b} > c$. +\end{proposition} + +We are now ready to proof the main result, formalized as Theorem ref{theo.Theobald2}, that for some $\lambda$ the ridge regression estimator yields a lower MSE than the ML regression estimator. + +\begin{theorem} *(Theorem 2 of \citealp{Theo1974*}) label{theo.Theobald2} +\\ +There exists $\lambda > 0$ such that $\mbox{MSE}[\hat{\bbeta}(\lambda)] < \mbox{MSE}[\hat{\bbeta}(0)] = \mbox{MSE}[\hat{\bbeta}]$. +\end{theorem} + +\begin{proof} +The second order moment matrix of the ridge estimator is: +\begin{align*} +\mathbf{M} (\lambda) & := & \mathbb{E} [ (\hat{\bbeta}(\lambda) - \bbeta) (\hat{\bbeta} (\lambda) - \bbeta)^{\top} ] +\\ +& = \mathbb{E} \{ \hat{\bbeta}(\lambda) [\hat{\bbeta}(\lambda)]^{\top} \} - \mathbb{E} [ \hat{\bbeta}(\lambda) ] \{ \mathbb{E} [ \hat{\bbeta}(\lambda) ] \}^{\top} + \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \{ \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \}^{\top} +\\ +& = \mbox{Var}[ \hat{\bbeta}(\lambda) ] + \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \{ \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \}^{\top}. +\end{align*} +Then: +\begin{align*} +\mathbf{M} ( 0 ) - \mathbf{M}(\lambda) & = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] +\\ +& & - (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \bbeta \bbeta^{\top} (\mathbf{W}_{\lambda} -\mathbf{I}_{pp})^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} +\\ +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \bbeta \bbeta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \bbeta \bbeta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& = \lambda [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \bbeta \bbeta^{\top} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top}. +\end{align*} +This is positive definite if and only if $ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \bbeta \bbeta^{\top} \succ 0$. Hereto it suffices to show that $2 \, \sigma^2 \, \mathbf{I}_{pp} - \lambda \bbeta \bbeta^{\top} \succ 0$. By Proposition ref{prop.Farebrother} this holds for $\lambda$ such that $2 \sigma^2 (\bbeta^{\top} \bbeta)^{-1} > \lambda$. For these $\lambda$, we thus have $\mathbf{M} ( 0 ) - \mathbf{M}(\lambda)$. Application of Theorem ref{theo.Theobald1} now concludes the proof. +\end{proof} +This result of cite{Theo1974} is generalized by cite{Fare1976} to the class of design matrices $\mathbf{X}$ with $\mbox{rank}(\mathbf{X}) < p$. +\\ +\\ +Theorem ref{theo.Theobald2} can be used to illustrate that the ridge regression estimator strikes a balance between the bias and variance. This is illustrated in the left panel of Figure ref{fig.MSEridge}. For small $\lambda$, the variance of the ridge estimator dominates the MSE. This may be understood when realizing that in this domain of $\lambda$ the ridge estimator is close to the unbiased ML regression estimator. For large $\lambda$, the variance vanishes and the bias dominates the MSE. For small enough values of $\lambda$, the decrease in variance of the ridge regression estimator exceeds the increase in its bias. As the MSE is the sum of these two, the MSE first decreases as $\lambda$ moves away from zero. In particular, as $\lambda = 0$ corresponds to the ML regression estimator, the ridge regression estimator yields a lower MSE for these values of $\lambda$. In the right panel of Figure ref{fig.MSEridge} $\mbox{MSE}[ \hat{\bbeta}(\lambda)] < \mbox{MSE}[ \hat{\bbeta}(0)]$ for $\lambda < 7$ (roughly) and the ridge estimator outperforms the ML estimator. +# original latex figure with scale=0.45, angle=0 + +FIGURE: [MSE_ridge.eps, width=400 frac=1.0] Left panel: mean squared error, and its `bias' and `variance' parts, of the ridge regression estimator (for artificial data). Right panel: mean squared error of the ridge and ML estimator of the regression coefficient vector (for the same artificial data).} label{fig.MSEridge +# \afterpage{} +\\ +\\ +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +Besides another motivation behind the ridge regression estimator, the use of Theorem ref{theo.Theobald2} is limited. The optimal choice of $\lambda$ depends on the quantities $\bbeta$ and $\sigma^2$. These are unknown in practice. Then, the penalty parameter is chosen in a data-driven fashion by means of cross-validation (see Section ref{subsect.crossvalidation}). +\\ +\\ +\begin{remark} \mbox{ } +\\ +Theorem ref{theo.Theobald2} can also be used to conclude on the biasedness of the ridge regression estimator. The Gauss-Markov theorem \citep{Rao1973} states (under some assumptions) that the ML regression estimator is the best linear unbiased estimator (BLUE) with the smallest MSE. As the ridge regression estimator is a linear estimator and outperforms (in terms of MSE) this ML estimator, it must be biased (for it would otherwise refute the Gauss-Markov theorem). +\end{remark} + + + + +======= Constrained estimation} label{sect.constrainedEstimation ======= +The ad-hoc fix of cite{Hoer1970} to super-collinearity of the design matrix (and, consequently the singularity of the matrix $\mathbf{X}^{\top} \mathbf{X}$) has been motivated post-hoc. +The ridge estimator minimizes the *ridge loss function*, which is defined as: + +!bt + +!bt +\begin{align} +\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \lambda \| \bbeta \|^2_2 +\, \, \, = \, \, \, \sum_{i=1}^n (Y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 + \lambda \sum_{j=1}^p \beta_j^2. label{form.ridgeLossFunction} +\end{align} +!et +!et +This loss function is the traditional sum-of-squares augmented with a *penalty*. The particular form of the penalty, $\lambda \| \bbeta \|^2_2$ is referred to as the *ridge penalty* and $\lambda$ as the *penalty parameter*. For $\lambda=0$, minimization of the ridge loss function yields the ML estimator. For any $\lambda > 0$, the ridge penalty contributes to the loss function, affecting its minimum and its location. The minimum of the sum-of-squares is well-known. The minimum of the ridge penalty is attained at $\bbeta = \mathbf{0}_{p}$ whenever $\lambda > 0$. The $\bbeta$ that minimizes $\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda)$ then balances the sum-of-squares and the penalty. The effect of the penalty in this balancing act is to shrink the regression coefficients towards zero, its minimum. In particular, the larger $\lambda$, the larger the contribution of the penalty to the loss function, the stronger the tendency to shrink non-zero regression coefficients to zero (and decrease the contribution of the penalty to the loss function). This motivates the name `penalty' as non-zero elements of $\bbeta$ increase (or penalize) the loss function. + + +To verify that the ridge estimator indeed minimizes the ridge loss function, proceed as usual. Take the derivative with respect to $\bbeta$: +\begin{align*} +\frac{\partial}{\partial \bbeta} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = -2 \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \bbeta) + 2 \, \lambda \, \mathbf{I}_{pp} \, \bbeta \, \, \, = \, \, \, -2 \, \mathbf{X}^{\top} \mathbf{Y} + 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}) \bbeta. +\end{align*} +Equate the derivative to zero and solve for $\bbeta$. This yields the ridge regression estimator. + +The ridge estimator is thus a stationary point of the ridge loss function. A stationary point corresponds to a minimum if the Hessian matrix with second order partial derivatives is positive definite. The Hessian of the ridge loss function is +\begin{align*} +\frac{\partial^2}{\partial \bbeta \, \partial \bbeta^{\top}} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}). +\end{align*} +This Hessian is the sum of the semi-positive definite matrix $\mathbf{X}^{\top} \mathbf{X}$ and the positive definite matrix $\lambda \, \mathbf{I}_{pp}$. Lemma 14.2.4 of cite{Harv2008} then states that the sum of these matrices is itself a positive definite matrix. Hence, the Hessian is positive definite and the ridge loss function has a stationary point at the ridge estimator, which is a minimum. + +The ridge regression estimator minimizes the ridge loss function. It rests to verify that it is a global minimum. To this end we introduce the concept of a convex function. As a prerequisite, a set $\mathcal{S} \subset \mathbb{R}^p$ is called *convex* if for all $\bbeta_1, \bbeta_2 \in \mathcal{S}$ their weighted average $\bbeta_{\theta} = (1 - \theta) \bbeta_1 + \theta \bbeta_2$ for all $\theta \in [0, 1]$ is itself an element of $\mathcal{S}$, thus $\bbeta_{\theta} \in \mathcal{S}$. If for all $\theta \in (0, 1)$, the weighted average $\bbeta_{\theta}$ is inside $\mathcal{S}$ and not on its boundary, the set is called *strict convex*. Examples of (strict) convex and nonconvex sets are depicted in Figure ref{fig.ridgeAsConstrainedEst}. A function $f(\cdot)$ is *(strict) convex* if the set $\{ y \, : \, y \geq f(\bbeta) \mbox{ for all } \bbeta \in \mathcal{S} \mbox{ for any convex } \mathcal{S} \}$, called the epigraph of $f(\cdot)$, is (strict) convex. Examples of (strict) convex and nonconvex functions are depicted in Figure ref{fig.ridgeAsConstrainedEst}. The ridge loss function is the sum of two parabola's: one at least convex and the other a strict convex function in $\bbeta$. The sum of convex and strict convex function is itself strict convex (confer Lemma 9.4.2 of \citealt{Flet2008}). The ridge loss function is thus strict convex. Theorem 9.4.1 of \citealt{Flet2008} then warrants, by the strict convexity of the ridge loss function, that the ridge estimator is a global minimum. +\\ +\\ +From the ridge loss function the limiting behavior of the variance of the ridge regression estimator can be understood. The ridge penalty with its minimum $\bbeta = \mathbf{0}_{p}$ does not involve data and, consequently, the variance of its minimum equals zero. With the ridge regression being a compromise between the ML estimator and the minimum of the penalty, so is its variance a compromise of their variances. As $\lambda$ tends to infinity, the ridge estimator and its variance converge to the minimum and the variance of the minimum, respectively. Hence, in the limit (large $\lambda$) the variance of the ridge regression estimator vanishes. Understandably, as the penalty now fully dominates the loss function and, consequently, it does no longer involve data (i.e. randomness). +# original latex figure with scale=0.38, angle=0 + +FIGURE: [convexSets.eps, width=400 frac=1.0] Top panels show examples of convex (left) and nonconvex (right) sets. Middle panels show examples of convex (left) and nonconvex (right) functions. The left bottom panel illustrates the ridge estimation as a constrained estimation problem. The ellipses represent the contours of the ML loss function, with the blue dot at the center the ML estimate. The circle is the ridge parameter constraint. The red dot is the ridge estimate. It is at the intersection of the ridge constraint and the smallest contour with a non-empty intersection with the constraint. The right bottom panel shows the data corresponding to Example ref{exam.overfitting}. The grey line represents the `true' relationship, while the black line the fitted one. } label{fig.ridgeAsConstrainedEst +\afterpage{} +\\ +\\ +Above it has been shown that the ridge estimator can be defined as: + +!bt + +!bt +\begin{align} label{form.ridgeEstViaPenEst} +\hat{\bbeta}(\lambda) & = \arg \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \lambda \| \bbeta \|^2_2. +\end{align} +!et +!et +This minimization problem can be reformulated into the following constrained optimization problem (illustrated in Figure ref{fig.ridgeAsConstrainedEst}): + +!bt + +!bt +\begin{align} label{form.constrEstProblemRidge} +\hat{\bbeta}(\lambda) & = \arg \min_{\| \bbeta \|_2^2 \leq c} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2, +\end{align} +!et +!et +for some suitable $c > 0$. The constrained optimization problem (ref{form.constrEstProblemRidge}) can be solved by means of the Karush-Kuhn-Tucker (KKT) multiplier method, which minimizes a function subject to inequality constraints. The KKT multiplier method states that, under some regularity conditions (all met here), there exists a constant $\nu \geq 0$, called the *multiplier*, such that the solution $\hat{\bbeta}(\nu)$ of the constrained minimization problem (ref{form.constrEstProblemRidge}) satisfies the so-called KKT conditions. The first KKT condition (referred to as the stationarity condition) demands that the gradient (with respect to $\bbeta$) of the Lagrangian associated with the minimization problem equals zero at the solution $\hat{\bbeta}(\nu)$. The Lagrangian for problem (ref{form.constrEstProblemRidge}) is: +\begin{align*} +\| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \nu ( \| \bbeta \|^2_2 - c). +\end{align*} +The second KKT condition (the complementarity condition) requires that $\nu (\| \hat{\bbeta}(\nu) \|_2^2 - c) = 0$. If $\nu = \lambda$ and $c = \| \hat{\bbeta}(\lambda) \|_2^2$, the ridge estimator $\bbeta (\lambda)$ satisfies both KKT conditions. Hence, both problems have the same solution when $c = \| \hat{\bbeta}(\lambda) \|_2^2$. +\\ +\\ +The relevance of viewing the ridge regression estimator as the solution to a constrained estimation problem becomes obvious when considering a typical threat to high-dimensional data analysis: overfitting. *Overfitting* refers to the phenomenon of modelling the noise rather than the signal. In case the true model is parsimonious (few covariates driving the response) and data on many covariates are available, it is likely that a linear combination of all covariates yields a higher likelihood than a combination of the few that are actually related to the response. As only the few covariates related to the response contain the signal, the model involving all covariates then cannot but explain more than the signal alone: it also models the error. Hence, it overfits the data. In high-dimensional settings overfitting is a real threat. The number of explanatory variables exceeds the number of observations. It is thus possible to form a linear combination of the covariates that perfectly explains the response, including the noise. + +Large estimates of regression coefficients are often an indication of overfitting. Augmentation of the estimation procedure with a constraint on the regression coefficients is a simple remedy to large parameter estimates. As a consequence it decreases the probability of overfitting. Overfitting is illustrated in the next example. + + +\begin{example} *(Overfitting)* label{exam.overfitting} +\\ +Consider an artificial data set comprising of ten observations on a response $Y_i$ and nine covariates $X_{i,j}$. All covariate data are sampled from the standard normal distribution: $X_{i,j} \sim \mathcal{N}(0, 1)$. The response is generated by $Y_i = X_{i,1} + \varepsilon_i$ with $\varepsilon_{i} \sim \mathcal{N}(0, 1/4)$. Hence, only the first covariate contributes to the response. + +The regression model $Y_i = \sum_{j=1}^9 X_{i,j} \beta_j+ \varepsilon_i$ is fitted to the artificial data using `R`. This yields the regression parameter estimates: +\begin{align*} +\hat{\bbeta}^{\top} & = (0.048, -2.386, -5.528, 6.243, -4.819, 0.760, -3.345, -4.748, 2.136). +\end{align*} +As $\bbeta^{\top} = (1, 0, \ldots, 0)$, many regression coefficient are clearly over-estimated. + +The fitted values $\widehat{Y}_i = \mathbf{X}_i \hat{\bbeta}$ are plotted against the values of the first covariates in the right bottom panel of Figure ref{fig.ridgeAsConstrainedEst}. As a reference the line $x=y$ is added, which represents the `true' model. The fitted model follows the `true' relationship. But it also captures the deviations from this line that represent the errors. +\end{example} + + + +======= Bayesian regression} label{sect:Bayes ======= +Ridge regression has a close connection to Bayesian linear regression. Bayesian linear regression assumes the parameters $\bbeta$ and $\sigma^2$ to be the random variables, while at the same time considering $\mathbf{X}$ and $\mathbf{Y}$ as fixed. Within the regression context, the conjugate priors of $\bbeta$ and $\sigma^2$ are: +\begin{align*} +\bbeta \, | \, \sigma^2 \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp}) \qquad \mbox{and} \qquad \sigma^2 \sim \mathcal{IG}(\alpha_0, \beta_0), +\end{align*} +where $\mathcal{IG}$ denotes the inverse Gamma distribution with shape parameter $\alpha_0$ and scale parameter $\beta_0$. The penalty parameter can be interpreted as the precision of the prior, determining how informative the prior should be. A smaller penalty (i.e. precision) corresponds to a wider prior, and a larger penalty to a more informative, concentrated prior (Figure ref{fig.ridgePriorOfBeta}). +# original latex figure with scale=0.45, angle=0 + +FIGURE: [ridgePriorOfBeta.eps, width=400 frac=1.0] Conjugate prior of the regression parameter $\bbeta$ for various choices of $\lambda$, the penalty parameters c.q. precision.} label{fig.ridgePriorOfBeta + +Under the assumption of the conjugate priors above, the joint posterior distribution of $\bbeta$ and $\sigma^2$ is then: +\begin{align*} +f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \bbeta, \sigma^2) \, f_{\beta}(\bbeta | \sigma^2) \, f_{\sigma}(\sigma^2) +\\ +& \propto & \sigma^{-n} \exp \Big[ - \frac{1}{2\sigma^2} ( \mathbf{Y} - \mathbf{X} \bbeta)^{\top} ( \mathbf{Y} - \mathbf{X} \bbeta) \Big] +\\ +& & \times \, \, \sigma^{-p} \exp \Big[ - \frac{1}{2\sigma^2} \lambda \bbeta^{\top} \bbeta \Big] \, \times \, \, [\sigma^2]^{-\alpha_0-1} \exp \Big[ - \frac{\beta_0}{2\sigma^2} \Big]. +\end{align*} +As +\begin{align*} +& & \hspace{-1.5cm} ( \mathbf{Y} - \mathbf{X} \bbeta)^{\top} ( \mathbf{Y} - \mathbf{X} \bbeta) + \lambda \bbeta^{\top} \bbeta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - +\bbeta^{\top} \mathbf{X}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} \bbeta + \bbeta^{\top} \mathbf{X}^{\top} \mathbf{X} \bbeta + \lambda \bbeta^{\top} \bbeta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& & - \, \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta + \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) +\hat{\bbeta} (\lambda) +\\ +& & - \, [ \hat{\bbeta} (\lambda) ]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta + \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& & + \, \big[ \bbeta - \hat{\bbeta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \bbeta - \hat{\bbeta}(\lambda) \big], +\end{align*} +the posterior distribution can be rewritten to: +\begin{align*} +f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +& \propto & g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +\end{align*} +with +\begin{align*} +g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \bbeta - \hat{\bbeta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \bbeta - \hat{\bbeta}(\lambda) \big] \Big\}. +\end{align*} +Then, clearly the conditional posterior mean of $\bbeta$ is $\mathbb{E}(\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\bbeta}(\lambda)$. Hence, the ridge regression estimator can be viewed as the Bayesian posterior mean estimator of $\bbeta$ when imposing a Gaussian prior on the regression parameter. + +With little extra work we may also obtain the conditional posterior of $\sigma^2$ from the joint posterior distribution: +\begin{align*} +f_{\sigma^2} (\sigma^2 \, | \, \bbeta, \mathbf{Y}, \mathbf{X}) & \propto & (\sigma^2)^{-[(n+p)/2 + \alpha_0 + 1]} +\exp [ - \frac{1}{2\sigma^2} ( \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + +\lambda \| \bbeta \|_2^2 + \beta_0) ], +\end{align*} +in which one can recognize the shape of an inverse gamma distribution. +\\ +\\ +A Bayesian estimator of a parameter $\ttheta$ is the estimator that minimizes the Bayes risk over a prior distribution of the parameter $\ttheta$. The Bayes risk is defined as $\int_{\ttheta} \mathbb{E} [(\hat{\ttheta} - \ttheta)^{\top} (\hat{\ttheta} - \ttheta)] \pi_{\ttheta}(\ttheta; \aalpha) d\ttheta$, where $\pi_{\ttheta}(\ttheta; \aalpha)$ is the prior distribution of $\ttheta$ with hyperparameter $\aalpha$. It is thus a weighted average of the Mean Squared Error, with weights specified through the prior. The Bayes risk is minimized by the mean posterior $\mathbb{E}_{\ttheta}(\ttheta \, | \, \mbox{data})$ (cf., e.g., \citealp{Bijm2017}). The Bayesian estimator of $\ttheta$ thus yields the smallest possible expected MSE, under the assumption of the employed prior. + +The Bayes risk of the ridge estimator over the normal prior $\bbeta \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp})$ is: +\begin{align*} +\mathbb{E}_{\bbeta} \{\mbox{MSE}[\hat{\bbeta}(\lambda)] +\, | \, \sigma^2, \mathbf{Y}, \mathbf{X} \} & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \mathbb{E}_{\bbeta} [ \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta] +\\ +& = \sigma^2 \, \big\{ \mbox{tr} \big[ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big] + +\lambda^{-1} \mbox{tr} [(\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})] \big\} +\\ +& = \sigma^2 \sum_{j=1}^p (d_{jj}^2 + \lambda)^{-1}, +\end{align*} +in which we have used *i)* the previously derived explicit expression (ref{form.ridgeMSE}) of the ridge estimator's MSE, *ii)* the expectation of the quadratic form of a multivariate random variable \citep{Math1992}, *iii)* the singular value decomposition of $\mathbf{X}$ with singular values $d_{jj}$, and *iv)* the fact that the trace of a square matrix equals the sum of its eigenvalues. As the ridge estimator coincides with the posterior mean, this is the minimal achievable MSE under a zero-centered normal prior with an uncorrelated and equivariant covariance matrix. + +Above the Bayes risk of the ridge estimator factorizes with respect to $\sigma^2$ and $\lambda$. Hence, the larger the hyperparameter $\lambda$ the lower the Bayes risk of the ridge estimator. In particular, its Bayes risk converges to zero as $\lambda \rightarrow \infty$. This can be understood as follows. The limit corresponds to an infinite precision of the prior, thus reducing the variance contribution to the MSE. Moreover, as the ridge estimator shrinks towards zero and the prior distribution of $\bbeta$ has a zero mean, the bias too vanishes as $\lambda \rightarrow \infty$. + +The calculation of the Bayes risk above relates the Bayesian and frequentist statements on the MSE of the ridge estimator. For the latter revisit Theorem ref{theo.Theobald2} of Section ref{sect:ridgeMSE}, which states the existence of a $\lambda$ such that the resulting ridge estimator has a superior MSE over that of the ML estimator. This result made no assumption on (the distribution of) $\bbeta$. In fact, it can be viewed as a statement of the MSE conditional on $\bbeta$. The Bayesian result integrates out the uncertainty - specified by the prior - in $\bbeta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. + + + +# The Bayesian result states that the ridge regression estimator minimizes the weighted MSE (over a normal prior on the parameter $\bbeta$). It thus integrates out the uncertainty in $\bbeta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. + +======= Degrees of freedom ======= + +The degrees of freedom consumed by ridge regression is calculated. The degrees of freedom may be used in combination with an information criterion to decide on the value of the penalty parameter. Recall from ordinary regression that: +\begin{align*} +\widehat{\mathbf{Y}} & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, \mathbf{H} \mathbf{Y}, +\end{align*} +where $\mathbf{H}$ is the hat matrix. The degrees of freedom used in the regression is then equal to $\mbox{tr}(\mathbf{H})$, the trace of $\mathbf{H}$. In particular, if $ \mathbf{X}$ is of full rank, i.e. $\mbox{rank}(\mathbf{X}) = p$, then $\mbox{tr}(\mathbf{H}) = p$. + +By analogy, the ridge-version of the hat matrix is: +\begin{align*} +\mathbf{H}(\lambda) & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}. +\end{align*} +Continuing this analogy, the degrees of freedom of ridge regression is given by the trace of the ridge hat matrix $\mathbf{H}(\lambda)$: +\begin{align*} +\mbox{tr}[ \mathbf{H}(\lambda)] & = \mbox{tr}[ \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} ] \, \, \, = \, \, \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}. +\end{align*} +The degrees of freedom consumed by ridge regression is monotone decreasing in $\lambda$. In particular: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mbox{tr}[ \mathbf{H}(\lambda)] & = 0. +\end{align*} +That is, in the limit no information from $\mathbf{X}$ is used. Indeed, $\bbeta$ is forced to equal $\mathbf{0}_{p}$ which is not derived from data. + + + +======= Efficient calculation ======= + +In the high-dimensional setting the number of covariates $p$ is large compared to the number of samples $n$. In a microarray experiment $p = 40000$ and $n= 100$ is not uncommon. To perform ridge regression in this context, the following expression needs to be evaluated numerically: +\begin{align*} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\end{align*} +For $p=40000$ this requires the inversion of a $40000 \times 40000$ dimensional matrix. This is not feasible on most desktop computers. However, there is a workaround. + +Revisit the singular value decomposition of $\mathbf{X} = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}$ and write $\mathbf{R}_x = \mathbf{U}_x \mathbf{D}_x$. As both $\mathbf{U}_x$ and $\mathbf{D}_x$ are $(n \times n)$-dimensional matrices, so is $\mathbf{R}_x$. Consequently, $\mathbf{X}$ is now decomposed as $\mathbf{X} = \mathbf{R}_x \mathbf{V}_x^{\top}$. The ridge estimator can be rewritten in terms of $\mathbf{R}_x$ and $\mathbf{V}_x$: +\begin{align*} +\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{R}_x^{\top} \mathbf{Y}. +\end{align*} +Hence, the reformulated ridge estimator involves the inversion of an $(n \times n)$-dimensional matrix. With $n= 100$ this is feasible on most standard computers. + +cite{Hast2004} point out that the number of computation operations reduces from $\mathcal{O}(p^3)$ to $\mathcal{O}(p n^2)$. In addition, they point out that this computational short-cut can be used in combination with other loss functions, for instance that of standard generalized linear models. +\\ +\\ +Avoidance of the inversion of the $(p \times p)$-dimensional matrix may be achieved in an other way. Hereto one needs the Woodbury identity. Let $\mathbf{A}$, $\mathbf{U}$ and $\mathbf{V}$ be $(p \times p)$-, $(p \times n)$- and $(n \times p)$-dimensional matrices, respectively. +The (simplified form of the) Woodbury identity then is: +\begin{align*} +(\mathbf{A} + \mathbf{U} \mathbf{V})^{-1} & = \mathbf{A}^{-1} - \mathbf{A}^{-1} \mathbf{U} (\mathbf{I}_{nn} + \mathbf{V} \mathbf{A}^{-1} \mathbf{U})^{-1} \mathbf{V} \mathbf{A}^{-1}. +\end{align*} +Application of the Woodbury identity to the matrix inverse in the ridge estimator of the regression parameter gives: +\begin{align*} +(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} & = \lambda^{-1} \mathbf{I}_{pp} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X}. +\end{align*} +This gives: +\begin{align*} +(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} & = \lambda^{-1} \mathbf{X}^{\top} \mathbf{Y} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \lambda^{-1} \mathbf{X}^{\top} \left[ \mathbf{Y} - \lambda^{-1} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{Y} \right]. +\end{align*} +The inversion of the $(p \times p)$-dimensional matrix $\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X}$ is thus replaced by that of the $(n \times n)$-dimensional matrix $\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top}$. In addition, this expression of the ridge regression estimator avoids the singular value decomposition of $\mathbf{X}$, which may in some cases introduce additional numerical errors (e.g. at the level of machine precision). + + + +======= Choice of the penalty parameter ======= + +Throughout the introduction of ridge regression and the subsequent discussion of its properties the penalty parameter is considered known or `given'. In practice, it is unknown and the user needs to make an informed decision on its value. Several strategies to facilitate such a decision are presented. + +===== Information criterion ===== + +A popular strategy is to choose a penalty parameter that yields a good but parsimonious model. Information criteria measure the balance between model fit and model complexity. Here we present the Aikaike's information criterion (AIC), but many other criteria have been presented in the literature (e.g. \citealp{Akai1974}, \citealp{Schw1978}). The AIC measures model fit by the log-likelihood and model complexity is measured by the number of parameters used by the model. The number of model parameters in regular regression simply corresponds to the number of covariates in the model. Or, by the degrees of freedom consumed by the model, which is equivalent to the trace of the hat matrix. For ridge regression it thus seems natural to define model complexity +analogously by the trace of the ridge hat matrix. This yields the AIC for the linear regression model with ridge estimates: +\begin{align*} +\mbox{AIC}(\lambda) & = 2 \, p - 2 \log(\hat{L}) +\\ +& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\bbeta}(\lambda), \hat{\sigma}^2(\lambda)]\} +\\ +& = 2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda} ++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\bbeta}(\lambda)]^2. +\end{align*} +The value of $\lambda$ which minimizes $\mbox{AIC}(\lambda)$ corresponds to the `optimal' balance of model complexity and overfitting. + +Information criteria guide the decision process when having to decide among various different models. Different models use different sets of explanatory variables to explain the behaviour of the response variable. In that sense, the use of information criteria for the deciding on the ridge penalty parameter may be considered inappropriate: ridge regression uses the same set of explanatory variables irrespective of the value of the penalty parameter. Moreover, often ridge regression is employed to predict a response and not to provide an insightful explanatory model. The latter need not yield the best predictions. Finally, empirically we observe that the AIC often does not show an optimum *inside* the domain of the ridge penalty parameter. Henceforth, we refrain from the use of the AIC (or any of its relatives) in determining the optimal ridge penalty parameter. + + +===== Cross-validation} label{subsect.crossvalidation ===== +Instead of choosing the penalty parameter to balance model fit with model complexity, cross-validation requires it (i.e. the penalty parameter) to yield a model with good prediction performance. Commonly, this performance is evaluated on novel data. Novel data need not be easy to come by and one has to make do with the data at hand. The setting of `original' and novel data is then mimicked by sample splitting: the data set is divided into two (groups of samples). One of these two data sets, called the *training set*, plays the role of `original' data on which the model is built. The second of these data sets, called the *test set*, plays the role of the `novel' data and is used to evaluate the prediction performance (often operationalized as the log-likelihood or the prediction error) of the model built on the training data set. This procedure (model building and prediction evaluation on training and test set, respectively) is done for a collection of possible penalty parameter choices. The penalty parameter that yields the model with the best prediction performance is to be preferred. The thus obtained performance evaluation depends on the actual split of the data set. To remove this dependence the data set is split many times into a training and test set. For each split the model parameters are estimated for all choices of $\lambda$ using the training data and estimated parameters are evaluated on the corresponding test set. The penalty parameter that on average over the test sets performs best (in some sense) is then selected. + +When the repetitive splitting of the data set is done randomly, samples may accidently end up in a fast majority of the splits in either training or test set. Such samples may have an unbalanced influence on either model building or prediction evaluation. To avoid this $k$-fold cross-validation structures the data splitting. The samples are divided into $k$ more or less equally sized exhaustive and mutually exclusive subsets. In turn (at each split) one of these subsets plays the role of the test set while the union of the remaining subsets constitutes the training set. Such a splitting warrants a balanced representation of each sample in both training and test set over the splits. Still the division into the $k$ subsets involves a degree of randomness. This may be fully excluded when choosing $k=n$. This particular case is referred to as leave-one-out cross-validation (LOOCV). For illustration purposes the LOOCV procedure is detailed fully below: +\begin{compactitem} +\item[0)] Define a range of interest for the penalty parameter. + +\item[1)] Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively. + +\item[2)] Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set. This yields: +\begin{align*} +\hat{\bbeta}_{-i}(\lambda) & = ( \mathbf{X}_{-i, \ast}^{\top} +\mathbf{X}_{-i, \ast} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{-i, \ast}^{\top} \mathbf{Y}_{-i} +\end{align*} +and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$. + +\item[3)] Evaluate the prediction performance of these models on the test set by $\log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\bbeta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)|$, possibly squared. + +\item[4)] Repeat steps 1) to 3) such that each sample plays the role of the test set once. + +\item[5)] Average the prediction performances of the test sets at each grid point of the penalty parameter: +\begin{align*} +\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\bbeta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\end{align*} +The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. + +\item[6)] The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. +\end{compactitem} +The procedure is straightforwardly adopted to $k$-fold cross-validation, a different criterion, and different estimators. + +In the LOOCV procedure above resampling can be avoided when the prediction performance is measured by Allen's PRESS (Predicted Residual Error Sum of Squares) statistic \citep{Alle1974}. For then, the LOOCV prediction performance can be expressed analytically in terms of the known quantities derived from the design matrix and response (as pointed out but not detailed in \citealt{Golu1979}). Define the optimal penalty parameter to minimize Allen's PRESS statistic: +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)]^2. +\end{align*} +To derive an analytic expression for the right-hand side first +rewrite $(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1}$ by means of the Woodbury identity as: +\begin{align*} +(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} - \mathbf{X}_{i, \ast}^{\top} \mathbf{X}_{i, \ast})^{-1} +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}]^{-1} +\\ +& & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\end{align*} +with $\mathbf{H}_{ii}(\lambda) = \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. +Furthermore, $\mathbf{X}_{-i}^{\top} \mathbf{Y}_{-i} = \mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i$. Substitute both in the leave-one-out ridge regression estimator and manipulate: +\begin{align*} +\hat{\bbeta}_{- i}(\lambda) & = (\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{- i, \ast}^{\top} \mathbf{Y}_{- i} +\\ +& = \{(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}\} +\\ +& & \times (\mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i ) +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} Y_i +\\ +& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii} (\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}^{\top} \mathbf{Y} +\\ +& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} Y_i +\\ +& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i +\\ +& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) +\\ +& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{H}_{ii} (\lambda) Y_i +\\ +& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \{ [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{H}_{ii}(\lambda) Y_i \} +\\ +& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ]. +\end{align*} +The latter enables the reformulation of the prediction error as: +\begin{align*} +Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda) & = Y_i - \mathbf{X}_{i, \ast} \{ \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda)] \} +\\ +& = Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ] +\\ +& = Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{H}_{ii} (\lambda)[ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ] +\\ +& = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast}^{\top} \hat{\bbeta}(\lambda) ], +\end{align*} +which in turn results in the re-expression of Allen's PRESS statistic: +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} & = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)]^2 +\, \, \, = \, \, \, \arg \min_{\lambda} \tfrac{1}{n} \| \mathbf{B}(\lambda) [\mathbf{I}_{nn} - \mathbf{H}(\lambda)] \mathbf{Y} \|_ F^2, +\end{align*} +where $\mathbf{B}(\lambda)$ is diagonal with $[\mathbf{B}(\lambda)]_{ii} = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1}$. Hence, the prediction performance for a given $\lambda$ can be assessed directly from the ridge hat matrix and the response vector without the recalculation of the $n$ leave-one-out ridge estimators. Computationally, this is a considerable gain. + + + +======= Simulations ======= + +Simulations are presented that illustrate properties of the ridge estimator not discussed explicitly in the previous sections of this chapter. + +===== Role of the variance of the covariates} label{ridge:covariateVariances ===== +In many applications of high-dimensional data the covariates are standardized prior to the execution of the ridge regression. Before we discuss whether this is appropriate, we first illustrate the effect of ridge penalization on covariates with distinct variances using simulated data. + +The simulation involves one response to be (ridge) regressed on fifty covariates. Data (with $n=1000$) for the covariates, denoted $\mathbf{X}$, are drawn from a multivariate normal distribution: $\mathbf{X} \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{\Sigma})$ with $\mathbf{\Sigma}$ diagonal and $(\mathbf{\Sigma})_{jj} = j / 10$. From this the response is generated through $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\bbeta = \mathbf{1}_{50}$ and $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{I}_{50 \times 50})$. + +With the simulated data at hand the ridge regression estimates of $\bbeta$ are evaluated for a large grid of the penalty parameter $\lambda$. The resulting ridge regularization paths of the regression coefficients are plotted (Figure ref{fig.effectOfRidge_effectOfVariances}). All paths start ($\lambda=0$) close to one and vanish as $\lambda \rightarrow \infty$. However, ridge regularization paths of regression coefficients corresponding to covariates with a large variance dominate those with a low variance. + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [covariatesWithDifferentVariance.eps, width=400 frac=1.0] Top panel: Ridge regularization paths for coefficients of the 50 uncorrelated covariates with distinct variances. Color and line type indicated the grouping of the covariates by their variance. Bottom panels: Graphical illustration of the effect of a covariate's variance on the ridge estimator. The grey circle depicts the ridge parameter constraint. The dashed black ellipsoids are the level sets of the least squares loss function. The red dot is the ridge regression estimate. Left and right panels represent the cases with equal and unequal, respectively, variances of the covariates.} label{fig.effectOfRidge_effectOfVariances + + + + + +Ridge regression's preference of covariates with a large variance can intuitively be understood as follows. First note that the ridge regression estimator now can be written as: +\begin{align*} +\bbeta (\lambda) & = [ \mbox{Var}(\mathbf{X}) + \lambda \mathbf{I}_{50 \times 50}]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) +\\ +& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma [ \mbox{Var}(\mathbf{X}) ]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) +\\ +& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma \bbeta. +\end{align*} +Plug in the employed parametrization of $\mathbf{\Sigma}$, which gives: +\begin{align*} +[\bbeta (\lambda)]_j & = \frac{j}{j + 50 \lambda} \, (\bbeta)_j. +\end{align*} +Hence, the larger the covariate's variance (corresponding to the larger $j$), the larger its ridge regression coefficient estimate. Ridge regression thus prefers (among a set of covariates with comparable effect sizes) those with larger variances. + + + +The reformulation of ridge penalized estimation as a constrained estimation problem offers a geometrical interpretation of this phenomenon. Let $p=2$ and the design matrix $\mathbf{X}$ be orthogonal, while both covariates contribute equally to the response. Contrast the cases with $\mbox{Var}(X_1) \approx \mbox{Var}(X_2)$ and $\mbox{Var}(X_1) \gg \mbox{Var}(X_2)$. The level sets of the least squares loss function associated with the former case are circular, while that of the latter are strongly ellipsoidal (see Figure ref{fig.effectOfRidge_effectOfVariances}). The diameters along the principal axes (that - due to the orthogonality of $\mathbf{X}$ - are parallel to that of the $\beta_1$- and $\beta_2$-axes) of both circle and ellipsoid are reciprocals of the variance of the covariates. When the variances of both covariates are equal, the level sets of the loss function expand equally fast along both axis. With the two covariates having the same regression coefficient, the point of these level sets closest to the parameter constraint is to be found on the line $\beta_1 = \beta_2$ (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). Consequently, the ridge regression estimate satisfies $\hat{\beta}_1 (\lambda) \approx \hat{\beta}_2(\lambda)$. With unequal variances between the covariates, the ellipsoidal level sets of the loss function have diameters of rather different sizes. In particular, along the $\beta_1$-axis it is narrow (as $\mbox{Var}(X_1)$ is large), and - vice versa - wide along the $\beta_2$-axis. Consequently, the point of these level sets closest to the circular parameter constraint will be closer to the $\beta_1$- than to the $\beta_2$-axis (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). For the ridge estimates of the regression parameter this implies $0 \ll \hat{\beta}_1 (\lambda) < 1$ and $0 < \hat{\beta}_2 (\lambda) \ll 1$. Hence, the covariate with a larger variance yields the larger ridge regression estimate. + + +Should one thus standardize the covariates prior to ridge regression analysis? When dealing with gene expression data from microarrays, the data have been subjected to a series of pre-processing steps (e.g. quality control, background correction, within- and between-normalization). The purpose of these steps is to make the expression levels of genes comparable both within and between hybridizations. The preprocessing should thus be considered an inherent part of the measurement. As such it is to be done independently of whatever down-stream analysis is to follow and further tinkering with the data is preferably to be avoided (as it may mess up the `comparable-ness' of the expression levels as achieved by the preprocessing). For other data types different considerations may apply. + +Among the considerations to decide on standardization of the covariates, one should also include the fact that ridge estimates prior and posterior to scaling do not simply differ by a factor. To see this assume that the covariates have been centered. Scaling of the covariates amounts to post-multiplication of the design matrix by a $(p \times p)$-dimensional diagonal matrix $\mathbf{A}$ with the reciprocals of the covariates' scale estimates on its diagonal \citep{Sard2008}. Hence, the ridge estimator (for the rescaled data) is then given by: +\begin{align*} +\min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \mathbf{A} \bbeta \|_2^2 + \lambda \| \bbeta \|_ 2^2. +\end{align*} +Apply the change-of-variable $\ggamma = \mathbf{A} \bbeta$ and obtain: +\begin{align*} +\min_{\ggamma} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \lambda \| \mathbf{A}^{-1} \ggamma \|_ 2^2 & = \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \sum_{j=1}^p \lambda [(\mathbf{A})_{jj}]^{-2} \gamma_j^2. +\end{align*} +Effectively, the scaling is equivalent to covariate-wise penalization. The `scaled' ridge estimator may then be derived along the same lines as before in Section ref{sect.constrainedEstimation}: +\begin{align*} +\hat{\bbeta}^{\mbox{{\tiny (scaled)}}} (\lambda) & = \mathbf{A}^{-1} \hat{\ggamma} (\lambda) +\, \, \, = \, \, \, \mathbf{A}^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{A}^{-2})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\end{align*} +In general, this is unequal to the ridge estimator without the rescaling of the columns of the design matrix. Moreover, it should be clear that $\hat{\bbeta}^{\mbox{{\tiny (scaled)}}} (\lambda) \not= \mathbf{A} \hat{\bbeta}(\lambda)$. + +===== Ridge regression and collinearity} label{sect:collinearCovariates ===== +Initially, ridge regression was motivated as an ad-hoc fix of (super)-collinear covariates in order to obtain a well-defined estimator. We now study the effect of this ad-hoc fix on the regression coefficient estimates of collinear covariates. In particular, their ridge regularization paths are contrasted to those of `non-collinear' covariates. + +To this end, we consider a simulation in which one response is regressed on 50 covariates. The data of these covariates, stored in a design matrix denoted $\mathbf{X}$, are sampled from a multivariate normal distribution, with mean zero and a $5 \times 5$ blocked covariance matrix: +\begin{align*} +\mathbf{\Sigma} & = \left( +\begin{array}{ccccc} +\mathbf{\Sigma}_{11} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{22} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{\Sigma}_{33} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{44} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{55} +\end{array} +\right) +\end{align*} +with +\begin{align*} +\mathbf{\Sigma}_{kk} & = \frac{k-1}{5} \, \mathbf{1}_{10 \times 10} + \frac{6-k}{5} \, \mathbf{I}_{10 \times 10}. +\end{align*} +The data of the response variable $\mathbf{Y}$ are then obtained through: $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, with $\vvarepsilon \sim \mathcal{N}( \mathbf{0}_{n}, \mathbf{I}_{nn})$ and $\bbeta = \mathbf{1}_{50}$. Hence, all covariates contribute equally to the response. Would the columns of $\mathbf{X}$ be orthogonal, little difference in the ridge estimates of the regression coefficients is expected. + +The results of this simulation study with sample size $n=1000$ are presented in Figure ref{fig.ridgeEstimates_correlatedCovariates}. All 50 regularization paths start close to one as $\lambda$ is small and converge to zero as $\lambda \rightarrow \infty$. But the paths of covariates of the same block of the covariance matrix $\mathbf{\Sigma}$ quickly group, with those corresponding to a block with larger off-diagonal elements above those with smaller ones. Thus, ridge regression prefers (i.e. shrinks less) coefficient estimates of strongly positively correlated covariates. + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [correlatedCovariates.eps, width=400 frac=1.0] Left panel: Ridge regularization paths for coefficients of the 50 covariates, with various degree of collinearity but equal variance. Color and line type correspond to the five blocks of the covariate matrix $\mathbf{\Sigma}$. Right panel: Graphical illustration of the effect of the collinearity among covariates on the ridge estimator. The solid and dotted grey circles depict the ridge parameter constraint for the collinear and orthogonal cases, respectively. The dashed black ellipsoids are the level sets of the sum-of-squares squares loss function. The red dot and violet diamond are the ridge regression for the positive collinear and orthogonal case, respectively.} label{fig.ridgeEstimates_correlatedCovariates +# \afterpage{} + + + +Intuitive understanding of the observed behaviour may be obtained from the $p=2$ case. Let $U$, $V$ and $\varepsilon$ be independent random variables with zero mean. Define $X_1 = U + V$, $X_2 = U - V$, and $Y = \beta_1 X_1 + \beta_2 X_2 + \varepsilon$ with $\beta_1$ and $\beta_2$ constants. Hence, $\mathbb{E}(Y) = 0$. Then: +\begin{align*} +# Y & = \beta_1 X_1 + \beta_2 X_2 + \varepsilon +# \\ +# & = \beta_1 (U + V) + \beta_2 (U - V) + \varepsilon +# \\ +Y & = (\beta_1 + \beta_2) U + (\beta_1 - \beta_2) V + \varepsilon +\\ +& = \gamma_u U + \gamma_v V + \varepsilon +\end{align*} +and $\mbox{Cor}(X_{1}, X_{2}) = [\mbox{Var}(U) - \mbox{Var}(V)] / [ \mbox{Var}(U) + \mbox{Var}(V) ]$. +# \begin{align*} +# \mbox{Cor}(X_{1}, X_{2}) +# & = \frac{ \mbox{Cov}(X_{1}, X_{2}) }{ \sqrt{\mbox{Var}(X_{1})} \sqrt{\mbox{Var}(X_{2})} } +# \\ +# & = \frac{ \mbox{Cov}(U + V, U - V) }{ \sqrt{\mbox{Var}(U + V)} \sqrt{\mbox{Var}(U - V)} } +# \\ +# & = +# \frac{ \mbox{Var}(U) - \mbox{Var}(V) }{ \mbox{Var}(U) + \mbox{Var}(V) }. +# \end{align*} +The random variables $X_1$ and $X_2$ are strongly positively correlated if $\mbox{Var}(U) \gg \mbox{Var}(V)$. + +The ridge regression estimator associated with regression of $Y$ on $U$ and $V$ is: +\begin{align*} +\ggamma(\lambda) & = \left( +\begin{array}{rr} +\mbox{Var}(U) + \lambda & 0 +\\ +0 & \mbox{Var}(V) + \lambda +\end{array} +\right)^{-1} +\left( +\begin{array}{r} +\mbox{Cov}(U, Y) +\\ +\mbox{Cov}(V, Y) +\end{array} +\right). +\end{align*} +For large enough $\lambda$ +\begin{align*} +\ggamma(\lambda) +# & & \frac{1}{\lambda} \left( \begin{array}{rr} \mbox{Var}(U) & 0 \\ 0 \mbox{Var}(V) \end{array} \right) \left( \begin{array}{rr} \mbox{Var}(U) & 0 +# \\ +# 0 & \mbox{Var}(V) +# \end{array} +# \right)^{-1} +# \left( +# \begin{array}{r} +# \mbox{Cov}(U, Y) +# \\ +# \mbox{Cov}(V, Y) +# \end{array} +# \right) +# \\ +& \approx \frac{1}{\lambda} +\left( +\begin{array}{rr} +\mbox{Var}(U) & 0 +\\ +0 & \mbox{Var}(V) +\end{array} +\right) +\left( +\begin{array}{r} +\beta_1 + \beta_2 +\\ +\beta_1 - \beta_2 +\end{array} +\right). +\end{align*} +When $\mbox{Var}(U) \gg \mbox{Var}(V)$ and $\beta_1 \approx \beta_2$, the ridge estimate of $\gamma_v$ vanishes for large $\lambda$. Hence, ridge regression prefers positively covariates with similar effect sizes. + + +This phenomenon too can be explained geometrically. For the illustration consider ridge estimation with $\lambda=1$ of the linear model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\bbeta = (3, 3)^{\top}$, $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_2, \mathbf{I}_{22})$ and the columns of $\mathbf{X}$ strongly and positively collinear. The level sets of the sum-of-squares loss, $\| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2$, are plotted in the right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}. Recall that the ridge estimate is found by looking for the smallest loss level set that hits the ridge contraint. The sought-for estimate is then the point of intersection between this level set and the constraint, and - for the case at hand - is on the $x=y$-line. This is no different from the case with orthogonal $\mathbf{X}$ columns. Yet their estimates differ, even though the same $\lambda$ is applied. The difference is to due to fact that the radius of the ridge constraint depends on $\lambda$, $\mathbf{X}$ and $\mathbf{Y}$. This is immediate from the fact that the radius of the constraint equals $\| \hat{\bbeta}(\lambda) \|_2^2$ (see Section ref{sect.constrainedEstimation}). To study the effect of $\mathbf{X}$ on the radius, we remove its dependence on $\mathbf{Y}$ by considering its expectation, which is: +\begin{align*} +\mathbb{E}[ \| \hat{\bbeta}(\lambda) \|_2^2 ] & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta}]^{\top} \, (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta} \} \nonumber % \, \, \, = \, \, \, \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) \nonumber +\\ +# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +# \\ +# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X})(\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +# \\ +& = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +\\ +& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \big\} + \bbeta^{\top} \mathbf{X}^{ \top} \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \, \mathbf{X} \bbeta. % label{form.expectationPenalty} +\end{align*} +In the last step we have used $\mathbf{Y} \sim \mathcal{N}( \mathbf{X} \bbeta, \sigma^2 \mathbf{I}_{pp})$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ is $\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}$ (cf. \citealp{Math1992}). The expression for the expectation of the radius of the ridge constraint can now be evaluated for the orthogonal $\mathbf{X}$ and the strongly, positively collinear $\mathbf{X}$. It turns out that the latter is larger than the former. This results in a larger ridge constraint. For the larger ridge constraint there is a smaller level set that hits it first. The point of intersection, still on the $x=y$-line, is now thus closer to $\bbeta$ and further from the origin (cf. right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}). The resulting estimate is thus larger than that from the orthogonal case. + +The above needs some attenuation. Among others it depends on: *i)* the number of covariates in each block, *ii)* the size of the effects, i.e. regression coefficients of each covariate, and +*iii)* the degree of collinearity. Possibly, there are more factors influencing the behaviour of the ridge estimator presented in this subsection. + +This behaviour of ridge regression is to be understood when using (say) gene expression data to predict a certain clinical outcome. Genes work in concert to fulfil a certain function in the cell. Consequently, one expects their expression levels to be correlated. Indeed, gene expression studies exhibit many co-expressed genes, that is, genes with correlating transcript levels. + + + + +======= Illustration} label{sect.ridgeRegressionDataIllustration ======= +The application of ridge regression to actual data aims to illustrate its use in practice. + +===== MCM7 expression regulation by microRNAs ===== + +Recently, a new class of RNA was discovered, referred to as microRNA. MicroRNAs are non-coding, single stranded RNAs of approximately 22 nucleotides. Like mRNAs, microRNAs are encoded in and transcribed from the DNA. MicroRNAs play an important role in the regulatory mechanism of the cell. MicroRNAs down-regulate gene expression by either of two post-transcriptional mechanisms: mRNA cleavage or transcriptional repression. This depends on the degree of complementarity between the microRNA and the target. Perfect or nearly perfect complementarity of the mRNA to the microRNA will lead to cleavage and degradation of the target mRNA. Imperfect complementarity will repress the productive translation and reduction in protein levels without affecting the mRNA levels. +A single microRNA can bind to and regulate many different mRNA targets. Conversely, several microRNAs can bind to and cooperatively control a single mRNA target (\citealp{Bart2004}; \citealp{Esqu2006}; \citealp{Kim2006}). + +In this illustration we wish to confirm the regulation of mRNA expression by microRNAs in an independent data set. We cherry pick an arbitrary finding from literature reported in cite{Ambs2008}, which focusses on the microRNA regulation of the MCM7 gene in prostate cancer. The MCM7 gene is involved in DNA replication \citep{Tye1999}, a cellular process often derailed in cancer. Furthermore, MCM7 interacts with the tumor-suppressor gene RB1 \citep{Ster1998}. Several studies indeed confirm the involvement of MCM7 in prostate cancer \citep{Padm2004}. And recently, it has been reported that in prostate cancer MCM7 may be regulated by microRNAs \citep{Ambs2008}. + +We here assess whether the MCM7 down-regulation by microRNAs can be observed in a data set other than the one upon which the microRNA-regulation of MCM7 claim has been based. To this end we download from the Gene Expression Omnibus (GEO) a prostate cancer data set (presented by \citealp{Wang2009GeneNetworks}). This data set (with GEO identifier: GSE20161) has both mRNA and microRNA profiles for all samples available. The preprocessed (as detailed in \citealp{Wang2009GeneNetworks}) data are downloaded and require only minor further manipulations to suit our purpose. These manipulations comprise *i)* averaging of duplicated profiles of several samples, *ii)* gene- and mir-wise zero-centering of the expression data, *iii)* averaging the expression levels of the probes that interrogate MCM7. Eventually, this leaves 90 profiles each comprising of 735 microRNA expression measurements. + + + +\lstinputlisting{prostate_ridgeAnalysis.r} + + +With this prostate data set at hand we now investigate whether MCM7 is regulated by microRNAs. Hereto we fit a linear regression model regressing the expression levels of MCM7 onto those of the microRNAs. As the number of microRNAs exceeds the number of samples, ordinary least squares fails and we resort to the ridge estimator of the regression coefficients. First, an informed choice of the penalty parameter is made through maximization of the LOOCV log-likelihood, resulting in $\lambda_{\mbox{{\tiny opt}}} = 1812.826$. Having decided on the value of the to-be-employed penalty parameter, the ridge regression estimator can now readily be evaluated. The thus fitted model allows for the evaluation of microRNA-regulation of MCM7. E.g., by the proportion of variation of the MCM7 expression levels by the microRNAs as expressed in coefficient of determination: $R^2 = 0.4492$. Alternatively, but closely related, observed expression levels may be related to the linear predictor of the MCM7 expression levels: $\hat{\mathbf{Y}}(\lambda_{\mbox{{\tiny opt}}}) = \mathbf{X} \hat{\bbeta} (\lambda_{\mbox{{\tiny opt}}})$. The Spearman correlation of response and predictor equals 0.6295. A visual inspection is provided by the left panel of Figure ref{fig.RidgeProstateExample}. Note the difference in scale of the $x$- and $y$-axes. This is due to the fact that the regression coefficients have been estimated in penalized fashion, consequently shrinking estimates of the regression coefficients towards zero leading to small estimates and in turn compressing the range of the linear prediction. The above suggests there is indeed association between the microRNA expression levels and those of MCM7. + + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [prostate_ridgeFit.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values. Right panel: Histogram of the ridge regression coefficient estimates.} label{fig.RidgeProstateExample + + +The overall aim of this illustration was to assess whether microRNA-regulation of MCM7 could also be observed in this prostate cancer data set. In this endeavour the dogma (stating this regulation should be negative) has nowhere been used. A first simple assessment of the validity of this dogma studies the signs of the estimated regression coefficients. The ridge regression estimate has 394 out of the 735 microRNA probes with a negative coefficient. Hence, a small majority has a sign in line with the `microRNA $\downarrow$ mRNA' dogma. When, in addition, taking the size of these coefficients into account (Figure ref{fig.RidgeProstateExample}, right panel), the negative regression coefficient estimates do not substantially differ from their positive counterparts (as can be witnessed from their almost symmetrical distribution around zero). Hence, the value of the `microRNA $\downarrow$ mRNA' dogma is not confirmed by this ridge regression analysis of the MCM7-regulation by microRNAs. Nor is it refuted. + + +The implementation of ridge regression in the {\tt penalized}-package offers the possibility to fully obey the dogma on negative regulation of mRNA expression by microRNAs. This requires all regression coefficients to be negative. Incorporation of the requirement into the ridge estimation augments the constrained estimation problem with an additional constraint: +\begin{align*} % label{form.constrEstProblemRidge} +\hat{\bbeta}(\lambda) & = \arg \min_{\| \bbeta \|_2^2 \leq c (\lambda) \atop \beta_j \leq 0 \, \mbox{{\tiny for all $j$}}} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. +\end{align*} +With the additional non-positivity constraint on the parameters, there is no explicit solution for the estimator. The ridge estimate of the regression parameters is then found by numerical optimization using e.g. the Newton-Raphson algorithm or a gradient descent approach. The next listing gives the R-code for ridge estimation with the non-positivity constraint of the linear regression model. + +\lstinputlisting{prostate_ridgeAnalysisWithConstraint.r} + +The linear regression model linking MCM7 expression to that of the microRNAs is fitted by ridge regression while simultaneously obeying the `negative regulation of mRNA by microRNA'-dogma to the prostate cancer data. In the resulting model 401 out of 735 microRNA probes have a nonzero (and negative) coefficient. There is a large overlap in microRNAs with a negative coefficient between those from this and the previous fit. The models are also compared in terms of their fit to the data. The Spearman rank correlation coefficient between response and predictor for the model without positive regression coefficients equals 0.679 and its coefficient of determination 0.524 (confer the left panel of ref{fig.RidgeProstateExample_constrainedAnalysis} for a visualization). This is a slight improvement upon the unconstrained ridge estimated model. The improvement may be small but it should be kept in mind that the number of parameters used by both models is 401 (for the model without positive regression coefficients) vs. 735. Hence, with close to half the number of parameters the dogma-obeying model gives a somewhat better description of the data. This may suggest that there is some value in the dogma as inclusion of this prior information leads to a more parsimonious model without any loss in fit. + + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [prostate_ridgeFitConstrained.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values (with the non-positive constraint on the parameters in place). Right panel: Histogram of the ridge regression coefficient estimates (from the non-positivity constrained analysis).} label{fig.RidgeProstateExample_constrainedAnalysis + + + +The dogma-obeying model selects 401 microRNAs that aid in the explanation of the variation in the gene expression levels of MCM7. There is an active field of research, called *target prediction*, trying to identify which microRNAs target the mRNA of which genes. Within {\tt R} there is a collection of packages that provide the target prediction of known microRNAs. The packages differ on the method (e.g. experimental or sequence comparison) that has been used to arrive at the prediction. These target predictions may be used to evaluate the value of the found 401 microRNAs. Ideally, there would be a substantial amount of overlap. The {\tt R}-script that loads the target predictions and does the comparison is below. + +\lstinputlisting{prostate_link2predictionTools.r} + + +|-----------------------------------------------------------------| +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-7pt} +# #endif | | | +|---------r--------------------r--------------------r-------------| +| # #if FORMAT in ("latex", "pdflatex") +\vspace{3pt} +# #endif | $\hat{\beta}_j = 0$ | $\hat{\beta}_j < 0$ | +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif | | | +| microRNA not target | 323 | 390 | +| microRNA target | 11 | 11 | +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-9pt} +# #endif | | | +|-----------------------------------------------------------------| + + +\caption{Cross-tabulation of the microRNAs being a potential target of MCM7 vs. the value of its regression coefficient in the dogma-obeying model.} label{table.mir2mcm7prediction} + + +With knowledge available on each microRNA whether it is predicted (by at least one target prediction package) to be a potential target of MCM7, it may be cross-tabulated against its corresponding regression coefficient estimate in the dogma-obeying model being equal to zero or not. Table ref{table.mir2mcm7prediction} contains the result. Somewhat superfluous considering the data, we may test whether the targets of MCM7 are overrepresented in the group of strictly negatively estimated regression coefficients. The corresponding chi-squared test (with Yates' continuity correction) yields the test statistic $\chi^2 = 0.0478$ with a $p$-value equal to 0.827. Hence, there is no enrichment among the 401 microRNAS of those that have been predicted to target MCM7. This may seem worrisome. However, the microRNAs have been selected for their predictive power of the expression levels of MCM7. Variable selection has not been a criterion (although the sign constraint implies selection). Moreover, criticism on the value of the microRNA target prediction has been accumulating in recent years. + + +======= Conclusion ======= + +We discussed ridge regression as a modification of linear regression to overcome the empirical non-identifiability of the latter when confronted with high-dimensional data. The means to this end was the addition of a (ridge) penalty to the sum-of-squares loss function of the linear regression model, which turned out to be equivalent to constraining the parameter domain. This warranted the identification of the regression coefficients, but came at the cost of introducing bias in the estimates. Several properties of ridge regression like moments, MSE, and its Bayesian interpretation have been reviewed. Finally, its behaviour and use have been illustrated in simulation and omics data. + + +======= Exercises ======= +\begin{question}[^footnote2][^footnote2]: This exercise is freely rendered from cite{Drap1998} \mbox{ } +\\ +Find the ridge regression solution for the data below for a general value of $\lambda$ and for the straight line model $Y = \beta_0 + \beta_1 X + \varepsilon$ (only apply the ridge penalty to the slope parameter, not to the intercept). Show that when $\lambda$ is chosen as 0.4, the ridge solution fit is $\hat{Y} = 40 + 1.75 X$. Data: $\mathbf{X}^{\top} = (X_1, X_2, \ldots, X_{8})^{\top} = (-2, -1, -1, -1, 0, 1, 2, 2)^{\top}$, and $\mathbf{Y}^{\top} = (Y_1, Y_2, \ldots, Y_{8})^{\top} = (35, 40, 36, 38, 40, 43, 45, 43)^{\top}$. +\end{question} + +\begin{question}[^footnote3][^footnote3]: This exercise is freely rendered from cite{Hast2009, but can be found in many other places. The original source is unknown to the author.} label{question:ridgeAugmentation} \mbox{ } +\\ +Show that the ridge regression estimates can be obtained by ordinary least squares regression on an augmented data set. We augment the centered matrix $\mathbf{X}$ with $p$ additional row $\sqrt{\lambda} \mathbf{I}$, and augment $\mathbf{y}$ with $p$ zeros. +\end{question} + +\begin{question} label{question.ridgeResidualsProjection} \mbox{ } +\\ +The coefficients $\bbeta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, are estimated by $\hat{\bbeta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. The associated fitted values then given by $\widehat{\mathbf{Y}} = \mathbf{X} \, \hat{\bbeta} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} = \mathbf{H} \mathbf{Y}$, where $\mathbf{H} =\mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}$ referred to as the hat matrix. The matrix $\mathbf{P}$ is a projection matrix and satisfies $\mathbf{H} = \mathbf{H}^ 2$. Hence, linear regression projects the response $\mathbf{Y}$ onto the vector space spanned by the columns of $\mathbf{Y}$. Consequently, the residuals $\hat{\vvarepsilon}$ and $\hat{\mathbf{Y}}$ are orthogonal. Now consider the ridge estimator of the regression coefficients: $\hat{\bbeta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. Let $\hat{\mathbf{Y}}(\lambda) = \mathbf{X} \hat{\bbeta}(\lambda)$ be the vector of associated fitted values. + +\begin{compactitem} +\item[*a)*] Show that the matrix $\mathbf{H}(\lambda) = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}$, associated with ridge regression, is not a projection matrix (for any $\lambda > 0$). + +\item[*b)*] Show that the `ridge fit' $\widehat{\mathbf{Y}}(\lambda)$ is not orthogonal to the associated `ridge residuals' $\hat{\vvarepsilon}(\lambda)$ (for any $\lambda > 0$). + +\item[*c)*] Derive the distribution of the `ridge residuals'. +\end{compactitem} +\end{question} + + +\begin{question} label{question.MSEofLinearPredictor} \mbox{ } +\\ +Recall that there exists $\lambda > 0$ such that $MSE(\hat{\bbeta}) > MSE[\hat{\bbeta}(\lambda)]$. Verify that this carries over to the linear predictor. That is, there exists a $\lambda > 0$ such that$MSE(\widehat{\mathbf{Y}}) = MSE(\mathbf{X} \hat{\bbeta}) > MSE[\mathbf{X}\hat{\bbeta}(\lambda)]$. +\end{question} + +\begin{question} \mbox{ } +\\ +Consider a 3-gene pathway. Expression levels of these three genes have been measured in an observational study involving hundred individuals. In order to assess how the expression levels of gene A are affect by that of genes B and C, a medical researcher fits the +\begin{align*} +Y_{i}^{(A)} &= \beta_b Y_{i}^{(B)} + \beta_c Y_{i}^{(C)} + +\varepsilon_{i}, +\end{align*} +with $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. This model fitted by means of ridge regression, but with a separate penalty parameter, $\lambda_{2,b}$ and $\lambda_{2,c}$, for the two regression coefficient, $\beta_b$ and $\beta_c$, respectively. + +\begin{compactitem} +\item[*a)*] Write down the ridge penalized loss function employed by the researcher. + +\item[*b)*] Does a different choice of penalty parameter for the second regression coefficient affect the estimation of the first regression coefficient? Motivate your answer. + +\item[*c)*] The researcher decides that the second covariate $Y_i^{(C)}$ is irrelevant. Instead of removing the covariate from model, the researcher decides to set $\lambda_{2,c} = +\infty$. Show that this results in the same ridge estimate for $\beta_b$ as when fitting (again by means of ridge regression) the model without the second covariate. +\end{compactitem} +\end{question} + +\begin{question} \mbox{ } +\\ +The expression levels of the $j$-the gene are explained by a linear regression model in terms of those of all other genes. Consider the following two ridge regression estimators of the regression parameter of this model, defined as: +\begin{align*} +\arg \max_\bbeta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \bbeta_j)^2 + \lambda \| \bbeta_j \|_ 2^2 +\quad +\mbox{and} \quad +\arg \max_\bbeta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \bbeta_j)^2 + n \lambda \| \bbeta_j \|_ 2^2. +\end{align*} +Which do you prefer? Motivate. +\end{question} + + + + +========= Generalizing ridge regression ========= label{chap:genRidge} +# ======= Generalized ridge regression ======= + +The expos\'{e} on ridge regression may be generalized in many ways. Among others different generalized linear models may be considered (confer Section ref{sect.ridgeLogistic}). In this section we stick to the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with the usual assumptions, but fit it in weighted fashion and generalize the common, spherical penalty. The loss function corresponding to this scenario is: + +!bt + +!bt +\begin{align} label{form:generalizedRidgeLoss} +(\mathbf{Y} - \mathbf{X} \bbeta)^{\top} \mathbf{W} (\mathbf{Y} - \mathbf{X} \bbeta) + (\bbeta - \bbeta_0)^{\top} \mathbf{\Delta} +(\bbeta - \bbeta_0), +\end{align} +!et +!et +which comprises a weighted least squares criterion and a generalized ridge penalty. In this $\mathbf{W}$ is a $(n \times n)$-dimensional, diagonal matrix with $(\mathbf{W})_{ii} \in [0,1]$ representing the weight of the $i$-th observation. The penalty is now a quadratic form with penalty parameter $\mathbf{\Delta}$, a $(p \times p)$-dimensional, positive definite, symmetric matrix. When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, one regains the spherical penalty of `regular ridge regression'. This penalty shrinks each element of the regression parameter $\bbeta$ equally along the unit vectors $\mathbf{e}_j$. Generalizing $\mathbf{\Delta}$ to the class of symmetric, positive definite matrices $\mathcal{S}_{++}$ allows for *i)* different penalization per regression parameter, and *ii)* joint (or correlated) shrinkage among the elements of $\bbeta$. The penalty parameter $\mathbf{\Delta}$ determines the speed and direction of shrinkage. The $p$-dimensional column vector $\bbeta_0$ is a user-specified, non-random target towards which $\bbeta$ is shrunken as the penalty parameter increases. When recasting generalized ridge estimation as a constrained estimation problem, the implications of the penalty may be visualized (Figure ref{fig:generalizedRidgeIllustration}, left panel). The generalized ridge penalty is a quadratic form centered around $\bbeta_0$. In Figure ref{fig:generalizedRidgeIllustration} the parameter constraint clearly is ellipsoidal (and not spherical). Moreover, the center of this ellipsoid is not at zero. + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [generalizedRidgeIllustration_penalty.eps, width=400 frac=1.0] Left panel: the contours of the likelihood (grey solid ellipsoids) and the parameter constraint implied by the generalized penalty (black dashed ellipsoid. Right panel: generalized (fat coloured lines) and `regular' (thin coloured lines) regularization paths of four regression coefficients. The dotted grey (straight) lines indicated the targets towards the generalized ridge penalty shrinks regression coefficient estimates.} label{fig:generalizedRidgeIllustration + + +The addition of the generalized ridge penalty to the sum-of-squares ensures the existence of a unique regression estimator in the face of super-collinearity. The generalized penalty is a non-degenerated quadratic form in $\bbeta$ due to the positive definiteness of the matrix $\mathbf{\Delta}$. As it is non-degenerate, it is strictly convex. Consequently, the generalized ridge regression loss function (ref{form:generalizedRidgeLoss}), being the sum of a convex and strictly convex function, is also strictly convex. This warrants the existence of a unique global minimum and, thereby, a unique estimator. + + +Like for the `regular' ridge loss function (ref{form.ridgeLossFunction}), there is an explicit expression for the optimum of the generalized ridge loss function (ref{form:generalizedRidgeLoss}). To see this, obtain the estimating equation of $\bbeta$ through equating its derivative with respect to $\bbeta$ to zero: +\begin{align*} +2 \mathbf{X}^{\top} \mathbf{W} \mathbf{Y} - 2 \mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta - 2 \mathbf{\Delta} \bbeta + 2 \mathbf{\Delta} \bbeta_0 & = \mathbf{0}_{p}. +\end{align*} +This is solved by: + +!bt + +!bt +\begin{align} +\hat{\bbeta}(\mathbf{\Delta}) & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{Y} + \mathbf{\Delta} \bbeta_0). +\end{align} +!et +!et +Clearly, this reduces to the `regular' ridge estimator by setting $\mathbf{W} = \mathbf{I}_{nn}$, $\bbeta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$. The effects of the generalized ridge penalty on the estimates can be seen in the regularization paths of the estimates. Figure ref{fig:generalizedRidgeIllustration} (right panel) contains an example of the regularization paths for coefficients of a linear regression model with four explanatory variables. Most striking is the limiting behaviour of the estimates of $\beta_3$ and $\beta_4$ for large values of the penalty parameter $\lambda$: they convergence to a non-zero value (as was specified by a nonzero $\beta_0$). More subtle is the (temporary) convergence of the regularization paths of the estimates of $\beta_2$ and $\beta_3$. That of $\beta_2$ is pulled away from zero (its true value and approximately its unpenalized estimate) towards the estimate of $\beta_3$. In the regularization path of $\beta_3$ this can be observed in a delayed convergence to its nonzero target value (for comparison consider that of $\beta_4$). For reference the corresponding regularization paths of the `regular' ridge estimates (as thinner lines of the same colour) are included in Figure ref{fig:generalizedRidgeIllustration}. + + + +\begin{example} *Fused ridge estimation* label{example.gRidge2orthonormalDesign} +\\ +An example of a generalized ridge penalty is the *fused ridge penalty*. Consider the standard linear model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$. The fused ridge estimator of $\bbeta$ then minimizes: + +!bt + +!bt +\begin{align} label{form:fusedRidgeLoss} +\| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2. +\end{align} +!et +!et +The penalty in the loss function above can be written as a generalized ridge penalty: +\begin{align*} +\lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2 & = \bbeta^{\top} +\left( +\begin{array}{rrrrrr} +\lambda & -\lambda & 0 & \ldots & \ldots & 0 +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +-\lambda & 2 \lambda & -\lambda & \ddots & & \vdots +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +0 & -\lambda & 2 \lambda & \ddots & \ddots & \vdots +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +\vdots & \ddots & \ddots & \ddots & \ddots & 0 +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +\vdots & & \ddots & \ddots & \ddots & -\lambda +# # #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +0 & \ldots & \ldots & 0 & -\lambda & \lambda +\end{array}\right) +\bbeta. +\end{align*} +The matrix $\mathbf{\Delta}$ employed above is semi-positive definite and therefore the loss function (ref{form:fusedRidgeLoss}) is not strictly convex. Hence, often a regular ridge penalty $\| \bbeta \|_2^2$ is added (with its own penalty parameter). + +To illustrate the effect of the fused ridge penalty on the estimation of the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, let $\beta_j = \phi_{0,1}(z_j)$ with $z_j =-30 + \tfrac{6}{50} j$ for $j=1, \ldots, 500$. Sample the elements of the design matrix $\mathbf{X}$ and those of the error vector $\vvarepsilon$ from the standard normal distribution, then form the response $\mathbf{Y}$ from the linear model. The regression parameter is estimated through fused ridge loss minimization with $\lambda=1000$. The estimate is shown in Figure ref{fig:fusedRidgeIllustration} (red line). For reference the figure includes the true $\bbeta$ (black line) and the `regular ridge' estimate with $\lambda=1$ (blue line). Clearly, the fused ridge estimate yields a nice smooth vector of $\bbeta$ estimates +\end{example} + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [fusedRidgeIllustration_simulation.eps, width=400 frac=1.0] Left panel: illustration of the fused ridge estimator (in simulation). The true parameter $\bbeta$ and its ridge and fused ridge estimates against their spatial order. Right panel: Ridge vs. fused ridge estimates of the DNA copy effect on KRAS expression levels. The dashed, grey vertical bar indicates the location of the KRAS gene. } label{fig:fusedRidgeIllustration + + + + +======= Moments ======= + +The expectation and variance of $\hat{\bbeta}(\mathbf{\Delta})$ are obtained through application of the same matrix algebra and expectation and covariance rules used in the derivation of their counterparts of the `regular' ridge regression estimator. This leads to: +\begin{align*} +\mathbb{E}[\hat{\bbeta}(\mathbf{\Delta})] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta + \mathbf{\Delta} \bbeta_0), +\\ +\mbox{Var}[\hat{\bbeta}(\mathbf{\Delta})] & = \sigma^2 (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}. +\end{align*} +From these expressions similar limiting behaviour as for the `regular' ridge regression case can be deduced. To this end let $\mathbf{V}_{\delta} \mathbf{D}_{\delta} \mathbf{V}_{\delta}^{\top}$ be the eigendecomposition of $\mathbf{\Delta}$ and $d_{\delta,j} = (\mathbf{D}_{\delta})_{jj}$. Furthermore, define (with some abuse of notation) $\lim_{\mathbf{\Delta} \rightarrow \infty}$ as the limit of all $d_{\delta,j}$ simultaneously tending to infinity. Then, $\lim_{\mathbf{\Delta} \rightarrow \infty} \mathbb{E}[\hat{\bbeta}(\mathbf{\Delta})] = \bbeta_0$ and $\lim_{\mathbf{\Delta} \rightarrow \infty} \mbox{Var}[\hat{\bbeta}(\mathbf{\Delta})] = \mathbf{0}_{pp}$. + + +\begin{example} \mbox{ } label{example.gRidge2orthonormalDesign} +\\ +Let $\mathbf{X}$ be an $(n \times p)$-dimensional, orthonormal design matrix. Contrast the regular and generalized ridge regression estimator, the latter with $\mathbf{W} = \mathbf{I}_{pp}$, $\bbeta_0 = \mathbf{0}_p$ and $\mathbf{\Delta} = \lambda \mathbf{R}$ where $\mathbf{R} = (1-\rho) \mathbf{I}_{pp} + \rho \mathbf{1}_{pp}$ for $\rho \in (-(p-1)^{-1}, 1)$. For $\rho =0$ the two estimators coincide. The variance of the generalized ridge regression estimator then is $\mbox{Var}[ \hat{\bbeta}(\mathbf{\Delta})] = (\mathbf{I}_{pp} + \mathbf{\Delta})^{-2}$. The efficiency of this estimator, measured by its generalized variance, is: +\begin{align*} +\det \{ \mbox{Var}[ \hat{\bbeta}(\mathbf{\Delta})] \} & = \{ [1 + \lambda + (p-1) \rho] (1 + \lambda-\rho)^{p-1} \}^{-2}. +\end{align*} +This efficiency attains its minimum at $\rho = 0$. In the present case, the regular ridge regression estimator is thus more efficient than its generalized counterpart. +\end{example} + + +\begin{example} *(MSE with perfect target)* +\\ +Set $\bbeta_0 = \bbeta$, i.e. the target is equal to the true value of the regression parameter. Then: +\begin{align*} +\mathbb{E}[\hat{\bbeta}(\Delta)] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \Delta)^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta + \Delta \bbeta) \, \, \, = \, \, \, \bbeta. +\end{align*} +Hence, irrespective of the choice of $\Delta$, the generalized ridge is then unbiased. Thus: +\begin{align*} +\mbox{MSE}[\hat{\bbeta}(\Delta)] & = \mbox{tr} \{ +\mbox{Var}[\hat{\bbeta}(\Delta)] \} +\\ +& = \mbox{tr}[ \sigma^{2} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}] +\\ +& = \sigma^2 \mbox{tr}[ \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-2}]. +\end{align*} +When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, this MSE is smaller than that of the ML regression estimator, irrespective of the choice of $\lambda$. +\end{example} + + +======= The Bayesian connection} label{sect:genRidgeBayes ======= +This generalized ridge estimator can, like the regular ridge estimator, be viewed as a Bayesian estimator. It requires to replace the conjugate prior on $\bbeta$ by a more general normal law, $\bbeta \sim \mathcal{N}(\bbeta_0, \sigma^2 \mathbf{\Delta}^{-1})$, but retains the gamma prior on $\sigma^2$. The joint posterior distribution of $\bbeta$ and $\sigma^2$ is then obtained analogously (the details are left as Exercise ref{question.generalizedRidgeAndBayes}) to Section ref{sect:Bayes}: +\begin{align*} +f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \bbeta, \sigma^2) \, f_{\beta}(\bbeta | \sigma^2) \, f_{\sigma}(\sigma^2) +\\ +& \propto & g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +\end{align*} +with +\begin{align*} +g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \bbeta - \hat{\bbeta}(\mathbf{\Delta}) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta}) \big[ \bbeta - \hat{\bbeta}(\mathbf{\Delta}) \big] \Big\}. +\end{align*} +This implies $\mathbb{E}(\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\bbeta}(\mathbf{\Delta})$. Hence, the generalized ridge regression estimator too can be viewed as the Bayesian posterior mean estimator of $\bbeta$ when imposing a multivariate Gaussian prior on the regression parameter. + + + + + +======= Application ======= + +An illustration involving omics data can be found in the explanation of a gene's expression levels in terms of its DNA copy number. The latter is simply the number of gene copies encoded in the DNA. For instance, for most genes on the autosomal chromosomes the DNA copy number is two, as there is a single gene copy on each chromosome and autosomal chromosomes come in pairs. Alternatively, in males the copy number is one for genes that map to the X or Y chromosome, while in females it is zero for genes on the Y chromosome. In cancer the DNA replication process has often been compromised leading to a (partially) reshuffled and aberrated DNA. Consequently, the cancer cell may exhibit gene copy numbers well over a hundred for classic oncogenes. A faulted replication process does - of course - not nicely follow the boundaries of gene encoding regions. This causes contiguous genes to commonly share aberrated copy numbers. With genes being transcribed from the DNA and a higher DNA copy number implying an enlarged availability of the gene's template, the latter is expected to lead to elevated expression levels. Intuitively, one expects this effect to be localized (a so-called *cis*-effect), but some suggest that aberrations elsewhere in the DNA may directly affect the expression levels of distant genes (referred to as a *trans*-effect). + +The *cis*- and *trans*-effects of DNA copy aberrations on the expression levels of the KRAS oncogene in colorectal cancer are investigated. Data of both molecular levels from the TCGA (The Cancer Genome Atlas) repository are downloaded \citep{TCGA2012colon}. The gene expression data are limited to that of KRAS, while for the DNA copy number data only that of chromosome 12, which harbors KRAS, is retained. This leaves genomic profiles of 195 samples comprising 927 aberrations. Both molecular data types are zero centered feature-wise. Moreover, the data are limited to ten - conveniently chosen? - samples. The KRAS expression levels are explained by the DNA copy number aberrations through the linear regression model. The model is fitted by means of ridge regression, with $\lambda \mathbf{\Delta}$ and $\mathbf{\Delta} = \mathbf{I}_{pp}$ and a single-banded $\mathbf{\Delta}$ with unit diagonal and the elements of the first off-diagonal equal to the arbitrary value of $-0.4$. The latter choice appeals to the spatial structure of the genome and encourages similar regression estimates for contiguous DNA copy numbers. The penalty parameter is chosen by means of leave-one-out cross-validation using the squared error loss. + +\lstinputlisting{fusedRidge_KRASexample.r} + +The right panel of Figure ref{fig:fusedRidgeIllustration} shows the ridge regression estimate with both choices of $\mathbf{\Delta}$ and optimal penalty parameters plotted against the chromosomal order. The location of KRAS is indicated by a vertical dashed bar. The ordinary ridge regression estimates show a minor peak at the location of KRAS but is otherwise more or less flat. In the generalized ridge estimates the peak at KRAS is emphasized. Moreover, the region close to KRAS exhibits clearly elevated estimates, suggesting co-abberated DNA. For the remainder the generalized ridge estimates portray a flat surface, with the exception of a single downward spike away from KRAS. Such negative effects are biologically nonsensible (more gene templates leading to reduced expression levels?). On the whole the generalized ridge estimates point towards the *cis*-effect as the dominant genomic regulation mechanism of KRAS expression. The isolated spike may suggest the presence of a *trans*-effect, but its sign is biological nonsensible and the spike is fully absent in the ordinary ridge estimates. This leads us to ignore the possibility of a genomic *trans*-effect on KRAS expression levels in colorectal cancer. + +The sample selection demands justification. It yields a clear illustrate-able difference between the ordinary and ridge estimates. When all samples are left in, the *cis*-effect is clearly present, discernable from both estimates that yield a virtually similar profile. + + +======= Generalized ridge regression ======= + +What is generally referred to as `generalized ridge regression' (cf. \citealp{Hoer1970, Hemm1975}) is the particular case of loss function (ref{form:generalizedRidgeLoss}) in which $\mathbf{W} = \mathbf{I}_{nn}$, $\bbeta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \mathbf{V}_{x} \mathbf{\Lambda} \mathbf{V}_x^{\top}$, where $\mathbf{V}_x$ is obtained from the singular value decomposition of $\mathbf{X}$ (i.e., $\mathbf{X} = \mathbf{U}_{x} \mathbf{D}_x \mathbf{V}_x^{\top}$ with its constituents endowed with the usual interpretation) and $\mathbf{\Lambda}$ a positive definite diagonal matrix. This gives the estimator: +\begin{align*} +\hat{\bbeta}(\mathbf{\Lambda}) & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x \mathbf{U}_x \mathbf{Y}. +\end{align*} +From this last expression it becomes clear how this estimator generalizes the `regular ridge estimator'. The latter shrinks all eigenvalues, irrespectively of their size, in the same manner through a common penalty parameter. The `generalized ridge estimator', through differing penalty parameters (i.e. the diagonal elements of $\mathbf{\Lambda}$), shrinks them individually. + +The generalized ridge estimator coincides with the Bayesian linear regression estimator with the normal prior $\mathcal{N}[\mathbf{0}_p, (\mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1}]$ on the regression parameter $\bbeta$ (and preserving the inverse gamma prior on the error variance). Assume $\mathbf{X}$ to be of full column rank and choose $\mathbf{\Lambda} = g^{-1} \mathbf{D}_x^2$ with $g$ a positive scalar. The prior on $\bbeta$ then - assuming $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits - reduces to Zellner's $g$-prior: $\bbeta \sim \mathcal{N}[\mathbf{0}_p, g (\mathbf{X}^{\top} \mathbf{X})^{-1}]$ \citep{Zell1986}. The corresponding estimator of the regression coefficient is: $\hat{\bbeta}(g) = g (1+g)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$, which is proportional to the unpenalized ordinary least squares estimator of $\bbeta$. + +For convenience of notation in the analysis of the generalized ridge estimator the linear regression model is usually rewritten as: +\begin{align*} +\mathbf{Y} & = \mathbf{X} \bbeta + \vvarepsilon \, \, \, = \, \, \, +\mathbf{X} \mathbf{V}_x \mathbf{V}_x^{\top} \bbeta + \vvarepsilon \, \, \, = \, \, \, \tilde{\mathbf{X}} \aalpha + \vvarepsilon, +\end{align*} +with $\tilde{\mathbf{X}} = \mathbf{X} \mathbf{V}_x = \mathbf{U}_x +\mathbf{D}_x$ (and thus $\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} = \mathbf{D}_x^ 2$) and $\aalpha = \mathbf{V}_x^{\top} \bbeta$ with loss function $(\mathbf{Y} - \tilde{\mathbf{X}} \aalpha)^{\top} (\mathbf{Y} - \tilde{\mathbf{X}} \aalpha) + \aalpha^{\top} \mathbf{\Lambda} \aalpha$. In the notation above the generalized ridge estimator is then: +\begin{align*} +\hat{\aalpha}(\mathbf{\Lambda}) & = (\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} + \mathbf{\Lambda})^ {-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y} = (\mathbf{D}_x^ 2 + \mathbf{\Lambda})^{-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y}, +\end{align*} +from which one obtains $\hat{\bbeta}(\mathbf{\Lambda}) = \mathbf{V}_x \hat{\aalpha}(\mathbf{\Lambda})$. Using $\mathbb{E}[\hat{\aalpha}(\mathbf{\Lambda})] = (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 \aalpha$ and $\mbox{Var}[\hat{\aalpha}(\mathbf{\Lambda})] = \sigma^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1}$, the MSE for the generalized ridge estimator can be written as: +\begin{align*} +\mbox{MSE}[\hat{\aalpha}(\mathbf{\Lambda})] & = \sum_{j=1}^p ( \sigma^2 d_{x,j}^2 + \alpha_j^2 \lambda_{j}^2 ) (d_{x,j}^2 + \lambda_{j} )^{-2}, +\end{align*} +where $d_{x,j} = (\mathbf{D}_x)_{jj}$ and $\lambda_j = (\mathbf{\Lambda})_{jj}$. Having $\aalpha$ and $\sigma^ 2$ available, it is easily seen (equate the derivative w.r.t. $\lambda_j$ to zero and solve) that the MSE of $\hat{\aalpha}(\mathbf{\Lambda})$ is minimized by $\lambda_j = \sigma^2 / \alpha_j^2$ for all $j$. With $\aalpha$ and $\sigma^2$ unknown, cite{Hoer1970} suggest an iterative procedure to estimate the $\lambda_j$'s. Initiate the procedure with the OLS estimates of $\aalpha$ and $\sigma^2$, followed by sequentially updating the $\lambda_j$'s and +the estimates of $\aalpha$ and $\sigma^2$. An analytic expression of the limit of this procedure exists (\citealp{Hemm1975}). This limit, however, still depends on the observed $\mathbf{Y}$ and as such it does not necessarily yield the minimal attainable value of the MSE. This limit may nonetheless still yield a potential gain in MSE. This is investigated in cite{Lawl1981}. Under a variety of cases it seems to indeed outperform the OLS estimator, but there are exceptions. + +======= Conclusion ======= + +To conclude: a note of caution. The generalized ridge penalty is extremely flexible. It can incorporate any prior knowledge on the parameter values (through specification of $\bbeta_0$) and the relations among these parameters (via $\mathbf{\Delta}$). While a pilot study or literature may provide a suggestion for $\bbeta_0$, it is less obvious how to choose an informative $\mathbf{\Delta}$ (although a spatial structure is a nice exception). In general, exact knowledge on the parameters should not be incorporated implicitly via the penalty (read: prior) but preferably be used explicitly in the model - the likelihood - itself. Though this may be the viewpoint of a prudent frequentist and a subjective Bayesian might disagree. + + +======= Exercises ======= +\begin{question} \mbox{ } +\\ +Consider the linear regression model $Y_i = \beta_1 X_{i,1} + \beta_2 X_{i,2} + \varepsilon_i$ for $i=1, \ldots, n$. +Suppose estimates of the regression parameters $(\beta_1, \beta_2)$ of this model are obtained through the minimization of the sum-of-squares augmented with a ridge-type penalty: +\begin{align*} +\Big[ \sum_{i=1}^n (Y_i - \beta_1 X_{i,1} - \beta_2 X_{i,2})^2 \Big] + \lambda (\beta_1^2 + \beta_2^2 + 2 \nu \beta_1 \beta_2), +\end{align*} +with penalty parameters $\lambda \in \mathbb{R}_{> 0}$ and $\nu \in (-1, 1)$. + +\begin{compactitem} +\item[*a)*] Sketch (for both $\nu=0$ and $\nu=0.9$) the shape of the parameter constraint induced by the penalty above and describe in words the qualitative difference between both shapes. + +\item[*b)*] When $\nu = -1$ and $\lambda \rightarrow \infty$ the estimates of $\beta_1$ and $\beta_2$ (resulting from minimization of the penalized loss function above) converge towards each other: +$\lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1) = \lim_{\lambda \rightarrow \infty} \hat{\beta}_2(\lambda, -1)$. Motivated by this observation a data scientists incorporates the equality constraint $\beta_1 = \beta = \beta_2$ explicitly into the model, and s/he estimates the `joint regression parameter' $\beta$ through the minimization (with respect to $\beta$) of: +\begin{align*} +\Big[ \sum_{i=1}^n (Y_i - \beta X_{i,1} - \beta X_{i,2})^2 \Big] + \delta \beta^2, +\end{align*} +with penalty parameter $\delta \in \mathbb{R}_{> 0}$. The data scientist is surprised to find that resulting estimate $\hat{\beta}(\delta)$ does not have the same limiting (in the penalty parameter) behavior as the $\hat{\beta}_1(\lambda, -1)$, i.e. $\lim_{\delta \rightarrow \infty} \hat{\beta} (\delta) \not= \lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1)$. Explain the misconception of the data scientist. + +\item[*c)*] Assume that *i)* $n \gg 2$, *ii)* the unpenalized estimates $(\hat{\beta}_1(0, 0), \hat{\beta}_2(0, 0))$ equal $(-2,2)$, and *iii)* that the two covariates $X_1$ and $X_2$ are zero-centered, have equal variance, and are strongly negatively correlated. Consider $(\hat{\beta}_1(\lambda, \nu), \hat{\beta}_2(\lambda, \nu))$ for both $\nu=-0.9$ and $\nu=0.9$. For which value of $\nu$ do you expect the sum of the absolute value of the estimates to be largest? *Hint:* Distinguish between small and large values of $\lambda$ and think geometrically! +\end{compactitem} +\end{question} + +\begin{question} label{question.generalizedRidgeAndBayes} \mbox{ } +\\ +Consider the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \mathbf{I}_{pp})$. Assume $\bbeta \sim \mathcal{N}(\bbeta_0, \sigma^2 \mathbf{\Delta}^{-1})$ with $\bbeta_0 \in \mathbb{R}^p$ and $\mathbf{\Delta} \succ 0$ and a gamma prior on the error variance. Verify (i.e., work out the details of the derivation) that the posterior mean coincides with the generalized ridge estimator defined as: +\begin{align*} +\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{Y} + \mathbf{\Delta} \bbeta_0). +\end{align*} +\end{question} + + + +\begin{question} label{question.ridgeAR1penalty} \mbox{ } +\\ +The ridge penalty may be interpreted as a multivariate normal prior on the regression coefficients: $\bbeta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \mathbf{I}_{pp})$. Different priors may be considered. In case the covariates are spatially related in some sense (e.g. genomically), it may of interest to assume a first-order autoregressive prior: $\bbeta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \SSigma_A)$, in which $\SSigma_A$ is a $p \times p$-correlation matrix with $(\SSigma_A)_{j_1, j_2} = \rho^{ | j_1 - j_2 | } $ for some correlation coefficient $\rho \in [0, 1)$. Hence, +\begin{align*} +\SSigma_A \, \, \, = \, \, \, +\left( +\begin{array}{cccc} +1 & \rho & \ldots & \rho^{p-1} +\\ +\rho & 1 & \ldots & \rho^{p-2} +\\ +\vdots & \vdots & \ddots & \vdots +\\ +\rho^{p-1} & \rho^{p-2} & \ldots & 1 +\end{array} +\right). +\end{align*} + +\begin{compactitem} +\item[*a)*] The penalized loss function associated with this AR(1) prior is: +\begin{align*} +\mathcal{L}(\bbeta; \lambda, \SSigma_A) & = \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda \bbeta^{\top} \SSigma_A^{-1} \bbeta. +\end{align*} +Find the minimizer of this loss function. + +\item[*b)*] What is the effect of $\rho$ on the ridge estimates? Contrast this to the effect of $\lambda$. Illustrate this on (simulated) data. + +\item[*c)*] Instead of an AR(1) prior assume a prior with a uniform correlation between the elements of $\bbeta$. That is, replace $\SSigma_A$ by $\SSigma_U$, given by: +\begin{align*} +\SSigma_U \, \, \, = \, \, \, \left( +\begin{array}{ccc} +1 & \rho & \rho +\\ +\rho & 1 & \rho +\\ +\rho & \rho & 1 +\end{array} +\right) +\end{align*} +\\ +Investigate (again on data) the effect of changing from the AR(1) to the uniform prior on the ridge regression estimates. +\end{compactitem} +\end{question} + + + + +========= Ridge logistic regression ========= label{sect.ridgeLogistic} +# ======= Ridge logistic regression ======= + +Ridge penalized estimation is not limited to the standard linear regression model, but may be used to estimate (virtually) any model. Here we illustrate how it may be used to fit the logistic regression model. To this end we first recap this model and the (unpenalized) maximum likelihood estimation of its parameters. After which the model is estimated by means of ridge penalized maximum likelihood, which will turn out to be a relatively straightforward modification of unpenalized estimation. + +======= Logistic regression ======= + +The logistic regression model explains a binary response variable (through some transformation) by a linear combination of a set of covariates (as in the linear regression model). Denote this response of the $i$-th sample by $Y_i$ with $Y_i \in \{ 0, 1 \}$ for $i=1, \ldots, n$. The $n$-dimensional column vector $\mathbf{Y}$ stacks these $n$ responses. For each sample information on the $p$ explanatory variables $X_{i,1}, \ldots, X_{i,p}$ is available. In row vector form this information is denoted $\mathbf{X}_{i,\ast} = (X_{i,1}, \ldots, X_{i,p})$. Or, in short, $\mathbf{X}_i$ when the context tolerates no confusion. The $(n \times p)$-dimensional matrix $\mathbf{X}$ aggregates these vectors, such that $\mathbf{X}_i$ is the $i$-th row vector. + +The binary response cannot be modelled as in the linear model like $Y_i = \mathbf{X}_i \bbeta + \varepsilon_i$. With each element of $\mathbf{X}_i$ and $\bbeta$ assuming a value in $\mathbb{R}$, the linear predictor is not restricted to the domain of the response. This is resolved by modeling $p_i = P(Y_i = 1)$ instead. Still the linear predictor may exceed the domain of the response ($p_i \in [0,1]$). Hence, a transformation is applied to map $p_i$ to $\mathbb{R}$, the range of the linear predictor. +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logistic_effectOfBeta0.eps, width=400 frac=1.0] Top row, left panel: the response curve for various choices of the intercept $\beta_0$. Top row, right panel: the response curve for various choices of the regression coefficent $\beta_1$. Bottom row, left panel: the responce curve for various choices of the link function. Bottom panel, right panel: observations, fits and their deviations. } label{fig.logisticRidge_modelAndFitIllustration +The transformation associated with the logistic regression model is the logarithm of the odds, with the odds defined as: $\mbox{*odds*} = P(\mbox{succes}) / P(\mbox{failure}) = p_i/ (1-p_i)$. The logistic model is then written as $\log[ p_i / (1-p_i)] = \mathbf{X}_i \bbeta$ for all $i$. Or, expressed in terms of the response: +\begin{align*} +p_i & = P(Y_i = 1) \, \, \, = \, \, \, g^{-1}(\mathbf{X}_i; \bbeta) \, \, \, = \, \, \,\frac{\exp(\mathbf{X}_i \bbeta) }{1 + \exp(\mathbf{X}_i \bbeta) }. +\end{align*} +The function $g(\cdot; \cdot)$ is called the *link function*. It links the response to the explanatory variables. The one above is called the logistic link function. Or short, logit. The regression parameters have tangible interpretations. When the first covariate represents the intercept, i.e. $X_{i,j} = 1$ for all $i$, then $\beta_1$ determines where the link function equals a half when all other covariates fail to contribute to the linear predictor (i.e. where $P (Y_i = 1 \, | \, \mathbf{X}_{i}) = 0.5$ when $\mathbf{X}_{i} \bbeta = \beta_1$). This is illustrated in the top-left panel of Figure ref{fig.logisticRidge_modelAndFitIllustration} for various choices of the intercept. On the other hand, the regression parameters are directly related to the odds ratio: $\mbox{*odds ratio*} = \mbox{odds}(X_{i,j}+1) / \mbox{odds}(X_{i,j}) = \exp(\beta_j)$. Hence, the effect of a unit change in the $j$-th covariate on the odds ratio is $\exp(\beta_j)$ (see Figure ref{fig.logisticRidge_modelAndFitIllustration}, top-right panel). Other link functions (depicted in Figure ref{fig.logisticRidge_modelAndFitIllustration}, bottom-left panel) are common, e.g. the *probit*: $p_i = \Phi_{0,1}(\mathbf{X}_i \bbeta)$; the *cloglog*: $p_i = \frac{1}{\pi} \arctan(\mathbf{X}_i \bbeta) + \frac{1}{2}$; the *Cauchit*: $p_i = \exp[ - \exp(\mathbf{X}_i \bbeta)]$. All these link function are invertible. Irrespective of the choice of the link function, the binary data are thus modelled as $Y_i \sim \mathcal{B}[g^{-1}(\mathbf{X}_i; \bbeta), 1]$. That is, as a single draw from the Binomial distribution with success probability $g^{-1}(\mathbf{X}_i; \bbeta)$. + + +Let us now estimate the parameter of the logistic regression model by means of the maximum likelihood method. The likelihood of the experiment is then: +\begin{align*} +L(\mathbf{Y} \, | \, \mathbf{X}; \bbeta) & = \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i}. +\end{align*} +After taking the logarithm and some ready algebra, the log-likelihood is found to be: +\begin{align*} +\mathcal{L}(\mathbf{Y} \, | \, \mathbf{X}; \bbeta) & = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \bbeta - \log [ 1 + \exp(\mathbf{X}_i \bbeta) ] \big\}. +\end{align*} +Differentiate the log-likelihood with respect to $\bbeta$, equate it zero, and obtain the estimating equation for $\bbeta$: + +!bt + +!bt +\begin{align} label{form:logisticRidge_estimatingEquationOfBeta} +\frac{\partial \mathcal{L}}{\partial \bbeta } & = \sum_{i=1}^n \Big[ Y_i - \frac{\exp(\mathbf{X}_i \bbeta)}{ 1 + \exp(\mathbf{X}_i \bbeta)} \Big] \mathbf{X}_i^{\top} \, \, \, = \, \, \, \mathbf{0}_p. +\end{align} +!et +!et +The ML estimate of $\bbeta$ strikes a (weighted by the $\mathbf{X}_i$) balance between observation and model. Put differently (and illustrated in the bottom-right panel of Figure ref{fig.logisticRidge_modelAndFitIllustration}), a curve is fit through data by minimizing the distance between them: at the ML estimate of $\bbeta$ a weighted average of their deviations is zero. + +The maximum likelihood estimate of $\bbeta$ is evaluated by solving Equation (ref{form:logisticRidge_estimatingEquationOfBeta}) with respect to $\bbeta$ by means of the Newton-Raphson algorithm. The Newton-Raphson algorithm iteratively finds the zeros of a smooth enough function $f(\cdot)$. Let $x_0$ denote an initial guess of the zero. Then, approximate $f(\cdot)$ around $x_0$ by means of a first order Taylor series: $f(x) \approx x_0 + (x - x_0) \, (d f / d x) |_{x=x_0}$. Solve this for $x$ and obtain: $x = x_0 - [ (d f / d x) |_{x=x_0} ]^{-1} f(x_0)$. Let $x_1$ be the solution for $x$, use this as the new guess and repeat the above until convergence. When the function $f(\cdot)$ has multiple arguments, is vector-valued and denoted by $\vec{\mathbf{f}}$, and the Taylor approximation becomes: $\vec{\mathbf{f}}(\mathbf{x}) \approx \mathbf{x}_0 + J \vec{\mathbf{f}} \big|_{\mathbf{x}=\mathbf{x}_0} (\mathbf{x} - \mathbf{x}_0)$ with +\begin{align*} +J \vec{\mathbf{f}} = \left( +\begin{array}{llll} +\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \ldots & \frac{\partial f_1}{\partial x_p} +\\ +\frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \ldots & \frac{\partial f_2}{\partial x_p} +\\ +\vdots & \vdots & \ddots & \vdots +\\ +\frac{\partial f_q}{\partial x_1} & \frac{\partial f_q}{\partial x_2} & \ldots & \frac{\partial f_q}{\partial x_p} +\end{array} +\right), +\end{align*} +the Jacobi matrix. An update of $x_0$ is now readily constructed by solving (the approximation for) $\vec{\mathbf{f}}(\mathbf{x}) = \mathbf{0}$ for $\mathbf{x}$. + +When applied here to the maximum likelihood estimation of the regression parameter $\bbeta$ of the logistic regression model, the Newton-Raphson update is: +\begin{align*} +\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} - \Big( \frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} \Big)^{-1} \Big|_{\bbeta = \hat{\bbeta}^{\mbox{{\tiny old}}} } \, \, \frac{\partial \mathcal{L}}{\partial \bbeta } \Big|_{\bbeta = \hat{\bbeta}^{\mbox{{\tiny old}}} } +\end{align*} +where the Hessian of the log-likelihood equals: +\begin{align*} +\frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} & = - \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \bbeta)}{ [1 + \exp(\mathbf{X}_i \bbeta)]^2} \mathbf{X}_i^{\top} \mathbf{X}_i. +\end{align*} +Iterative application of this updating formula converges to the ML estimate of $\bbeta$. + +The Newton-Raphson algorithm is often reformulated as an iteratively re-weighted least squares algorithm. Hereto, first write the gradient and Hessian in matrix notation: +\begin{align*} +\frac{\partial \mathcal{L}}{\partial \bbeta } \, \, \, = \, \, \, \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta)] +& \mbox{ and } & +\frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} \, \, \, = \, \, \, - \mathbf{X}^{\top} \mathbf{W} \mathbf{X}, +\end{align*} +where $\vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta) = [g^{-1}( \mathbf{X}_{1, \ast}; \bbeta), \ldots, g^{-1}( \mathbf{X}_{n, \ast}; \bbeta)]^{\top}$ with $g^{-1}(\cdot; \cdot) = \exp(\cdot; \cdot) / [1 + \exp(\cdot; \cdot)]$ and $\mathbf{W}$ diagonal with $(\mathbf{W})_{ii} = \exp(\mathbf{X}_i \hat{\bbeta}^{\mbox{{\scriptsize old}}} ) [ 1 + \exp(\mathbf{X}_i \hat{\bbeta}^{\mbox{{\scriptsize old}}} ) ]^{-2}$. The updating formula of the estimate then becomes: +\begin{align*} +\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} + (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] +\\ +& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \} +\\ +& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, +\end{align*} +where $\mathbf{Z} = \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \}$. The Newton-Raphson update is thus the solution to the following weighted least squares problem: +\begin{align*} +\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \arg \min_{\bbeta} (\mathbf{Z} - \mathbf{X} \bbeta)^{\top} \mathbf{W} (\mathbf{Z} - \mathbf{X} \bbeta). +\end{align*} +Effectively, at each iteration the *adjusted response* $\mathbf{Z}$ is regressed on the covariates that comprise $\mathbf{X}$. For more on logistic regression confer the monograph of cite{Hosm2013}. + + +======= Ridge estimation ======= + +High-dimensionally, the linear predictor $\mathbf{X} \bbeta$ may be uniquely defined, but the maximum likelihood estimate of the logistic regression parameter is not. Assume $p > n$ and an estimate $\hat{\bbeta}$ available. Due to the high-dimensionality, the null space of $\mathbf{X}$ is non-trivial. Hence, let $\ggamma \in \mbox{null}(\mbox{span}(\mathbf{X}))$. Then: $\mathbf{X} \hat{\bbeta} = \mathbf{X} \hat{\bbeta} + \mathbf{X} \ggamma = \mathbf{X} (\hat{\bbeta} + \ggamma)$. As the null space is a $p-n$-dimensional subspace, $\ggamma$ need not equal zero. Hence, an infinite number of estimates of the logistic regression parameter exists that yield the same log-likelihood. Augmentation of the loss function with a ridge penalty resolves the matter, as their sum is strictly concave in $\bbeta$ (not convex as a maximum rather than a minimum is sought here) and thereby has a unique maximum. + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticPenLL2ridgePenalty.eps, width=400 frac=1.0] Top row, left panel: contour plot of the penalized log-likelihood of a logistic regression model with the ridge constraint (red line). Top row, right panel: the regularization paths of the ridge estimator of the logistic regression parameter. Bottom row, left panel: variance of the ridge estimator of the logistic regression parameter against the logarithm of the penalty parameter. Bottom panel, right panel: the predicted success probability versus the linear predictor for various choices of the penalty parameter. } label{fig.logisticRidge_effectOfPenalization +\mbox{ } + + +Ridge maximum likelihood estimates of the logistic model parameters are found by the maximization of the ridge penalized loglikelihood (cf. \citealt{Scha1984,LeCe1992}): +\begin{align*} +\mathcal{L}^{\mbox{{\tiny pen}}}(\mathbf{Y}, \mathbf{X}; \bbeta, \lambda) & = \mathcal{L} (\mathbf{Y}, \mathbf{X}; \bbeta) - \tfrac{1}{2} \lambda \| \bbeta \|_2^2 +\\ +& = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \bbeta - \log [ 1 + \exp(\mathbf{X}_i \bbeta) ] \big\} - \tfrac{1}{2} \lambda \bbeta^{\top} \bbeta, +\end{align*} +where the second summand is the ridge penalty (the sum of the square of the elements of $\bbeta$) with $\lambda$ the penalty parameter. Note that as in Section ref{sect.constrainedEstimation} maximization of this penalized loss function can be reformulated as a constrained estimation problem. This is illustrated by the top left panel of Figure ref{fig.logisticRidge_effectOfPenalization}, which depicts the contours (black lines) of the log-likelihood and the spherical domain of the parameter (red line). The optimization of the above loss function proceeds, due to the differentiability of the penalty, fully analogous to the unpenalized case and uses the Newton-Raphson algorithm for solving the (penalized) estimating equation. Hence, the unpenalized ML estimation procedure is modified straightforwardly by replacing gradient and Hessian by their `penalized' counterparts: +\begin{align*} +\frac{\partial \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \bbeta } \, \, \, = \, \, \, \frac{\partial \mathcal{L}}{\partial \bbeta } - \lambda \bbeta & \mbox{ and } & \frac{\partial^2 \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \bbeta \partial \bbeta^{\top}} \, \, \, = \, \, \, \frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} - \lambda \mathbf{I}_{pp}. +\end{align*} +With these at hand, the Newton-Raphson algorithm is (again) reformulated as an iteratively re-weighted least squares algorithm with the updating step changes accordingly to: +\begin{align*} +\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \{ \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}}) ] - \lambda \bbeta^{\mbox{{\scriptsize old}}} \} +\\ +& = \mathbf{V}^{-1} \mathbf{V} \hat{\bbeta}^{\mbox{{\scriptsize old}}} - \lambda \mathbf{V}^{-1} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] +\\ +& = \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \} +\\ +& = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, +\end{align*} +where $\mathbf{V} = \mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp}$ and $\mathbf{W}$ and $\mathbf{Z}$ as before. Hence, use this to update the estimate of $\bbeta$ until convergence, which yields the desired ridge ML estimate. + +Obviously, the ridge estimate of the logistic regression parameter tends to zero as $\lambda \rightarrow \infty$. Now consider a linear predictor with an intercept that is left unpenalized. When $\lambda$ tends to infinity, all regression coefficients but the intercept vanish. The intercept is left to model the success probability. Hence, in this case $\lim_{\lambda \rightarrow \infty} \hat{\beta}_0 (\lambda) = \log [ \tfrac{1}{n} \sum_{i=1}^n Y_i / \tfrac{1}{n} \sum_{i=1}^n (1-Y_i)]$. + + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticRidge_effectOnPrediction_lambda0.eps, width=400 frac=1.0] The realized design as scatter plot ($X_1$ vs $X_2$ overlayed by the success (\textcolor{red}{RED}) and failure regions (\textcolor{green}{GREEN}) for various choices of the penalty parameter: $\lambda = 0$ (top row, left panel), $\lambda = 10$ (top row, right panel) $\lambda = 40$ (bottom row, left panel), $\lambda = 100$ (bottom row, right panel).} label{fig.logisticRidge_effectOnPrediction + + + +The effect of the ridge penalty on parameter estimates propagates to the predictor $\hat{p}_i$. The linear predictor of the linear regression model involving the ridge estimator $\mathbf{X}_i \hat{\bbeta}(\lambda)$ shrinks towards a common value for each $i$, leading to a scale difference between observation and predictor (as seen before in Section ref{sect.ridgeRegressionDataIllustration}). This behaviour transfers to the ridge logistic regression predictor, as is illustrated on simulated data. The dimension and sample size of these data are $p=2$ and $n=200$, respectively. The covariate data are drawn from the standard normal, while that of the response is sampled from a Bernoulli distribution with success probability $P(Y_i=1) = \exp(2 X_{i,1} - 2 X_{i,2}) / [ 1 + \exp(2 X_{i,1} - 2 X_{i,2})]$. The logistic regression model is estimated from these data by means of ridge penalized likelihood maximization with various choices of the penalty parameter. The bottom right plot in Figure ref{fig.logisticRidge_effectOfPenalization} shows the predicted success probability versus the linear predictor for various choices of the penalty parameter. Larger values of the penalty parameter $\lambda$ flatten the slope of this curve. Consequently, for larger $\lambda$ more excessive values of the covariates are needed to achieve the same predicted success probability as those obtained with smaller $\lambda$ at more moderate covariate values. The implications for the resulting classification may become clearer when studying the effect of the penalty parameter on the `failure' and `success regions' respectively defined by: +\\ +\indent $\{(x_1, x_2) : P({\color{green}{\mathbf{Y=0}}} \, | \, X_1=x_1, X_2=x_2, \hat{\bbeta}(\lambda)) > 0.75 \}$, +\\ +\indent $\{(x_1, x_2) : P({\color{red}{\mathbf{Y=1}}} \, | \, X_1=x_1, X_2=x_2, \hat{\bbeta}(\lambda)) > 0.75 \}$. +\\ +This separates the design space in a light red (`failure') and light green (`success') domain. The white bar between them is the domain where samples cannot be classified with high enough certainty. As $\lambda$ grows, so does the white area that separates the failure and success regions. Hence, as stronger penalization shrinks the logistic regression parameter estimate towards zero, it produces a predictor that is less outspoken in its class assignments. + + + + +======= Moments ======= + +The $1^{\mbox{{\tiny st}}}$ and $2^{\mbox{{\tiny nd}}}$ order moment of the ridge ML parameter of the logistic model may be approximated by the final update of the Newton-Raphson estimate. Assume the one-to-last update $\hat{\bbeta}^{\mbox{{\scriptsize old}}}$ to be non-random and proceed as for the ridge estimator of the linear regression model parameter to arrive at: +# \begin{align*} +# & & \hat{\bbeta}^{\mbox{{\scriptsize new}}} = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + 2 \lambda \mathbf{I}_{p \times p} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{z} +# \end{align*} +\begin{align*} +\mathbb{E} \big( \hat{\bbeta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbb{E}( \mathbf{Z}), +\\ +\mbox{Var} \big( \hat{\bbeta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \big[ \mbox{Var} ( \mathbf{Z} ) \big] \mathbf{W} \mathbf{X} [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}, +\end{align*} +with +\begin{align*} +\mathbb{E}(\mathbf{Z}) & = \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbb{E}(\mathbf{Y}) - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \}, +\\ +\mbox{Var}(\mathbf{Z}) & = \mathbf{W}^{-1} \mbox{Var}(\mathbf{Y}) \mathbf{W}^{-1} = \mathbf{W}^{-1}, +\end{align*} +where the identity $\mbox{Var}(\mathbf{Y}) = \mathbf{W}$ follows from the variance of a Binomial distributed random variable. From these expressions similar properties as for the ridge ML estimate of the regression parameter of the linear model may be deduced. For instance, the ridge ML estimate of the logistic regression parameter converges to zero as the penalty parameter tends to infinity (confer the top right panel of Figure ref{fig.logisticRidge_effectOfPenalization}). Similarly, their variances vanish as $\lambda \rightarrow \infty$ (illustrated in the bottom left panel of Figure ref{fig.logisticRidge_effectOfPenalization}). + +======= The Bayesian connection ======= + +All penalized estimators can be formulated as Bayesian estimators, including the ridge logistic estimator. In particular, ridge estimators correspond to Bayesian estimators with a multivariate normal prior on the regression coefficients. Thus, assume $\bbeta \sim \mathcal{N}(\mathbf{0}_p, \mathbf{\Delta}^{-1})$. The posterior distribution of $\bbeta$ then is: +\begin{align*} +f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X}) & \propto & \Big\{ \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i} \Big\} \exp( - \tfrac{1}{2} \bbeta \mathbf{\Delta} \bbeta). +\end{align*} +This does not coincide with any standard distribution. But, under appropriate conditions, the posterior distribution is asymptotically normal. This invites a (multivariate) normal approximation to the posterior distribution above. The Laplace's method provides (cf. \citealp{Bish2006}). + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticRidge_laplace2posterior.eps, width=400 frac=1.0] Right panel: Laplace approximation to the posterior density of the Bayesian logistic regression parameter.} label{fig.logisticRidge_MSEandLaplace2posterior + +Laplace's method *i)* centers the normal approximation at the mode of the posterior, and *ii)* chooses the covariance to match the curvature of the posterior at the mode. The posterior mode is the location of the maximum of the posterior distribution. The location of this maximum coincides with that of the logarithm of the posterior. The latter is the log-likelihood augmented with a ridge penalty. Hence, the posterior mode, which is taken as the mean of the approximating Gaussian, coincides +with the ridge logistic estimator. For the covariance of the approximating Gaussian, the logarithm of the posterior is approximated by a second order Taylor series around the posterior mode and limited to second order terms: +\begin{align*} +\log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] & \propto & +\left. \log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] +\right|_{\bbeta = \hat{\bbeta}_{\mbox{{\tiny MAP}}}} +\\ +& & + \tfrac{1}{2} (\bbeta - \hat{\bbeta}_{\mbox{{\tiny MAP}}})^{\top} \left. \frac{\partial^2}{\partial \bbeta \partial \bbeta^{\top}} +\log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] + \right|_{\bbeta = \hat{\bbeta}_{\mbox{{\tiny MAP}}}} (\bbeta - \hat{\bbeta}_{\mbox{{\tiny MAP}}})^{\top}, +\end{align*} +in which the first order term cancels as the derivative of $f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})$ with respect to $\bbeta$ vanishes at the posterior mode - its maximum. Take the exponential of this approximation and match its arguments to that of a multivariate Gaussian $\exp[-\tfrac{1}{2} (\bbeta - \mmu_{\beta})^{\top} \mathbf{\Sigma}_{\bbeta}^{-1} (\bbeta - \mmu_{\beta})]$. The covariance of the sought Gaussian approximation is thus the inverse of the Hessian of the negative penalized log-likelihood. Put together the posterior is approximated by: +\begin{align*} +\bbeta \, | \, \mathbf{Y}, \mathbf{X} \sim \mathcal{N} \Big( \hat{\bbeta}_{\mbox{{\tiny MAP}}}, \Big\{ \mathbf{\Delta} + \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \bbeta)}{ [1+ \exp(\mathbf{X}_i \bbeta) ]^2 } \mathbf{X}_i \mathbf{X}_i^{\top} \Big\}^{-1} \Big). +\end{align*} +The Gaussian approximation is convenient but need not be good. Fortunately, the Bernstein-Von Mises Theorem \citep{VdVa2000} tells it is very accurate when the model is regular, the prior smooth, and the sample size sufficiently large. The quality of the approximation for an artificial example data set is shown in Figure ref{fig.logisticRidge_MSEandLaplace2posterior}. + +======= Penalty parameter selection ======= + +As before the penalty parameter may be chosen through $K$-fold cross-validation. For the $K=n$ case cite{Meij2013} describe a computationally efficient approximation of the leave-one-out cross-validated loglikelihood. It is based on the exact evaluation of the LOOCV loss, discussed in Section ref{subsect.crossvalidation}, that avoided resampling. The approach of cite{Meij2013} hinges upon the first-order Taylor expansion of the left-out penalized loglikelihood of the left-out estimate $\hat{\bbeta}_{-i} (\lambda)$ around $\hat{\bbeta} (\lambda)$, which yields an approximation of the former: +\begin{align*} +\hat{\bbeta}_{-i} (\lambda) & \approx \hat{\bbeta} (\lambda) - +\left( \left. \frac{\partial^2 \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \bbeta \partial \bbeta^{\top}} \right|_{\bbeta = \hat{\bbeta}(\lambda)} \right)^{-1} \left. \frac{\partial \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \bbeta } \right|_{\bbeta = \hat{\bbeta}(\lambda)} +\\ +& = \hat{\bbeta} (\lambda) + (\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \{ \mathbf{X}_{- i, \ast}^{\top} [\mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\bbeta}(\lambda))] - \lambda \hat{\bbeta}(\lambda) \}. +\end{align*} +This approximation involves the inverse of a $p \times p$ dimensional matrix, which amounts to the evaluation of $n$ such inverses for the LOOCV loss. As in Section ref{subsect.crossvalidation} this may be avoided. Rewrite both the gradient and the Hessian of the left-out loglikelihood in the approximation of the preceding display: +\begin{align*} +\mathbf{X}_{-i, \ast}^{\top} \{ \mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\bbeta}(\lambda)]\} - \lambda \hat{\bbeta}(\lambda) +& = \mathbf{X}^{\top} \{ \mathbf{Y} - \vec{\mathbf{g}}^{-1}[\mathbf{X}; \hat{\bbeta}(\lambda)]\} - \lambda \hat{\bbeta}(\lambda) +- \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)]\} +\\ +& = - \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)]\} +\end{align*} +and +\begin{align*} +(\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} + \mathbf{W}_{ii} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} +\\ +& & \qquad \qquad \qquad \qquad \qquad \quad [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}, +\end{align*} +where the Woodbury identity has been used and now $\mathbf{H}_{ii}(\lambda) = \mathbf{W}_{ii} \mathbf{X}_{i, \ast}(\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. Substitute both in the approximation of the left-out ridge logistic regression estimator and manipulate as in Section ref{subsect.crossvalidation} to obtain: +\begin{align*} +\hat{\bbeta}_{- i}(\lambda) & \approx \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - g^{-1}(\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)) ]. +\end{align*} +Hence, the leave-one-out cross-validated loglikelihood $\sum_{i=1}^n \mathcal{L} [Y_i \, | \, \mathbf{X}_{i, \ast}, \hat{\bbeta}_{-i}(\lambda)]$ can now be evaluated by means of a single inverse of a $p \times p$ dimensional matrix and some matrix multiplications. For the performance of this approximation in terms of accuracy and speed confer cite{Meij2013}. + + + + + + +======= Application ======= + +The ridge logistic regression is used here to explain the status (dead or alive) of ovarian cancer samples at the close of the study from gene expression data at baseline. Data stem from the TCGA study \citep{TCGA2011ovarian}, which measured gene expression by means of sequencing technology. Available are 295 samples with both status and transcriptomic profiles. These profiles are composed of 19990 transcript reads. The sequencing data, being representative of the mRNA transcript count, is heavily skewed. cite{Zwie2014} show that a simple transformation of the data prior to model building generally yields a better model than tailor-made approaches. Motivated by this observation the data were - to accommodate the zero counts - $\mbox{asinh}$-transformed. The logistic regression model is then fitted in ridge penalized fashion, leaving the intercept unpenalized. The ridge penalty parameter is chosen through 10-fold cross-validation minimizing the cross-validated error. R-code, and that for the sequel of this example, is to be found below. + +\lstinputlisting{ridgeLogistic_ovarianExample.r} + +The fit of the resulting model is studied. Hereto the fitted linear predictor $\mathbf{X} \hat{\bbeta}(\lambda_{\mbox{\tiny opt}})$ is plotted against the status (Figure ref{fig:logisticRidge_ovarianExample}, left panel). The plot shows some overlap between the boxes, but also a clear separation. The latter suggests gene expression at baseline thus enables us to distinguish surviving from the to-be-diseased ovarian cancer patients. Ideally, a decision rule based on the linear predictor can be formulated to predict an individual's outcome. + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [ridgeLogistic_ovarianExample_fit.ps, width=400 frac=1.0] Left panel: Box plot of the status vs. the fitted linear predictor using the full data set. Right panel: Box plot of the status vs. the linear prediction in the left-out samples of the 10 folds.} label{fig:logisticRidge_ovarianExample + + +The fit, however, is evaluated on the samples that have been used to build the model. This gives no insight on the model's predictive performance on novel samples. A replication of the study is generally costly and comparable data sets need not be at hand. A common workaround is to evaluate the predictive performance on the same data \citep{Subr2010}. This requires to put several samples aside for performance evaluation while the remainder is used for model building. The left-out sample may accidently be chosen to yield an exaggerated (either dramatically poor or overly optimistic) performance. This is avoided through the repetition of this exercise, leaving (groups of) samples out one at the time. The left-out performance evaluations are then averaged and believed to be representative of the predictive performance of the model on novel samples. +Note that, effectively, as the model building involves cross-validation and so does the performance evaluation, a double cross-validation loop is applied. This procedure is applied with a ten-fold split in both loops. Denote the outer folds by $f = 1, \ldots, 10$. Then, $\mathbf{X}_{f}$ and $\mathbf{X}_{-f}$ represent the design matrix of the samples comprising fold $f$ and that of the remaining samples, respectively. Define $\mathbf{Y}_{f}$ and $\mathbf{Y}_{-f}$ similarly. The linear prediction for the left-out fold $f$ is then $\mathbf{X}_{f} \hat{\bbeta}_{-f} (\lambda_{\mbox{{\tiny opt, -f}}})$. For reference to the fit, this is compared to $\mathbf{Y}_{f}$ visually by means of a boxplot as used above (see Figure ref{fig:logisticRidge_ovarianExample}, right panel). The boxes overlap almost perfectly. Hence, little to nothing remains of the predictive power suggested by the boxplot of the fit. The fit may thus give a reasonable description of the data at hand, but it extrapolates poorly to new samples. + + +======= Conclusion ======= + +To deal with response variables other than continuous ones, ridge logistic regression was discussed. High-dimensionally, the empirical identifiability problem then persists. Again, penalization came to the rescue: the ridge penalty may be combined with other link functions than the identity. Properties of ridge regression were shown to carry over to its logistic equivalent. + + + + + + +======= Exercises ======= +\begin{question} label{question.ridgeCoxNewtonRaphson} \mbox{ } +\\ +Consider an experiment involving $n$ cancer samples. For each sample $i$ the transcriptome of its tumor has been profiled and is denoted $\mathbf{X}_{i} = (X_{i1}, \ldots, X_{ip})^{\top}$ where $X_{ij}$ represents the gene $j=1, \ldots, p$ in sample $i$. Additionally, the overall survival data, $(Y_i, c_i)$ for $i=1, \ldots,n$ of these samples is available. In this $Y_i$ denotes the survival time of sample $i$ and $c_i$ the event indicator with $c_i = 0$ and $c_i = 1$ representing non- and censoredness, respectively. You may ignore the possibility of ties in the remainder. + +\begin{compactitem} +\item[*a)*] Write down the Cox proportional regression model that links overall survival times (as the response variable) to the expression levels. + +\item[*b)*] Specify its loss function for penalized maximum partial (!) likelihood estimation of the parameters. Penalization is via the ridge penalty. + +\item[*c)*] From this loss function, derive the estimation equation for the Cox regression coefficients. + +\item[*d)*] Describe (in words) how you would find the `ridge ML estimate'. +\end{compactitem} +\end{question} + +\begin{question} \mbox{ } +\\ +Download the {\tt multtest} package from BioConductor: +\\ +\indent {\tt > source("http://www.bioconductor.org/biocLite.R")} +\\ +\indent {\tt > biocLite("multtest")} +\\ +Activate the library and load leukemia data from the package: +\\ +\indent {\tt > library(multtest)} +\\ +\indent {\tt > data(golub)} +\\ +The objects {\tt golub} and {\tt golub.cl} are now available. The matrix-object {\tt golub} contains the expression profiles of 38 leukemia patients. Each profile comprises expression levels of 3051 genes. The numeric-object {\tt golub.cl} is an indicator variable for the leukemia type (AML or ALL) of the patient. + +\begin{compactitem} +\item[*a)*] Relate the leukemia subtype and the gene expression levels by a logistic regression model. Fit this model by means of penalized maximum likelihood, employing the ridge penalty with penalty parameter $\lambda=1$. This is implemented in the {\tt penalized}-packages available from {\tt CRAN}. *Note:* center (gene-wise) the expression levels around zero. + +\item[*b)*] Obtain the fits from the regression model. The fit is almost perfect. Could this be due to overfitting the data? Alternatively, could it be that the biological information in the gene expression levels indeed determines the leukemia subtype almost perfectly? + +\item[*c)*] To discern between the two explanations for the almost perfect fit, randomly shuffle the subtypes. Refit the logistic regression model and obtain the fits. On the basis of this and the previous fit, which explanation is more plausible? + +\item[*d)*] Compare the fit of the logistic model with different penalty parameters, say $\lambda = 1$ and $\lambda = 1000$. How does $\lambda$ influence the possibility of overfitting the data? + +\item[*e)*] Describe what you would do to prevent overfitting. +\end{compactitem} +\end{question} + + +\begin{question} label{question.ridgeLogisticBreastData} \mbox{ } +\\ +Download the {\tt breastCancerNKI} package from BioConductor: +\\ +\indent {\tt > source("http://www.bioconductor.org/biocLite.R")} +\\ +\indent {\tt > biocLite("breastCancerNKI")} +\\ +Activate the library and load leukemia data from the package: +\\ +\indent {\tt > library(breastCancerNKI)} +\\ +\indent {\tt > data(nki)} +\\ +The eset-object {\tt nki} is now available. It contains the expression profiles of 337 breast cancer patients. Each profile comprises expression levels of 24481 genes. Extract the expression data from the object, remove all genes with missing values, center the gene expression gene-wise around zero, and limit the data set to the first thousand genes. The reduction of the gene dimensionality is only for computational speed. +\\ +\indent {\tt X <- exprs(nki) } +\\ +\indent {\tt X <- X[-which(rowSums(is.na(X)) > 0),] } +\\ +\indent {\tt X <- apply(X[1:1000,], 1, function(X){ X - mean(X) }) }. +\\ +Furthermore, extract the estrogen receptor status (short: ER status), an important prognostic indicator for breast cancer. +\\ +\indent {\tt Y <- pData(nki)[,8] } + + +\begin{compactitem} +\item[*a)*] Relate the ER status and the gene expression levels by a logistic regression model, which is fitted by means of ridge penalized maximum likelihood. First, find the optimal value of the penalty parameter of $\lambda$ by means of cross-validation. This is implemented in {\tt optL2}-function of the {\tt penalized}-package available from {\tt CRAN}. + +\item[*b)*] Evaluate whether the cross-validated likelihood indeed attains a maximum at the optimal value of $\lambda$. This can be done with the {\tt profL2}-function of the {\tt penalized}-package available from {\tt CRAN}. + +\item[*c)*] Investigate the sensitivity of the penalty parameter selection with respect to the choice of the cross-validation fold. + +\item[*d)*] Does the optimal lambda produce a reasonable fit? +\end{compactitem} +\end{question} + + + + + + From 5d287fa028a21f10344280cb51d463e45fd0813f Mon Sep 17 00:00:00 2001 From: mhjensen Date: Wed, 5 Sep 2018 15:39:52 +0200 Subject: [PATCH 05/12] Typo in ridge matrix in exercise 4 of hw2 --- doc/Projects/2018/hw2/html/hw2-bs.html | 4 +- doc/Projects/2018/hw2/html/hw2.html | 4 +- .../2018/hw2/ipynb/ipynb-hw2-src.tar.gz | Bin 207 -> 207 bytes doc/Projects/2018/hw2/pdf/hw2.p.tex | 4 +- doc/Projects/2018/hw2/pdf/hw2.pdf | Bin 198096 -> 198097 bytes doc/Projects/2018/hw2/pdf/hw2.tex | 4 +- doc/src/Projects/2018/Exercises/hw2.do.txt | 2 +- doc/src/Regression/Regression.do.txt | 2205 ++++++++++++++++- doc/src/Regression/out.txt | 1147 ++++----- 9 files changed, 2716 insertions(+), 654 deletions(-) diff --git a/doc/Projects/2018/hw2/html/hw2-bs.html b/doc/Projects/2018/hw2/html/hw2-bs.html index 7e1d9a41e..5ff410dfb 100644 --- a/doc/Projects/2018/hw2/html/hw2-bs.html +++ b/doc/Projects/2018/hw2/html/hw2-bs.html @@ -114,7 +114,7 @@ MathJax.Hub.Config({

    Department of Physics, University of Oslo, Norway

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    Sep 3, 2018

    +

    Sep 5, 2018


    @@ -145,7 +145,7 @@ y = 5*x<

    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
    2. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      3. -
    3. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
      4. +
    4. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})+\lambda\hat{I})^{-1} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    5. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    6. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
    diff --git a/doc/Projects/2018/hw2/html/hw2.html b/doc/Projects/2018/hw2/html/hw2.html index e68f82db2..66482871e 100644 --- a/doc/Projects/2018/hw2/html/hw2.html +++ b/doc/Projects/2018/hw2/html/hw2.html @@ -81,7 +81,7 @@ MathJax.Hub.Config({
    Department of Physics, University of Oslo, Norway

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    Exercise 4

    @@ -110,7 +110,7 @@ y = 5*x<
    1. Write your own code for the Ridge method (see chapter 3.4 of Hastie et al., equations (3.43) and (3.44)) and compute the parametrization for different values of \( \lambda \). Compare and analyze your results with those from exercise 2. Study the dependence on \( \lambda \) while also varying the strength of the noise in your expression for \( y(x) \).
    2. Repeat the above but using the functionality of scikit-learn. Compare your code with the results from scikit-learn. Remember to run with the same random numbers for generating \( x \) and \( y \).
      3. -
    3. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})^{-1}+\lambda\hat{I} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
      4. +
    4. Our next step is to study the variance of the parameters \( \beta_1 \) and \( \beta_2 \) (assuming that we are parametrizing our function with a second-order polynomial. We will use standard linear regression and the Ridge regression. You can now opt for either writing your own function that calculates the variance of these paramaters (recall that this is equal to the diagonal elements of the matrix \( (\hat{X}^T\hat{X})+\lambda\hat{I})^{-1} \)) or use the functionality of scikit-learn and compute their variances. Discuss the results of these variances as functions of \( \lambda \). In particular, try to link your discussion with the discussion in Hastie et al. and their figure 3.11.
    5. Repeat the previous step but add now the Lasso method, see equation (3.53) of Hastie et al.. Discuss your results and compare with standard regression and the Ridge regression results. You can write your own code or use the functionality of scikit-learn.
    6. Finally, using scikit-learn or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
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zzWY6GUMIJCmu~9{)sIa-fF_CHG$;e22xyd|zSh9RCzcCRyN1$48h6%V1bMVi9Dc(Sc#hfK+BO0-UQ z1m!eAo{=buMbEV}yC_+%<%`Z$zT1vtYyr9LFqwA?umjkg^MohDA33-pf6+{dqAIY{(&2$pj^oxhwhb4Lwv`fm|8Eldg{Nv$3CsWXL1ma`x-DC>Z{$kAO77HMVAcNG^Nk#$C=e$uAra%asixw!}QU$nP}BAYV>7dxbE z^3kBkrXoOp#Hiu)YN_A}(Q~m!snViqrU*a~N3Khvos}ogfRrdkn8g;Q>gq0Y>7~hS2EI)|>JJ1erZ*fD2r zy9INr8-olS|EF>2e!}@bk3+2h%|uuHKgS^$gXYr8STgF|&metIo7c&0NQqAnZEG=Z z79)X+^6b=>>gW)WUiqHKJ;+?-q)c=sD0VBwx4ruUBM1c3lwi=8!Q7)0Tj`@kfFi8eW!oJ=-3c_r)t z2^N4@108||Ay*VgP}2g1MFTp2^XW)aj3oUfDaq;kLzE#w2JMy#F*HFlY{CZYz!v17 z3de8&EwBr%&;}>44~MV~M{o+=&|&4T;xZ4N&;@%?fK}*$b?Ai^=!1S(g8`^O5lS#x z2tzOoBQOeMFm4qqaXw+)-b77W^-|Q7^>!UKZ8d(PX6)Q+RN1~))U5sT8#U)tyNjB4 zdVYvnaH?0M7M n$, the vector $\beta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = +\mathbf{X}^{\top} \mathbf{0}_{p} = \mathbf{0}_{n}$, the solution of the normal equations is: + +!bt +\begin{align*} +\hat{\beta} & = ( \mathbf{X}^{\top} \mathbf{X})^{-} \mathbf{X}^{\top} \mathbf{Y} + \mathbf{v} \qquad \mbox{for all } \mathbf{v} \in V, +\end{align*} +!et + +where $\mathbf{A}^{-}$ denotes the Moore-Penrose inverse of the matrix $\mathbf{A}$, which is defined as: + +!bt +\begin{align*} +\mathbf{A}^{-} & = \sum_{j=1}^p \nu_j^{-1} \, I_{\{ \nu_j \not= 0 \} } \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. +\end{align*} +!et + +The solution of the normal equations is thus only determined up to an +element from a non-trivial space $V$, and there is no unique estimator +of the regression parameter. + +To obtain an estimate of the regression parameter $\beta$ when +$\mathbf{X}$ is (close to) super-collinearity, cite{Hoer1970} proposed +an ad-hoc fix to resolve the (almost) singularity of +$\mathbf{X}^{\top} \mathbf{X}$. Simply replace $\mathbf{X}^{\top} +\mathbf{X}$ by $\mathbf{X}^{\top} \mathbf{X} + \lambda +\mathbf{I}_{pp}$ with $\lambda \in [0, \infty)$. The scalar $\lambda$ +is a tuning parameter, henceforth called the *penalty parameter*. + + + +Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Then, for (say) $\lambda = 1$: +!bt +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} & = \left( +\begin{array}{rrr} +5 & 2 & 2 +\\ +2 & 7 & -4 +\\ +2 & -4 & 7 +\end{array} \right). +\end{align*} +!et +The eigenvalues of this matrix are 11, 7, and 1. Hence, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ has no zero eigenvalue and its inverse is well-defined. + +With the ad-hoc fix for the singularity of $\mathbf{X}^{\top} \mathbf{X}$, cite{Hoer1970} proceed to define the *ridge regression estimator*: + + +!bt +\begin{align} label{form.ridgeRegressionEstimator} +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}, +\end{align} +!et + +for $\lambda \in [0, \infty)$. Clearly, this is - for $\lambda$ +strictly positive - a well-defined estimator, even if $\mathbf{X}$ is +high-dimensional. However, each choice of $\lambda$ leads to a +different ridge regression estimate. The set of all ridge regression +estimates $\{ \hat{\beta}(\lambda) \, : \, \lambda \in [0, \infty) \}$ +is called the *solution* or *regularization path* of the ridge +estimator. + + + + +Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Suppose that the corresponding response vector is $\mathbf{Y} = (1.3, -0.5, 2.6, 0.9)^{\top}$. +The ridge regression estimates for, e.g. $\lambda = 1, 2$, and $10$ are then: +\begin{align*} +\hat{\beta}(1) & = (0.614, 0.548, 0.066)^{\top}, +\\ +\hat{\beta}(2) & = (0.537, 0.490, 0.048)^{\top}, +\\ +\hat{\beta}(10) & = (0.269, 0.267, 0.002)^{\top}. +\end{align*} +The full solution path of the ridge estimator is plotted in Figure ref{fig.ridgeSolPathPlusVar}. + + +Having obtained an estimate of the regression parameter $\beta$, one can define the fit $\widehat{\mathbf{Y}}$. It is defined analogous to the standard case: + +!bt +\begin{align*} +\widehat{\mathbf{Y}}(\lambda) & = \mathbf{X} \hat{\beta}(\lambda) +\, \, \, = \, \, \, \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\, \, \, := \, \, \, \mathbf{H}(\lambda) \mathbf{Y}. +\end{align*} +!et + +Previously, when using the ML estimator, the fit could be understood as a projection of $\mathbf{Y}$ onto the subspace spanned by the columns of $\mathbf{X}$. The fit $\widehat{\mathbf{Y}}(\lambda)$ corresponding to the ridge estimator is not a projection of $\mathbf{Y}$ onto $\mathbf{X}$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$a). Consequently, the `ridge residuals' $\mathbf{Y} - \widehat{\mathbf{Y}}(\lambda)$ are not orthogonal to the fit $\widehat{\mathbf{Y}}(\lambda)$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$b). + + +======= Eigenvalue shrinkage ======= + +The effect of the ridge penalty may also studied from the perspective of singular values. Let the singular value decomposition of the $(n \times p)$-dimensional design matrix $\mathbf{X}$ be: + +!bt +\begin{align*} +\mathbf{X} & = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}, +\end{align*} +!et + +where $\mathbf{D}_x$ an $(n \times n)$-dimensional diagonal matrix +with the singular values, $\mathbf{U}_x$ an $(n \times n)$-dimensional +matrix with columns containing the left singular vectors (denoted +$\mathbf{u}_i$), and $\mathbf{V}_x$ a $(p \times n)$-dimensional +matrix with columns containing the right singular vectors (denoted +$\mathbf{v}_i$). The columns of $\mathbf{U}_x$ and $\mathbf{V}_x$ are +orthogonal: $\mathbf{U}_x^{\top} \mathbf{U}_x = \mathbf{I}_{nn} = +\mathbf{V}_x^{\top} \mathbf{V}_x$. + +The OLS estimator can then be rewritten in terms of the SVD-matrices as: + + +\begin{align*} +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}, +\end{align*} +where $\mathbf{D}_x^{-2} \mathbf{D}_x$ is not simplified further to emphasize the effect of the ridge penalty. Similarly, the ridge estimator can be rewritten in terms of the SVD-matrices as: +\begin{align*} +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. +\end{align*} +Combining the two results and writing $(\mathbf{D}_x)_{jj} = d_{x,jj}$ we have: +\begin{align*} +d_{x,jj}^{-1} & \geq & \frac{d_{x,jj}}{d_{x,jj}^2 + \lambda} \qquad \mbox{ for all } \lambda > 0. +\end{align*} +Thus, the ridge penalty shrinks the singular values. +\\ +\\ +Return to the problem of the super-collinearity of $\mathbf{X}$ in the high-dimensional setting ($p > n$). The super-collinearity implies the singularity of $\mathbf{X}^{\top} \mathbf{X}$ and prevents the calculation of the OLS estimator of the regression coefficients. However, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ is non-singular, with inverse: +\begin{align*} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} & = \sum_{j=1}^p (d_{x,jj}^2 + \lambda)^{-1} \mathbf{v}_j \mathbf{v}_j^{\top}. +\end{align*} +The right-hand side is well-defined for $\lambda > 0$. + + +===== Principal components regression ===== + +Principal component regression is a close relative to ridge regression that can also be applied in a high-dimensional context. Principal components regression explains the response not by the covariates themselves but by linear combinations of the covariates as defined by the principal components of $\mathbf{X}$. Let $\mathbf{U} \mathbf{D} \mathbf{V}^{\top}$ be the singular value decomposition of $\mathbf{X}$. The $i$-th principal component of $\mathbf{X}$ is then $\mathbf{X} \mathbf{v}_i$, henceforth denoted $\mathbf{z}_i$. Let $\mathbf{Z}_k$ be the matrix of the first $k$ principal components, i.e. $\mathbf{Z}_k = \mathbf{X} \mathbf{V}_k$ where $\mathbf{V}_k$ contains the first $k$ right singular vectors as columns. Principal components regression then amounts to regressing the response $\mathbf{Y}$ onto $\mathbf{Z}_{k}$, that is, it fits the model $\mathbf{Y} = \mathbf{Z}_k \ggamma + \vvarepsilon$. The least squares estimator of $\ggamma$ then is (with some abuse of notation): +\begin{align*} +\hat{\ggamma} & = (\mathbf{Z}_k^{\top} \mathbf{Z}_k)^{-1} \mathbf{Z}_k^{\top} \mathbf{Y} \, \, \, = \, \, \, (\mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{X} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{U} \mathbf{D} \mathbf{V}^{\top} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} +\\ +& = (\mathbf{I}_{kn} \mathbf{D}^2 \mathbf{I}_{nk})^{-1} \mathbf{I}_{kn} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} +\\ +& = \mathbf{D}_k^{-2} \widetilde{\mathbf{D}}_k \mathbf{U}^{\top} \mathbf{Y} \, \, \, = \, \, \ \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}, +\end{align*} +where $\mathbf{D}_k$ and $\widetilde{\mathbf{D}}_k$ are submatrices of $\mathbf{D}$. The matrix $\mathbf{D}_k$ is obtained from $\mathbf{D}$ by removal of the last $n-p$ rows and columsn, while for $\widetilde{\mathbf{D}}_k$ only the last $n-k$ rows are dropped. Similarly, $\mathbf{I}_{kn}$ and $\mathbf{I}_{nk}$ are obtained from $\mathbf{I}_{nn}$ by removal of the last $n-k$ rows and columns, respectively. The principal component regression estimator of $\beta$ then is $\hat{\beta}_{\mbox{{\tiny pcr}}} = \mathbf{V}_k \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}$. When $k$ is set equal to the column rank of $\mathbf{X}$, and thus to the rank of $\mathbf{X}^{\top} \mathbf{X}$, the +principal component regression estimator $\hat{\beta}_{\mbox{{\tiny pcr}}} = (\mathbf{X}^{\top} \mathbf{X})^- \mathbf{X}^{\top} \mathbf{Y}$, where $\mathbf{A}^-$ denotes the Moore-Penrose inverse of matrix $\mathbf{A}$. + +The relation between ridge and principal component regression becomes clear when their corresponding estimators are written in terms of the singular value decomposition of $\mathbf{X}$: +\begin{align*} +\hat{\beta}_{\mbox{{\tiny pcr}}} & = \mathbf{V}_x (\mathbf{I}_{nk} \mathbf{D}_x \mathbf{I}_{kn})^{-1} \mathbf{U}_x^{\top} \mathbf{Y}, +\\ +\hat{\beta} (\lambda) & = \mathbf{V}_x (\mathbf{D}_x^2 + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. +\end{align*} +Both operate on the singular values of the design matrix. But where principal component regression thresholds the singular values of $\mathbf{X}$, ridge regression shrinks them (depending on their size). Hence, one applies a discrete map on the singular values while the other a continuous one. + + + + + +======= Moments ======= + +The first two moments of the ridge regression estimator are derived. Next the performance of the ridge regression estimator is studied in terms of the mean squared error, which combines the first two moments. + + +===== Expectation ===== + +The left panel of Figure ref{fig.ridgeSolPathPlusVar} shows ridge estimates of the regression parameters converging to zero as the penalty parameter tends to infinity. This behaviour of the ridge estimator does not depend on the specifics of the data set. To see this study the expectation of the ridge estimator: +\begin{align*} +\mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \big] +\\ +& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \big] +\\ +& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \big] +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \mathbb{E} ( \hat{\beta} ) +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta. +\end{align*} +Clearly, $\mathbb{E} \big[ \hat{\beta}(\lambda) \big] \not= \beta$ for any $\lambda > 0$. Hence, the ridge estimator is biased. + +From the expression above it is clear that the expectation of the ridge estimator vanishes as $\lambda$ tends to infinity: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta \, \, \, = \, \, \, \mathbf{0}_{p}. +\end{align*} +Hence, all regression coefficients are shrunken towards zero as the penalty parameter increases. This also holds for $\mathbf{X}$ with $p > n$. Furthermore, this behaviour is not strictly monotone in $\lambda$: $\lambda_{a} > \lambda_b$ does not necessarily imply $|\hat{\beta}_j (\lambda_a) | < |\hat{\beta}_j (\lambda_b) |$. Upon close inspection this can be witnessed from the ridge solution path of $\beta_3$ in Figure ref{fig.ridgeSolPathPlusVar}. + + +\begin{example} *Orthonormal design matrix* label{example.orthoronormalDesign} +\\ +Consider an orthonormal design matrix $\mathbf{X}$, i.e.: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} & = \mathbf{I}_{pp} \, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1}. +\end{align*} +An example of an orthonormal design matrix would be: +\begin{align*} +\mathbf{X} & = \frac{1}{2} \left( +\begin{array}{rr} +-1 & -1 +\\ +-1 & 1 +\\ +1 & -1 +\\ +1 & 1 +\end{array} \right). +\end{align*} +This design matrix is orthonormal as $\mathbf{X}^{\top} \mathbf{X} = \mathbf{I}_{22}$, which is easily verified: +\begin{align*} +\mathbf{X}^{\top} \mathbf{X} & = \frac{1}{4} +\left( +\begin{array}{rrrr} +-1 & -1 & 1 & 1 +\\ +-1 & 1 & -1 & 1 +\end{array} \right) +\left( +\begin{array}{rr} +-1 & -1 +\\ +-1 & 1 +\\ +1 & -1 +\\ +1 & 1 +\end{array} \right) \, \, \, = \, \, \, +\frac{1}{4} +\left( +\begin{array}{rr} +4 & 0 +\\ +0 & 4 +\end{array} \right) \, \, \, = \, \, \, \mathbf{I}_{22}. +\end{align*} +In case of an orthonormal design matrix the relation between the OLS and ridge estimator is: +\begin{align*} +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\, \, \, = \, \, \, (\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (1 + \lambda)^{-1} \mathbf{I}_{pp} \mathbf{X}^{\top} \mathbf{Y} +\qquad \, \, = \, \, \, (1 + \lambda)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (1 + \lambda)^{-1} \hat{\beta}. +\end{align*} +Hence, the ridge estimator scales the OLS estimator by a factor. When taking the expectation on both sides, it is evident that the ridge estimator converges to zero as $\lambda \rightarrow \infty$. +\end{example} + + + +===== Variance ===== + +As for the ML estimate of the regression parameter $\beta$ of model (ref{form.linRegressionModelinMatrix}), we derive the second moment of the ridge estimator. Hereto define: +\begin{align*} +\mathbf{W}_{\lambda} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X}. +\end{align*} +Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\beta}(\lambda)$ can be expressed as $\mathbf{W}_{\lambda} \hat{\beta}$ for: +\begin{align*} +\mathbf{W}_{\lambda} \hat{\beta} & = \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ] \}^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \hat{\beta}(\lambda). +\end{align*} +The linear operator $\mathbf{W}_{\lambda}$ thus transforms the ML estimator of the regression parameter into the ridge estimator. + +It is now easily seen that: +\begin{align*} +\mbox{Var}[ \hat{\beta}(\lambda) ] & = \mbox{Var}[ \mathbf{W}_{\lambda} \hat{\beta} ] \qquad \qquad \, \, \, \, \, \, = \, \, \, \mathbf{W}_{\lambda} \mbox{Var}[\hat{\beta} ] \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}, +\end{align*} +in which we have used $\mbox{Var}(\mathbf{A} \mathbf{Y}) = \mathbf{A} \mbox{Var}( \mathbf{Y}) \mathbf{A}^{\top}$ for a non-random matrix $\mathbf{A}$, the fact that $\mathbf{W}_{\lambda}$ is non-random, and $ \mbox{Var}[\hat{\beta} ] = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}$. + + +Like the expectation the variance of the ridge estimator vanishes as $\lambda$ tends to infinity: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mbox{Var} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \mathbf{0}_{pp}. +\end{align*} +Hence, the variance of the ridge regression coefficient estimates decreases towards zero as the penalty parameter becomes large. This is illustrated in the right panel of Figure ref{fig.ridgeSolPathPlusVar} for the data of Example ref{example.supercollinearity}. + + +With an explicit expression of the variance of the ridge estimator at hand, we can compare it to that of the OLS estimator: +\begin{align*} +\mbox{Var}[ \hat{\beta} ] - \mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 [(\mathbf{X}^{\top} \mathbf{X})^{-1} - \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] +\\ +& = \sigma^2 \mathbf{W}_{\lambda} \{ [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ] (\mathbf{X}^{\top} \mathbf{X})^{-1} [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ]^{\top} - (\mathbf{X}^{\top} \mathbf{X})^{-1} \} \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} +\\ +& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}. +\end{align*} +The difference is non-negative definite as each component in the matrix product is non-negative definite. Hence, the variance of the ML estimator exceeds (in the positive definite ordering) that of the ridge estimator: + +!bt + +!bt +\begin{align} label{form.VarInequalityMLandRidge} +\mbox{Var}[ \hat{\beta} ] & \succeq & \mbox{Var}[ \hat{\beta}(\lambda) ], +\end{align} +!et +!et +with the inequality being strict if $\lambda > 0$. In other words, the variance of the ML estimator is larger than that of the ridge estimator (in the sense that their difference is non-negative definite). The variance inequality (ref{form.VarInequalityMLandRidge}) can be interpreted in terms of the stochastic behaviour of the estimator. This is illustrated by the next example. + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [varMLellipses.eps, width=400 frac=1.0] Level sets of the distribution of the ML (left panel) and ridge (right panel) regression estimators.} label{fig.varOLSandRidge +# \afterpage{} + + +\begin{example} *Variance comparison* +\\ +Consider the design matrix: +\begin{align*} +\mathbf{X} & = \left( +\begin{array}{rr} +-1 & 2 +\\ +0 & 1 +\\ +2 & -1 +\\ +1 & 0 +\end{array} \right). +\end{align*} +The variances of the ML and ridge (with $\lambda=1$) estimates of the regression coefficients then are: +\begin{align*} +\mbox{Var}(\hat{\beta}) & = \sigma^2 \left( +\begin{array}{rr} +0.3 & 0.2 +\\ +0.2 & 0.3 +\end{array} \right) +\qquad \mbox{and} \qquad +\mbox{Var}[\hat{\beta}(\lambda)] \, \, \, = \, \, \, \sigma^2 \left( +\begin{array}{rr} +0.1524 & 0.0698 +\\ +0.0698 & 0.1524 +\end{array} \right). +\end{align*} +These variances can be used to construct levels sets of the distribution of the estimates. The level sets that contain 50\%, 75\% and 95\% of the distribution of the ML and ridge estimates are plotted in Figure ref{fig.varOLSandRidge}. In line with inequality (ref{form.VarInequalityMLandRidge}) the level sets of the ridge estimate are smaller than that of the ML estimate: it thus varies less. +\end{example} + + + + +\begin{contexample}_ref{example.orthoronormalDesign_} *Orthonormal design matrix (continued)* +\\ +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{Var}[ \hat{\beta} ] = \sigma^2 \mathbf{I}_{pp}$ and +\begin{align*} +\mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} +\, \, \, = \, \, \, \sigma^2 [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{I}_{pp} \{ [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \}^{\top} +\, \, \, = \, \, \, \sigma^2 (1 + \lambda )^{-2} \mathbf{I}_{pp} . +\end{align*} +As the penalty parameter $\lambda$ is non-negative the former exceeds the latter. In particular, this expression vanishes as $\lambda \rightarrow \infty$. +\end{contexample} + +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +The full distribution of the ridge regression estimator is now known. The estimator, $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ is a linear estimator, linear in $\mathbf{Y}$. As $\mathbf{Y}$ is normally distributed, so is $\hat{\beta}(\lambda)$. Moreover, the normal distribution is fully characterized by its first two moments, which are available. Hence: +\begin{align*} +\hat{\beta}(\lambda) & \sim & \mathcal{N} \big( (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X} \, \beta, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top} \big). +\end{align*} +Given $\lambda$ and $\beta$, the random behavior of the estimator is thus known. + + + +===== Mean squared error} label{sect:ridgeMSE ===== +Previously, we motivated the ridge estimator as an ad hoc solution to collinearity. An alternative motivation comes from studying the Mean Squared Error (MSE) of the ridge regression estimator: for a suitable choice of $\lambda$ the ridge regression estimator may outperform the ML regression estimator in terms of the MSE. Before we prove this, we first derive the MSE of the ridge estimator and quote some auxiliary results. + +Recall that (in general) for any estimator of a parameter $\theta$: +\begin{align*} +\mbox{MSE}( \hat{\theta} ) & = \mathbb{E} [ ( \hat{\theta} - \theta)^2 ] +\, \, \, = \, \, \, \mbox{Var}( \hat{ \theta} ) + [\mbox{Bias} ( \hat{\theta} )]^2. +\end{align*} +Hence, the MSE is a measure of the quality of the estimator. + +The MSE of the ridge estimator is: + +!bt + +!bt +\begin{align} +\mbox{MSE}[\hat{\beta}(\lambda)] & = \mathbb{E} [ (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta)^{\top} \, (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta) ] \nonumber +\\ +& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber +\\ +& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) + \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) \nonumber +\\ +& & - \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) + \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) + \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) \nonumber +\\ +& & - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber +\\ +& = \mathbb{E} [ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) ] \nonumber +\\ +& & - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta + \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \beta + \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta \nonumber +\\ +& & - \beta^{\top} \, \mathbf{W}_{\lambda} \, \beta - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta + \beta^{\top} \beta \nonumber +\\ +& = \mathbb{E} \big\{ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta \nonumber +\\ +& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta. label{form.ridgeMSE} +\end{align} +!et +!et +In the last step we have used $\hat{\beta} \sim \mathcal{N}( \beta, \sigma^2 \, [\mathbf{X}^{\top} \mathbf{X}]^{-1} )$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ for some nonrandom symmetric positive definite matrix $\LLambda$ is (cf. \citealt{Math1992}): +\begin{align*} +\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) & = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}, +\end{align*} +of course replacing $\vvarepsilon$ by $\hat{\beta}$ in this expectation. The first summand in the final derived expression for $\mbox{MSE}[\hat{\beta}(\lambda)]$ is the sum of the variances of the ridge estimator, while the second summand can be thought of the ``squared bias'' of the ridge estimator. In particular, $\lim_{\lambda \rightarrow \infty} \mbox{MSE}[\hat{\beta}(\lambda)] = \beta^{\top} \beta$, which is the squared biased for an estimator that equals zero (as does the ridge estimator in the limit). + + + +\begin{example} *Orthonormal design matrix* +\\ +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{MSE}[ \hat{\beta} ] = p \, \sigma^2$ and +\begin{align*} +\mbox{MSE}[ \hat{\beta}(\lambda) ] & = \frac{p \, \sigma^2}{ (1+ \lambda)^{2}} + \frac{\lambda^2}{ (1+ \lambda)^{2}} \beta^{\top} \beta. +\end{align*} +The latter achieves its minimum at: $\lambda = p \sigma^2 / \beta^{\top} \beta$. +\end{example} + + + +The following theorem and proposition are required for the proof of the main result. + +\begin{theorem} *(Theorem 1 of \citealp{Theo1974*}) label{theo.Theobald1} +\\ +Let $\hat{\ttheta}_1$ and $\hat{\ttheta}_2$ be (different) estimators of $\ttheta$ with second order moments: +\begin{align*} +\mathbf{M}_k & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta) (\hat{\ttheta}_k - \ttheta)^{\top} ] \qquad \mbox{for } k=1,2, +\end{align*} +and +\begin{align*} +\mbox{MSE}(\hat{\ttheta}_k) & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta)^{\top} \mathbf{A} (\hat{\ttheta}_k - \ttheta) ] \qquad \mbox{for } k=1,2, +\end{align*} +where $\mathbf{A} \succeq 0$. Then, $\mathbf{M}_1 - \mathbf{M}_2 \succeq 0$ if and only if $\mbox{MSE}(\hat{\ttheta}_1) - \mbox{MSE}(\hat{\ttheta}_2) \geq 0$ for all $\mathbf{A} \succeq 0$. +\end{theorem} + + +\begin{proposition} *(\citealp{Fare1976*}) label{prop.Farebrother} +\\ +Let $\mathbf{A}$ be a $(p \times p)$-dimensional, positive definite matrix, $\mathbf{b}$ be a nonzero $p$ dimensional vector, and $c \in \mathbb{R}_+$. Then, $c \mathbf{A} - \mathbf{b} \mathbf{b}^{\top} \succ 0$ if and only if $\mathbf{b}^{\top} \mathbf{A}^{-1} \mathbf{b} > c$. +\end{proposition} + +We are now ready to proof the main result, formalized as Theorem ref{theo.Theobald2}, that for some $\lambda$ the ridge regression estimator yields a lower MSE than the ML regression estimator. + +\begin{theorem} *(Theorem 2 of \citealp{Theo1974*}) label{theo.Theobald2} +\\ +There exists $\lambda > 0$ such that $\mbox{MSE}[\hat{\beta}(\lambda)] < \mbox{MSE}[\hat{\beta}(0)] = \mbox{MSE}[\hat{\beta}]$. +\end{theorem} + +\begin{proof} +The second order moment matrix of the ridge estimator is: +\begin{align*} +\mathbf{M} (\lambda) & := & \mathbb{E} [ (\hat{\beta}(\lambda) - \beta) (\hat{\beta} (\lambda) - \beta)^{\top} ] +\\ +& = \mathbb{E} \{ \hat{\beta}(\lambda) [\hat{\beta}(\lambda)]^{\top} \} - \mathbb{E} [ \hat{\beta}(\lambda) ] \{ \mathbb{E} [ \hat{\beta}(\lambda) ] \}^{\top} + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top} +\\ +& = \mbox{Var}[ \hat{\beta}(\lambda) ] + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top}. +\end{align*} +Then: +\begin{align*} +\mathbf{M} ( 0 ) - \mathbf{M}(\lambda) & = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] +\\ +& & - (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \beta \beta^{\top} (\mathbf{W}_{\lambda} -\mathbf{I}_{pp})^{\top} +\\ +& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} +\\ +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +\\ +& = \lambda [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top}. +\end{align*} +This is positive definite if and only if $ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} \succ 0$. Hereto it suffices to show that $2 \, \sigma^2 \, \mathbf{I}_{pp} - \lambda \beta \beta^{\top} \succ 0$. By Proposition ref{prop.Farebrother} this holds for $\lambda$ such that $2 \sigma^2 (\beta^{\top} \beta)^{-1} > \lambda$. For these $\lambda$, we thus have $\mathbf{M} ( 0 ) - \mathbf{M}(\lambda)$. Application of Theorem ref{theo.Theobald1} now concludes the proof. +\end{proof} +This result of cite{Theo1974} is generalized by cite{Fare1976} to the class of design matrices $\mathbf{X}$ with $\mbox{rank}(\mathbf{X}) < p$. +\\ +\\ +Theorem ref{theo.Theobald2} can be used to illustrate that the ridge regression estimator strikes a balance between the bias and variance. This is illustrated in the left panel of Figure ref{fig.MSEridge}. For small $\lambda$, the variance of the ridge estimator dominates the MSE. This may be understood when realizing that in this domain of $\lambda$ the ridge estimator is close to the unbiased ML regression estimator. For large $\lambda$, the variance vanishes and the bias dominates the MSE. For small enough values of $\lambda$, the decrease in variance of the ridge regression estimator exceeds the increase in its bias. As the MSE is the sum of these two, the MSE first decreases as $\lambda$ moves away from zero. In particular, as $\lambda = 0$ corresponds to the ML regression estimator, the ridge regression estimator yields a lower MSE for these values of $\lambda$. In the right panel of Figure ref{fig.MSEridge} $\mbox{MSE}[ \hat{\beta}(\lambda)] < \mbox{MSE}[ \hat{\beta}(0)]$ for $\lambda < 7$ (roughly) and the ridge estimator outperforms the ML estimator. +# original latex figure with scale=0.45, angle=0 + +FIGURE: [MSE_ridge.eps, width=400 frac=1.0] Left panel: mean squared error, and its `bias' and `variance' parts, of the ridge regression estimator (for artificial data). Right panel: mean squared error of the ridge and ML estimator of the regression coefficient vector (for the same artificial data).} label{fig.MSEridge +# \afterpage{} +\\ +\\ +# #if FORMAT in ("latex", "pdflatex") +\noindent +# #endif +Besides another motivation behind the ridge regression estimator, the use of Theorem ref{theo.Theobald2} is limited. The optimal choice of $\lambda$ depends on the quantities $\beta$ and $\sigma^2$. These are unknown in practice. Then, the penalty parameter is chosen in a data-driven fashion by means of cross-validation (see Section ref{subsect.crossvalidation}). +\\ +\\ +\begin{remark} \mbox{ } +\\ +Theorem ref{theo.Theobald2} can also be used to conclude on the biasedness of the ridge regression estimator. The Gauss-Markov theorem \citep{Rao1973} states (under some assumptions) that the ML regression estimator is the best linear unbiased estimator (BLUE) with the smallest MSE. As the ridge regression estimator is a linear estimator and outperforms (in terms of MSE) this ML estimator, it must be biased (for it would otherwise refute the Gauss-Markov theorem). +\end{remark} + + + + +======= Constrained estimation} label{sect.constrainedEstimation ======= +The ad-hoc fix of cite{Hoer1970} to super-collinearity of the design matrix (and, consequently the singularity of the matrix $\mathbf{X}^{\top} \mathbf{X}$) has been motivated post-hoc. +The ridge estimator minimizes the *ridge loss function*, which is defined as: + +!bt + +!bt +\begin{align} +\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2 +\, \, \, = \, \, \, \sum_{i=1}^n (Y_i - \mathbf{X}_{i\ast} \, \beta)^2 + \lambda \sum_{j=1}^p \beta_j^2. label{form.ridgeLossFunction} +\end{align} +!et +!et +This loss function is the traditional sum-of-squares augmented with a *penalty*. The particular form of the penalty, $\lambda \| \beta \|^2_2$ is referred to as the *ridge penalty* and $\lambda$ as the *penalty parameter*. For $\lambda=0$, minimization of the ridge loss function yields the ML estimator. For any $\lambda > 0$, the ridge penalty contributes to the loss function, affecting its minimum and its location. The minimum of the sum-of-squares is well-known. The minimum of the ridge penalty is attained at $\beta = \mathbf{0}_{p}$ whenever $\lambda > 0$. The $\beta$ that minimizes $\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda)$ then balances the sum-of-squares and the penalty. The effect of the penalty in this balancing act is to shrink the regression coefficients towards zero, its minimum. In particular, the larger $\lambda$, the larger the contribution of the penalty to the loss function, the stronger the tendency to shrink non-zero regression coefficients to zero (and decrease the contribution of the penalty to the loss function). This motivates the name `penalty' as non-zero elements of $\beta$ increase (or penalize) the loss function. + + +To verify that the ridge estimator indeed minimizes the ridge loss function, proceed as usual. Take the derivative with respect to $\beta$: +\begin{align*} +\frac{\partial}{\partial \beta} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = -2 \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \beta) + 2 \, \lambda \, \mathbf{I}_{pp} \, \beta \, \, \, = \, \, \, -2 \, \mathbf{X}^{\top} \mathbf{Y} + 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}) \beta. +\end{align*} +Equate the derivative to zero and solve for $\beta$. This yields the ridge regression estimator. + +The ridge estimator is thus a stationary point of the ridge loss function. A stationary point corresponds to a minimum if the Hessian matrix with second order partial derivatives is positive definite. The Hessian of the ridge loss function is +\begin{align*} +\frac{\partial^2}{\partial \beta \, \partial \beta^{\top}} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}). +\end{align*} +This Hessian is the sum of the semi-positive definite matrix $\mathbf{X}^{\top} \mathbf{X}$ and the positive definite matrix $\lambda \, \mathbf{I}_{pp}$. Lemma 14.2.4 of cite{Harv2008} then states that the sum of these matrices is itself a positive definite matrix. Hence, the Hessian is positive definite and the ridge loss function has a stationary point at the ridge estimator, which is a minimum. + +The ridge regression estimator minimizes the ridge loss function. It rests to verify that it is a global minimum. To this end we introduce the concept of a convex function. As a prerequisite, a set $\mathcal{S} \subset \mathbb{R}^p$ is called *convex* if for all $\beta_1, \beta_2 \in \mathcal{S}$ their weighted average $\beta_{\theta} = (1 - \theta) \beta_1 + \theta \beta_2$ for all $\theta \in [0, 1]$ is itself an element of $\mathcal{S}$, thus $\beta_{\theta} \in \mathcal{S}$. If for all $\theta \in (0, 1)$, the weighted average $\beta_{\theta}$ is inside $\mathcal{S}$ and not on its boundary, the set is called *strict convex*. Examples of (strict) convex and nonconvex sets are depicted in Figure ref{fig.ridgeAsConstrainedEst}. A function $f(\cdot)$ is *(strict) convex* if the set $\{ y \, : \, y \geq f(\beta) \mbox{ for all } \beta \in \mathcal{S} \mbox{ for any convex } \mathcal{S} \}$, called the epigraph of $f(\cdot)$, is (strict) convex. Examples of (strict) convex and nonconvex functions are depicted in Figure ref{fig.ridgeAsConstrainedEst}. The ridge loss function is the sum of two parabola's: one at least convex and the other a strict convex function in $\beta$. The sum of convex and strict convex function is itself strict convex (confer Lemma 9.4.2 of \citealt{Flet2008}). The ridge loss function is thus strict convex. Theorem 9.4.1 of \citealt{Flet2008} then warrants, by the strict convexity of the ridge loss function, that the ridge estimator is a global minimum. +\\ +\\ +From the ridge loss function the limiting behavior of the variance of the ridge regression estimator can be understood. The ridge penalty with its minimum $\beta = \mathbf{0}_{p}$ does not involve data and, consequently, the variance of its minimum equals zero. With the ridge regression being a compromise between the ML estimator and the minimum of the penalty, so is its variance a compromise of their variances. As $\lambda$ tends to infinity, the ridge estimator and its variance converge to the minimum and the variance of the minimum, respectively. Hence, in the limit (large $\lambda$) the variance of the ridge regression estimator vanishes. Understandably, as the penalty now fully dominates the loss function and, consequently, it does no longer involve data (i.e. randomness). +# original latex figure with scale=0.38, angle=0 + +FIGURE: [convexSets.eps, width=400 frac=1.0] Top panels show examples of convex (left) and nonconvex (right) sets. Middle panels show examples of convex (left) and nonconvex (right) functions. The left bottom panel illustrates the ridge estimation as a constrained estimation problem. The ellipses represent the contours of the ML loss function, with the blue dot at the center the ML estimate. The circle is the ridge parameter constraint. The red dot is the ridge estimate. It is at the intersection of the ridge constraint and the smallest contour with a non-empty intersection with the constraint. The right bottom panel shows the data corresponding to Example ref{exam.overfitting}. The grey line represents the `true' relationship, while the black line the fitted one. } label{fig.ridgeAsConstrainedEst +\afterpage{} +\\ +\\ +Above it has been shown that the ridge estimator can be defined as: + +!bt + +!bt +\begin{align} label{form.ridgeEstViaPenEst} +\hat{\beta}(\lambda) & = \arg \min_{\beta} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2. +\end{align} +!et +!et +This minimization problem can be reformulated into the following constrained optimization problem (illustrated in Figure ref{fig.ridgeAsConstrainedEst}): + +!bt + +!bt +\begin{align} label{form.constrEstProblemRidge} +\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2, +\end{align} +!et +!et +for some suitable $c > 0$. The constrained optimization problem (ref{form.constrEstProblemRidge}) can be solved by means of the Karush-Kuhn-Tucker (KKT) multiplier method, which minimizes a function subject to inequality constraints. The KKT multiplier method states that, under some regularity conditions (all met here), there exists a constant $\nu \geq 0$, called the *multiplier*, such that the solution $\hat{\beta}(\nu)$ of the constrained minimization problem (ref{form.constrEstProblemRidge}) satisfies the so-called KKT conditions. The first KKT condition (referred to as the stationarity condition) demands that the gradient (with respect to $\beta$) of the Lagrangian associated with the minimization problem equals zero at the solution $\hat{\beta}(\nu)$. The Lagrangian for problem (ref{form.constrEstProblemRidge}) is: +\begin{align*} +\| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \nu ( \| \beta \|^2_2 - c). +\end{align*} +The second KKT condition (the complementarity condition) requires that $\nu (\| \hat{\beta}(\nu) \|_2^2 - c) = 0$. If $\nu = \lambda$ and $c = \| \hat{\beta}(\lambda) \|_2^2$, the ridge estimator $\beta (\lambda)$ satisfies both KKT conditions. Hence, both problems have the same solution when $c = \| \hat{\beta}(\lambda) \|_2^2$. +\\ +\\ +The relevance of viewing the ridge regression estimator as the solution to a constrained estimation problem becomes obvious when considering a typical threat to high-dimensional data analysis: overfitting. *Overfitting* refers to the phenomenon of modelling the noise rather than the signal. In case the true model is parsimonious (few covariates driving the response) and data on many covariates are available, it is likely that a linear combination of all covariates yields a higher likelihood than a combination of the few that are actually related to the response. As only the few covariates related to the response contain the signal, the model involving all covariates then cannot but explain more than the signal alone: it also models the error. Hence, it overfits the data. In high-dimensional settings overfitting is a real threat. The number of explanatory variables exceeds the number of observations. It is thus possible to form a linear combination of the covariates that perfectly explains the response, including the noise. + +Large estimates of regression coefficients are often an indication of overfitting. Augmentation of the estimation procedure with a constraint on the regression coefficients is a simple remedy to large parameter estimates. As a consequence it decreases the probability of overfitting. Overfitting is illustrated in the next example. + + +\begin{example} *(Overfitting)* label{exam.overfitting} +\\ +Consider an artificial data set comprising of ten observations on a response $Y_i$ and nine covariates $X_{i,j}$. All covariate data are sampled from the standard normal distribution: $X_{i,j} \sim \mathcal{N}(0, 1)$. The response is generated by $Y_i = X_{i,1} + \varepsilon_i$ with $\varepsilon_{i} \sim \mathcal{N}(0, 1/4)$. Hence, only the first covariate contributes to the response. + +The regression model $Y_i = \sum_{j=1}^9 X_{i,j} \beta_j+ \varepsilon_i$ is fitted to the artificial data using `R`. This yields the regression parameter estimates: +\begin{align*} +\hat{\beta}^{\top} & = (0.048, -2.386, -5.528, 6.243, -4.819, 0.760, -3.345, -4.748, 2.136). +\end{align*} +As $\beta^{\top} = (1, 0, \ldots, 0)$, many regression coefficient are clearly over-estimated. + +The fitted values $\widehat{Y}_i = \mathbf{X}_i \hat{\beta}$ are plotted against the values of the first covariates in the right bottom panel of Figure ref{fig.ridgeAsConstrainedEst}. As a reference the line $x=y$ is added, which represents the `true' model. The fitted model follows the `true' relationship. But it also captures the deviations from this line that represent the errors. +\end{example} + + + +======= Bayesian regression} label{sect:Bayes ======= +Ridge regression has a close connection to Bayesian linear regression. Bayesian linear regression assumes the parameters $\beta$ and $\sigma^2$ to be the random variables, while at the same time considering $\mathbf{X}$ and $\mathbf{Y}$ as fixed. Within the regression context, the conjugate priors of $\beta$ and $\sigma^2$ are: +\begin{align*} +\beta \, | \, \sigma^2 \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp}) \qquad \mbox{and} \qquad \sigma^2 \sim \mathcal{IG}(\alpha_0, \beta_0), +\end{align*} +where $\mathcal{IG}$ denotes the inverse Gamma distribution with shape parameter $\alpha_0$ and scale parameter $\beta_0$. The penalty parameter can be interpreted as the precision of the prior, determining how informative the prior should be. A smaller penalty (i.e. precision) corresponds to a wider prior, and a larger penalty to a more informative, concentrated prior (Figure ref{fig.ridgePriorOfBeta}). +# original latex figure with scale=0.45, angle=0 + +FIGURE: [ridgePriorOfBeta.eps, width=400 frac=1.0] Conjugate prior of the regression parameter $\beta$ for various choices of $\lambda$, the penalty parameters c.q. precision.} label{fig.ridgePriorOfBeta + +Under the assumption of the conjugate priors above, the joint posterior distribution of $\beta$ and $\sigma^2$ is then: +\begin{align*} +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) +\\ +& \propto & \sigma^{-n} \exp \Big[ - \frac{1}{2\sigma^2} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) \Big] +\\ +& & \times \, \, \sigma^{-p} \exp \Big[ - \frac{1}{2\sigma^2} \lambda \beta^{\top} \beta \Big] \, \times \, \, [\sigma^2]^{-\alpha_0-1} \exp \Big[ - \frac{\beta_0}{2\sigma^2} \Big]. +\end{align*} +As +\begin{align*} +& & \hspace{-1.5cm} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) + \lambda \beta^{\top} \beta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - +\beta^{\top} \mathbf{X}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} \beta + \beta^{\top} \mathbf{X}^{\top} \mathbf{X} \beta + \lambda \beta^{\top} \beta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& & - \, \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) +\hat{\beta} (\lambda) +\\ +& & - \, [ \hat{\beta} (\lambda) ]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta +\\ +& = \mathbf{Y}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& & + \, \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big], +\end{align*} +the posterior distribution can be rewritten to: +\begin{align*} +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +\end{align*} +with +\begin{align*} +g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big] \Big\}. +\end{align*} +Then, clearly the conditional posterior mean of $\beta$ is $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\lambda)$. Hence, the ridge regression estimator can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a Gaussian prior on the regression parameter. + +With little extra work we may also obtain the conditional posterior of $\sigma^2$ from the joint posterior distribution: +\begin{align*} +f_{\sigma^2} (\sigma^2 \, | \, \beta, \mathbf{Y}, \mathbf{X}) & \propto & (\sigma^2)^{-[(n+p)/2 + \alpha_0 + 1]} +\exp [ - \frac{1}{2\sigma^2} ( \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + +\lambda \| \beta \|_2^2 + \beta_0) ], +\end{align*} +in which one can recognize the shape of an inverse gamma distribution. +\\ +\\ +A Bayesian estimator of a parameter $\ttheta$ is the estimator that minimizes the Bayes risk over a prior distribution of the parameter $\ttheta$. The Bayes risk is defined as $\int_{\ttheta} \mathbb{E} [(\hat{\ttheta} - \ttheta)^{\top} (\hat{\ttheta} - \ttheta)] \pi_{\ttheta}(\ttheta; \aalpha) d\ttheta$, where $\pi_{\ttheta}(\ttheta; \aalpha)$ is the prior distribution of $\ttheta$ with hyperparameter $\aalpha$. It is thus a weighted average of the Mean Squared Error, with weights specified through the prior. The Bayes risk is minimized by the mean posterior $\mathbb{E}_{\ttheta}(\ttheta \, | \, \mbox{data})$ (cf., e.g., \citealp{Bijm2017}). The Bayesian estimator of $\ttheta$ thus yields the smallest possible expected MSE, under the assumption of the employed prior. + +The Bayes risk of the ridge estimator over the normal prior $\beta \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp})$ is: +\begin{align*} +\mathbb{E}_{\beta} \{\mbox{MSE}[\hat{\beta}(\lambda)] +\, | \, \sigma^2, \mathbf{Y}, \mathbf{X} \} & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \mathbb{E}_{\beta} [ \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta] +\\ +& = \sigma^2 \, \big\{ \mbox{tr} \big[ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big] + +\lambda^{-1} \mbox{tr} [(\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})] \big\} +\\ +& = \sigma^2 \sum_{j=1}^p (d_{jj}^2 + \lambda)^{-1}, +\end{align*} +in which we have used *i)* the previously derived explicit expression (ref{form.ridgeMSE}) of the ridge estimator's MSE, *ii)* the expectation of the quadratic form of a multivariate random variable \citep{Math1992}, *iii)* the singular value decomposition of $\mathbf{X}$ with singular values $d_{jj}$, and *iv)* the fact that the trace of a square matrix equals the sum of its eigenvalues. As the ridge estimator coincides with the posterior mean, this is the minimal achievable MSE under a zero-centered normal prior with an uncorrelated and equivariant covariance matrix. + +Above the Bayes risk of the ridge estimator factorizes with respect to $\sigma^2$ and $\lambda$. Hence, the larger the hyperparameter $\lambda$ the lower the Bayes risk of the ridge estimator. In particular, its Bayes risk converges to zero as $\lambda \rightarrow \infty$. This can be understood as follows. The limit corresponds to an infinite precision of the prior, thus reducing the variance contribution to the MSE. Moreover, as the ridge estimator shrinks towards zero and the prior distribution of $\beta$ has a zero mean, the bias too vanishes as $\lambda \rightarrow \infty$. + +The calculation of the Bayes risk above relates the Bayesian and frequentist statements on the MSE of the ridge estimator. For the latter revisit Theorem ref{theo.Theobald2} of Section ref{sect:ridgeMSE}, which states the existence of a $\lambda$ such that the resulting ridge estimator has a superior MSE over that of the ML estimator. This result made no assumption on (the distribution of) $\beta$. In fact, it can be viewed as a statement of the MSE conditional on $\beta$. The Bayesian result integrates out the uncertainty - specified by the prior - in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. + + + +# The Bayesian result states that the ridge regression estimator minimizes the weighted MSE (over a normal prior on the parameter $\beta$). It thus integrates out the uncertainty in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. + +======= Degrees of freedom ======= + +The degrees of freedom consumed by ridge regression is calculated. The degrees of freedom may be used in combination with an information criterion to decide on the value of the penalty parameter. Recall from ordinary regression that: +\begin{align*} +\widehat{\mathbf{Y}} & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, \mathbf{H} \mathbf{Y}, +\end{align*} +where $\mathbf{H}$ is the hat matrix. The degrees of freedom used in the regression is then equal to $\mbox{tr}(\mathbf{H})$, the trace of $\mathbf{H}$. In particular, if $ \mathbf{X}$ is of full rank, i.e. $\mbox{rank}(\mathbf{X}) = p$, then $\mbox{tr}(\mathbf{H}) = p$. + +By analogy, the ridge-version of the hat matrix is: +\begin{align*} +\mathbf{H}(\lambda) & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}. +\end{align*} +Continuing this analogy, the degrees of freedom of ridge regression is given by the trace of the ridge hat matrix $\mathbf{H}(\lambda)$: +\begin{align*} +\mbox{tr}[ \mathbf{H}(\lambda)] & = \mbox{tr}[ \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} ] \, \, \, = \, \, \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}. +\end{align*} +The degrees of freedom consumed by ridge regression is monotone decreasing in $\lambda$. In particular: +\begin{align*} +\lim_{\lambda \rightarrow \infty} \mbox{tr}[ \mathbf{H}(\lambda)] & = 0. +\end{align*} +That is, in the limit no information from $\mathbf{X}$ is used. Indeed, $\beta$ is forced to equal $\mathbf{0}_{p}$ which is not derived from data. + + + +======= Efficient calculation ======= + +In the high-dimensional setting the number of covariates $p$ is large compared to the number of samples $n$. In a microarray experiment $p = 40000$ and $n= 100$ is not uncommon. To perform ridge regression in this context, the following expression needs to be evaluated numerically: +\begin{align*} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\end{align*} +For $p=40000$ this requires the inversion of a $40000 \times 40000$ dimensional matrix. This is not feasible on most desktop computers. However, there is a workaround. + +Revisit the singular value decomposition of $\mathbf{X} = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}$ and write $\mathbf{R}_x = \mathbf{U}_x \mathbf{D}_x$. As both $\mathbf{U}_x$ and $\mathbf{D}_x$ are $(n \times n)$-dimensional matrices, so is $\mathbf{R}_x$. Consequently, $\mathbf{X}$ is now decomposed as $\mathbf{X} = \mathbf{R}_x \mathbf{V}_x^{\top}$. The ridge estimator can be rewritten in terms of $\mathbf{R}_x$ and $\mathbf{V}_x$: +\begin{align*} +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{R}_x^{\top} \mathbf{Y}. +\end{align*} +Hence, the reformulated ridge estimator involves the inversion of an $(n \times n)$-dimensional matrix. With $n= 100$ this is feasible on most standard computers. + +cite{Hast2004} point out that the number of computation operations reduces from $\mathcal{O}(p^3)$ to $\mathcal{O}(p n^2)$. In addition, they point out that this computational short-cut can be used in combination with other loss functions, for instance that of standard generalized linear models. +\\ +\\ +Avoidance of the inversion of the $(p \times p)$-dimensional matrix may be achieved in an other way. Hereto one needs the Woodbury identity. Let $\mathbf{A}$, $\mathbf{U}$ and $\mathbf{V}$ be $(p \times p)$-, $(p \times n)$- and $(n \times p)$-dimensional matrices, respectively. +The (simplified form of the) Woodbury identity then is: +\begin{align*} +(\mathbf{A} + \mathbf{U} \mathbf{V})^{-1} & = \mathbf{A}^{-1} - \mathbf{A}^{-1} \mathbf{U} (\mathbf{I}_{nn} + \mathbf{V} \mathbf{A}^{-1} \mathbf{U})^{-1} \mathbf{V} \mathbf{A}^{-1}. +\end{align*} +Application of the Woodbury identity to the matrix inverse in the ridge estimator of the regression parameter gives: +\begin{align*} +(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} & = \lambda^{-1} \mathbf{I}_{pp} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X}. +\end{align*} +This gives: +\begin{align*} +(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} & = \lambda^{-1} \mathbf{X}^{\top} \mathbf{Y} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = \lambda^{-1} \mathbf{X}^{\top} \left[ \mathbf{Y} - \lambda^{-1} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{Y} \right]. +\end{align*} +The inversion of the $(p \times p)$-dimensional matrix $\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X}$ is thus replaced by that of the $(n \times n)$-dimensional matrix $\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top}$. In addition, this expression of the ridge regression estimator avoids the singular value decomposition of $\mathbf{X}$, which may in some cases introduce additional numerical errors (e.g. at the level of machine precision). + + + +======= Choice of the penalty parameter ======= + +Throughout the introduction of ridge regression and the subsequent discussion of its properties the penalty parameter is considered known or `given'. In practice, it is unknown and the user needs to make an informed decision on its value. Several strategies to facilitate such a decision are presented. + +===== Information criterion ===== + +A popular strategy is to choose a penalty parameter that yields a good but parsimonious model. Information criteria measure the balance between model fit and model complexity. Here we present the Aikaike's information criterion (AIC), but many other criteria have been presented in the literature (e.g. \citealp{Akai1974}, \citealp{Schw1978}). The AIC measures model fit by the log-likelihood and model complexity is measured by the number of parameters used by the model. The number of model parameters in regular regression simply corresponds to the number of covariates in the model. Or, by the degrees of freedom consumed by the model, which is equivalent to the trace of the hat matrix. For ridge regression it thus seems natural to define model complexity +analogously by the trace of the ridge hat matrix. This yields the AIC for the linear regression model with ridge estimates: +\begin{align*} +\mbox{AIC}(\lambda) & = 2 \, p - 2 \log(\hat{L}) +\\ +& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\} +\\ +& = 2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda} ++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2. +\end{align*} +The value of $\lambda$ which minimizes $\mbox{AIC}(\lambda)$ corresponds to the `optimal' balance of model complexity and overfitting. + +Information criteria guide the decision process when having to decide among various different models. Different models use different sets of explanatory variables to explain the behaviour of the response variable. In that sense, the use of information criteria for the deciding on the ridge penalty parameter may be considered inappropriate: ridge regression uses the same set of explanatory variables irrespective of the value of the penalty parameter. Moreover, often ridge regression is employed to predict a response and not to provide an insightful explanatory model. The latter need not yield the best predictions. Finally, empirically we observe that the AIC often does not show an optimum *inside* the domain of the ridge penalty parameter. Henceforth, we refrain from the use of the AIC (or any of its relatives) in determining the optimal ridge penalty parameter. + + +===== Cross-validation} label{subsect.crossvalidation ===== +Instead of choosing the penalty parameter to balance model fit with model complexity, cross-validation requires it (i.e. the penalty parameter) to yield a model with good prediction performance. Commonly, this performance is evaluated on novel data. Novel data need not be easy to come by and one has to make do with the data at hand. The setting of `original' and novel data is then mimicked by sample splitting: the data set is divided into two (groups of samples). One of these two data sets, called the *training set*, plays the role of `original' data on which the model is built. The second of these data sets, called the *test set*, plays the role of the `novel' data and is used to evaluate the prediction performance (often operationalized as the log-likelihood or the prediction error) of the model built on the training data set. This procedure (model building and prediction evaluation on training and test set, respectively) is done for a collection of possible penalty parameter choices. The penalty parameter that yields the model with the best prediction performance is to be preferred. The thus obtained performance evaluation depends on the actual split of the data set. To remove this dependence the data set is split many times into a training and test set. For each split the model parameters are estimated for all choices of $\lambda$ using the training data and estimated parameters are evaluated on the corresponding test set. The penalty parameter that on average over the test sets performs best (in some sense) is then selected. + +When the repetitive splitting of the data set is done randomly, samples may accidently end up in a fast majority of the splits in either training or test set. Such samples may have an unbalanced influence on either model building or prediction evaluation. To avoid this $k$-fold cross-validation structures the data splitting. The samples are divided into $k$ more or less equally sized exhaustive and mutually exclusive subsets. In turn (at each split) one of these subsets plays the role of the test set while the union of the remaining subsets constitutes the training set. Such a splitting warrants a balanced representation of each sample in both training and test set over the splits. Still the division into the $k$ subsets involves a degree of randomness. This may be fully excluded when choosing $k=n$. This particular case is referred to as leave-one-out cross-validation (LOOCV). For illustration purposes the LOOCV procedure is detailed fully below: +\begin{compactitem} +\item[0)] Define a range of interest for the penalty parameter. + +\item[1)] Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively. + +\item[2)] Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set. This yields: +\begin{align*} +\hat{\beta}_{-i}(\lambda) & = ( \mathbf{X}_{-i, \ast}^{\top} +\mathbf{X}_{-i, \ast} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{-i, \ast}^{\top} \mathbf{Y}_{-i} +\end{align*} +and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$. + +\item[3)] Evaluate the prediction performance of these models on the test set by $\log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)|$, possibly squared. + +\item[4)] Repeat steps 1) to 3) such that each sample plays the role of the test set once. + +\item[5)] Average the prediction performances of the test sets at each grid point of the penalty parameter: +\begin{align*} +\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\end{align*} +The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. + +\item[6)] The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. +\end{compactitem} +The procedure is straightforwardly adopted to $k$-fold cross-validation, a different criterion, and different estimators. + +In the LOOCV procedure above resampling can be avoided when the prediction performance is measured by Allen's PRESS (Predicted Residual Error Sum of Squares) statistic \citep{Alle1974}. For then, the LOOCV prediction performance can be expressed analytically in terms of the known quantities derived from the design matrix and response (as pointed out but not detailed in \citealt{Golu1979}). Define the optimal penalty parameter to minimize Allen's PRESS statistic: +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. +\end{align*} +To derive an analytic expression for the right-hand side first +rewrite $(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1}$ by means of the Woodbury identity as: +\begin{align*} +(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} - \mathbf{X}_{i, \ast}^{\top} \mathbf{X}_{i, \ast})^{-1} +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}]^{-1} +\\ +& & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\end{align*} +with $\mathbf{H}_{ii}(\lambda) = \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. +Furthermore, $\mathbf{X}_{-i}^{\top} \mathbf{Y}_{-i} = \mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i$. Substitute both in the leave-one-out ridge regression estimator and manipulate: +\begin{align*} +\hat{\beta}_{- i}(\lambda) & = (\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{- i, \ast}^{\top} \mathbf{Y}_{- i} +\\ +& = \{(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} ++ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}\} +\\ +& & \times (\mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i ) +\\ +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} Y_i +\\ +& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii} (\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}^{\top} \mathbf{Y} +\\ +& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} +\mathbf{X}_{i, \ast}^{\top} Y_i +\\ +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i +\\ +& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) +\\ +& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{H}_{ii} (\lambda) Y_i +\\ +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \{ [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii}(\lambda) Y_i \} +\\ +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ]. +\end{align*} +The latter enables the reformulation of the prediction error as: +\begin{align*} +Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda) & = Y_i - \mathbf{X}_{i, \ast} \{ \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda)] \} +\\ +& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] +\\ +& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii} (\lambda)[ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] +\\ +& = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast}^{\top} \hat{\beta}(\lambda) ], +\end{align*} +which in turn results in the re-expression of Allen's PRESS statistic: +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} & = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2 +\, \, \, = \, \, \, \arg \min_{\lambda} \tfrac{1}{n} \| \mathbf{B}(\lambda) [\mathbf{I}_{nn} - \mathbf{H}(\lambda)] \mathbf{Y} \|_ F^2, +\end{align*} +where $\mathbf{B}(\lambda)$ is diagonal with $[\mathbf{B}(\lambda)]_{ii} = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1}$. Hence, the prediction performance for a given $\lambda$ can be assessed directly from the ridge hat matrix and the response vector without the recalculation of the $n$ leave-one-out ridge estimators. Computationally, this is a considerable gain. + + + +======= Simulations ======= + +Simulations are presented that illustrate properties of the ridge estimator not discussed explicitly in the previous sections of this chapter. + +===== Role of the variance of the covariates} label{ridge:covariateVariances ===== +In many applications of high-dimensional data the covariates are standardized prior to the execution of the ridge regression. Before we discuss whether this is appropriate, we first illustrate the effect of ridge penalization on covariates with distinct variances using simulated data. + +The simulation involves one response to be (ridge) regressed on fifty covariates. Data (with $n=1000$) for the covariates, denoted $\mathbf{X}$, are drawn from a multivariate normal distribution: $\mathbf{X} \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{\Sigma})$ with $\mathbf{\Sigma}$ diagonal and $(\mathbf{\Sigma})_{jj} = j / 10$. From this the response is generated through $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = \mathbf{1}_{50}$ and $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{I}_{50 \times 50})$. + +With the simulated data at hand the ridge regression estimates of $\beta$ are evaluated for a large grid of the penalty parameter $\lambda$. The resulting ridge regularization paths of the regression coefficients are plotted (Figure ref{fig.effectOfRidge_effectOfVariances}). All paths start ($\lambda=0$) close to one and vanish as $\lambda \rightarrow \infty$. However, ridge regularization paths of regression coefficients corresponding to covariates with a large variance dominate those with a low variance. + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [covariatesWithDifferentVariance.eps, width=400 frac=1.0] Top panel: Ridge regularization paths for coefficients of the 50 uncorrelated covariates with distinct variances. Color and line type indicated the grouping of the covariates by their variance. Bottom panels: Graphical illustration of the effect of a covariate's variance on the ridge estimator. The grey circle depicts the ridge parameter constraint. The dashed black ellipsoids are the level sets of the least squares loss function. The red dot is the ridge regression estimate. Left and right panels represent the cases with equal and unequal, respectively, variances of the covariates.} label{fig.effectOfRidge_effectOfVariances + + + + + +Ridge regression's preference of covariates with a large variance can intuitively be understood as follows. First note that the ridge regression estimator now can be written as: +\begin{align*} +\beta (\lambda) & = [ \mbox{Var}(\mathbf{X}) + \lambda \mathbf{I}_{50 \times 50}]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) +\\ +& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma [ \mbox{Var}(\mathbf{X}) ]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) +\\ +& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma \beta. +\end{align*} +Plug in the employed parametrization of $\mathbf{\Sigma}$, which gives: +\begin{align*} +[\beta (\lambda)]_j & = \frac{j}{j + 50 \lambda} \, (\beta)_j. +\end{align*} +Hence, the larger the covariate's variance (corresponding to the larger $j$), the larger its ridge regression coefficient estimate. Ridge regression thus prefers (among a set of covariates with comparable effect sizes) those with larger variances. + + + +The reformulation of ridge penalized estimation as a constrained estimation problem offers a geometrical interpretation of this phenomenon. Let $p=2$ and the design matrix $\mathbf{X}$ be orthogonal, while both covariates contribute equally to the response. Contrast the cases with $\mbox{Var}(X_1) \approx \mbox{Var}(X_2)$ and $\mbox{Var}(X_1) \gg \mbox{Var}(X_2)$. The level sets of the least squares loss function associated with the former case are circular, while that of the latter are strongly ellipsoidal (see Figure ref{fig.effectOfRidge_effectOfVariances}). The diameters along the principal axes (that - due to the orthogonality of $\mathbf{X}$ - are parallel to that of the $\beta_1$- and $\beta_2$-axes) of both circle and ellipsoid are reciprocals of the variance of the covariates. When the variances of both covariates are equal, the level sets of the loss function expand equally fast along both axis. With the two covariates having the same regression coefficient, the point of these level sets closest to the parameter constraint is to be found on the line $\beta_1 = \beta_2$ (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). Consequently, the ridge regression estimate satisfies $\hat{\beta}_1 (\lambda) \approx \hat{\beta}_2(\lambda)$. With unequal variances between the covariates, the ellipsoidal level sets of the loss function have diameters of rather different sizes. In particular, along the $\beta_1$-axis it is narrow (as $\mbox{Var}(X_1)$ is large), and - vice versa - wide along the $\beta_2$-axis. Consequently, the point of these level sets closest to the circular parameter constraint will be closer to the $\beta_1$- than to the $\beta_2$-axis (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). For the ridge estimates of the regression parameter this implies $0 \ll \hat{\beta}_1 (\lambda) < 1$ and $0 < \hat{\beta}_2 (\lambda) \ll 1$. Hence, the covariate with a larger variance yields the larger ridge regression estimate. + + +Should one thus standardize the covariates prior to ridge regression analysis? When dealing with gene expression data from microarrays, the data have been subjected to a series of pre-processing steps (e.g. quality control, background correction, within- and between-normalization). The purpose of these steps is to make the expression levels of genes comparable both within and between hybridizations. The preprocessing should thus be considered an inherent part of the measurement. As such it is to be done independently of whatever down-stream analysis is to follow and further tinkering with the data is preferably to be avoided (as it may mess up the `comparable-ness' of the expression levels as achieved by the preprocessing). For other data types different considerations may apply. + +Among the considerations to decide on standardization of the covariates, one should also include the fact that ridge estimates prior and posterior to scaling do not simply differ by a factor. To see this assume that the covariates have been centered. Scaling of the covariates amounts to post-multiplication of the design matrix by a $(p \times p)$-dimensional diagonal matrix $\mathbf{A}$ with the reciprocals of the covariates' scale estimates on its diagonal \citep{Sard2008}. Hence, the ridge estimator (for the rescaled data) is then given by: +\begin{align*} +\min_{\beta} \| \mathbf{Y} - \mathbf{X} \mathbf{A} \beta \|_2^2 + \lambda \| \beta \|_ 2^2. +\end{align*} +Apply the change-of-variable $\ggamma = \mathbf{A} \beta$ and obtain: +\begin{align*} +\min_{\ggamma} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \lambda \| \mathbf{A}^{-1} \ggamma \|_ 2^2 & = \min_{\beta} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \sum_{j=1}^p \lambda [(\mathbf{A})_{jj}]^{-2} \gamma_j^2. +\end{align*} +Effectively, the scaling is equivalent to covariate-wise penalization. The `scaled' ridge estimator may then be derived along the same lines as before in Section ref{sect.constrainedEstimation}: +\begin{align*} +\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) & = \mathbf{A}^{-1} \hat{\ggamma} (\lambda) +\, \, \, = \, \, \, \mathbf{A}^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{A}^{-2})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\end{align*} +In general, this is unequal to the ridge estimator without the rescaling of the columns of the design matrix. Moreover, it should be clear that $\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) \not= \mathbf{A} \hat{\beta}(\lambda)$. + +===== Ridge regression and collinearity} label{sect:collinearCovariates ===== +Initially, ridge regression was motivated as an ad-hoc fix of (super)-collinear covariates in order to obtain a well-defined estimator. We now study the effect of this ad-hoc fix on the regression coefficient estimates of collinear covariates. In particular, their ridge regularization paths are contrasted to those of `non-collinear' covariates. + +To this end, we consider a simulation in which one response is regressed on 50 covariates. The data of these covariates, stored in a design matrix denoted $\mathbf{X}$, are sampled from a multivariate normal distribution, with mean zero and a $5 \times 5$ blocked covariance matrix: +\begin{align*} +\mathbf{\Sigma} & = \left( +\begin{array}{ccccc} +\mathbf{\Sigma}_{11} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{22} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{\Sigma}_{33} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{44} & \mathbf{0}_{10 \times 10} +\\ +\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} +& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{55} +\end{array} +\right) +\end{align*} +with +\begin{align*} +\mathbf{\Sigma}_{kk} & = \frac{k-1}{5} \, \mathbf{1}_{10 \times 10} + \frac{6-k}{5} \, \mathbf{I}_{10 \times 10}. +\end{align*} +The data of the response variable $\mathbf{Y}$ are then obtained through: $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, with $\vvarepsilon \sim \mathcal{N}( \mathbf{0}_{n}, \mathbf{I}_{nn})$ and $\beta = \mathbf{1}_{50}$. Hence, all covariates contribute equally to the response. Would the columns of $\mathbf{X}$ be orthogonal, little difference in the ridge estimates of the regression coefficients is expected. + +The results of this simulation study with sample size $n=1000$ are presented in Figure ref{fig.ridgeEstimates_correlatedCovariates}. All 50 regularization paths start close to one as $\lambda$ is small and converge to zero as $\lambda \rightarrow \infty$. But the paths of covariates of the same block of the covariance matrix $\mathbf{\Sigma}$ quickly group, with those corresponding to a block with larger off-diagonal elements above those with smaller ones. Thus, ridge regression prefers (i.e. shrinks less) coefficient estimates of strongly positively correlated covariates. + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [correlatedCovariates.eps, width=400 frac=1.0] Left panel: Ridge regularization paths for coefficients of the 50 covariates, with various degree of collinearity but equal variance. Color and line type correspond to the five blocks of the covariate matrix $\mathbf{\Sigma}$. Right panel: Graphical illustration of the effect of the collinearity among covariates on the ridge estimator. The solid and dotted grey circles depict the ridge parameter constraint for the collinear and orthogonal cases, respectively. The dashed black ellipsoids are the level sets of the sum-of-squares squares loss function. The red dot and violet diamond are the ridge regression for the positive collinear and orthogonal case, respectively.} label{fig.ridgeEstimates_correlatedCovariates +# \afterpage{} + + + +Intuitive understanding of the observed behaviour may be obtained from the $p=2$ case. Let $U$, $V$ and $\varepsilon$ be independent random variables with zero mean. Define $X_1 = U + V$, $X_2 = U - V$, and $Y = \beta_1 X_1 + \beta_2 X_2 + \varepsilon$ with $\beta_1$ and $\beta_2$ constants. Hence, $\mathbb{E}(Y) = 0$. Then: +\begin{align*} +# Y & = \beta_1 X_1 + \beta_2 X_2 + \varepsilon +# \\ +# & = \beta_1 (U + V) + \beta_2 (U - V) + \varepsilon +# \\ +Y & = (\beta_1 + \beta_2) U + (\beta_1 - \beta_2) V + \varepsilon +\\ +& = \gamma_u U + \gamma_v V + \varepsilon +\end{align*} +and $\mbox{Cor}(X_{1}, X_{2}) = [\mbox{Var}(U) - \mbox{Var}(V)] / [ \mbox{Var}(U) + \mbox{Var}(V) ]$. +# \begin{align*} +# \mbox{Cor}(X_{1}, X_{2}) +# & = \frac{ \mbox{Cov}(X_{1}, X_{2}) }{ \sqrt{\mbox{Var}(X_{1})} \sqrt{\mbox{Var}(X_{2})} } +# \\ +# & = \frac{ \mbox{Cov}(U + V, U - V) }{ \sqrt{\mbox{Var}(U + V)} \sqrt{\mbox{Var}(U - V)} } +# \\ +# & = +# \frac{ \mbox{Var}(U) - \mbox{Var}(V) }{ \mbox{Var}(U) + \mbox{Var}(V) }. +# \end{align*} +The random variables $X_1$ and $X_2$ are strongly positively correlated if $\mbox{Var}(U) \gg \mbox{Var}(V)$. + +The ridge regression estimator associated with regression of $Y$ on $U$ and $V$ is: +\begin{align*} +\ggamma(\lambda) & = \left( +\begin{array}{rr} +\mbox{Var}(U) + \lambda & 0 +\\ +0 & \mbox{Var}(V) + \lambda +\end{array} +\right)^{-1} +\left( +\begin{array}{r} +\mbox{Cov}(U, Y) +\\ +\mbox{Cov}(V, Y) +\end{array} +\right). +\end{align*} +For large enough $\lambda$ +\begin{align*} +\ggamma(\lambda) +# & & \frac{1}{\lambda} \left( \begin{array}{rr} \mbox{Var}(U) & 0 \\ 0 \mbox{Var}(V) \end{array} \right) \left( \begin{array}{rr} \mbox{Var}(U) & 0 +# \\ +# 0 & \mbox{Var}(V) +# \end{array} +# \right)^{-1} +# \left( +# \begin{array}{r} +# \mbox{Cov}(U, Y) +# \\ +# \mbox{Cov}(V, Y) +# \end{array} +# \right) +# \\ +& \approx \frac{1}{\lambda} +\left( +\begin{array}{rr} +\mbox{Var}(U) & 0 +\\ +0 & \mbox{Var}(V) +\end{array} +\right) +\left( +\begin{array}{r} +\beta_1 + \beta_2 +\\ +\beta_1 - \beta_2 +\end{array} +\right). +\end{align*} +When $\mbox{Var}(U) \gg \mbox{Var}(V)$ and $\beta_1 \approx \beta_2$, the ridge estimate of $\gamma_v$ vanishes for large $\lambda$. Hence, ridge regression prefers positively covariates with similar effect sizes. + + +This phenomenon too can be explained geometrically. For the illustration consider ridge estimation with $\lambda=1$ of the linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = (3, 3)^{\top}$, $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_2, \mathbf{I}_{22})$ and the columns of $\mathbf{X}$ strongly and positively collinear. The level sets of the sum-of-squares loss, $\| \mathbf{Y} - \mathbf{X} \beta \|_2^2$, are plotted in the right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}. Recall that the ridge estimate is found by looking for the smallest loss level set that hits the ridge contraint. The sought-for estimate is then the point of intersection between this level set and the constraint, and - for the case at hand - is on the $x=y$-line. This is no different from the case with orthogonal $\mathbf{X}$ columns. Yet their estimates differ, even though the same $\lambda$ is applied. The difference is to due to fact that the radius of the ridge constraint depends on $\lambda$, $\mathbf{X}$ and $\mathbf{Y}$. This is immediate from the fact that the radius of the constraint equals $\| \hat{\beta}(\lambda) \|_2^2$ (see Section ref{sect.constrainedEstimation}). To study the effect of $\mathbf{X}$ on the radius, we remove its dependence on $\mathbf{Y}$ by considering its expectation, which is: +\begin{align*} +\mathbb{E}[ \| \hat{\beta}(\lambda) \|_2^2 ] & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta}]^{\top} \, (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \} \nonumber % \, \, \, = \, \, \, \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) \nonumber +\\ +# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +# \\ +# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X})(\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +# \\ +& = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber +\\ +& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \big\} + \beta^{\top} \mathbf{X}^{ \top} \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \, \mathbf{X} \beta. % label{form.expectationPenalty} +\end{align*} +In the last step we have used $\mathbf{Y} \sim \mathcal{N}( \mathbf{X} \beta, \sigma^2 \mathbf{I}_{pp})$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ is $\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}$ (cf. \citealp{Math1992}). The expression for the expectation of the radius of the ridge constraint can now be evaluated for the orthogonal $\mathbf{X}$ and the strongly, positively collinear $\mathbf{X}$. It turns out that the latter is larger than the former. This results in a larger ridge constraint. For the larger ridge constraint there is a smaller level set that hits it first. The point of intersection, still on the $x=y$-line, is now thus closer to $\beta$ and further from the origin (cf. right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}). The resulting estimate is thus larger than that from the orthogonal case. + +The above needs some attenuation. Among others it depends on: *i)* the number of covariates in each block, *ii)* the size of the effects, i.e. regression coefficients of each covariate, and +*iii)* the degree of collinearity. Possibly, there are more factors influencing the behaviour of the ridge estimator presented in this subsection. + +This behaviour of ridge regression is to be understood when using (say) gene expression data to predict a certain clinical outcome. Genes work in concert to fulfil a certain function in the cell. Consequently, one expects their expression levels to be correlated. Indeed, gene expression studies exhibit many co-expressed genes, that is, genes with correlating transcript levels. + + + + +======= Illustration} label{sect.ridgeRegressionDataIllustration ======= +The application of ridge regression to actual data aims to illustrate its use in practice. + +===== MCM7 expression regulation by microRNAs ===== + +Recently, a new class of RNA was discovered, referred to as microRNA. MicroRNAs are non-coding, single stranded RNAs of approximately 22 nucleotides. Like mRNAs, microRNAs are encoded in and transcribed from the DNA. MicroRNAs play an important role in the regulatory mechanism of the cell. MicroRNAs down-regulate gene expression by either of two post-transcriptional mechanisms: mRNA cleavage or transcriptional repression. This depends on the degree of complementarity between the microRNA and the target. Perfect or nearly perfect complementarity of the mRNA to the microRNA will lead to cleavage and degradation of the target mRNA. Imperfect complementarity will repress the productive translation and reduction in protein levels without affecting the mRNA levels. +A single microRNA can bind to and regulate many different mRNA targets. Conversely, several microRNAs can bind to and cooperatively control a single mRNA target (\citealp{Bart2004}; \citealp{Esqu2006}; \citealp{Kim2006}). + +In this illustration we wish to confirm the regulation of mRNA expression by microRNAs in an independent data set. We cherry pick an arbitrary finding from literature reported in cite{Ambs2008}, which focusses on the microRNA regulation of the MCM7 gene in prostate cancer. The MCM7 gene is involved in DNA replication \citep{Tye1999}, a cellular process often derailed in cancer. Furthermore, MCM7 interacts with the tumor-suppressor gene RB1 \citep{Ster1998}. Several studies indeed confirm the involvement of MCM7 in prostate cancer \citep{Padm2004}. And recently, it has been reported that in prostate cancer MCM7 may be regulated by microRNAs \citep{Ambs2008}. + +We here assess whether the MCM7 down-regulation by microRNAs can be observed in a data set other than the one upon which the microRNA-regulation of MCM7 claim has been based. To this end we download from the Gene Expression Omnibus (GEO) a prostate cancer data set (presented by \citealp{Wang2009GeneNetworks}). This data set (with GEO identifier: GSE20161) has both mRNA and microRNA profiles for all samples available. The preprocessed (as detailed in \citealp{Wang2009GeneNetworks}) data are downloaded and require only minor further manipulations to suit our purpose. These manipulations comprise *i)* averaging of duplicated profiles of several samples, *ii)* gene- and mir-wise zero-centering of the expression data, *iii)* averaging the expression levels of the probes that interrogate MCM7. Eventually, this leaves 90 profiles each comprising of 735 microRNA expression measurements. + + + +\lstinputlisting{prostate_ridgeAnalysis.r} + + +With this prostate data set at hand we now investigate whether MCM7 is regulated by microRNAs. Hereto we fit a linear regression model regressing the expression levels of MCM7 onto those of the microRNAs. As the number of microRNAs exceeds the number of samples, ordinary least squares fails and we resort to the ridge estimator of the regression coefficients. First, an informed choice of the penalty parameter is made through maximization of the LOOCV log-likelihood, resulting in $\lambda_{\mbox{{\tiny opt}}} = 1812.826$. Having decided on the value of the to-be-employed penalty parameter, the ridge regression estimator can now readily be evaluated. The thus fitted model allows for the evaluation of microRNA-regulation of MCM7. E.g., by the proportion of variation of the MCM7 expression levels by the microRNAs as expressed in coefficient of determination: $R^2 = 0.4492$. Alternatively, but closely related, observed expression levels may be related to the linear predictor of the MCM7 expression levels: $\hat{\mathbf{Y}}(\lambda_{\mbox{{\tiny opt}}}) = \mathbf{X} \hat{\beta} (\lambda_{\mbox{{\tiny opt}}})$. The Spearman correlation of response and predictor equals 0.6295. A visual inspection is provided by the left panel of Figure ref{fig.RidgeProstateExample}. Note the difference in scale of the $x$- and $y$-axes. This is due to the fact that the regression coefficients have been estimated in penalized fashion, consequently shrinking estimates of the regression coefficients towards zero leading to small estimates and in turn compressing the range of the linear prediction. The above suggests there is indeed association between the microRNA expression levels and those of MCM7. + + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [prostate_ridgeFit.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values. Right panel: Histogram of the ridge regression coefficient estimates.} label{fig.RidgeProstateExample + + +The overall aim of this illustration was to assess whether microRNA-regulation of MCM7 could also be observed in this prostate cancer data set. In this endeavour the dogma (stating this regulation should be negative) has nowhere been used. A first simple assessment of the validity of this dogma studies the signs of the estimated regression coefficients. The ridge regression estimate has 394 out of the 735 microRNA probes with a negative coefficient. Hence, a small majority has a sign in line with the `microRNA $\downarrow$ mRNA' dogma. When, in addition, taking the size of these coefficients into account (Figure ref{fig.RidgeProstateExample}, right panel), the negative regression coefficient estimates do not substantially differ from their positive counterparts (as can be witnessed from their almost symmetrical distribution around zero). Hence, the value of the `microRNA $\downarrow$ mRNA' dogma is not confirmed by this ridge regression analysis of the MCM7-regulation by microRNAs. Nor is it refuted. + + +The implementation of ridge regression in the {\tt penalized}-package offers the possibility to fully obey the dogma on negative regulation of mRNA expression by microRNAs. This requires all regression coefficients to be negative. Incorporation of the requirement into the ridge estimation augments the constrained estimation problem with an additional constraint: +\begin{align*} % label{form.constrEstProblemRidge} +\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c (\lambda) \atop \beta_j \leq 0 \, \mbox{{\tiny for all $j$}}} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. +\end{align*} +With the additional non-positivity constraint on the parameters, there is no explicit solution for the estimator. The ridge estimate of the regression parameters is then found by numerical optimization using e.g. the Newton-Raphson algorithm or a gradient descent approach. The next listing gives the R-code for ridge estimation with the non-positivity constraint of the linear regression model. + +\lstinputlisting{prostate_ridgeAnalysisWithConstraint.r} + +The linear regression model linking MCM7 expression to that of the microRNAs is fitted by ridge regression while simultaneously obeying the `negative regulation of mRNA by microRNA'-dogma to the prostate cancer data. In the resulting model 401 out of 735 microRNA probes have a nonzero (and negative) coefficient. There is a large overlap in microRNAs with a negative coefficient between those from this and the previous fit. The models are also compared in terms of their fit to the data. The Spearman rank correlation coefficient between response and predictor for the model without positive regression coefficients equals 0.679 and its coefficient of determination 0.524 (confer the left panel of ref{fig.RidgeProstateExample_constrainedAnalysis} for a visualization). This is a slight improvement upon the unconstrained ridge estimated model. The improvement may be small but it should be kept in mind that the number of parameters used by both models is 401 (for the model without positive regression coefficients) vs. 735. Hence, with close to half the number of parameters the dogma-obeying model gives a somewhat better description of the data. This may suggest that there is some value in the dogma as inclusion of this prior information leads to a more parsimonious model without any loss in fit. + + + +# original latex figure with scale=0.45, angle=0 + +FIGURE: [prostate_ridgeFitConstrained.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values (with the non-positive constraint on the parameters in place). Right panel: Histogram of the ridge regression coefficient estimates (from the non-positivity constrained analysis).} label{fig.RidgeProstateExample_constrainedAnalysis + + + +The dogma-obeying model selects 401 microRNAs that aid in the explanation of the variation in the gene expression levels of MCM7. There is an active field of research, called *target prediction*, trying to identify which microRNAs target the mRNA of which genes. Within {\tt R} there is a collection of packages that provide the target prediction of known microRNAs. The packages differ on the method (e.g. experimental or sequence comparison) that has been used to arrive at the prediction. These target predictions may be used to evaluate the value of the found 401 microRNAs. Ideally, there would be a substantial amount of overlap. The {\tt R}-script that loads the target predictions and does the comparison is below. + +\lstinputlisting{prostate_link2predictionTools.r} + + +|-----------------------------------------------------------------| +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-7pt} +# #endif | | | +|---------r--------------------r--------------------r-------------| +| # #if FORMAT in ("latex", "pdflatex") +\vspace{3pt} +# #endif | $\hat{\beta}_j = 0$ | $\hat{\beta}_j < 0$ | +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif | | | +| microRNA not target | 323 | 390 | +| microRNA target | 11 | 11 | +| # #if FORMAT in ("latex", "pdflatex") +\vspace{-9pt} +# #endif | | | +|-----------------------------------------------------------------| + + +\caption{Cross-tabulation of the microRNAs being a potential target of MCM7 vs. the value of its regression coefficient in the dogma-obeying model.} label{table.mir2mcm7prediction} + + +With knowledge available on each microRNA whether it is predicted (by at least one target prediction package) to be a potential target of MCM7, it may be cross-tabulated against its corresponding regression coefficient estimate in the dogma-obeying model being equal to zero or not. Table ref{table.mir2mcm7prediction} contains the result. Somewhat superfluous considering the data, we may test whether the targets of MCM7 are overrepresented in the group of strictly negatively estimated regression coefficients. The corresponding chi-squared test (with Yates' continuity correction) yields the test statistic $\chi^2 = 0.0478$ with a $p$-value equal to 0.827. Hence, there is no enrichment among the 401 microRNAS of those that have been predicted to target MCM7. This may seem worrisome. However, the microRNAs have been selected for their predictive power of the expression levels of MCM7. Variable selection has not been a criterion (although the sign constraint implies selection). Moreover, criticism on the value of the microRNA target prediction has been accumulating in recent years. + + +======= Conclusion ======= + +We discussed ridge regression as a modification of linear regression to overcome the empirical non-identifiability of the latter when confronted with high-dimensional data. The means to this end was the addition of a (ridge) penalty to the sum-of-squares loss function of the linear regression model, which turned out to be equivalent to constraining the parameter domain. This warranted the identification of the regression coefficients, but came at the cost of introducing bias in the estimates. Several properties of ridge regression like moments, MSE, and its Bayesian interpretation have been reviewed. Finally, its behaviour and use have been illustrated in simulation and omics data. + + +======= Exercises ======= +\begin{question}[^footnote2][^footnote2]: This exercise is freely rendered from cite{Drap1998} \mbox{ } +\\ +Find the ridge regression solution for the data below for a general value of $\lambda$ and for the straight line model $Y = \beta_0 + \beta_1 X + \varepsilon$ (only apply the ridge penalty to the slope parameter, not to the intercept). Show that when $\lambda$ is chosen as 0.4, the ridge solution fit is $\hat{Y} = 40 + 1.75 X$. Data: $\mathbf{X}^{\top} = (X_1, X_2, \ldots, X_{8})^{\top} = (-2, -1, -1, -1, 0, 1, 2, 2)^{\top}$, and $\mathbf{Y}^{\top} = (Y_1, Y_2, \ldots, Y_{8})^{\top} = (35, 40, 36, 38, 40, 43, 45, 43)^{\top}$. +\end{question} + +\begin{question}[^footnote3][^footnote3]: This exercise is freely rendered from cite{Hast2009, but can be found in many other places. The original source is unknown to the author.} label{question:ridgeAugmentation} \mbox{ } +\\ +Show that the ridge regression estimates can be obtained by ordinary least squares regression on an augmented data set. We augment the centered matrix $\mathbf{X}$ with $p$ additional row $\sqrt{\lambda} \mathbf{I}$, and augment $\mathbf{y}$ with $p$ zeros. +\end{question} + +\begin{question} label{question.ridgeResidualsProjection} \mbox{ } +\\ +The coefficients $\beta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, are estimated by $\hat{\beta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. The associated fitted values then given by $\widehat{\mathbf{Y}} = \mathbf{X} \, \hat{\beta} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} = \mathbf{H} \mathbf{Y}$, where $\mathbf{H} =\mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}$ referred to as the hat matrix. The matrix $\mathbf{P}$ is a projection matrix and satisfies $\mathbf{H} = \mathbf{H}^ 2$. Hence, linear regression projects the response $\mathbf{Y}$ onto the vector space spanned by the columns of $\mathbf{Y}$. Consequently, the residuals $\hat{\vvarepsilon}$ and $\hat{\mathbf{Y}}$ are orthogonal. Now consider the ridge estimator of the regression coefficients: $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. Let $\hat{\mathbf{Y}}(\lambda) = \mathbf{X} \hat{\beta}(\lambda)$ be the vector of associated fitted values. + +\begin{compactitem} +\item[*a)*] Show that the matrix $\mathbf{H}(\lambda) = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}$, associated with ridge regression, is not a projection matrix (for any $\lambda > 0$). + +\item[*b)*] Show that the `ridge fit' $\widehat{\mathbf{Y}}(\lambda)$ is not orthogonal to the associated `ridge residuals' $\hat{\vvarepsilon}(\lambda)$ (for any $\lambda > 0$). + +\item[*c)*] Derive the distribution of the `ridge residuals'. +\end{compactitem} +\end{question} + + +\begin{question} label{question.MSEofLinearPredictor} \mbox{ } +\\ +Recall that there exists $\lambda > 0$ such that $MSE(\hat{\beta}) > MSE[\hat{\beta}(\lambda)]$. Verify that this carries over to the linear predictor. That is, there exists a $\lambda > 0$ such that$MSE(\widehat{\mathbf{Y}}) = MSE(\mathbf{X} \hat{\beta}) > MSE[\mathbf{X}\hat{\beta}(\lambda)]$. +\end{question} + +\begin{question} \mbox{ } +\\ +Consider a 3-gene pathway. Expression levels of these three genes have been measured in an observational study involving hundred individuals. In order to assess how the expression levels of gene A are affect by that of genes B and C, a medical researcher fits the +\begin{align*} +Y_{i}^{(A)} &= \beta_b Y_{i}^{(B)} + \beta_c Y_{i}^{(C)} + +\varepsilon_{i}, +\end{align*} +with $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. This model fitted by means of ridge regression, but with a separate penalty parameter, $\lambda_{2,b}$ and $\lambda_{2,c}$, for the two regression coefficient, $\beta_b$ and $\beta_c$, respectively. + +\begin{compactitem} +\item[*a)*] Write down the ridge penalized loss function employed by the researcher. + +\item[*b)*] Does a different choice of penalty parameter for the second regression coefficient affect the estimation of the first regression coefficient? Motivate your answer. + +\item[*c)*] The researcher decides that the second covariate $Y_i^{(C)}$ is irrelevant. Instead of removing the covariate from model, the researcher decides to set $\lambda_{2,c} = +\infty$. Show that this results in the same ridge estimate for $\beta_b$ as when fitting (again by means of ridge regression) the model without the second covariate. +\end{compactitem} +\end{question} + +\begin{question} \mbox{ } +\\ +The expression levels of the $j$-the gene are explained by a linear regression model in terms of those of all other genes. Consider the following two ridge regression estimators of the regression parameter of this model, defined as: +\begin{align*} +\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + \lambda \| \beta_j \|_ 2^2 +\quad +\mbox{and} \quad +\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + n \lambda \| \beta_j \|_ 2^2. +\end{align*} +Which do you prefer? Motivate. +\end{question} + + + + +========= Generalizing ridge regression ========= label{chap:genRidge} +# ======= Generalized ridge regression ======= + +The expos\'{e} on ridge regression may be generalized in many ways. Among others different generalized linear models may be considered (confer Section ref{sect.ridgeLogistic}). In this section we stick to the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with the usual assumptions, but fit it in weighted fashion and generalize the common, spherical penalty. The loss function corresponding to this scenario is: + +!bt + +!bt +\begin{align} label{form:generalizedRidgeLoss} +(\mathbf{Y} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Y} - \mathbf{X} \beta) + (\beta - \beta_0)^{\top} \mathbf{\Delta} +(\beta - \beta_0), +\end{align} +!et +!et +which comprises a weighted least squares criterion and a generalized ridge penalty. In this $\mathbf{W}$ is a $(n \times n)$-dimensional, diagonal matrix with $(\mathbf{W})_{ii} \in [0,1]$ representing the weight of the $i$-th observation. The penalty is now a quadratic form with penalty parameter $\mathbf{\Delta}$, a $(p \times p)$-dimensional, positive definite, symmetric matrix. When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, one regains the spherical penalty of `regular ridge regression'. This penalty shrinks each element of the regression parameter $\beta$ equally along the unit vectors $\mathbf{e}_j$. Generalizing $\mathbf{\Delta}$ to the class of symmetric, positive definite matrices $\mathcal{S}_{++}$ allows for *i)* different penalization per regression parameter, and *ii)* joint (or correlated) shrinkage among the elements of $\beta$. The penalty parameter $\mathbf{\Delta}$ determines the speed and direction of shrinkage. The $p$-dimensional column vector $\beta_0$ is a user-specified, non-random target towards which $\beta$ is shrunken as the penalty parameter increases. When recasting generalized ridge estimation as a constrained estimation problem, the implications of the penalty may be visualized (Figure ref{fig:generalizedRidgeIllustration}, left panel). The generalized ridge penalty is a quadratic form centered around $\beta_0$. In Figure ref{fig:generalizedRidgeIllustration} the parameter constraint clearly is ellipsoidal (and not spherical). Moreover, the center of this ellipsoid is not at zero. + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [generalizedRidgeIllustration_penalty.eps, width=400 frac=1.0] Left panel: the contours of the likelihood (grey solid ellipsoids) and the parameter constraint implied by the generalized penalty (black dashed ellipsoid. Right panel: generalized (fat coloured lines) and `regular' (thin coloured lines) regularization paths of four regression coefficients. The dotted grey (straight) lines indicated the targets towards the generalized ridge penalty shrinks regression coefficient estimates.} label{fig:generalizedRidgeIllustration + + +The addition of the generalized ridge penalty to the sum-of-squares ensures the existence of a unique regression estimator in the face of super-collinearity. The generalized penalty is a non-degenerated quadratic form in $\beta$ due to the positive definiteness of the matrix $\mathbf{\Delta}$. As it is non-degenerate, it is strictly convex. Consequently, the generalized ridge regression loss function (ref{form:generalizedRidgeLoss}), being the sum of a convex and strictly convex function, is also strictly convex. This warrants the existence of a unique global minimum and, thereby, a unique estimator. + + +Like for the `regular' ridge loss function (ref{form.ridgeLossFunction}), there is an explicit expression for the optimum of the generalized ridge loss function (ref{form:generalizedRidgeLoss}). To see this, obtain the estimating equation of $\beta$ through equating its derivative with respect to $\beta$ to zero: +\begin{align*} +2 \mathbf{X}^{\top} \mathbf{W} \mathbf{Y} - 2 \mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta - 2 \mathbf{\Delta} \beta + 2 \mathbf{\Delta} \beta_0 & = \mathbf{0}_{p}. +\end{align*} +This is solved by: + +!bt + +!bt +\begin{align} +\hat{\beta}(\mathbf{\Delta}) & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{Y} + \mathbf{\Delta} \beta_0). +\end{align} +!et +!et +Clearly, this reduces to the `regular' ridge estimator by setting $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$. The effects of the generalized ridge penalty on the estimates can be seen in the regularization paths of the estimates. Figure ref{fig:generalizedRidgeIllustration} (right panel) contains an example of the regularization paths for coefficients of a linear regression model with four explanatory variables. Most striking is the limiting behaviour of the estimates of $\beta_3$ and $\beta_4$ for large values of the penalty parameter $\lambda$: they convergence to a non-zero value (as was specified by a nonzero $\beta_0$). More subtle is the (temporary) convergence of the regularization paths of the estimates of $\beta_2$ and $\beta_3$. That of $\beta_2$ is pulled away from zero (its true value and approximately its unpenalized estimate) towards the estimate of $\beta_3$. In the regularization path of $\beta_3$ this can be observed in a delayed convergence to its nonzero target value (for comparison consider that of $\beta_4$). For reference the corresponding regularization paths of the `regular' ridge estimates (as thinner lines of the same colour) are included in Figure ref{fig:generalizedRidgeIllustration}. + + + +\begin{example} *Fused ridge estimation* label{example.gRidge2orthonormalDesign} +\\ +An example of a generalized ridge penalty is the *fused ridge penalty*. Consider the standard linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$. The fused ridge estimator of $\beta$ then minimizes: + +!bt + +!bt +\begin{align} label{form:fusedRidgeLoss} +\| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2. +\end{align} +!et +!et +The penalty in the loss function above can be written as a generalized ridge penalty: +\begin{align*} +\lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2 & = \beta^{\top} +\left( +\begin{array}{rrrrrr} +\lambda & -\lambda & 0 & \ldots & \ldots & 0 +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +-\lambda & 2 \lambda & -\lambda & \ddots & & \vdots +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +0 & -\lambda & 2 \lambda & \ddots & \ddots & \vdots +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +\vdots & \ddots & \ddots & \ddots & \ddots & 0 +# #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +\vdots & & \ddots & \ddots & \ddots & -\lambda +# # #if FORMAT in ("latex", "pdflatex") +\vspace{-4pt} +# #endif +\\ +0 & \ldots & \ldots & 0 & -\lambda & \lambda +\end{array}\right) +\beta. +\end{align*} +The matrix $\mathbf{\Delta}$ employed above is semi-positive definite and therefore the loss function (ref{form:fusedRidgeLoss}) is not strictly convex. Hence, often a regular ridge penalty $\| \beta \|_2^2$ is added (with its own penalty parameter). + +To illustrate the effect of the fused ridge penalty on the estimation of the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, let $\beta_j = \phi_{0,1}(z_j)$ with $z_j =-30 + \tfrac{6}{50} j$ for $j=1, \ldots, 500$. Sample the elements of the design matrix $\mathbf{X}$ and those of the error vector $\vvarepsilon$ from the standard normal distribution, then form the response $\mathbf{Y}$ from the linear model. The regression parameter is estimated through fused ridge loss minimization with $\lambda=1000$. The estimate is shown in Figure ref{fig:fusedRidgeIllustration} (red line). For reference the figure includes the true $\beta$ (black line) and the `regular ridge' estimate with $\lambda=1$ (blue line). Clearly, the fused ridge estimate yields a nice smooth vector of $\beta$ estimates +\end{example} + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [fusedRidgeIllustration_simulation.eps, width=400 frac=1.0] Left panel: illustration of the fused ridge estimator (in simulation). The true parameter $\beta$ and its ridge and fused ridge estimates against their spatial order. Right panel: Ridge vs. fused ridge estimates of the DNA copy effect on KRAS expression levels. The dashed, grey vertical bar indicates the location of the KRAS gene. } label{fig:fusedRidgeIllustration + + + + +======= Moments ======= + +The expectation and variance of $\hat{\beta}(\mathbf{\Delta})$ are obtained through application of the same matrix algebra and expectation and covariance rules used in the derivation of their counterparts of the `regular' ridge regression estimator. This leads to: +\begin{align*} +\mathbb{E}[\hat{\beta}(\mathbf{\Delta})] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \mathbf{\Delta} \beta_0), +\\ +\mbox{Var}[\hat{\beta}(\mathbf{\Delta})] & = \sigma^2 (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}. +\end{align*} +From these expressions similar limiting behaviour as for the `regular' ridge regression case can be deduced. To this end let $\mathbf{V}_{\delta} \mathbf{D}_{\delta} \mathbf{V}_{\delta}^{\top}$ be the eigendecomposition of $\mathbf{\Delta}$ and $d_{\delta,j} = (\mathbf{D}_{\delta})_{jj}$. Furthermore, define (with some abuse of notation) $\lim_{\mathbf{\Delta} \rightarrow \infty}$ as the limit of all $d_{\delta,j}$ simultaneously tending to infinity. Then, $\lim_{\mathbf{\Delta} \rightarrow \infty} \mathbb{E}[\hat{\beta}(\mathbf{\Delta})] = \beta_0$ and $\lim_{\mathbf{\Delta} \rightarrow \infty} \mbox{Var}[\hat{\beta}(\mathbf{\Delta})] = \mathbf{0}_{pp}$. + + +\begin{example} \mbox{ } label{example.gRidge2orthonormalDesign} +\\ +Let $\mathbf{X}$ be an $(n \times p)$-dimensional, orthonormal design matrix. Contrast the regular and generalized ridge regression estimator, the latter with $\mathbf{W} = \mathbf{I}_{pp}$, $\beta_0 = \mathbf{0}_p$ and $\mathbf{\Delta} = \lambda \mathbf{R}$ where $\mathbf{R} = (1-\rho) \mathbf{I}_{pp} + \rho \mathbf{1}_{pp}$ for $\rho \in (-(p-1)^{-1}, 1)$. For $\rho =0$ the two estimators coincide. The variance of the generalized ridge regression estimator then is $\mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] = (\mathbf{I}_{pp} + \mathbf{\Delta})^{-2}$. The efficiency of this estimator, measured by its generalized variance, is: +\begin{align*} +\det \{ \mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] \} & = \{ [1 + \lambda + (p-1) \rho] (1 + \lambda-\rho)^{p-1} \}^{-2}. +\end{align*} +This efficiency attains its minimum at $\rho = 0$. In the present case, the regular ridge regression estimator is thus more efficient than its generalized counterpart. +\end{example} + + +\begin{example} *(MSE with perfect target)* +\\ +Set $\beta_0 = \beta$, i.e. the target is equal to the true value of the regression parameter. Then: +\begin{align*} +\mathbb{E}[\hat{\beta}(\Delta)] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \Delta)^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \Delta \beta) \, \, \, = \, \, \, \beta. +\end{align*} +Hence, irrespective of the choice of $\Delta$, the generalized ridge is then unbiased. Thus: +\begin{align*} +\mbox{MSE}[\hat{\beta}(\Delta)] & = \mbox{tr} \{ +\mbox{Var}[\hat{\beta}(\Delta)] \} +\\ +& = \mbox{tr}[ \sigma^{2} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}] +\\ +& = \sigma^2 \mbox{tr}[ \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-2}]. +\end{align*} +When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, this MSE is smaller than that of the ML regression estimator, irrespective of the choice of $\lambda$. +\end{example} + + +======= The Bayesian connection} label{sect:genRidgeBayes ======= +This generalized ridge estimator can, like the regular ridge estimator, be viewed as a Bayesian estimator. It requires to replace the conjugate prior on $\beta$ by a more general normal law, $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$, but retains the gamma prior on $\sigma^2$. The joint posterior distribution of $\beta$ and $\sigma^2$ is then obtained analogously (the details are left as Exercise ref{question.generalizedRidgeAndBayes}) to Section ref{sect:Bayes}: +\begin{align*} +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) +\\ +& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +\end{align*} +with +\begin{align*} +g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta}) \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big] \Big\}. +\end{align*} +This implies $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\mathbf{\Delta})$. Hence, the generalized ridge regression estimator too can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a multivariate Gaussian prior on the regression parameter. + + + + + +======= Application ======= + +An illustration involving omics data can be found in the explanation of a gene's expression levels in terms of its DNA copy number. The latter is simply the number of gene copies encoded in the DNA. For instance, for most genes on the autosomal chromosomes the DNA copy number is two, as there is a single gene copy on each chromosome and autosomal chromosomes come in pairs. Alternatively, in males the copy number is one for genes that map to the X or Y chromosome, while in females it is zero for genes on the Y chromosome. In cancer the DNA replication process has often been compromised leading to a (partially) reshuffled and aberrated DNA. Consequently, the cancer cell may exhibit gene copy numbers well over a hundred for classic oncogenes. A faulted replication process does - of course - not nicely follow the boundaries of gene encoding regions. This causes contiguous genes to commonly share aberrated copy numbers. With genes being transcribed from the DNA and a higher DNA copy number implying an enlarged availability of the gene's template, the latter is expected to lead to elevated expression levels. Intuitively, one expects this effect to be localized (a so-called *cis*-effect), but some suggest that aberrations elsewhere in the DNA may directly affect the expression levels of distant genes (referred to as a *trans*-effect). + +The *cis*- and *trans*-effects of DNA copy aberrations on the expression levels of the KRAS oncogene in colorectal cancer are investigated. Data of both molecular levels from the TCGA (The Cancer Genome Atlas) repository are downloaded \citep{TCGA2012colon}. The gene expression data are limited to that of KRAS, while for the DNA copy number data only that of chromosome 12, which harbors KRAS, is retained. This leaves genomic profiles of 195 samples comprising 927 aberrations. Both molecular data types are zero centered feature-wise. Moreover, the data are limited to ten - conveniently chosen? - samples. The KRAS expression levels are explained by the DNA copy number aberrations through the linear regression model. The model is fitted by means of ridge regression, with $\lambda \mathbf{\Delta}$ and $\mathbf{\Delta} = \mathbf{I}_{pp}$ and a single-banded $\mathbf{\Delta}$ with unit diagonal and the elements of the first off-diagonal equal to the arbitrary value of $-0.4$. The latter choice appeals to the spatial structure of the genome and encourages similar regression estimates for contiguous DNA copy numbers. The penalty parameter is chosen by means of leave-one-out cross-validation using the squared error loss. + +\lstinputlisting{fusedRidge_KRASexample.r} + +The right panel of Figure ref{fig:fusedRidgeIllustration} shows the ridge regression estimate with both choices of $\mathbf{\Delta}$ and optimal penalty parameters plotted against the chromosomal order. The location of KRAS is indicated by a vertical dashed bar. The ordinary ridge regression estimates show a minor peak at the location of KRAS but is otherwise more or less flat. In the generalized ridge estimates the peak at KRAS is emphasized. Moreover, the region close to KRAS exhibits clearly elevated estimates, suggesting co-abberated DNA. For the remainder the generalized ridge estimates portray a flat surface, with the exception of a single downward spike away from KRAS. Such negative effects are biologically nonsensible (more gene templates leading to reduced expression levels?). On the whole the generalized ridge estimates point towards the *cis*-effect as the dominant genomic regulation mechanism of KRAS expression. The isolated spike may suggest the presence of a *trans*-effect, but its sign is biological nonsensible and the spike is fully absent in the ordinary ridge estimates. This leads us to ignore the possibility of a genomic *trans*-effect on KRAS expression levels in colorectal cancer. + +The sample selection demands justification. It yields a clear illustrate-able difference between the ordinary and ridge estimates. When all samples are left in, the *cis*-effect is clearly present, discernable from both estimates that yield a virtually similar profile. + + +======= Generalized ridge regression ======= + +What is generally referred to as `generalized ridge regression' (cf. \citealp{Hoer1970, Hemm1975}) is the particular case of loss function (ref{form:generalizedRidgeLoss}) in which $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \mathbf{V}_{x} \mathbf{\Lambda} \mathbf{V}_x^{\top}$, where $\mathbf{V}_x$ is obtained from the singular value decomposition of $\mathbf{X}$ (i.e., $\mathbf{X} = \mathbf{U}_{x} \mathbf{D}_x \mathbf{V}_x^{\top}$ with its constituents endowed with the usual interpretation) and $\mathbf{\Lambda}$ a positive definite diagonal matrix. This gives the estimator: +\begin{align*} +\hat{\beta}(\mathbf{\Lambda}) & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\\ +& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x \mathbf{Y} +\\ +& = \mathbf{V}_x (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x \mathbf{U}_x \mathbf{Y}. +\end{align*} +From this last expression it becomes clear how this estimator generalizes the `regular ridge estimator'. The latter shrinks all eigenvalues, irrespectively of their size, in the same manner through a common penalty parameter. The `generalized ridge estimator', through differing penalty parameters (i.e. the diagonal elements of $\mathbf{\Lambda}$), shrinks them individually. + +The generalized ridge estimator coincides with the Bayesian linear regression estimator with the normal prior $\mathcal{N}[\mathbf{0}_p, (\mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1}]$ on the regression parameter $\beta$ (and preserving the inverse gamma prior on the error variance). Assume $\mathbf{X}$ to be of full column rank and choose $\mathbf{\Lambda} = g^{-1} \mathbf{D}_x^2$ with $g$ a positive scalar. The prior on $\beta$ then - assuming $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits - reduces to Zellner's $g$-prior: $\beta \sim \mathcal{N}[\mathbf{0}_p, g (\mathbf{X}^{\top} \mathbf{X})^{-1}]$ \citep{Zell1986}. The corresponding estimator of the regression coefficient is: $\hat{\beta}(g) = g (1+g)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$, which is proportional to the unpenalized ordinary least squares estimator of $\beta$. + +For convenience of notation in the analysis of the generalized ridge estimator the linear regression model is usually rewritten as: +\begin{align*} +\mathbf{Y} & = \mathbf{X} \beta + \vvarepsilon \, \, \, = \, \, \, +\mathbf{X} \mathbf{V}_x \mathbf{V}_x^{\top} \beta + \vvarepsilon \, \, \, = \, \, \, \tilde{\mathbf{X}} \aalpha + \vvarepsilon, +\end{align*} +with $\tilde{\mathbf{X}} = \mathbf{X} \mathbf{V}_x = \mathbf{U}_x +\mathbf{D}_x$ (and thus $\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} = \mathbf{D}_x^ 2$) and $\aalpha = \mathbf{V}_x^{\top} \beta$ with loss function $(\mathbf{Y} - \tilde{\mathbf{X}} \aalpha)^{\top} (\mathbf{Y} - \tilde{\mathbf{X}} \aalpha) + \aalpha^{\top} \mathbf{\Lambda} \aalpha$. In the notation above the generalized ridge estimator is then: +\begin{align*} +\hat{\aalpha}(\mathbf{\Lambda}) & = (\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} + \mathbf{\Lambda})^ {-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y} = (\mathbf{D}_x^ 2 + \mathbf{\Lambda})^{-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y}, +\end{align*} +from which one obtains $\hat{\beta}(\mathbf{\Lambda}) = \mathbf{V}_x \hat{\aalpha}(\mathbf{\Lambda})$. Using $\mathbb{E}[\hat{\aalpha}(\mathbf{\Lambda})] = (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 \aalpha$ and $\mbox{Var}[\hat{\aalpha}(\mathbf{\Lambda})] = \sigma^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1}$, the MSE for the generalized ridge estimator can be written as: +\begin{align*} +\mbox{MSE}[\hat{\aalpha}(\mathbf{\Lambda})] & = \sum_{j=1}^p ( \sigma^2 d_{x,j}^2 + \alpha_j^2 \lambda_{j}^2 ) (d_{x,j}^2 + \lambda_{j} )^{-2}, +\end{align*} +where $d_{x,j} = (\mathbf{D}_x)_{jj}$ and $\lambda_j = (\mathbf{\Lambda})_{jj}$. Having $\aalpha$ and $\sigma^ 2$ available, it is easily seen (equate the derivative w.r.t. $\lambda_j$ to zero and solve) that the MSE of $\hat{\aalpha}(\mathbf{\Lambda})$ is minimized by $\lambda_j = \sigma^2 / \alpha_j^2$ for all $j$. With $\aalpha$ and $\sigma^2$ unknown, cite{Hoer1970} suggest an iterative procedure to estimate the $\lambda_j$'s. Initiate the procedure with the OLS estimates of $\aalpha$ and $\sigma^2$, followed by sequentially updating the $\lambda_j$'s and +the estimates of $\aalpha$ and $\sigma^2$. An analytic expression of the limit of this procedure exists (\citealp{Hemm1975}). This limit, however, still depends on the observed $\mathbf{Y}$ and as such it does not necessarily yield the minimal attainable value of the MSE. This limit may nonetheless still yield a potential gain in MSE. This is investigated in cite{Lawl1981}. Under a variety of cases it seems to indeed outperform the OLS estimator, but there are exceptions. + +======= Conclusion ======= + +To conclude: a note of caution. The generalized ridge penalty is extremely flexible. It can incorporate any prior knowledge on the parameter values (through specification of $\beta_0$) and the relations among these parameters (via $\mathbf{\Delta}$). While a pilot study or literature may provide a suggestion for $\beta_0$, it is less obvious how to choose an informative $\mathbf{\Delta}$ (although a spatial structure is a nice exception). In general, exact knowledge on the parameters should not be incorporated implicitly via the penalty (read: prior) but preferably be used explicitly in the model - the likelihood - itself. Though this may be the viewpoint of a prudent frequentist and a subjective Bayesian might disagree. + + +======= Exercises ======= +\begin{question} \mbox{ } +\\ +Consider the linear regression model $Y_i = \beta_1 X_{i,1} + \beta_2 X_{i,2} + \varepsilon_i$ for $i=1, \ldots, n$. +Suppose estimates of the regression parameters $(\beta_1, \beta_2)$ of this model are obtained through the minimization of the sum-of-squares augmented with a ridge-type penalty: +\begin{align*} +\Big[ \sum_{i=1}^n (Y_i - \beta_1 X_{i,1} - \beta_2 X_{i,2})^2 \Big] + \lambda (\beta_1^2 + \beta_2^2 + 2 \nu \beta_1 \beta_2), +\end{align*} +with penalty parameters $\lambda \in \mathbb{R}_{> 0}$ and $\nu \in (-1, 1)$. + +\begin{compactitem} +\item[*a)*] Sketch (for both $\nu=0$ and $\nu=0.9$) the shape of the parameter constraint induced by the penalty above and describe in words the qualitative difference between both shapes. + +\item[*b)*] When $\nu = -1$ and $\lambda \rightarrow \infty$ the estimates of $\beta_1$ and $\beta_2$ (resulting from minimization of the penalized loss function above) converge towards each other: +$\lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1) = \lim_{\lambda \rightarrow \infty} \hat{\beta}_2(\lambda, -1)$. Motivated by this observation a data scientists incorporates the equality constraint $\beta_1 = \beta = \beta_2$ explicitly into the model, and s/he estimates the `joint regression parameter' $\beta$ through the minimization (with respect to $\beta$) of: +\begin{align*} +\Big[ \sum_{i=1}^n (Y_i - \beta X_{i,1} - \beta X_{i,2})^2 \Big] + \delta \beta^2, +\end{align*} +with penalty parameter $\delta \in \mathbb{R}_{> 0}$. The data scientist is surprised to find that resulting estimate $\hat{\beta}(\delta)$ does not have the same limiting (in the penalty parameter) behavior as the $\hat{\beta}_1(\lambda, -1)$, i.e. $\lim_{\delta \rightarrow \infty} \hat{\beta} (\delta) \not= \lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1)$. Explain the misconception of the data scientist. + +\item[*c)*] Assume that *i)* $n \gg 2$, *ii)* the unpenalized estimates $(\hat{\beta}_1(0, 0), \hat{\beta}_2(0, 0))$ equal $(-2,2)$, and *iii)* that the two covariates $X_1$ and $X_2$ are zero-centered, have equal variance, and are strongly negatively correlated. Consider $(\hat{\beta}_1(\lambda, \nu), \hat{\beta}_2(\lambda, \nu))$ for both $\nu=-0.9$ and $\nu=0.9$. For which value of $\nu$ do you expect the sum of the absolute value of the estimates to be largest? *Hint:* Distinguish between small and large values of $\lambda$ and think geometrically! +\end{compactitem} +\end{question} + +\begin{question} label{question.generalizedRidgeAndBayes} \mbox{ } +\\ +Consider the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \mathbf{I}_{pp})$. Assume $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$ with $\beta_0 \in \mathbb{R}^p$ and $\mathbf{\Delta} \succ 0$ and a gamma prior on the error variance. Verify (i.e., work out the details of the derivation) that the posterior mean coincides with the generalized ridge estimator defined as: +\begin{align*} +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{Y} + \mathbf{\Delta} \beta_0). +\end{align*} +\end{question} + + + +\begin{question} label{question.ridgeAR1penalty} \mbox{ } +\\ +The ridge penalty may be interpreted as a multivariate normal prior on the regression coefficients: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \mathbf{I}_{pp})$. Different priors may be considered. In case the covariates are spatially related in some sense (e.g. genomically), it may of interest to assume a first-order autoregressive prior: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \SSigma_A)$, in which $\SSigma_A$ is a $p \times p$-correlation matrix with $(\SSigma_A)_{j_1, j_2} = \rho^{ | j_1 - j_2 | } $ for some correlation coefficient $\rho \in [0, 1)$. Hence, +\begin{align*} +\SSigma_A \, \, \, = \, \, \, +\left( +\begin{array}{cccc} +1 & \rho & \ldots & \rho^{p-1} +\\ +\rho & 1 & \ldots & \rho^{p-2} +\\ +\vdots & \vdots & \ddots & \vdots +\\ +\rho^{p-1} & \rho^{p-2} & \ldots & 1 +\end{array} +\right). +\end{align*} + +\begin{compactitem} +\item[*a)*] The penalized loss function associated with this AR(1) prior is: +\begin{align*} +\mathcal{L}(\beta; \lambda, \SSigma_A) & = \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \beta^{\top} \SSigma_A^{-1} \beta. +\end{align*} +Find the minimizer of this loss function. + +\item[*b)*] What is the effect of $\rho$ on the ridge estimates? Contrast this to the effect of $\lambda$. Illustrate this on (simulated) data. + +\item[*c)*] Instead of an AR(1) prior assume a prior with a uniform correlation between the elements of $\beta$. That is, replace $\SSigma_A$ by $\SSigma_U$, given by: +\begin{align*} +\SSigma_U \, \, \, = \, \, \, \left( +\begin{array}{ccc} +1 & \rho & \rho +\\ +\rho & 1 & \rho +\\ +\rho & \rho & 1 +\end{array} +\right) +\end{align*} +\\ +Investigate (again on data) the effect of changing from the AR(1) to the uniform prior on the ridge regression estimates. +\end{compactitem} +\end{question} + + + + +========= Ridge logistic regression ========= label{sect.ridgeLogistic} +# ======= Ridge logistic regression ======= + +Ridge penalized estimation is not limited to the standard linear regression model, but may be used to estimate (virtually) any model. Here we illustrate how it may be used to fit the logistic regression model. To this end we first recap this model and the (unpenalized) maximum likelihood estimation of its parameters. After which the model is estimated by means of ridge penalized maximum likelihood, which will turn out to be a relatively straightforward modification of unpenalized estimation. + +======= Logistic regression ======= + +The logistic regression model explains a binary response variable (through some transformation) by a linear combination of a set of covariates (as in the linear regression model). Denote this response of the $i$-th sample by $Y_i$ with $Y_i \in \{ 0, 1 \}$ for $i=1, \ldots, n$. The $n$-dimensional column vector $\mathbf{Y}$ stacks these $n$ responses. For each sample information on the $p$ explanatory variables $X_{i,1}, \ldots, X_{i,p}$ is available. In row vector form this information is denoted $\mathbf{X}_{i,\ast} = (X_{i,1}, \ldots, X_{i,p})$. Or, in short, $\mathbf{X}_i$ when the context tolerates no confusion. The $(n \times p)$-dimensional matrix $\mathbf{X}$ aggregates these vectors, such that $\mathbf{X}_i$ is the $i$-th row vector. + +The binary response cannot be modelled as in the linear model like $Y_i = \mathbf{X}_i \beta + \varepsilon_i$. With each element of $\mathbf{X}_i$ and $\beta$ assuming a value in $\mathbb{R}$, the linear predictor is not restricted to the domain of the response. This is resolved by modeling $p_i = P(Y_i = 1)$ instead. Still the linear predictor may exceed the domain of the response ($p_i \in [0,1]$). Hence, a transformation is applied to map $p_i$ to $\mathbb{R}$, the range of the linear predictor. +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logistic_effectOfBeta0.eps, width=400 frac=1.0] Top row, left panel: the response curve for various choices of the intercept $\beta_0$. Top row, right panel: the response curve for various choices of the regression coefficent $\beta_1$. Bottom row, left panel: the responce curve for various choices of the link function. Bottom panel, right panel: observations, fits and their deviations. } label{fig.logisticRidge_modelAndFitIllustration +The transformation associated with the logistic regression model is the logarithm of the odds, with the odds defined as: $\mbox{*odds*} = P(\mbox{succes}) / P(\mbox{failure}) = p_i/ (1-p_i)$. The logistic model is then written as $\log[ p_i / (1-p_i)] = \mathbf{X}_i \beta$ for all $i$. Or, expressed in terms of the response: +\begin{align*} +p_i & = P(Y_i = 1) \, \, \, = \, \, \, g^{-1}(\mathbf{X}_i; \beta) \, \, \, = \, \, \,\frac{\exp(\mathbf{X}_i \beta) }{1 + \exp(\mathbf{X}_i \beta) }. +\end{align*} +The function $g(\cdot; \cdot)$ is called the *link function*. It links the response to the explanatory variables. The one above is called the logistic link function. Or short, logit. The regression parameters have tangible interpretations. When the first covariate represents the intercept, i.e. $X_{i,j} = 1$ for all $i$, then $\beta_1$ determines where the link function equals a half when all other covariates fail to contribute to the linear predictor (i.e. where $P (Y_i = 1 \, | \, \mathbf{X}_{i}) = 0.5$ when $\mathbf{X}_{i} \beta = \beta_1$). This is illustrated in the top-left panel of Figure ref{fig.logisticRidge_modelAndFitIllustration} for various choices of the intercept. On the other hand, the regression parameters are directly related to the odds ratio: $\mbox{*odds ratio*} = \mbox{odds}(X_{i,j}+1) / \mbox{odds}(X_{i,j}) = \exp(\beta_j)$. Hence, the effect of a unit change in the $j$-th covariate on the odds ratio is $\exp(\beta_j)$ (see Figure ref{fig.logisticRidge_modelAndFitIllustration}, top-right panel). Other link functions (depicted in Figure ref{fig.logisticRidge_modelAndFitIllustration}, bottom-left panel) are common, e.g. the *probit*: $p_i = \Phi_{0,1}(\mathbf{X}_i \beta)$; the *cloglog*: $p_i = \frac{1}{\pi} \arctan(\mathbf{X}_i \beta) + \frac{1}{2}$; the *Cauchit*: $p_i = \exp[ - \exp(\mathbf{X}_i \beta)]$. All these link function are invertible. Irrespective of the choice of the link function, the binary data are thus modelled as $Y_i \sim \mathcal{B}[g^{-1}(\mathbf{X}_i; \beta), 1]$. That is, as a single draw from the Binomial distribution with success probability $g^{-1}(\mathbf{X}_i; \beta)$. + + +Let us now estimate the parameter of the logistic regression model by means of the maximum likelihood method. The likelihood of the experiment is then: +\begin{align*} +L(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i}. +\end{align*} +After taking the logarithm and some ready algebra, the log-likelihood is found to be: +\begin{align*} +\mathcal{L}(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\}. +\end{align*} +Differentiate the log-likelihood with respect to $\beta$, equate it zero, and obtain the estimating equation for $\beta$: + +!bt + +!bt +\begin{align} label{form:logisticRidge_estimatingEquationOfBeta} +\frac{\partial \mathcal{L}}{\partial \beta } & = \sum_{i=1}^n \Big[ Y_i - \frac{\exp(\mathbf{X}_i \beta)}{ 1 + \exp(\mathbf{X}_i \beta)} \Big] \mathbf{X}_i^{\top} \, \, \, = \, \, \, \mathbf{0}_p. +\end{align} +!et +!et +The ML estimate of $\beta$ strikes a (weighted by the $\mathbf{X}_i$) balance between observation and model. Put differently (and illustrated in the bottom-right panel of Figure ref{fig.logisticRidge_modelAndFitIllustration}), a curve is fit through data by minimizing the distance between them: at the ML estimate of $\beta$ a weighted average of their deviations is zero. + +The maximum likelihood estimate of $\beta$ is evaluated by solving Equation (ref{form:logisticRidge_estimatingEquationOfBeta}) with respect to $\beta$ by means of the Newton-Raphson algorithm. The Newton-Raphson algorithm iteratively finds the zeros of a smooth enough function $f(\cdot)$. Let $x_0$ denote an initial guess of the zero. Then, approximate $f(\cdot)$ around $x_0$ by means of a first order Taylor series: $f(x) \approx x_0 + (x - x_0) \, (d f / d x) |_{x=x_0}$. Solve this for $x$ and obtain: $x = x_0 - [ (d f / d x) |_{x=x_0} ]^{-1} f(x_0)$. Let $x_1$ be the solution for $x$, use this as the new guess and repeat the above until convergence. When the function $f(\cdot)$ has multiple arguments, is vector-valued and denoted by $\vec{\mathbf{f}}$, and the Taylor approximation becomes: $\vec{\mathbf{f}}(\mathbf{x}) \approx \mathbf{x}_0 + J \vec{\mathbf{f}} \big|_{\mathbf{x}=\mathbf{x}_0} (\mathbf{x} - \mathbf{x}_0)$ with +\begin{align*} +J \vec{\mathbf{f}} = \left( +\begin{array}{llll} +\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \ldots & \frac{\partial f_1}{\partial x_p} +\\ +\frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \ldots & \frac{\partial f_2}{\partial x_p} +\\ +\vdots & \vdots & \ddots & \vdots +\\ +\frac{\partial f_q}{\partial x_1} & \frac{\partial f_q}{\partial x_2} & \ldots & \frac{\partial f_q}{\partial x_p} +\end{array} +\right), +\end{align*} +the Jacobi matrix. An update of $x_0$ is now readily constructed by solving (the approximation for) $\vec{\mathbf{f}}(\mathbf{x}) = \mathbf{0}$ for $\mathbf{x}$. + +When applied here to the maximum likelihood estimation of the regression parameter $\beta$ of the logistic regression model, the Newton-Raphson update is: +\begin{align*} +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} - \Big( \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \Big)^{-1} \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } \, \, \frac{\partial \mathcal{L}}{\partial \beta } \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } +\end{align*} +where the Hessian of the log-likelihood equals: +\begin{align*} +\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} & = - \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1 + \exp(\mathbf{X}_i \beta)]^2} \mathbf{X}_i^{\top} \mathbf{X}_i. +\end{align*} +Iterative application of this updating formula converges to the ML estimate of $\beta$. + +The Newton-Raphson algorithm is often reformulated as an iteratively re-weighted least squares algorithm. Hereto, first write the gradient and Hessian in matrix notation: +\begin{align*} +\frac{\partial \mathcal{L}}{\partial \beta } \, \, \, = \, \, \, \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta)] +& \mbox{ and } & +\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, - \mathbf{X}^{\top} \mathbf{W} \mathbf{X}, +\end{align*} +where $\vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta) = [g^{-1}( \mathbf{X}_{1, \ast}; \beta), \ldots, g^{-1}( \mathbf{X}_{n, \ast}; \beta)]^{\top}$ with $g^{-1}(\cdot; \cdot) = \exp(\cdot; \cdot) / [1 + \exp(\cdot; \cdot)]$ and $\mathbf{W}$ diagonal with $(\mathbf{W})_{ii} = \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) [ 1 + \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) ]^{-2}$. The updating formula of the estimate then becomes: +\begin{align*} +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] +\\ +& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} +\\ +& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, +\end{align*} +where $\mathbf{Z} = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}$. The Newton-Raphson update is thus the solution to the following weighted least squares problem: +\begin{align*} +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \arg \min_{\beta} (\mathbf{Z} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Z} - \mathbf{X} \beta). +\end{align*} +Effectively, at each iteration the *adjusted response* $\mathbf{Z}$ is regressed on the covariates that comprise $\mathbf{X}$. For more on logistic regression confer the monograph of cite{Hosm2013}. + + +======= Ridge estimation ======= + +High-dimensionally, the linear predictor $\mathbf{X} \beta$ may be uniquely defined, but the maximum likelihood estimate of the logistic regression parameter is not. Assume $p > n$ and an estimate $\hat{\beta}$ available. Due to the high-dimensionality, the null space of $\mathbf{X}$ is non-trivial. Hence, let $\ggamma \in \mbox{null}(\mbox{span}(\mathbf{X}))$. Then: $\mathbf{X} \hat{\beta} = \mathbf{X} \hat{\beta} + \mathbf{X} \ggamma = \mathbf{X} (\hat{\beta} + \ggamma)$. As the null space is a $p-n$-dimensional subspace, $\ggamma$ need not equal zero. Hence, an infinite number of estimates of the logistic regression parameter exists that yield the same log-likelihood. Augmentation of the loss function with a ridge penalty resolves the matter, as their sum is strictly concave in $\beta$ (not convex as a maximum rather than a minimum is sought here) and thereby has a unique maximum. + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticPenLL2ridgePenalty.eps, width=400 frac=1.0] Top row, left panel: contour plot of the penalized log-likelihood of a logistic regression model with the ridge constraint (red line). Top row, right panel: the regularization paths of the ridge estimator of the logistic regression parameter. Bottom row, left panel: variance of the ridge estimator of the logistic regression parameter against the logarithm of the penalty parameter. Bottom panel, right panel: the predicted success probability versus the linear predictor for various choices of the penalty parameter. } label{fig.logisticRidge_effectOfPenalization +\mbox{ } + + +Ridge maximum likelihood estimates of the logistic model parameters are found by the maximization of the ridge penalized loglikelihood (cf. \citealt{Scha1984,LeCe1992}): +\begin{align*} +\mathcal{L}^{\mbox{{\tiny pen}}}(\mathbf{Y}, \mathbf{X}; \beta, \lambda) & = \mathcal{L} (\mathbf{Y}, \mathbf{X}; \beta) - \tfrac{1}{2} \lambda \| \beta \|_2^2 +\\ +& = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\} - \tfrac{1}{2} \lambda \beta^{\top} \beta, +\end{align*} +where the second summand is the ridge penalty (the sum of the square of the elements of $\beta$) with $\lambda$ the penalty parameter. Note that as in Section ref{sect.constrainedEstimation} maximization of this penalized loss function can be reformulated as a constrained estimation problem. This is illustrated by the top left panel of Figure ref{fig.logisticRidge_effectOfPenalization}, which depicts the contours (black lines) of the log-likelihood and the spherical domain of the parameter (red line). The optimization of the above loss function proceeds, due to the differentiability of the penalty, fully analogous to the unpenalized case and uses the Newton-Raphson algorithm for solving the (penalized) estimating equation. Hence, the unpenalized ML estimation procedure is modified straightforwardly by replacing gradient and Hessian by their `penalized' counterparts: +\begin{align*} +\frac{\partial \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta } \, \, \, = \, \, \, \frac{\partial \mathcal{L}}{\partial \beta } - \lambda \beta & \mbox{ and } & \frac{\partial^2 \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} - \lambda \mathbf{I}_{pp}. +\end{align*} +With these at hand, the Newton-Raphson algorithm is (again) reformulated as an iteratively re-weighted least squares algorithm with the updating step changes accordingly to: +\begin{align*} +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \{ \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}}) ] - \lambda \beta^{\mbox{{\scriptsize old}}} \} +\\ +& = \mathbf{V}^{-1} \mathbf{V} \hat{\beta}^{\mbox{{\scriptsize old}}} - \lambda \mathbf{V}^{-1} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] +\\ +& = \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} +\\ +& = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, +\end{align*} +where $\mathbf{V} = \mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp}$ and $\mathbf{W}$ and $\mathbf{Z}$ as before. Hence, use this to update the estimate of $\beta$ until convergence, which yields the desired ridge ML estimate. + +Obviously, the ridge estimate of the logistic regression parameter tends to zero as $\lambda \rightarrow \infty$. Now consider a linear predictor with an intercept that is left unpenalized. When $\lambda$ tends to infinity, all regression coefficients but the intercept vanish. The intercept is left to model the success probability. Hence, in this case $\lim_{\lambda \rightarrow \infty} \hat{\beta}_0 (\lambda) = \log [ \tfrac{1}{n} \sum_{i=1}^n Y_i / \tfrac{1}{n} \sum_{i=1}^n (1-Y_i)]$. + + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticRidge_effectOnPrediction_lambda0.eps, width=400 frac=1.0] The realized design as scatter plot ($X_1$ vs $X_2$ overlayed by the success (\textcolor{red}{RED}) and failure regions (\textcolor{green}{GREEN}) for various choices of the penalty parameter: $\lambda = 0$ (top row, left panel), $\lambda = 10$ (top row, right panel) $\lambda = 40$ (bottom row, left panel), $\lambda = 100$ (bottom row, right panel).} label{fig.logisticRidge_effectOnPrediction + + + +The effect of the ridge penalty on parameter estimates propagates to the predictor $\hat{p}_i$. The linear predictor of the linear regression model involving the ridge estimator $\mathbf{X}_i \hat{\beta}(\lambda)$ shrinks towards a common value for each $i$, leading to a scale difference between observation and predictor (as seen before in Section ref{sect.ridgeRegressionDataIllustration}). This behaviour transfers to the ridge logistic regression predictor, as is illustrated on simulated data. The dimension and sample size of these data are $p=2$ and $n=200$, respectively. The covariate data are drawn from the standard normal, while that of the response is sampled from a Bernoulli distribution with success probability $P(Y_i=1) = \exp(2 X_{i,1} - 2 X_{i,2}) / [ 1 + \exp(2 X_{i,1} - 2 X_{i,2})]$. The logistic regression model is estimated from these data by means of ridge penalized likelihood maximization with various choices of the penalty parameter. The bottom right plot in Figure ref{fig.logisticRidge_effectOfPenalization} shows the predicted success probability versus the linear predictor for various choices of the penalty parameter. Larger values of the penalty parameter $\lambda$ flatten the slope of this curve. Consequently, for larger $\lambda$ more excessive values of the covariates are needed to achieve the same predicted success probability as those obtained with smaller $\lambda$ at more moderate covariate values. The implications for the resulting classification may become clearer when studying the effect of the penalty parameter on the `failure' and `success regions' respectively defined by: +\\ +\indent $\{(x_1, x_2) : P({\color{green}{\mathbf{Y=0}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$, +\\ +\indent $\{(x_1, x_2) : P({\color{red}{\mathbf{Y=1}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$. +\\ +This separates the design space in a light red (`failure') and light green (`success') domain. The white bar between them is the domain where samples cannot be classified with high enough certainty. As $\lambda$ grows, so does the white area that separates the failure and success regions. Hence, as stronger penalization shrinks the logistic regression parameter estimate towards zero, it produces a predictor that is less outspoken in its class assignments. + + + + +======= Moments ======= + +The $1^{\mbox{{\tiny st}}}$ and $2^{\mbox{{\tiny nd}}}$ order moment of the ridge ML parameter of the logistic model may be approximated by the final update of the Newton-Raphson estimate. Assume the one-to-last update $\hat{\beta}^{\mbox{{\scriptsize old}}}$ to be non-random and proceed as for the ridge estimator of the linear regression model parameter to arrive at: +# \begin{align*} +# & & \hat{\beta}^{\mbox{{\scriptsize new}}} = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + 2 \lambda \mathbf{I}_{p \times p} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{z} +# \end{align*} +\begin{align*} +\mathbb{E} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbb{E}( \mathbf{Z}), +\\ +\mbox{Var} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \big[ \mbox{Var} ( \mathbf{Z} ) \big] \mathbf{W} \mathbf{X} [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}, +\end{align*} +with +\begin{align*} +\mathbb{E}(\mathbf{Z}) & = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbb{E}(\mathbf{Y}) - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}, +\\ +\mbox{Var}(\mathbf{Z}) & = \mathbf{W}^{-1} \mbox{Var}(\mathbf{Y}) \mathbf{W}^{-1} = \mathbf{W}^{-1}, +\end{align*} +where the identity $\mbox{Var}(\mathbf{Y}) = \mathbf{W}$ follows from the variance of a Binomial distributed random variable. From these expressions similar properties as for the ridge ML estimate of the regression parameter of the linear model may be deduced. For instance, the ridge ML estimate of the logistic regression parameter converges to zero as the penalty parameter tends to infinity (confer the top right panel of Figure ref{fig.logisticRidge_effectOfPenalization}). Similarly, their variances vanish as $\lambda \rightarrow \infty$ (illustrated in the bottom left panel of Figure ref{fig.logisticRidge_effectOfPenalization}). + +======= The Bayesian connection ======= + +All penalized estimators can be formulated as Bayesian estimators, including the ridge logistic estimator. In particular, ridge estimators correspond to Bayesian estimators with a multivariate normal prior on the regression coefficients. Thus, assume $\beta \sim \mathcal{N}(\mathbf{0}_p, \mathbf{\Delta}^{-1})$. The posterior distribution of $\beta$ then is: +\begin{align*} +f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X}) & \propto & \Big\{ \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i} \Big\} \exp( - \tfrac{1}{2} \beta \mathbf{\Delta} \beta). +\end{align*} +This does not coincide with any standard distribution. But, under appropriate conditions, the posterior distribution is asymptotically normal. This invites a (multivariate) normal approximation to the posterior distribution above. The Laplace's method provides (cf. \citealp{Bish2006}). + +# original latex figure with scale=0.40, angle=0 + +FIGURE: [logisticRidge_laplace2posterior.eps, width=400 frac=1.0] Right panel: Laplace approximation to the posterior density of the Bayesian logistic regression parameter.} label{fig.logisticRidge_MSEandLaplace2posterior + +Laplace's method *i)* centers the normal approximation at the mode of the posterior, and *ii)* chooses the covariance to match the curvature of the posterior at the mode. The posterior mode is the location of the maximum of the posterior distribution. The location of this maximum coincides with that of the logarithm of the posterior. The latter is the log-likelihood augmented with a ridge penalty. Hence, the posterior mode, which is taken as the mean of the approximating Gaussian, coincides +with the ridge logistic estimator. For the covariance of the approximating Gaussian, the logarithm of the posterior is approximated by a second order Taylor series around the posterior mode and limited to second order terms: +\begin{align*} +\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] & \propto & +\left. \log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] +\right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} +\\ +& & + \tfrac{1}{2} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top} \left. \frac{\partial^2}{\partial \beta \partial \beta^{\top}} +\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] + \right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top}, +\end{align*} +in which the first order term cancels as the derivative of $f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})$ with respect to $\beta$ vanishes at the posterior mode - its maximum. Take the exponential of this approximation and match its arguments to that of a multivariate Gaussian $\exp[-\tfrac{1}{2} (\beta - \mmu_{\beta})^{\top} \mathbf{\Sigma}_{\beta}^{-1} (\beta - \mmu_{\beta})]$. The covariance of the sought Gaussian approximation is thus the inverse of the Hessian of the negative penalized log-likelihood. Put together the posterior is approximated by: +\begin{align*} +\beta \, | \, \mathbf{Y}, \mathbf{X} \sim \mathcal{N} \Big( \hat{\beta}_{\mbox{{\tiny MAP}}}, \Big\{ \mathbf{\Delta} + \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1+ \exp(\mathbf{X}_i \beta) ]^2 } \mathbf{X}_i \mathbf{X}_i^{\top} \Big\}^{-1} \Big). +\end{align*} +The Gaussian approximation is convenient but need not be good. Fortunately, the Bernstein-Von Mises Theorem \citep{VdVa2000} tells it is very accurate when the model is regular, the prior smooth, and the sample size sufficiently large. The quality of the approximation for an artificial example data set is shown in Figure ref{fig.logisticRidge_MSEandLaplace2posterior}. + +======= Penalty parameter selection ======= + +As before the penalty parameter may be chosen through $K$-fold cross-validation. For the $K=n$ case cite{Meij2013} describe a computationally efficient approximation of the leave-one-out cross-validated loglikelihood. It is based on the exact evaluation of the LOOCV loss, discussed in Section ref{subsect.crossvalidation}, that avoided resampling. The approach of cite{Meij2013} hinges upon the first-order Taylor expansion of the left-out penalized loglikelihood of the left-out estimate $\hat{\beta}_{-i} (\lambda)$ around $\hat{\beta} (\lambda)$, which yields an approximation of the former: +\begin{align*} +\hat{\beta}_{-i} (\lambda) & \approx \hat{\beta} (\lambda) - +\left( \left. \frac{\partial^2 \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \right|_{\beta = \hat{\beta}(\lambda)} \right)^{-1} \left. \frac{\partial \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta } \right|_{\beta = \hat{\beta}(\lambda)} +\\ +& = \hat{\beta} (\lambda) + (\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \{ \mathbf{X}_{- i, \ast}^{\top} [\mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda))] - \lambda \hat{\beta}(\lambda) \}. +\end{align*} +This approximation involves the inverse of a $p \times p$ dimensional matrix, which amounts to the evaluation of $n$ such inverses for the LOOCV loss. As in Section ref{subsect.crossvalidation} this may be avoided. Rewrite both the gradient and the Hessian of the left-out loglikelihood in the approximation of the preceding display: +\begin{align*} +\mathbf{X}_{-i, \ast}^{\top} \{ \mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) +& = \mathbf{X}^{\top} \{ \mathbf{Y} - \vec{\mathbf{g}}^{-1}[\mathbf{X}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) +- \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} +\\ +& = - \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} +\end{align*} +and +\begin{align*} +(\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} + \mathbf{W}_{ii} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} +\\ +& & \qquad \qquad \qquad \qquad \qquad \quad [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}, +\end{align*} +where the Woodbury identity has been used and now $\mathbf{H}_{ii}(\lambda) = \mathbf{W}_{ii} \mathbf{X}_{i, \ast}(\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. Substitute both in the approximation of the left-out ridge logistic regression estimator and manipulate as in Section ref{subsect.crossvalidation} to obtain: +\begin{align*} +\hat{\beta}_{- i}(\lambda) & \approx \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - g^{-1}(\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)) ]. +\end{align*} +Hence, the leave-one-out cross-validated loglikelihood $\sum_{i=1}^n \mathcal{L} [Y_i \, | \, \mathbf{X}_{i, \ast}, \hat{\beta}_{-i}(\lambda)]$ can now be evaluated by means of a single inverse of a $p \times p$ dimensional matrix and some matrix multiplications. For the performance of this approximation in terms of accuracy and speed confer cite{Meij2013}. + + + + diff --git a/doc/src/Regression/out.txt b/doc/src/Regression/out.txt index e7253cd8d..95ddc0d91 100644 --- a/doc/src/Regression/out.txt +++ b/doc/src/Regression/out.txt @@ -1,17 +1,4 @@ -# #ifdef LATEX2DOCONCE -This is the result of the doconce latex2doconce program. -The translation from LaTeX is just a helper. The text must -be carefully examined! (Be prepared that some text might also -be lost in the translation - in seldom cases.) - -## search for CHECK to see if auto editing was correct - -# #endif - -========= Ridge regression ========= label{chap:ridgeRegression} -\pagenumbering{arabic} -# \setcounter{page} High-throughput techniques measure many characteristics of a single sample simultaneously. The number of characteristics $p$ measured may easily exceed ten thousand. In most medical studies the number of samples $n$ involved often falls behind the number of characteristics measured, i.e: $p > n$. The resulting $(n \times p)$-dimensional data matrix $\mathbf{X}$: \begin{align*} \mathbf{X} & = @@ -35,213 +22,316 @@ X_{n,1} & \ldots & X_{n,p} \end{align*} from such a study contains a larger number of covariates than samples. When $p > n$ the data matrix $\mathbf{X}$ is said to be *high-dimensional*. -In this chapter we adopt the traditional statistical notation of the data matrix. An alternative notation would be $\mathbf{X}^{\top}$ (rather than $\mathbf{X}$), which is employed in the field of (statistical) bioinformatics. In $\mathbf{X}^{\top}$ the rows comprise the samples rather than the covariates. The case for the bioinformatics notation stems from practical arguments. A spreadsheet is designed to have more rows than columns. In case $p > n$ the traditional notation yields a spreadsheet with more columns than rows. When $p > 10000$ the conventional display is impractical. In this chapter we stick to the conventional statistical notation of the data matrix as all mathematical expressions involving $\mathbf{X}$ are then in line with those of standard textbooks on regression. +In this chapter we adopt the traditional statistical notation of the +data matrix. An alternative notation would be $\mathbf{X}^{\top}$ +(rather than $\mathbf{X}$), which is employed in the field of +(statistical) bioinformatics. In $\mathbf{X}^{\top}$ the rows comprise +the samples rather than the covariates. The case for the +bioinformatics notation stems from practical arguments. A spreadsheet +is designed to have more rows than columns. In case $p > n$ the +traditional notation yields a spreadsheet with more columns than +rows. When $p > 10000$ the conventional display is impractical. In +this chapter we stick to the conventional statistical notation of the +data matrix as all mathematical expressions involving $\mathbf{X}$ are +then in line with those of standard textbooks on regression. -The information contained in $\mathbf{X}$ is often used to explain a particular property of the samples involved. In applications in molecular biology $\mathbf{X}$ may contain microRNA expression data from which the expression levels of a gene are to be described. When the gene's expression levels are denoted by $\mathbf{Y} = (Y_{1}, \ldots, Y_n)^{\top}$, the aim is to find the linear relation $Y_i = \mathbf{X}_{i, \ast} \bbeta$ from the data at hand by means of regression analysis. Regression is however frustrated by the high-dimensionality of $\mathbf{X}$ (illustrated in Section ref{sect.ridgeRegression} and at the end of Section ref{sect.constrainedEstimation}). These notes discuss how regression may be modified to accommodate the high-dimensionality of $\mathbf{X}$. First, however, `standard' linear regression is recaputilated. +The information contained in $\mathbf{X}$ is often used to explain a +particular property of the samples involved. In applications in +molecular biology $\mathbf{X}$ may contain microRNA expression data +from which the expression levels of a gene are to be described. When +the gene's expression levels are denoted by $\mathbf{Y} = (Y_{1}, +\ldots, Y_n)^{\top}$, the aim is to find the linear relation $Y_i = +\mathbf{X}_{i, \ast} \beta$ from the data at hand by means of +regression analysis. Regression is however frustrated by the +high-dimensionality of $\mathbf{X}$ (illustrated in Section +ref{sect.ridgeRegression} and at the end of Section +ref{sect.constrainedEstimation}). These notes discuss how regression +may be modified to accommodate the high-dimensionality of +$\mathbf{X}$. First, however, `standard' linear regression is +recaputilated. ======= Linear regression ======= -Consider an experiment in which $p$ characteristics of $n$ samples are measured. The data from this experiment are denoted $\mathbf{X}$, with $\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design matrix*. Additional information of the samples is available in the form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is generally referred to as the *response variable*. The aim of regression analysis is to explain $\mathbf{Y}$ in terms of $\mathbf{X}$ through a functional relationship like $Y_i = f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of $f(\cdot)$ is available, it is common to assume a linear relationship between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to the *linear regression model*: +Consider an experiment in which $p$ characteristics of $n$ samples are +measured. The data from this experiment are denoted $\mathbf{X}$, with +$\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design +matrix*. Additional information of the samples is available in the +form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is +generally referred to as the *response variable*. The aim of +regression analysis is to explain $\mathbf{Y}$ in terms of +$\mathbf{X}$ through a functional relationship like $Y_i = +f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of +$f(\cdot)$ is available, it is common to assume a linear relationship +between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to +the *linear regression model*: -!bt !bt \begin{align} -Y_{i} & = \mathbf{X}_{i,\ast} \, \bbeta + \varepsilon_i label{form.linRegressionModel} +Y_{i} & = \mathbf{X}_{i,\ast} \, \beta + \varepsilon_i label{form.linRegressionModel} \\ \nonumber & = \beta_1 \, X_{i,1} + \ldots + \beta_{p} \, X_{i, p} + \varepsilon_i. \end{align} !et -!et -In model (ref{form.linRegressionModel}) $\bbeta = (\beta_1, \ldots, \beta_p)^{\top}$ is the *regression parameter*. The parameter $\beta_j$, $j=1, \ldots, p$, represents the -effect size of covariate $j$ on the response. That is, for each unit change in covariate $j$ (while keeping the other covariates fixed) the observed change in the response is equal to $\beta_j$. -The second summand on the right-hand side of the model, $\varepsilon_i$, is referred to as the error. It represents the part of the response not explained by the functional part $\mathbf{X}_{i,\ast} \, \bbeta$ of the model (ref{form.linRegressionModel}). In contrast to the functional part, which is considered to be systematic (i.e. non-random), the error is assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal $Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= \mathbf{X}_{i_2,\ast}$. To complete the formulation of model (ref{form.linRegressionModel}) we need to specify the probability distribution of $\varepsilon_i$. It is assumed that $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$ and the $\varepsilon_{i}$ are independent, i.e.: -\begin{align*} -\mbox{Cov}(\varepsilon_{i_1}, \varepsilon_{i_2}) & = -\left\{ -\begin{array}{lcc} -\sigma^2 & \mbox{if} & i_1 = i_2, -\\ -0 & \mbox{if} & i_1 \not= i_2. -\end{array} -\right. -\end{align*} -The randomness of $\varepsilon_i$ implies that $\mathbf{Y}_i$ is also a random variable. In particular, $\mathbf{Y}_i$ is normally distributed, because $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$ and $\mathbf{X}_{i,\ast} \, \bbeta$ is a non-random scalar. To specify the parameters of the distribution of $\mathbf{Y}_i$ we need to calculate its first two moments. Its expectation equals: -\begin{align*} -\mathbb{E}(Y_i) & = \mathbb{E}(\mathbf{X}_{i, \ast} \, \bbeta) + \mathbb{E}(\varepsilon_i) \, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \bbeta, -\end{align*} -while its variance is: -\begin{align*} -\mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i - \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - [\mathbb{E}(Y_i)]^2 -# \\ -# & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \bbeta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 -\\ -& = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \bbeta)^2 + 2 \varepsilon_i \mathbf{X}_{i, \ast} \, \bbeta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 -\\ -# & = ( \mathbf{X}_{i, \ast} \, \bbeta)^2 + 2 \mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \bbeta + \mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \bbeta)^2 -# \\ -& = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, \mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. -\end{align*} -Hence, $Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \bbeta, \sigma^2)$. This formulation (in terms of the normal distribution) is equivalent to the formulation of model -(ref{form.linRegressionModel}), as both capture the assumptions involved: the linearity of the functional part and the normality of the error. + +In model (ref{form.linRegressionModel}) $\beta = (\beta_1, \ldots, +\beta_p)^{\top}$ is the *regression parameter*. The parameter +$\beta_j$, $j=1, \ldots, p$, represents the effect size of covariate +$j$ on the response. That is, for each unit change in covariate $j$ +(while keeping the other covariates fixed) the observed change in the +response is equal to $\beta_j$. The second summand on the right-hand +side of the model, $\varepsilon_i$, is referred to as the error. It +represents the part of the response not explained by the functional +part $\mathbf{X}_{i,\ast} \, \beta$ of the model +(ref{form.linRegressionModel}). In contrast to the functional part, +which is considered to be systematic (i.e. non-random), the error is +assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal +$Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= +\mathbf{X}_{i_2,\ast}$. To complete the formulation of model +(ref{form.linRegressionModel}) we need to specify the probability +distribution of $\varepsilon_i$. It is assumed that $\varepsilon_i +\sim \mathcal{N}(0, \sigma^2)$ and the $\varepsilon_{i}$ are +independent, i.e.: \begin{align*} \mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} The randomness of $\varepsilon_i$ implies that +$\mathbf{Y}_i$ is also a random variable. In particular, +$\mathbf{Y}_i$ is normally distributed, because $\varepsilon_i \sim +\mathcal{N}(0, \sigma^2)$ and $\mathbf{X}_{i,\ast} \, \beta$ is a +non-random scalar. To specify the parameters of the distribution of +$\mathbf{Y}_i$ we need to calculate its first two moments. Its +expectation equals: \begin{align*} \mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, \end{align*} while +its variance is: \begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 # \\ # & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ # & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 # +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. \end{align*} +Hence, $Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, +\sigma^2)$. This formulation (in terms of the normal distribution) is +equivalent to the formulation of model (ref{form.linRegressionModel}), +as both capture the assumptions involved: the linearity of the +functional part and the normality of the error. Model (ref{form.linRegressionModel}) is often written in a more condensed matrix form: -!bt !bt \begin{align} -\mathbf{Y} & = \mathbf{X} \, \bbeta + \vvarepsilon, label{form.linRegressionModelinMatrix} +\mathbf{Y} & = \mathbf{X} \, \beta + \vvarepsilon, label{form.linRegressionModelinMatrix} \end{align} !et -!et -where $\vvarepsilon = (\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)^{\top}$ and distributed as $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{p}, \sigma^2 \mathbf{I}_{nn})$. As above model (ref{form.linRegressionModelinMatrix}) can be expressed as a multivariate normal distribution: $\mathbf{Y} \sim \mathcal{N}(\mathbf{X} \, \bbeta, \sigma^2 \mathbf{I}_{nn})$. +where $\vvarepsilon = (\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)^{\top}$ and distributed as $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{p}, \sigma^2 \mathbf{I}_{nn})$. As above model (ref{form.linRegressionModelinMatrix}) can be expressed as a multivariate normal distribution: $\mathbf{Y} \sim \mathcal{N}(\mathbf{X} \, \beta, \sigma^2 \mathbf{I}_{nn})$. -Model (ref{form.linRegressionModelinMatrix}) is a so-called hierarchical model. This terminology emphasizes that $\mathbf{X}$ and $\mathbf{Y}$ are not on a par, they play different roles in the model. The former is used to explain the latter. In model (ref{form.linRegressionModel}) $\mathbf{X}$ is referred as the *explanatory* or *independent* variable, while the variable $\mathbf{Y}$ is generally referred to as the *response* or *dependent* variable. +Model (ref{form.linRegressionModelinMatrix}) is a so-called +hierarchical model. This terminology emphasizes that $\mathbf{X}$ and +$\mathbf{Y}$ are not on a par, they play different roles in the +model. The former is used to explain the latter. In model +(ref{form.linRegressionModel}) $\mathbf{X}$ is referred as the +*explanatory* or *independent* variable, while the variable +$\mathbf{Y}$ is generally referred to as the *response* or *dependent* +variable. -The covariates, the columns of $\mathbf{X}$, may themselves be random. To apply the linear model they are temporarily assumed fixed. The linear regression model is then to be interpreted as $\mathbf{Y} \, | \, \mathbf{X} \sim \mathcal{N}(\mathbf{X} \, \bbeta, \sigma^2 \mathbf{I}_{nn})$ +The covariates, the columns of $\mathbf{X}$, may themselves be +random. To apply the linear model they are temporarily assumed +fixed. The linear regression model is then to be interpreted as +$\mathbf{Y} \, | \, \mathbf{X} \sim \mathcal{N}(\mathbf{X} \, \beta, +\sigma^2 \mathbf{I}_{nn})$ -\begin{example} *Methylation of a tumor-suppressor gene* -\\ -Consider a study which measures the gene expression levels of a tumor-suppressor genes (TSG) and two methylation markers (MM1 and MM2) on 67 samples. A methylation marker is a gene that promotes methylation. Methylation refers to attachment of a methyl group to a nucleotide of the DNA. In case this attachment takes place in or close by the promotor region of a gene, this complicates the transcription of the gene. Methylation may down-regulate a gene. This mechanism also works in the reverse direction: removal of methyl groups may up-regulate a gene. A tumor-suppressor gene is a gene that halts the progression of the cell towards a cancerous state. -The medical question associated with these data: do the expression levels methylation markers affect the expression levels of the tumor-suppressor gene? To answer this question we may formulate the following linear regression model: +The linear regression model (ref{form.linRegressionModel}) involves the unknown parameters: $\beta$ and $\sigma^2$, which need to be learned from the data. The parameters of the regression model, $\beta$ and $\sigma^2$ are estimated by means of likelihood maximization. Recall that $Y_i \sim \mathcal{N}( \mathbf{X}_{i,\ast} \, \beta, \sigma^2)$ with corresponding density: $ f_{Y_i}(y_i) = (2 \, \pi \, \sigma^2)^{-1/2} \, \exp[ - (y_i - \mathbf{X}_{i\ast} \, \beta)^2 / 2 \sigma^2 ]$. The likelihood thus is: + + +!bt \begin{align*} -Y_{i, `{\footnotesize tsg`}} & = \beta_0 + \beta_{`{\footnotesize mm1`}} X_{i, `{\footnotesize mm1`}} -+ \beta_{`{\footnotesize mm2`}} X_{i, `{\footnotesize mm2`}} + \varepsilon_i, -\end{align*} -with $i = 1, \ldots, 67$ and $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. The interest focusses on $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$. A non-zero value of at least one of these two regression parameters indicates that there is a linear association between the expression levels of the tumor-suppressor gene and that of the methylation markers. - -Prior knowledge from biology suggests that the $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$ are both non-positive. High expression levels of the methylation markers lead to hyper-methylation, in turn inhibiting the transcription of the tumor-suppressor gene. Vice versa, low expression levels of MM1 and MM2 are (via hypo-methylation) associated with high expression levels of TSG. Hence, a negative concordant effect between MM1 and MM2 (on one side) and TSG (on the other) is expected. Of course, the methylation markers may affect expression levels of other genes that in turn regulate the tumor-suppressor gene. The regression parameters $\beta_{`{\footnotesize mm1`}}$ and $\beta_{`{\footnotesize mm2`}}$ then reflect the indirect effect of the methylation markers on the expression levels of the tumor suppressor gene. -\end{example} - - -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -The linear regression model (ref{form.linRegressionModel}) involves the unknown parameters: $\bbeta$ and $\sigma^2$, which need to be learned from the data. The parameters of the regression model, $\bbeta$ and $\sigma^2$ are estimated by means of likelihood maximization. Recall that $Y_i \sim \mathcal{N}( \mathbf{X}_{i,\ast} \, \bbeta, \sigma^2)$ with corresponding density: $ f_{Y_i}(y_i) = (2 \, \pi \, \sigma^2)^{-1/2} \, \exp[ - (y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 / 2 \sigma^2 ]$. The likelihood thus is: -\begin{align*} -L(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = +L(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = # f_{\mathbf{Y}}(\mathbf{y}) \, \, \, = \, \, \, \prod_{i=1}^n f_{Y_i}(y_i) \, \, \, = \, \, \, -\prod_{i=1}^n \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, \exp[ - (Y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2 / 2 \sigma^2 ], +\prod_{i=1}^n \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, \exp[ - (Y_i - \mathbf{X}_{i, \ast} \, \beta)^2 / 2 \sigma^2 ], \end{align*} -in which the independence of the observations has been used. Because of the concavity of the logarithm, the maximization of the likelihood coincides with the maximum of the logarithm of the likelihood (called the log-likelihood). Hence, to obtain maximum likelihood (ML) estimates of the parameter it is equivalent to find the maximum of the log-likelihood. The log-likelihood is: +!et + +in which the independence of the observations has been used. Because +of the concavity of the logarithm, the maximization of the likelihood +coincides with the maximum of the logarithm of the likelihood (called +the log-likelihood). Hence, to obtain maximum likelihood (ML) +estimates of the parameter it is equivalent to find the maximum of the +log-likelihood. The log-likelihood is: + +!bt \begin{align*} -\mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = -\log[ L(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) ] +\mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = +\log[ L(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) ] \, \, \, = \, \, \, % \log \Big[ \prod_{i=1}^n f_{Y_i}(y_i) \Big] # \, \, \, = \, \, \, \sum_{i=1}^n \log[ f_{Y_i}(y_i) ] # \\ -# & = \sum_{i=1}^n [ -\log(\sqrt{2 \, \pi} \, \sigma) - (y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 / 2 \sigma^2 ] +# & = \sum_{i=1}^n [ -\log(\sqrt{2 \, \pi} \, \sigma) - (y_i - \mathbf{X}_{i\ast} \, \beta)^2 / 2 \sigma^2 ] # \\ # & = --n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \sum_{i=1}^n (y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2. +-n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \sum_{i=1}^n (y_i - \mathbf{X}_{i, \ast} \, \beta)^2. \end{align*} -After noting that $\sum_{i=1}^n (Y_i - \mathbf{X}_{i, \ast} \, \bbeta)^2 = \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 \, \, \, = \, \, \, (\mathbf{Y} - \mathbf{X} \, \bbeta)^{\top} \, (\mathbf{Y} - \mathbf{X} \, \bbeta)$, the log-likelihood can be written as: +After noting that $\sum_{i=1}^n (Y_i - \mathbf{X}_{i, \ast} \, \beta)^2 = \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 \, \, \, = \, \, \, (\mathbf{Y} - \mathbf{X} \, \beta)^{\top} \, (\mathbf{Y} - \mathbf{X} \, \beta)$, the log-likelihood can be written as: \begin{align*} -\mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = -n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \, \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. +\mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = -n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \, \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. \end{align*} -In order to find the maximum of the log-likelihood, take its derivate with respect to $\bbeta$: -\begin{align*} -\frac{\partial }{\partial \, \beta} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = - \frac{1}{ 2 \sigma^2} \, \frac{\partial }{\partial \, \beta} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 \, \, \, = \, \, \, \frac{1}{\sigma^2} \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \, \bbeta). -\end{align*} -Equate this derivative to zero gives the estimating equation for $\bbeta$: +!et +In order to find the maximum of the log-likelihood, take its derivate with respect to $\beta$: !bt +\begin{align*} +\frac{\partial }{\partial \, \beta} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = - \frac{1}{ 2 \sigma^2} \, \frac{\partial }{\partial \, \beta} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 \, \, \, = \, \, \, \frac{1}{\sigma^2} \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \, \beta). +\end{align*} +!et + +Equate this derivative to zero gives the estimating equation for $\beta$: !bt \begin{align} label{form.normalEquation} -\mathbf{X}^{\top} \mathbf{X} \, \bbeta & = \mathbf{X}^{\top} \mathbf{Y}. +\mathbf{X}^{\top} \mathbf{X} \, \beta & = \mathbf{X}^{\top} \mathbf{Y}. \end{align} !et -!et -Equation (ref{form.normalEquation}) is called to the *normal equation*. Pre-multiplication of both sides of the normal equation by $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ now yields the ML estimator of the regression parameter: $\hat{\bbeta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}$, in which it is assumed that $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ is well-defined. + +Equation (ref{form.normalEquation}) is called to the *normal equation*. Pre-multiplication of both sides of the normal equation by $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ now yields the ML estimator of the regression parameter: $\hat{\beta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}$, in which it is assumed that $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ is well-defined. Along the same lines one obtains the ML estimator of the residual variance. Take the partial derivative of the log-likelihood with respect to $\sigma^2$: -\begin{align*} -\frac{\partial }{\partial \, \sigma} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \bbeta, \sigma^2) & = - \frac{n}{\sigma} + \frac{1}{\sigma^3} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. -\end{align*} -Equate the right-hand side to zero and solve for $\sigma^2$ to find $\hat{\sigma}^2 = \tfrac{1}{n} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2$. In this expression $\bbeta$ is unknown and the ML estimate of $\bbeta$ is plugged-in. -\\ -\\ -With explicit expressions of the ML estimators at hand, we can study their properties. The expectation of the ML estimator of the regression parameter $\bbeta$ is: -\begin{align*} -\mathbb{E}(\hat{\bbeta}) & = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] -\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbb{E}[ \mathbf{Y}] -# \\ -\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{X} \, \bbeta -\, \, \, \, \, = \, \, \, \bbeta. -\end{align*} -Hence, the ML estimator of the regression coefficients is unbiased. -The variance of the ML estimator of $\bbeta$ is: +!bt \begin{align*} -\mbox{Var}(\hat{\bbeta}) & = \mathbb{E} \{ [\hat{\bbeta} - \mathbb{E}(\hat{\bbeta})] [\hat{\bbeta} - \mathbb{E}(\hat{\bbeta})]^{\top} \} +\frac{\partial }{\partial \, \sigma} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = - \frac{n}{\sigma} + \frac{1}{\sigma^3} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. +\end{align*} +!et + + +Equate the right-hand side to zero and solve for $\sigma^2$ to find +$\hat{\sigma}^2 = \tfrac{1}{n} \| \mathbf{Y} - \mathbf{X} \, \beta +\|^2_2$. In this expression $\beta$ is unknown and the ML estimate of +$\beta$ is plugged-in. \\ \\ With explicit expressions of the ML +estimators at hand, we can study their properties. The expectation of +the ML estimator of the regression parameter $\beta$ is: + +!bt +\begin{align*} +\mathbb{E}(\hat{\beta}) & = \mathbb{E}[ +(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] +\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, +\mathbf{X}^{\top} \mathbb{E}[ \mathbf{Y}] # \\ \, \, \, = \, \, \, +(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{X} \, +\beta \, \, \, \, \, = \, \, \, \beta. +\end{align*} +!et + +Hence, the ML +estimator of the regression coefficients is unbiased. + +The variance of the ML estimator of $\beta$ is: + +!bt +\begin{align*} +\mbox{Var}(\hat{\beta}) & = \mathbb{E} \{ [\hat{\beta} - \mathbb{E}(\hat{\beta})] [\hat{\beta} - \mathbb{E}(\hat{\beta})]^{\top} \} \\ -& = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \bbeta] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \bbeta]^{\top} \} +& = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \beta] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \beta]^{\top} \} \\ -# & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}]^{\top} \} - \bbeta \, \bbeta^{\top} +# & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}]^{\top} \} - \beta \, \beta^{\top} # \\ -# & = \mathbb{E} \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \, \mathbf{Y}^{\top} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \} - \bbeta \, \bbeta^{\top} +# & = \mathbb{E} \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \, \mathbf{Y}^{\top} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \} - \beta \, \beta^{\top} # \\ -& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{\top} \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{\top} \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} \\ -& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \{ \mathbf{X} \, \bbeta \, \bbeta^{\top} \, \mathbf{X}^{\top} + \SSigma \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \{ \mathbf{X} \, \beta \, \beta^{\top} \, \mathbf{X}^{\top} + \SSigma \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} # \\ -# & = (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \bbeta \, \bbeta^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} +# & = (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \beta \, \beta^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} # \\ -# & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \bbeta \bbeta^T +# & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \beta \beta^T \\ -& = \bbeta \, \bbeta^{\top} + \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \bbeta \, \bbeta^{\top} +& = \beta \, \beta^{\top} + \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} \, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1}, \end{align*} -in which we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{\top}) = \mathbf{X} \, \bbeta \, \bbeta^{\top} \, \mathbf{X}^{\top} + \sigma^2 \, \mathbf{I}_{nn}$. From $\mbox{Var}(\hat{\bbeta}) = \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1}$, one obtains an estimate of the variance of the estimate of the $j$-th regression coefficient: $\hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ [(\mathbf{X}^{\top} \mathbf{X})^{-1}]_{jj} }$. This may be used to construct a confidence interval for the estimates or test the hypothesis $H_0: \beta_j = 0$. In the latter $\hat{\sigma}^2$ should not be the maximum likelihood estimator, as it is biased. It is then to be replaced by the residual sum-of-squares divided by $n-p$ rather than $n$. -\\ -\\ -The prediction of $Y_i$, denoted $\widehat{Y}_i$, is the expected value of $Y_i$ according the linear regression model (with its parameters replaced by their estimates). The prediction of $Y_i$ thus equals $\mathbb{E}(Y_i; \hat{\bbeta}, \hat{\sigma}^2) = \mathbf{X}_{i, \ast} \hat{\bbeta}$. In matrix notation the prediction is: +!et + +in which we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{\top}) += \mathbf{X} \, \beta \, \beta^{\top} \, \mathbf{X}^{\top} + \sigma^2 +\, \mathbf{I}_{nn}$. From $\mbox{Var}(\hat{\beta}) = \sigma^2 \, +(\mathbf{X}^{\top} \mathbf{X})^{-1}$, one obtains an estimate of the +variance of the estimate of the $j$-th regression coefficient: +$\hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ +[(\mathbf{X}^{\top} \mathbf{X})^{-1}]_{jj} }$. This may be used to +construct a confidence interval for the estimates or test the +hypothesis $H_0: \beta_j = 0$. In the latter $\hat{\sigma}^2$ should +not be the maximum likelihood estimator, as it is biased. It is then +to be replaced by the residual sum-of-squares divided by $n-p$ rather +than $n$. + + +The prediction of $Y_i$, denoted $\widehat{Y}_i$, is the expected +value of $Y_i$ according the linear regression model (with its +parameters replaced by their estimates). The prediction of $Y_i$ thus +equals $\mathbb{E}(Y_i; \hat{\beta}, \hat{\sigma}^2) = \mathbf{X}_{i, +\ast} \hat{\beta}$. In matrix notation the prediction is: + +!bt \begin{align*} -\widehat{\mathbf{Y}} & = \mathbf{X} \, \hat{\bbeta} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, := \, \, \, \mathbf{H} \mathbf{Y}, +\widehat{\mathbf{Y}} & = \mathbf{X} \, \hat{\beta} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, := \, \, \, \mathbf{H} \mathbf{Y}, \end{align*} -where $\mathbf{H}$ is the *hat matrix*, as it `puts the hat' on $\mathbf{Y}$. Note that the hat matrix is a projection matrix, i.e. $\mathbf{H}^2 = \mathbf{H}$ for +!et + +where $\mathbf{H}$ is the *hat matrix*, as it `puts the hat' on +$\mathbf{Y}$. Note that the hat matrix is a projection matrix, +i.e. $\mathbf{H}^2 = \mathbf{H}$ for + +!bt \begin{align*} \mathbf{H}^2 & = \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}. \end{align*} -Thus, the prediction $\widehat{\mathbf{Y}}$ is an orthogonal projection of $\mathbf{Y}$ onto the space spanned by the columns of $\mathbf{X}$. +!et -With $\widehat{\bbeta}$ available, an estimate of the errors $\hat{\varepsilon}_i$, dubbed the *residuals* are obtained via: +Thus, the prediction $\widehat{\mathbf{Y}}$ is an orthogonal +projection of $\mathbf{Y}$ onto the space spanned by the columns of +$\mathbf{X}$. + +With $\widehat{\beta}$ available, an estimate of the errors +$\hat{\varepsilon}_i$, dubbed the *residuals* are obtained via: + +!bt \begin{align*} -\hat{\vvarepsilon} & = \mathbf{Y} - \widehat{\mathbf{Y}} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, \hat{\bbeta} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, [ \mathbf{I} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} ] \, \mathbf{Y}. +\hat{\vvarepsilon} & = \mathbf{Y} - \widehat{\mathbf{Y}} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, \hat{\beta} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, [ \mathbf{I} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} ] \, \mathbf{Y}. \end{align*} -Thus, the residuals are a projection of $\mathbf{Y}$ onto the orthogonal complement of the space spanned by the columns of $\mathbf{X}$. The residuals are to be used in diagnostics, e.g. checking of the normality assumption by means of a normal probability plot. -\\ -\\ -For more on the linear regression model confer the monograph of cite{Drap1998}. +!et + +Thus, the residuals are a projection of $\mathbf{Y}$ onto the +orthogonal complement of the space spanned by the columns of +$\mathbf{X}$. The residuals are to be used in diagnostics, +e.g. checking of the normality assumption by means of a normal +probability plot. -======= Ridge regression} label{sect.ridgeRegression ======= -When the design matrix is high-dimensional, the covariates (the columns of $\mathbf{X}$) are super-collinear. Recall *collinearity* in regression analysis refers to the event of two (or multiple) covariates being highly linearly related. Consequently, the subspace spanned by collinear covariates may not be (or close to not being) of full rank. When the subspace (onto which $\mathbf{Y}$ is projected) is (close to) rank deficient, it is (almost) impossible to separate the contribution of the individual covariates. The uncertainty with respect to the covariate responsible for the variation explained in $\mathbf{Y}$ is often reflected in the fit of the linear regression model to data by a large error of the estimates of the regression parameters corresponding to the collinear covariates. +======= Ridge regression ======= -\begin{example} \mbox{ } \\ -The flotillins (the FLOT-1 and FLOT-2 genes) have been observed to regulate the proto-oncogene ERBB2 *in vitro* \citep{Pust2013}. One may wish to corroborate this *in vivo*. To this end we use gene expression data of a breast cancer study, available as a Bioconductor package: {\tt breastCancerVDX}. From this study the expression levels of probes interrogating the FLOT-1 and ERBB2 genes are retrieved. For clarity of the illustration the FLOT-2 gene is ignored. After centering, the expression levels of the first ERBB2 probe are regressed on those of the four FLOT-1 probes. The R-code below carries out the data retrieval and analysis. -\lstinputlisting{introExampleForNotes.r} -Prior to the regression analysis, we first assess whether there is collinearity among the FLOT-1 probes through evaluation of the correlation matrix. This reveals a strong correlation ($\hat{\rho} = 0.91$) between the second and third probe. All other cross-correlations do not exceed the 0.20 (in an absolute sense). Hence, there is collinearity among the columns of the design matrix in the to-be-performed regression analysis. -\verbatiminput{introExampleRegressionOutput.txt} -The output of the regression analysis above shows the first probe to be significantly associated to the expression levels of ERBB2. The collinearity of the second and third probe reveals itself in the standard errors of the effect size: for these probes the standard error is much larger than those of the other two probes. This reflects the uncertainty in the estimates. Regression analysis has difficulty to decide to which covariate the explained proportion of variation in the response should be attributed. The large standard error of these effect sizes propagates to the testing as the Wald test statistic is the ratio of the estimated effect size and its standard error. Collinear covariates are thus less likely to pass the significance threshold. -\end{example} - -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -The case of two (or multiple) covariates being perfectly linearly dependent is referred as *super-collinearity*. The rank of a high-dimensional design matrix is maximally equal to $n$: $\mbox{rank}(\mathbf{X}) \leq n$. Consequently, the dimension of subspace spanned by the columns of $\mathbf{X}$ is smaller than or equal to $n$. As $p > n$, this implies that columns of $\mathbf{X}$ are linearly dependent. Put differently, a high-dimensional $\mathbf{X}$ suffers from super-collinearity. +When the design matrix is high-dimensional, the covariates (the +columns of $\mathbf{X}$) are super-collinear. Recall *collinearity* in +regression analysis refers to the event of two (or multiple) +covariates being highly linearly related. Consequently, the subspace +spanned by collinear covariates may not be (or close to not being) of +full rank. When the subspace (onto which $\mathbf{Y}$ is projected) +is (close to) rank deficient, it is (almost) impossible to separate +the contribution of the individual covariates. The uncertainty with +respect to the covariate responsible for the variation explained in +$\mathbf{Y}$ is often reflected in the fit of the linear regression +model to data by a large error of the estimates of the regression +parameters corresponding to the collinear covariates. - -\begin{example} *Super-collinearity* label{example.supercollinearity} -\\ Consider the design matrix: + +!bt \begin{align*} \mathbf{X} & = \left( \begin{array}{rrr} @@ -254,18 +344,23 @@ Consider the design matrix: 1 & 1 & 0 \end{array} \right) \end{align*} -The columns of $\mathbf{X}$ are linearly dependent: the first column is the row-wise sum of the other two columns. The rank (more correct, the column rank) of a matrix is the dimension of space spanned by the column vectors. Hence, the rank of $\mathbf{X}$ is equal to the number of linearly independent columns: $\mbox{rank}(\mathbf{X}) = 2$. -\end{example} +!et + +The columns of $\mathbf{X}$ are linearly dependent: the first column +is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of space spanned by the +column vectors. Hence, the rank of $\mathbf{X}$ is equal to the number +of linearly independent columns: $\mbox{rank}(\mathbf{X}) = 2$. + + -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif Super-collinearity of an $(n \times p)$-dimensional design matrix $\mathbf{X}$ implies[^footnote1][^footnote1]: If the (column) rank of $\mathbf{X$ is smaller than $p$, there exists a non-trivial $\mathbf{v} \in \mathbb{R}^p$ such that $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. Multiplication of this inequality by $\mathbf{X}^{\top}$ yields $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. As $\mathbf{v} \not= \mathbf{0}_{p}$, this implies that $\mathbf{X}^{\top} \mathbf{X}$ is not invertible.} that the rank of the $(p \times p)$-dimensional matrix $\mathbf{X}^{\top} \mathbf{X}$ is smaller than $p$, and, consequently, it is singular. A square matrix that does not have an inverse is called *singular*. A matrix $\mathbf{A}$ is singular if and only if its determinant is zero: $\mbox{det}(\mathbf{A}) = 0$. -\begin{example} *Singularity* label{example.singular} -\\ + + Consider the matrix $\mathbf{A}$ given by: +!bt \begin{align*} \mathbf{A} & = \left( \begin{array}{rr} @@ -274,70 +369,98 @@ Consider the matrix $\mathbf{A}$ given by: 2 & 4 \end{array} \right) \end{align*} +!et Clearly, $\mbox{det}(\mathbf{A}) = a_{11} a_{22} - a_{12} a_{21} = 1 \times 4 - 2 \times 2 = 0$. Hence, $\mathbf{A}$ is singular and its inverse is undefined. -\end{example} -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -As $\mbox{det}(\mathbf{A})$ is equal to the product of the eigenvalues $\nu_j$ of $\mathbf{A}$, the matrix $\mathbf{A}$ is singular if one (or more) of the eigenvalues of $\mathbf{A}$ is zero. To see this, consider the spectral decomposition of $\mathbf{A}$: +As $\mbox{det}(\mathbf{A})$ is equal to the product of the eigenvalues +$\nu_j$ of $\mathbf{A}$, the matrix $\mathbf{A}$ is singular if one +(or more) of the eigenvalues of $\mathbf{A}$ is zero. To see this, +consider the spectral decomposition of $\mathbf{A}$: + +!bt \begin{align*} \mathbf{A} & = \sum_{j=1}^p \nu_j \, \mathbf{v}_j \, \mathbf{v}_j^{\top}, \end{align*} +!et where $\mathbf{v}_j$ is the eigenvector corresponding to $\nu_j$. The inverse of $\mathbf{A}$ is then: +!bt \begin{align*} \mathbf{A}^{-1} & = \sum_{j=1}^p \nu_j^{-1} \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. \end{align*} +!et + The right-hand side is undefined if $\nu_j =0$ for any $j$. -\begin{contexample}_ref{example.singular_} \hspace{3pt}*Singularity (continued)* -\\ -Revisit Example ref{example.singular}. Matrix $\mathbf{A}$ has eigenvalues $\nu_1 =5$ and $\nu_2=0$. According to the spectral decomposition, the inverse of $\mathbf{A}$ is: +Matrix $\mathbf{A}$ has eigenvalues $\nu_1 =5$ and $\nu_2=0$. According to the spectral decomposition, the inverse of $\mathbf{A}$ is: +!bt \begin{align*} \mathbf{A}^{-1} & = \frac{1}{5} \, \mathbf{v}_1 \, \mathbf{v}_1^{\top} + \frac{1}{0} \, \mathbf{v}_2 \, \mathbf{v}_2^{\top}. \end{align*} +!et + This expression is undefined as we divide by zero in the second summand on the right-hand side. -\end{contexample} -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -In summary, the columns of a high-dimensional design matrix $\mathbf{X}$ are linearly dependent and this super-collinearity causes $\mathbf{X}^{\top} \mathbf{X}$ to be singular. Now recall the ML estimator of the parameter of the linear regression model: -!bt + +In summary, the columns of a high-dimensional design matrix +$\mathbf{X}$ are linearly dependent and this super-collinearity causes +$\mathbf{X}^{\top} \mathbf{X}$ to be singular. Now recall the ML +estimator of the parameter of the linear regression model: !bt \begin{align} -\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}. \end{align} !et -!et -This estimator is only well-defined if $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits. Hence, when $\mathbf{X}$ is high-dimensional the regression parameter $\bbeta$ cannot be estimated. -\\ -\\ + + +This estimator is only well-defined if $(\mathbf{X}^{\top} +\mathbf{X})^{-1}$ exits. Hence, when $\mathbf{X}$ is high-dimensional +the regression parameter $\beta$ cannot be estimated. + + Above only the practical consequence of high-dimensionality is presented: the expression $( \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ cannot be evaluated numerically. But the problem arising from the high-dimensionality of the data is more fundamental. To appreciate this, consider the normal equations: + +!bt \begin{align*} -\mathbf{X}^{\top} \mathbf{X} \bbeta & = \mathbf{X}^{\top} \mathbf{Y}. +\mathbf{X}^{\top} \mathbf{X} \beta & = \mathbf{X}^{\top} \mathbf{Y}. \end{align*} -The matrix $\mathbf{X}^{\top} \mathbf{X}$ is of rank $n$, while $\bbeta$ is a vector of length $p$. Hence, while there are $p$ unknowns, the system of linear equations from which these are to be solved effectively comprises $n$ degrees of freedom. If $p > n$, the vector $\bbeta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = +!et + +The matrix $\mathbf{X}^{\top} \mathbf{X}$ is of rank $n$, while $\beta$ is a vector of length $p$. Hence, while there are $p$ unknowns, the system of linear equations from which these are to be solved effectively comprises $n$ degrees of freedom. If $p > n$, the vector $\beta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{X}^{\top} \mathbf{0}_{p} = \mathbf{0}_{n}$, the solution of the normal equations is: + +!bt \begin{align*} -\hat{\bbeta} & = ( \mathbf{X}^{\top} \mathbf{X})^{-} \mathbf{X}^{\top} \mathbf{Y} + \mathbf{v} \qquad \mbox{for all } \mathbf{v} \in V, +\hat{\beta} & = ( \mathbf{X}^{\top} \mathbf{X})^{-} \mathbf{X}^{\top} \mathbf{Y} + \mathbf{v} \qquad \mbox{for all } \mathbf{v} \in V, \end{align*} +!et + where $\mathbf{A}^{-}$ denotes the Moore-Penrose inverse of the matrix $\mathbf{A}$, which is defined as: + +!bt \begin{align*} \mathbf{A}^{-} & = \sum_{j=1}^p \nu_j^{-1} \, I_{\{ \nu_j \not= 0 \} } \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. \end{align*} -The solution of the normal equations is thus only determined up to an element from a non-trivial space $V$, and there is no unique estimator of the regression parameter. -\\ -\\ -To obtain an estimate of the regression parameter $\bbeta$ when $\mathbf{X}$ is (close to) super-collinearity, cite{Hoer1970} proposed an ad-hoc fix to resolve the (almost) singularity of $\mathbf{X}^{\top} \mathbf{X}$. Simply replace $\mathbf{X}^{\top} \mathbf{X}$ by $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ with $\lambda \in [0, \infty)$. The scalar $\lambda$ is a tuning parameter, henceforth called the *penalty parameter*. +!et + +The solution of the normal equations is thus only determined up to an +element from a non-trivial space $V$, and there is no unique estimator +of the regression parameter. + +To obtain an estimate of the regression parameter $\beta$ when +$\mathbf{X}$ is (close to) super-collinearity, cite{Hoer1970} proposed +an ad-hoc fix to resolve the (almost) singularity of +$\mathbf{X}^{\top} \mathbf{X}$. Simply replace $\mathbf{X}^{\top} +\mathbf{X}$ by $\mathbf{X}^{\top} \mathbf{X} + \lambda +\mathbf{I}_{pp}$ with $\lambda \in [0, \infty)$. The scalar $\lambda$ +is a tuning parameter, henceforth called the *penalty parameter*. + -\begin{contexample}_ref{example.supercollinearity_} \hspace{3pt}*Super-collinearity (continued)* -\\ Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Then, for (say) $\lambda = 1$: +!bt \begin{align*} \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} & = \left( \begin{array}{rrr} @@ -348,69 +471,78 @@ Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.sup 2 & -4 & 7 \end{array} \right). \end{align*} +!et The eigenvalues of this matrix are 11, 7, and 1. Hence, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ has no zero eigenvalue and its inverse is well-defined. -\end{contexample} -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif With the ad-hoc fix for the singularity of $\mathbf{X}^{\top} \mathbf{X}$, cite{Hoer1970} proceed to define the *ridge regression estimator*: -!bt !bt \begin{align} label{form.ridgeRegressionEstimator} -\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}, +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}, \end{align} !et -!et -for $\lambda \in [0, \infty)$. Clearly, this is - for $\lambda$ strictly positive - a well-defined estimator, even if $\mathbf{X}$ is high-dimensional. However, each choice of $\lambda$ leads to a different ridge regression estimate. The set of all ridge regression estimates $\{ \hat{\bbeta}(\lambda) \, : \, \lambda \in [0, \infty) \}$ is called the *solution* or *regularization path* of the ridge estimator. + +for $\lambda \in [0, \infty)$. Clearly, this is - for $\lambda$ +strictly positive - a well-defined estimator, even if $\mathbf{X}$ is +high-dimensional. However, each choice of $\lambda$ leads to a +different ridge regression estimate. The set of all ridge regression +estimates $\{ \hat{\beta}(\lambda) \, : \, \lambda \in [0, \infty) \}$ +is called the *solution* or *regularization path* of the ridge +estimator. + -\begin{contexample}_ref{example.supercollinearity_} \hspace{3pt}*Super-collinearity (continued)* -\\ Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Suppose that the corresponding response vector is $\mathbf{Y} = (1.3, -0.5, 2.6, 0.9)^{\top}$. The ridge regression estimates for, e.g. $\lambda = 1, 2$, and $10$ are then: \begin{align*} -\hat{\bbeta}(1) & = (0.614, 0.548, 0.066)^{\top}, +\hat{\beta}(1) & = (0.614, 0.548, 0.066)^{\top}, \\ -\hat{\bbeta}(2) & = (0.537, 0.490, 0.048)^{\top}, +\hat{\beta}(2) & = (0.537, 0.490, 0.048)^{\top}, \\ -\hat{\bbeta}(10) & = (0.269, 0.267, 0.002)^{\top}. +\hat{\beta}(10) & = (0.269, 0.267, 0.002)^{\top}. \end{align*} The full solution path of the ridge estimator is plotted in Figure ref{fig.ridgeSolPathPlusVar}. -# original latex figure with scale=0.45, angle=0 -FIGURE: [ridgeSolutionPath_logLambda.eps, width=400 frac=1.0] Solution path of the ridge estimator and its variance. The left panel shows the solution path of the ridge estimator for the data of Example ref{example.supercollinearity}. In the right panel the corresponding variance of the ridge estimator is plotted against the (logarithm of the) penalty parameter.} label{fig.ridgeSolPathPlusVar -# \afterpage{} -\end{contexample} +Having obtained an estimate of the regression parameter $\beta$, one can define the fit $\widehat{\mathbf{Y}}$. It is defined analogous to the standard case: -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -Having obtained an estimate of the regression parameter $\bbeta$, one can define the fit $\widehat{\mathbf{Y}}$. It is defined analogous to the standard case: +!bt \begin{align*} -\widehat{\mathbf{Y}}(\lambda) & = \mathbf{X} \hat{\bbeta}(\lambda) +\widehat{\mathbf{Y}}(\lambda) & = \mathbf{X} \hat{\beta}(\lambda) \, \, \, = \, \, \, \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, := \, \, \, \mathbf{H}(\lambda) \mathbf{Y}. \end{align*} +!et + Previously, when using the ML estimator, the fit could be understood as a projection of $\mathbf{Y}$ onto the subspace spanned by the columns of $\mathbf{X}$. The fit $\widehat{\mathbf{Y}}(\lambda)$ corresponding to the ridge estimator is not a projection of $\mathbf{Y}$ onto $\mathbf{X}$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$a). Consequently, the `ridge residuals' $\mathbf{Y} - \widehat{\mathbf{Y}}(\lambda)$ are not orthogonal to the fit $\widehat{\mathbf{Y}}(\lambda)$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$b). ======= Eigenvalue shrinkage ======= The effect of the ridge penalty may also studied from the perspective of singular values. Let the singular value decomposition of the $(n \times p)$-dimensional design matrix $\mathbf{X}$ be: + +!bt \begin{align*} \mathbf{X} & = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}, \end{align*} -where $\mathbf{D}_x$ an $(n \times n)$-dimensional diagonal matrix with the singular values, -$\mathbf{U}_x$ an $(n \times n)$-dimensional matrix with columns containing the left singular vectors (denoted $\mathbf{u}_i$), and $\mathbf{V}_x$ a $(p \times n)$-dimensional matrix with columns containing the right singular vectors (denoted $\mathbf{v}_i$). The columns of $\mathbf{U}_x$ and $\mathbf{V}_x$ are orthogonal: $\mathbf{U}_x^{\top} \mathbf{U}_x = \mathbf{I}_{nn} = \mathbf{V}_x^{\top} \mathbf{V}_x$. +!et + +where $\mathbf{D}_x$ an $(n \times n)$-dimensional diagonal matrix +with the singular values, $\mathbf{U}_x$ an $(n \times n)$-dimensional +matrix with columns containing the left singular vectors (denoted +$\mathbf{u}_i$), and $\mathbf{V}_x$ a $(p \times n)$-dimensional +matrix with columns containing the right singular vectors (denoted +$\mathbf{v}_i$). The columns of $\mathbf{U}_x$ and $\mathbf{V}_x$ are +orthogonal: $\mathbf{U}_x^{\top} \mathbf{U}_x = \mathbf{I}_{nn} = +\mathbf{V}_x^{\top} \mathbf{V}_x$. The OLS estimator can then be rewritten in terms of the SVD-matrices as: + + \begin{align*} -\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} \\ @@ -422,7 +554,7 @@ The OLS estimator can then be rewritten in terms of the SVD-matrices as: \end{align*} where $\mathbf{D}_x^{-2} \mathbf{D}_x$ is not simplified further to emphasize the effect of the ridge penalty. Similarly, the ridge estimator can be rewritten in terms of the SVD-matrices as: \begin{align*} -\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} \\ @@ -458,14 +590,14 @@ Principal component regression is a close relative to ridge regression that can \\ & = \mathbf{D}_k^{-2} \widetilde{\mathbf{D}}_k \mathbf{U}^{\top} \mathbf{Y} \, \, \, = \, \, \ \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}, \end{align*} -where $\mathbf{D}_k$ and $\widetilde{\mathbf{D}}_k$ are submatrices of $\mathbf{D}$. The matrix $\mathbf{D}_k$ is obtained from $\mathbf{D}$ by removal of the last $n-p$ rows and columsn, while for $\widetilde{\mathbf{D}}_k$ only the last $n-k$ rows are dropped. Similarly, $\mathbf{I}_{kn}$ and $\mathbf{I}_{nk}$ are obtained from $\mathbf{I}_{nn}$ by removal of the last $n-k$ rows and columns, respectively. The principal component regression estimator of $\bbeta$ then is $\hat{\bbeta}_{\mbox{{\tiny pcr}}} = \mathbf{V}_k \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}$. When $k$ is set equal to the column rank of $\mathbf{X}$, and thus to the rank of $\mathbf{X}^{\top} \mathbf{X}$, the -principal component regression estimator $\hat{\bbeta}_{\mbox{{\tiny pcr}}} = (\mathbf{X}^{\top} \mathbf{X})^- \mathbf{X}^{\top} \mathbf{Y}$, where $\mathbf{A}^-$ denotes the Moore-Penrose inverse of matrix $\mathbf{A}$. +where $\mathbf{D}_k$ and $\widetilde{\mathbf{D}}_k$ are submatrices of $\mathbf{D}$. The matrix $\mathbf{D}_k$ is obtained from $\mathbf{D}$ by removal of the last $n-p$ rows and columsn, while for $\widetilde{\mathbf{D}}_k$ only the last $n-k$ rows are dropped. Similarly, $\mathbf{I}_{kn}$ and $\mathbf{I}_{nk}$ are obtained from $\mathbf{I}_{nn}$ by removal of the last $n-k$ rows and columns, respectively. The principal component regression estimator of $\beta$ then is $\hat{\beta}_{\mbox{{\tiny pcr}}} = \mathbf{V}_k \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}$. When $k$ is set equal to the column rank of $\mathbf{X}$, and thus to the rank of $\mathbf{X}^{\top} \mathbf{X}$, the +principal component regression estimator $\hat{\beta}_{\mbox{{\tiny pcr}}} = (\mathbf{X}^{\top} \mathbf{X})^- \mathbf{X}^{\top} \mathbf{Y}$, where $\mathbf{A}^-$ denotes the Moore-Penrose inverse of matrix $\mathbf{A}$. The relation between ridge and principal component regression becomes clear when their corresponding estimators are written in terms of the singular value decomposition of $\mathbf{X}$: \begin{align*} -\hat{\bbeta}_{\mbox{{\tiny pcr}}} & = \mathbf{V}_x (\mathbf{I}_{nk} \mathbf{D}_x \mathbf{I}_{kn})^{-1} \mathbf{U}_x^{\top} \mathbf{Y}, +\hat{\beta}_{\mbox{{\tiny pcr}}} & = \mathbf{V}_x (\mathbf{I}_{nk} \mathbf{D}_x \mathbf{I}_{kn})^{-1} \mathbf{U}_x^{\top} \mathbf{Y}, \\ -\hat{\bbeta} (\lambda) & = \mathbf{V}_x (\mathbf{D}_x^2 + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. +\hat{\beta} (\lambda) & = \mathbf{V}_x (\mathbf{D}_x^2 + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. \end{align*} Both operate on the singular values of the design matrix. But where principal component regression thresholds the singular values of $\mathbf{X}$, ridge regression shrinks them (depending on their size). Hence, one applies a discrete map on the singular values while the other a continuous one. @@ -482,21 +614,21 @@ The first two moments of the ridge regression estimator are derived. Next the pe The left panel of Figure ref{fig.ridgeSolPathPlusVar} shows ridge estimates of the regression parameters converging to zero as the penalty parameter tends to infinity. This behaviour of the ridge estimator does not depend on the specifics of the data set. To see this study the expectation of the ridge estimator: \begin{align*} -\mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \big] +\mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \big] \\ & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \big] \\ -& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta} \big] +& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \big] \\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \mathbb{E} ( \hat{\bbeta} ) +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \mathbb{E} ( \hat{\beta} ) \\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \bbeta. +& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta. \end{align*} -Clearly, $\mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] \not= \bbeta$ for any $\lambda > 0$. Hence, the ridge estimator is biased. +Clearly, $\mathbb{E} \big[ \hat{\beta}(\lambda) \big] \not= \beta$ for any $\lambda > 0$. Hence, the ridge estimator is biased. From the expression above it is clear that the expectation of the ridge estimator vanishes as $\lambda$ tends to infinity: \begin{align*} -\lim_{\lambda \rightarrow \infty} \mathbb{E} \big[ \hat{\bbeta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \bbeta \, \, \, = \, \, \, \mathbf{0}_{p}. +\lim_{\lambda \rightarrow \infty} \mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta \, \, \, = \, \, \, \mathbf{0}_{p}. \end{align*} Hence, all regression coefficients are shrunken towards zero as the penalty parameter increases. This also holds for $\mathbf{X}$ with $p > n$. Furthermore, this behaviour is not strictly monotone in $\lambda$: $\lambda_{a} > \lambda_b$ does not necessarily imply $|\hat{\beta}_j (\lambda_a) | < |\hat{\beta}_j (\lambda_b) |$. Upon close inspection this can be witnessed from the ridge solution path of $\beta_3$ in Figure ref{fig.ridgeSolPathPlusVar}. @@ -549,13 +681,13 @@ This design matrix is orthonormal as $\mathbf{X}^{\top} \mathbf{X} = \mathbf{I}_ \end{align*} In case of an orthonormal design matrix the relation between the OLS and ridge estimator is: \begin{align*} -\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, (\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = (1 + \lambda)^{-1} \mathbf{I}_{pp} \mathbf{X}^{\top} \mathbf{Y} \qquad \, \, = \, \, \, (1 + \lambda)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ -& = (1 + \lambda)^{-1} \hat{\bbeta}. +& = (1 + \lambda)^{-1} \hat{\beta}. \end{align*} Hence, the ridge estimator scales the OLS estimator by a factor. When taking the expectation on both sides, it is evident that the ridge estimator converges to zero as $\lambda \rightarrow \infty$. \end{example} @@ -564,13 +696,13 @@ Hence, the ridge estimator scales the OLS estimator by a factor. When taking the ===== Variance ===== -As for the ML estimate of the regression parameter $\bbeta$ of model (ref{form.linRegressionModelinMatrix}), we derive the second moment of the ridge estimator. Hereto define: +As for the ML estimate of the regression parameter $\beta$ of model (ref{form.linRegressionModelinMatrix}), we derive the second moment of the ridge estimator. Hereto define: \begin{align*} \mathbf{W}_{\lambda} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X}. \end{align*} -Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\bbeta}(\lambda)$ can be expressed as $\mathbf{W}_{\lambda} \hat{\bbeta}$ for: +Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\beta}(\lambda)$ can be expressed as $\mathbf{W}_{\lambda} \hat{\beta}$ for: \begin{align*} -\mathbf{W}_{\lambda} \hat{\bbeta} & = \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\mathbf{W}_{\lambda} \hat{\beta} & = \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ] \}^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ @@ -578,29 +710,29 @@ Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\bbeta}(\lambda)$ can be \\ & = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ -& = \hat{\bbeta}(\lambda). +& = \hat{\beta}(\lambda). \end{align*} The linear operator $\mathbf{W}_{\lambda}$ thus transforms the ML estimator of the regression parameter into the ridge estimator. It is now easily seen that: \begin{align*} -\mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \mbox{Var}[ \mathbf{W}_{\lambda} \hat{\bbeta} ] \qquad \qquad \, \, \, \, \, \, = \, \, \, \mathbf{W}_{\lambda} \mbox{Var}[\hat{\bbeta} ] \mathbf{W}_{\lambda}^{\top} +\mbox{Var}[ \hat{\beta}(\lambda) ] & = \mbox{Var}[ \mathbf{W}_{\lambda} \hat{\beta} ] \qquad \qquad \, \, \, \, \, \, = \, \, \, \mathbf{W}_{\lambda} \mbox{Var}[\hat{\beta} ] \mathbf{W}_{\lambda}^{\top} \\ & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}, \end{align*} -in which we have used $\mbox{Var}(\mathbf{A} \mathbf{Y}) = \mathbf{A} \mbox{Var}( \mathbf{Y}) \mathbf{A}^{\top}$ for a non-random matrix $\mathbf{A}$, the fact that $\mathbf{W}_{\lambda}$ is non-random, and $ \mbox{Var}[\hat{\bbeta} ] = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}$. +in which we have used $\mbox{Var}(\mathbf{A} \mathbf{Y}) = \mathbf{A} \mbox{Var}( \mathbf{Y}) \mathbf{A}^{\top}$ for a non-random matrix $\mathbf{A}$, the fact that $\mathbf{W}_{\lambda}$ is non-random, and $ \mbox{Var}[\hat{\beta} ] = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}$. Like the expectation the variance of the ridge estimator vanishes as $\lambda$ tends to infinity: \begin{align*} -\lim_{\lambda \rightarrow \infty} \mbox{Var} \big[ \hat{\bbeta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \mathbf{0}_{pp}. +\lim_{\lambda \rightarrow \infty} \mbox{Var} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \mathbf{0}_{pp}. \end{align*} Hence, the variance of the ridge regression coefficient estimates decreases towards zero as the penalty parameter becomes large. This is illustrated in the right panel of Figure ref{fig.ridgeSolPathPlusVar} for the data of Example ref{example.supercollinearity}. With an explicit expression of the variance of the ridge estimator at hand, we can compare it to that of the OLS estimator: \begin{align*} -\mbox{Var}[ \hat{\bbeta} ] - \mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \sigma^2 [(\mathbf{X}^{\top} \mathbf{X})^{-1} - \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] +\mbox{Var}[ \hat{\beta} ] - \mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 [(\mathbf{X}^{\top} \mathbf{X})^{-1} - \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] \\ & = \sigma^2 \mathbf{W}_{\lambda} \{ [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ] (\mathbf{X}^{\top} \mathbf{X})^{-1} [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ]^{\top} - (\mathbf{X}^{\top} \mathbf{X})^{-1} \} \mathbf{W}_{\lambda}^{\top} \\ @@ -614,7 +746,7 @@ The difference is non-negative definite as each component in the matrix product !bt \begin{align} label{form.VarInequalityMLandRidge} -\mbox{Var}[ \hat{\bbeta} ] & \succeq & \mbox{Var}[ \hat{\bbeta}(\lambda) ], +\mbox{Var}[ \hat{\beta} ] & \succeq & \mbox{Var}[ \hat{\beta}(\lambda) ], \end{align} !et !et @@ -643,14 +775,14 @@ Consider the design matrix: \end{align*} The variances of the ML and ridge (with $\lambda=1$) estimates of the regression coefficients then are: \begin{align*} -\mbox{Var}(\hat{\bbeta}) & = \sigma^2 \left( +\mbox{Var}(\hat{\beta}) & = \sigma^2 \left( \begin{array}{rr} 0.3 & 0.2 \\ 0.2 & 0.3 \end{array} \right) \qquad \mbox{and} \qquad -\mbox{Var}[\hat{\bbeta}(\lambda)] \, \, \, = \, \, \, \sigma^2 \left( +\mbox{Var}[\hat{\beta}(\lambda)] \, \, \, = \, \, \, \sigma^2 \left( \begin{array}{rr} 0.1524 & 0.0698 \\ @@ -665,9 +797,9 @@ These variances can be used to construct levels sets of the distribution of the \begin{contexample}_ref{example.orthoronormalDesign_} *Orthonormal design matrix (continued)* \\ -Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{Var}[ \hat{\bbeta} ] = \sigma^2 \mathbf{I}_{pp}$ and +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{Var}[ \hat{\beta} ] = \sigma^2 \mathbf{I}_{pp}$ and \begin{align*} -\mbox{Var}[ \hat{\bbeta}(\lambda) ] & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} +\mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \sigma^2 [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{I}_{pp} \{ [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \}^{\top} \, \, \, = \, \, \, \sigma^2 (1 + \lambda )^{-2} \mathbf{I}_{pp} . \end{align*} @@ -677,11 +809,11 @@ As the penalty parameter $\lambda$ is non-negative the former exceeds the latter # #if FORMAT in ("latex", "pdflatex") \noindent # #endif -The full distribution of the ridge regression estimator is now known. The estimator, $\hat{\bbeta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ is a linear estimator, linear in $\mathbf{Y}$. As $\mathbf{Y}$ is normally distributed, so is $\hat{\bbeta}(\lambda)$. Moreover, the normal distribution is fully characterized by its first two moments, which are available. Hence: +The full distribution of the ridge regression estimator is now known. The estimator, $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ is a linear estimator, linear in $\mathbf{Y}$. As $\mathbf{Y}$ is normally distributed, so is $\hat{\beta}(\lambda)$. Moreover, the normal distribution is fully characterized by its first two moments, which are available. Hence: \begin{align*} -\hat{\bbeta}(\lambda) & \sim & \mathcal{N} \big( (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X} \, \bbeta, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top} \big). +\hat{\beta}(\lambda) & \sim & \mathcal{N} \big( (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X} \, \beta, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top} \big). \end{align*} -Given $\lambda$ and $\bbeta$, the random behavior of the estimator is thus known. +Given $\lambda$ and $\beta$, the random behavior of the estimator is thus known. @@ -701,43 +833,43 @@ The MSE of the ridge estimator is: !bt \begin{align} -\mbox{MSE}[\hat{\bbeta}(\lambda)] & = \mathbb{E} [ (\mathbf{W}_{\lambda} \, \hat{\bbeta} - \bbeta)^{\top} \, (\mathbf{W}_{\lambda} \, \hat{\bbeta} - \bbeta) ] \nonumber +\mbox{MSE}[\hat{\beta}(\lambda)] & = \mathbb{E} [ (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta)^{\top} \, (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta) ] \nonumber \\ -& = \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta) + \mathbb{E} ( \bbeta^{\top} \bbeta) \nonumber +& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber \\ -& = \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta) + \mathbb{E} ( \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta ) \nonumber +& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) + \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) \nonumber \\ -& & - \mathbb{E} ( \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta ) + \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\bbeta}) + \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta) \nonumber +& & - \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) + \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) + \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) \nonumber \\ -& & - \mathbb{E} ( \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\bbeta}) - \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta) + \mathbb{E} ( \bbeta^{\top} \bbeta) \nonumber +& & - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber \\ -& = \mathbb{E} [ ( \hat{\bbeta} - \bbeta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\bbeta} - \bbeta) ] \nonumber +& = \mathbb{E} [ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) ] \nonumber \\ -& & - \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \bbeta + \bbeta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \bbeta + \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \bbeta \nonumber +& & - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta + \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \beta + \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta \nonumber \\ -& & - \bbeta^{\top} \, \mathbf{W}_{\lambda} \, \bbeta - \bbeta^{\top} \mathbf{W}_{\lambda}^{\top} \, \bbeta + \bbeta^{\top} \bbeta \nonumber +& & - \beta^{\top} \, \mathbf{W}_{\lambda} \, \beta - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta + \beta^{\top} \beta \nonumber \\ -& = \mathbb{E} \big\{ ( \hat{\bbeta} - \bbeta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\bbeta} - \bbeta) \big\} + \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta \nonumber +& = \mathbb{E} \big\{ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta \nonumber \\ -& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta. label{form.ridgeMSE} +& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta. label{form.ridgeMSE} \end{align} !et !et -In the last step we have used $\hat{\bbeta} \sim \mathcal{N}( \bbeta, \sigma^2 \, [\mathbf{X}^{\top} \mathbf{X}]^{-1} )$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ for some nonrandom symmetric positive definite matrix $\LLambda$ is (cf. \citealt{Math1992}): +In the last step we have used $\hat{\beta} \sim \mathcal{N}( \beta, \sigma^2 \, [\mathbf{X}^{\top} \mathbf{X}]^{-1} )$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ for some nonrandom symmetric positive definite matrix $\LLambda$ is (cf. \citealt{Math1992}): \begin{align*} \mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) & = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}, \end{align*} -of course replacing $\vvarepsilon$ by $\hat{\bbeta}$ in this expectation. The first summand in the final derived expression for $\mbox{MSE}[\hat{\bbeta}(\lambda)]$ is the sum of the variances of the ridge estimator, while the second summand can be thought of the ``squared bias'' of the ridge estimator. In particular, $\lim_{\lambda \rightarrow \infty} \mbox{MSE}[\hat{\bbeta}(\lambda)] = \bbeta^{\top} \bbeta$, which is the squared biased for an estimator that equals zero (as does the ridge estimator in the limit). +of course replacing $\vvarepsilon$ by $\hat{\beta}$ in this expectation. The first summand in the final derived expression for $\mbox{MSE}[\hat{\beta}(\lambda)]$ is the sum of the variances of the ridge estimator, while the second summand can be thought of the ``squared bias'' of the ridge estimator. In particular, $\lim_{\lambda \rightarrow \infty} \mbox{MSE}[\hat{\beta}(\lambda)] = \beta^{\top} \beta$, which is the squared biased for an estimator that equals zero (as does the ridge estimator in the limit). \begin{example} *Orthonormal design matrix* \\ -Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{MSE}[ \hat{\bbeta} ] = p \, \sigma^2$ and +Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{MSE}[ \hat{\beta} ] = p \, \sigma^2$ and \begin{align*} -\mbox{MSE}[ \hat{\bbeta}(\lambda) ] & = \frac{p \, \sigma^2}{ (1+ \lambda)^{2}} + \frac{\lambda^2}{ (1+ \lambda)^{2}} \bbeta^{\top} \bbeta. +\mbox{MSE}[ \hat{\beta}(\lambda) ] & = \frac{p \, \sigma^2}{ (1+ \lambda)^{2}} + \frac{\lambda^2}{ (1+ \lambda)^{2}} \beta^{\top} \beta. \end{align*} -The latter achieves its minimum at: $\lambda = p \sigma^2 / \bbeta^{\top} \bbeta$. +The latter achieves its minimum at: $\lambda = p \sigma^2 / \beta^{\top} \beta$. \end{example} @@ -767,40 +899,40 @@ We are now ready to proof the main result, formalized as Theorem ref{theo.Theoba \begin{theorem} *(Theorem 2 of \citealp{Theo1974*}) label{theo.Theobald2} \\ -There exists $\lambda > 0$ such that $\mbox{MSE}[\hat{\bbeta}(\lambda)] < \mbox{MSE}[\hat{\bbeta}(0)] = \mbox{MSE}[\hat{\bbeta}]$. +There exists $\lambda > 0$ such that $\mbox{MSE}[\hat{\beta}(\lambda)] < \mbox{MSE}[\hat{\beta}(0)] = \mbox{MSE}[\hat{\beta}]$. \end{theorem} \begin{proof} The second order moment matrix of the ridge estimator is: \begin{align*} -\mathbf{M} (\lambda) & := & \mathbb{E} [ (\hat{\bbeta}(\lambda) - \bbeta) (\hat{\bbeta} (\lambda) - \bbeta)^{\top} ] +\mathbf{M} (\lambda) & := & \mathbb{E} [ (\hat{\beta}(\lambda) - \beta) (\hat{\beta} (\lambda) - \beta)^{\top} ] \\ -& = \mathbb{E} \{ \hat{\bbeta}(\lambda) [\hat{\bbeta}(\lambda)]^{\top} \} - \mathbb{E} [ \hat{\bbeta}(\lambda) ] \{ \mathbb{E} [ \hat{\bbeta}(\lambda) ] \}^{\top} + \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \{ \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \}^{\top} +& = \mathbb{E} \{ \hat{\beta}(\lambda) [\hat{\beta}(\lambda)]^{\top} \} - \mathbb{E} [ \hat{\beta}(\lambda) ] \{ \mathbb{E} [ \hat{\beta}(\lambda) ] \}^{\top} + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top} \\ -& = \mbox{Var}[ \hat{\bbeta}(\lambda) ] + \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \{ \mathbb{E} [\hat{\bbeta} (\lambda) - \bbeta)] \}^{\top}. +& = \mbox{Var}[ \hat{\beta}(\lambda) ] + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top}. \end{align*} Then: \begin{align*} \mathbf{M} ( 0 ) - \mathbf{M}(\lambda) & = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] \\ -& & - (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \bbeta \bbeta^{\top} (\mathbf{W}_{\lambda} -\mathbf{I}_{pp})^{\top} +& & - (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \beta \beta^{\top} (\mathbf{W}_{\lambda} -\mathbf{I}_{pp})^{\top} \\ & = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} \\ -& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \bbeta \bbeta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} \\ & = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} \\ -& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \bbeta \bbeta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} +& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} \\ -& = \lambda [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \bbeta \bbeta^{\top} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top}. +& = \lambda [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top}. \end{align*} -This is positive definite if and only if $ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \bbeta \bbeta^{\top} \succ 0$. Hereto it suffices to show that $2 \, \sigma^2 \, \mathbf{I}_{pp} - \lambda \bbeta \bbeta^{\top} \succ 0$. By Proposition ref{prop.Farebrother} this holds for $\lambda$ such that $2 \sigma^2 (\bbeta^{\top} \bbeta)^{-1} > \lambda$. For these $\lambda$, we thus have $\mathbf{M} ( 0 ) - \mathbf{M}(\lambda)$. Application of Theorem ref{theo.Theobald1} now concludes the proof. +This is positive definite if and only if $ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} \succ 0$. Hereto it suffices to show that $2 \, \sigma^2 \, \mathbf{I}_{pp} - \lambda \beta \beta^{\top} \succ 0$. By Proposition ref{prop.Farebrother} this holds for $\lambda$ such that $2 \sigma^2 (\beta^{\top} \beta)^{-1} > \lambda$. For these $\lambda$, we thus have $\mathbf{M} ( 0 ) - \mathbf{M}(\lambda)$. Application of Theorem ref{theo.Theobald1} now concludes the proof. \end{proof} This result of cite{Theo1974} is generalized by cite{Fare1976} to the class of design matrices $\mathbf{X}$ with $\mbox{rank}(\mathbf{X}) < p$. \\ \\ -Theorem ref{theo.Theobald2} can be used to illustrate that the ridge regression estimator strikes a balance between the bias and variance. This is illustrated in the left panel of Figure ref{fig.MSEridge}. For small $\lambda$, the variance of the ridge estimator dominates the MSE. This may be understood when realizing that in this domain of $\lambda$ the ridge estimator is close to the unbiased ML regression estimator. For large $\lambda$, the variance vanishes and the bias dominates the MSE. For small enough values of $\lambda$, the decrease in variance of the ridge regression estimator exceeds the increase in its bias. As the MSE is the sum of these two, the MSE first decreases as $\lambda$ moves away from zero. In particular, as $\lambda = 0$ corresponds to the ML regression estimator, the ridge regression estimator yields a lower MSE for these values of $\lambda$. In the right panel of Figure ref{fig.MSEridge} $\mbox{MSE}[ \hat{\bbeta}(\lambda)] < \mbox{MSE}[ \hat{\bbeta}(0)]$ for $\lambda < 7$ (roughly) and the ridge estimator outperforms the ML estimator. +Theorem ref{theo.Theobald2} can be used to illustrate that the ridge regression estimator strikes a balance between the bias and variance. This is illustrated in the left panel of Figure ref{fig.MSEridge}. For small $\lambda$, the variance of the ridge estimator dominates the MSE. This may be understood when realizing that in this domain of $\lambda$ the ridge estimator is close to the unbiased ML regression estimator. For large $\lambda$, the variance vanishes and the bias dominates the MSE. For small enough values of $\lambda$, the decrease in variance of the ridge regression estimator exceeds the increase in its bias. As the MSE is the sum of these two, the MSE first decreases as $\lambda$ moves away from zero. In particular, as $\lambda = 0$ corresponds to the ML regression estimator, the ridge regression estimator yields a lower MSE for these values of $\lambda$. In the right panel of Figure ref{fig.MSEridge} $\mbox{MSE}[ \hat{\beta}(\lambda)] < \mbox{MSE}[ \hat{\beta}(0)]$ for $\lambda < 7$ (roughly) and the ridge estimator outperforms the ML estimator. # original latex figure with scale=0.45, angle=0 FIGURE: [MSE_ridge.eps, width=400 frac=1.0] Left panel: mean squared error, and its `bias' and `variance' parts, of the ridge regression estimator (for artificial data). Right panel: mean squared error of the ridge and ML estimator of the regression coefficient vector (for the same artificial data).} label{fig.MSEridge @@ -810,7 +942,7 @@ FIGURE: [MSE_ridge.eps, width=400 frac=1.0] Left panel: mean squared error, and # #if FORMAT in ("latex", "pdflatex") \noindent # #endif -Besides another motivation behind the ridge regression estimator, the use of Theorem ref{theo.Theobald2} is limited. The optimal choice of $\lambda$ depends on the quantities $\bbeta$ and $\sigma^2$. These are unknown in practice. Then, the penalty parameter is chosen in a data-driven fashion by means of cross-validation (see Section ref{subsect.crossvalidation}). +Besides another motivation behind the ridge regression estimator, the use of Theorem ref{theo.Theobald2} is limited. The optimal choice of $\lambda$ depends on the quantities $\beta$ and $\sigma^2$. These are unknown in practice. Then, the penalty parameter is chosen in a data-driven fashion by means of cross-validation (see Section ref{subsect.crossvalidation}). \\ \\ \begin{remark} \mbox{ } @@ -829,30 +961,30 @@ The ridge estimator minimizes the *ridge loss function*, which is defined as: !bt \begin{align} -\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \lambda \| \bbeta \|^2_2 -\, \, \, = \, \, \, \sum_{i=1}^n (Y_i - \mathbf{X}_{i\ast} \, \bbeta)^2 + \lambda \sum_{j=1}^p \beta_j^2. label{form.ridgeLossFunction} +\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2 +\, \, \, = \, \, \, \sum_{i=1}^n (Y_i - \mathbf{X}_{i\ast} \, \beta)^2 + \lambda \sum_{j=1}^p \beta_j^2. label{form.ridgeLossFunction} \end{align} !et !et -This loss function is the traditional sum-of-squares augmented with a *penalty*. The particular form of the penalty, $\lambda \| \bbeta \|^2_2$ is referred to as the *ridge penalty* and $\lambda$ as the *penalty parameter*. For $\lambda=0$, minimization of the ridge loss function yields the ML estimator. For any $\lambda > 0$, the ridge penalty contributes to the loss function, affecting its minimum and its location. The minimum of the sum-of-squares is well-known. The minimum of the ridge penalty is attained at $\bbeta = \mathbf{0}_{p}$ whenever $\lambda > 0$. The $\bbeta$ that minimizes $\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda)$ then balances the sum-of-squares and the penalty. The effect of the penalty in this balancing act is to shrink the regression coefficients towards zero, its minimum. In particular, the larger $\lambda$, the larger the contribution of the penalty to the loss function, the stronger the tendency to shrink non-zero regression coefficients to zero (and decrease the contribution of the penalty to the loss function). This motivates the name `penalty' as non-zero elements of $\bbeta$ increase (or penalize) the loss function. +This loss function is the traditional sum-of-squares augmented with a *penalty*. The particular form of the penalty, $\lambda \| \beta \|^2_2$ is referred to as the *ridge penalty* and $\lambda$ as the *penalty parameter*. For $\lambda=0$, minimization of the ridge loss function yields the ML estimator. For any $\lambda > 0$, the ridge penalty contributes to the loss function, affecting its minimum and its location. The minimum of the sum-of-squares is well-known. The minimum of the ridge penalty is attained at $\beta = \mathbf{0}_{p}$ whenever $\lambda > 0$. The $\beta$ that minimizes $\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda)$ then balances the sum-of-squares and the penalty. The effect of the penalty in this balancing act is to shrink the regression coefficients towards zero, its minimum. In particular, the larger $\lambda$, the larger the contribution of the penalty to the loss function, the stronger the tendency to shrink non-zero regression coefficients to zero (and decrease the contribution of the penalty to the loss function). This motivates the name `penalty' as non-zero elements of $\beta$ increase (or penalize) the loss function. -To verify that the ridge estimator indeed minimizes the ridge loss function, proceed as usual. Take the derivative with respect to $\bbeta$: +To verify that the ridge estimator indeed minimizes the ridge loss function, proceed as usual. Take the derivative with respect to $\beta$: \begin{align*} -\frac{\partial}{\partial \bbeta} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = -2 \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \bbeta) + 2 \, \lambda \, \mathbf{I}_{pp} \, \bbeta \, \, \, = \, \, \, -2 \, \mathbf{X}^{\top} \mathbf{Y} + 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}) \bbeta. +\frac{\partial}{\partial \beta} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = -2 \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \beta) + 2 \, \lambda \, \mathbf{I}_{pp} \, \beta \, \, \, = \, \, \, -2 \, \mathbf{X}^{\top} \mathbf{Y} + 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}) \beta. \end{align*} -Equate the derivative to zero and solve for $\bbeta$. This yields the ridge regression estimator. +Equate the derivative to zero and solve for $\beta$. This yields the ridge regression estimator. The ridge estimator is thus a stationary point of the ridge loss function. A stationary point corresponds to a minimum if the Hessian matrix with second order partial derivatives is positive definite. The Hessian of the ridge loss function is \begin{align*} -\frac{\partial^2}{\partial \bbeta \, \partial \bbeta^{\top}} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\bbeta; \lambda) & = 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}). +\frac{\partial^2}{\partial \beta \, \partial \beta^{\top}} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}). \end{align*} This Hessian is the sum of the semi-positive definite matrix $\mathbf{X}^{\top} \mathbf{X}$ and the positive definite matrix $\lambda \, \mathbf{I}_{pp}$. Lemma 14.2.4 of cite{Harv2008} then states that the sum of these matrices is itself a positive definite matrix. Hence, the Hessian is positive definite and the ridge loss function has a stationary point at the ridge estimator, which is a minimum. -The ridge regression estimator minimizes the ridge loss function. It rests to verify that it is a global minimum. To this end we introduce the concept of a convex function. As a prerequisite, a set $\mathcal{S} \subset \mathbb{R}^p$ is called *convex* if for all $\bbeta_1, \bbeta_2 \in \mathcal{S}$ their weighted average $\bbeta_{\theta} = (1 - \theta) \bbeta_1 + \theta \bbeta_2$ for all $\theta \in [0, 1]$ is itself an element of $\mathcal{S}$, thus $\bbeta_{\theta} \in \mathcal{S}$. If for all $\theta \in (0, 1)$, the weighted average $\bbeta_{\theta}$ is inside $\mathcal{S}$ and not on its boundary, the set is called *strict convex*. Examples of (strict) convex and nonconvex sets are depicted in Figure ref{fig.ridgeAsConstrainedEst}. A function $f(\cdot)$ is *(strict) convex* if the set $\{ y \, : \, y \geq f(\bbeta) \mbox{ for all } \bbeta \in \mathcal{S} \mbox{ for any convex } \mathcal{S} \}$, called the epigraph of $f(\cdot)$, is (strict) convex. Examples of (strict) convex and nonconvex functions are depicted in Figure ref{fig.ridgeAsConstrainedEst}. The ridge loss function is the sum of two parabola's: one at least convex and the other a strict convex function in $\bbeta$. The sum of convex and strict convex function is itself strict convex (confer Lemma 9.4.2 of \citealt{Flet2008}). The ridge loss function is thus strict convex. Theorem 9.4.1 of \citealt{Flet2008} then warrants, by the strict convexity of the ridge loss function, that the ridge estimator is a global minimum. +The ridge regression estimator minimizes the ridge loss function. It rests to verify that it is a global minimum. To this end we introduce the concept of a convex function. As a prerequisite, a set $\mathcal{S} \subset \mathbb{R}^p$ is called *convex* if for all $\beta_1, \beta_2 \in \mathcal{S}$ their weighted average $\beta_{\theta} = (1 - \theta) \beta_1 + \theta \beta_2$ for all $\theta \in [0, 1]$ is itself an element of $\mathcal{S}$, thus $\beta_{\theta} \in \mathcal{S}$. If for all $\theta \in (0, 1)$, the weighted average $\beta_{\theta}$ is inside $\mathcal{S}$ and not on its boundary, the set is called *strict convex*. Examples of (strict) convex and nonconvex sets are depicted in Figure ref{fig.ridgeAsConstrainedEst}. A function $f(\cdot)$ is *(strict) convex* if the set $\{ y \, : \, y \geq f(\beta) \mbox{ for all } \beta \in \mathcal{S} \mbox{ for any convex } \mathcal{S} \}$, called the epigraph of $f(\cdot)$, is (strict) convex. Examples of (strict) convex and nonconvex functions are depicted in Figure ref{fig.ridgeAsConstrainedEst}. The ridge loss function is the sum of two parabola's: one at least convex and the other a strict convex function in $\beta$. The sum of convex and strict convex function is itself strict convex (confer Lemma 9.4.2 of \citealt{Flet2008}). The ridge loss function is thus strict convex. Theorem 9.4.1 of \citealt{Flet2008} then warrants, by the strict convexity of the ridge loss function, that the ridge estimator is a global minimum. \\ \\ -From the ridge loss function the limiting behavior of the variance of the ridge regression estimator can be understood. The ridge penalty with its minimum $\bbeta = \mathbf{0}_{p}$ does not involve data and, consequently, the variance of its minimum equals zero. With the ridge regression being a compromise between the ML estimator and the minimum of the penalty, so is its variance a compromise of their variances. As $\lambda$ tends to infinity, the ridge estimator and its variance converge to the minimum and the variance of the minimum, respectively. Hence, in the limit (large $\lambda$) the variance of the ridge regression estimator vanishes. Understandably, as the penalty now fully dominates the loss function and, consequently, it does no longer involve data (i.e. randomness). +From the ridge loss function the limiting behavior of the variance of the ridge regression estimator can be understood. The ridge penalty with its minimum $\beta = \mathbf{0}_{p}$ does not involve data and, consequently, the variance of its minimum equals zero. With the ridge regression being a compromise between the ML estimator and the minimum of the penalty, so is its variance a compromise of their variances. As $\lambda$ tends to infinity, the ridge estimator and its variance converge to the minimum and the variance of the minimum, respectively. Hence, in the limit (large $\lambda$) the variance of the ridge regression estimator vanishes. Understandably, as the penalty now fully dominates the loss function and, consequently, it does no longer involve data (i.e. randomness). # original latex figure with scale=0.38, angle=0 FIGURE: [convexSets.eps, width=400 frac=1.0] Top panels show examples of convex (left) and nonconvex (right) sets. Middle panels show examples of convex (left) and nonconvex (right) functions. The left bottom panel illustrates the ridge estimation as a constrained estimation problem. The ellipses represent the contours of the ML loss function, with the blue dot at the center the ML estimate. The circle is the ridge parameter constraint. The red dot is the ridge estimate. It is at the intersection of the ridge constraint and the smallest contour with a non-empty intersection with the constraint. The right bottom panel shows the data corresponding to Example ref{exam.overfitting}. The grey line represents the `true' relationship, while the black line the fitted one. } label{fig.ridgeAsConstrainedEst @@ -865,7 +997,7 @@ Above it has been shown that the ridge estimator can be defined as: !bt \begin{align} label{form.ridgeEstViaPenEst} -\hat{\bbeta}(\lambda) & = \arg \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \lambda \| \bbeta \|^2_2. +\hat{\beta}(\lambda) & = \arg \min_{\beta} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2. \end{align} !et !et @@ -875,15 +1007,15 @@ This minimization problem can be reformulated into the following constrained opt !bt \begin{align} label{form.constrEstProblemRidge} -\hat{\bbeta}(\lambda) & = \arg \min_{\| \bbeta \|_2^2 \leq c} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2, +\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2, \end{align} !et !et -for some suitable $c > 0$. The constrained optimization problem (ref{form.constrEstProblemRidge}) can be solved by means of the Karush-Kuhn-Tucker (KKT) multiplier method, which minimizes a function subject to inequality constraints. The KKT multiplier method states that, under some regularity conditions (all met here), there exists a constant $\nu \geq 0$, called the *multiplier*, such that the solution $\hat{\bbeta}(\nu)$ of the constrained minimization problem (ref{form.constrEstProblemRidge}) satisfies the so-called KKT conditions. The first KKT condition (referred to as the stationarity condition) demands that the gradient (with respect to $\bbeta$) of the Lagrangian associated with the minimization problem equals zero at the solution $\hat{\bbeta}(\nu)$. The Lagrangian for problem (ref{form.constrEstProblemRidge}) is: +for some suitable $c > 0$. The constrained optimization problem (ref{form.constrEstProblemRidge}) can be solved by means of the Karush-Kuhn-Tucker (KKT) multiplier method, which minimizes a function subject to inequality constraints. The KKT multiplier method states that, under some regularity conditions (all met here), there exists a constant $\nu \geq 0$, called the *multiplier*, such that the solution $\hat{\beta}(\nu)$ of the constrained minimization problem (ref{form.constrEstProblemRidge}) satisfies the so-called KKT conditions. The first KKT condition (referred to as the stationarity condition) demands that the gradient (with respect to $\beta$) of the Lagrangian associated with the minimization problem equals zero at the solution $\hat{\beta}(\nu)$. The Lagrangian for problem (ref{form.constrEstProblemRidge}) is: \begin{align*} -\| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2 + \nu ( \| \bbeta \|^2_2 - c). +\| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \nu ( \| \beta \|^2_2 - c). \end{align*} -The second KKT condition (the complementarity condition) requires that $\nu (\| \hat{\bbeta}(\nu) \|_2^2 - c) = 0$. If $\nu = \lambda$ and $c = \| \hat{\bbeta}(\lambda) \|_2^2$, the ridge estimator $\bbeta (\lambda)$ satisfies both KKT conditions. Hence, both problems have the same solution when $c = \| \hat{\bbeta}(\lambda) \|_2^2$. +The second KKT condition (the complementarity condition) requires that $\nu (\| \hat{\beta}(\nu) \|_2^2 - c) = 0$. If $\nu = \lambda$ and $c = \| \hat{\beta}(\lambda) \|_2^2$, the ridge estimator $\beta (\lambda)$ satisfies both KKT conditions. Hence, both problems have the same solution when $c = \| \hat{\beta}(\lambda) \|_2^2$. \\ \\ The relevance of viewing the ridge regression estimator as the solution to a constrained estimation problem becomes obvious when considering a typical threat to high-dimensional data analysis: overfitting. *Overfitting* refers to the phenomenon of modelling the noise rather than the signal. In case the true model is parsimonious (few covariates driving the response) and data on many covariates are available, it is likely that a linear combination of all covariates yields a higher likelihood than a combination of the few that are actually related to the response. As only the few covariates related to the response contain the signal, the model involving all covariates then cannot but explain more than the signal alone: it also models the error. Hence, it overfits the data. In high-dimensional settings overfitting is a real threat. The number of explanatory variables exceeds the number of observations. It is thus possible to form a linear combination of the covariates that perfectly explains the response, including the noise. @@ -897,81 +1029,81 @@ Consider an artificial data set comprising of ten observations on a response $Y_ The regression model $Y_i = \sum_{j=1}^9 X_{i,j} \beta_j+ \varepsilon_i$ is fitted to the artificial data using `R`. This yields the regression parameter estimates: \begin{align*} -\hat{\bbeta}^{\top} & = (0.048, -2.386, -5.528, 6.243, -4.819, 0.760, -3.345, -4.748, 2.136). +\hat{\beta}^{\top} & = (0.048, -2.386, -5.528, 6.243, -4.819, 0.760, -3.345, -4.748, 2.136). \end{align*} -As $\bbeta^{\top} = (1, 0, \ldots, 0)$, many regression coefficient are clearly over-estimated. +As $\beta^{\top} = (1, 0, \ldots, 0)$, many regression coefficient are clearly over-estimated. -The fitted values $\widehat{Y}_i = \mathbf{X}_i \hat{\bbeta}$ are plotted against the values of the first covariates in the right bottom panel of Figure ref{fig.ridgeAsConstrainedEst}. As a reference the line $x=y$ is added, which represents the `true' model. The fitted model follows the `true' relationship. But it also captures the deviations from this line that represent the errors. +The fitted values $\widehat{Y}_i = \mathbf{X}_i \hat{\beta}$ are plotted against the values of the first covariates in the right bottom panel of Figure ref{fig.ridgeAsConstrainedEst}. As a reference the line $x=y$ is added, which represents the `true' model. The fitted model follows the `true' relationship. But it also captures the deviations from this line that represent the errors. \end{example} ======= Bayesian regression} label{sect:Bayes ======= -Ridge regression has a close connection to Bayesian linear regression. Bayesian linear regression assumes the parameters $\bbeta$ and $\sigma^2$ to be the random variables, while at the same time considering $\mathbf{X}$ and $\mathbf{Y}$ as fixed. Within the regression context, the conjugate priors of $\bbeta$ and $\sigma^2$ are: +Ridge regression has a close connection to Bayesian linear regression. Bayesian linear regression assumes the parameters $\beta$ and $\sigma^2$ to be the random variables, while at the same time considering $\mathbf{X}$ and $\mathbf{Y}$ as fixed. Within the regression context, the conjugate priors of $\beta$ and $\sigma^2$ are: \begin{align*} -\bbeta \, | \, \sigma^2 \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp}) \qquad \mbox{and} \qquad \sigma^2 \sim \mathcal{IG}(\alpha_0, \beta_0), +\beta \, | \, \sigma^2 \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp}) \qquad \mbox{and} \qquad \sigma^2 \sim \mathcal{IG}(\alpha_0, \beta_0), \end{align*} where $\mathcal{IG}$ denotes the inverse Gamma distribution with shape parameter $\alpha_0$ and scale parameter $\beta_0$. The penalty parameter can be interpreted as the precision of the prior, determining how informative the prior should be. A smaller penalty (i.e. precision) corresponds to a wider prior, and a larger penalty to a more informative, concentrated prior (Figure ref{fig.ridgePriorOfBeta}). # original latex figure with scale=0.45, angle=0 -FIGURE: [ridgePriorOfBeta.eps, width=400 frac=1.0] Conjugate prior of the regression parameter $\bbeta$ for various choices of $\lambda$, the penalty parameters c.q. precision.} label{fig.ridgePriorOfBeta +FIGURE: [ridgePriorOfBeta.eps, width=400 frac=1.0] Conjugate prior of the regression parameter $\beta$ for various choices of $\lambda$, the penalty parameters c.q. precision.} label{fig.ridgePriorOfBeta -Under the assumption of the conjugate priors above, the joint posterior distribution of $\bbeta$ and $\sigma^2$ is then: +Under the assumption of the conjugate priors above, the joint posterior distribution of $\beta$ and $\sigma^2$ is then: \begin{align*} -f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \bbeta, \sigma^2) \, f_{\beta}(\bbeta | \sigma^2) \, f_{\sigma}(\sigma^2) +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) \\ -& \propto & \sigma^{-n} \exp \Big[ - \frac{1}{2\sigma^2} ( \mathbf{Y} - \mathbf{X} \bbeta)^{\top} ( \mathbf{Y} - \mathbf{X} \bbeta) \Big] +& \propto & \sigma^{-n} \exp \Big[ - \frac{1}{2\sigma^2} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) \Big] \\ -& & \times \, \, \sigma^{-p} \exp \Big[ - \frac{1}{2\sigma^2} \lambda \bbeta^{\top} \bbeta \Big] \, \times \, \, [\sigma^2]^{-\alpha_0-1} \exp \Big[ - \frac{\beta_0}{2\sigma^2} \Big]. +& & \times \, \, \sigma^{-p} \exp \Big[ - \frac{1}{2\sigma^2} \lambda \beta^{\top} \beta \Big] \, \times \, \, [\sigma^2]^{-\alpha_0-1} \exp \Big[ - \frac{\beta_0}{2\sigma^2} \Big]. \end{align*} As \begin{align*} -& & \hspace{-1.5cm} ( \mathbf{Y} - \mathbf{X} \bbeta)^{\top} ( \mathbf{Y} - \mathbf{X} \bbeta) + \lambda \bbeta^{\top} \bbeta +& & \hspace{-1.5cm} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) + \lambda \beta^{\top} \beta \\ & = \mathbf{Y}^{\top} \mathbf{Y} - -\bbeta^{\top} \mathbf{X}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} \bbeta + \bbeta^{\top} \mathbf{X}^{\top} \mathbf{X} \bbeta + \lambda \bbeta^{\top} \bbeta +\beta^{\top} \mathbf{X}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} \beta + \beta^{\top} \mathbf{X}^{\top} \mathbf{X} \beta + \lambda \beta^{\top} \beta \\ -& = \mathbf{Y}^{\top} \mathbf{Y} - \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ -& & - \, \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta + \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta +& & - \, \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta \\ -& = \mathbf{Y}^{\top} \mathbf{Y} - \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) -\hat{\bbeta} (\lambda) +& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) +\hat{\beta} (\lambda) \\ -& & - \, [ \hat{\bbeta} (\lambda) ]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta + \bbeta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \bbeta +& & - \, [ \hat{\beta} (\lambda) ]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta \\ & = \mathbf{Y}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ -& & + \, \big[ \bbeta - \hat{\bbeta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \bbeta - \hat{\bbeta}(\lambda) \big], +& & + \, \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big], \end{align*} the posterior distribution can be rewritten to: \begin{align*} -f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) -& \propto & g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) +& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) \end{align*} with \begin{align*} -g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) -& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \bbeta - \hat{\bbeta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \bbeta - \hat{\bbeta}(\lambda) \big] \Big\}. +g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big] \Big\}. \end{align*} -Then, clearly the conditional posterior mean of $\bbeta$ is $\mathbb{E}(\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\bbeta}(\lambda)$. Hence, the ridge regression estimator can be viewed as the Bayesian posterior mean estimator of $\bbeta$ when imposing a Gaussian prior on the regression parameter. +Then, clearly the conditional posterior mean of $\beta$ is $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\lambda)$. Hence, the ridge regression estimator can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a Gaussian prior on the regression parameter. With little extra work we may also obtain the conditional posterior of $\sigma^2$ from the joint posterior distribution: \begin{align*} -f_{\sigma^2} (\sigma^2 \, | \, \bbeta, \mathbf{Y}, \mathbf{X}) & \propto & (\sigma^2)^{-[(n+p)/2 + \alpha_0 + 1]} -\exp [ - \frac{1}{2\sigma^2} ( \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + -\lambda \| \bbeta \|_2^2 + \beta_0) ], +f_{\sigma^2} (\sigma^2 \, | \, \beta, \mathbf{Y}, \mathbf{X}) & \propto & (\sigma^2)^{-[(n+p)/2 + \alpha_0 + 1]} +\exp [ - \frac{1}{2\sigma^2} ( \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + +\lambda \| \beta \|_2^2 + \beta_0) ], \end{align*} in which one can recognize the shape of an inverse gamma distribution. \\ \\ A Bayesian estimator of a parameter $\ttheta$ is the estimator that minimizes the Bayes risk over a prior distribution of the parameter $\ttheta$. The Bayes risk is defined as $\int_{\ttheta} \mathbb{E} [(\hat{\ttheta} - \ttheta)^{\top} (\hat{\ttheta} - \ttheta)] \pi_{\ttheta}(\ttheta; \aalpha) d\ttheta$, where $\pi_{\ttheta}(\ttheta; \aalpha)$ is the prior distribution of $\ttheta$ with hyperparameter $\aalpha$. It is thus a weighted average of the Mean Squared Error, with weights specified through the prior. The Bayes risk is minimized by the mean posterior $\mathbb{E}_{\ttheta}(\ttheta \, | \, \mbox{data})$ (cf., e.g., \citealp{Bijm2017}). The Bayesian estimator of $\ttheta$ thus yields the smallest possible expected MSE, under the assumption of the employed prior. -The Bayes risk of the ridge estimator over the normal prior $\bbeta \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp})$ is: +The Bayes risk of the ridge estimator over the normal prior $\beta \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp})$ is: \begin{align*} -\mathbb{E}_{\bbeta} \{\mbox{MSE}[\hat{\bbeta}(\lambda)] -\, | \, \sigma^2, \mathbf{Y}, \mathbf{X} \} & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \mathbb{E}_{\bbeta} [ \bbeta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \bbeta] +\mathbb{E}_{\beta} \{\mbox{MSE}[\hat{\beta}(\lambda)] +\, | \, \sigma^2, \mathbf{Y}, \mathbf{X} \} & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \mathbb{E}_{\beta} [ \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta] \\ & = \sigma^2 \, \big\{ \mbox{tr} \big[ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big] + \lambda^{-1} \mbox{tr} [(\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})] \big\} @@ -980,13 +1112,13 @@ The Bayes risk of the ridge estimator over the normal prior $\bbeta \sim \mathca \end{align*} in which we have used *i)* the previously derived explicit expression (ref{form.ridgeMSE}) of the ridge estimator's MSE, *ii)* the expectation of the quadratic form of a multivariate random variable \citep{Math1992}, *iii)* the singular value decomposition of $\mathbf{X}$ with singular values $d_{jj}$, and *iv)* the fact that the trace of a square matrix equals the sum of its eigenvalues. As the ridge estimator coincides with the posterior mean, this is the minimal achievable MSE under a zero-centered normal prior with an uncorrelated and equivariant covariance matrix. -Above the Bayes risk of the ridge estimator factorizes with respect to $\sigma^2$ and $\lambda$. Hence, the larger the hyperparameter $\lambda$ the lower the Bayes risk of the ridge estimator. In particular, its Bayes risk converges to zero as $\lambda \rightarrow \infty$. This can be understood as follows. The limit corresponds to an infinite precision of the prior, thus reducing the variance contribution to the MSE. Moreover, as the ridge estimator shrinks towards zero and the prior distribution of $\bbeta$ has a zero mean, the bias too vanishes as $\lambda \rightarrow \infty$. +Above the Bayes risk of the ridge estimator factorizes with respect to $\sigma^2$ and $\lambda$. Hence, the larger the hyperparameter $\lambda$ the lower the Bayes risk of the ridge estimator. In particular, its Bayes risk converges to zero as $\lambda \rightarrow \infty$. This can be understood as follows. The limit corresponds to an infinite precision of the prior, thus reducing the variance contribution to the MSE. Moreover, as the ridge estimator shrinks towards zero and the prior distribution of $\beta$ has a zero mean, the bias too vanishes as $\lambda \rightarrow \infty$. -The calculation of the Bayes risk above relates the Bayesian and frequentist statements on the MSE of the ridge estimator. For the latter revisit Theorem ref{theo.Theobald2} of Section ref{sect:ridgeMSE}, which states the existence of a $\lambda$ such that the resulting ridge estimator has a superior MSE over that of the ML estimator. This result made no assumption on (the distribution of) $\bbeta$. In fact, it can be viewed as a statement of the MSE conditional on $\bbeta$. The Bayesian result integrates out the uncertainty - specified by the prior - in $\bbeta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. +The calculation of the Bayes risk above relates the Bayesian and frequentist statements on the MSE of the ridge estimator. For the latter revisit Theorem ref{theo.Theobald2} of Section ref{sect:ridgeMSE}, which states the existence of a $\lambda$ such that the resulting ridge estimator has a superior MSE over that of the ML estimator. This result made no assumption on (the distribution of) $\beta$. In fact, it can be viewed as a statement of the MSE conditional on $\beta$. The Bayesian result integrates out the uncertainty - specified by the prior - in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. -# The Bayesian result states that the ridge regression estimator minimizes the weighted MSE (over a normal prior on the parameter $\bbeta$). It thus integrates out the uncertainty in $\bbeta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. +# The Bayesian result states that the ridge regression estimator minimizes the weighted MSE (over a normal prior on the parameter $\beta$). It thus integrates out the uncertainty in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. ======= Degrees of freedom ======= @@ -1008,7 +1140,7 @@ The degrees of freedom consumed by ridge regression is monotone decreasing in $\ \begin{align*} \lim_{\lambda \rightarrow \infty} \mbox{tr}[ \mathbf{H}(\lambda)] & = 0. \end{align*} -That is, in the limit no information from $\mathbf{X}$ is used. Indeed, $\bbeta$ is forced to equal $\mathbf{0}_{p}$ which is not derived from data. +That is, in the limit no information from $\mathbf{X}$ is used. Indeed, $\beta$ is forced to equal $\mathbf{0}_{p}$ which is not derived from data. @@ -1022,7 +1154,7 @@ For $p=40000$ this requires the inversion of a $40000 \times 40000$ dimensional Revisit the singular value decomposition of $\mathbf{X} = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}$ and write $\mathbf{R}_x = \mathbf{U}_x \mathbf{D}_x$. As both $\mathbf{U}_x$ and $\mathbf{D}_x$ are $(n \times n)$-dimensional matrices, so is $\mathbf{R}_x$. Consequently, $\mathbf{X}$ is now decomposed as $\mathbf{X} = \mathbf{R}_x \mathbf{V}_x^{\top}$. The ridge estimator can be rewritten in terms of $\mathbf{R}_x$ and $\mathbf{V}_x$: \begin{align*} -\hat{\bbeta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} \\ @@ -1067,10 +1199,10 @@ analogously by the trace of the ridge hat matrix. This yields the AIC for the li \begin{align*} \mbox{AIC}(\lambda) & = 2 \, p - 2 \log(\hat{L}) \\ -& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\bbeta}(\lambda), \hat{\sigma}^2(\lambda)]\} +& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\} \\ & = 2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda} -+ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\bbeta}(\lambda)]^2. ++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2. \end{align*} The value of $\lambda$ which minimizes $\mbox{AIC}(\lambda)$ corresponds to the `optimal' balance of model complexity and overfitting. @@ -1088,19 +1220,19 @@ When the repetitive splitting of the data set is done randomly, samples may acc \item[2)] Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set. This yields: \begin{align*} -\hat{\bbeta}_{-i}(\lambda) & = ( \mathbf{X}_{-i, \ast}^{\top} +\hat{\beta}_{-i}(\lambda) & = ( \mathbf{X}_{-i, \ast}^{\top} \mathbf{X}_{-i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{-i, \ast}^{\top} \mathbf{Y}_{-i} \end{align*} and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$. -\item[3)] Evaluate the prediction performance of these models on the test set by $\log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\bbeta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)|$, possibly squared. +\item[3)] Evaluate the prediction performance of these models on the test set by $\log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)|$, possibly squared. \item[4)] Repeat steps 1) to 3) such that each sample plays the role of the test set once. \item[5)] Average the prediction performances of the test sets at each grid point of the penalty parameter: \begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\bbeta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. \end{align*} The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. @@ -1110,7 +1242,7 @@ The procedure is straightforwardly adopted to $k$-fold cross-validation, a diffe In the LOOCV procedure above resampling can be avoided when the prediction performance is measured by Allen's PRESS (Predicted Residual Error Sum of Squares) statistic \citep{Alle1974}. For then, the LOOCV prediction performance can be expressed analytically in terms of the known quantities derived from the design matrix and response (as pointed out but not detailed in \citealt{Golu1979}). Define the optimal penalty parameter to minimize Allen's PRESS statistic: \begin{align*} -\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)]^2. +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. \end{align*} To derive an analytic expression for the right-hand side first rewrite $(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1}$ by means of the Woodbury identity as: @@ -1130,7 +1262,7 @@ rewrite $(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf with $\mathbf{H}_{ii}(\lambda) = \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. Furthermore, $\mathbf{X}_{-i}^{\top} \mathbf{Y}_{-i} = \mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i$. Substitute both in the leave-one-out ridge regression estimator and manipulate: \begin{align*} -\hat{\bbeta}_{- i}(\lambda) & = (\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{- i, \ast}^{\top} \mathbf{Y}_{- i} +\hat{\beta}_{- i}(\lambda) & = (\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{- i, \ast}^{\top} \mathbf{Y}_{- i} \\ & = \{(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} @@ -1148,29 +1280,29 @@ Furthermore, $\mathbf{X}_{-i}^{\top} \mathbf{Y}_{-i} = \mathbf{X}^{\top} \mathbf \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} Y_i \\ -& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i \\ -& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) +& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) \\ & & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{H}_{ii} (\lambda) Y_i \\ -& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \{ [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{H}_{ii}(\lambda) Y_i \} +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \{ [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii}(\lambda) Y_i \} \\ -& = \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ]. +& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ]. \end{align*} The latter enables the reformulation of the prediction error as: \begin{align*} -Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda) & = Y_i - \mathbf{X}_{i, \ast} \{ \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda)] \} +Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda) & = Y_i - \mathbf{X}_{i, \ast} \{ \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda)] \} \\ -& = Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ] +& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] \\ -& = Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) + \mathbf{H}_{ii} (\lambda)[ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}(\lambda) ] +& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii} (\lambda)[ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] \\ -& = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast}^{\top} \hat{\bbeta}(\lambda) ], +& = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast}^{\top} \hat{\beta}(\lambda) ], \end{align*} which in turn results in the re-expression of Allen's PRESS statistic: \begin{align*} -\lambda_{\mbox{{\tiny opt}}} & = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\bbeta}_{-i}(\lambda)]^2 +\lambda_{\mbox{{\tiny opt}}} & = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2 \, \, \, = \, \, \, \arg \min_{\lambda} \tfrac{1}{n} \| \mathbf{B}(\lambda) [\mathbf{I}_{nn} - \mathbf{H}(\lambda)] \mathbf{Y} \|_ F^2, \end{align*} where $\mathbf{B}(\lambda)$ is diagonal with $[\mathbf{B}(\lambda)]_{ii} = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1}$. Hence, the prediction performance for a given $\lambda$ can be assessed directly from the ridge hat matrix and the response vector without the recalculation of the $n$ leave-one-out ridge estimators. Computationally, this is a considerable gain. @@ -1184,9 +1316,9 @@ Simulations are presented that illustrate properties of the ridge estimator not ===== Role of the variance of the covariates} label{ridge:covariateVariances ===== In many applications of high-dimensional data the covariates are standardized prior to the execution of the ridge regression. Before we discuss whether this is appropriate, we first illustrate the effect of ridge penalization on covariates with distinct variances using simulated data. -The simulation involves one response to be (ridge) regressed on fifty covariates. Data (with $n=1000$) for the covariates, denoted $\mathbf{X}$, are drawn from a multivariate normal distribution: $\mathbf{X} \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{\Sigma})$ with $\mathbf{\Sigma}$ diagonal and $(\mathbf{\Sigma})_{jj} = j / 10$. From this the response is generated through $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\bbeta = \mathbf{1}_{50}$ and $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{I}_{50 \times 50})$. +The simulation involves one response to be (ridge) regressed on fifty covariates. Data (with $n=1000$) for the covariates, denoted $\mathbf{X}$, are drawn from a multivariate normal distribution: $\mathbf{X} \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{\Sigma})$ with $\mathbf{\Sigma}$ diagonal and $(\mathbf{\Sigma})_{jj} = j / 10$. From this the response is generated through $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = \mathbf{1}_{50}$ and $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{I}_{50 \times 50})$. -With the simulated data at hand the ridge regression estimates of $\bbeta$ are evaluated for a large grid of the penalty parameter $\lambda$. The resulting ridge regularization paths of the regression coefficients are plotted (Figure ref{fig.effectOfRidge_effectOfVariances}). All paths start ($\lambda=0$) close to one and vanish as $\lambda \rightarrow \infty$. However, ridge regularization paths of regression coefficients corresponding to covariates with a large variance dominate those with a low variance. +With the simulated data at hand the ridge regression estimates of $\beta$ are evaluated for a large grid of the penalty parameter $\lambda$. The resulting ridge regularization paths of the regression coefficients are plotted (Figure ref{fig.effectOfRidge_effectOfVariances}). All paths start ($\lambda=0$) close to one and vanish as $\lambda \rightarrow \infty$. However, ridge regularization paths of regression coefficients corresponding to covariates with a large variance dominate those with a low variance. # original latex figure with scale=0.45, angle=0 @@ -1199,15 +1331,15 @@ FIGURE: [covariatesWithDifferentVariance.eps, width=400 frac=1.0] Top panel: Rid Ridge regression's preference of covariates with a large variance can intuitively be understood as follows. First note that the ridge regression estimator now can be written as: \begin{align*} -\bbeta (\lambda) & = [ \mbox{Var}(\mathbf{X}) + \lambda \mathbf{I}_{50 \times 50}]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) +\beta (\lambda) & = [ \mbox{Var}(\mathbf{X}) + \lambda \mathbf{I}_{50 \times 50}]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) \\ & = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma [ \mbox{Var}(\mathbf{X}) ]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) \\ -& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma \bbeta. +& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma \beta. \end{align*} Plug in the employed parametrization of $\mathbf{\Sigma}$, which gives: \begin{align*} -[\bbeta (\lambda)]_j & = \frac{j}{j + 50 \lambda} \, (\bbeta)_j. +[\beta (\lambda)]_j & = \frac{j}{j + 50 \lambda} \, (\beta)_j. \end{align*} Hence, the larger the covariate's variance (corresponding to the larger $j$), the larger its ridge regression coefficient estimate. Ridge regression thus prefers (among a set of covariates with comparable effect sizes) those with larger variances. @@ -1220,18 +1352,18 @@ Should one thus standardize the covariates prior to ridge regression analysis? Among the considerations to decide on standardization of the covariates, one should also include the fact that ridge estimates prior and posterior to scaling do not simply differ by a factor. To see this assume that the covariates have been centered. Scaling of the covariates amounts to post-multiplication of the design matrix by a $(p \times p)$-dimensional diagonal matrix $\mathbf{A}$ with the reciprocals of the covariates' scale estimates on its diagonal \citep{Sard2008}. Hence, the ridge estimator (for the rescaled data) is then given by: \begin{align*} -\min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \mathbf{A} \bbeta \|_2^2 + \lambda \| \bbeta \|_ 2^2. +\min_{\beta} \| \mathbf{Y} - \mathbf{X} \mathbf{A} \beta \|_2^2 + \lambda \| \beta \|_ 2^2. \end{align*} -Apply the change-of-variable $\ggamma = \mathbf{A} \bbeta$ and obtain: +Apply the change-of-variable $\ggamma = \mathbf{A} \beta$ and obtain: \begin{align*} -\min_{\ggamma} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \lambda \| \mathbf{A}^{-1} \ggamma \|_ 2^2 & = \min_{\bbeta} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \sum_{j=1}^p \lambda [(\mathbf{A})_{jj}]^{-2} \gamma_j^2. +\min_{\ggamma} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \lambda \| \mathbf{A}^{-1} \ggamma \|_ 2^2 & = \min_{\beta} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \sum_{j=1}^p \lambda [(\mathbf{A})_{jj}]^{-2} \gamma_j^2. \end{align*} Effectively, the scaling is equivalent to covariate-wise penalization. The `scaled' ridge estimator may then be derived along the same lines as before in Section ref{sect.constrainedEstimation}: \begin{align*} -\hat{\bbeta}^{\mbox{{\tiny (scaled)}}} (\lambda) & = \mathbf{A}^{-1} \hat{\ggamma} (\lambda) +\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) & = \mathbf{A}^{-1} \hat{\ggamma} (\lambda) \, \, \, = \, \, \, \mathbf{A}^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{A}^{-2})^{-1} \mathbf{X}^{\top} \mathbf{Y}. \end{align*} -In general, this is unequal to the ridge estimator without the rescaling of the columns of the design matrix. Moreover, it should be clear that $\hat{\bbeta}^{\mbox{{\tiny (scaled)}}} (\lambda) \not= \mathbf{A} \hat{\bbeta}(\lambda)$. +In general, this is unequal to the ridge estimator without the rescaling of the columns of the design matrix. Moreover, it should be clear that $\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) \not= \mathbf{A} \hat{\beta}(\lambda)$. ===== Ridge regression and collinearity} label{sect:collinearCovariates ===== Initially, ridge regression was motivated as an ad-hoc fix of (super)-collinear covariates in order to obtain a well-defined estimator. We now study the effect of this ad-hoc fix on the regression coefficient estimates of collinear covariates. In particular, their ridge regularization paths are contrasted to those of `non-collinear' covariates. @@ -1261,7 +1393,7 @@ with \begin{align*} \mathbf{\Sigma}_{kk} & = \frac{k-1}{5} \, \mathbf{1}_{10 \times 10} + \frac{6-k}{5} \, \mathbf{I}_{10 \times 10}. \end{align*} -The data of the response variable $\mathbf{Y}$ are then obtained through: $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, with $\vvarepsilon \sim \mathcal{N}( \mathbf{0}_{n}, \mathbf{I}_{nn})$ and $\bbeta = \mathbf{1}_{50}$. Hence, all covariates contribute equally to the response. Would the columns of $\mathbf{X}$ be orthogonal, little difference in the ridge estimates of the regression coefficients is expected. +The data of the response variable $\mathbf{Y}$ are then obtained through: $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, with $\vvarepsilon \sim \mathcal{N}( \mathbf{0}_{n}, \mathbf{I}_{nn})$ and $\beta = \mathbf{1}_{50}$. Hence, all covariates contribute equally to the response. Would the columns of $\mathbf{X}$ be orthogonal, little difference in the ridge estimates of the regression coefficients is expected. The results of this simulation study with sample size $n=1000$ are presented in Figure ref{fig.ridgeEstimates_correlatedCovariates}. All 50 regularization paths start close to one as $\lambda$ is small and converge to zero as $\lambda \rightarrow \infty$. But the paths of covariates of the same block of the covariance matrix $\mathbf{\Sigma}$ quickly group, with those corresponding to a block with larger off-diagonal elements above those with smaller ones. Thus, ridge regression prefers (i.e. shrinks less) coefficient estimates of strongly positively correlated covariates. @@ -1346,9 +1478,9 @@ For large enough $\lambda$ When $\mbox{Var}(U) \gg \mbox{Var}(V)$ and $\beta_1 \approx \beta_2$, the ridge estimate of $\gamma_v$ vanishes for large $\lambda$. Hence, ridge regression prefers positively covariates with similar effect sizes. -This phenomenon too can be explained geometrically. For the illustration consider ridge estimation with $\lambda=1$ of the linear model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\bbeta = (3, 3)^{\top}$, $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_2, \mathbf{I}_{22})$ and the columns of $\mathbf{X}$ strongly and positively collinear. The level sets of the sum-of-squares loss, $\| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2$, are plotted in the right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}. Recall that the ridge estimate is found by looking for the smallest loss level set that hits the ridge contraint. The sought-for estimate is then the point of intersection between this level set and the constraint, and - for the case at hand - is on the $x=y$-line. This is no different from the case with orthogonal $\mathbf{X}$ columns. Yet their estimates differ, even though the same $\lambda$ is applied. The difference is to due to fact that the radius of the ridge constraint depends on $\lambda$, $\mathbf{X}$ and $\mathbf{Y}$. This is immediate from the fact that the radius of the constraint equals $\| \hat{\bbeta}(\lambda) \|_2^2$ (see Section ref{sect.constrainedEstimation}). To study the effect of $\mathbf{X}$ on the radius, we remove its dependence on $\mathbf{Y}$ by considering its expectation, which is: +This phenomenon too can be explained geometrically. For the illustration consider ridge estimation with $\lambda=1$ of the linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = (3, 3)^{\top}$, $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_2, \mathbf{I}_{22})$ and the columns of $\mathbf{X}$ strongly and positively collinear. The level sets of the sum-of-squares loss, $\| \mathbf{Y} - \mathbf{X} \beta \|_2^2$, are plotted in the right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}. Recall that the ridge estimate is found by looking for the smallest loss level set that hits the ridge contraint. The sought-for estimate is then the point of intersection between this level set and the constraint, and - for the case at hand - is on the $x=y$-line. This is no different from the case with orthogonal $\mathbf{X}$ columns. Yet their estimates differ, even though the same $\lambda$ is applied. The difference is to due to fact that the radius of the ridge constraint depends on $\lambda$, $\mathbf{X}$ and $\mathbf{Y}$. This is immediate from the fact that the radius of the constraint equals $\| \hat{\beta}(\lambda) \|_2^2$ (see Section ref{sect.constrainedEstimation}). To study the effect of $\mathbf{X}$ on the radius, we remove its dependence on $\mathbf{Y}$ by considering its expectation, which is: \begin{align*} -\mathbb{E}[ \| \hat{\bbeta}(\lambda) \|_2^2 ] & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta}]^{\top} \, (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\bbeta} \} \nonumber % \, \, \, = \, \, \, \mathbb{E} ( \hat{\bbeta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\bbeta} ) \nonumber +\mathbb{E}[ \| \hat{\beta}(\lambda) \|_2^2 ] & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta}]^{\top} \, (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \} \nonumber % \, \, \, = \, \, \, \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) \nonumber \\ # & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber # \\ @@ -1358,10 +1490,10 @@ This phenomenon too can be explained geometrically. For the illustration conside (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber \\ & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{X} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \big\} + \bbeta^{\top} \mathbf{X}^{ \top} \mathbf{X} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \, \mathbf{X} \bbeta. % label{form.expectationPenalty} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \big\} + \beta^{\top} \mathbf{X}^{ \top} \mathbf{X} +(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \, \mathbf{X} \beta. % label{form.expectationPenalty} \end{align*} -In the last step we have used $\mathbf{Y} \sim \mathcal{N}( \mathbf{X} \bbeta, \sigma^2 \mathbf{I}_{pp})$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ is $\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}$ (cf. \citealp{Math1992}). The expression for the expectation of the radius of the ridge constraint can now be evaluated for the orthogonal $\mathbf{X}$ and the strongly, positively collinear $\mathbf{X}$. It turns out that the latter is larger than the former. This results in a larger ridge constraint. For the larger ridge constraint there is a smaller level set that hits it first. The point of intersection, still on the $x=y$-line, is now thus closer to $\bbeta$ and further from the origin (cf. right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}). The resulting estimate is thus larger than that from the orthogonal case. +In the last step we have used $\mathbf{Y} \sim \mathcal{N}( \mathbf{X} \beta, \sigma^2 \mathbf{I}_{pp})$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ is $\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}$ (cf. \citealp{Math1992}). The expression for the expectation of the radius of the ridge constraint can now be evaluated for the orthogonal $\mathbf{X}$ and the strongly, positively collinear $\mathbf{X}$. It turns out that the latter is larger than the former. This results in a larger ridge constraint. For the larger ridge constraint there is a smaller level set that hits it first. The point of intersection, still on the $x=y$-line, is now thus closer to $\beta$ and further from the origin (cf. right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}). The resulting estimate is thus larger than that from the orthogonal case. The above needs some attenuation. Among others it depends on: *i)* the number of covariates in each block, *ii)* the size of the effects, i.e. regression coefficients of each covariate, and *iii)* the degree of collinearity. Possibly, there are more factors influencing the behaviour of the ridge estimator presented in this subsection. @@ -1388,7 +1520,7 @@ We here assess whether the MCM7 down-regulation by microRNAs can be observed in \lstinputlisting{prostate_ridgeAnalysis.r} -With this prostate data set at hand we now investigate whether MCM7 is regulated by microRNAs. Hereto we fit a linear regression model regressing the expression levels of MCM7 onto those of the microRNAs. As the number of microRNAs exceeds the number of samples, ordinary least squares fails and we resort to the ridge estimator of the regression coefficients. First, an informed choice of the penalty parameter is made through maximization of the LOOCV log-likelihood, resulting in $\lambda_{\mbox{{\tiny opt}}} = 1812.826$. Having decided on the value of the to-be-employed penalty parameter, the ridge regression estimator can now readily be evaluated. The thus fitted model allows for the evaluation of microRNA-regulation of MCM7. E.g., by the proportion of variation of the MCM7 expression levels by the microRNAs as expressed in coefficient of determination: $R^2 = 0.4492$. Alternatively, but closely related, observed expression levels may be related to the linear predictor of the MCM7 expression levels: $\hat{\mathbf{Y}}(\lambda_{\mbox{{\tiny opt}}}) = \mathbf{X} \hat{\bbeta} (\lambda_{\mbox{{\tiny opt}}})$. The Spearman correlation of response and predictor equals 0.6295. A visual inspection is provided by the left panel of Figure ref{fig.RidgeProstateExample}. Note the difference in scale of the $x$- and $y$-axes. This is due to the fact that the regression coefficients have been estimated in penalized fashion, consequently shrinking estimates of the regression coefficients towards zero leading to small estimates and in turn compressing the range of the linear prediction. The above suggests there is indeed association between the microRNA expression levels and those of MCM7. +With this prostate data set at hand we now investigate whether MCM7 is regulated by microRNAs. Hereto we fit a linear regression model regressing the expression levels of MCM7 onto those of the microRNAs. As the number of microRNAs exceeds the number of samples, ordinary least squares fails and we resort to the ridge estimator of the regression coefficients. First, an informed choice of the penalty parameter is made through maximization of the LOOCV log-likelihood, resulting in $\lambda_{\mbox{{\tiny opt}}} = 1812.826$. Having decided on the value of the to-be-employed penalty parameter, the ridge regression estimator can now readily be evaluated. The thus fitted model allows for the evaluation of microRNA-regulation of MCM7. E.g., by the proportion of variation of the MCM7 expression levels by the microRNAs as expressed in coefficient of determination: $R^2 = 0.4492$. Alternatively, but closely related, observed expression levels may be related to the linear predictor of the MCM7 expression levels: $\hat{\mathbf{Y}}(\lambda_{\mbox{{\tiny opt}}}) = \mathbf{X} \hat{\beta} (\lambda_{\mbox{{\tiny opt}}})$. The Spearman correlation of response and predictor equals 0.6295. A visual inspection is provided by the left panel of Figure ref{fig.RidgeProstateExample}. Note the difference in scale of the $x$- and $y$-axes. This is due to the fact that the regression coefficients have been estimated in penalized fashion, consequently shrinking estimates of the regression coefficients towards zero leading to small estimates and in turn compressing the range of the linear prediction. The above suggests there is indeed association between the microRNA expression levels and those of MCM7. @@ -1402,7 +1534,7 @@ The overall aim of this illustration was to assess whether microRNA-regulation o The implementation of ridge regression in the {\tt penalized}-package offers the possibility to fully obey the dogma on negative regulation of mRNA expression by microRNAs. This requires all regression coefficients to be negative. Incorporation of the requirement into the ridge estimation augments the constrained estimation problem with an additional constraint: \begin{align*} % label{form.constrEstProblemRidge} -\hat{\bbeta}(\lambda) & = \arg \min_{\| \bbeta \|_2^2 \leq c (\lambda) \atop \beta_j \leq 0 \, \mbox{{\tiny for all $j$}}} \| \mathbf{Y} - \mathbf{X} \, \bbeta \|^2_2. +\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c (\lambda) \atop \beta_j \leq 0 \, \mbox{{\tiny for all $j$}}} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. \end{align*} With the additional non-positivity constraint on the parameters, there is no explicit solution for the estimator. The ridge estimate of the regression parameters is then found by numerical optimization using e.g. the Newton-Raphson algorithm or a gradient descent approach. The next listing gives the R-code for ridge estimation with the non-positivity constraint of the linear regression model. @@ -1466,7 +1598,7 @@ Show that the ridge regression estimates can be obtained by ordinary least squar \begin{question} label{question.ridgeResidualsProjection} \mbox{ } \\ -The coefficients $\bbeta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, are estimated by $\hat{\bbeta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. The associated fitted values then given by $\widehat{\mathbf{Y}} = \mathbf{X} \, \hat{\bbeta} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} = \mathbf{H} \mathbf{Y}$, where $\mathbf{H} =\mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}$ referred to as the hat matrix. The matrix $\mathbf{P}$ is a projection matrix and satisfies $\mathbf{H} = \mathbf{H}^ 2$. Hence, linear regression projects the response $\mathbf{Y}$ onto the vector space spanned by the columns of $\mathbf{Y}$. Consequently, the residuals $\hat{\vvarepsilon}$ and $\hat{\mathbf{Y}}$ are orthogonal. Now consider the ridge estimator of the regression coefficients: $\hat{\bbeta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. Let $\hat{\mathbf{Y}}(\lambda) = \mathbf{X} \hat{\bbeta}(\lambda)$ be the vector of associated fitted values. +The coefficients $\beta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, are estimated by $\hat{\beta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. The associated fitted values then given by $\widehat{\mathbf{Y}} = \mathbf{X} \, \hat{\beta} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} = \mathbf{H} \mathbf{Y}$, where $\mathbf{H} =\mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}$ referred to as the hat matrix. The matrix $\mathbf{P}$ is a projection matrix and satisfies $\mathbf{H} = \mathbf{H}^ 2$. Hence, linear regression projects the response $\mathbf{Y}$ onto the vector space spanned by the columns of $\mathbf{Y}$. Consequently, the residuals $\hat{\vvarepsilon}$ and $\hat{\mathbf{Y}}$ are orthogonal. Now consider the ridge estimator of the regression coefficients: $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. Let $\hat{\mathbf{Y}}(\lambda) = \mathbf{X} \hat{\beta}(\lambda)$ be the vector of associated fitted values. \begin{compactitem} \item[*a)*] Show that the matrix $\mathbf{H}(\lambda) = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}$, associated with ridge regression, is not a projection matrix (for any $\lambda > 0$). @@ -1480,7 +1612,7 @@ The coefficients $\bbeta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \begin{question} label{question.MSEofLinearPredictor} \mbox{ } \\ -Recall that there exists $\lambda > 0$ such that $MSE(\hat{\bbeta}) > MSE[\hat{\bbeta}(\lambda)]$. Verify that this carries over to the linear predictor. That is, there exists a $\lambda > 0$ such that$MSE(\widehat{\mathbf{Y}}) = MSE(\mathbf{X} \hat{\bbeta}) > MSE[\mathbf{X}\hat{\bbeta}(\lambda)]$. +Recall that there exists $\lambda > 0$ such that $MSE(\hat{\beta}) > MSE[\hat{\beta}(\lambda)]$. Verify that this carries over to the linear predictor. That is, there exists a $\lambda > 0$ such that$MSE(\widehat{\mathbf{Y}}) = MSE(\mathbf{X} \hat{\beta}) > MSE[\mathbf{X}\hat{\beta}(\lambda)]$. \end{question} \begin{question} \mbox{ } @@ -1506,10 +1638,10 @@ with $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. This model fitted by means o \\ The expression levels of the $j$-the gene are explained by a linear regression model in terms of those of all other genes. Consider the following two ridge regression estimators of the regression parameter of this model, defined as: \begin{align*} -\arg \max_\bbeta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \bbeta_j)^2 + \lambda \| \bbeta_j \|_ 2^2 +\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + \lambda \| \beta_j \|_ 2^2 \quad \mbox{and} \quad -\arg \max_\bbeta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \bbeta_j)^2 + n \lambda \| \bbeta_j \|_ 2^2. +\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + n \lambda \| \beta_j \|_ 2^2. \end{align*} Which do you prefer? Motivate. \end{question} @@ -1520,30 +1652,30 @@ Which do you prefer? Motivate. ========= Generalizing ridge regression ========= label{chap:genRidge} # ======= Generalized ridge regression ======= -The expos\'{e} on ridge regression may be generalized in many ways. Among others different generalized linear models may be considered (confer Section ref{sect.ridgeLogistic}). In this section we stick to the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with the usual assumptions, but fit it in weighted fashion and generalize the common, spherical penalty. The loss function corresponding to this scenario is: +The expos\'{e} on ridge regression may be generalized in many ways. Among others different generalized linear models may be considered (confer Section ref{sect.ridgeLogistic}). In this section we stick to the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with the usual assumptions, but fit it in weighted fashion and generalize the common, spherical penalty. The loss function corresponding to this scenario is: !bt !bt \begin{align} label{form:generalizedRidgeLoss} -(\mathbf{Y} - \mathbf{X} \bbeta)^{\top} \mathbf{W} (\mathbf{Y} - \mathbf{X} \bbeta) + (\bbeta - \bbeta_0)^{\top} \mathbf{\Delta} -(\bbeta - \bbeta_0), +(\mathbf{Y} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Y} - \mathbf{X} \beta) + (\beta - \beta_0)^{\top} \mathbf{\Delta} +(\beta - \beta_0), \end{align} !et !et -which comprises a weighted least squares criterion and a generalized ridge penalty. In this $\mathbf{W}$ is a $(n \times n)$-dimensional, diagonal matrix with $(\mathbf{W})_{ii} \in [0,1]$ representing the weight of the $i$-th observation. The penalty is now a quadratic form with penalty parameter $\mathbf{\Delta}$, a $(p \times p)$-dimensional, positive definite, symmetric matrix. When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, one regains the spherical penalty of `regular ridge regression'. This penalty shrinks each element of the regression parameter $\bbeta$ equally along the unit vectors $\mathbf{e}_j$. Generalizing $\mathbf{\Delta}$ to the class of symmetric, positive definite matrices $\mathcal{S}_{++}$ allows for *i)* different penalization per regression parameter, and *ii)* joint (or correlated) shrinkage among the elements of $\bbeta$. The penalty parameter $\mathbf{\Delta}$ determines the speed and direction of shrinkage. The $p$-dimensional column vector $\bbeta_0$ is a user-specified, non-random target towards which $\bbeta$ is shrunken as the penalty parameter increases. When recasting generalized ridge estimation as a constrained estimation problem, the implications of the penalty may be visualized (Figure ref{fig:generalizedRidgeIllustration}, left panel). The generalized ridge penalty is a quadratic form centered around $\bbeta_0$. In Figure ref{fig:generalizedRidgeIllustration} the parameter constraint clearly is ellipsoidal (and not spherical). Moreover, the center of this ellipsoid is not at zero. +which comprises a weighted least squares criterion and a generalized ridge penalty. In this $\mathbf{W}$ is a $(n \times n)$-dimensional, diagonal matrix with $(\mathbf{W})_{ii} \in [0,1]$ representing the weight of the $i$-th observation. The penalty is now a quadratic form with penalty parameter $\mathbf{\Delta}$, a $(p \times p)$-dimensional, positive definite, symmetric matrix. When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, one regains the spherical penalty of `regular ridge regression'. This penalty shrinks each element of the regression parameter $\beta$ equally along the unit vectors $\mathbf{e}_j$. Generalizing $\mathbf{\Delta}$ to the class of symmetric, positive definite matrices $\mathcal{S}_{++}$ allows for *i)* different penalization per regression parameter, and *ii)* joint (or correlated) shrinkage among the elements of $\beta$. The penalty parameter $\mathbf{\Delta}$ determines the speed and direction of shrinkage. The $p$-dimensional column vector $\beta_0$ is a user-specified, non-random target towards which $\beta$ is shrunken as the penalty parameter increases. When recasting generalized ridge estimation as a constrained estimation problem, the implications of the penalty may be visualized (Figure ref{fig:generalizedRidgeIllustration}, left panel). The generalized ridge penalty is a quadratic form centered around $\beta_0$. In Figure ref{fig:generalizedRidgeIllustration} the parameter constraint clearly is ellipsoidal (and not spherical). Moreover, the center of this ellipsoid is not at zero. # original latex figure with scale=0.40, angle=0 FIGURE: [generalizedRidgeIllustration_penalty.eps, width=400 frac=1.0] Left panel: the contours of the likelihood (grey solid ellipsoids) and the parameter constraint implied by the generalized penalty (black dashed ellipsoid. Right panel: generalized (fat coloured lines) and `regular' (thin coloured lines) regularization paths of four regression coefficients. The dotted grey (straight) lines indicated the targets towards the generalized ridge penalty shrinks regression coefficient estimates.} label{fig:generalizedRidgeIllustration -The addition of the generalized ridge penalty to the sum-of-squares ensures the existence of a unique regression estimator in the face of super-collinearity. The generalized penalty is a non-degenerated quadratic form in $\bbeta$ due to the positive definiteness of the matrix $\mathbf{\Delta}$. As it is non-degenerate, it is strictly convex. Consequently, the generalized ridge regression loss function (ref{form:generalizedRidgeLoss}), being the sum of a convex and strictly convex function, is also strictly convex. This warrants the existence of a unique global minimum and, thereby, a unique estimator. +The addition of the generalized ridge penalty to the sum-of-squares ensures the existence of a unique regression estimator in the face of super-collinearity. The generalized penalty is a non-degenerated quadratic form in $\beta$ due to the positive definiteness of the matrix $\mathbf{\Delta}$. As it is non-degenerate, it is strictly convex. Consequently, the generalized ridge regression loss function (ref{form:generalizedRidgeLoss}), being the sum of a convex and strictly convex function, is also strictly convex. This warrants the existence of a unique global minimum and, thereby, a unique estimator. -Like for the `regular' ridge loss function (ref{form.ridgeLossFunction}), there is an explicit expression for the optimum of the generalized ridge loss function (ref{form:generalizedRidgeLoss}). To see this, obtain the estimating equation of $\bbeta$ through equating its derivative with respect to $\bbeta$ to zero: +Like for the `regular' ridge loss function (ref{form.ridgeLossFunction}), there is an explicit expression for the optimum of the generalized ridge loss function (ref{form:generalizedRidgeLoss}). To see this, obtain the estimating equation of $\beta$ through equating its derivative with respect to $\beta$ to zero: \begin{align*} -2 \mathbf{X}^{\top} \mathbf{W} \mathbf{Y} - 2 \mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta - 2 \mathbf{\Delta} \bbeta + 2 \mathbf{\Delta} \bbeta_0 & = \mathbf{0}_{p}. +2 \mathbf{X}^{\top} \mathbf{W} \mathbf{Y} - 2 \mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta - 2 \mathbf{\Delta} \beta + 2 \mathbf{\Delta} \beta_0 & = \mathbf{0}_{p}. \end{align*} This is solved by: @@ -1551,29 +1683,29 @@ This is solved by: !bt \begin{align} -\hat{\bbeta}(\mathbf{\Delta}) & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{Y} + \mathbf{\Delta} \bbeta_0). +\hat{\beta}(\mathbf{\Delta}) & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{Y} + \mathbf{\Delta} \beta_0). \end{align} !et !et -Clearly, this reduces to the `regular' ridge estimator by setting $\mathbf{W} = \mathbf{I}_{nn}$, $\bbeta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$. The effects of the generalized ridge penalty on the estimates can be seen in the regularization paths of the estimates. Figure ref{fig:generalizedRidgeIllustration} (right panel) contains an example of the regularization paths for coefficients of a linear regression model with four explanatory variables. Most striking is the limiting behaviour of the estimates of $\beta_3$ and $\beta_4$ for large values of the penalty parameter $\lambda$: they convergence to a non-zero value (as was specified by a nonzero $\beta_0$). More subtle is the (temporary) convergence of the regularization paths of the estimates of $\beta_2$ and $\beta_3$. That of $\beta_2$ is pulled away from zero (its true value and approximately its unpenalized estimate) towards the estimate of $\beta_3$. In the regularization path of $\beta_3$ this can be observed in a delayed convergence to its nonzero target value (for comparison consider that of $\beta_4$). For reference the corresponding regularization paths of the `regular' ridge estimates (as thinner lines of the same colour) are included in Figure ref{fig:generalizedRidgeIllustration}. +Clearly, this reduces to the `regular' ridge estimator by setting $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$. The effects of the generalized ridge penalty on the estimates can be seen in the regularization paths of the estimates. Figure ref{fig:generalizedRidgeIllustration} (right panel) contains an example of the regularization paths for coefficients of a linear regression model with four explanatory variables. Most striking is the limiting behaviour of the estimates of $\beta_3$ and $\beta_4$ for large values of the penalty parameter $\lambda$: they convergence to a non-zero value (as was specified by a nonzero $\beta_0$). More subtle is the (temporary) convergence of the regularization paths of the estimates of $\beta_2$ and $\beta_3$. That of $\beta_2$ is pulled away from zero (its true value and approximately its unpenalized estimate) towards the estimate of $\beta_3$. In the regularization path of $\beta_3$ this can be observed in a delayed convergence to its nonzero target value (for comparison consider that of $\beta_4$). For reference the corresponding regularization paths of the `regular' ridge estimates (as thinner lines of the same colour) are included in Figure ref{fig:generalizedRidgeIllustration}. \begin{example} *Fused ridge estimation* label{example.gRidge2orthonormalDesign} \\ -An example of a generalized ridge penalty is the *fused ridge penalty*. Consider the standard linear model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$. The fused ridge estimator of $\bbeta$ then minimizes: +An example of a generalized ridge penalty is the *fused ridge penalty*. Consider the standard linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$. The fused ridge estimator of $\beta$ then minimizes: !bt !bt \begin{align} label{form:fusedRidgeLoss} -\| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2. +\| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2. \end{align} !et !et The penalty in the loss function above can be written as a generalized ridge penalty: \begin{align*} -\lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2 & = \bbeta^{\top} +\lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2 & = \beta^{\top} \left( \begin{array}{rrrrrr} \lambda & -\lambda & 0 & \ldots & \ldots & 0 @@ -1603,36 +1735,36 @@ The penalty in the loss function above can be written as a generalized ridge pen \\ 0 & \ldots & \ldots & 0 & -\lambda & \lambda \end{array}\right) -\bbeta. +\beta. \end{align*} -The matrix $\mathbf{\Delta}$ employed above is semi-positive definite and therefore the loss function (ref{form:fusedRidgeLoss}) is not strictly convex. Hence, often a regular ridge penalty $\| \bbeta \|_2^2$ is added (with its own penalty parameter). +The matrix $\mathbf{\Delta}$ employed above is semi-positive definite and therefore the loss function (ref{form:fusedRidgeLoss}) is not strictly convex. Hence, often a regular ridge penalty $\| \beta \|_2^2$ is added (with its own penalty parameter). -To illustrate the effect of the fused ridge penalty on the estimation of the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$, let $\beta_j = \phi_{0,1}(z_j)$ with $z_j =-30 + \tfrac{6}{50} j$ for $j=1, \ldots, 500$. Sample the elements of the design matrix $\mathbf{X}$ and those of the error vector $\vvarepsilon$ from the standard normal distribution, then form the response $\mathbf{Y}$ from the linear model. The regression parameter is estimated through fused ridge loss minimization with $\lambda=1000$. The estimate is shown in Figure ref{fig:fusedRidgeIllustration} (red line). For reference the figure includes the true $\bbeta$ (black line) and the `regular ridge' estimate with $\lambda=1$ (blue line). Clearly, the fused ridge estimate yields a nice smooth vector of $\bbeta$ estimates +To illustrate the effect of the fused ridge penalty on the estimation of the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, let $\beta_j = \phi_{0,1}(z_j)$ with $z_j =-30 + \tfrac{6}{50} j$ for $j=1, \ldots, 500$. Sample the elements of the design matrix $\mathbf{X}$ and those of the error vector $\vvarepsilon$ from the standard normal distribution, then form the response $\mathbf{Y}$ from the linear model. The regression parameter is estimated through fused ridge loss minimization with $\lambda=1000$. The estimate is shown in Figure ref{fig:fusedRidgeIllustration} (red line). For reference the figure includes the true $\beta$ (black line) and the `regular ridge' estimate with $\lambda=1$ (blue line). Clearly, the fused ridge estimate yields a nice smooth vector of $\beta$ estimates \end{example} # original latex figure with scale=0.40, angle=0 -FIGURE: [fusedRidgeIllustration_simulation.eps, width=400 frac=1.0] Left panel: illustration of the fused ridge estimator (in simulation). The true parameter $\bbeta$ and its ridge and fused ridge estimates against their spatial order. Right panel: Ridge vs. fused ridge estimates of the DNA copy effect on KRAS expression levels. The dashed, grey vertical bar indicates the location of the KRAS gene. } label{fig:fusedRidgeIllustration +FIGURE: [fusedRidgeIllustration_simulation.eps, width=400 frac=1.0] Left panel: illustration of the fused ridge estimator (in simulation). The true parameter $\beta$ and its ridge and fused ridge estimates against their spatial order. Right panel: Ridge vs. fused ridge estimates of the DNA copy effect on KRAS expression levels. The dashed, grey vertical bar indicates the location of the KRAS gene. } label{fig:fusedRidgeIllustration ======= Moments ======= -The expectation and variance of $\hat{\bbeta}(\mathbf{\Delta})$ are obtained through application of the same matrix algebra and expectation and covariance rules used in the derivation of their counterparts of the `regular' ridge regression estimator. This leads to: +The expectation and variance of $\hat{\beta}(\mathbf{\Delta})$ are obtained through application of the same matrix algebra and expectation and covariance rules used in the derivation of their counterparts of the `regular' ridge regression estimator. This leads to: \begin{align*} -\mathbb{E}[\hat{\bbeta}(\mathbf{\Delta})] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta + \mathbf{\Delta} \bbeta_0), +\mathbb{E}[\hat{\beta}(\mathbf{\Delta})] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \mathbf{\Delta} \beta_0), \\ -\mbox{Var}[\hat{\bbeta}(\mathbf{\Delta})] & = \sigma^2 (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}. +\mbox{Var}[\hat{\beta}(\mathbf{\Delta})] & = \sigma^2 (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}. \end{align*} -From these expressions similar limiting behaviour as for the `regular' ridge regression case can be deduced. To this end let $\mathbf{V}_{\delta} \mathbf{D}_{\delta} \mathbf{V}_{\delta}^{\top}$ be the eigendecomposition of $\mathbf{\Delta}$ and $d_{\delta,j} = (\mathbf{D}_{\delta})_{jj}$. Furthermore, define (with some abuse of notation) $\lim_{\mathbf{\Delta} \rightarrow \infty}$ as the limit of all $d_{\delta,j}$ simultaneously tending to infinity. Then, $\lim_{\mathbf{\Delta} \rightarrow \infty} \mathbb{E}[\hat{\bbeta}(\mathbf{\Delta})] = \bbeta_0$ and $\lim_{\mathbf{\Delta} \rightarrow \infty} \mbox{Var}[\hat{\bbeta}(\mathbf{\Delta})] = \mathbf{0}_{pp}$. +From these expressions similar limiting behaviour as for the `regular' ridge regression case can be deduced. To this end let $\mathbf{V}_{\delta} \mathbf{D}_{\delta} \mathbf{V}_{\delta}^{\top}$ be the eigendecomposition of $\mathbf{\Delta}$ and $d_{\delta,j} = (\mathbf{D}_{\delta})_{jj}$. Furthermore, define (with some abuse of notation) $\lim_{\mathbf{\Delta} \rightarrow \infty}$ as the limit of all $d_{\delta,j}$ simultaneously tending to infinity. Then, $\lim_{\mathbf{\Delta} \rightarrow \infty} \mathbb{E}[\hat{\beta}(\mathbf{\Delta})] = \beta_0$ and $\lim_{\mathbf{\Delta} \rightarrow \infty} \mbox{Var}[\hat{\beta}(\mathbf{\Delta})] = \mathbf{0}_{pp}$. \begin{example} \mbox{ } label{example.gRidge2orthonormalDesign} \\ -Let $\mathbf{X}$ be an $(n \times p)$-dimensional, orthonormal design matrix. Contrast the regular and generalized ridge regression estimator, the latter with $\mathbf{W} = \mathbf{I}_{pp}$, $\bbeta_0 = \mathbf{0}_p$ and $\mathbf{\Delta} = \lambda \mathbf{R}$ where $\mathbf{R} = (1-\rho) \mathbf{I}_{pp} + \rho \mathbf{1}_{pp}$ for $\rho \in (-(p-1)^{-1}, 1)$. For $\rho =0$ the two estimators coincide. The variance of the generalized ridge regression estimator then is $\mbox{Var}[ \hat{\bbeta}(\mathbf{\Delta})] = (\mathbf{I}_{pp} + \mathbf{\Delta})^{-2}$. The efficiency of this estimator, measured by its generalized variance, is: +Let $\mathbf{X}$ be an $(n \times p)$-dimensional, orthonormal design matrix. Contrast the regular and generalized ridge regression estimator, the latter with $\mathbf{W} = \mathbf{I}_{pp}$, $\beta_0 = \mathbf{0}_p$ and $\mathbf{\Delta} = \lambda \mathbf{R}$ where $\mathbf{R} = (1-\rho) \mathbf{I}_{pp} + \rho \mathbf{1}_{pp}$ for $\rho \in (-(p-1)^{-1}, 1)$. For $\rho =0$ the two estimators coincide. The variance of the generalized ridge regression estimator then is $\mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] = (\mathbf{I}_{pp} + \mathbf{\Delta})^{-2}$. The efficiency of this estimator, measured by its generalized variance, is: \begin{align*} -\det \{ \mbox{Var}[ \hat{\bbeta}(\mathbf{\Delta})] \} & = \{ [1 + \lambda + (p-1) \rho] (1 + \lambda-\rho)^{p-1} \}^{-2}. +\det \{ \mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] \} & = \{ [1 + \lambda + (p-1) \rho] (1 + \lambda-\rho)^{p-1} \}^{-2}. \end{align*} This efficiency attains its minimum at $\rho = 0$. In the present case, the regular ridge regression estimator is thus more efficient than its generalized counterpart. \end{example} @@ -1640,14 +1772,14 @@ This efficiency attains its minimum at $\rho = 0$. In the present case, the regu \begin{example} *(MSE with perfect target)* \\ -Set $\bbeta_0 = \bbeta$, i.e. the target is equal to the true value of the regression parameter. Then: +Set $\beta_0 = \beta$, i.e. the target is equal to the true value of the regression parameter. Then: \begin{align*} -\mathbb{E}[\hat{\bbeta}(\Delta)] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \Delta)^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \bbeta + \Delta \bbeta) \, \, \, = \, \, \, \bbeta. +\mathbb{E}[\hat{\beta}(\Delta)] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \Delta)^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \Delta \beta) \, \, \, = \, \, \, \beta. \end{align*} Hence, irrespective of the choice of $\Delta$, the generalized ridge is then unbiased. Thus: \begin{align*} -\mbox{MSE}[\hat{\bbeta}(\Delta)] & = \mbox{tr} \{ -\mbox{Var}[\hat{\bbeta}(\Delta)] \} +\mbox{MSE}[\hat{\beta}(\Delta)] & = \mbox{tr} \{ +\mbox{Var}[\hat{\beta}(\Delta)] \} \\ & = \mbox{tr}[ \sigma^{2} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}] \\ @@ -1658,19 +1790,19 @@ When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, this MSE is smaller than that ======= The Bayesian connection} label{sect:genRidgeBayes ======= -This generalized ridge estimator can, like the regular ridge estimator, be viewed as a Bayesian estimator. It requires to replace the conjugate prior on $\bbeta$ by a more general normal law, $\bbeta \sim \mathcal{N}(\bbeta_0, \sigma^2 \mathbf{\Delta}^{-1})$, but retains the gamma prior on $\sigma^2$. The joint posterior distribution of $\bbeta$ and $\sigma^2$ is then obtained analogously (the details are left as Exercise ref{question.generalizedRidgeAndBayes}) to Section ref{sect:Bayes}: +This generalized ridge estimator can, like the regular ridge estimator, be viewed as a Bayesian estimator. It requires to replace the conjugate prior on $\beta$ by a more general normal law, $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$, but retains the gamma prior on $\sigma^2$. The joint posterior distribution of $\beta$ and $\sigma^2$ is then obtained analogously (the details are left as Exercise ref{question.generalizedRidgeAndBayes}) to Section ref{sect:Bayes}: \begin{align*} -f_{\bbeta, \sigma^2} (\bbeta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \bbeta, \sigma^2) \, f_{\beta}(\bbeta | \sigma^2) \, f_{\sigma}(\sigma^2) +f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) \\ -& \propto & g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, +& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) \end{align*} with \begin{align*} -g_{\bbeta} (\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) -& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \bbeta - \hat{\bbeta}(\mathbf{\Delta}) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta}) \big[ \bbeta - \hat{\bbeta}(\mathbf{\Delta}) \big] \Big\}. +g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) +& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta}) \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big] \Big\}. \end{align*} -This implies $\mathbb{E}(\bbeta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\bbeta}(\mathbf{\Delta})$. Hence, the generalized ridge regression estimator too can be viewed as the Bayesian posterior mean estimator of $\bbeta$ when imposing a multivariate Gaussian prior on the regression parameter. +This implies $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\mathbf{\Delta})$. Hence, the generalized ridge regression estimator too can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a multivariate Gaussian prior on the regression parameter. @@ -1691,9 +1823,9 @@ The sample selection demands justification. It yields a clear illustrate-able di ======= Generalized ridge regression ======= -What is generally referred to as `generalized ridge regression' (cf. \citealp{Hoer1970, Hemm1975}) is the particular case of loss function (ref{form:generalizedRidgeLoss}) in which $\mathbf{W} = \mathbf{I}_{nn}$, $\bbeta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \mathbf{V}_{x} \mathbf{\Lambda} \mathbf{V}_x^{\top}$, where $\mathbf{V}_x$ is obtained from the singular value decomposition of $\mathbf{X}$ (i.e., $\mathbf{X} = \mathbf{U}_{x} \mathbf{D}_x \mathbf{V}_x^{\top}$ with its constituents endowed with the usual interpretation) and $\mathbf{\Lambda}$ a positive definite diagonal matrix. This gives the estimator: +What is generally referred to as `generalized ridge regression' (cf. \citealp{Hoer1970, Hemm1975}) is the particular case of loss function (ref{form:generalizedRidgeLoss}) in which $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \mathbf{V}_{x} \mathbf{\Lambda} \mathbf{V}_x^{\top}$, where $\mathbf{V}_x$ is obtained from the singular value decomposition of $\mathbf{X}$ (i.e., $\mathbf{X} = \mathbf{U}_{x} \mathbf{D}_x \mathbf{V}_x^{\top}$ with its constituents endowed with the usual interpretation) and $\mathbf{\Lambda}$ a positive definite diagonal matrix. This gives the estimator: \begin{align*} -\hat{\bbeta}(\mathbf{\Lambda}) & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y} +\hat{\beta}(\mathbf{\Lambda}) & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y} \\ & = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x \mathbf{Y} \\ @@ -1701,19 +1833,19 @@ What is generally referred to as `generalized ridge regression' (cf. \citealp{Ho \end{align*} From this last expression it becomes clear how this estimator generalizes the `regular ridge estimator'. The latter shrinks all eigenvalues, irrespectively of their size, in the same manner through a common penalty parameter. The `generalized ridge estimator', through differing penalty parameters (i.e. the diagonal elements of $\mathbf{\Lambda}$), shrinks them individually. -The generalized ridge estimator coincides with the Bayesian linear regression estimator with the normal prior $\mathcal{N}[\mathbf{0}_p, (\mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1}]$ on the regression parameter $\bbeta$ (and preserving the inverse gamma prior on the error variance). Assume $\mathbf{X}$ to be of full column rank and choose $\mathbf{\Lambda} = g^{-1} \mathbf{D}_x^2$ with $g$ a positive scalar. The prior on $\bbeta$ then - assuming $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits - reduces to Zellner's $g$-prior: $\bbeta \sim \mathcal{N}[\mathbf{0}_p, g (\mathbf{X}^{\top} \mathbf{X})^{-1}]$ \citep{Zell1986}. The corresponding estimator of the regression coefficient is: $\hat{\bbeta}(g) = g (1+g)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$, which is proportional to the unpenalized ordinary least squares estimator of $\bbeta$. +The generalized ridge estimator coincides with the Bayesian linear regression estimator with the normal prior $\mathcal{N}[\mathbf{0}_p, (\mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1}]$ on the regression parameter $\beta$ (and preserving the inverse gamma prior on the error variance). Assume $\mathbf{X}$ to be of full column rank and choose $\mathbf{\Lambda} = g^{-1} \mathbf{D}_x^2$ with $g$ a positive scalar. The prior on $\beta$ then - assuming $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits - reduces to Zellner's $g$-prior: $\beta \sim \mathcal{N}[\mathbf{0}_p, g (\mathbf{X}^{\top} \mathbf{X})^{-1}]$ \citep{Zell1986}. The corresponding estimator of the regression coefficient is: $\hat{\beta}(g) = g (1+g)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$, which is proportional to the unpenalized ordinary least squares estimator of $\beta$. For convenience of notation in the analysis of the generalized ridge estimator the linear regression model is usually rewritten as: \begin{align*} -\mathbf{Y} & = \mathbf{X} \bbeta + \vvarepsilon \, \, \, = \, \, \, -\mathbf{X} \mathbf{V}_x \mathbf{V}_x^{\top} \bbeta + \vvarepsilon \, \, \, = \, \, \, \tilde{\mathbf{X}} \aalpha + \vvarepsilon, +\mathbf{Y} & = \mathbf{X} \beta + \vvarepsilon \, \, \, = \, \, \, +\mathbf{X} \mathbf{V}_x \mathbf{V}_x^{\top} \beta + \vvarepsilon \, \, \, = \, \, \, \tilde{\mathbf{X}} \aalpha + \vvarepsilon, \end{align*} with $\tilde{\mathbf{X}} = \mathbf{X} \mathbf{V}_x = \mathbf{U}_x -\mathbf{D}_x$ (and thus $\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} = \mathbf{D}_x^ 2$) and $\aalpha = \mathbf{V}_x^{\top} \bbeta$ with loss function $(\mathbf{Y} - \tilde{\mathbf{X}} \aalpha)^{\top} (\mathbf{Y} - \tilde{\mathbf{X}} \aalpha) + \aalpha^{\top} \mathbf{\Lambda} \aalpha$. In the notation above the generalized ridge estimator is then: +\mathbf{D}_x$ (and thus $\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} = \mathbf{D}_x^ 2$) and $\aalpha = \mathbf{V}_x^{\top} \beta$ with loss function $(\mathbf{Y} - \tilde{\mathbf{X}} \aalpha)^{\top} (\mathbf{Y} - \tilde{\mathbf{X}} \aalpha) + \aalpha^{\top} \mathbf{\Lambda} \aalpha$. In the notation above the generalized ridge estimator is then: \begin{align*} \hat{\aalpha}(\mathbf{\Lambda}) & = (\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} + \mathbf{\Lambda})^ {-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y} = (\mathbf{D}_x^ 2 + \mathbf{\Lambda})^{-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y}, \end{align*} -from which one obtains $\hat{\bbeta}(\mathbf{\Lambda}) = \mathbf{V}_x \hat{\aalpha}(\mathbf{\Lambda})$. Using $\mathbb{E}[\hat{\aalpha}(\mathbf{\Lambda})] = (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 \aalpha$ and $\mbox{Var}[\hat{\aalpha}(\mathbf{\Lambda})] = \sigma^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1}$, the MSE for the generalized ridge estimator can be written as: +from which one obtains $\hat{\beta}(\mathbf{\Lambda}) = \mathbf{V}_x \hat{\aalpha}(\mathbf{\Lambda})$. Using $\mathbb{E}[\hat{\aalpha}(\mathbf{\Lambda})] = (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 \aalpha$ and $\mbox{Var}[\hat{\aalpha}(\mathbf{\Lambda})] = \sigma^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1}$, the MSE for the generalized ridge estimator can be written as: \begin{align*} \mbox{MSE}[\hat{\aalpha}(\mathbf{\Lambda})] & = \sum_{j=1}^p ( \sigma^2 d_{x,j}^2 + \alpha_j^2 \lambda_{j}^2 ) (d_{x,j}^2 + \lambda_{j} )^{-2}, \end{align*} @@ -1722,7 +1854,7 @@ the estimates of $\aalpha$ and $\sigma^2$. An analytic expression of the limit o ======= Conclusion ======= -To conclude: a note of caution. The generalized ridge penalty is extremely flexible. It can incorporate any prior knowledge on the parameter values (through specification of $\bbeta_0$) and the relations among these parameters (via $\mathbf{\Delta}$). While a pilot study or literature may provide a suggestion for $\bbeta_0$, it is less obvious how to choose an informative $\mathbf{\Delta}$ (although a spatial structure is a nice exception). In general, exact knowledge on the parameters should not be incorporated implicitly via the penalty (read: prior) but preferably be used explicitly in the model - the likelihood - itself. Though this may be the viewpoint of a prudent frequentist and a subjective Bayesian might disagree. +To conclude: a note of caution. The generalized ridge penalty is extremely flexible. It can incorporate any prior knowledge on the parameter values (through specification of $\beta_0$) and the relations among these parameters (via $\mathbf{\Delta}$). While a pilot study or literature may provide a suggestion for $\beta_0$, it is less obvious how to choose an informative $\mathbf{\Delta}$ (although a spatial structure is a nice exception). In general, exact knowledge on the parameters should not be incorporated implicitly via the penalty (read: prior) but preferably be used explicitly in the model - the likelihood - itself. Though this may be the viewpoint of a prudent frequentist and a subjective Bayesian might disagree. ======= Exercises ======= @@ -1751,9 +1883,9 @@ with penalty parameter $\delta \in \mathbb{R}_{> 0}$. The data scientist is surp \begin{question} label{question.generalizedRidgeAndBayes} \mbox{ } \\ -Consider the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsilon$ with $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \mathbf{I}_{pp})$. Assume $\bbeta \sim \mathcal{N}(\bbeta_0, \sigma^2 \mathbf{\Delta}^{-1})$ with $\bbeta_0 \in \mathbb{R}^p$ and $\mathbf{\Delta} \succ 0$ and a gamma prior on the error variance. Verify (i.e., work out the details of the derivation) that the posterior mean coincides with the generalized ridge estimator defined as: +Consider the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \mathbf{I}_{pp})$. Assume $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$ with $\beta_0 \in \mathbb{R}^p$ and $\mathbf{\Delta} \succ 0$ and a gamma prior on the error variance. Verify (i.e., work out the details of the derivation) that the posterior mean coincides with the generalized ridge estimator defined as: \begin{align*} -\hat{\bbeta} & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{Y} + \mathbf{\Delta} \bbeta_0). +\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{Y} + \mathbf{\Delta} \beta_0). \end{align*} \end{question} @@ -1761,7 +1893,7 @@ Consider the linear regression model $\mathbf{Y} = \mathbf{X} \bbeta + \vvarepsi \begin{question} label{question.ridgeAR1penalty} \mbox{ } \\ -The ridge penalty may be interpreted as a multivariate normal prior on the regression coefficients: $\bbeta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \mathbf{I}_{pp})$. Different priors may be considered. In case the covariates are spatially related in some sense (e.g. genomically), it may of interest to assume a first-order autoregressive prior: $\bbeta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \SSigma_A)$, in which $\SSigma_A$ is a $p \times p$-correlation matrix with $(\SSigma_A)_{j_1, j_2} = \rho^{ | j_1 - j_2 | } $ for some correlation coefficient $\rho \in [0, 1)$. Hence, +The ridge penalty may be interpreted as a multivariate normal prior on the regression coefficients: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \mathbf{I}_{pp})$. Different priors may be considered. In case the covariates are spatially related in some sense (e.g. genomically), it may of interest to assume a first-order autoregressive prior: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \SSigma_A)$, in which $\SSigma_A$ is a $p \times p$-correlation matrix with $(\SSigma_A)_{j_1, j_2} = \rho^{ | j_1 - j_2 | } $ for some correlation coefficient $\rho \in [0, 1)$. Hence, \begin{align*} \SSigma_A \, \, \, = \, \, \, \left( @@ -1780,13 +1912,13 @@ The ridge penalty may be interpreted as a multivariate normal prior on the regre \begin{compactitem} \item[*a)*] The penalized loss function associated with this AR(1) prior is: \begin{align*} -\mathcal{L}(\bbeta; \lambda, \SSigma_A) & = \| \mathbf{Y} - \mathbf{X} \bbeta \|_2^2 + \lambda \bbeta^{\top} \SSigma_A^{-1} \bbeta. +\mathcal{L}(\beta; \lambda, \SSigma_A) & = \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \beta^{\top} \SSigma_A^{-1} \beta. \end{align*} Find the minimizer of this loss function. \item[*b)*] What is the effect of $\rho$ on the ridge estimates? Contrast this to the effect of $\lambda$. Illustrate this on (simulated) data. -\item[*c)*] Instead of an AR(1) prior assume a prior with a uniform correlation between the elements of $\bbeta$. That is, replace $\SSigma_A$ by $\SSigma_U$, given by: +\item[*c)*] Instead of an AR(1) prior assume a prior with a uniform correlation between the elements of $\beta$. That is, replace $\SSigma_A$ by $\SSigma_U$, given by: \begin{align*} \SSigma_U \, \, \, = \, \, \, \left( \begin{array}{ccc} @@ -1815,38 +1947,38 @@ Ridge penalized estimation is not limited to the standard linear regression mode The logistic regression model explains a binary response variable (through some transformation) by a linear combination of a set of covariates (as in the linear regression model). Denote this response of the $i$-th sample by $Y_i$ with $Y_i \in \{ 0, 1 \}$ for $i=1, \ldots, n$. The $n$-dimensional column vector $\mathbf{Y}$ stacks these $n$ responses. For each sample information on the $p$ explanatory variables $X_{i,1}, \ldots, X_{i,p}$ is available. In row vector form this information is denoted $\mathbf{X}_{i,\ast} = (X_{i,1}, \ldots, X_{i,p})$. Or, in short, $\mathbf{X}_i$ when the context tolerates no confusion. The $(n \times p)$-dimensional matrix $\mathbf{X}$ aggregates these vectors, such that $\mathbf{X}_i$ is the $i$-th row vector. -The binary response cannot be modelled as in the linear model like $Y_i = \mathbf{X}_i \bbeta + \varepsilon_i$. With each element of $\mathbf{X}_i$ and $\bbeta$ assuming a value in $\mathbb{R}$, the linear predictor is not restricted to the domain of the response. This is resolved by modeling $p_i = P(Y_i = 1)$ instead. Still the linear predictor may exceed the domain of the response ($p_i \in [0,1]$). Hence, a transformation is applied to map $p_i$ to $\mathbb{R}$, the range of the linear predictor. +The binary response cannot be modelled as in the linear model like $Y_i = \mathbf{X}_i \beta + \varepsilon_i$. With each element of $\mathbf{X}_i$ and $\beta$ assuming a value in $\mathbb{R}$, the linear predictor is not restricted to the domain of the response. This is resolved by modeling $p_i = P(Y_i = 1)$ instead. Still the linear predictor may exceed the domain of the response ($p_i \in [0,1]$). Hence, a transformation is applied to map $p_i$ to $\mathbb{R}$, the range of the linear predictor. # original latex figure with scale=0.40, angle=0 FIGURE: [logistic_effectOfBeta0.eps, width=400 frac=1.0] Top row, left panel: the response curve for various choices of the intercept $\beta_0$. Top row, right panel: the response curve for various choices of the regression coefficent $\beta_1$. Bottom row, left panel: the responce curve for various choices of the link function. Bottom panel, right panel: observations, fits and their deviations. } label{fig.logisticRidge_modelAndFitIllustration -The transformation associated with the logistic regression model is the logarithm of the odds, with the odds defined as: $\mbox{*odds*} = P(\mbox{succes}) / P(\mbox{failure}) = p_i/ (1-p_i)$. The logistic model is then written as $\log[ p_i / (1-p_i)] = \mathbf{X}_i \bbeta$ for all $i$. Or, expressed in terms of the response: +The transformation associated with the logistic regression model is the logarithm of the odds, with the odds defined as: $\mbox{*odds*} = P(\mbox{succes}) / P(\mbox{failure}) = p_i/ (1-p_i)$. The logistic model is then written as $\log[ p_i / (1-p_i)] = \mathbf{X}_i \beta$ for all $i$. Or, expressed in terms of the response: \begin{align*} -p_i & = P(Y_i = 1) \, \, \, = \, \, \, g^{-1}(\mathbf{X}_i; \bbeta) \, \, \, = \, \, \,\frac{\exp(\mathbf{X}_i \bbeta) }{1 + \exp(\mathbf{X}_i \bbeta) }. +p_i & = P(Y_i = 1) \, \, \, = \, \, \, g^{-1}(\mathbf{X}_i; \beta) \, \, \, = \, \, \,\frac{\exp(\mathbf{X}_i \beta) }{1 + \exp(\mathbf{X}_i \beta) }. \end{align*} -The function $g(\cdot; \cdot)$ is called the *link function*. It links the response to the explanatory variables. The one above is called the logistic link function. Or short, logit. The regression parameters have tangible interpretations. When the first covariate represents the intercept, i.e. $X_{i,j} = 1$ for all $i$, then $\beta_1$ determines where the link function equals a half when all other covariates fail to contribute to the linear predictor (i.e. where $P (Y_i = 1 \, | \, \mathbf{X}_{i}) = 0.5$ when $\mathbf{X}_{i} \bbeta = \beta_1$). This is illustrated in the top-left panel of Figure ref{fig.logisticRidge_modelAndFitIllustration} for various choices of the intercept. On the other hand, the regression parameters are directly related to the odds ratio: $\mbox{*odds ratio*} = \mbox{odds}(X_{i,j}+1) / \mbox{odds}(X_{i,j}) = \exp(\beta_j)$. Hence, the effect of a unit change in the $j$-th covariate on the odds ratio is $\exp(\beta_j)$ (see Figure ref{fig.logisticRidge_modelAndFitIllustration}, top-right panel). Other link functions (depicted in Figure ref{fig.logisticRidge_modelAndFitIllustration}, bottom-left panel) are common, e.g. the *probit*: $p_i = \Phi_{0,1}(\mathbf{X}_i \bbeta)$; the *cloglog*: $p_i = \frac{1}{\pi} \arctan(\mathbf{X}_i \bbeta) + \frac{1}{2}$; the *Cauchit*: $p_i = \exp[ - \exp(\mathbf{X}_i \bbeta)]$. All these link function are invertible. Irrespective of the choice of the link function, the binary data are thus modelled as $Y_i \sim \mathcal{B}[g^{-1}(\mathbf{X}_i; \bbeta), 1]$. That is, as a single draw from the Binomial distribution with success probability $g^{-1}(\mathbf{X}_i; \bbeta)$. +The function $g(\cdot; \cdot)$ is called the *link function*. It links the response to the explanatory variables. The one above is called the logistic link function. Or short, logit. The regression parameters have tangible interpretations. When the first covariate represents the intercept, i.e. $X_{i,j} = 1$ for all $i$, then $\beta_1$ determines where the link function equals a half when all other covariates fail to contribute to the linear predictor (i.e. where $P (Y_i = 1 \, | \, \mathbf{X}_{i}) = 0.5$ when $\mathbf{X}_{i} \beta = \beta_1$). This is illustrated in the top-left panel of Figure ref{fig.logisticRidge_modelAndFitIllustration} for various choices of the intercept. On the other hand, the regression parameters are directly related to the odds ratio: $\mbox{*odds ratio*} = \mbox{odds}(X_{i,j}+1) / \mbox{odds}(X_{i,j}) = \exp(\beta_j)$. Hence, the effect of a unit change in the $j$-th covariate on the odds ratio is $\exp(\beta_j)$ (see Figure ref{fig.logisticRidge_modelAndFitIllustration}, top-right panel). Other link functions (depicted in Figure ref{fig.logisticRidge_modelAndFitIllustration}, bottom-left panel) are common, e.g. the *probit*: $p_i = \Phi_{0,1}(\mathbf{X}_i \beta)$; the *cloglog*: $p_i = \frac{1}{\pi} \arctan(\mathbf{X}_i \beta) + \frac{1}{2}$; the *Cauchit*: $p_i = \exp[ - \exp(\mathbf{X}_i \beta)]$. All these link function are invertible. Irrespective of the choice of the link function, the binary data are thus modelled as $Y_i \sim \mathcal{B}[g^{-1}(\mathbf{X}_i; \beta), 1]$. That is, as a single draw from the Binomial distribution with success probability $g^{-1}(\mathbf{X}_i; \beta)$. Let us now estimate the parameter of the logistic regression model by means of the maximum likelihood method. The likelihood of the experiment is then: \begin{align*} -L(\mathbf{Y} \, | \, \mathbf{X}; \bbeta) & = \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i}. +L(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i}. \end{align*} After taking the logarithm and some ready algebra, the log-likelihood is found to be: \begin{align*} -\mathcal{L}(\mathbf{Y} \, | \, \mathbf{X}; \bbeta) & = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \bbeta - \log [ 1 + \exp(\mathbf{X}_i \bbeta) ] \big\}. +\mathcal{L}(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\}. \end{align*} -Differentiate the log-likelihood with respect to $\bbeta$, equate it zero, and obtain the estimating equation for $\bbeta$: +Differentiate the log-likelihood with respect to $\beta$, equate it zero, and obtain the estimating equation for $\beta$: !bt !bt \begin{align} label{form:logisticRidge_estimatingEquationOfBeta} -\frac{\partial \mathcal{L}}{\partial \bbeta } & = \sum_{i=1}^n \Big[ Y_i - \frac{\exp(\mathbf{X}_i \bbeta)}{ 1 + \exp(\mathbf{X}_i \bbeta)} \Big] \mathbf{X}_i^{\top} \, \, \, = \, \, \, \mathbf{0}_p. +\frac{\partial \mathcal{L}}{\partial \beta } & = \sum_{i=1}^n \Big[ Y_i - \frac{\exp(\mathbf{X}_i \beta)}{ 1 + \exp(\mathbf{X}_i \beta)} \Big] \mathbf{X}_i^{\top} \, \, \, = \, \, \, \mathbf{0}_p. \end{align} !et !et -The ML estimate of $\bbeta$ strikes a (weighted by the $\mathbf{X}_i$) balance between observation and model. Put differently (and illustrated in the bottom-right panel of Figure ref{fig.logisticRidge_modelAndFitIllustration}), a curve is fit through data by minimizing the distance between them: at the ML estimate of $\bbeta$ a weighted average of their deviations is zero. +The ML estimate of $\beta$ strikes a (weighted by the $\mathbf{X}_i$) balance between observation and model. Put differently (and illustrated in the bottom-right panel of Figure ref{fig.logisticRidge_modelAndFitIllustration}), a curve is fit through data by minimizing the distance between them: at the ML estimate of $\beta$ a weighted average of their deviations is zero. -The maximum likelihood estimate of $\bbeta$ is evaluated by solving Equation (ref{form:logisticRidge_estimatingEquationOfBeta}) with respect to $\bbeta$ by means of the Newton-Raphson algorithm. The Newton-Raphson algorithm iteratively finds the zeros of a smooth enough function $f(\cdot)$. Let $x_0$ denote an initial guess of the zero. Then, approximate $f(\cdot)$ around $x_0$ by means of a first order Taylor series: $f(x) \approx x_0 + (x - x_0) \, (d f / d x) |_{x=x_0}$. Solve this for $x$ and obtain: $x = x_0 - [ (d f / d x) |_{x=x_0} ]^{-1} f(x_0)$. Let $x_1$ be the solution for $x$, use this as the new guess and repeat the above until convergence. When the function $f(\cdot)$ has multiple arguments, is vector-valued and denoted by $\vec{\mathbf{f}}$, and the Taylor approximation becomes: $\vec{\mathbf{f}}(\mathbf{x}) \approx \mathbf{x}_0 + J \vec{\mathbf{f}} \big|_{\mathbf{x}=\mathbf{x}_0} (\mathbf{x} - \mathbf{x}_0)$ with +The maximum likelihood estimate of $\beta$ is evaluated by solving Equation (ref{form:logisticRidge_estimatingEquationOfBeta}) with respect to $\beta$ by means of the Newton-Raphson algorithm. The Newton-Raphson algorithm iteratively finds the zeros of a smooth enough function $f(\cdot)$. Let $x_0$ denote an initial guess of the zero. Then, approximate $f(\cdot)$ around $x_0$ by means of a first order Taylor series: $f(x) \approx x_0 + (x - x_0) \, (d f / d x) |_{x=x_0}$. Solve this for $x$ and obtain: $x = x_0 - [ (d f / d x) |_{x=x_0} ]^{-1} f(x_0)$. Let $x_1$ be the solution for $x$, use this as the new guess and repeat the above until convergence. When the function $f(\cdot)$ has multiple arguments, is vector-valued and denoted by $\vec{\mathbf{f}}$, and the Taylor approximation becomes: $\vec{\mathbf{f}}(\mathbf{x}) \approx \mathbf{x}_0 + J \vec{\mathbf{f}} \big|_{\mathbf{x}=\mathbf{x}_0} (\mathbf{x} - \mathbf{x}_0)$ with \begin{align*} J \vec{\mathbf{f}} = \left( \begin{array}{llll} @@ -1862,40 +1994,40 @@ J \vec{\mathbf{f}} = \left( \end{align*} the Jacobi matrix. An update of $x_0$ is now readily constructed by solving (the approximation for) $\vec{\mathbf{f}}(\mathbf{x}) = \mathbf{0}$ for $\mathbf{x}$. -When applied here to the maximum likelihood estimation of the regression parameter $\bbeta$ of the logistic regression model, the Newton-Raphson update is: +When applied here to the maximum likelihood estimation of the regression parameter $\beta$ of the logistic regression model, the Newton-Raphson update is: \begin{align*} -\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} - \Big( \frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} \Big)^{-1} \Big|_{\bbeta = \hat{\bbeta}^{\mbox{{\tiny old}}} } \, \, \frac{\partial \mathcal{L}}{\partial \bbeta } \Big|_{\bbeta = \hat{\bbeta}^{\mbox{{\tiny old}}} } +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} - \Big( \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \Big)^{-1} \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } \, \, \frac{\partial \mathcal{L}}{\partial \beta } \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } \end{align*} where the Hessian of the log-likelihood equals: \begin{align*} -\frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} & = - \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \bbeta)}{ [1 + \exp(\mathbf{X}_i \bbeta)]^2} \mathbf{X}_i^{\top} \mathbf{X}_i. +\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} & = - \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1 + \exp(\mathbf{X}_i \beta)]^2} \mathbf{X}_i^{\top} \mathbf{X}_i. \end{align*} -Iterative application of this updating formula converges to the ML estimate of $\bbeta$. +Iterative application of this updating formula converges to the ML estimate of $\beta$. The Newton-Raphson algorithm is often reformulated as an iteratively re-weighted least squares algorithm. Hereto, first write the gradient and Hessian in matrix notation: \begin{align*} -\frac{\partial \mathcal{L}}{\partial \bbeta } \, \, \, = \, \, \, \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta)] +\frac{\partial \mathcal{L}}{\partial \beta } \, \, \, = \, \, \, \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta)] & \mbox{ and } & -\frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} \, \, \, = \, \, \, - \mathbf{X}^{\top} \mathbf{W} \mathbf{X}, +\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, - \mathbf{X}^{\top} \mathbf{W} \mathbf{X}, \end{align*} -where $\vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta) = [g^{-1}( \mathbf{X}_{1, \ast}; \bbeta), \ldots, g^{-1}( \mathbf{X}_{n, \ast}; \bbeta)]^{\top}$ with $g^{-1}(\cdot; \cdot) = \exp(\cdot; \cdot) / [1 + \exp(\cdot; \cdot)]$ and $\mathbf{W}$ diagonal with $(\mathbf{W})_{ii} = \exp(\mathbf{X}_i \hat{\bbeta}^{\mbox{{\scriptsize old}}} ) [ 1 + \exp(\mathbf{X}_i \hat{\bbeta}^{\mbox{{\scriptsize old}}} ) ]^{-2}$. The updating formula of the estimate then becomes: +where $\vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta) = [g^{-1}( \mathbf{X}_{1, \ast}; \beta), \ldots, g^{-1}( \mathbf{X}_{n, \ast}; \beta)]^{\top}$ with $g^{-1}(\cdot; \cdot) = \exp(\cdot; \cdot) / [1 + \exp(\cdot; \cdot)]$ and $\mathbf{W}$ diagonal with $(\mathbf{W})_{ii} = \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) [ 1 + \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) ]^{-2}$. The updating formula of the estimate then becomes: \begin{align*} -\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} + (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \\ -& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \} +& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} \\ & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, \end{align*} -where $\mathbf{Z} = \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \}$. The Newton-Raphson update is thus the solution to the following weighted least squares problem: +where $\mathbf{Z} = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}$. The Newton-Raphson update is thus the solution to the following weighted least squares problem: \begin{align*} -\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \arg \min_{\bbeta} (\mathbf{Z} - \mathbf{X} \bbeta)^{\top} \mathbf{W} (\mathbf{Z} - \mathbf{X} \bbeta). +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \arg \min_{\beta} (\mathbf{Z} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Z} - \mathbf{X} \beta). \end{align*} Effectively, at each iteration the *adjusted response* $\mathbf{Z}$ is regressed on the covariates that comprise $\mathbf{X}$. For more on logistic regression confer the monograph of cite{Hosm2013}. ======= Ridge estimation ======= -High-dimensionally, the linear predictor $\mathbf{X} \bbeta$ may be uniquely defined, but the maximum likelihood estimate of the logistic regression parameter is not. Assume $p > n$ and an estimate $\hat{\bbeta}$ available. Due to the high-dimensionality, the null space of $\mathbf{X}$ is non-trivial. Hence, let $\ggamma \in \mbox{null}(\mbox{span}(\mathbf{X}))$. Then: $\mathbf{X} \hat{\bbeta} = \mathbf{X} \hat{\bbeta} + \mathbf{X} \ggamma = \mathbf{X} (\hat{\bbeta} + \ggamma)$. As the null space is a $p-n$-dimensional subspace, $\ggamma$ need not equal zero. Hence, an infinite number of estimates of the logistic regression parameter exists that yield the same log-likelihood. Augmentation of the loss function with a ridge penalty resolves the matter, as their sum is strictly concave in $\bbeta$ (not convex as a maximum rather than a minimum is sought here) and thereby has a unique maximum. +High-dimensionally, the linear predictor $\mathbf{X} \beta$ may be uniquely defined, but the maximum likelihood estimate of the logistic regression parameter is not. Assume $p > n$ and an estimate $\hat{\beta}$ available. Due to the high-dimensionality, the null space of $\mathbf{X}$ is non-trivial. Hence, let $\ggamma \in \mbox{null}(\mbox{span}(\mathbf{X}))$. Then: $\mathbf{X} \hat{\beta} = \mathbf{X} \hat{\beta} + \mathbf{X} \ggamma = \mathbf{X} (\hat{\beta} + \ggamma)$. As the null space is a $p-n$-dimensional subspace, $\ggamma$ need not equal zero. Hence, an infinite number of estimates of the logistic regression parameter exists that yield the same log-likelihood. Augmentation of the loss function with a ridge penalty resolves the matter, as their sum is strictly concave in $\beta$ (not convex as a maximum rather than a minimum is sought here) and thereby has a unique maximum. # original latex figure with scale=0.40, angle=0 @@ -1905,25 +2037,25 @@ FIGURE: [logisticPenLL2ridgePenalty.eps, width=400 frac=1.0] Top row, left panel Ridge maximum likelihood estimates of the logistic model parameters are found by the maximization of the ridge penalized loglikelihood (cf. \citealt{Scha1984,LeCe1992}): \begin{align*} -\mathcal{L}^{\mbox{{\tiny pen}}}(\mathbf{Y}, \mathbf{X}; \bbeta, \lambda) & = \mathcal{L} (\mathbf{Y}, \mathbf{X}; \bbeta) - \tfrac{1}{2} \lambda \| \bbeta \|_2^2 +\mathcal{L}^{\mbox{{\tiny pen}}}(\mathbf{Y}, \mathbf{X}; \beta, \lambda) & = \mathcal{L} (\mathbf{Y}, \mathbf{X}; \beta) - \tfrac{1}{2} \lambda \| \beta \|_2^2 \\ -& = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \bbeta - \log [ 1 + \exp(\mathbf{X}_i \bbeta) ] \big\} - \tfrac{1}{2} \lambda \bbeta^{\top} \bbeta, +& = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\} - \tfrac{1}{2} \lambda \beta^{\top} \beta, \end{align*} -where the second summand is the ridge penalty (the sum of the square of the elements of $\bbeta$) with $\lambda$ the penalty parameter. Note that as in Section ref{sect.constrainedEstimation} maximization of this penalized loss function can be reformulated as a constrained estimation problem. This is illustrated by the top left panel of Figure ref{fig.logisticRidge_effectOfPenalization}, which depicts the contours (black lines) of the log-likelihood and the spherical domain of the parameter (red line). The optimization of the above loss function proceeds, due to the differentiability of the penalty, fully analogous to the unpenalized case and uses the Newton-Raphson algorithm for solving the (penalized) estimating equation. Hence, the unpenalized ML estimation procedure is modified straightforwardly by replacing gradient and Hessian by their `penalized' counterparts: +where the second summand is the ridge penalty (the sum of the square of the elements of $\beta$) with $\lambda$ the penalty parameter. Note that as in Section ref{sect.constrainedEstimation} maximization of this penalized loss function can be reformulated as a constrained estimation problem. This is illustrated by the top left panel of Figure ref{fig.logisticRidge_effectOfPenalization}, which depicts the contours (black lines) of the log-likelihood and the spherical domain of the parameter (red line). The optimization of the above loss function proceeds, due to the differentiability of the penalty, fully analogous to the unpenalized case and uses the Newton-Raphson algorithm for solving the (penalized) estimating equation. Hence, the unpenalized ML estimation procedure is modified straightforwardly by replacing gradient and Hessian by their `penalized' counterparts: \begin{align*} -\frac{\partial \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \bbeta } \, \, \, = \, \, \, \frac{\partial \mathcal{L}}{\partial \bbeta } - \lambda \bbeta & \mbox{ and } & \frac{\partial^2 \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \bbeta \partial \bbeta^{\top}} \, \, \, = \, \, \, \frac{\partial^2 \mathcal{L}}{\partial \bbeta \partial \bbeta^{\top}} - \lambda \mathbf{I}_{pp}. +\frac{\partial \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta } \, \, \, = \, \, \, \frac{\partial \mathcal{L}}{\partial \beta } - \lambda \beta & \mbox{ and } & \frac{\partial^2 \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} - \lambda \mathbf{I}_{pp}. \end{align*} With these at hand, the Newton-Raphson algorithm is (again) reformulated as an iteratively re-weighted least squares algorithm with the updating step changes accordingly to: \begin{align*} -\hat{\bbeta}^{\mbox{{\scriptsize new}}} & = \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \{ \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}}) ] - \lambda \bbeta^{\mbox{{\scriptsize old}}} \} +\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \{ \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}}) ] - \lambda \beta^{\mbox{{\scriptsize old}}} \} \\ -& = \mathbf{V}^{-1} \mathbf{V} \hat{\bbeta}^{\mbox{{\scriptsize old}}} - \lambda \mathbf{V}^{-1} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] +& = \mathbf{V}^{-1} \mathbf{V} \hat{\beta}^{\mbox{{\scriptsize old}}} - \lambda \mathbf{V}^{-1} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \\ -& = \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \} +& = \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} \\ & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, \end{align*} -where $\mathbf{V} = \mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp}$ and $\mathbf{W}$ and $\mathbf{Z}$ as before. Hence, use this to update the estimate of $\bbeta$ until convergence, which yields the desired ridge ML estimate. +where $\mathbf{V} = \mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp}$ and $\mathbf{W}$ and $\mathbf{Z}$ as before. Hence, use this to update the estimate of $\beta$ until convergence, which yields the desired ridge ML estimate. Obviously, the ridge estimate of the logistic regression parameter tends to zero as $\lambda \rightarrow \infty$. Now consider a linear predictor with an intercept that is left unpenalized. When $\lambda$ tends to infinity, all regression coefficients but the intercept vanish. The intercept is left to model the success probability. Hence, in this case $\lim_{\lambda \rightarrow \infty} \hat{\beta}_0 (\lambda) = \log [ \tfrac{1}{n} \sum_{i=1}^n Y_i / \tfrac{1}{n} \sum_{i=1}^n (1-Y_i)]$. @@ -1934,11 +2066,11 @@ FIGURE: [logisticRidge_effectOnPrediction_lambda0.eps, width=400 frac=1.0] The r -The effect of the ridge penalty on parameter estimates propagates to the predictor $\hat{p}_i$. The linear predictor of the linear regression model involving the ridge estimator $\mathbf{X}_i \hat{\bbeta}(\lambda)$ shrinks towards a common value for each $i$, leading to a scale difference between observation and predictor (as seen before in Section ref{sect.ridgeRegressionDataIllustration}). This behaviour transfers to the ridge logistic regression predictor, as is illustrated on simulated data. The dimension and sample size of these data are $p=2$ and $n=200$, respectively. The covariate data are drawn from the standard normal, while that of the response is sampled from a Bernoulli distribution with success probability $P(Y_i=1) = \exp(2 X_{i,1} - 2 X_{i,2}) / [ 1 + \exp(2 X_{i,1} - 2 X_{i,2})]$. The logistic regression model is estimated from these data by means of ridge penalized likelihood maximization with various choices of the penalty parameter. The bottom right plot in Figure ref{fig.logisticRidge_effectOfPenalization} shows the predicted success probability versus the linear predictor for various choices of the penalty parameter. Larger values of the penalty parameter $\lambda$ flatten the slope of this curve. Consequently, for larger $\lambda$ more excessive values of the covariates are needed to achieve the same predicted success probability as those obtained with smaller $\lambda$ at more moderate covariate values. The implications for the resulting classification may become clearer when studying the effect of the penalty parameter on the `failure' and `success regions' respectively defined by: +The effect of the ridge penalty on parameter estimates propagates to the predictor $\hat{p}_i$. The linear predictor of the linear regression model involving the ridge estimator $\mathbf{X}_i \hat{\beta}(\lambda)$ shrinks towards a common value for each $i$, leading to a scale difference between observation and predictor (as seen before in Section ref{sect.ridgeRegressionDataIllustration}). This behaviour transfers to the ridge logistic regression predictor, as is illustrated on simulated data. The dimension and sample size of these data are $p=2$ and $n=200$, respectively. The covariate data are drawn from the standard normal, while that of the response is sampled from a Bernoulli distribution with success probability $P(Y_i=1) = \exp(2 X_{i,1} - 2 X_{i,2}) / [ 1 + \exp(2 X_{i,1} - 2 X_{i,2})]$. The logistic regression model is estimated from these data by means of ridge penalized likelihood maximization with various choices of the penalty parameter. The bottom right plot in Figure ref{fig.logisticRidge_effectOfPenalization} shows the predicted success probability versus the linear predictor for various choices of the penalty parameter. Larger values of the penalty parameter $\lambda$ flatten the slope of this curve. Consequently, for larger $\lambda$ more excessive values of the covariates are needed to achieve the same predicted success probability as those obtained with smaller $\lambda$ at more moderate covariate values. The implications for the resulting classification may become clearer when studying the effect of the penalty parameter on the `failure' and `success regions' respectively defined by: \\ -\indent $\{(x_1, x_2) : P({\color{green}{\mathbf{Y=0}}} \, | \, X_1=x_1, X_2=x_2, \hat{\bbeta}(\lambda)) > 0.75 \}$, +\indent $\{(x_1, x_2) : P({\color{green}{\mathbf{Y=0}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$, \\ -\indent $\{(x_1, x_2) : P({\color{red}{\mathbf{Y=1}}} \, | \, X_1=x_1, X_2=x_2, \hat{\bbeta}(\lambda)) > 0.75 \}$. +\indent $\{(x_1, x_2) : P({\color{red}{\mathbf{Y=1}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$. \\ This separates the design space in a light red (`failure') and light green (`success') domain. The white bar between them is the domain where samples cannot be classified with high enough certainty. As $\lambda$ grows, so does the white area that separates the failure and success regions. Hence, as stronger penalization shrinks the logistic regression parameter estimate towards zero, it produces a predictor that is less outspoken in its class assignments. @@ -1947,18 +2079,18 @@ This separates the design space in a light red (`failure') and light green (`suc ======= Moments ======= -The $1^{\mbox{{\tiny st}}}$ and $2^{\mbox{{\tiny nd}}}$ order moment of the ridge ML parameter of the logistic model may be approximated by the final update of the Newton-Raphson estimate. Assume the one-to-last update $\hat{\bbeta}^{\mbox{{\scriptsize old}}}$ to be non-random and proceed as for the ridge estimator of the linear regression model parameter to arrive at: +The $1^{\mbox{{\tiny st}}}$ and $2^{\mbox{{\tiny nd}}}$ order moment of the ridge ML parameter of the logistic model may be approximated by the final update of the Newton-Raphson estimate. Assume the one-to-last update $\hat{\beta}^{\mbox{{\scriptsize old}}}$ to be non-random and proceed as for the ridge estimator of the linear regression model parameter to arrive at: # \begin{align*} -# & & \hat{\bbeta}^{\mbox{{\scriptsize new}}} = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + 2 \lambda \mathbf{I}_{p \times p} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{z} +# & & \hat{\beta}^{\mbox{{\scriptsize new}}} = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + 2 \lambda \mathbf{I}_{p \times p} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{z} # \end{align*} \begin{align*} -\mathbb{E} \big( \hat{\bbeta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbb{E}( \mathbf{Z}), +\mathbb{E} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbb{E}( \mathbf{Z}), \\ -\mbox{Var} \big( \hat{\bbeta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \big[ \mbox{Var} ( \mathbf{Z} ) \big] \mathbf{W} \mathbf{X} [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}, +\mbox{Var} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \big[ \mbox{Var} ( \mathbf{Z} ) \big] \mathbf{W} \mathbf{X} [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}, \end{align*} with \begin{align*} -\mathbb{E}(\mathbf{Z}) & = \{ \mathbf{X} \hat{\bbeta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbb{E}(\mathbf{Y}) - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \bbeta^{\mbox{{\scriptsize old}}})] \}, +\mathbb{E}(\mathbf{Z}) & = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbb{E}(\mathbf{Y}) - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}, \\ \mbox{Var}(\mathbf{Z}) & = \mathbf{W}^{-1} \mbox{Var}(\mathbf{Y}) \mathbf{W}^{-1} = \mathbf{W}^{-1}, \end{align*} @@ -1966,9 +2098,9 @@ where the identity $\mbox{Var}(\mathbf{Y}) = \mathbf{W}$ follows from the varian ======= The Bayesian connection ======= -All penalized estimators can be formulated as Bayesian estimators, including the ridge logistic estimator. In particular, ridge estimators correspond to Bayesian estimators with a multivariate normal prior on the regression coefficients. Thus, assume $\bbeta \sim \mathcal{N}(\mathbf{0}_p, \mathbf{\Delta}^{-1})$. The posterior distribution of $\bbeta$ then is: +All penalized estimators can be formulated as Bayesian estimators, including the ridge logistic estimator. In particular, ridge estimators correspond to Bayesian estimators with a multivariate normal prior on the regression coefficients. Thus, assume $\beta \sim \mathcal{N}(\mathbf{0}_p, \mathbf{\Delta}^{-1})$. The posterior distribution of $\beta$ then is: \begin{align*} -f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X}) & \propto & \Big\{ \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i} \Big\} \exp( - \tfrac{1}{2} \bbeta \mathbf{\Delta} \bbeta). +f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X}) & \propto & \Big\{ \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i} \Big\} \exp( - \tfrac{1}{2} \beta \mathbf{\Delta} \beta). \end{align*} This does not coincide with any standard distribution. But, under appropriate conditions, the posterior distribution is asymptotically normal. This invites a (multivariate) normal approximation to the posterior distribution above. The Laplace's method provides (cf. \citealp{Bish2006}). @@ -1979,36 +2111,36 @@ FIGURE: [logisticRidge_laplace2posterior.eps, width=400 frac=1.0] Right panel: L Laplace's method *i)* centers the normal approximation at the mode of the posterior, and *ii)* chooses the covariance to match the curvature of the posterior at the mode. The posterior mode is the location of the maximum of the posterior distribution. The location of this maximum coincides with that of the logarithm of the posterior. The latter is the log-likelihood augmented with a ridge penalty. Hence, the posterior mode, which is taken as the mean of the approximating Gaussian, coincides with the ridge logistic estimator. For the covariance of the approximating Gaussian, the logarithm of the posterior is approximated by a second order Taylor series around the posterior mode and limited to second order terms: \begin{align*} -\log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] & \propto & -\left. \log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] -\right|_{\bbeta = \hat{\bbeta}_{\mbox{{\tiny MAP}}}} +\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] & \propto & +\left. \log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] +\right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} \\ -& & + \tfrac{1}{2} (\bbeta - \hat{\bbeta}_{\mbox{{\tiny MAP}}})^{\top} \left. \frac{\partial^2}{\partial \bbeta \partial \bbeta^{\top}} -\log[f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})] - \right|_{\bbeta = \hat{\bbeta}_{\mbox{{\tiny MAP}}}} (\bbeta - \hat{\bbeta}_{\mbox{{\tiny MAP}}})^{\top}, +& & + \tfrac{1}{2} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top} \left. \frac{\partial^2}{\partial \beta \partial \beta^{\top}} +\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] + \right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top}, \end{align*} -in which the first order term cancels as the derivative of $f_{\bbeta}(\bbeta \, | \, \mathbf{Y}, \mathbf{X})$ with respect to $\bbeta$ vanishes at the posterior mode - its maximum. Take the exponential of this approximation and match its arguments to that of a multivariate Gaussian $\exp[-\tfrac{1}{2} (\bbeta - \mmu_{\beta})^{\top} \mathbf{\Sigma}_{\bbeta}^{-1} (\bbeta - \mmu_{\beta})]$. The covariance of the sought Gaussian approximation is thus the inverse of the Hessian of the negative penalized log-likelihood. Put together the posterior is approximated by: +in which the first order term cancels as the derivative of $f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})$ with respect to $\beta$ vanishes at the posterior mode - its maximum. Take the exponential of this approximation and match its arguments to that of a multivariate Gaussian $\exp[-\tfrac{1}{2} (\beta - \mmu_{\beta})^{\top} \mathbf{\Sigma}_{\beta}^{-1} (\beta - \mmu_{\beta})]$. The covariance of the sought Gaussian approximation is thus the inverse of the Hessian of the negative penalized log-likelihood. Put together the posterior is approximated by: \begin{align*} -\bbeta \, | \, \mathbf{Y}, \mathbf{X} \sim \mathcal{N} \Big( \hat{\bbeta}_{\mbox{{\tiny MAP}}}, \Big\{ \mathbf{\Delta} + \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \bbeta)}{ [1+ \exp(\mathbf{X}_i \bbeta) ]^2 } \mathbf{X}_i \mathbf{X}_i^{\top} \Big\}^{-1} \Big). +\beta \, | \, \mathbf{Y}, \mathbf{X} \sim \mathcal{N} \Big( \hat{\beta}_{\mbox{{\tiny MAP}}}, \Big\{ \mathbf{\Delta} + \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1+ \exp(\mathbf{X}_i \beta) ]^2 } \mathbf{X}_i \mathbf{X}_i^{\top} \Big\}^{-1} \Big). \end{align*} The Gaussian approximation is convenient but need not be good. Fortunately, the Bernstein-Von Mises Theorem \citep{VdVa2000} tells it is very accurate when the model is regular, the prior smooth, and the sample size sufficiently large. The quality of the approximation for an artificial example data set is shown in Figure ref{fig.logisticRidge_MSEandLaplace2posterior}. ======= Penalty parameter selection ======= -As before the penalty parameter may be chosen through $K$-fold cross-validation. For the $K=n$ case cite{Meij2013} describe a computationally efficient approximation of the leave-one-out cross-validated loglikelihood. It is based on the exact evaluation of the LOOCV loss, discussed in Section ref{subsect.crossvalidation}, that avoided resampling. The approach of cite{Meij2013} hinges upon the first-order Taylor expansion of the left-out penalized loglikelihood of the left-out estimate $\hat{\bbeta}_{-i} (\lambda)$ around $\hat{\bbeta} (\lambda)$, which yields an approximation of the former: +As before the penalty parameter may be chosen through $K$-fold cross-validation. For the $K=n$ case cite{Meij2013} describe a computationally efficient approximation of the leave-one-out cross-validated loglikelihood. It is based on the exact evaluation of the LOOCV loss, discussed in Section ref{subsect.crossvalidation}, that avoided resampling. The approach of cite{Meij2013} hinges upon the first-order Taylor expansion of the left-out penalized loglikelihood of the left-out estimate $\hat{\beta}_{-i} (\lambda)$ around $\hat{\beta} (\lambda)$, which yields an approximation of the former: \begin{align*} -\hat{\bbeta}_{-i} (\lambda) & \approx \hat{\bbeta} (\lambda) - -\left( \left. \frac{\partial^2 \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \bbeta \partial \bbeta^{\top}} \right|_{\bbeta = \hat{\bbeta}(\lambda)} \right)^{-1} \left. \frac{\partial \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \bbeta } \right|_{\bbeta = \hat{\bbeta}(\lambda)} +\hat{\beta}_{-i} (\lambda) & \approx \hat{\beta} (\lambda) - +\left( \left. \frac{\partial^2 \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \right|_{\beta = \hat{\beta}(\lambda)} \right)^{-1} \left. \frac{\partial \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta } \right|_{\beta = \hat{\beta}(\lambda)} \\ -& = \hat{\bbeta} (\lambda) + (\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \{ \mathbf{X}_{- i, \ast}^{\top} [\mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\bbeta}(\lambda))] - \lambda \hat{\bbeta}(\lambda) \}. +& = \hat{\beta} (\lambda) + (\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \{ \mathbf{X}_{- i, \ast}^{\top} [\mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda))] - \lambda \hat{\beta}(\lambda) \}. \end{align*} This approximation involves the inverse of a $p \times p$ dimensional matrix, which amounts to the evaluation of $n$ such inverses for the LOOCV loss. As in Section ref{subsect.crossvalidation} this may be avoided. Rewrite both the gradient and the Hessian of the left-out loglikelihood in the approximation of the preceding display: \begin{align*} -\mathbf{X}_{-i, \ast}^{\top} \{ \mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\bbeta}(\lambda)]\} - \lambda \hat{\bbeta}(\lambda) -& = \mathbf{X}^{\top} \{ \mathbf{Y} - \vec{\mathbf{g}}^{-1}[\mathbf{X}; \hat{\bbeta}(\lambda)]\} - \lambda \hat{\bbeta}(\lambda) -- \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)]\} +\mathbf{X}_{-i, \ast}^{\top} \{ \mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) +& = \mathbf{X}^{\top} \{ \mathbf{Y} - \vec{\mathbf{g}}^{-1}[\mathbf{X}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) +- \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} \\ -& = - \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)]\} +& = - \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} \end{align*} and \begin{align*} @@ -2018,124 +2150,9 @@ and \end{align*} where the Woodbury identity has been used and now $\mathbf{H}_{ii}(\lambda) = \mathbf{W}_{ii} \mathbf{X}_{i, \ast}(\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. Substitute both in the approximation of the left-out ridge logistic regression estimator and manipulate as in Section ref{subsect.crossvalidation} to obtain: \begin{align*} -\hat{\bbeta}_{- i}(\lambda) & \approx \hat{\bbeta}(\lambda) - (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - g^{-1}(\mathbf{X}_{i, \ast}; \hat{\bbeta}(\lambda)) ]. +\hat{\beta}_{- i}(\lambda) & \approx \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - g^{-1}(\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)) ]. \end{align*} -Hence, the leave-one-out cross-validated loglikelihood $\sum_{i=1}^n \mathcal{L} [Y_i \, | \, \mathbf{X}_{i, \ast}, \hat{\bbeta}_{-i}(\lambda)]$ can now be evaluated by means of a single inverse of a $p \times p$ dimensional matrix and some matrix multiplications. For the performance of this approximation in terms of accuracy and speed confer cite{Meij2013}. - - - - - - -======= Application ======= - -The ridge logistic regression is used here to explain the status (dead or alive) of ovarian cancer samples at the close of the study from gene expression data at baseline. Data stem from the TCGA study \citep{TCGA2011ovarian}, which measured gene expression by means of sequencing technology. Available are 295 samples with both status and transcriptomic profiles. These profiles are composed of 19990 transcript reads. The sequencing data, being representative of the mRNA transcript count, is heavily skewed. cite{Zwie2014} show that a simple transformation of the data prior to model building generally yields a better model than tailor-made approaches. Motivated by this observation the data were - to accommodate the zero counts - $\mbox{asinh}$-transformed. The logistic regression model is then fitted in ridge penalized fashion, leaving the intercept unpenalized. The ridge penalty parameter is chosen through 10-fold cross-validation minimizing the cross-validated error. R-code, and that for the sequel of this example, is to be found below. - -\lstinputlisting{ridgeLogistic_ovarianExample.r} - -The fit of the resulting model is studied. Hereto the fitted linear predictor $\mathbf{X} \hat{\bbeta}(\lambda_{\mbox{\tiny opt}})$ is plotted against the status (Figure ref{fig:logisticRidge_ovarianExample}, left panel). The plot shows some overlap between the boxes, but also a clear separation. The latter suggests gene expression at baseline thus enables us to distinguish surviving from the to-be-diseased ovarian cancer patients. Ideally, a decision rule based on the linear predictor can be formulated to predict an individual's outcome. - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [ridgeLogistic_ovarianExample_fit.ps, width=400 frac=1.0] Left panel: Box plot of the status vs. the fitted linear predictor using the full data set. Right panel: Box plot of the status vs. the linear prediction in the left-out samples of the 10 folds.} label{fig:logisticRidge_ovarianExample - - -The fit, however, is evaluated on the samples that have been used to build the model. This gives no insight on the model's predictive performance on novel samples. A replication of the study is generally costly and comparable data sets need not be at hand. A common workaround is to evaluate the predictive performance on the same data \citep{Subr2010}. This requires to put several samples aside for performance evaluation while the remainder is used for model building. The left-out sample may accidently be chosen to yield an exaggerated (either dramatically poor or overly optimistic) performance. This is avoided through the repetition of this exercise, leaving (groups of) samples out one at the time. The left-out performance evaluations are then averaged and believed to be representative of the predictive performance of the model on novel samples. -Note that, effectively, as the model building involves cross-validation and so does the performance evaluation, a double cross-validation loop is applied. This procedure is applied with a ten-fold split in both loops. Denote the outer folds by $f = 1, \ldots, 10$. Then, $\mathbf{X}_{f}$ and $\mathbf{X}_{-f}$ represent the design matrix of the samples comprising fold $f$ and that of the remaining samples, respectively. Define $\mathbf{Y}_{f}$ and $\mathbf{Y}_{-f}$ similarly. The linear prediction for the left-out fold $f$ is then $\mathbf{X}_{f} \hat{\bbeta}_{-f} (\lambda_{\mbox{{\tiny opt, -f}}})$. For reference to the fit, this is compared to $\mathbf{Y}_{f}$ visually by means of a boxplot as used above (see Figure ref{fig:logisticRidge_ovarianExample}, right panel). The boxes overlap almost perfectly. Hence, little to nothing remains of the predictive power suggested by the boxplot of the fit. The fit may thus give a reasonable description of the data at hand, but it extrapolates poorly to new samples. - - -======= Conclusion ======= - -To deal with response variables other than continuous ones, ridge logistic regression was discussed. High-dimensionally, the empirical identifiability problem then persists. Again, penalization came to the rescue: the ridge penalty may be combined with other link functions than the identity. Properties of ridge regression were shown to carry over to its logistic equivalent. - - - - - - -======= Exercises ======= -\begin{question} label{question.ridgeCoxNewtonRaphson} \mbox{ } -\\ -Consider an experiment involving $n$ cancer samples. For each sample $i$ the transcriptome of its tumor has been profiled and is denoted $\mathbf{X}_{i} = (X_{i1}, \ldots, X_{ip})^{\top}$ where $X_{ij}$ represents the gene $j=1, \ldots, p$ in sample $i$. Additionally, the overall survival data, $(Y_i, c_i)$ for $i=1, \ldots,n$ of these samples is available. In this $Y_i$ denotes the survival time of sample $i$ and $c_i$ the event indicator with $c_i = 0$ and $c_i = 1$ representing non- and censoredness, respectively. You may ignore the possibility of ties in the remainder. - -\begin{compactitem} -\item[*a)*] Write down the Cox proportional regression model that links overall survival times (as the response variable) to the expression levels. - -\item[*b)*] Specify its loss function for penalized maximum partial (!) likelihood estimation of the parameters. Penalization is via the ridge penalty. - -\item[*c)*] From this loss function, derive the estimation equation for the Cox regression coefficients. - -\item[*d)*] Describe (in words) how you would find the `ridge ML estimate'. -\end{compactitem} -\end{question} - -\begin{question} \mbox{ } -\\ -Download the {\tt multtest} package from BioConductor: -\\ -\indent {\tt > source("http://www.bioconductor.org/biocLite.R")} -\\ -\indent {\tt > biocLite("multtest")} -\\ -Activate the library and load leukemia data from the package: -\\ -\indent {\tt > library(multtest)} -\\ -\indent {\tt > data(golub)} -\\ -The objects {\tt golub} and {\tt golub.cl} are now available. The matrix-object {\tt golub} contains the expression profiles of 38 leukemia patients. Each profile comprises expression levels of 3051 genes. The numeric-object {\tt golub.cl} is an indicator variable for the leukemia type (AML or ALL) of the patient. - -\begin{compactitem} -\item[*a)*] Relate the leukemia subtype and the gene expression levels by a logistic regression model. Fit this model by means of penalized maximum likelihood, employing the ridge penalty with penalty parameter $\lambda=1$. This is implemented in the {\tt penalized}-packages available from {\tt CRAN}. *Note:* center (gene-wise) the expression levels around zero. - -\item[*b)*] Obtain the fits from the regression model. The fit is almost perfect. Could this be due to overfitting the data? Alternatively, could it be that the biological information in the gene expression levels indeed determines the leukemia subtype almost perfectly? - -\item[*c)*] To discern between the two explanations for the almost perfect fit, randomly shuffle the subtypes. Refit the logistic regression model and obtain the fits. On the basis of this and the previous fit, which explanation is more plausible? - -\item[*d)*] Compare the fit of the logistic model with different penalty parameters, say $\lambda = 1$ and $\lambda = 1000$. How does $\lambda$ influence the possibility of overfitting the data? - -\item[*e)*] Describe what you would do to prevent overfitting. -\end{compactitem} -\end{question} - - -\begin{question} label{question.ridgeLogisticBreastData} \mbox{ } -\\ -Download the {\tt breastCancerNKI} package from BioConductor: -\\ -\indent {\tt > source("http://www.bioconductor.org/biocLite.R")} -\\ -\indent {\tt > biocLite("breastCancerNKI")} -\\ -Activate the library and load leukemia data from the package: -\\ -\indent {\tt > library(breastCancerNKI)} -\\ -\indent {\tt > data(nki)} -\\ -The eset-object {\tt nki} is now available. It contains the expression profiles of 337 breast cancer patients. Each profile comprises expression levels of 24481 genes. Extract the expression data from the object, remove all genes with missing values, center the gene expression gene-wise around zero, and limit the data set to the first thousand genes. The reduction of the gene dimensionality is only for computational speed. -\\ -\indent {\tt X <- exprs(nki) } -\\ -\indent {\tt X <- X[-which(rowSums(is.na(X)) > 0),] } -\\ -\indent {\tt X <- apply(X[1:1000,], 1, function(X){ X - mean(X) }) }. -\\ -Furthermore, extract the estrogen receptor status (short: ER status), an important prognostic indicator for breast cancer. -\\ -\indent {\tt Y <- pData(nki)[,8] } - - -\begin{compactitem} -\item[*a)*] Relate the ER status and the gene expression levels by a logistic regression model, which is fitted by means of ridge penalized maximum likelihood. First, find the optimal value of the penalty parameter of $\lambda$ by means of cross-validation. This is implemented in {\tt optL2}-function of the {\tt penalized}-package available from {\tt CRAN}. - -\item[*b)*] Evaluate whether the cross-validated likelihood indeed attains a maximum at the optimal value of $\lambda$. This can be done with the {\tt profL2}-function of the {\tt penalized}-package available from {\tt CRAN}. - -\item[*c)*] Investigate the sensitivity of the penalty parameter selection with respect to the choice of the cross-validation fold. - -\item[*d)*] Does the optimal lambda produce a reasonable fit? -\end{compactitem} -\end{question} +Hence, the leave-one-out cross-validated loglikelihood $\sum_{i=1}^n \mathcal{L} [Y_i \, | \, \mathbf{X}_{i, \ast}, \hat{\beta}_{-i}(\lambda)]$ can now be evaluated by means of a single inverse of a $p \times p$ dimensional matrix and some matrix multiplications. For the performance of this approximation in terms of accuracy and speed confer cite{Meij2013}. From 1b59ce5a0ca2751461ec29cf4523039fee4fb902 Mon Sep 17 00:00:00 2001 From: mhjensen Date: Wed, 5 Sep 2018 16:46:51 +0200 Subject: [PATCH 06/12] added hint to exercise 5 --- doc/Projects/2018/hw2/html/hw2-bs.html | 7 +++++-- doc/Projects/2018/hw2/html/hw2.html | 7 +++++-- .../2018/hw2/ipynb/ipynb-hw2-src.tar.gz | Bin 207 -> 207 bytes doc/Projects/2018/hw2/pdf/hw2.p.tex | 6 ++++-- doc/Projects/2018/hw2/pdf/hw2.pdf | Bin 198097 -> 198515 bytes doc/Projects/2018/hw2/pdf/hw2.tex | 6 ++++-- doc/src/Projects/2018/Exercises/hw2.do.txt | 6 ++++-- 7 files changed, 22 insertions(+), 10 deletions(-) diff --git a/doc/Projects/2018/hw2/html/hw2-bs.html b/doc/Projects/2018/hw2/html/hw2-bs.html index 5ff410dfb..2cac701d8 100644 --- a/doc/Projects/2018/hw2/html/hw2-bs.html +++ b/doc/Projects/2018/hw2/html/hw2-bs.html @@ -171,12 +171,15 @@ Discuss these quantities as functions of the variable \( \lambda \) in the Ridge

    Using the singular value decomposition, show that the variance of the direction vector -\( \hat{z}_i=\hat{X}\hat{v}_i \) is equal to (equation (3.49) of Hastie et al.) +\( \hat{z}_i=\hat{X}\hat{v}_i=\hat{u}_1d_1 \) is equal to (equation (3.49) of Hastie et al.) $$ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, $$ -where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where \( d_i \) are the singular values of the matrix \( \hat{X} \). In Hastie et al, the matrix elements of \( X \) are centered. The consequence is that the mean values of for example \( \hat{u}_i \) are zero. + +

    +Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise.

    diff --git a/doc/Projects/2018/hw2/html/hw2.html b/doc/Projects/2018/hw2/html/hw2.html index 66482871e..102ba1e39 100644 --- a/doc/Projects/2018/hw2/html/hw2.html +++ b/doc/Projects/2018/hw2/html/hw2.html @@ -136,12 +136,15 @@ Discuss these quantities as functions of the variable \( \lambda \) in the Ridge

    Using the singular value decomposition, show that the variance of the direction vector -\( \hat{z}_i=\hat{X}\hat{v}_i \) is equal to (equation (3.49) of Hastie et al.) +\( \hat{z}_i=\hat{X}\hat{v}_i=\hat{u}_1d_1 \) is equal to (equation (3.49) of Hastie et al.) $$ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, $$ -where \( d_i \) are the singular values of the matrix \( \hat{X} \). Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where \( d_i \) are the singular values of the matrix \( \hat{X} \). In Hastie et al, the matrix elements of \( X \) are centered. The consequence is that the mean values of for example \( \hat{u}_i \) are zero. + +

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    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018


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    General linear models

    +

    Regression analysis, overarching aims II

    -Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \hat{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \hat{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.

    -Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is -$$ -y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_i x_i^j+\epsilon_i, -$$ - -where \( \epsilon_i \) is the error in our approximation. +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \).

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    Rewriting the fitting procedure as a linear algebra problem

    +

    General linear models

    -For every set of values \( y_i,x_i \) we have thus the corresponding set of equations +Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \hat{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \hat{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function. + +

    +Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is $$ -\begin{align*} -y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ -y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ -y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\ -\end{align*} +y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_i x_i^j+\epsilon_i, $$ + +where \( \epsilon_i \) is the error in our approximation. + +

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    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem

    -Defining the vectors +For every set of values \( y_i,x_i \) we have thus the corresponding set of equations $$ -\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, -$$ - -and -$$ -\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, -$$ - -and -$$ -\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, -$$ - -and the matrix -$$ -\hat{X}= -\begin{bmatrix} -1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ -1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ -1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ -\end{bmatrix} -$$ - -we can rewrite our equations as -$$ -\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. +\begin{align*} +y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ +y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ +y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\ +\end{align*} $$

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    Generalizing the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    -We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. -For every set of values \( y_i,x_i \) we can then generalize the equations to +Defining the vectors $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, +$$ + +and +$$ +\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, +$$ + +and +$$ +\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, +$$ + +and the matrix +$$ +\hat{X}= +\begin{bmatrix} +1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ +1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ +1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ +\end{bmatrix} +$$ + +we can rewrite our equations as +$$ +\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. $$

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    -We redefine in turn the matrix \( \hat{X} \) as +We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. +For every set of values \( y_i,x_i \) we can then generalize the equations to $$ -\hat{X}= -\begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ -\end{bmatrix} +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ - -and without loss of generality we rewrite again our equations as -$$ -\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. -$$ - -The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?

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    Optimizing our parameters

    +

    Generalizing the fitting procedure as a linear algebra problem

    -We have defined the matrix \( \hat{X} \) +We redefine in turn the matrix \( \hat{X} \) as $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\hat{X}= +\begin{bmatrix} +x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ +x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ +x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +\end{bmatrix} $$ + +and without loss of generality we rewrite again our equations as +$$ +\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. +$$ + +The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?

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    Optimizing our parameters, more details

    +

    Optimizing our parameters

    -We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as +We have defined the matrix \( \hat{X} \) $$ -\hat{\tilde{y}}= \hat{X}\hat{\beta}, -$$ - -and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely -$$ -Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), -$$ - -or using the matrix \( \hat{X} \) as -$$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$

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    Interpretations and optimizing our parameters

    +

    Optimizing our parameters, more details

    -The function +We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as $$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), +\hat{\tilde{y}}= \hat{X}\hat{\beta}, $$ -can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value +and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely $$ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, +Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), $$ -where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. - -

    -In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring +or using the matrix \( \hat{X} \) as $$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). $$ - -which results in -$$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, -$$ - -or in a matrix-vector form as -$$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). -$$ - -

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    -We can rewrite +The function $$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), $$ -as +can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value $$ -\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, +y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, $$ -and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution +where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. + +

    +In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring $$ -\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +$$ + +which results in +$$ +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, +$$ + +or in a matrix-vector form as +$$ +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). $$

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    -The residuals \( \hat{\epsilon} \) are in turn given by +We can rewrite $$ -\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), $$ -and with +as $$ -\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, $$ -we have +and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution $$ -\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. $$ -meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. -

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    The \( \chi^2 \) function

    +

    Interpretations and optimizing our parameters

    - -

    -Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. - -

    -Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as +The residuals \( \hat{\epsilon} \) are in turn given by $$ -\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right), +\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, $$ -where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements. +and with +$$ +\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +we have +$$ +\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.

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    -In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring +Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. + +

    +Introducing the standard deviation \( \sigma_i \) for each measurement \( y_i \), we define now the \( \chi^2 \) function as $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, +\chi^2(\hat{\beta})=\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\left(\hat{y}-\hat{\tilde{y}}\right)^T\frac{1}{\hat{\Sigma^2}}\left(\hat{y}-\hat{\tilde{y}}\right), $$ -which results in -$$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, -$$ +where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements. -or in a matrix-vector form as -$$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right). -$$ - -where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \). +

    @@ -255,7 +279,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix
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    • diff --git a/doc/pub/Regression/html/._Regression-bs014.html b/doc/pub/Regression/html/._Regression-bs014.html index 896a5038a..bbde82e75 100644 --- a/doc/pub/Regression/html/._Regression-bs014.html +++ b/doc/pub/Regression/html/._Regression-bs014.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -209,20 +238,22 @@ MathJax.Hub.Config({

    -We can rewrite +In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\hat{\beta}) \) by requiring $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right), +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, $$ -as +which results in $$ -\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, $$ -and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution +or in a matrix-vector form as $$ -\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. +\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right). $$ + +where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \hat{b} \) with elements \( b_i = y_i/\sigma_i \).

    @@ -253,7 +284,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs015.html b/doc/pub/Regression/html/._Regression-bs015.html index dd2dbedfa..9bb6354df 100644 --- a/doc/pub/Regression/html/._Regression-bs015.html +++ b/doc/pub/Regression/html/._Regression-bs015.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -209,24 +238,19 @@ MathJax.Hub.Config({

    -If we then introduce the matrix +We can rewrite $$ -\hat{H} = \left(\hat{A}^T\hat{A}\right)^{-1}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{A}^T\left( \hat{b}-\hat{A}\hat{\beta}\right), $$ -we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \)) +as $$ -\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} +\hat{A}^T\hat{b} = \hat{A}^T\hat{A}\hat{\beta}, $$ -We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) +and if the matrix \( \hat{A}^T\hat{A} \) is invertible we have the solution $$ -\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, -$$ - -resulting in -$$ -\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! +\hat{\beta} =\left(\hat{A}^T\hat{A}\right)^{-1}\hat{A}^T\hat{b}. $$ @@ -258,7 +282,7 @@ $$

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    • diff --git a/doc/pub/Regression/html/._Regression-bs016.html b/doc/pub/Regression/html/._Regression-bs016.html index 6a8765bee..b3aeb045a 100644 --- a/doc/pub/Regression/html/._Regression-bs016.html +++ b/doc/pub/Regression/html/._Regression-bs016.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -207,19 +236,26 @@ MathJax.Hub.Config({

    -The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write + +

    +If we then introduce the matrix $$ -y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. +\hat{H} = \left(\hat{A}^T\hat{A}\right)^{-1}, $$ -By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by +we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \hat{H} \) are \( h_{ij} \)) $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, +\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} $$ -and +We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise) $$ -\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. +\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, +$$ + +resulting in +$$ +\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! $$

    @@ -251,7 +287,7 @@ $$
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    - -

    -For a linear fit we don't need to invert a matrix!! -Defining +The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write $$ -\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. $$ - +By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by $$ -\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, $$ +and $$ -\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +\frac{\partial \chi^2(\hat{\beta})}{\partial \beta_0} = -2\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. $$ - -$$ -\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, -$$ - -$$ -\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, -$$ - -we obtain -$$ -\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, -$$ - -$$ -\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. -$$ - -

    -This approach (different linear and non-linear regression) suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below. Or using Lasso and Ridge regression. See below.

    @@ -273,7 +280,7 @@ This approach (different linear and non-linear regression) suffers often from bo
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    Simple regression model

    -We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens +

    The \( \chi^2 \) function

    +
    +
    +

    +

    +For a linear fit we don't need to invert a matrix!! +Defining +$$ +\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, +$$ + + +$$ +\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, +$$ + +$$ +\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), +$$ + +$$ +\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, +$$ + +$$ +\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, +$$ + +we obtain +$$ +\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, +$$ + +$$ +\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. +$$ - -

    import numpy as np
-print(np.c_[np.array([1,2,3]), np.array([4,5,6])])
-print(np.c_[np.array([[1,2,3]]), 0, 0, np.array([[4,5,6]])])
-

    +This approach (different linear and non-linear regression) suffers often from both being underdetermined and overdetermined in the unknown coefficients \( \beta_i \). A better approach is to use the Singular Value Decomposition (SVD) method discussed below. Or using Lasso and Ridge regression. See below. +

    +
    - -
    # Importing various packages
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(beta)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression')
-plt.show()
-
    -

    -We see that, as expected, a linear fit gives a seemingly (from the graph) good representation of the data.

    @@ -266,7 +302,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r

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    Simple regression model, now using scikit-learn

    - +

    Simple regression model

    +We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    -We can repeat the above algorithm using scikit-learn as follows + + +

    import numpy as np
+print(np.c_[np.array([1,2,3]), np.array([4,5,6])])
+print(np.c_[np.array([[1,2,3]]), 0, 0, np.array([[4,5,6]])])
+

    @@ -214,23 +248,27 @@ We can repeat the above algorithm using scikit-learn as follows from random import random, seed import numpy as np import matplotlib.pyplot as plt -from sklearn.linear_model import LinearRegression x = 2*np.random.rand(100,1) y = 4+3*x+np.random.randn(100,1) -linreg = LinearRegression() -linreg.fit(x,y) + +xb = np.c_[np.ones((100,1)), x] +beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y) xnew = np.array([[0],[2]]) -ypredict = linreg.predict(xnew) +xbnew = np.c_[np.ones((2,1)), xnew] +ypredict = xbnew.dot(beta) plt.plot(xnew, ypredict, "r-") plt.plot(x, y ,'ro') plt.axis([0,2.0,0, 15.0]) plt.xlabel(r'$x$') plt.ylabel(r'$y$') -plt.title(r'Random numbers ') +plt.title(r'Linear Regression') plt.show() +

    +We see that, as expected, a linear fit gives a seemingly (from the graph) good representation of the data. +

    @@ -257,7 +295,7 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs020.html b/doc/pub/Regression/html/._Regression-bs020.html index 420ad0471..4cbf19b9e 100644 --- a/doc/pub/Regression/html/._Regression-bs020.html +++ b/doc/pub/Regression/html/._Regression-bs020.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,69 +232,32 @@ MathJax.Hub.Config({ -

    Simple linear regression model using scikit-learn

    +

    Simple regression model, now using scikit-learn

    -We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. -What follows is a simple Python code where we have defined function \( y \) in terms of the variable \( x \). Both are defined as vectors of dimension \( 1\times 100 \). The entries to the vector \( \hat{x} \) are given by random numbers generated with a uniform distribution with entries \( x_i \in [0,1] \) (more about probability distribution functions later). These values are then used to define a function \( y(x) \) (tabulated again as a vector) with a linear dependence on \( x \) plus a random noise added via the normal distribution. - -

    -The Numpy functions are imported used the import numpy as np -statement and the random number generator for the uniform distribution -is called using the function np.random.rand(), where we specificy -that we want \( 100 \) random variables. Using Numpy we define -automatically an array with the specified number of elements, \( 100 \) in -our case. With the Numpy function randn() we can compute random -numbers with the normal distribution (mean value \( \mu \) equal to zero and -variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear -dependence as function of \( x \) - -$$ -y = 2x+N(0,1), -$$ - -

    -where \( N(0,1) \) represents random numbers generated by the normal -distribution. From scikit-learn we import then the -LinearRegression functionality and make a prediction \( \tilde{y} = -\alpha + \beta x \) using the function fit(x,y). We call the set of -data \( (\hat{x},\hat{y}) \) for our training data. The Python package -scikit-learn has also a functionality which extracts the above -fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will -distinguish between training data and test data. - -

    -For plotting we use the Python package -matplotlib which produces publication -quality figures. Feel free to explore the extensive -gallery of examples. In -this example we plot our original values of \( x \) and \( y \) as well as the -prediction ypredict (\( \tilde{y} \)), which attempts at fitting our -data with a straight line. - -

    -The Python code follows here. +We can repeat the above algorithm using scikit-learn as follows 

    # Importing various packages
+from random import random, seed
 import numpy as np
 import matplotlib.pyplot as plt
 from sklearn.linear_model import LinearRegression
 
-x = np.random.rand(100,1)
-y = 2*x+np.random.randn(100,1)
+x = 2*np.random.rand(100,1)
+y = 4+3*x+np.random.randn(100,1)
 linreg = LinearRegression()
 linreg.fit(x,y)
-xnew = np.array([[0],[1]])
+xnew = np.array([[0],[2]])
 ypredict = linreg.predict(xnew)
 
 plt.plot(xnew, ypredict, "r-")
 plt.plot(x, y ,'ro')
-plt.axis([0,1.0,0, 5.0])
+plt.axis([0,2.0,0, 15.0])
 plt.xlabel(r'$x$')
 plt.ylabel(r'$y$')
-plt.title(r'Simple Linear Regression')
+plt.title(r'Random numbers ')
 plt.show()

    @@ -294,7 +286,7 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs021.html b/doc/pub/Regression/html/._Regression-bs021.html index c1f975f55..a768c9fd4 100644 --- a/doc/pub/Regression/html/._Regression-bs021.html +++ b/doc/pub/Regression/html/._Regression-bs021.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,25 +232,71 @@ MathJax.Hub.Config({ -

    Simple linear regression model

    +

    Simple linear regression model using scikit-learn

    -This example serves several aims. It allows us to demonstrate several -aspects of data analysis and later machine learning algorithms. The -immediate visualization shows that our linear fit is not -impressive. It goes through the data points, but there are many -outliers which are not reproduced by our linear regression. We could -now play around with this small program and change for example the -factor in front of \( x \) and the normal distribution. Try to change the -function \( y \) to +We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. +What follows is a simple Python code where we have defined function \( y \) in terms of the variable \( x \). Both are defined as vectors of dimension \( 1\times 100 \). The entries to the vector \( \hat{x} \) are given by random numbers generated with a uniform distribution with entries \( x_i \in [0,1] \) (more about probability distribution functions later). These values are then used to define a function \( y(x) \) (tabulated again as a vector) with a linear dependence on \( x \) plus a random noise added via the normal distribution. + +

    +The Numpy functions are imported used the import numpy as np +statement and the random number generator for the uniform distribution +is called using the function np.random.rand(), where we specificy +that we want \( 100 \) random variables. Using Numpy we define +automatically an array with the specified number of elements, \( 100 \) in +our case. With the Numpy function randn() we can compute random +numbers with the normal distribution (mean value \( \mu \) equal to zero and +variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear +dependence as function of \( x \) $$ -y = 10x+0.01 \times N(0,1), +y = 2x+N(0,1), $$

    -where \( x \) is defined as before. +where \( N(0,1) \) represents random numbers generated by the normal +distribution. From scikit-learn we import then the +LinearRegression functionality and make a prediction \( \tilde{y} = +\alpha + \beta x \) using the function fit(x,y). We call the set of +data \( (\hat{x},\hat{y}) \) for our training data. The Python package +scikit-learn has also a functionality which extracts the above +fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will +distinguish between training data and test data. +

    +For plotting we use the Python package +matplotlib which produces publication +quality figures. Feel free to explore the extensive +gallery of examples. In +this example we plot our original values of \( x \) and \( y \) as well as the +prediction ypredict (\( \tilde{y} \)), which attempts at fitting our +data with a straight line. + +

    +The Python code follows here. +

    + + +

    # Importing various packages
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 2*x+np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+xnew = np.array([[0],[1]])
+ypredict = linreg.predict(xnew)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,1.0,0, 5.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Simple Linear Regression')
+plt.show()
+

    @@ -248,7 +323,7 @@ where \( x \) is defined as before.

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    Less noise

    +

    Simple linear regression model

    -Does the fit look better? Indeed, by -reducing the role of the normal distribution we see immediately that -our linear prediction seemingly reproduces better the training -set. However, this testing 'by the eye' is obviouly not satisfactory in the -long run. Here we have only defined the training data and our model, and -have not discussed a more rigorous approach to the cost function. +This example serves several aims. It allows us to demonstrate several +aspects of data analysis and later machine learning algorithms. The +immediate visualization shows that our linear fit is not +impressive. It goes through the data points, but there are many +outliers which are not reproduced by our linear regression. We could +now play around with this small program and change for example the +factor in front of \( x \) and the normal distribution. Try to change the +function \( y \) to + +$$ +y = 10x+0.01 \times N(0,1), +$$ + +

    +where \( x \) is defined as before.

    @@ -239,7 +277,7 @@ have not discussed a more rigorous approach to the cost function.

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    How to study our fits

    +

    Less noise

    -We need more rigorous criteria in defining whether we have succeeded or -not in modeling our training data. You will be surprised to see that -many scientists seldomly venture beyond this 'by the eye' approach. A -standard approach for the cost function is the so-called \( \chi^2 \) -function - -$$ \chi^2 = \frac{1}{n} -\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, -$$ - -

    -where \( \sigma_i^2 \) is the variance (to be defined later) of the entry -\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves -however the aim of scaling the equations and make the cost function -dimensionless. +Does the fit look better? Indeed, by +reducing the role of the normal distribution we see immediately that +our linear prediction seemingly reproduces better the training +set. However, this testing 'by the eye' is obviouly not satisfactory in the +long run. Here we have only defined the training data and our model, and +have not discussed a more rigorous approach to the cost function.

    @@ -248,7 +268,7 @@ dimensionless.

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    Minimizing the cost function

    +

    How to study our fits

    -Minimizing the cost function is a central aspect of -our discussions to come. Finding its minima as function of the model -parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring -theme in these series of lectures. Essentially all machine learning -algorithms we will discuss center around the minimization of the -chosen cost function. This depends in turn on our specific -model for describing the data, a typical situation in supervised -learning. Automatizing the search for the minima of the cost function is a -central ingredient in all algorithms. Typical methods which are -employed are various variants of gradient methods. These will be -discussed in more detail later. Again, you'll be surprised to hear that -many practitioners minimize the above function ''by the eye', popularly dubbed as -'chi by the eye'. That is, change a parameter and see (visually and numerically) that -the \( \chi^2 \) function becomes smaller. +We need more rigorous criteria in defining whether we have succeeded or +not in modeling our training data. You will be surprised to see that +many scientists seldomly venture beyond this 'by the eye' approach. A +standard approach for the cost function is the so-called \( \chi^2 \) +function + +$$ \chi^2 = \frac{1}{n} +\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, +$$ + +

    +where \( \sigma_i^2 \) is the variance (to be defined later) of the entry +\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves +however the aim of scaling the equations and make the cost function +dimensionless.

    @@ -247,7 +277,7 @@ the \( \chi^2 \) function becomes smaller.

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    Relative error

    +

    Minimizing the cost function

    -There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define -the relative error as - -$$ -\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. -$$ - -We can modify easily the above Python code and plot the relative error instead -

    - - -

    import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 5*x+0.01*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-
-plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
-plt.axis([0,1.0,0.0, 0.5])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
-plt.title(r'Relative error')
-plt.show()
-
    -

    -Depending on the parameter in front of the normal distribution, we may -have a small or larger relative error. Try to play around with -different training data sets and study (graphically) the value of the -relative error. +Minimizing the cost function is a central aspect of +our discussions to come. Finding its minima as function of the model +parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring +theme in these series of lectures. Essentially all machine learning +algorithms we will discuss center around the minimization of the +chosen cost function. This depends in turn on our specific +model for describing the data, a typical situation in supervised +learning. Automatizing the search for the minima of the cost function is a +central ingredient in all algorithms. Typical methods which are +employed are various variants of gradient methods. These will be +discussed in more detail later. Again, you'll be surprised to hear that +many practitioners minimize the above function ''by the eye', popularly dubbed as +'chi by the eye'. That is, change a parameter and see (visually and numerically) that +the \( \chi^2 \) function becomes smaller.

    @@ -266,7 +276,7 @@ relative error.

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    The richness of scikit-learn

    +

    Relative error

    -As mentioned above, scikit-learn has an impressive functionality. -We can for example extract the values of \( \alpha \) and \( \beta \) and -their error estimates, or the variance and standard deviation and many -other properties from the statistical data analysis. +There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define +the relative error as -

    -Here we show an -example of the functionality of scikit-learn. +$$ +\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. +$$ + +We can modify easily the above Python code and plot the relative error instead

    -

    import numpy as np 
-import matplotlib.pyplot as plt 
-from sklearn.linear_model import LinearRegression 
-from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
 
 x = np.random.rand(100,1)
-y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+y = 5*x+0.01*np.random.randn(100,1)
 linreg = LinearRegression()
 linreg.fit(x,y)
 ypredict = linreg.predict(x)
-print('The intercept alpha: \n', linreg.intercept_)
-print('Coefficient beta : \n', linreg.coef_)
-# The mean squared error                               
-print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
-# Explained variance score: 1 is perfect prediction                                 
-print('Variance score: %.2f' % r2_score(y, ypredict))
-# Mean squared log error                                                        
-print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
-# Mean absolute error                                                           
-print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
-plt.plot(x, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0.0,1.0,1.5, 7.0])
+
+plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
+plt.axis([0,1.0,0.0, 0.5])
 plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression fit ')
+plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
+plt.title(r'Relative error')
 plt.show()
 
    
+
    
+Depending on the parameter in front of the normal distribution, we may
+have a small or larger relative error. Try to play around with
+different training data sets and study (graphically) the value of the
+relative error.
+
 
    
 
    
 
@@ -271,7 +295,7 @@ plt.show()
   
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diff --git a/doc/pub/Regression/html/._Regression-bs027.html b/doc/pub/Regression/html/._Regression-bs027.html
index d68ae908b..79b422d69 100644
--- a/doc/pub/Regression/html/._Regression-bs027.html
+++ b/doc/pub/Regression/html/._Regression-bs027.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,20 +232,48 @@ MathJax.Hub.Config({
 
 
 
-
    Functions in scikit-learn 
    
+
    The richness of scikit-learn 
    
 
 
    
-The function coef gives us the parameter \( \beta \) of our fit while intercept yields 
-\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as
-$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
-\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
-$$
+As mentioned above, scikit-learn has an impressive functionality.
+We can for example extract the values of \( \alpha \) and \( \beta \) and
+their error estimates, or the variance and standard deviation and many
+other properties from the statistical data analysis.
 
 
    
-The smaller the value, the better the fit. Ideally we would like to
-have an MSE equal zero.  The attentive reader has probably recognized
-this function as being similar to the \( \chi^2 \) function defined above.
+Here we show an
+example of the functionality of scikit-learn.
+
    
 
+
+
    import numpy as np 
+import matplotlib.pyplot as plt 
+from sklearn.linear_model import LinearRegression 
+from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
+x = np.random.rand(100,1)
+y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+print('The intercept alpha: \n', linreg.intercept_)
+print('Coefficient beta : \n', linreg.coef_)
+# The mean squared error                               
+print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
+# Explained variance score: 1 is perfect prediction                                 
+print('Variance score: %.2f' % r2_score(y, ypredict))
+# Mean squared log error                                                        
+print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
+# Mean absolute error                                                           
+print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
+plt.plot(x, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0.0,1.0,1.5, 7.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Linear Regression fit ')
+plt.show()
+
    
 
    
 
    
 
@@ -243,7 +300,7 @@ this function as being similar to the \( \chi^2 \) function defined above.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs028.html b/doc/pub/Regression/html/._Regression-bs028.html
index 39f493b62..79c5f693a 100644
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@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,26 +232,19 @@ MathJax.Hub.Config({
 
 
 
-
    Other functions in  scikit-learn 
    
+
    Functions in scikit-learn 
    
 
 
    
-The r2score function computes \( R^2 \), the coefficient of
-determination. It provides a measure of how well future samples are
-likely to be predicted by the model. Best possible score is 1.0 and it
-can be negative (because the model can be arbitrarily worse). A
-constant model that always predicts the expected value of \( \hat{y} \),
-disregarding the input features, would get a \( R^2 \) score of \( 0.0 \).
+The function coef gives us the parameter \( \beta \) of our fit while intercept yields 
+\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as
+$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
+\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
+$$
 
 
    
-If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
-$$
-R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
-$$
-
-where we have defined the mean value  of \( \hat{y} \) as
-$$
-\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
-$$
+The smaller the value, the better the fit. Ideally we would like to
+have an MSE equal zero.  The attentive reader has probably recognized
+this function as being similar to the \( \chi^2 \) function defined above.
 
 
    
 
    
@@ -249,6 +271,8 @@ $$
   
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diff --git a/doc/pub/Regression/html/._Regression-bs029.html b/doc/pub/Regression/html/._Regression-bs029.html
index 6a182d377..33cd41ef8 100644
--- a/doc/pub/Regression/html/._Regression-bs029.html
+++ b/doc/pub/Regression/html/._Regression-bs029.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,27 +232,26 @@ MathJax.Hub.Config({
 
 
 
-
    The mean absolute error and other functions in  scikit-learn 
    
+
    Other functions in  scikit-learn 
    
 
 
    
-Another quantity will meet again in our discussions of regression analysis is 
- mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error.
-The MAE is defined as follows
-$$
-\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|.
-$$
-
-Finally we present the 
-squared logarithmic (quadratic) error
-$$
-\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2,
-$$
+The r2score function computes \( R^2 \), the coefficient of
+determination. It provides a measure of how well future samples are
+likely to be predicted by the model. Best possible score is 1.0 and it
+can be negative (because the model can be arbitrarily worse). A
+constant model that always predicts the expected value of \( \hat{y} \),
+disregarding the input features, would get a \( R^2 \) score of \( 0.0 \).
 
 
    
-where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error
-estimate is best to use when targets having exponential growth, such
-as population counts, average sales of a commodity over a span of
-years etc.
+If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as
+$$
+R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
+$$
+
+where we have defined the mean value  of \( \hat{y} \) as
+$$
+\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
+$$
 
 
    
 
    
@@ -249,6 +277,9 @@ years etc.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs030.html b/doc/pub/Regression/html/._Regression-bs030.html
index a0e4dfa92..69b76d122 100644
--- a/doc/pub/Regression/html/._Regression-bs030.html
+++ b/doc/pub/Regression/html/._Regression-bs030.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,50 +232,27 @@ MathJax.Hub.Config({
 
 
 
-
    Cubic polynomial in  scikit-learn 
    
+
    The mean absolute error and other functions in  scikit-learn 
    
 
 
    
-We will discuss in more
-detail these and other functions in the various lectures.  We conclude this part with another example. Instead of 
-a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn.
+Another quantity will meet again in our discussions of regression analysis is 
+ mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error.
+The MAE is defined as follows
+$$
+\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|.
+$$
+
+Finally we present the 
+squared logarithmic (quadratic) error
+$$
+\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2,
+$$
 
 
    
-
-
-
    import matplotlib.pyplot as plt
-import numpy as np
-import random
-from sklearn.linear_model import Ridge
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.pipeline import make_pipeline
-from sklearn.linear_model import LinearRegression
-
-x=np.linspace(0.02,0.98,200)
-noise = np.asarray(random.sample((range(200)),200))
-y=x**3*noise
-yn=x**3*100
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = LinearRegression()
-clf3.fit(X,y)
-
-Xplot=poly3.fit_transform(x[:,np.newaxis])
-poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
-plt.plot(x,yn, color='red', label="True Cubic")
-plt.scatter(x, y, label='Data', color='orange', s=15)
-plt.legend()
-plt.show()
-
-def error(a):
-    for i in y:
-        err=(y-yn)/yn
-    return abs(np.sum(err))/len(err)
-
-print (error(y))
-
    
-
    
-Similarly, using R, we can perform similar studies. 
-(more details on R will be inserted later).
+where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error
+estimate is best to use when targets having exponential growth, such
+as population counts, average sales of a commodity over a span of
+years etc.
 
 
    
 
    
@@ -271,6 +277,10 @@ Similarly, using R, we can perform similar studies.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs031.html b/doc/pub/Regression/html/._Regression-bs031.html
index 3ad7cabb6..3033d8063 100644
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@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,47 +232,51 @@ MathJax.Hub.Config({
 
 
 
-
    Simple regression model with gradient descent  
    
-Add info about the equations, play around with different learning rates
+
    Cubic polynomial in  scikit-learn 
    
+
+
    
+We will discuss in more
+detail these and other functions in the various lectures.  We conclude this part with another example. Instead of 
+a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. 
+Add description of the various python commands.
+
 
    
 
 
-
    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
+
    import matplotlib.pyplot as plt
 import numpy as np
-import matplotlib.pyplot as plt
+import random
+from sklearn.linear_model import Ridge
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.pipeline import make_pipeline
+from sklearn.linear_model import LinearRegression
 
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
+x=np.linspace(0.02,0.98,200)
+noise = np.asarray(random.sample((range(200)),200))
+y=x**3*noise
+yn=x**3*100
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = LinearRegression()
+clf3.fit(X,y)
 
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
+Xplot=poly3.fit_transform(x[:,np.newaxis])
+poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
+plt.plot(x,yn, color='red', label="True Cubic")
+plt.scatter(x, y, label='Data', color='orange', s=15)
+plt.legend()
 plt.show()
+
+def error(a):
+    for i in y:
+        err=(y-yn)/yn
+    return abs(np.sum(err))/len(err)
+
+print (error(y))
 
    
+
    
+Using R, we can perform similar studies.
+
 
    
 
    
 
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diff --git a/doc/pub/Regression/html/._Regression-bs032.html b/doc/pub/Regression/html/._Regression-bs032.html
index cae569fdf..0f9b89474 100644
--- a/doc/pub/Regression/html/._Regression-bs032.html
+++ b/doc/pub/Regression/html/._Regression-bs032.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,8 +232,7 @@ MathJax.Hub.Config({
 
 
 
-
    Simple regression model with stochastic gradient descent  
    
-Add info about the equations, play around with different learning rates
+
    Polynomial Regression 
    
 
    
 
 
@@ -213,17 +241,24 @@ Add info about the equations, play around with different learning rates
 from random import random, seed
 import numpy as np
 import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
 
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
+m = 100
+x = 2*np.random.rand(m,1)+4.
+y = 4+3*x*x+ +x-np.random.randn(m,1)
 
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
+xb = np.c_[np.ones((m,1)), x]
+theta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
+xnew = np.array([[0],[2]])
+xbnew = np.c_[np.ones((2,1)), xnew]
+ypredict = xbnew.dot(theta)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,2.0,0, 15.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Random numbers ')
+plt.show()
 
    
 
    
 
    
@@ -246,6 +281,12 @@ sgdreg.fit(x,y.
   
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diff --git a/doc/pub/Regression/html/._Regression-bs033.html b/doc/pub/Regression/html/._Regression-bs033.html
index 21186180b..d55114e2c 100644
--- a/doc/pub/Regression/html/._Regression-bs033.html
+++ b/doc/pub/Regression/html/._Regression-bs033.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -201,36 +230,32 @@ MathJax.Hub.Config({
 
     
     
     
     
 
 
-
+
+
+
    Linking the regression analysis with a statistical interpretation 
    
 
-
    Polynomial Regression 
    
 
    
+Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way.
 
-
-
    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
+
    
+It is assumed that \( \varepsilon_i
+\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are
+independent, i.e.: 
+$$
+\begin{align*} 
+\mbox{Cov}(\varepsilon_{i_1},
+\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if}
+& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2.  \end{array} \right.
+\end{align*} 
+$$
 
-m = 100
-x = 2*np.random.rand(m,1)+4.
-y = 4+3*x*x+ +x-np.random.randn(m,1)
+The randomness of \( \varepsilon_i \) implies that
+\( \mathbf{Y}_i \) is also a random variable. In particular,
+\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim
+\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a
+non-random scalar. To specify the parameters of the distribution of
+\( \mathbf{Y}_i \) we need to calculate its first two moments.
 
-xb = np.c_[np.ones((m,1)), x]
-theta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    
 
    
 
    
 
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diff --git a/doc/pub/Regression/html/._Regression-bs034.html b/doc/pub/Regression/html/._Regression-bs034.html
index 364fa6719..a5c44e236 100644
--- a/doc/pub/Regression/html/._Regression-bs034.html
+++ b/doc/pub/Regression/html/._Regression-bs034.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -203,84 +232,37 @@ MathJax.Hub.Config({
 
 
 
-
    Ridge and Lasso Regression 
    
+
    Expectation value and variance 
    
 
 
    
+Its expectation equals: 
+$$
+\begin{align*} 
+\mathbb{E}(Y_i) & =
+\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i)
+\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, 
+\end{align*} 
+$$
 
-
-
    import matplotlib.pyplot as plt
-import numpy as np
-from sklearn import linear_model
-from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score
+while
+its variance is 
+$$
+\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i
+- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) -
+[\mathbb{E}(Y_i)]^2  \\  & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \,
+\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ &
+= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i
+\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i,
+\ast} \, \beta)^2 \\  & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2
+\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta +
+\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 
+\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \,
+\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2.  
+\end{align*}
+$$
 
-#creating data with random noise
-x=np.arange(50)
+Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \).
 
-delta=np.random.uniform(-2.5,2.5, size=(50))
-np.random.shuffle(delta)
-y =0.5*x+5+delta
-
-#arranging data into 2x50 matrix
-a=np.array(x) #inputs
-b=np.array(y) #outputs
-
-#Split into training and test
-X_train=a[:37, np.newaxis]
-X_test=a[37:, np.newaxis]
-y_train=b[:37]
-y_test=b[37:]
-
-print ("X_train: ", X_train.shape)
-print ("y_train: ", y_train.shape)
-print ("X_test: ", X_test.shape)
-print ("y_test: ", y_test.shape)
-
-print ("------------------------------------")
-
-print ("Ordinary Least Squares")
-#Add Ordinary Least Squares fit
-reg=LinearRegression()
-reg.fit(X_train, y_train)
-pred=reg.predict(X_test)
-print ("Prediction Shape: ", pred.shape)
-
-print('Coefficients: \n', reg.coef_)
-# The mean squared error
-print("Mean squared error: %.2f"
-      % mean_squared_error(y_test, pred))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(y_test, pred))
-
-#plot
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, pred, color='black', label="Fit Line")
-plt.legend()
-plt.show()
-
-print ("------------------------------------")
-
-print ("Ridge Regression")
-
-ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
-ridge.fit(X_train,y_train)
-print ("Ridge Coefficient: ",ridge.coef_)
-print ("Ridge Intercept: ", ridge.intercept_)
-#Look into graphing with Ridge fit
-
-print ("------------------------------------")
-
-print ("Lasso")
-lasso=linear_model.Lasso(alpha=0.1)
-lasso.fit(X_train,y_train)
-predl=lasso.predict(X_test)
-print("Lasso Coefficient: ", lasso.coef_)
-print("Lasso Intercept: ", lasso.intercept_)
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, predl, color='blue', label="Lasso")
-plt.legend()
-plt.show()
-
    
 
    
 
    
 
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diff --git a/doc/pub/Regression/html/._Regression-bs035.html b/doc/pub/Regression/html/._Regression-bs035.html
index 476608042..3902c8b08 100644
--- a/doc/pub/Regression/html/._Regression-bs035.html
+++ b/doc/pub/Regression/html/._Regression-bs035.html
@@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source
 
 
 
@@ -152,42 +173,50 @@ MathJax.Hub.Config({
         Contents 
         
       
@@ -207,16 +236,20 @@ MathJax.Hub.Config({
 
    
 
    
 
     
-How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
-$$
-\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
-$$
-
    
-
    
-
 
 
    
-Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue
+A general
+\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal
+matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal
+matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality
+\( m \times m \) and the last dimensionality \( n\times n \). 
+We have then 
+$$ 
+\hat{A} = \hat{U}\hat{D}\hat{V}^T 
+$$
+
    + +

    @@ -236,6 +269,15 @@ Add codes and discuss this in connection with lasso and ridge, show example wher

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    • diff --git a/doc/pub/Regression/html/._Regression-bs036.html b/doc/pub/Regression/html/._Regression-bs036.html index b5551cea1..a2a01b043 100644 --- a/doc/pub/Regression/html/._Regression-bs036.html +++ b/doc/pub/Regression/html/._Regression-bs036.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,11 +232,84 @@ MathJax.Hub.Config({ -

    Lasso and Ridge regression

    +

    Code examples for Ridge and Lasso Regression

    -Discuss the mathematics here + +

    import matplotlib.pyplot as plt
+import numpy as np
+from sklearn import linear_model
+from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score
+
+#creating data with random noise
+x=np.arange(50)
+
+delta=np.random.uniform(-2.5,2.5, size=(50))
+np.random.shuffle(delta)
+y =0.5*x+5+delta
+
+#arranging data into 2x50 matrix
+a=np.array(x) #inputs
+b=np.array(y) #outputs
+
+#Split into training and test
+X_train=a[:37, np.newaxis]
+X_test=a[37:, np.newaxis]
+y_train=b[:37]
+y_test=b[37:]
+
+print ("X_train: ", X_train.shape)
+print ("y_train: ", y_train.shape)
+print ("X_test: ", X_test.shape)
+print ("y_test: ", y_test.shape)
+
+print ("------------------------------------")
+
+print ("Ordinary Least Squares")
+#Add Ordinary Least Squares fit
+reg=LinearRegression()
+reg.fit(X_train, y_train)
+pred=reg.predict(X_test)
+print ("Prediction Shape: ", pred.shape)
+
+print('Coefficients: \n', reg.coef_)
+# The mean squared error
+print("Mean squared error: %.2f"
+      % mean_squared_error(y_test, pred))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(y_test, pred))
+
+#plot
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, pred, color='black', label="Fit Line")
+plt.legend()
+plt.show()
+
+print ("------------------------------------")
+
+print ("Ridge Regression")
+
+ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
+ridge.fit(X_train,y_train)
+print ("Ridge Coefficient: ",ridge.coef_)
+print ("Ridge Intercept: ", ridge.intercept_)
+#Look into graphing with Ridge fit
+
+print ("------------------------------------")
+
+print ("Lasso")
+lasso=linear_model.Lasso(alpha=0.1)
+lasso.fit(X_train,y_train)
+predl=lasso.predict(X_test)
+print("Lasso Coefficient: ", lasso.coef_)
+print("Lasso Intercept: ", lasso.intercept_)
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, predl, color='blue', label="Lasso")
+plt.legend()
+plt.show()
+

    @@ -225,6 +327,14 @@ Discuss the mathematics here

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    • diff --git a/doc/pub/Regression/html/._Regression-bs037.html b/doc/pub/Regression/html/._Regression-bs037.html index 1a3204fc2..d7790e7d6 100644 --- a/doc/pub/Regression/html/._Regression-bs037.html +++ b/doc/pub/Regression/html/._Regression-bs037.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -203,14 +232,55 @@ MathJax.Hub.Config({ -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    From standard regression to Ridge regressions

    -Add examples on classification problems +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. + +

    diff --git a/doc/pub/Regression/html/._Regression-bs038.html b/doc/pub/Regression/html/._Regression-bs038.html new file mode 100644 index 000000000..cca079168 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs038.html @@ -0,0 +1,307 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fixing the singularity

    + +

    +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\tag{1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs039.html b/doc/pub/Regression/html/._Regression-bs039.html new file mode 100644 index 000000000..aa8c96dc3 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs039.html @@ -0,0 +1,392 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs040.html b/doc/pub/Regression/html/._Regression-bs040.html new file mode 100644 index 000000000..e62c176d7 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs040.html @@ -0,0 +1,304 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fitting vs. predicting when data is in the model class

    + +

    +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs041.html b/doc/pub/Regression/html/._Regression-bs041.html new file mode 100644 index 000000000..0449005ac --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs041.html @@ -0,0 +1,291 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs042.html b/doc/pub/Regression/html/._Regression-bs042.html new file mode 100644 index 000000000..ff3f3130a --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs042.html @@ -0,0 +1,365 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs043.html b/doc/pub/Regression/html/._Regression-bs043.html new file mode 100644 index 000000000..b45774529 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs043.html @@ -0,0 +1,327 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs044.html b/doc/pub/Regression/html/._Regression-bs044.html new file mode 100644 index 000000000..1ad5ddece --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs044.html @@ -0,0 +1,278 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Lasso regression

    + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs045.html b/doc/pub/Regression/html/._Regression-bs045.html new file mode 100644 index 000000000..59fd87d05 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs045.html @@ -0,0 +1,275 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    + + +
    + +

     

     

     

    + + + + +

    Logistic regression

    + +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/Regression-bs.html b/doc/pub/Regression/html/Regression-bs.html index 9560a3323..c92c12f8a 100644 --- a/doc/pub/Regression/html/Regression-bs.html +++ b/doc/pub/Regression/html/Regression-bs.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents @@ -222,7 +251,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018


    @@ -246,7 +275,7 @@ MathJax.Hub.Config({

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    • diff --git a/doc/pub/Regression/html/Regression-reveal.html b/doc/pub/Regression/html/Regression-reveal.html index d5aaa438c..13e0c45e3 100644 --- a/doc/pub/Regression/html/Regression-reveal.html +++ b/doc/pub/Regression/html/Regression-reveal.html @@ -148,7 +148,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

     
    -

    Aug 24, 2018

    +

    Sep 6, 2018


    @@ -182,7 +182,35 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)

    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    +
    + + +
    +

    General linear models

    @@ -204,7 +232,7 @@ where \( \epsilon_i \) is the error in our approximation.

    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -225,7 +253,7 @@ $$

    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -275,7 +303,7 @@ $$

    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -299,7 +327,7 @@ $$

    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -330,7 +358,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo

    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -353,7 +381,7 @@ $$

    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -382,7 +410,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -430,7 +458,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -461,7 +489,7 @@ $$

    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -494,7 +522,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -517,7 +545,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -548,7 +576,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -577,7 +605,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -613,7 +641,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -642,7 +670,7 @@ $$

    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -698,7 +726,7 @@ This approach (different linear and non-linear regression) suffers often from bo

    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -738,7 +766,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r

    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -770,7 +798,7 @@ plt.show()

    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -841,7 +869,7 @@ plt.show()

    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -865,7 +893,7 @@ where \( x \) is defined as before.

    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -878,7 +906,7 @@ have not discussed a more rigorous approach to the cost function.

    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -902,7 +930,7 @@ dimensionless.

    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -923,7 +951,7 @@ the \( \chi^2 \) function becomes smaller.

    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -965,7 +993,7 @@ relative error.

    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -1011,7 +1039,7 @@ plt.show()

    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -1030,7 +1058,7 @@ this function as being similar to the \( \chi^2 \) function defined above.

    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -1058,7 +1086,7 @@ $$

    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1087,12 +1115,13 @@ years etc.

    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1129,84 +1158,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.

    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -
    - - -
    -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -
    - - -
    -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1238,7 +1195,94 @@ plt.show()

    -

    Ridge and Lasso Regression

    +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +

     
    +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ +

     
    + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +

    + + +
    +

    Expectation value and variance

    + +

    +Its expectation equals: +

     
    +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ +

     
    + +while +its variance is +

     
    +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ +

     
    + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). +

    + + +
    +

    The singular value decompostion

    +
    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +

     
    +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

     
    +

    +
    + + +
    +

    Code examples for Ridge and Lasso Regression

    @@ -1320,37 +1364,403 @@ plt.show()

    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -

     
    -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

     
    -

    +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +

     
    +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ +

     
    + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +

     
    +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ +

     
    + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.

    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +

     
    +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\tag{1} +\end{align} +$$ +

     
    + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +

     
    +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ +

     
    + +where \( \hat{I} \) is the identity matrix.

    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +
    + + +
    +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +

    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +

    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +

    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +

    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    +

    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. +

    + + +
    +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +

    1. Do better fits lead to better predictions?
      2. +

    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    +

    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. +

    + + +
    +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +
    + + +
    +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +
    + + +
    +

    Lasso regression

    +
    + + +
    +

    Logistic regression

    diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html index 9b2167152..9b2031654 100644 --- a/doc/pub/Regression/html/Regression-solarized.html +++ b/doc/pub/Regression/html/Regression-solarized.html @@ -61,17 +61,14 @@ div { text-align: justify; text-justify: inter-word; } @@ -175,7 +196,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018












    @@ -205,7 +226,38 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)











    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    + + +

    +









    + +

    General linear models

    @@ -226,7 +278,7 @@ where \( \epsilon_i \) is the error in our approximation.











    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -246,7 +298,7 @@ $$











    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -287,7 +339,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -310,7 +362,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -338,7 +390,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo











    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -360,7 +412,7 @@ $$











    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -384,7 +436,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -423,7 +475,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -449,7 +501,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -477,7 +529,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -500,7 +552,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -528,7 +580,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -554,7 +606,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -585,7 +637,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -609,7 +661,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -655,7 +707,7 @@ This approach (different linear and non-linear regression) suffers often from bo











    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -695,7 +747,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r











    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -726,7 +778,7 @@ plt.show()











    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -794,7 +846,7 @@ plt.show()











    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -816,7 +868,7 @@ where \( x \) is defined as before.











    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -829,7 +881,7 @@ have not discussed a more rigorous approach to the cost function.











    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -851,7 +903,7 @@ dimensionless.











    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -872,7 +924,7 @@ the \( \chi^2 \) function becomes smaller.











    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -912,7 +964,7 @@ relative error.











    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -957,7 +1009,7 @@ plt.show()











    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -974,7 +1026,7 @@ this function as being similar to the \( \chi^2 \) function defined above.











    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -998,7 +1050,7 @@ $$











    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1023,12 +1075,13 @@ years etc.











    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1065,82 +1118,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.











    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -

    -









    - -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -

    -









    - -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1168,10 +1151,92 @@ plt.ylabel(r'$y$') plt.title(r'Random numbers ') plt.show()

    +

    + + +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +











    -

    Ridge and Lasso Regression

    +

    Expectation value and variance

    + +

    +Its expectation equals: +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ + +while +its variance is +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). + +

    +









    + +

    The singular value decompostion

    +
    + +

    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

    + + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1252,38 +1317,390 @@ plt.show()











    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

    - +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.











    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix.











    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +









    + +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \).

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +









    + +

    Lasso regression

    + +

    +









    + +

    Logistic regression

    diff --git a/doc/pub/Regression/html/Regression.html b/doc/pub/Regression/html/Regression.html index 7e1a6bbe5..fd86d2474 100644 --- a/doc/pub/Regression/html/Regression.html +++ b/doc/pub/Regression/html/Regression.html @@ -66,17 +66,14 @@ div { text-align: justify; text-justify: inter-word; } @@ -180,7 +201,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Aug 24, 2018

    +

    Sep 6, 2018












    @@ -210,7 +231,38 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)











    -

    General linear models

    +

    Regression analysis, overarching aims II

    +
    + +

    + +

    +Consider an experiment in which \( p \) characteristics of \( n \) samples are +measured. The data from this experiment are denoted \( \mathbf{X} \), with +\( \mathbf{X} \) as above. The matrix \( \mathbf{X} \) is called the design +matrix. Additional information of the samples is available in the +form of \( \mathbf{Y} \) (also as above). The variable \( \mathbf{Y} \) is +generally referred to as the response variable. The aim of +regression analysis is to explain \( \mathbf{Y} \) in terms of +\( \mathbf{X} \) through a functional relationship like \( Y_i = +f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of +\( f(\cdot) \) is available, it is common to assume a linear relationship +between \( \mathbf{X} \) and \( \mathbf{Y} \). This assumption gives rise to +the linear regression model where \( \beta = (\beta_1, \ldots, +\beta_p)^{\top} \) is the regression parameter. The parameter +\( \beta_j \), \( j=1, \ldots, p \), represents the effect size of covariate +\( j \) on the response. That is, for each unit change in covariate \( j \) +(while keeping the other covariates fixed) the observed change in the +response is equal to \( \beta_j \). + + +

    + + +

    +









    + +

    General linear models

    @@ -231,7 +283,7 @@ where \( \epsilon_i \) is the error in our approximation.











    -

    Rewriting the fitting procedure as a linear algebra problem

    +

    Rewriting the fitting procedure as a linear algebra problem

    @@ -251,7 +303,7 @@ $$











    -

    Rewriting the fitting procedure as a linear algebra problem, follows

    +

    Rewriting the fitting procedure as a linear algebra problem, follows

    @@ -292,7 +344,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -315,7 +367,7 @@ $$











    -

    Generalizing the fitting procedure as a linear algebra problem

    +

    Generalizing the fitting procedure as a linear algebra problem

    @@ -343,7 +395,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo











    -

    Optimizing our parameters

    +

    Optimizing our parameters

    @@ -365,7 +417,7 @@ $$











    -

    Optimizing our parameters, more details

    +

    Optimizing our parameters, more details

    @@ -389,7 +441,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -428,7 +480,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -454,7 +506,7 @@ $$











    -

    Interpretations and optimizing our parameters

    +

    Interpretations and optimizing our parameters

    @@ -482,7 +534,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -505,7 +557,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -533,7 +585,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -559,7 +611,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -590,7 +642,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -614,7 +666,7 @@ $$











    -

    The \( \chi^2 \) function

    +

    The \( \chi^2 \) function

    @@ -660,7 +712,7 @@ This approach (different linear and non-linear regression) suffers often from bo











    -

    Simple regression model

    +

    Simple regression model

    We are now ready to write our first program which aims at solving the above linear regression equations. We start with data we have produced ourselves, in this case normally distributed random numbers along the \( x \)-axis. These numbers define then the value of a function \( y(x)=4+3x+N(0,1) \). Thereafter we order the \( x \) values and employ our linear regression algorithm to set up the best fit. Here we find it useful to use the numpy function \( c\_ \) arrays where arrays are stacked along their last axis after being upgraded to at least two dimensions with ones post-pended to the shape. The following examples help in understanding what happens

    @@ -700,7 +752,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r











    -

    Simple regression model, now using scikit-learn

    +

    Simple regression model, now using scikit-learn

    We can repeat the above algorithm using scikit-learn as follows @@ -731,7 +783,7 @@ plt.show()











    -

    Simple linear regression model using scikit-learn

    +

    Simple linear regression model using scikit-learn

    We start with perhaps our simplest possible example, using scikit-learn to perform linear regression analysis on a data set produced by us. @@ -799,7 +851,7 @@ plt.show()











    -

    Simple linear regression model

    +

    Simple linear regression model

    This example serves several aims. It allows us to demonstrate several @@ -821,7 +873,7 @@ where \( x \) is defined as before.











    -

    Less noise

    +

    Less noise

    Does the fit look better? Indeed, by @@ -834,7 +886,7 @@ have not discussed a more rigorous approach to the cost function.











    -

    How to study our fits

    +

    How to study our fits

    We need more rigorous criteria in defining whether we have succeeded or @@ -856,7 +908,7 @@ dimensionless.











    -

    Minimizing the cost function

    +

    Minimizing the cost function

    Minimizing the cost function is a central aspect of @@ -877,7 +929,7 @@ the \( \chi^2 \) function becomes smaller.











    -

    Relative error

    +

    Relative error

    There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define @@ -917,7 +969,7 @@ relative error.











    -

    The richness of scikit-learn

    +

    The richness of scikit-learn

    As mentioned above, scikit-learn has an impressive functionality. @@ -962,7 +1014,7 @@ plt.show()











    -

    Functions in scikit-learn

    +

    Functions in scikit-learn

    The function coef gives us the parameter \( \beta \) of our fit while intercept yields @@ -979,7 +1031,7 @@ this function as being similar to the \( \chi^2 \) function defined above.











    -

    Other functions in scikit-learn

    +

    Other functions in scikit-learn

    The r2score function computes \( R^2 \), the coefficient of @@ -1003,7 +1055,7 @@ $$











    -

    The mean absolute error and other functions in scikit-learn

    +

    The mean absolute error and other functions in scikit-learn

    Another quantity will meet again in our discussions of regression analysis is @@ -1028,12 +1080,13 @@ years etc.











    -

    Cubic polynomial in scikit-learn

    +

    Cubic polynomial in scikit-learn

    We will discuss in more detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. +Add description of the various python commands.

    @@ -1070,82 +1123,12 @@ plt.show() print (error(y))

    -Similarly, using R, we can perform similar studies. -(more details on R will be inserted later). +Using R, we can perform similar studies.











    -

    Simple regression model with gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-theta = np.random.randn(2,1)
-
-eta = 0.1
-Niterations = 1000
-m = 100
-
-for iter in range(Niterations):
-    gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)
-    theta -= eta*gradients
-
-print(theta)
-xnew = np.array([[0],[2]])
-xbnew = np.c_[np.ones((2,1)), xnew]
-ypredict = xbnew.dot(theta)
-ypredict2 = xbnew.dot(theta_linreg)
-plt.plot(xnew, ypredict, "r-")
-plt.plot(xnew, ypredict2, "b-")
-plt.plot(x, y ,'ro')
-plt.axis([0,2.0,0, 15.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Random numbers ')
-plt.show()
-
    -

    -









    - -

    Simple regression model with stochastic gradient descent

    -Add info about the equations, play around with different learning rates -

    - - -

    # Importing various packages
-from math import exp, sqrt
-from random import random, seed
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import SGDRegressor
-
-x = 2*np.random.rand(100,1)
-y = 4+3*x+np.random.randn(100,1)
-
-xb = np.c_[np.ones((100,1)), x]
-theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
-print(theta_linreg)
-sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
-sgdreg.fit(x,y.ravel())
-print(sgdreg.intercept_, sgdreg.coef_)
-
    -

    -









    - -

    Polynomial Regression

    +

    Polynomial Regression

    @@ -1173,10 +1156,92 @@ plt.ylabel(r plt.title(r'Random numbers ') plt.show()

    +

    + + +

    Linking the regression analysis with a statistical interpretation

    + +

    +Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. + +

    +It is assumed that \( \varepsilon_i +\sim \mathcal{N}(0, \sigma^2) \) and the \( \varepsilon_{i} \) are +independent, i.e.: +$$ +\begin{align*} +\mbox{Cov}(\varepsilon_{i_1}, +\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} +& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. +\end{align*} +$$ + +The randomness of \( \varepsilon_i \) implies that +\( \mathbf{Y}_i \) is also a random variable. In particular, +\( \mathbf{Y}_i \) is normally distributed, because \( \varepsilon_i \sim +\mathcal{N}(0, \sigma^2) \) and \( \mathbf{X}_{i,\ast} \, \beta \) is a +non-random scalar. To specify the parameters of the distribution of +\( \mathbf{Y}_i \) we need to calculate its first two moments. +











    -

    Ridge and Lasso Regression

    +

    Expectation value and variance

    + +

    +Its expectation equals: +$$ +\begin{align*} +\mathbb{E}(Y_i) & = +\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) +\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, +\end{align*} +$$ + +while +its variance is +$$ +\begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i +- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - +[\mathbb{E}(Y_i)]^2 \\ & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, +\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & += \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i +\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, +\ast} \, \beta)^2 \\ & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 +\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + +\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 +\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, +\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. +\end{align*} +$$ + +Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \). + +

    +









    + +

    The singular value decompostion

    +
    + +

    + +

    +A general +\( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal +matrix \( \hat{D} \) of dimensionality \( n\times n \) and two orthognal +matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality +\( m \times m \) and the last dimensionality \( n\times n \). +We have then +$$ +\hat{A} = \hat{U}\hat{D}\hat{V}^T +$$ +

    + + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1257,38 +1322,390 @@ plt.show()











    -

    The singular value decompostion

    -
    - -

    -How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then -$$ -\hat{A} = \hat{U}\hat{\Sigma}\hat{V} -$$ -

    - +

    From standard regression to Ridge regressions

    -Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.











    -

    Lasso and Ridge regression

    +

    Fixing the singularity

    -Discuss the mathematics here +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix.











    -

    Logistic regression

    -Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors +

    A second-order polynomial with Ridge and Lasso

    +

    + + +

    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    +

    +









    + +

    Fitting vs. predicting when data is in the model class

    -Add examples on classification problems +We start by considering the case +\( f(x)=2x \).

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    The code

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
+#Linear Filename
+#filename_test=Title+"pred-linear.pdf"
+#Tenth Order Filename
+#filename_test=Title+"pred-o10.pdf"
+#plt.savefig(filename_test)
+#plt.ylim((-6,12))
+
    +

    +









    + +

    Lasso regression

    + +

    +









    + +

    Logistic regression

    diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index a4a85095f..af7e36ea4 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -10,7 +10,7 @@ " \n", "**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University\n", "\n", - "Date: **Aug 24, 2018**\n", + "Date: **Sep 6, 2018**\n", "\n", "Copyright 1999-2018, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license\n", "\n", @@ -33,6 +33,28 @@ "\n", "\n", "\n", + "## Regression analysis, overarching aims II\n", + "\n", + "\n", + "Consider an experiment in which $p$ characteristics of $n$ samples are\n", + "measured. The data from this experiment are denoted $\\mathbf{X}$, with\n", + "$\\mathbf{X}$ as above. The matrix $\\mathbf{X}$ is called the *design\n", + "matrix*. Additional information of the samples is available in the\n", + "form of $\\mathbf{Y}$ (also as above). The variable $\\mathbf{Y}$ is\n", + "generally referred to as the *response variable*. The aim of\n", + "regression analysis is to explain $\\mathbf{Y}$ in terms of\n", + "$\\mathbf{X}$ through a functional relationship like $Y_i =\n", + "f(\\mathbf{X}_{i,\\ast})$. When no prior knowledge on the form of\n", + "$f(\\cdot)$ is available, it is common to assume a linear relationship\n", + "between $\\mathbf{X}$ and $\\mathbf{Y}$. This assumption gives rise to\n", + "the *linear regression model* where $\\beta = (\\beta_1, \\ldots,\n", + "\\beta_p)^{\\top}$ is the *regression parameter*. The parameter\n", + "$\\beta_j$, $j=1, \\ldots, p$, represents the effect size of covariate\n", + "$j$ on the response. That is, for each unit change in covariate $j$\n", + "(while keeping the other covariates fixed) the observed change in the\n", + "response is equal to $\\beta_j$.\n", + "\n", + "\n", "\n", "## General linear models\n", "Before we proceed let us study a case from linear algebra where we aim at fitting a set of data $\\hat{y}=[y_0,y_1,\\dots,y_{n-1}]$. We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables $\\hat{x}=[x_0,x_1,\\dots,x_{n-1}]$, that is $y_i = y(x_i)$ with $i=0,1,2,\\dots,n-1$. The variables $x_i$ could represent physical quantities like time, temperature, position etc. We assume that $y(x)$ is a smooth function. \n", @@ -897,19 +919,10 @@ { "cell_type": "code", "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "[[1 4]\n", - " [2 5]\n", - " [3 6]]\n", - "[[1 2 3 0 0 4 5 6]]\n" - ] - } - ], + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np\n", "print(np.c_[np.array([1,2,3]), np.array([4,5,6])])\n", @@ -918,20 +931,11 @@ }, { "cell_type": "code", - "execution_count": 15, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "%matplotlib inline\n", "\n", @@ -941,7 +945,7 @@ "import matplotlib.pyplot as plt\n", "\n", "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+0.01*np.random.randn(100,1)\n", + "y = 4+3*x+np.random.randn(100,1)\n", "\n", "xb = np.c_[np.ones((100,1)), x]\n", "beta = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", @@ -975,20 +979,11 @@ }, { "cell_type": "code", - "execution_count": 17, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "# Importing various packages\n", "from random import random, seed\n", @@ -997,7 +992,7 @@ "from sklearn.linear_model import LinearRegression\n", "\n", "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+0.01*np.random.randn(100,1)\n", + "y = 4+3*x+np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "xnew = np.array([[0],[2]])\n", @@ -1068,20 +1063,11 @@ }, { "cell_type": "code", - "execution_count": 11, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "# Importing various packages\n", "import numpy as np\n", @@ -1216,27 +1202,18 @@ }, { "cell_type": "code", - "execution_count": 19, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", "from sklearn.linear_model import LinearRegression\n", "\n", "x = np.random.rand(100,1)\n", - "y = 5*x+np.random.randn(100,1)\n", + "y = 5*x+0.01*np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "ypredict = linreg.predict(x)\n", @@ -1272,34 +1249,11 @@ }, { "cell_type": "code", - "execution_count": 21, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "The intercept alpha: \n", - " [2.00137415]\n", - "Coefficient beta : \n", - " [[4.99698001]]\n", - "Mean squared error: 0.00\n", - "Variance score: 1.00\n", - "Mean squared log error: 0.00\n", - "Mean absolute error: 0.01\n" - ] - }, - { - "data": { - "image/png": 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- "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np \n", "import matplotlib.pyplot as plt \n", @@ -1307,7 +1261,7 @@ "from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error\n", "\n", "x = np.random.rand(100,1)\n", - "y = 2.0+ 5*x+0.01*np.random.randn(100,1)\n", + "y = 2.0+ 5*x+0.5*np.random.randn(100,1)\n", "linreg = LinearRegression()\n", "linreg.fit(x,y)\n", "ypredict = linreg.predict(x)\n", @@ -1446,32 +1400,17 @@ "\n", "We will discuss in more\n", "detail these and other functions in the various lectures. We conclude this part with another example. Instead of \n", - "a linear $x$-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn." + "a linear $x$-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. \n", + "Add description of the various python commands." ] }, { "cell_type": "code", - "execution_count": 22, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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\n", 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    " - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "0.004999999999999991\n" - ] - } - ], + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", @@ -1509,110 +1448,22 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "Similarly, using **R**, we can perform similar studies. \n", - "(more details on **R** will be inserted later).\n", + "Using **R**, we can perform similar studies. \n", "\n", "\n", "\n", "\n", "\n", "\n", - "\n", - "\n", - "\n", - "\n", - "\n", - "\n", - "## Simple regression model with gradient descent\n", - "Add info about the equations, play around with different learning rates" - ] - }, - { - "cell_type": "code", - "execution_count": 8, - "metadata": {}, - "outputs": [], - "source": [ - "# Importing various packages\n", - "from math import exp, sqrt\n", - "from random import random, seed\n", - "import numpy as np\n", - "import matplotlib.pyplot as plt\n", - "\n", - "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+np.random.randn(100,1)\n", - "\n", - "xb = np.c_[np.ones((100,1)), x]\n", - "theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", - "print(theta_linreg)\n", - "theta = np.random.randn(2,1)\n", - "\n", - "eta = 0.1\n", - "Niterations = 1000\n", - "m = 100\n", - "\n", - "for iter in range(Niterations):\n", - " gradients = 2.0/m*xb.T.dot(xb.dot(theta)-y)\n", - " theta -= eta*gradients\n", - "\n", - "print(theta)\n", - "xnew = np.array([[0],[2]])\n", - "xbnew = np.c_[np.ones((2,1)), xnew]\n", - "ypredict = xbnew.dot(theta)\n", - "ypredict2 = xbnew.dot(theta_linreg)\n", - "plt.plot(xnew, ypredict, \"r-\")\n", - "plt.plot(xnew, ypredict2, \"b-\")\n", - "plt.plot(x, y ,'ro')\n", - "plt.axis([0,2.0,0, 15.0])\n", - "plt.xlabel(r'$x$')\n", - "plt.ylabel(r'$y$')\n", - "plt.title(r'Random numbers ')\n", - "plt.show()" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Simple regression model with stochastic gradient descent\n", - "Add info about the equations, play around with different learning rates" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": {}, - "outputs": [], - "source": [ - "# Importing various packages\n", - "from math import exp, sqrt\n", - "from random import random, seed\n", - "import numpy as np\n", - "import matplotlib.pyplot as plt\n", - "from sklearn.linear_model import SGDRegressor\n", - "\n", - "x = 2*np.random.rand(100,1)\n", - "y = 4+3*x+np.random.randn(100,1)\n", - "\n", - "xb = np.c_[np.ones((100,1)), x]\n", - "theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)\n", - "print(theta_linreg)\n", - "sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)\n", - "sgdreg.fit(x,y.ravel())\n", - "print(sgdreg.intercept_, sgdreg.coef_)" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ "## Polynomial Regression" ] }, { "cell_type": "code", - "execution_count": 10, - "metadata": {}, + "execution_count": 8, + "metadata": { + "collapsed": false + }, "outputs": [], "source": [ "# Importing various packages\n", @@ -1644,72 +1495,131 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Ridge and Lasso Regression" + "\n", + "## Linking the regression analysis with a statistical interpretation\n", + "\n", + "Before we proceed, and to link with our discussions of Bayesian statistics to come, it is useful the derive the standard regression analysis equations using a statistical interpretation. This allows us also to derive quantities like the variance and other expectation values in a rather straightforward way. \n", + "\n", + "It is assumed that $\\varepsilon_i\n", + "\\sim \\mathcal{N}(0, \\sigma^2)$ and the $\\varepsilon_{i}$ are\n", + "independent, i.e.:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \n", + "\\mbox{Cov}(\\varepsilon_{i_1},\n", + "\\varepsilon_{i_2}) & = \\left\\{ \\begin{array}{lcc} \\sigma^2 & \\mbox{if}\n", + "& i_1 = i_2, \\\\ 0 & \\mbox{if} & i_1 \\not= i_2. \\end{array} \\right.\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The randomness of $\\varepsilon_i$ implies that\n", + "$\\mathbf{Y}_i$ is also a random variable. In particular,\n", + "$\\mathbf{Y}_i$ is normally distributed, because $\\varepsilon_i \\sim\n", + "\\mathcal{N}(0, \\sigma^2)$ and $\\mathbf{X}_{i,\\ast} \\, \\beta$ is a\n", + "non-random scalar. To specify the parameters of the distribution of\n", + "$\\mathbf{Y}_i$ we need to calculate its first two moments. \n", + "\n", + "## Expectation value and variance\n", + "\n", + "Its expectation equals:" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \n", + "\\mathbb{E}(Y_i) & =\n", + "\\mathbb{E}(\\mathbf{X}_{i, \\ast} \\, \\beta) + \\mathbb{E}(\\varepsilon_i)\n", + "\\, \\, \\, = \\, \\, \\, \\mathbf{X}_{i, \\ast} \\, \\beta, \n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "while\n", + "its variance is" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*} \\mbox{Var}(Y_i) & = \\mathbb{E} \\{ [Y_i\n", + "- \\mathbb{E}(Y_i)]^2 \\} \\, \\, \\, = \\, \\, \\, \\mathbb{E} ( Y_i^2 ) -\n", + "[\\mathbb{E}(Y_i)]^2 \\\\ & = \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\,\n", + "\\beta + \\varepsilon_i )^2] - ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 \\\\ &\n", + "= \\mathbb{E} [ ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 + 2 \\varepsilon_i\n", + "\\mathbf{X}_{i, \\ast} \\, \\beta + \\varepsilon_i^2 ] - ( \\mathbf{X}_{i,\n", + "\\ast} \\, \\beta)^2 \\\\ & = ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 + 2\n", + "\\mathbb{E}(\\varepsilon_i) \\mathbf{X}_{i, \\ast} \\, \\beta +\n", + "\\mathbb{E}(\\varepsilon_i^2 ) - ( \\mathbf{X}_{i, \\ast} \\, \\beta)^2 \n", + "\\\\ & = \\mathbb{E}(\\varepsilon_i^2 ) \\, \\, \\, = \\, \\, \\,\n", + "\\mbox{Var}(\\varepsilon_i) \\, \\, \\, = \\, \\, \\, \\sigma^2. \n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Hence, $Y_i \\sim \\mathcal{N}( \\mathbf{X}_{i, \\ast} \\, \\beta, \\sigma^2)$. \n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "## The singular value decompostion\n", + "\n", + "\n", + "A general\n", + "$m\\times n$ matrix $\\hat{A}$ can be written in terms of a diagonal\n", + "matrix $\\hat{D}$ of dimensionality $n\\times n$ and two orthognal\n", + "matrices $\\hat{U}$ and $\\hat{V}$, where the first has dimensionality\n", + "$m \\times m$ and the last dimensionality $n\\times n$. \n", + "We have then" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\hat{A} = \\hat{U}\\hat{D}\\hat{V}^T\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Code examples for Ridge and Lasso Regression" ] }, { "cell_type": "code", - "execution_count": 1, - "metadata": {}, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "X_train: (37, 1)\n", - "y_train: (37,)\n", - "X_test: (13, 1)\n", - "y_test: (13,)\n", - "------------------------------------\n", - "Ordinary Least Squares\n", - "Prediction Shape: (13,)\n", - "Coefficients: \n", - " [0.54090544]\n", - "Mean squared error: 4.92\n", - "Variance score: 0.11\n" - ] - }, - { - "name": "stderr", - "output_type": "stream", - "text": [ - "/usr/local/lib/python3.7/site-packages/sklearn/linear_model/base.py:509: RuntimeWarning: internal gelsd driver lwork query error, required iwork dimension not returned. This is likely the result of LAPACK bug 0038, fixed in LAPACK 3.2.2 (released July 21, 2010). Falling back to 'gelss' driver.\n", - " linalg.lstsq(X, y)\n" - ] - }, - { - "data": { - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "------------------------------------\n", - "Ridge Regression\n", - "Ridge Coefficient: [0.54089262]\n", - "Ridge Intercept: 4.6076494432587065\n", - "------------------------------------\n", - "Lasso\n", - "Lasso Coefficient: [0.54002825]\n", - "Lasso Intercept: 4.623208095009552\n" - ] - }, - { - "data": { - "text/plain": [ - "
    " - ] - }, - "metadata": {}, - "output_type": "display_data" - } - ], + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", @@ -1789,8 +1699,14 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## The singular value decompostion\n", - "How can we use the singular value decomposition to find the parameters $\\beta_j$? More details will come. We first note that a general $m\\times n$ matrix $\\hat{A}$ can be written in terms of a diagonal matrix $\\hat{\\Sigma}$ of dimensionality $n\\times n$ and two orthognal matrices $\\hat{U}$ and $\\hat{V}$, where the first has dimensionality $m \\times n$ and the last dimensionality $n\\times n$. We have then" + "## From standard regression to Ridge regressions\n", + "\n", + "One of the typical problems we encounter with linear regression, in particular \n", + "when the matrix $\\hat{X}$ (our so-called design matrix) is high-dimensional, \n", + "are problems with near singular or singular matrices. The column vectors of $\\hat{X}$ \n", + "may be linearly dependent, normally referred to as super-collinearity. \n", + "This means that the matrix may be rank deficient and it is basically impossible to \n", + "to model the data using linear regression. As an example, consider the matrix" ] }, { @@ -1798,7 +1714,18 @@ "metadata": {}, "source": [ "$$\n", - "\\hat{A} = \\hat{U}\\hat{\\Sigma}\\hat{V}\n", + "\\begin{align*}\n", + "\\mathbf{X} & = \\left[\n", + "\\begin{array}{rrr}\n", + "1 & -1 & 2\n", + "\\\\\n", + "1 & 0 & 1\n", + "\\\\\n", + "1 & 2 & -1\n", + "\\\\\n", + "1 & 1 & 0\n", + "\\end{array} \\right]\n", + "\\end{align*}\n", "$$" ] }, @@ -1806,39 +1733,414 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "Add codes and discuss this in connection with lasso and ridge, show example where the standard inversion of a matrix fails and where SVD comes to rescue\n", + "The columns of $\\hat{X}$ are linearly dependent. We se this easily since the \n", + "the first column is the row-wise sum of the other two columns. The rank (more correct,\n", + "the column rank) of a matrix is the dimension of the space spanned by the\n", + "column vectors. Hence, the rank of $\\mathbf{X}$ is equal to the number\n", + "of linearly independent columns. In this particular case the matrix has rank 2.\n", + "\n", + "Super-collinearity of an $(n \\times p)$-dimensional design matrix $\\mathbf{X}$ implies\n", + "that the inverse of the matrix $\\hat{X}^T\\hat{x}$ (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\\hat{X} & = \\left[\n", + "\\begin{array}{rr}\n", + "1 & -1\n", + "\\\\\n", + "1 & -1\n", + "\\end{array} \\right].\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "We see easily that $\\mbox{det}(\\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \\times (-1) - 1 \\times (-1) = 0$. Hence, $\\mathbf{X}$ is singular and its inverse is undefined.\n", + "This is equivalent to saying that the matrix $\\hat{X}$ has at least an eigenvalue which is zero.\n", + "\n", + "## Fixing the singularity\n", + "\n", + "If our design matrix $\\hat{X}$ which enters the linear regression problem" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "
    \n", + "\n", + "$$\n", + "\\begin{equation}\n", + "\\hat{\\beta} = (\\hat{X}^{T} \\hat{X})^{-1} \\hat{X}^{T} \\hat{y},\n", + "\\label{_auto1} \\tag{1}\n", + "\\end{equation}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "has linearly dependent column vectors, we will not be able to compute the inverse\n", + "of $\\hat{X}^T\\hat{X}$ and we cannot find the parameters (estimators) $\\beta_i$. \n", + "The estimators are only well-defined if $(\\hat{X}^{T}\\hat{X})^{-1}$ exits. \n", + "This is more likely to happen when the matrix $\\hat{X}$ is high-dimensional. In this case it is likely to encounter a situation where \n", + "the regression parameters $\\beta_i$ cannot be estimated.\n", + "\n", + "The *ad hoc* approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\hat{X}^{T} \\hat{X} \\rightarrow \\hat{X}^{T} \\hat{X}+\\lambda \\hat{I},\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "where $\\hat{I}$ is the identity matrix.\n", "\n", "\n", - "## Lasso and Ridge regression\n", "\n", - "Discuss the mathematics here\n", + "## A second-order polynomial with Ridge and Lasso" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "from sklearn.linear_model import Ridge\n", + "from sklearn.metrics import r2_score\n", "\n", - "## Logistic regression\n", - "Add discussion about classification versus regression, show examples of more than two cases and why regression is not the best approach. Motivate for k-nearest neighbors\n", + "np.random.seed(4155)\n", "\n", - "Add examples on classification problems" + "n_samples = 100\n", + "\n", + "x = np.random.rand(n_samples,1)\n", + "y = 5*x*x + 0.1*np.random.rand(n_samples,1)\n", + "\n", + "# Centering x and y.\n", + "x_ = x - np.mean(x)\n", + "y_ = y - np.mean(y) # beta_0 = mean(y)\n", + "\n", + "X = np.c_[np.ones((n_samples,1)), x, x**2]\n", + "X_ = np.c_[x_, x_**2]\n", + "\n", + "\n", + "### 1.\n", + "lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]\n", + "num_values = len(lmb_values)\n", + "\n", + "## Ridge-regression of centered and not centered data\n", + "beta_ridge = np.zeros((3,num_values))\n", + "beta_ridge_centered = np.zeros((3,num_values))\n", + "\n", + "I3 = np.eye(3)\n", + "I2 = np.eye(2)\n", + "\n", + "for i,lmb in enumerate(lmb_values):\n", + " beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()\n", + " beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()\n", + "\n", + "# sett beta_0 = np.mean(y)\n", + "beta_ridge_centered[0,:] = np.mean(y)\n", + "\n", + "## OLS (ordinary least squares) solution \n", + "beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y\n", + "\n", + "## Evaluate the models\n", + "pred_ls = X @ beta_ls\n", + "pred_ridge = X @ beta_ridge\n", + "pred_ridge_centered = X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]\n", + "\n", + "## Plot the results\n", + "\n", + "# Sorting\n", + "sort_ind = np.argsort(x[:,0])\n", + "\n", + "x_plot = x[sort_ind,0]\n", + "x_centered_plot = x_[sort_ind,0]\n", + "\n", + "pred_ls_plot = pred_ls[sort_ind,0]\n", + "pred_ridge_plot = pred_ridge[sort_ind,:]\n", + "pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]\n", + "\n", + "# Plott not centered\n", + "plt.plot(x_plot,pred_ls_plot,label='ls')\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x,y,'ro')\n", + "\n", + "plt.title('linear regression on un-centered data')\n", + "plt.legend()\n", + "\n", + "# Plott centered\n", + "plt.figure()\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x_,y,'ro')\n", + "\n", + "plt.title('linear regression on centered data')\n", + "plt.legend()\n", + "\n", + "\n", + "# 2.\n", + "\n", + "pred_ridge_scikit = np.zeros((n_samples,num_values))\n", + "for i,lmb in enumerate(lmb_values):\n", + " pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X\n", + "\n", + "plt.figure()\n", + "\n", + "plt.plot(x_plot,pred_ls_plot,label='ls')\n", + "\n", + "for i in range(num_values):\n", + " plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])\n", + "\n", + "plt.plot(x,y,'ro')\n", + "plt.legend()\n", + "plt.title('linear regression using scikit')\n", + "\n", + "plt.show()\n", + "\n", + "### R2-score of the results\n", + "for i in range(num_values):\n", + " print('lambda = %g'%lmb_values[i])\n", + " print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))\n", + " print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))\n", + " print('r2 for own, centered: %g\\n'%r2_score(y,pred_ridge_centered[:,i]))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Fitting vs. predicting when data is in the model class\n", + "\n", + "We start by considering the case\n", + "$f(x)=2x$.\n", + "\n", + "Then the data is clearly generated by a model that is contained within\n", + "all three model classes we are using to make predictions (linear\n", + "models, third order polynomials, and tenth order polynomials).\n", + "\n", + "Run the code for the following cases:\n", + "\n", + "1. For $f(x)=2x$ , $Ntrain=10$ and $\\sigma =0$ (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when $x \\in [0,1]$ . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?\n", + "\n", + "2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?\n", + "\n", + "3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example $x \\in [0,1.2]$ ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?\n", + "\n", + "4. Repeat the above for $f(x)=2x$ , $Ntrain=10$ , and $\\sigma=1$ . What changes?\n", + "\n", + "Repeat the exercises above for $f(x)=2x$ , $Ntrain=100$ , and $\\sigma=1$ . What changes?\n", + "Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.\n", + "\n", + "\n", + "## Fitting versus predicting when data is not in the model class\n", + "\n", + "Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider $f(x)=2x-10x^5+15x^{10}$ . Notice that the for linear and third-order polynomial the true model $f(x)$ is not contained in model class.\n", + "\n", + "1. Do better fits lead to better predictions?\n", + "\n", + "2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points $Ntrain$ and $\\sigma$?\n", + "\n", + "Summarize what you think you learned about the relationship of knowing the true model class and predictive power.\n", + "\n", + "## The code" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import numpy as np\n", + "import sklearn as sk\n", + "from sklearn import datasets, linear_model\n", + "from sklearn.preprocessing import PolynomialFeatures\n", + "\n", + "import matplotlib as mpl\n", + "from matplotlib import pyplot as plt\n", + "\n", + "%matplotlib notebook\n", + "\n", + "# The Training Data\n", + "\n", + "N_train=100\n", + "\n", + "sigma_train=1;\n", + "\n", + "# Train on integers\n", + "x=np.linspace(0.05,0.95,N_train)\n", + "# Draw random noise\n", + "s = sigma_train*np.random.randn(N_train)\n", + "\n", + "#linear\n", + "y=2*x+s\n", + "\n", + "#Tenth Order\n", + "#y=2*x-10*x**5+15*x**10+s\n", + "\n", + "p1=plt.plot(x,y, \"o\",ms=15, label='Training')\n", + "\n", + "#Linear Regression\n", + "# Create linear regression object\n", + "clf = linear_model.LinearRegression()\n", + "\n", + "# Train the model using the training sets\n", + "clf.fit(x[:, np.newaxis], y)\n", + "# The coefficients\n", + "\n", + "xplot=np.linspace(0.02,0.98,200)\n", + "linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')\n", + "\n", + "#Polynomial Regression\n", + "\n", + "\n", + "poly3 = PolynomialFeatures(degree=3)\n", + "X = poly3.fit_transform(x[:,np.newaxis])\n", + "clf3 = linear_model.LinearRegression()\n", + "clf3.fit(X,y)\n", + "\n", + "\n", + "Xplot=poly3.fit_transform(xplot[:,np.newaxis])\n", + "poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')\n", + "\n", + "\n", + "\n", + "#poly5 = PolynomialFeatures(degree=5)\n", + "#X = poly5.fit_transform(x[:,np.newaxis])\n", + "#clf5 = linear_model.LinearRegression()\n", + "#clf5.fit(X,y)\n", + "\n", + "#Xplot=poly5.fit_transform(xplot[:,np.newaxis])\n", + "#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)\n", + "\n", + "poly10 = PolynomialFeatures(degree=10)\n", + "X = poly10.fit_transform(x[:,np.newaxis])\n", + "clf10 = linear_model.LinearRegression()\n", + "clf10.fit(X,y)\n", + "\n", + "Xplot=poly10.fit_transform(xplot[:,np.newaxis])\n", + "poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')\n", + "\n", + "axes = plt.gca()\n", + "axes.set_ylim([-7,7])\n", + "\n", + "handles, labels=axes.get_legend_handles_labels()\n", + "plt.legend(handles,labels, loc='lower center')\n", + "plt.xlabel(\"$x$\")\n", + "plt.ylabel(\"$y$\")\n", + "Title=\"$N=$\"+str(N_train)+\", $\\sigma=$\"+str(sigma_train)\n", + "plt.title(Title+\" (train)\")\n", + "plt.tight_layout()\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "## Generating test data" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "# Generate Test Data\n", + "\n", + "#Number of test data\n", + "N_test=20\n", + "\n", + "sigma_test=sigma_train\n", + "\n", + "max_x=1.2\n", + "x_test=max_x*np.random.random(N_test)\n", + "# Draw random noise\n", + "s_test = sigma_test*np.random.randn(N_test)\n", + "\n", + "#Linear\n", + "y_test=2*x_test+s_test\n", + "#Tenth order\n", + "#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", + "\n", + "#Make design matrices for prediction\n", + "x_plot=np.linspace(0,max_x, 200)\n", + "X3 = poly3.fit_transform(x_plot[:,np.newaxis])\n", + "X10 = poly10.fit_transform(x_plot[:,np.newaxis])\n", + "\n", + "%matplotlib notebook\n", + "\n", + "fig = plt.figure() \n", + "p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')\n", + "p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')\n", + "p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')\n", + "p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')\n", + "\n", + "\n", + "plt.legend(loc=2)\n", + "plt.xlabel('$x$')\n", + "plt.ylabel('$y$')\n", + "plt.legend(loc='best')\n", + "plt.title(Title+\" (pred.)\")\n", + "plt.tight_layout()\n", + "plt.show()\n", + "\n", + "#Linear Filename\n", + "#filename_test=Title+\"pred-linear.pdf\"\n", + "#Tenth Order Filename\n", + "#filename_test=Title+\"pred-o10.pdf\"\n", + "#plt.savefig(filename_test)\n", + "#plt.ylim((-6,12))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Lasso regression\n", + "\n", + "\n", + "## Logistic regression" ] } ], - "metadata": { - "kernelspec": { - "display_name": "Python 3", - "language": "python", - "name": "python3" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 3 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython3", - "version": "3.7.0" - } - }, + "metadata": {}, "nbformat": 4, "nbformat_minor": 2 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None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
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  • Code examples for Ridge and Lasso Regression
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  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
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  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -272,7 +300,7 @@ A regression model aims at finding a likelihood function \( p(y\vert \hat{x}) \)
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    • diff --git a/doc/pub/Regression/html/._Regression-bs002.html b/doc/pub/Regression/html/._Regression-bs002.html index 523f9d68c..a514bd2ec 100644 --- a/doc/pub/Regression/html/._Regression-bs002.html +++ b/doc/pub/Regression/html/._Regression-bs002.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
    • +
  • Generating test data
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  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -279,7 +307,7 @@ response is equal to \( \beta_j \).
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    • diff --git a/doc/pub/Regression/html/._Regression-bs003.html b/doc/pub/Regression/html/._Regression-bs003.html index d18ec8f83..572119bce 100644 --- a/doc/pub/Regression/html/._Regression-bs003.html +++ b/doc/pub/Regression/html/._Regression-bs003.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -270,7 +298,7 @@ where \( \epsilon_i \) is the error in our approximation.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs004.html b/doc/pub/Regression/html/._Regression-bs004.html index 643f337c1..de10c4d9c 100644 --- a/doc/pub/Regression/html/._Regression-bs004.html +++ b/doc/pub/Regression/html/._Regression-bs004.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
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  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -270,7 +298,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs005.html b/doc/pub/Regression/html/._Regression-bs005.html index 79e79e7b3..acae72078 100644 --- a/doc/pub/Regression/html/._Regression-bs005.html +++ b/doc/pub/Regression/html/._Regression-bs005.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -292,7 +320,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs006.html b/doc/pub/Regression/html/._Regression-bs006.html index 0666eaf89..afe7d5b85 100644 --- a/doc/pub/Regression/html/._Regression-bs006.html +++ b/doc/pub/Regression/html/._Regression-bs006.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -275,7 +303,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs007.html b/doc/pub/Regression/html/._Regression-bs007.html index 1cbc6bb56..dc27b195c 100644 --- a/doc/pub/Regression/html/._Regression-bs007.html +++ b/doc/pub/Regression/html/._Regression-bs007.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
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  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
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    • -
  • Lasso regression
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  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -281,7 +309,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • A second-order polynomial with Ridge and Lasso
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  • Lasso regression
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  • An example code without the model assessment part
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -276,7 +304,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs009.html b/doc/pub/Regression/html/._Regression-bs009.html index 2c4747519..9bda0b2aa 100644 --- a/doc/pub/Regression/html/._Regression-bs009.html +++ b/doc/pub/Regression/html/._Regression-bs009.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
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  • Generating test data
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  • Lasso regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • An example code without the model assessment part
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -279,7 +307,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs010.html b/doc/pub/Regression/html/._Regression-bs010.html index 90e7a3478..d7076e786 100644 --- a/doc/pub/Regression/html/._Regression-bs010.html +++ b/doc/pub/Regression/html/._Regression-bs010.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
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  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -295,7 +323,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs011.html b/doc/pub/Regression/html/._Regression-bs011.html index 94864f6a6..f40aab856 100644 --- a/doc/pub/Regression/html/._Regression-bs011.html +++ b/doc/pub/Regression/html/._Regression-bs011.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • Various steps in cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -282,7 +310,7 @@ $$
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    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
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    • -
  • Lasso regression
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -284,7 +312,7 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
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    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
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  • Lasso regression
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • A second-order polynomial with Ridge and Lasso
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -279,7 +307,7 @@ where the matrix \( \hat{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as
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  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -284,7 +312,7 @@ where we have defined the matrix \( \hat{A} =\hat{X}/\hat{\Sigma} \) with matrix
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    • diff --git a/doc/pub/Regression/html/._Regression-bs015.html b/doc/pub/Regression/html/._Regression-bs015.html index 9bb6354df..0d26f1e4c 100644 --- a/doc/pub/Regression/html/._Regression-bs015.html +++ b/doc/pub/Regression/html/._Regression-bs015.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
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  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
    • -
  • Generating test data
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  • Lasso regression
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  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Predicted Residual Error Sum of Squares
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  • Bootstrap
    • @@ -282,7 +310,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs016.html b/doc/pub/Regression/html/._Regression-bs016.html index b3aeb045a..a53ae1b6b 100644 --- a/doc/pub/Regression/html/._Regression-bs016.html +++ b/doc/pub/Regression/html/._Regression-bs016.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
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  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • Various steps in cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -287,7 +315,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs017.html b/doc/pub/Regression/html/._Regression-bs017.html index a258a9d2c..9eca50aa1 100644 --- a/doc/pub/Regression/html/._Regression-bs017.html +++ b/doc/pub/Regression/html/._Regression-bs017.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
    • +
  • Generating test data
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  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • Various steps in cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -280,7 +308,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs018.html b/doc/pub/Regression/html/._Regression-bs018.html index cc9cfc97e..ea0076b77 100644 --- a/doc/pub/Regression/html/._Regression-bs018.html +++ b/doc/pub/Regression/html/._Regression-bs018.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
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  • The singular value decompostion
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    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
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  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -302,7 +330,7 @@ This approach (different linear and non-linear regression) suffers often from bo
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    • diff --git a/doc/pub/Regression/html/._Regression-bs019.html b/doc/pub/Regression/html/._Regression-bs019.html index 26f66fce1..5e38d1e83 100644 --- a/doc/pub/Regression/html/._Regression-bs019.html +++ b/doc/pub/Regression/html/._Regression-bs019.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
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  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
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  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
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  • An example code without the model assessment part
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  • Generating test data
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  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling approaches can be computationally expensive
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  • Cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
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  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -295,7 +323,7 @@ We see that, as expected, a linear fit gives a seemingly (from the graph) good r
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    • diff --git a/doc/pub/Regression/html/._Regression-bs020.html b/doc/pub/Regression/html/._Regression-bs020.html index 4cbf19b9e..188ed0aa4 100644 --- a/doc/pub/Regression/html/._Regression-bs020.html +++ b/doc/pub/Regression/html/._Regression-bs020.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
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  • From standard regression to Ridge regressions
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  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
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  • How can we effectively evaluate the various models?
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  • Code examples for Ridge and Lasso Regression
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  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
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  • Log-likelihood
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  • Cross-validation
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  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -286,7 +314,7 @@ plt.show()
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    • diff --git a/doc/pub/Regression/html/._Regression-bs021.html b/doc/pub/Regression/html/._Regression-bs021.html index a768c9fd4..b02ce24d9 100644 --- a/doc/pub/Regression/html/._Regression-bs021.html +++ b/doc/pub/Regression/html/._Regression-bs021.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -323,7 +351,7 @@ plt.show()
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    • diff --git a/doc/pub/Regression/html/._Regression-bs022.html b/doc/pub/Regression/html/._Regression-bs022.html index dc088d321..ef3e8f482 100644 --- a/doc/pub/Regression/html/._Regression-bs022.html +++ b/doc/pub/Regression/html/._Regression-bs022.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -277,7 +305,7 @@ where \( x \) is defined as before.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs023.html b/doc/pub/Regression/html/._Regression-bs023.html index 193801b1c..e2730e41b 100644 --- a/doc/pub/Regression/html/._Regression-bs023.html +++ b/doc/pub/Regression/html/._Regression-bs023.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -268,7 +296,7 @@ have not discussed a more rigorous approach to the cost function.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs024.html b/doc/pub/Regression/html/._Regression-bs024.html index fe69a6e2a..cd61ec01b 100644 --- a/doc/pub/Regression/html/._Regression-bs024.html +++ b/doc/pub/Regression/html/._Regression-bs024.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -277,7 +305,7 @@ dimensionless.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs025.html b/doc/pub/Regression/html/._Regression-bs025.html index a1e2a076e..31f5d6002 100644 --- a/doc/pub/Regression/html/._Regression-bs025.html +++ b/doc/pub/Regression/html/._Regression-bs025.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -276,7 +304,7 @@ the \( \chi^2 \) function becomes smaller.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs026.html b/doc/pub/Regression/html/._Regression-bs026.html index 2b050bf0f..f72d47d9f 100644 --- a/doc/pub/Regression/html/._Regression-bs026.html +++ b/doc/pub/Regression/html/._Regression-bs026.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -295,7 +323,7 @@ relative error.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs027.html b/doc/pub/Regression/html/._Regression-bs027.html index 79b422d69..402165acb 100644 --- a/doc/pub/Regression/html/._Regression-bs027.html +++ b/doc/pub/Regression/html/._Regression-bs027.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
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    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -300,7 +328,7 @@ plt.show()
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    • diff --git a/doc/pub/Regression/html/._Regression-bs028.html b/doc/pub/Regression/html/._Regression-bs028.html index 79c5f693a..56e72568c 100644 --- a/doc/pub/Regression/html/._Regression-bs028.html +++ b/doc/pub/Regression/html/._Regression-bs028.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
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  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -272,7 +300,7 @@ this function as being similar to the \( \chi^2 \) function defined above.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs029.html b/doc/pub/Regression/html/._Regression-bs029.html index 33cd41ef8..a17909406 100644 --- a/doc/pub/Regression/html/._Regression-bs029.html +++ b/doc/pub/Regression/html/._Regression-bs029.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
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  • Fitting vs. predicting when data is in the model class
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  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -279,7 +307,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs030.html b/doc/pub/Regression/html/._Regression-bs030.html index 69b76d122..c26af9a7b 100644 --- a/doc/pub/Regression/html/._Regression-bs030.html +++ b/doc/pub/Regression/html/._Regression-bs030.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -280,7 +308,7 @@ years etc.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs031.html b/doc/pub/Regression/html/._Regression-bs031.html index 3033d8063..ff865ccb9 100644 --- a/doc/pub/Regression/html/._Regression-bs031.html +++ b/doc/pub/Regression/html/._Regression-bs031.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -303,7 +331,7 @@ Using R, we can perform similar studies.
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    • diff --git a/doc/pub/Regression/html/._Regression-bs032.html b/doc/pub/Regression/html/._Regression-bs032.html index 0f9b89474..1101eac0f 100644 --- a/doc/pub/Regression/html/._Regression-bs032.html +++ b/doc/pub/Regression/html/._Regression-bs032.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -286,7 +314,7 @@ plt.show()
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    • diff --git a/doc/pub/Regression/html/._Regression-bs033.html b/doc/pub/Regression/html/._Regression-bs033.html index d55114e2c..a9b1394ed 100644 --- a/doc/pub/Regression/html/._Regression-bs033.html +++ b/doc/pub/Regression/html/._Regression-bs033.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
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  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
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  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
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  • Cross-validation
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  • Computationally expensive
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  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -282,7 +310,7 @@ non-random scalar. To specify the parameters of the distribution of
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    • diff --git a/doc/pub/Regression/html/._Regression-bs034.html b/doc/pub/Regression/html/._Regression-bs034.html index a5c44e236..c5be7672b 100644 --- a/doc/pub/Regression/html/._Regression-bs034.html +++ b/doc/pub/Regression/html/._Regression-bs034.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -289,7 +317,7 @@ Hence, \( Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, \sigma^2) \).
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    • diff --git a/doc/pub/Regression/html/._Regression-bs035.html b/doc/pub/Regression/html/._Regression-bs035.html index 3902c8b08..7da384969 100644 --- a/doc/pub/Regression/html/._Regression-bs035.html +++ b/doc/pub/Regression/html/._Regression-bs035.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -277,7 +305,7 @@ $$
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    • diff --git a/doc/pub/Regression/html/._Regression-bs036.html b/doc/pub/Regression/html/._Regression-bs036.html index a2a01b043..b1e3b9849 100644 --- a/doc/pub/Regression/html/._Regression-bs036.html +++ b/doc/pub/Regression/html/._Regression-bs036.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,84 +260,54 @@ MathJax.Hub.Config({ -

    Code examples for Ridge and Lasso Regression

    +

    From standard regression to Ridge regressions

    +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ - -

    import matplotlib.pyplot as plt
-import numpy as np
-from sklearn import linear_model
-from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score
+
    
+The columns of \( \hat{X} \) are linearly dependent. We se this easily since the 
+the first column is the row-wise sum of the other two columns. The rank (more correct,
+the column rank) of a matrix is the dimension of the space spanned by the
+column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number
+of linearly independent columns. In this particular case the matrix has rank 2.
 
-#creating data with random noise
-x=np.arange(50)
+
    
+Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies
+that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this
+$$
+\begin{align*}
+\hat{X} & =  \left[
+\begin{array}{rr}
+1 & -1
+\\
+1 & -1
+\end{array} \right].
+\end{align*}
+$$
 
-delta=np.random.uniform(-2.5,2.5, size=(50))
-np.random.shuffle(delta)
-y =0.5*x+5+delta
+We see easily that  \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined.
+This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.
 
-#arranging data into 2x50 matrix
-a=np.array(x) #inputs
-b=np.array(y) #outputs
-
-#Split into training and test
-X_train=a[:37, np.newaxis]
-X_test=a[37:, np.newaxis]
-y_train=b[:37]
-y_test=b[37:]
-
-print ("X_train: ", X_train.shape)
-print ("y_train: ", y_train.shape)
-print ("X_test: ", X_test.shape)
-print ("y_test: ", y_test.shape)
-
-print ("------------------------------------")
-
-print ("Ordinary Least Squares")
-#Add Ordinary Least Squares fit
-reg=LinearRegression()
-reg.fit(X_train, y_train)
-pred=reg.predict(X_test)
-print ("Prediction Shape: ", pred.shape)
-
-print('Coefficients: \n', reg.coef_)
-# The mean squared error
-print("Mean squared error: %.2f"
-      % mean_squared_error(y_test, pred))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(y_test, pred))
-
-#plot
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, pred, color='black', label="Fit Line")
-plt.legend()
-plt.show()
-
-print ("------------------------------------")
-
-print ("Ridge Regression")
-
-ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
-ridge.fit(X_train,y_train)
-print ("Ridge Coefficient: ",ridge.coef_)
-print ("Ridge Intercept: ", ridge.intercept_)
-#Look into graphing with Ridge fit
-
-print ("------------------------------------")
-
-print ("Lasso")
-lasso=linear_model.Lasso(alpha=0.1)
-lasso.fit(X_train,y_train)
-predl=lasso.predict(X_test)
-print("Lasso Coefficient: ", lasso.coef_)
-print("Lasso Intercept: ", lasso.intercept_)
-plt.scatter(X_test,y_test,color='green', label="Training Data")
-plt.plot(X_test, predl, color='blue', label="Lasso")
-plt.legend()
-plt.show()
-

    @@ -335,6 +333,8 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs037.html b/doc/pub/Regression/html/._Regression-bs037.html index d7790e7d6..a01473605 100644 --- a/doc/pub/Regression/html/._Regression-bs037.html +++ b/doc/pub/Regression/html/._Regression-bs037.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,53 +260,30 @@ MathJax.Hub.Config({ -

    From standard regression to Ridge regressions

    +

    Fixing the singularity

    -One of the typical problems we encounter with linear regression, in particular -when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, -are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) -may be linearly dependent, normally referred to as super-collinearity. -This means that the matrix may be rank deficient and it is basically impossible to -to model the data using linear regression. As an example, consider the matrix +If our design matrix \( \hat{X} \) which enters the linear regression problem $$ -\begin{align*} -\mathbf{X} & = \left[ -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right] -\end{align*} +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\tag{1} +\end{align} $$ +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. +

    -The columns of \( \hat{X} \) are linearly dependent. We se this easily since the -the first column is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of the space spanned by the -column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number -of linearly independent columns. In this particular case the matrix has rank 2. - -

    -Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies -that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change $$ -\begin{align*} -\hat{X} & = \left[ -\begin{array}{rr} -1 & -1 -\\ -1 & -1 -\end{array} \right]. -\end{align*} +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, $$ -We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. -This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. +where \( \hat{I} \) is the identity matrix.

    @@ -304,6 +309,9 @@ This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigen

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    • diff --git a/doc/pub/Regression/html/._Regression-bs038.html b/doc/pub/Regression/html/._Regression-bs038.html index cca079168..1d99dea3a 100644 --- a/doc/pub/Regression/html/._Regression-bs038.html +++ b/doc/pub/Regression/html/._Regression-bs038.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,30 +260,29 @@ MathJax.Hub.Config({ -

    Fixing the singularity

    +

    Fitting vs. predicting when data is in the model class

    -If our design matrix \( \hat{X} \) which enters the linear regression problem -$$ -\begin{align} -\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, -\tag{1} -\end{align} -$$ - -has linearly dependent column vectors, we will not be able to compute the inverse -of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). -The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. -This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where -the regression parameters \( \beta_i \) cannot be estimated. +We start by considering the case +\( f(x)=2x \).

    -The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change -$$ -\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, -$$ +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). -where \( \hat{I} \) is the identity matrix. +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.

    @@ -280,6 +307,10 @@ where \( \hat{I} \) is the identity matrix.

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    • diff --git a/doc/pub/Regression/html/._Regression-bs039.html b/doc/pub/Regression/html/._Regression-bs039.html index aa8c96dc3..82cb86d33 100644 --- a/doc/pub/Regression/html/._Regression-bs039.html +++ b/doc/pub/Regression/html/._Regression-bs039.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,117 +260,18 @@ MathJax.Hub.Config({ -

    A second-order polynomial with Ridge and Lasso

    +

    Fitting versus predicting when data is not in the model class

    +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. - -

    import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import Ridge
-from sklearn.metrics import r2_score
+
      
+
    1.  Do better fits lead to better predictions?
      2. 
+
    2.  What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points  \( Ntrain \)  and  \( \sigma \)?
      3. 
+
    
 
-np.random.seed(4155)
+Summarize what you think you learned about the relationship of knowing the true model class and predictive power.
 
-n_samples = 100
-
-x = np.random.rand(n_samples,1)
-y = 5*x*x + 0.1*np.random.rand(n_samples,1)
-
-# Centering  x and y.
-x_ = x - np.mean(x)
-y_ = y - np.mean(y) # beta_0 = mean(y)
-
-X = np.c_[np.ones((n_samples,1)), x, x**2]
-X_ = np.c_[x_, x_**2]
-
-
-### 1.
-lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
-num_values = len(lmb_values)
-
-## Ridge-regression of centered and not centered data
-beta_ridge = np.zeros((3,num_values))
-beta_ridge_centered = np.zeros((3,num_values))
-
-I3 = np.eye(3)
-I2 = np.eye(2)
-
-for i,lmb in enumerate(lmb_values):
-    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
-    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
-
-# sett beta_0 = np.mean(y)
-beta_ridge_centered[0,:] = np.mean(y)
-
-## OLS (ordinary least squares) solution 
-beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
-
-## Evaluate the models
-pred_ls = X @ beta_ls
-pred_ridge =  X @ beta_ridge
-pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
-
-## Plot the results
-
-# Sorting
-sort_ind = np.argsort(x[:,0])
-
-x_plot = x[sort_ind,0]
-x_centered_plot = x_[sort_ind,0]
-
-pred_ls_plot = pred_ls[sort_ind,0]
-pred_ridge_plot = pred_ridge[sort_ind,:]
-pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
-
-# Plott not centered
-plt.plot(x_plot,pred_ls_plot,label='ls')
-
-for i in range(num_values):
-    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
-
-plt.plot(x,y,'ro')
-
-plt.title('linear regression on un-centered data')
-plt.legend()
-
-# Plott centered
-plt.figure()
-
-for i in range(num_values):
-    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
-
-plt.plot(x_,y,'ro')
-
-plt.title('linear regression on centered data')
-plt.legend()
-
-
-# 2.
-
-pred_ridge_scikit =  np.zeros((n_samples,num_values))
-for i,lmb in enumerate(lmb_values):
-    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
-
-plt.figure()
-
-plt.plot(x_plot,pred_ls_plot,label='ls')
-
-for i in range(num_values):
-    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
-
-plt.plot(x,y,'ro')
-plt.legend()
-plt.title('linear regression using scikit')
-
-plt.show()
-
-### R2-score of the results
-for i in range(num_values):
-    print('lambda = %g'%lmb_values[i])
-    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
-    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
-    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
-

    @@ -365,6 +294,11 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs040.html b/doc/pub/Regression/html/._Regression-bs040.html index e62c176d7..d42992185 100644 --- a/doc/pub/Regression/html/._Regression-bs040.html +++ b/doc/pub/Regression/html/._Regression-bs040.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,30 +260,93 @@ MathJax.Hub.Config({ -

    Fitting vs. predicting when data is in the model class

    +

    An example code without the model assessment part

    -We start by considering the case -\( f(x)=2x \). -

    -Then the data is clearly generated by a model that is contained within -all three model classes we are using to make predictions (linear -models, third order polynomials, and tenth order polynomials). + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
 
-
    
-Run the code for the following cases:
+import matplotlib as mpl
+from matplotlib import pyplot as plt
 
-
      
-
    1.  For  \( f(x)=2x \) ,  \( Ntrain=10 \)  and \( \sigma =0 \)  (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when  \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. 
-
    2.  Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. 
-
    3.  Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example  \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of  x  where you trained the model? How about points outside the original training data set?
      4. 
-
    4.  Repeat the above for  \( f(x)=2x \) ,  \( Ntrain=10 \) , and  \( \sigma=1 \) . What changes?
      5. 
-
    
+%matplotlib notebook
 
-Repeat the exercises above for  \( f(x)=2x \) ,  \( Ntrain=100 \) , and  \( \sigma=1 \) . What changes?
-Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.
+# The Training Data
 
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+

    @@ -277,6 +368,12 @@ Summarize what you have learned about the relationship between model complexity

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    • diff --git a/doc/pub/Regression/html/._Regression-bs041.html b/doc/pub/Regression/html/._Regression-bs041.html index 0449005ac..1c8a8d8a9 100644 --- a/doc/pub/Regression/html/._Regression-bs041.html +++ b/doc/pub/Regression/html/._Regression-bs041.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -230,20 +258,51 @@ MathJax.Hub.Config({

     

     

     

    - - -

    Fitting versus predicting when data is not in the model class

    + +

    Generating test data

    -Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. -

      -
    1. Do better fits lead to better predictions?
      2. -
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. -
    + +
    # Generate Test Data
 
-Summarize what you think you learned about the relationship of knowing the true model class and predictive power.
+#Number of test data
+N_test=20
 
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+

    @@ -264,6 +323,13 @@ Summarize what you think you learned about the relationship of knowing the true

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    • diff --git a/doc/pub/Regression/html/._Regression-bs042.html b/doc/pub/Regression/html/._Regression-bs042.html index ff3f3130a..ae514c7af 100644 --- a/doc/pub/Regression/html/._Regression-bs042.html +++ b/doc/pub/Regression/html/._Regression-bs042.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -232,93 +260,14 @@ MathJax.Hub.Config({ -

    The code

    +

    How can we effectively evaluate the various models?

    +In Ridge regression and the subsequent discussion of its properties +the bias or penalty parameter is considered known or `given'. In +practice, it is unknown and the user needs to make an informed +decision on its value. How do we do that? Much of the same considerations apply to the Lasso method. - -

    import numpy as np
-import sklearn as sk
-from sklearn import datasets, linear_model
-from sklearn.preprocessing import PolynomialFeatures
-
-import matplotlib as mpl
-from matplotlib import pyplot as plt
-
-%matplotlib notebook
-
-# The Training Data
-
-N_train=100
-
-sigma_train=1;
-
-# Train on integers
-x=np.linspace(0.05,0.95,N_train)
-# Draw random noise
-s = sigma_train*np.random.randn(N_train)
-
-#linear
-y=2*x+s
-
-#Tenth Order
-#y=2*x-10*x**5+15*x**10+s
-
-p1=plt.plot(x,y, "o",ms=15, label='Training')
-
-#Linear Regression
-# Create linear regression object
-clf = linear_model.LinearRegression()
-
-# Train the model using the training sets
-clf.fit(x[:, np.newaxis], y)
-# The coefficients
-
-xplot=np.linspace(0.02,0.98,200)
-linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
-
-#Polynomial Regression
-
-
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = linear_model.LinearRegression()
-clf3.fit(X,y)
-
-
-Xplot=poly3.fit_transform(xplot[:,np.newaxis])
-poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
-
-
-
-#poly5 = PolynomialFeatures(degree=5)
-#X = poly5.fit_transform(x[:,np.newaxis])
-#clf5 = linear_model.LinearRegression()
-#clf5.fit(X,y)
-
-#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
-#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
-
-poly10 = PolynomialFeatures(degree=10)
-X = poly10.fit_transform(x[:,np.newaxis])
-clf10 = linear_model.LinearRegression()
-clf10.fit(X,y)
-
-Xplot=poly10.fit_transform(xplot[:,np.newaxis])
-poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
-
-axes = plt.gca()
-axes.set_ylim([-7,7])
-
-handles, labels=axes.get_legend_handles_labels()
-plt.legend(handles,labels, loc='lower center')
-plt.xlabel("$x$")
-plt.ylabel("$y$")
-Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
-plt.title(Title+" (train)")
-plt.tight_layout()
-plt.show()
-

    @@ -338,6 +287,14 @@ plt.show()

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    • diff --git a/doc/pub/Regression/html/._Regression-bs043.html b/doc/pub/Regression/html/._Regression-bs043.html index b45774529..656700822 100644 --- a/doc/pub/Regression/html/._Regression-bs043.html +++ b/doc/pub/Regression/html/._Regression-bs043.html @@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -207,16 +227,24 @@ MathJax.Hub.Config({
  • Linking the regression analysis with a statistical interpretation
  • Expectation value and variance
  • The singular value decompostion
    • -
  • Code examples for Ridge and Lasso Regression
    • -
  • From standard regression to Ridge regressions
    • -
  • Fixing the singularity
    • -
  • A second-order polynomial with Ridge and Lasso
    • -
  • Fitting vs. predicting when data is in the model class
    • -
  • Fitting versus predicting when data is not in the model class
    • -
  • The code
    • -
  • Generating test data
    • -
  • Lasso regression
    • -
  • Logistic regression
    • +
  • From standard regression to Ridge regressions
    • +
  • Fixing the singularity
    • +
  • Fitting vs. predicting when data is in the model class
    • +
  • Fitting versus predicting when data is not in the model class
    • +
  • An example code without the model assessment part
    • +
  • Generating test data
    • +
  • How can we effectively evaluate the various models?
    • +
  • Code examples for Ridge and Lasso Regression
    • +
  • A second-order polynomial with Ridge and Lasso
    • +
  • Resampling methods
    • +
  • Resampling approaches can be computationally expensive
    • +
  • Log-likelihood
    • +
  • Cross-validation
    • +
  • Computationally expensive
    • +
  • Various steps in cross-validation
    • +
  • How to set up the cross-validation for Ridge and/or Lasso
    • +
  • Predicted Residual Error Sum of Squares
    • +
  • Bootstrap
    • @@ -230,57 +258,85 @@ MathJax.Hub.Config({

     

     

     

    - + + +

    Code examples for Ridge and Lasso Regression

    -

    Generating test data

    -

    # Generate Test Data
+
    import matplotlib.pyplot as plt
+import numpy as np
+from sklearn import linear_model
+from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score
 
-#Number of test data
-N_test=20
+#creating data with random noise
+x=np.arange(50)
 
-sigma_test=sigma_train
+delta=np.random.uniform(-2.5,2.5, size=(50))
+np.random.shuffle(delta)
+y =0.5*x+5+delta
 
-max_x=1.2
-x_test=max_x*np.random.random(N_test)
-# Draw random noise
-s_test = sigma_test*np.random.randn(N_test)
+#arranging data into 2x50 matrix
+a=np.array(x) #inputs
+b=np.array(y) #outputs
 
-#Linear
-y_test=2*x_test+s_test
-#Tenth order
-#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+#Split into training and test
+X_train=a[:37, np.newaxis]
+X_test=a[37:, np.newaxis]
+y_train=b[:37]
+y_test=b[37:]
 
-#Make design matrices for prediction
-x_plot=np.linspace(0,max_x, 200)
-X3 = poly3.fit_transform(x_plot[:,np.newaxis])
-X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+print ("X_train: ", X_train.shape)
+print ("y_train: ", y_train.shape)
+print ("X_test: ", X_test.shape)
+print ("y_test: ", y_test.shape)
 
-%matplotlib notebook
+print ("------------------------------------")
 
-fig = plt.figure() 
-p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
-p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
-p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
-p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+print ("Ordinary Least Squares")
+#Add Ordinary Least Squares fit
+reg=LinearRegression()
+reg.fit(X_train, y_train)
+pred=reg.predict(X_test)
+print ("Prediction Shape: ", pred.shape)
 
+print('Coefficients: \n', reg.coef_)
+# The mean squared error
+print("Mean squared error: %.2f"
+      % mean_squared_error(y_test, pred))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(y_test, pred))
 
-plt.legend(loc=2)
-plt.xlabel('$x$')
-plt.ylabel('$y$')
-plt.legend(loc='best')
-plt.title(Title+" (pred.)")
-plt.tight_layout()
+#plot
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, pred, color='black', label="Fit Line")
+plt.legend()
 plt.show()
 
-#Linear Filename
-#filename_test=Title+"pred-linear.pdf"
-#Tenth Order Filename
-#filename_test=Title+"pred-o10.pdf"
-#plt.savefig(filename_test)
-#plt.ylim((-6,12))
+print ("------------------------------------")
+
+print ("Ridge Regression")
+
+ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])
+ridge.fit(X_train,y_train)
+print ("Ridge Coefficient: ",ridge.coef_)
+print ("Ridge Intercept: ", ridge.intercept_)
+#Look into graphing with Ridge fit
+
+print ("------------------------------------")
+
+print ("Lasso")
+lasso=linear_model.Lasso(alpha=0.1)
+lasso.fit(X_train,y_train)
+predl=lasso.predict(X_test)
+print("Lasso Coefficient: ", lasso.coef_)
+print("Lasso Intercept: ", lasso.intercept_)
+plt.scatter(X_test,y_test,color='green', label="Training Data")
+plt.plot(X_test, predl, color='blue', label="Lasso")
+plt.legend()
+plt.show()
 
    
 
    
 
    
@@ -300,6 +356,15 @@ plt.show()
   
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diff --git a/doc/pub/Regression/html/._Regression-bs044.html b/doc/pub/Regression/html/._Regression-bs044.html
index 1ad5ddece..10e005345 100644
--- a/doc/pub/Regression/html/._Regression-bs044.html
+++ b/doc/pub/Regression/html/._Regression-bs044.html
@@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source
                '___sec32'),
               ('Expectation value and variance', 2, None, '___sec33'),
               ('The singular value decompostion', 2, None, '___sec34'),
-              ('Code examples for Ridge and Lasso Regression',
-               2,
-               None,
-               '___sec35'),
               ('From standard regression to Ridge regressions',
                2,
                None,
-               '___sec36'),
-              ('Fixing the singularity', 2, None, '___sec37'),
-              ('A second-order polynomial with Ridge and Lasso',
-               2,
-               None,
-               '___sec38'),
+               '___sec35'),
+              ('Fixing the singularity', 2, None, '___sec36'),
               ('Fitting vs. predicting when data is in the model class',
                2,
                None,
-               '___sec39'),
+               '___sec37'),
               ('Fitting versus predicting when data is not in the model class',
                2,
                None,
-               '___sec40'),
-              ('The code', 2, None, '___sec41'),
-              ('Generating test data', 2, None, '___sec42'),
-              ('Lasso regression', 2, None, '___sec43'),
-              ('Logistic regression', 2, None, '___sec44')]}
+               '___sec38'),
+              ('An example code without the model assessment part',
+               2,
+               None,
+               '___sec39'),
+              ('Generating test data', 2, None, '___sec40'),
+              ('How can we effectively evaluate the various models?',
+               2,
+               None,
+               '___sec41'),
+              ('Code examples for Ridge and Lasso Regression',
+               2,
+               None,
+               '___sec42'),
+              ('A second-order polynomial with Ridge and Lasso',
+               2,
+               None,
+               '___sec43'),
+              ('Resampling methods', 2, None, '___sec44'),
+              ('Resampling approaches can be computationally expensive',
+               2,
+               None,
+               '___sec45'),
+              ('Log-likelihood', 2, None, '___sec46'),
+              ('Cross-validation', 2, None, '___sec47'),
+              ('Computationally expensive', 2, None, '___sec48'),
+              ('Various steps in cross-validation', 2, None, '___sec49'),
+              ('How to set up the cross-validation for Ridge and/or Lasso',
+               2,
+               None,
+               '___sec50'),
+              ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'),
+              ('Bootstrap', 2, None, '___sec52')]}
 end of tocinfo -->
 
 
@@ -207,16 +227,24 @@ MathJax.Hub.Config({
       
  • Linking the regression analysis with a statistical interpretation
    • 
       
  • Expectation value and variance
    • 
       
  • The singular value decompostion
    • 
-      
  • Code examples for Ridge and Lasso Regression
    • 
-      
  • From standard regression to Ridge regressions
    • 
-      
  • Fixing the singularity
    • 
-      
  • A second-order polynomial with Ridge and Lasso
    • 
-      
  • Fitting vs. predicting when data is in the model class
    • 
-      
  • Fitting versus predicting when data is not in the model class
    • 
-      
  • The code
    • 
-      
  • Generating test data
    • 
-      
  • Lasso regression
    • 
-      
  • Logistic regression
    • 
+      
  • From standard regression to Ridge regressions
    • 
+      
  • Fixing the singularity
    • 
+      
  • Fitting vs. predicting when data is in the model class
    • 
+      
  • Fitting versus predicting when data is not in the model class
    • 
+      
  • An example code without the model assessment part
    • 
+      
  • Generating test data
    • 
+      
  • How can we effectively evaluate the various models?
    • 
+      
  • Code examples for Ridge and Lasso Regression
    • 
+      
  • A second-order polynomial with Ridge and Lasso
    • 
+      
  • Resampling methods
    • 
+      
  • Resampling approaches can be computationally expensive
    • 
+      
  • Log-likelihood
    • 
+      
  • Cross-validation
    • 
+      
  • Computationally expensive
    • 
+      
  • Various steps in cross-validation
    • 
+      
  • How to set up the cross-validation for Ridge and/or Lasso
    • 
+      
  • Predicted Residual Error Sum of Squares
    • 
+      
  • Bootstrap
    • 
 
         
       
@@ -232,8 +260,117 @@ MathJax.Hub.Config({
 
 
 
-
    Lasso regression 
    
+
    A second-order polynomial with Ridge and Lasso 
    
+
    
 
+
+
    import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import Ridge
+from sklearn.metrics import r2_score
+
+np.random.seed(4155)
+
+n_samples = 100
+
+x = np.random.rand(n_samples,1)
+y = 5*x*x + 0.1*np.random.rand(n_samples,1)
+
+# Centering  x and y.
+x_ = x - np.mean(x)
+y_ = y - np.mean(y) # beta_0 = mean(y)
+
+X = np.c_[np.ones((n_samples,1)), x, x**2]
+X_ = np.c_[x_, x_**2]
+
+
+### 1.
+lmb_values = [1e-4, 1e-3, 1e-2, 10, 1e2, 1e4]
+num_values = len(lmb_values)
+
+## Ridge-regression of centered and not centered data
+beta_ridge = np.zeros((3,num_values))
+beta_ridge_centered = np.zeros((3,num_values))
+
+I3 = np.eye(3)
+I2 = np.eye(2)
+
+for i,lmb in enumerate(lmb_values):
+    beta_ridge[:,i] = (np.linalg.inv( X.T @ X + lmb*I3) @ X.T @ y).flatten()
+    beta_ridge_centered[1:,i] = (np.linalg.inv( X_.T @ X_ + lmb*I2) @ X_.T @ y_).flatten()
+
+# sett beta_0 = np.mean(y)
+beta_ridge_centered[0,:] = np.mean(y)
+
+## OLS (ordinary least squares) solution 
+beta_ls = np.linalg.inv( X.T @ X ) @ X.T @ y
+
+## Evaluate the models
+pred_ls = X @ beta_ls
+pred_ridge =  X @ beta_ridge
+pred_ridge_centered =  X_ @ beta_ridge_centered[1:] + beta_ridge_centered[0,:]
+
+## Plot the results
+
+# Sorting
+sort_ind = np.argsort(x[:,0])
+
+x_plot = x[sort_ind,0]
+x_centered_plot = x_[sort_ind,0]
+
+pred_ls_plot = pred_ls[sort_ind,0]
+pred_ridge_plot = pred_ridge[sort_ind,:]
+pred_ridge_centered_plot = pred_ridge_centered[sort_ind,:]
+
+# Plott not centered
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+
+plt.title('linear regression on un-centered data')
+plt.legend()
+
+# Plott centered
+plt.figure()
+
+for i in range(num_values):
+    plt.plot(x_centered_plot,pred_ridge_centered_plot[:,i],label='ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x_,y,'ro')
+
+plt.title('linear regression on centered data')
+plt.legend()
+
+
+# 2.
+
+pred_ridge_scikit =  np.zeros((n_samples,num_values))
+for i,lmb in enumerate(lmb_values):
+    pred_ridge_scikit[:,i] = (Ridge(alpha=lmb,fit_intercept=False).fit(X,y).predict(X)).flatten() # fit_intercept=False fordi bias er allerede i X
+
+plt.figure()
+
+plt.plot(x_plot,pred_ls_plot,label='ls')
+
+for i in range(num_values):
+    plt.plot(x_plot,pred_ridge_scikit[sort_ind,i],label='scikit-ridge, lmb=%g'%lmb_values[i])
+
+plt.plot(x,y,'ro')
+plt.legend()
+plt.title('linear regression using scikit')
+
+plt.show()
+
+### R2-score of the results
+for i in range(num_values):
+    print('lambda = %g'%lmb_values[i])
+    print('r2 for scikit: %g'%r2_score(y,pred_ridge_scikit[:,i]))
+    print('r2 for own code, not centered: %g'%r2_score(y,pred_ridge[:,i]))
+    print('r2 for own, centered: %g\n'%r2_score(y,pred_ridge_centered[:,i]))
+
    
 
    
 
    
 
@@ -251,6 +388,14 @@ MathJax.Hub.Config({
   
  • 44
    • 
   
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    • 
   
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    • 
+  
  • 47
    • 
+  
  • 48
    • 
+  
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+  
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+  
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+  
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+  
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+  
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    • 
 
 
diff --git a/doc/pub/Regression/html/._Regression-bs045.html b/doc/pub/Regression/html/._Regression-bs045.html
index 59fd87d05..d8001720b 100644
--- a/doc/pub/Regression/html/._Regression-bs045.html
+++ b/doc/pub/Regression/html/._Regression-bs045.html
@@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source
                '___sec32'),
               ('Expectation value and variance', 2, None, '___sec33'),
               ('The singular value decompostion', 2, None, '___sec34'),
-              ('Code examples for Ridge and Lasso Regression',
-               2,
-               None,
-               '___sec35'),
               ('From standard regression to Ridge regressions',
                2,
                None,
-               '___sec36'),
-              ('Fixing the singularity', 2, None, '___sec37'),
-              ('A second-order polynomial with Ridge and Lasso',
-               2,
-               None,
-               '___sec38'),
+               '___sec35'),
+              ('Fixing the singularity', 2, None, '___sec36'),
               ('Fitting vs. predicting when data is in the model class',
                2,
                None,
-               '___sec39'),
+               '___sec37'),
               ('Fitting versus predicting when data is not in the model class',
                2,
                None,
-               '___sec40'),
-              ('The code', 2, None, '___sec41'),
-              ('Generating test data', 2, None, '___sec42'),
-              ('Lasso regression', 2, None, '___sec43'),
-              ('Logistic regression', 2, None, '___sec44')]}
+               '___sec38'),
+              ('An example code without the model assessment part',
+               2,
+               None,
+               '___sec39'),
+              ('Generating test data', 2, None, '___sec40'),
+              ('How can we effectively evaluate the various models?',
+               2,
+               None,
+               '___sec41'),
+              ('Code examples for Ridge and Lasso Regression',
+               2,
+               None,
+               '___sec42'),
+              ('A second-order polynomial with Ridge and Lasso',
+               2,
+               None,
+               '___sec43'),
+              ('Resampling methods', 2, None, '___sec44'),
+              ('Resampling approaches can be computationally expensive',
+               2,
+               None,
+               '___sec45'),
+              ('Log-likelihood', 2, None, '___sec46'),
+              ('Cross-validation', 2, None, '___sec47'),
+              ('Computationally expensive', 2, None, '___sec48'),
+              ('Various steps in cross-validation', 2, None, '___sec49'),
+              ('How to set up the cross-validation for Ridge and/or Lasso',
+               2,
+               None,
+               '___sec50'),
+              ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'),
+              ('Bootstrap', 2, None, '___sec52')]}
 end of tocinfo -->
 
 
@@ -207,16 +227,24 @@ MathJax.Hub.Config({
       
  • Linking the regression analysis with a statistical interpretation
    • 
       
  • Expectation value and variance
    • 
       
  • The singular value decompostion
    • 
-      
  • Code examples for Ridge and Lasso Regression
    • 
-      
  • From standard regression to Ridge regressions
    • 
-      
  • Fixing the singularity
    • 
-      
  • A second-order polynomial with Ridge and Lasso
    • 
-      
  • Fitting vs. predicting when data is in the model class
    • 
-      
  • Fitting versus predicting when data is not in the model class
    • 
-      
  • The code
    • 
-      
  • Generating test data
    • 
-      
  • Lasso regression
    • 
-      
  • Logistic regression
    • 
+      
  • From standard regression to Ridge regressions
    • 
+      
  • Fixing the singularity
    • 
+      
  • Fitting vs. predicting when data is in the model class
    • 
+      
  • Fitting versus predicting when data is not in the model class
    • 
+      
  • An example code without the model assessment part
    • 
+      
  • Generating test data
    • 
+      
  • How can we effectively evaluate the various models?
    • 
+      
  • Code examples for Ridge and Lasso Regression
    • 
+      
  • A second-order polynomial with Ridge and Lasso
    • 
+      
  • Resampling methods
    • 
+      
  • Resampling approaches can be computationally expensive
    • 
+      
  • Log-likelihood
    • 
+      
  • Cross-validation
    • 
+      
  • Computationally expensive
    • 
+      
  • Various steps in cross-validation
    • 
+      
  • How to set up the cross-validation for Ridge and/or Lasso
    • 
+      
  • Predicted Residual Error Sum of Squares
    • 
+      
  • Bootstrap
    • 
 
         
       
@@ -232,8 +260,25 @@ MathJax.Hub.Config({
 
 
 
-
    Logistic regression 
    
+
    Resampling methods 
    
+
    
+
    
+
     
+Resampling methods are an indispensable tool in modern
+statistics. They involve repeatedly drawing samples from a training
+set and refitting a model of interest on each sample in order to
+obtain additional information about the fitted model. For example, in
+order to estimate the variability of a linear regression fit, we can
+repeatedly draw different samples from the training data, fit a linear
+regression to each new sample, and then examine the extent to which
+the resulting fits differ. Such an approach may allow us to obtain
+information that would not be available from fitting the model only
+once using the original training sample.
+
    
+
    
 
+
+
    
 
    
 
 
    
 
 
diff --git a/doc/pub/Regression/html/Regression-bs.html b/doc/pub/Regression/html/Regression-bs.html
index c92c12f8a..bd71ca567 100644
--- a/doc/pub/Regression/html/Regression-bs.html
+++ b/doc/pub/Regression/html/Regression-bs.html
@@ -110,31 +110,51 @@ Automatically generated HTML file from DocOnce source
                '___sec32'),
               ('Expectation value and variance', 2, None, '___sec33'),
               ('The singular value decompostion', 2, None, '___sec34'),
-              ('Code examples for Ridge and Lasso Regression',
-               2,
-               None,
-               '___sec35'),
               ('From standard regression to Ridge regressions',
                2,
                None,
-               '___sec36'),
-              ('Fixing the singularity', 2, None, '___sec37'),
-              ('A second-order polynomial with Ridge and Lasso',
-               2,
-               None,
-               '___sec38'),
+               '___sec35'),
+              ('Fixing the singularity', 2, None, '___sec36'),
               ('Fitting vs. predicting when data is in the model class',
                2,
                None,
-               '___sec39'),
+               '___sec37'),
               ('Fitting versus predicting when data is not in the model class',
                2,
                None,
-               '___sec40'),
-              ('The code', 2, None, '___sec41'),
-              ('Generating test data', 2, None, '___sec42'),
-              ('Lasso regression', 2, None, '___sec43'),
-              ('Logistic regression', 2, None, '___sec44')]}
+               '___sec38'),
+              ('An example code without the model assessment part',
+               2,
+               None,
+               '___sec39'),
+              ('Generating test data', 2, None, '___sec40'),
+              ('How can we effectively evaluate the various models?',
+               2,
+               None,
+               '___sec41'),
+              ('Code examples for Ridge and Lasso Regression',
+               2,
+               None,
+               '___sec42'),
+              ('A second-order polynomial with Ridge and Lasso',
+               2,
+               None,
+               '___sec43'),
+              ('Resampling methods', 2, None, '___sec44'),
+              ('Resampling approaches can be computationally expensive',
+               2,
+               None,
+               '___sec45'),
+              ('Log-likelihood', 2, None, '___sec46'),
+              ('Cross-validation', 2, None, '___sec47'),
+              ('Computationally expensive', 2, None, '___sec48'),
+              ('Various steps in cross-validation', 2, None, '___sec49'),
+              ('How to set up the cross-validation for Ridge and/or Lasso',
+               2,
+               None,
+               '___sec50'),
+              ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'),
+              ('Bootstrap', 2, None, '___sec52')]}
 end of tocinfo -->
 
 
@@ -207,16 +227,24 @@ MathJax.Hub.Config({
       
  • Linking the regression analysis with a statistical interpretation
    • 
       
  • Expectation value and variance
    • 
       
  • The singular value decompostion
    • 
-      
  • Code examples for Ridge and Lasso Regression
    • 
-      
  • From standard regression to Ridge regressions
    • 
-      
  • Fixing the singularity
    • 
-      
  • A second-order polynomial with Ridge and Lasso
    • 
-      
  • Fitting vs. predicting when data is in the model class
    • 
-      
  • Fitting versus predicting when data is not in the model class
    • 
-      
  • The code
    • 
-      
  • Generating test data
    • 
-      
  • Lasso regression
    • 
-      
  • Logistic regression
    • 
+      
  • From standard regression to Ridge regressions
    • 
+      
  • Fixing the singularity
    • 
+      
  • Fitting vs. predicting when data is in the model class
    • 
+      
  • Fitting versus predicting when data is not in the model class
    • 
+      
  • An example code without the model assessment part
    • 
+      
  • Generating test data
    • 
+      
  • How can we effectively evaluate the various models?
    • 
+      
  • Code examples for Ridge and Lasso Regression
    • 
+      
  • A second-order polynomial with Ridge and Lasso
    • 
+      
  • Resampling methods
    • 
+      
  • Resampling approaches can be computationally expensive
    • 
+      
  • Log-likelihood
    • 
+      
  • Cross-validation
    • 
+      
  • Computationally expensive
    • 
+      
  • Various steps in cross-validation
    • 
+      
  • How to set up the cross-validation for Ridge and/or Lasso
    • 
+      
  • Predicted Residual Error Sum of Squares
    • 
+      
  • Bootstrap
    • 
 
         
       
@@ -251,7 +279,7 @@ MathJax.Hub.Config({
 
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
    
 
    
 
    
-
    Sep 6, 2018
     
+
    Sep 7, 2018
     
 
    
 
    
 
@@ -275,7 +303,7 @@ MathJax.Hub.Config({
   
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    • 
   
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    • 
   
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    • 
-  
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diff --git a/doc/pub/Regression/html/Regression-reveal.html b/doc/pub/Regression/html/Regression-reveal.html
index 13e0c45e3..ee2d9bcc5 100644
--- a/doc/pub/Regression/html/Regression-reveal.html
+++ b/doc/pub/Regression/html/Regression-reveal.html
@@ -148,7 +148,7 @@ MathJax.Hub.Config({
 
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University
    
 
    
 
     
    
-
    Sep 6, 2018
     
+
    Sep 7, 2018
     
 
    
 
    
 
@@ -1282,7 +1282,287 @@ $$
 
 
 
    
-
    Code examples for Ridge and Lasso Regression 
    
+
    From standard regression to Ridge regressions 
    
+
+
    
+One of the typical problems we encounter with linear regression, in particular 
+when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, 
+are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) 
+may be linearly dependent, normally referred to as super-collinearity.  
+This means that the matrix may be rank deficient and it is basically impossible to 
+to model the data using linear regression. As an example, consider the matrix
+
     
    
+$$
+\begin{align*}
+\mathbf{X} & =  \left[
+\begin{array}{rrr}
+1 & -1 & 2
+\\
+1 & 0 & 1
+\\
+1 & 2  & -1
+\\
+1 & 1  & 0
+\end{array} \right]
+\end{align*}
+$$
+
     
    
+
+
    
+The columns of \( \hat{X} \) are linearly dependent. We se this easily since the 
+the first column is the row-wise sum of the other two columns. The rank (more correct,
+the column rank) of a matrix is the dimension of the space spanned by the
+column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number
+of linearly independent columns. In this particular case the matrix has rank 2.
+
+
    
+Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies
+that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this
+
     
    
+$$
+\begin{align*}
+\hat{X} & =  \left[
+\begin{array}{rr}
+1 & -1
+\\
+1 & -1
+\end{array} \right].
+\end{align*}
+$$
+
     
    
+
+We see easily that  \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined.
+This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.
+
    
+
+
+
    
+
    Fixing the singularity 
    
+
+
    
+If our design matrix \( \hat{X} \) which enters the linear regression problem
+
     
    
+$$
+\begin{align}
+\hat{\beta} & =  (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y},
+\tag{1}
+\end{align}
+$$
+
     
    
+
+has linearly dependent column vectors, we will not be able to compute the inverse
+of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). 
+The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. 
+This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where 
+the regression parameters \( \beta_i \) cannot be estimated.
+
+
    
+The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change
+
     
    
+$$
+\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I},
+$$
+
     
    
+
+where \( \hat{I} \) is the identity matrix.
+
    
+
+
+
    
+
    Fitting vs. predicting when data is in the model class 
    
+
+
    
+We start by considering the case
+\( f(x)=2x \).
+
+
    
+Then the data is clearly generated by a model that is contained within
+all three model classes we are using to make predictions (linear
+models, third order polynomials, and tenth order polynomials).
+
+
    
+Run the code for the following cases:
+
+
      
+
    1.  For  \( f(x)=2x \) ,  \( Ntrain=10 \)  and \( \sigma =0 \)  (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when  \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. 
+
    2.  Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. 
+
    3.  Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example  \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of  x  where you trained the model? How about points outside the original training data set?
      4. 
+
    4.  Repeat the above for  \( f(x)=2x \) ,  \( Ntrain=10 \) , and  \( \sigma=1 \) . What changes?
      5. 
+
    
+
    
+
+Repeat the exercises above for  \( f(x)=2x \) ,  \( Ntrain=100 \) , and  \( \sigma=1 \) . What changes?
+Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.
+
    
+
+
+
    
+
    Fitting versus predicting when data is not in the model class 
    
+
+
    
+Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model  \( f(x) \)  is not contained in model class.
+
+
      
+
    1.  Do better fits lead to better predictions?
      2. 
+
    2.  What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points  \( Ntrain \)  and  \( \sigma \)?
      3. 
+
    
+
    
+
+Summarize what you think you learned about the relationship of knowing the true model class and predictive power.
+
    
+
+
+
    
+
    An example code without the model assessment part 
    
+
+
    
+
+
+
    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    
+
    
+
+
+
    
+
    Generating test data 
    
+
    
+
+
+
    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
    
+
    
+
+
+
    
+
    How can we effectively evaluate the various models? 
    
+
+
    
+In Ridge regression and the subsequent discussion of its properties
+the bias or penalty parameter is considered known or `given'. In
+practice, it is unknown and the user needs to make an informed
+decision on its value. How do we do that?  Much of the same considerations apply to the Lasso method.
+
    
+
+
+
    
+
    Code examples for Ridge and Lasso Regression 
    
 
 
    
 
@@ -1364,94 +1644,7 @@ plt.show()
 
 
 
    
-
    From standard regression to Ridge regressions 
    
-
-
    
-One of the typical problems we encounter with linear regression, in particular 
-when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, 
-are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) 
-may be linearly dependent, normally referred to as super-collinearity.  
-This means that the matrix may be rank deficient and it is basically impossible to 
-to model the data using linear regression. As an example, consider the matrix
-
     
    
-$$
-\begin{align*}
-\mathbf{X} & =  \left[
-\begin{array}{rrr}
-1 & -1 & 2
-\\
-1 & 0 & 1
-\\
-1 & 2  & -1
-\\
-1 & 1  & 0
-\end{array} \right]
-\end{align*}
-$$
-
     
    
-
-
    
-The columns of \( \hat{X} \) are linearly dependent. We se this easily since the 
-the first column is the row-wise sum of the other two columns. The rank (more correct,
-the column rank) of a matrix is the dimension of the space spanned by the
-column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number
-of linearly independent columns. In this particular case the matrix has rank 2.
-
-
    
-Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies
-that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this
-
     
    
-$$
-\begin{align*}
-\hat{X} & =  \left[
-\begin{array}{rr}
-1 & -1
-\\
-1 & -1
-\end{array} \right].
-\end{align*}
-$$
-
     
    
-
-We see easily that  \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined.
-This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero.
-
    
-
-
-
    
-
    Fixing the singularity 
    
-
-
    
-If our design matrix \( \hat{X} \) which enters the linear regression problem
-
     
    
-$$
-\begin{align}
-\hat{\beta} & =  (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y},
-\tag{1}
-\end{align}
-$$
-
     
    
-
-has linearly dependent column vectors, we will not be able to compute the inverse
-of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). 
-The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. 
-This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where 
-the regression parameters \( \beta_i \) cannot be estimated.
-
-
    
-The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change
-
     
    
-$$
-\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I},
-$$
-
     
    
-
-where \( \hat{I} \) is the identity matrix.
-
    
-
-
-
    
-
    A second-order polynomial with Ridge and Lasso 
    
+
    A second-order polynomial with Ridge and Lasso 
    
 
    
 
 
@@ -1566,201 +1759,233 @@ plt.show()
 
 
 
    
-
    Fitting vs. predicting when data is in the model class 
    
+
    Resampling methods 
    
+
    
+
+
    
+Resampling methods are an indispensable tool in modern
+statistics. They involve repeatedly drawing samples from a training
+set and refitting a model of interest on each sample in order to
+obtain additional information about the fitted model. For example, in
+order to estimate the variability of a linear regression fit, we can
+repeatedly draw different samples from the training data, fit a linear
+regression to each new sample, and then examine the extent to which
+the resulting fits differ. Such an approach may allow us to obtain
+information that would not be available from fitting the model only
+once using the original training sample.
+
    
+
    
+
+
+
    
+
    Resampling approaches can be computationally expensive 
    
+
    
+
+
    
+Resampling approaches can be computationally expensive, because they
+involve fitting the same statistical method multiple times using
+different subsets of the training data. However, due to recent
+advances in computing power, the computational requirements of
+resampling methods generally are not prohibitive. In this chapter, we
+discuss two of the most commonly used resampling methods,
+cross-validation and the bootstrap. Both methods are important tools
+in the practical application of many statistical learning
+procedures. For example, cross-validation can be used to estimate the
+test error associated with a given statistical learning method in
+order to evaluate its performance, or to select the appropriate level
+of flexibility. The process of evaluating a model’s performance is
+known as model assessment, whereas the process of selecting the proper
+level of flexibility for a model is known as model selection. The
+bootstrap is widely used.
+
    
+
    
+
+
+
    
+
    Log-likelihood 
    
 
 
    
-We start by considering the case
-\( f(x)=2x \).
+A popular strategy is to choose a penalty parameter that yields a good
+but parsimonious model. Information criteria measure the balance
+between model fit and model complexity. One possibility is  Aikaike's
+information criterion (AIC).
+The AIC measures model fit by the log-likelihood
+and model complexity is measured by the number of parameters used by
+the model. The number of model parameters in regular regression simply
+corresponds to the number of covariates in the model. Or, by the
+degrees of freedom consumed by the model, which is equivalent to the
+trace of the hat matrix. For ridge regression it thus seems natural to
+define model complexity analogously by the trace of the ridge hat
+matrix. This yields the AIC for the linear regression model with ridge
+estimates:
+
+
     
    
+$$
+\begin{align*}
+\mbox{AIC}(\lambda) & =  2 \, p - 2 \log(\hat{L})
+\\
+& =  2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\}
+\\
+& =  2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}
++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] +  \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2.
+\end{align*}
+$$
+
     
    
+
+The value of \( \lambda \) which minimizes \( \mbox{AIC}(\lambda) \) corresponds to the `optimal' balance of model complexity and overfitting.
+
    
+
+
+
    
+
    Cross-validation 
    
 
 
    
-Then the data is clearly generated by a model that is contained within
-all three model classes we are using to make predictions (linear
-models, third order polynomials, and tenth order polynomials).
+Instead of choosing the penalty parameter to balance model fit with
+model complexity, cross-validation requires it (i.e. the penalty
+parameter) to yield a model with good prediction
+performance. Commonly, this performance is evaluated on novel
+data. Novel data need not be easy to come by and one has to make do
+with the data at hand. The setting of `original' and novel data is
+then mimicked by sample splitting: the data set is divided into two
+(groups of samples). One of these two data sets, called the training
+set, plays the role of `original' data on which the model is
+built. The second of these data sets, called the test set, plays the
+role of the `novel' data and is used to evaluate the prediction
+performance (often operationalized as the log-likelihood or the
+prediction error or its square or the R2 score) of the model built on the training data set. This
+procedure (model building and prediction evaluation on training and
+test set, respectively) is done for a collection of possible penalty
+parameter choices. The penalty parameter that yields the model with
+the best prediction performance is to be preferred. The thus obtained
+performance evaluation depends on the actual split of the data set. To
+remove this dependence the data set is split many times into a
+training and test set. For each split the model parameters are
+estimated for all choices of \( \lambda \) using the training data and
+estimated parameters are evaluated on the corresponding test set. The
+penalty parameter that on average over the test sets performs best (in
+some sense) is then selected.
+
    
+
+
+
    
+
    Computationally expensive 
    
 
 
    
-Run the code for the following cases:
+The validation set approach is conceptually simple and is easy to implement. But it has two potential drawbacks:
+
+
      
+
    •  The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.
      • 
+
    •  In the validation approach, only a subset of the observations, those that are included in the training set rather than in the validation set are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
      • 
+
    
+
    
+
+
+
    
+
    Various steps in cross-validation 
    
+
+
    
+When the repetitive splitting of the data set is done randomly,
+samples may accidently end up in a fast majority of the splits in
+either training or test set. Such samples may have an unbalanced
+influence on either model building or prediction evaluation. To avoid
+this \( k \)-fold cross-validation structures the data splitting. The
+samples are divided into \( k \) more or less equally sized exhaustive and
+mutually exclusive subsets. In turn (at each split) one of these
+subsets plays the role of the test set while the union of the
+remaining subsets constitutes the training set. Such a splitting
+warrants a balanced representation of each sample in both training and
+test set over the splits. Still the division into the \( k \) subsets
+involves a degree of randomness. This may be fully excluded when
+choosing \( k=n \). This particular case is referred to as leave-one-out
+cross-validation (LOOCV).
+
    
+
+
+
    
+
    How to set up the cross-validation for Ridge and/or Lasso 
    
 
 
      
-
    1.  For  \( f(x)=2x \) ,  \( Ntrain=10 \)  and \( \sigma =0 \)  (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when  \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. 
-
    2.  Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. 
-
    3.  Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example  \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of  x  where you trained the model? How about points outside the original training data set?
      4. 
-
    4.  Repeat the above for  \( f(x)=2x \) ,  \( Ntrain=10 \) , and  \( \sigma=1 \) . What changes?
      5. 
+
    5.  Define a range of interest for the penalty parameter.
      6. 
+
    6.  Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
      7. 
+
    7.  Fit the linear regression model by means of ridge estimation  for each \( \lambda \) in the grid using the training set as
      8. 
 
    
-
    
+
     
    
+$$
+\begin{align*}
+\hat{\beta}_{-i}(\lambda) & =  ( \hat{X}_{-i, \ast}^{\top}
+\hat{X}_{-i, \ast} + \lambda \hat{I}_{pp})^{-1}
+\hat{X}_{-i, \ast}^{\top} \hat{y}_{-i}
+\end{align*}
+$$
+
     
    
 
-Repeat the exercises above for  \( f(x)=2x \) ,  \( Ntrain=100 \) , and  \( \sigma=1 \) . What changes?
-Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.
-
    
-
-
-
    
-
    Fitting versus predicting when data is not in the model class 
    
-
-
    
-Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model  \( f(x) \)  is not contained in model class.
+and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \).
 
 
      
-
    1.  Do better fits lead to better predictions?
      2. 
-
    2.  What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points  \( Ntrain \)  and  \( \sigma \)?
      3. 
+
    3.  Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
      4. 
+
    4.  Repeat steps 1) to 3) such that each sample plays the role of the test set once.
      5. 
+
    5.  Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
      6. 
 
    
-
    
+
     
    
+$$
+\begin{align*}
+\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}.
+\end{align*}
+$$
+
     
    
 
-Summarize what you think you learned about the relationship of knowing the true model class and predictive power.
+The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.
+
+
      
+
    1.  The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
      2. 
+
    
 
    
 
 
 
    
-
    The code 
    
+
    Predicted Residual Error Sum of Squares 
    
+
    
+
+
    
+Another approach in the LOOCV scheme is to the use the so-called Predicted Residual Error Sum of Squares (PRESS).
 
 
    
+We can define the optimal penalty parameter to minimize
+
     
    
+$$
+\begin{align*}
+\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2.
+\end{align*}
+$$
+
     
    
 
-
-
    import numpy as np
-import sklearn as sk
-from sklearn import datasets, linear_model
-from sklearn.preprocessing import PolynomialFeatures
-
-import matplotlib as mpl
-from matplotlib import pyplot as plt
-
-%matplotlib notebook
-
-# The Training Data
-
-N_train=100
-
-sigma_train=1;
-
-# Train on integers
-x=np.linspace(0.05,0.95,N_train)
-# Draw random noise
-s = sigma_train*np.random.randn(N_train)
-
-#linear
-y=2*x+s
-
-#Tenth Order
-#y=2*x-10*x**5+15*x**10+s
-
-p1=plt.plot(x,y, "o",ms=15, label='Training')
-
-#Linear Regression
-# Create linear regression object
-clf = linear_model.LinearRegression()
-
-# Train the model using the training sets
-clf.fit(x[:, np.newaxis], y)
-# The coefficients
-
-xplot=np.linspace(0.02,0.98,200)
-linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
-
-#Polynomial Regression
-
-
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = linear_model.LinearRegression()
-clf3.fit(X,y)
-
-
-Xplot=poly3.fit_transform(xplot[:,np.newaxis])
-poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
-
-
-
-#poly5 = PolynomialFeatures(degree=5)
-#X = poly5.fit_transform(x[:,np.newaxis])
-#clf5 = linear_model.LinearRegression()
-#clf5.fit(X,y)
-
-#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
-#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
-
-poly10 = PolynomialFeatures(degree=10)
-X = poly10.fit_transform(x[:,np.newaxis])
-clf10 = linear_model.LinearRegression()
-clf10.fit(X,y)
-
-Xplot=poly10.fit_transform(xplot[:,np.newaxis])
-poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
-
-axes = plt.gca()
-axes.set_ylim([-7,7])
-
-handles, labels=axes.get_legend_handles_labels()
-plt.legend(handles,labels, loc='lower center')
-plt.xlabel("$x$")
-plt.ylabel("$y$")
-Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
-plt.title(Title+" (train)")
-plt.tight_layout()
-plt.show()
-
    
-
    
-
-
-
    
-
    Generating test data 
    
 
    
-
-
-
    # Generate Test Data
-
-#Number of test data
-N_test=20
-
-sigma_test=sigma_train
-
-max_x=1.2
-x_test=max_x*np.random.random(N_test)
-# Draw random noise
-s_test = sigma_test*np.random.randn(N_test)
-
-#Linear
-y_test=2*x_test+s_test
-#Tenth order
-#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
-
-#Make design matrices for prediction
-x_plot=np.linspace(0,max_x, 200)
-X3 = poly3.fit_transform(x_plot[:,np.newaxis])
-X10 = poly10.fit_transform(x_plot[:,np.newaxis])
-
-%matplotlib notebook
-
-fig = plt.figure() 
-p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
-p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
-p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
-p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
-
-
-plt.legend(loc=2)
-plt.xlabel('$x$')
-plt.ylabel('$y$')
-plt.legend(loc='best')
-plt.title(Title+" (pred.)")
-plt.tight_layout()
-plt.show()
-
-#Linear Filename
-#filename_test=Title+"pred-linear.pdf"
-#Tenth Order Filename
-#filename_test=Title+"pred-o10.pdf"
-#plt.savefig(filename_test)
-#plt.ylim((-6,12))
-
    
+The LOOCV prediction performance can be
+expressed analytically in terms of the known quantities derived from
+the design matrix and the parameters \( \beta \).
+
    -

    Lasso regression

    -
    +

    Bootstrap

    +
    + +

    +Bootstrapping is a nonparametric approach to statistical inference +that substitutes computation for more traditional distributional +assumptions and asymptotic results. Bootstrapping offers a number of +advantages: +

      +

    1. The bootstrap is quite general, although there are some cases in which it fails.
      2. -
      -

      Logistic regression

      +

    2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small.
      3. + +

    3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.
      4. +

    4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).
      5. +
    +
    diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html index 9b2031654..afaf59f99 100644 --- a/doc/pub/Regression/html/Regression-solarized.html +++ b/doc/pub/Regression/html/Regression-solarized.html @@ -130,31 +130,51 @@ div { text-align: justify; text-justify: inter-word; } '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -196,7 +216,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Sep 6, 2018

    +

    Sep 7, 2018












    @@ -1236,7 +1256,275 @@ $$











    -

    Code examples for Ridge and Lasso Regression

    +

    From standard regression to Ridge regressions

    + +

    +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. + +

    +









    + +

    Fixing the singularity

    + +

    +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix. + +

    +









    + +

    Fitting vs. predicting when data is in the model class

    + +

    +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    An example code without the model assessment part

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
    +

    +









    + +

    How can we effectively evaluate the various models?

    + +

    +In Ridge regression and the subsequent discussion of its properties +the bias or penalty parameter is considered known or `given'. In +practice, it is unknown and the user needs to make an informed +decision on its value. How do we do that? Much of the same considerations apply to the Lasso method. + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1317,86 +1605,7 @@ plt.show()











    -

    From standard regression to Ridge regressions

    - -

    -One of the typical problems we encounter with linear regression, in particular -when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, -are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) -may be linearly dependent, normally referred to as super-collinearity. -This means that the matrix may be rank deficient and it is basically impossible to -to model the data using linear regression. As an example, consider the matrix -$$ -\begin{align*} -\mathbf{X} & = \left[ -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right] -\end{align*} -$$ - -

    -The columns of \( \hat{X} \) are linearly dependent. We se this easily since the -the first column is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of the space spanned by the -column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number -of linearly independent columns. In this particular case the matrix has rank 2. - -

    -Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies -that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this -$$ -\begin{align*} -\hat{X} & = \left[ -\begin{array}{rr} -1 & -1 -\\ -1 & -1 -\end{array} \right]. -\end{align*} -$$ - -We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. -This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. - -

    -









    - -

    Fixing the singularity

    - -

    -If our design matrix \( \hat{X} \) which enters the linear regression problem -$$ -\begin{align} -\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, -\label{_auto1} -\end{align} -$$ - -has linearly dependent column vectors, we will not be able to compute the inverse -of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). -The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. -This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where -the regression parameters \( \beta_i \) cannot be estimated. - -

    -The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change -$$ -\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, -$$ - -where \( \hat{I} \) is the identity matrix. - -

    -









    - -

    A second-order polynomial with Ridge and Lasso

    +

    A second-order polynomial with Ridge and Lasso

    @@ -1510,197 +1719,227 @@ plt.show()











    -

    Fitting vs. predicting when data is in the model class

    - +

    Resampling methods

    +
    +

    -We start by considering the case -\( f(x)=2x \). +Resampling methods are an indispensable tool in modern +statistics. They involve repeatedly drawing samples from a training +set and refitting a model of interest on each sample in order to +obtain additional information about the fitted model. For example, in +order to estimate the variability of a linear regression fit, we can +repeatedly draw different samples from the training data, fit a linear +regression to each new sample, and then examine the extent to which +the resulting fits differ. Such an approach may allow us to obtain +information that would not be available from fitting the model only +once using the original training sample. +

    -

    -Then the data is clearly generated by a model that is contained within -all three model classes we are using to make predictions (linear -models, third order polynomials, and tenth order polynomials). - -

    -Run the code for the following cases: - -

      -
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. -
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. -
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. -
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. -
    - -Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? -Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.











    -

    Fitting versus predicting when data is not in the model class

    - +

    Resampling approaches can be computationally expensive

    +
    +

    -Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. +Resampling approaches can be computationally expensive, because they +involve fitting the same statistical method multiple times using +different subsets of the training data. However, due to recent +advances in computing power, the computational requirements of +resampling methods generally are not prohibitive. In this chapter, we +discuss two of the most commonly used resampling methods, +cross-validation and the bootstrap. Both methods are important tools +in the practical application of many statistical learning +procedures. For example, cross-validation can be used to estimate the +test error associated with a given statistical learning method in +order to evaluate its performance, or to select the appropriate level +of flexibility. The process of evaluating a model’s performance is +known as model assessment, whereas the process of selecting the proper +level of flexibility for a model is known as model selection. The +bootstrap is widely used. +

    -
      -
    1. Do better fits lead to better predictions?
      2. -
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. -
    - -Summarize what you think you learned about the relationship of knowing the true model class and predictive power.











    -

    The code

    +

    Log-likelihood

    +A popular strategy is to choose a penalty parameter that yields a good +but parsimonious model. Information criteria measure the balance +between model fit and model complexity. One possibility is Aikaike's +information criterion (AIC). +The AIC measures model fit by the log-likelihood +and model complexity is measured by the number of parameters used by +the model. The number of model parameters in regular regression simply +corresponds to the number of covariates in the model. Or, by the +degrees of freedom consumed by the model, which is equivalent to the +trace of the hat matrix. For ridge regression it thus seems natural to +define model complexity analogously by the trace of the ridge hat +matrix. This yields the AIC for the linear regression model with ridge +estimates: - -

    import numpy as np
-import sklearn as sk
-from sklearn import datasets, linear_model
-from sklearn.preprocessing import PolynomialFeatures
+$$
+\begin{align*}
+\mbox{AIC}(\lambda) & =  2 \, p - 2 \log(\hat{L})
+\\
+& =  2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\}
+\\
+& =  2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}
++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] +  \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2.
+\end{align*}
+$$
 
-import matplotlib as mpl
-from matplotlib import pyplot as plt
+The value of \( \lambda \) which minimizes \( \mbox{AIC}(\lambda) \) corresponds to the `optimal' balance of model complexity and overfitting.
 
-%matplotlib notebook
-
-# The Training Data
-
-N_train=100
-
-sigma_train=1;
-
-# Train on integers
-x=np.linspace(0.05,0.95,N_train)
-# Draw random noise
-s = sigma_train*np.random.randn(N_train)
-
-#linear
-y=2*x+s
-
-#Tenth Order
-#y=2*x-10*x**5+15*x**10+s
-
-p1=plt.plot(x,y, "o",ms=15, label='Training')
-
-#Linear Regression
-# Create linear regression object
-clf = linear_model.LinearRegression()
-
-# Train the model using the training sets
-clf.fit(x[:, np.newaxis], y)
-# The coefficients
-
-xplot=np.linspace(0.02,0.98,200)
-linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
-
-#Polynomial Regression
-
-
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = linear_model.LinearRegression()
-clf3.fit(X,y)
-
-
-Xplot=poly3.fit_transform(xplot[:,np.newaxis])
-poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
-
-
-
-#poly5 = PolynomialFeatures(degree=5)
-#X = poly5.fit_transform(x[:,np.newaxis])
-#clf5 = linear_model.LinearRegression()
-#clf5.fit(X,y)
-
-#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
-#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
-
-poly10 = PolynomialFeatures(degree=10)
-X = poly10.fit_transform(x[:,np.newaxis])
-clf10 = linear_model.LinearRegression()
-clf10.fit(X,y)
-
-Xplot=poly10.fit_transform(xplot[:,np.newaxis])
-poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
-
-axes = plt.gca()
-axes.set_ylim([-7,7])
-
-handles, labels=axes.get_legend_handles_labels()
-plt.legend(handles,labels, loc='lower center')
-plt.xlabel("$x$")
-plt.ylabel("$y$")
-Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
-plt.title(Title+" (train)")
-plt.tight_layout()
-plt.show()
-

    -

    Generating test data

    +

    Cross-validation

    +

    - - -

    # Generate Test Data
-
-#Number of test data
-N_test=20
-
-sigma_test=sigma_train
-
-max_x=1.2
-x_test=max_x*np.random.random(N_test)
-# Draw random noise
-s_test = sigma_test*np.random.randn(N_test)
-
-#Linear
-y_test=2*x_test+s_test
-#Tenth order
-#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
-
-#Make design matrices for prediction
-x_plot=np.linspace(0,max_x, 200)
-X3 = poly3.fit_transform(x_plot[:,np.newaxis])
-X10 = poly10.fit_transform(x_plot[:,np.newaxis])
-
-%matplotlib notebook
-
-fig = plt.figure() 
-p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
-p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
-p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
-p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
-
-
-plt.legend(loc=2)
-plt.xlabel('$x$')
-plt.ylabel('$y$')
-plt.legend(loc='best')
-plt.title(Title+" (pred.)")
-plt.tight_layout()
-plt.show()
-
-#Linear Filename
-#filename_test=Title+"pred-linear.pdf"
-#Tenth Order Filename
-#filename_test=Title+"pred-o10.pdf"
-#plt.savefig(filename_test)
-#plt.ylim((-6,12))
-
    -

    -









    - -

    Lasso regression

    +Instead of choosing the penalty parameter to balance model fit with +model complexity, cross-validation requires it (i.e. the penalty +parameter) to yield a model with good prediction +performance. Commonly, this performance is evaluated on novel +data. Novel data need not be easy to come by and one has to make do +with the data at hand. The setting of `original' and novel data is +then mimicked by sample splitting: the data set is divided into two +(groups of samples). One of these two data sets, called the training +set, plays the role of `original' data on which the model is +built. The second of these data sets, called the test set, plays the +role of the `novel' data and is used to evaluate the prediction +performance (often operationalized as the log-likelihood or the +prediction error or its square or the R2 score) of the model built on the training data set. This +procedure (model building and prediction evaluation on training and +test set, respectively) is done for a collection of possible penalty +parameter choices. The penalty parameter that yields the model with +the best prediction performance is to be preferred. The thus obtained +performance evaluation depends on the actual split of the data set. To +remove this dependence the data set is split many times into a +training and test set. For each split the model parameters are +estimated for all choices of \( \lambda \) using the training data and +estimated parameters are evaluated on the corresponding test set. The +penalty parameter that on average over the test sets performs best (in +some sense) is then selected.











    -

    Logistic regression

    +

    Computationally expensive

    + +

    +The validation set approach is conceptually simple and is easy to implement. But it has two potential drawbacks: + +

      +
    • The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.
      • +
    • In the validation approach, only a subset of the observations, those that are included in the training set rather than in the validation set are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
      • +
    + + + +

    Various steps in cross-validation

    + +

    +When the repetitive splitting of the data set is done randomly, +samples may accidently end up in a fast majority of the splits in +either training or test set. Such samples may have an unbalanced +influence on either model building or prediction evaluation. To avoid +this \( k \)-fold cross-validation structures the data splitting. The +samples are divided into \( k \) more or less equally sized exhaustive and +mutually exclusive subsets. In turn (at each split) one of these +subsets plays the role of the test set while the union of the +remaining subsets constitutes the training set. Such a splitting +warrants a balanced representation of each sample in both training and +test set over the splits. Still the division into the \( k \) subsets +involves a degree of randomness. This may be fully excluded when +choosing \( k=n \). This particular case is referred to as leave-one-out +cross-validation (LOOCV). + +

    + + +

    How to set up the cross-validation for Ridge and/or Lasso

    + +
      +
    1. Define a range of interest for the penalty parameter.
      2. +
    2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
      3. +
    3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set as
      4. +
    + +$$ +\begin{align*} +\hat{\beta}_{-i}(\lambda) & = ( \hat{X}_{-i, \ast}^{\top} +\hat{X}_{-i, \ast} + \lambda \hat{I}_{pp})^{-1} +\hat{X}_{-i, \ast}^{\top} \hat{y}_{-i} +\end{align*} +$$ + +and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \). + +
      +
    1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
      2. +
    2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
      3. +
    3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
      4. +
    + +$$ +\begin{align*} +\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\end{align*} +$$ + +The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. + +
      +
    1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
      2. +
    + +









    + +

    Predicted Residual Error Sum of Squares

    +
    + +

    +Another approach in the LOOCV scheme is to the use the so-called Predicted Residual Error Sum of Squares (PRESS). + +

    +We can define the optimal penalty parameter to minimize +$$ +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. +\end{align*} +$$ + +

    +The LOOCV prediction performance can be +expressed analytically in terms of the known quantities derived from +the design matrix and the parameters \( \beta \). +

    + + +

    +









    + +

    Bootstrap

    +
    + +

    +Bootstrapping is a nonparametric approach to statistical inference +that substitutes computation for more traditional distributional +assumptions and asymptotic results. Bootstrapping offers a number of +advantages: + +

      +
    1. The bootstrap is quite general, although there are some cases in which it fails.
      2. +
    2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small.
      3. +
    3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.
      4. +
    4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).
      5. +
    +
    + diff --git a/doc/pub/Regression/html/Regression.html b/doc/pub/Regression/html/Regression.html index fd86d2474..366f94425 100644 --- a/doc/pub/Regression/html/Regression.html +++ b/doc/pub/Regression/html/Regression.html @@ -135,31 +135,51 @@ div { text-align: justify; text-justify: inter-word; } '___sec32'), ('Expectation value and variance', 2, None, '___sec33'), ('The singular value decompostion', 2, None, '___sec34'), - ('Code examples for Ridge and Lasso Regression', - 2, - None, - '___sec35'), ('From standard regression to Ridge regressions', 2, None, - '___sec36'), - ('Fixing the singularity', 2, None, '___sec37'), - ('A second-order polynomial with Ridge and Lasso', - 2, - None, - '___sec38'), + '___sec35'), + ('Fixing the singularity', 2, None, '___sec36'), ('Fitting vs. predicting when data is in the model class', 2, None, - '___sec39'), + '___sec37'), ('Fitting versus predicting when data is not in the model class', 2, None, - '___sec40'), - ('The code', 2, None, '___sec41'), - ('Generating test data', 2, None, '___sec42'), - ('Lasso regression', 2, None, '___sec43'), - ('Logistic regression', 2, None, '___sec44')]} + '___sec38'), + ('An example code without the model assessment part', + 2, + None, + '___sec39'), + ('Generating test data', 2, None, '___sec40'), + ('How can we effectively evaluate the various models?', + 2, + None, + '___sec41'), + ('Code examples for Ridge and Lasso Regression', + 2, + None, + '___sec42'), + ('A second-order polynomial with Ridge and Lasso', + 2, + None, + '___sec43'), + ('Resampling methods', 2, None, '___sec44'), + ('Resampling approaches can be computationally expensive', + 2, + None, + '___sec45'), + ('Log-likelihood', 2, None, '___sec46'), + ('Cross-validation', 2, None, '___sec47'), + ('Computationally expensive', 2, None, '___sec48'), + ('Various steps in cross-validation', 2, None, '___sec49'), + ('How to set up the cross-validation for Ridge and/or Lasso', + 2, + None, + '___sec50'), + ('Predicted Residual Error Sum of Squares', 2, None, '___sec51'), + ('Bootstrap', 2, None, '___sec52')]} end of tocinfo --> @@ -201,7 +221,7 @@ MathJax.Hub.Config({
    [2] Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

    -

    Sep 6, 2018

    +

    Sep 7, 2018












    @@ -1241,7 +1261,275 @@ $$











    -

    Code examples for Ridge and Lasso Regression

    +

    From standard regression to Ridge regressions

    + +

    +One of the typical problems we encounter with linear regression, in particular +when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix +$$ +\begin{align*} +\mathbf{X} & = \left[ +\begin{array}{rrr} +1 & -1 & 2 +\\ +1 & 0 & 1 +\\ +1 & 2 & -1 +\\ +1 & 1 & 0 +\end{array} \right] +\end{align*} +$$ + +

    +The columns of \( \hat{X} \) are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the +column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number +of linearly independent columns. In this particular case the matrix has rank 2. + +

    +Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies +that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this +$$ +\begin{align*} +\hat{X} & = \left[ +\begin{array}{rr} +1 & -1 +\\ +1 & -1 +\end{array} \right]. +\end{align*} +$$ + +We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. +This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. + +

    +









    + +

    Fixing the singularity

    + +

    +If our design matrix \( \hat{X} \) which enters the linear regression problem +$$ +\begin{align} +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, +\label{_auto1} +\end{align} +$$ + +has linearly dependent column vectors, we will not be able to compute the inverse +of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). +The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. +This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters \( \beta_i \) cannot be estimated. + +

    +The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change +$$ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +$$ + +where \( \hat{I} \) is the identity matrix. + +

    +









    + +

    Fitting vs. predicting when data is in the model class

    + +

    +We start by considering the case +\( f(x)=2x \). + +

    +Then the data is clearly generated by a model that is contained within +all three model classes we are using to make predictions (linear +models, third order polynomials, and tenth order polynomials). + +

    +Run the code for the following cases: + +

      +
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. +
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. +
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. +
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. +
    + +Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? +Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well. + +

    +









    + +

    Fitting versus predicting when data is not in the model class

    + +

    +Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. + +

      +
    1. Do better fits lead to better predictions?
      2. +
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. +
    + +Summarize what you think you learned about the relationship of knowing the true model class and predictive power. + +

    +









    + +

    An example code without the model assessment part

    + +

    + + +

    import numpy as np
+import sklearn as sk
+from sklearn import datasets, linear_model
+from sklearn.preprocessing import PolynomialFeatures
+
+import matplotlib as mpl
+from matplotlib import pyplot as plt
+
+%matplotlib notebook
+
+# The Training Data
+
+N_train=100
+
+sigma_train=1;
+
+# Train on integers
+x=np.linspace(0.05,0.95,N_train)
+# Draw random noise
+s = sigma_train*np.random.randn(N_train)
+
+#linear
+y=2*x+s
+
+#Tenth Order
+#y=2*x-10*x**5+15*x**10+s
+
+p1=plt.plot(x,y, "o",ms=15, label='Training')
+
+#Linear Regression
+# Create linear regression object
+clf = linear_model.LinearRegression()
+
+# Train the model using the training sets
+clf.fit(x[:, np.newaxis], y)
+# The coefficients
+
+xplot=np.linspace(0.02,0.98,200)
+linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
+
+#Polynomial Regression
+
+
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = linear_model.LinearRegression()
+clf3.fit(X,y)
+
+
+Xplot=poly3.fit_transform(xplot[:,np.newaxis])
+poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
+
+
+
+#poly5 = PolynomialFeatures(degree=5)
+#X = poly5.fit_transform(x[:,np.newaxis])
+#clf5 = linear_model.LinearRegression()
+#clf5.fit(X,y)
+
+#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
+#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
+
+poly10 = PolynomialFeatures(degree=10)
+X = poly10.fit_transform(x[:,np.newaxis])
+clf10 = linear_model.LinearRegression()
+clf10.fit(X,y)
+
+Xplot=poly10.fit_transform(xplot[:,np.newaxis])
+poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
+
+axes = plt.gca()
+axes.set_ylim([-7,7])
+
+handles, labels=axes.get_legend_handles_labels()
+plt.legend(handles,labels, loc='lower center')
+plt.xlabel("$x$")
+plt.ylabel("$y$")
+Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
+plt.title(Title+" (train)")
+plt.tight_layout()
+plt.show()
+
    +

    + + +

    Generating test data

    +

    + + +

    # Generate Test Data
+
+#Number of test data
+N_test=20
+
+sigma_test=sigma_train
+
+max_x=1.2
+x_test=max_x*np.random.random(N_test)
+# Draw random noise
+s_test = sigma_test*np.random.randn(N_test)
+
+#Linear
+y_test=2*x_test+s_test
+#Tenth order
+#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
+
+#Make design matrices for prediction
+x_plot=np.linspace(0,max_x, 200)
+X3 = poly3.fit_transform(x_plot[:,np.newaxis])
+X10 = poly10.fit_transform(x_plot[:,np.newaxis])
+
+%matplotlib notebook
+
+fig = plt.figure() 
+p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
+p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
+p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
+p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
+
+
+plt.legend(loc=2)
+plt.xlabel('$x$')
+plt.ylabel('$y$')
+plt.legend(loc='best')
+plt.title(Title+" (pred.)")
+plt.tight_layout()
+plt.show()
+
    +

    +









    + +

    How can we effectively evaluate the various models?

    + +

    +In Ridge regression and the subsequent discussion of its properties +the bias or penalty parameter is considered known or `given'. In +practice, it is unknown and the user needs to make an informed +decision on its value. How do we do that? Much of the same considerations apply to the Lasso method. + +

    +









    + +

    Code examples for Ridge and Lasso Regression

    @@ -1322,86 +1610,7 @@ plt.show()











    -

    From standard regression to Ridge regressions

    - -

    -One of the typical problems we encounter with linear regression, in particular -when the matrix \( \hat{X} \) (our so-called design matrix) is high-dimensional, -are problems with near singular or singular matrices. The column vectors of \( \hat{X} \) -may be linearly dependent, normally referred to as super-collinearity. -This means that the matrix may be rank deficient and it is basically impossible to -to model the data using linear regression. As an example, consider the matrix -$$ -\begin{align*} -\mathbf{X} & = \left[ -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right] -\end{align*} -$$ - -

    -The columns of \( \hat{X} \) are linearly dependent. We se this easily since the -the first column is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of the space spanned by the -column vectors. Hence, the rank of \( \mathbf{X} \) is equal to the number -of linearly independent columns. In this particular case the matrix has rank 2. - -

    -Super-collinearity of an \( (n \times p) \)-dimensional design matrix \( \mathbf{X} \) implies -that the inverse of the matrix \( \hat{X}^T\hat{x} \) (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this -$$ -\begin{align*} -\hat{X} & = \left[ -\begin{array}{rr} -1 & -1 -\\ -1 & -1 -\end{array} \right]. -\end{align*} -$$ - -We see easily that \( \mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0 \). Hence, \( \mathbf{X} \) is singular and its inverse is undefined. -This is equivalent to saying that the matrix \( \hat{X} \) has at least an eigenvalue which is zero. - -

    -









    - -

    Fixing the singularity

    - -

    -If our design matrix \( \hat{X} \) which enters the linear regression problem -$$ -\begin{align} -\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, -\label{_auto1} -\end{align} -$$ - -has linearly dependent column vectors, we will not be able to compute the inverse -of \( \hat{X}^T\hat{X} \) and we cannot find the parameters (estimators) \( \beta_i \). -The estimators are only well-defined if \( (\hat{X}^{T}\hat{X})^{-1} \) exits. -This is more likely to happen when the matrix \( \hat{X} \) is high-dimensional. In this case it is likely to encounter a situation where -the regression parameters \( \beta_i \) cannot be estimated. - -

    -The ad hoc approach which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change -$$ -\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, -$$ - -where \( \hat{I} \) is the identity matrix. - -

    -









    - -

    A second-order polynomial with Ridge and Lasso

    +

    A second-order polynomial with Ridge and Lasso

    @@ -1515,197 +1724,227 @@ plt.show()











    -

    Fitting vs. predicting when data is in the model class

    - +

    Resampling methods

    +
    +

    -We start by considering the case -\( f(x)=2x \). +Resampling methods are an indispensable tool in modern +statistics. They involve repeatedly drawing samples from a training +set and refitting a model of interest on each sample in order to +obtain additional information about the fitted model. For example, in +order to estimate the variability of a linear regression fit, we can +repeatedly draw different samples from the training data, fit a linear +regression to each new sample, and then examine the extent to which +the resulting fits differ. Such an approach may allow us to obtain +information that would not be available from fitting the model only +once using the original training sample. +

    -

    -Then the data is clearly generated by a model that is contained within -all three model classes we are using to make predictions (linear -models, third order polynomials, and tenth order polynomials). - -

    -Run the code for the following cases: - -

      -
    1. For \( f(x)=2x \) , \( Ntrain=10 \) and \( \sigma =0 \) (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when \( x \in [0,1] \) . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?
      2. -
    2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?
      3. -
    3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example \( x \in [0,1.2] \) ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?
      4. -
    4. Repeat the above for \( f(x)=2x \) , \( Ntrain=10 \) , and \( \sigma=1 \) . What changes?
      5. -
    - -Repeat the exercises above for \( f(x)=2x \) , \( Ntrain=100 \) , and \( \sigma=1 \) . What changes? -Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.











    -

    Fitting versus predicting when data is not in the model class

    - +

    Resampling approaches can be computationally expensive

    +
    +

    -Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider \( f(x)=2x-10x^5+15x^{10} \) . Notice that the for linear and third-order polynomial the true model \( f(x) \) is not contained in model class. +Resampling approaches can be computationally expensive, because they +involve fitting the same statistical method multiple times using +different subsets of the training data. However, due to recent +advances in computing power, the computational requirements of +resampling methods generally are not prohibitive. In this chapter, we +discuss two of the most commonly used resampling methods, +cross-validation and the bootstrap. Both methods are important tools +in the practical application of many statistical learning +procedures. For example, cross-validation can be used to estimate the +test error associated with a given statistical learning method in +order to evaluate its performance, or to select the appropriate level +of flexibility. The process of evaluating a model’s performance is +known as model assessment, whereas the process of selecting the proper +level of flexibility for a model is known as model selection. The +bootstrap is widely used. +

    -
      -
    1. Do better fits lead to better predictions?
      2. -
    2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points \( Ntrain \) and \( \sigma \)?
      3. -
    - -Summarize what you think you learned about the relationship of knowing the true model class and predictive power.











    -

    The code

    +

    Log-likelihood

    +A popular strategy is to choose a penalty parameter that yields a good +but parsimonious model. Information criteria measure the balance +between model fit and model complexity. One possibility is Aikaike's +information criterion (AIC). +The AIC measures model fit by the log-likelihood +and model complexity is measured by the number of parameters used by +the model. The number of model parameters in regular regression simply +corresponds to the number of covariates in the model. Or, by the +degrees of freedom consumed by the model, which is equivalent to the +trace of the hat matrix. For ridge regression it thus seems natural to +define model complexity analogously by the trace of the ridge hat +matrix. This yields the AIC for the linear regression model with ridge +estimates: - -

    import numpy as np
-import sklearn as sk
-from sklearn import datasets, linear_model
-from sklearn.preprocessing import PolynomialFeatures
+$$
+\begin{align*}
+\mbox{AIC}(\lambda) & =  2 \, p - 2 \log(\hat{L})
+\\
+& =  2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\}
+\\
+& =  2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}
++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] +  \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2.
+\end{align*}
+$$
 
-import matplotlib as mpl
-from matplotlib import pyplot as plt
+The value of \( \lambda \) which minimizes \( \mbox{AIC}(\lambda) \) corresponds to the `optimal' balance of model complexity and overfitting.
 
-%matplotlib notebook
-
-# The Training Data
-
-N_train=100
-
-sigma_train=1;
-
-# Train on integers
-x=np.linspace(0.05,0.95,N_train)
-# Draw random noise
-s = sigma_train*np.random.randn(N_train)
-
-#linear
-y=2*x+s
-
-#Tenth Order
-#y=2*x-10*x**5+15*x**10+s
-
-p1=plt.plot(x,y, "o",ms=15, label='Training')
-
-#Linear Regression
-# Create linear regression object
-clf = linear_model.LinearRegression()
-
-# Train the model using the training sets
-clf.fit(x[:, np.newaxis], y)
-# The coefficients
-
-xplot=np.linspace(0.02,0.98,200)
-linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')
-
-#Polynomial Regression
-
-
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = linear_model.LinearRegression()
-clf3.fit(X,y)
-
-
-Xplot=poly3.fit_transform(xplot[:,np.newaxis])
-poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')
-
-
-
-#poly5 = PolynomialFeatures(degree=5)
-#X = poly5.fit_transform(x[:,np.newaxis])
-#clf5 = linear_model.LinearRegression()
-#clf5.fit(X,y)
-
-#Xplot=poly5.fit_transform(xplot[:,np.newaxis])
-#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)
-
-poly10 = PolynomialFeatures(degree=10)
-X = poly10.fit_transform(x[:,np.newaxis])
-clf10 = linear_model.LinearRegression()
-clf10.fit(X,y)
-
-Xplot=poly10.fit_transform(xplot[:,np.newaxis])
-poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')
-
-axes = plt.gca()
-axes.set_ylim([-7,7])
-
-handles, labels=axes.get_legend_handles_labels()
-plt.legend(handles,labels, loc='lower center')
-plt.xlabel("$x$")
-plt.ylabel("$y$")
-Title="$N=$"+str(N_train)+", $\sigma=$"+str(sigma_train)
-plt.title(Title+" (train)")
-plt.tight_layout()
-plt.show()
-

    -

    Generating test data

    +

    Cross-validation

    +

    - - -

    # Generate Test Data
-
-#Number of test data
-N_test=20
-
-sigma_test=sigma_train
-
-max_x=1.2
-x_test=max_x*np.random.random(N_test)
-# Draw random noise
-s_test = sigma_test*np.random.randn(N_test)
-
-#Linear
-y_test=2*x_test+s_test
-#Tenth order
-#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test
-
-#Make design matrices for prediction
-x_plot=np.linspace(0,max_x, 200)
-X3 = poly3.fit_transform(x_plot[:,np.newaxis])
-X10 = poly10.fit_transform(x_plot[:,np.newaxis])
-
-%matplotlib notebook
-
-fig = plt.figure() 
-p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')
-p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')
-p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')
-p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')
-
-
-plt.legend(loc=2)
-plt.xlabel('$x$')
-plt.ylabel('$y$')
-plt.legend(loc='best')
-plt.title(Title+" (pred.)")
-plt.tight_layout()
-plt.show()
-
-#Linear Filename
-#filename_test=Title+"pred-linear.pdf"
-#Tenth Order Filename
-#filename_test=Title+"pred-o10.pdf"
-#plt.savefig(filename_test)
-#plt.ylim((-6,12))
-
    -

    -









    - -

    Lasso regression

    +Instead of choosing the penalty parameter to balance model fit with +model complexity, cross-validation requires it (i.e. the penalty +parameter) to yield a model with good prediction +performance. Commonly, this performance is evaluated on novel +data. Novel data need not be easy to come by and one has to make do +with the data at hand. The setting of `original' and novel data is +then mimicked by sample splitting: the data set is divided into two +(groups of samples). One of these two data sets, called the training +set, plays the role of `original' data on which the model is +built. The second of these data sets, called the test set, plays the +role of the `novel' data and is used to evaluate the prediction +performance (often operationalized as the log-likelihood or the +prediction error or its square or the R2 score) of the model built on the training data set. This +procedure (model building and prediction evaluation on training and +test set, respectively) is done for a collection of possible penalty +parameter choices. The penalty parameter that yields the model with +the best prediction performance is to be preferred. The thus obtained +performance evaluation depends on the actual split of the data set. To +remove this dependence the data set is split many times into a +training and test set. For each split the model parameters are +estimated for all choices of \( \lambda \) using the training data and +estimated parameters are evaluated on the corresponding test set. The +penalty parameter that on average over the test sets performs best (in +some sense) is then selected.











    -

    Logistic regression

    +

    Computationally expensive

    + +

    +The validation set approach is conceptually simple and is easy to implement. But it has two potential drawbacks: + +

      +
    • The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.
      • +
    • In the validation approach, only a subset of the observations, those that are included in the training set rather than in the validation set are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
      • +
    + + + +

    Various steps in cross-validation

    + +

    +When the repetitive splitting of the data set is done randomly, +samples may accidently end up in a fast majority of the splits in +either training or test set. Such samples may have an unbalanced +influence on either model building or prediction evaluation. To avoid +this \( k \)-fold cross-validation structures the data splitting. The +samples are divided into \( k \) more or less equally sized exhaustive and +mutually exclusive subsets. In turn (at each split) one of these +subsets plays the role of the test set while the union of the +remaining subsets constitutes the training set. Such a splitting +warrants a balanced representation of each sample in both training and +test set over the splits. Still the division into the \( k \) subsets +involves a degree of randomness. This may be fully excluded when +choosing \( k=n \). This particular case is referred to as leave-one-out +cross-validation (LOOCV). + +

    + + +

    How to set up the cross-validation for Ridge and/or Lasso

    + +
      +
    1. Define a range of interest for the penalty parameter.
      2. +
    2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
      3. +
    3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set as
      4. +
    + +$$ +\begin{align*} +\hat{\beta}_{-i}(\lambda) & = ( \hat{X}_{-i, \ast}^{\top} +\hat{X}_{-i, \ast} + \lambda \hat{I}_{pp})^{-1} +\hat{X}_{-i, \ast}^{\top} \hat{y}_{-i} +\end{align*} +$$ + +and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \). + +
      +
    1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
      2. +
    2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
      3. +
    3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
      4. +
    + +$$ +\begin{align*} +\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\end{align*} +$$ + +The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. + +
      +
    1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
      2. +
    + +









    + +

    Predicted Residual Error Sum of Squares

    +
    + +

    +Another approach in the LOOCV scheme is to the use the so-called Predicted Residual Error Sum of Squares (PRESS). + +

    +We can define the optimal penalty parameter to minimize +$$ +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. +\end{align*} +$$ + +

    +The LOOCV prediction performance can be +expressed analytically in terms of the known quantities derived from +the design matrix and the parameters \( \beta \). +

    + + +

    +









    + +

    Bootstrap

    +
    + +

    +Bootstrapping is a nonparametric approach to statistical inference +that substitutes computation for more traditional distributional +assumptions and asymptotic results. Bootstrapping offers a number of +advantages: + +

      +
    1. The bootstrap is quite general, although there are some cases in which it fails.
      2. +
    2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small.
      3. +
    3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.
      4. +
    4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).
      5. +
    +
    + diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index af7e36ea4..e36087719 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -10,7 +10,7 @@ " \n", "**Morten Hjorth-Jensen**, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University\n", "\n", - "Date: **Sep 6, 2018**\n", + "Date: **Sep 7, 2018**\n", "\n", "Copyright 1999-2018, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license\n", "\n", @@ -1606,95 +1606,6 @@ "$$" ] }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "## Code examples for Ridge and Lasso Regression" - ] - }, - { - "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [], - "source": [ - "import matplotlib.pyplot as plt\n", - "import numpy as np\n", - "from sklearn import linear_model\n", - "from sklearn.linear_model import LinearRegression\n", - "from sklearn.metrics import mean_squared_error, r2_score\n", - "\n", - "#creating data with random noise\n", - "x=np.arange(50)\n", - "\n", - "delta=np.random.uniform(-2.5,2.5, size=(50))\n", - "np.random.shuffle(delta)\n", - "y =0.5*x+5+delta\n", - "\n", - "#arranging data into 2x50 matrix\n", - "a=np.array(x) #inputs\n", - "b=np.array(y) #outputs\n", - "\n", - "#Split into training and test\n", - "X_train=a[:37, np.newaxis]\n", - "X_test=a[37:, np.newaxis]\n", - "y_train=b[:37]\n", - "y_test=b[37:]\n", - "\n", - "print (\"X_train: \", X_train.shape)\n", - "print (\"y_train: \", y_train.shape)\n", - "print (\"X_test: \", X_test.shape)\n", - "print (\"y_test: \", y_test.shape)\n", - "\n", - "print (\"------------------------------------\")\n", - "\n", - "print (\"Ordinary Least Squares\")\n", - "#Add Ordinary Least Squares fit\n", - "reg=LinearRegression()\n", - "reg.fit(X_train, y_train)\n", - "pred=reg.predict(X_test)\n", - "print (\"Prediction Shape: \", pred.shape)\n", - "\n", - "print('Coefficients: \\n', reg.coef_)\n", - "# The mean squared error\n", - "print(\"Mean squared error: %.2f\"\n", - " % mean_squared_error(y_test, pred))\n", - "# Explained variance score: 1 is perfect prediction\n", - "print('Variance score: %.2f' % r2_score(y_test, pred))\n", - "\n", - "#plot\n", - "plt.scatter(X_test,y_test,color='green', label=\"Training Data\")\n", - "plt.plot(X_test, pred, color='black', label=\"Fit Line\")\n", - "plt.legend()\n", - "plt.show()\n", - "\n", - "print (\"------------------------------------\")\n", - "\n", - "print (\"Ridge Regression\")\n", - "\n", - "ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])\n", - "ridge.fit(X_train,y_train)\n", - "print (\"Ridge Coefficient: \",ridge.coef_)\n", - "print (\"Ridge Intercept: \", ridge.intercept_)\n", - "#Look into graphing with Ridge fit\n", - "\n", - "print (\"------------------------------------\")\n", - "\n", - "print (\"Lasso\")\n", - "lasso=linear_model.Lasso(alpha=0.1)\n", - "lasso.fit(X_train,y_train)\n", - "predl=lasso.predict(X_test)\n", - "print(\"Lasso Coefficient: \", lasso.coef_)\n", - "print(\"Lasso Intercept: \", lasso.intercept_)\n", - "plt.scatter(X_test,y_test,color='green', label=\"Training Data\")\n", - "plt.plot(X_test, predl, color='blue', label=\"Lasso\")\n", - "plt.legend()\n", - "plt.show()" - ] - }, { "cell_type": "markdown", "metadata": {}, @@ -1816,7 +1727,141 @@ "\n", "\n", "\n", - "## A second-order polynomial with Ridge and Lasso" + "\n", + "\n", + "## Fitting vs. predicting when data is in the model class\n", + "\n", + "We start by considering the case\n", + "$f(x)=2x$.\n", + "\n", + "Then the data is clearly generated by a model that is contained within\n", + "all three model classes we are using to make predictions (linear\n", + "models, third order polynomials, and tenth order polynomials).\n", + "\n", + "Run the code for the following cases:\n", + "\n", + "1. For $f(x)=2x$ , $Ntrain=10$ and $\\sigma =0$ (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when $x \\in [0,1]$ . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?\n", + "\n", + "2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?\n", + "\n", + "3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example $x \\in [0,1.2]$ ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?\n", + "\n", + "4. Repeat the above for $f(x)=2x$ , $Ntrain=10$ , and $\\sigma=1$ . What changes?\n", + "\n", + "Repeat the exercises above for $f(x)=2x$ , $Ntrain=100$ , and $\\sigma=1$ . What changes?\n", + "Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.\n", + "\n", + "\n", + "## Fitting versus predicting when data is not in the model class\n", + "\n", + "Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider $f(x)=2x-10x^5+15x^{10}$ . Notice that the for linear and third-order polynomial the true model $f(x)$ is not contained in model class.\n", + "\n", + "1. Do better fits lead to better predictions?\n", + "\n", + "2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points $Ntrain$ and $\\sigma$?\n", + "\n", + "Summarize what you think you learned about the relationship of knowing the true model class and predictive power.\n", + "\n", + "## An example code without the model assessment part" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import numpy as np\n", + "import sklearn as sk\n", + "from sklearn import datasets, linear_model\n", + "from sklearn.preprocessing import PolynomialFeatures\n", + "\n", + "import matplotlib as mpl\n", + "from matplotlib import pyplot as plt\n", + "\n", + "%matplotlib notebook\n", + "\n", + "# The Training Data\n", + "\n", + "N_train=100\n", + "\n", + "sigma_train=1;\n", + "\n", + "# Train on integers\n", + "x=np.linspace(0.05,0.95,N_train)\n", + "# Draw random noise\n", + "s = sigma_train*np.random.randn(N_train)\n", + "\n", + "#linear\n", + "y=2*x+s\n", + "\n", + "#Tenth Order\n", + "#y=2*x-10*x**5+15*x**10+s\n", + "\n", + "p1=plt.plot(x,y, \"o\",ms=15, label='Training')\n", + "\n", + "#Linear Regression\n", + "# Create linear regression object\n", + "clf = linear_model.LinearRegression()\n", + "\n", + "# Train the model using the training sets\n", + "clf.fit(x[:, np.newaxis], y)\n", + "# The coefficients\n", + "\n", + "xplot=np.linspace(0.02,0.98,200)\n", + "linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')\n", + "\n", + "#Polynomial Regression\n", + "\n", + "\n", + "poly3 = PolynomialFeatures(degree=3)\n", + "X = poly3.fit_transform(x[:,np.newaxis])\n", + "clf3 = linear_model.LinearRegression()\n", + "clf3.fit(X,y)\n", + "\n", + "\n", + "Xplot=poly3.fit_transform(xplot[:,np.newaxis])\n", + "poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')\n", + "\n", + "\n", + "\n", + "#poly5 = PolynomialFeatures(degree=5)\n", + "#X = poly5.fit_transform(x[:,np.newaxis])\n", + "#clf5 = linear_model.LinearRegression()\n", + "#clf5.fit(X,y)\n", + "\n", + "#Xplot=poly5.fit_transform(xplot[:,np.newaxis])\n", + "#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)\n", + "\n", + "poly10 = PolynomialFeatures(degree=10)\n", + "X = poly10.fit_transform(x[:,np.newaxis])\n", + "clf10 = linear_model.LinearRegression()\n", + "clf10.fit(X,y)\n", + "\n", + "Xplot=poly10.fit_transform(xplot[:,np.newaxis])\n", + "poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')\n", + "\n", + "axes = plt.gca()\n", + "axes.set_ylim([-7,7])\n", + "\n", + "handles, labels=axes.get_legend_handles_labels()\n", + "plt.legend(handles,labels, loc='lower center')\n", + "plt.xlabel(\"$x$\")\n", + "plt.ylabel(\"$y$\")\n", + "Title=\"$N=$\"+str(N_train)+\", $\\sigma=$\"+str(sigma_train)\n", + "plt.title(Title+\" (train)\")\n", + "plt.tight_layout()\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "\n", + "## Generating test data" ] }, { @@ -1826,6 +1871,157 @@ "collapsed": false }, "outputs": [], + "source": [ + "# Generate Test Data\n", + "\n", + "#Number of test data\n", + "N_test=20\n", + "\n", + "sigma_test=sigma_train\n", + "\n", + "max_x=1.2\n", + "x_test=max_x*np.random.random(N_test)\n", + "# Draw random noise\n", + "s_test = sigma_test*np.random.randn(N_test)\n", + "\n", + "#Linear\n", + "y_test=2*x_test+s_test\n", + "#Tenth order\n", + "#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", + "\n", + "#Make design matrices for prediction\n", + "x_plot=np.linspace(0,max_x, 200)\n", + "X3 = poly3.fit_transform(x_plot[:,np.newaxis])\n", + "X10 = poly10.fit_transform(x_plot[:,np.newaxis])\n", + "\n", + "%matplotlib notebook\n", + "\n", + "fig = plt.figure() \n", + "p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')\n", + "p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')\n", + "p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')\n", + "p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')\n", + "\n", + "\n", + "plt.legend(loc=2)\n", + "plt.xlabel('$x$')\n", + "plt.ylabel('$y$')\n", + "plt.legend(loc='best')\n", + "plt.title(Title+\" (pred.)\")\n", + "plt.tight_layout()\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## How can we effectively evaluate the various models?\n", + "\n", + "In Ridge regression and the subsequent discussion of its properties\n", + "the bias or penalty parameter is considered known or `given'. In\n", + "practice, it is unknown and the user needs to make an informed\n", + "decision on its value. How do we do that? Much of the same considerations apply to the Lasso method. \n", + "\n", + "## Code examples for Ridge and Lasso Regression" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "from sklearn import linear_model\n", + "from sklearn.linear_model import LinearRegression\n", + "from sklearn.metrics import mean_squared_error, r2_score\n", + "\n", + "#creating data with random noise\n", + "x=np.arange(50)\n", + "\n", + "delta=np.random.uniform(-2.5,2.5, size=(50))\n", + "np.random.shuffle(delta)\n", + "y =0.5*x+5+delta\n", + "\n", + "#arranging data into 2x50 matrix\n", + "a=np.array(x) #inputs\n", + "b=np.array(y) #outputs\n", + "\n", + "#Split into training and test\n", + "X_train=a[:37, np.newaxis]\n", + "X_test=a[37:, np.newaxis]\n", + "y_train=b[:37]\n", + "y_test=b[37:]\n", + "\n", + "print (\"X_train: \", X_train.shape)\n", + "print (\"y_train: \", y_train.shape)\n", + "print (\"X_test: \", X_test.shape)\n", + "print (\"y_test: \", y_test.shape)\n", + "\n", + "print (\"------------------------------------\")\n", + "\n", + "print (\"Ordinary Least Squares\")\n", + "#Add Ordinary Least Squares fit\n", + "reg=LinearRegression()\n", + "reg.fit(X_train, y_train)\n", + "pred=reg.predict(X_test)\n", + "print (\"Prediction Shape: \", pred.shape)\n", + "\n", + "print('Coefficients: \\n', reg.coef_)\n", + "# The mean squared error\n", + "print(\"Mean squared error: %.2f\"\n", + " % mean_squared_error(y_test, pred))\n", + "# Explained variance score: 1 is perfect prediction\n", + "print('Variance score: %.2f' % r2_score(y_test, pred))\n", + "\n", + "#plot\n", + "plt.scatter(X_test,y_test,color='green', label=\"Training Data\")\n", + "plt.plot(X_test, pred, color='black', label=\"Fit Line\")\n", + "plt.legend()\n", + "plt.show()\n", + "\n", + "print (\"------------------------------------\")\n", + "\n", + "print (\"Ridge Regression\")\n", + "\n", + "ridge=linear_model.RidgeCV(alphas=[0.1,1.0,10.0])\n", + "ridge.fit(X_train,y_train)\n", + "print (\"Ridge Coefficient: \",ridge.coef_)\n", + "print (\"Ridge Intercept: \", ridge.intercept_)\n", + "#Look into graphing with Ridge fit\n", + "\n", + "print (\"------------------------------------\")\n", + "\n", + "print (\"Lasso\")\n", + "lasso=linear_model.Lasso(alpha=0.1)\n", + "lasso.fit(X_train,y_train)\n", + "predl=lasso.predict(X_test)\n", + "print(\"Lasso Coefficient: \", lasso.coef_)\n", + "print(\"Lasso Intercept: \", lasso.intercept_)\n", + "plt.scatter(X_test,y_test,color='green', label=\"Training Data\")\n", + "plt.plot(X_test, predl, color='blue', label=\"Lasso\")\n", + "plt.legend()\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## A second-order polynomial with Ridge and Lasso" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", @@ -1939,204 +2135,229 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Fitting vs. predicting when data is in the model class\n", - "\n", - "We start by considering the case\n", - "$f(x)=2x$.\n", - "\n", - "Then the data is clearly generated by a model that is contained within\n", - "all three model classes we are using to make predictions (linear\n", - "models, third order polynomials, and tenth order polynomials).\n", - "\n", - "Run the code for the following cases:\n", - "\n", - "1. For $f(x)=2x$ , $Ntrain=10$ and $\\sigma =0$ (noiseless case), train the three classes of models (linear, third-order polynomial, and tenth order polynomial) for a training set when $x \\in [0,1]$ . Make graphs comparing fits for different order of polynomials. Which model fits the data the best?\n", - "\n", - "2. Do you think that the data that has the least error on the training set will also make the best predictions? Why or why not? Can you try to discuss and formalize your intuition? What can go right and what can go wrong?\n", - "\n", - "3. Check your answer by seeing how well your fits predict newly generated test data (including on data outside the range you fit on, for example $x \\in [0,1.2]$ ) using the code below. How well do you do on points in the range of x where you trained the model? How about points outside the original training data set?\n", - "\n", - "4. Repeat the above for $f(x)=2x$ , $Ntrain=10$ , and $\\sigma=1$ . What changes?\n", - "\n", - "Repeat the exercises above for $f(x)=2x$ , $Ntrain=100$ , and $\\sigma=1$ . What changes?\n", - "Summarize what you have learned about the relationship between model complexity (number of parameters), goodness of fit on training data, and the ability to predict well.\n", - "\n", - "\n", - "## Fitting versus predicting when data is not in the model class\n", - "\n", - "Thus far, we have considered the case where the data is generated using a model contained in the model class. Now consider $f(x)=2x-10x^5+15x^{10}$ . Notice that the for linear and third-order polynomial the true model $f(x)$ is not contained in model class.\n", - "\n", - "1. Do better fits lead to better predictions?\n", - "\n", - "2. What is the relationship between the true model for generating the data and the model class that has the most predictive power? How is this related to the model complexity? How does this depend on the number of data points $Ntrain$ and $\\sigma$?\n", - "\n", - "Summarize what you think you learned about the relationship of knowing the true model class and predictive power.\n", - "\n", - "## The code" - ] - }, - { - "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [], - "source": [ - "import numpy as np\n", - "import sklearn as sk\n", - "from sklearn import datasets, linear_model\n", - "from sklearn.preprocessing import PolynomialFeatures\n", - "\n", - "import matplotlib as mpl\n", - "from matplotlib import pyplot as plt\n", - "\n", - "%matplotlib notebook\n", - "\n", - "# The Training Data\n", - "\n", - "N_train=100\n", - "\n", - "sigma_train=1;\n", - "\n", - "# Train on integers\n", - "x=np.linspace(0.05,0.95,N_train)\n", - "# Draw random noise\n", - "s = sigma_train*np.random.randn(N_train)\n", - "\n", - "#linear\n", - "y=2*x+s\n", - "\n", - "#Tenth Order\n", - "#y=2*x-10*x**5+15*x**10+s\n", - "\n", - "p1=plt.plot(x,y, \"o\",ms=15, label='Training')\n", - "\n", - "#Linear Regression\n", - "# Create linear regression object\n", - "clf = linear_model.LinearRegression()\n", - "\n", - "# Train the model using the training sets\n", - "clf.fit(x[:, np.newaxis], y)\n", - "# The coefficients\n", - "\n", - "xplot=np.linspace(0.02,0.98,200)\n", - "linear_plot=plt.plot(xplot, clf.predict(xplot[:, np.newaxis]),label='Linear')\n", - "\n", - "#Polynomial Regression\n", - "\n", - "\n", - "poly3 = PolynomialFeatures(degree=3)\n", - "X = poly3.fit_transform(x[:,np.newaxis])\n", - "clf3 = linear_model.LinearRegression()\n", - "clf3.fit(X,y)\n", - "\n", - "\n", - "Xplot=poly3.fit_transform(xplot[:,np.newaxis])\n", - "poly3_plot=plt.plot(xplot, clf3.predict(Xplot), label='Poly 3')\n", + "## Resampling methods\n", + "Resampling methods are an indispensable tool in modern\n", + "statistics. They involve repeatedly drawing samples from a training\n", + "set and refitting a model of interest on each sample in order to\n", + "obtain additional information about the fitted model. For example, in\n", + "order to estimate the variability of a linear regression fit, we can\n", + "repeatedly draw different samples from the training data, fit a linear\n", + "regression to each new sample, and then examine the extent to which\n", + "the resulting fits differ. Such an approach may allow us to obtain\n", + "information that would not be available from fitting the model only\n", + "once using the original training sample.\n", "\n", "\n", "\n", - "#poly5 = PolynomialFeatures(degree=5)\n", - "#X = poly5.fit_transform(x[:,np.newaxis])\n", - "#clf5 = linear_model.LinearRegression()\n", - "#clf5.fit(X,y)\n", + "## Resampling approaches can be computationally expensive\n", + "Resampling approaches can be computationally expensive, because they\n", + "involve fitting the same statistical method multiple times using\n", + "different subsets of the training data. However, due to recent\n", + "advances in computing power, the computational requirements of\n", + "resampling methods generally are not prohibitive. In this chapter, we\n", + "discuss two of the most commonly used resampling methods,\n", + "cross-validation and the bootstrap. Both methods are important tools\n", + "in the practical application of many statistical learning\n", + "procedures. For example, cross-validation can be used to estimate the\n", + "test error associated with a given statistical learning method in\n", + "order to evaluate its performance, or to select the appropriate level\n", + "of flexibility. The process of evaluating a model’s performance is\n", + "known as model assessment, whereas the process of selecting the proper\n", + "level of flexibility for a model is known as model selection. The\n", + "bootstrap is widely used.\n", "\n", - "#Xplot=poly5.fit_transform(xplot[:,np.newaxis])\n", - "#plt.plot(xplot, clf5.predict(Xplot), 'r--',linewidth=1)\n", "\n", - "poly10 = PolynomialFeatures(degree=10)\n", - "X = poly10.fit_transform(x[:,np.newaxis])\n", - "clf10 = linear_model.LinearRegression()\n", - "clf10.fit(X,y)\n", "\n", - "Xplot=poly10.fit_transform(xplot[:,np.newaxis])\n", - "poly10_plot=plt.plot(xplot, clf10.predict(Xplot), label='Poly 10')\n", "\n", - "axes = plt.gca()\n", - "axes.set_ylim([-7,7])\n", + "## Log-likelihood\n", "\n", - "handles, labels=axes.get_legend_handles_labels()\n", - "plt.legend(handles,labels, loc='lower center')\n", - "plt.xlabel(\"$x$\")\n", - "plt.ylabel(\"$y$\")\n", - "Title=\"$N=$\"+str(N_train)+\", $\\sigma=$\"+str(sigma_train)\n", - "plt.title(Title+\" (train)\")\n", - "plt.tight_layout()\n", - "plt.show()" + "A popular strategy is to choose a penalty parameter that yields a good\n", + "but parsimonious model. Information criteria measure the balance\n", + "between model fit and model complexity. One possibility is Aikaike's\n", + "information criterion (AIC).\n", + "The AIC measures model fit by the log-likelihood\n", + "and model complexity is measured by the number of parameters used by\n", + "the model. The number of model parameters in regular regression simply\n", + "corresponds to the number of covariates in the model. Or, by the\n", + "degrees of freedom consumed by the model, which is equivalent to the\n", + "trace of the hat matrix. For ridge regression it thus seems natural to\n", + "define model complexity analogously by the trace of the ridge hat\n", + "matrix. This yields the AIC for the linear regression model with ridge\n", + "estimates:" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ + "$$\n", + "\\begin{align*}\n", + "\\mbox{AIC}(\\lambda) & = 2 \\, p - 2 \\log(\\hat{L})\n", + "\\\\\n", + "& = 2 \\, \\mbox{tr} [\\mathbf{H}(\\lambda)] - 2 \\log\\{L[\\hat{\\beta}(\\lambda), \\hat{\\sigma}^2(\\lambda)]\\}\n", + "\\\\\n", + "& = 2 \\, \\sum_{j=1}^p \\frac{d_{jj}^2}{d_{jj}^2 + \\lambda}\n", + "+ 2 n \\, \\log[\\sqrt{2 \\, \\pi} \\, \\hat{\\sigma}(\\lambda)] + \\frac{1}{\\hat{\\sigma}^2(\\lambda)} \\sum_{i=1}^n [y_i - \\mathbf{X}_{i, \\ast} \\, \\hat{\\beta}(\\lambda)]^2.\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The value of $\\lambda$ which minimizes $\\mbox{AIC}(\\lambda)$ corresponds to the `optimal' balance of model complexity and overfitting.\n", + "\n", + "\n", "\n", - "## Generating test data" - ] - }, - { - "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [], - "source": [ - "# Generate Test Data\n", + "## Cross-validation\n", "\n", - "#Number of test data\n", - "N_test=20\n", - "\n", - "sigma_test=sigma_train\n", - "\n", - "max_x=1.2\n", - "x_test=max_x*np.random.random(N_test)\n", - "# Draw random noise\n", - "s_test = sigma_test*np.random.randn(N_test)\n", - "\n", - "#Linear\n", - "y_test=2*x_test+s_test\n", - "#Tenth order\n", - "#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", - "\n", - "#Make design matrices for prediction\n", - "x_plot=np.linspace(0,max_x, 200)\n", - "X3 = poly3.fit_transform(x_plot[:,np.newaxis])\n", - "X10 = poly10.fit_transform(x_plot[:,np.newaxis])\n", - "\n", - "%matplotlib notebook\n", - "\n", - "fig = plt.figure() \n", - "p1=plt.plot(x_test,y_test.transpose(), 'o', ms=12, label='data')\n", - "p2=plt.plot(x_plot,clf.predict(x_plot[:,np.newaxis]), label='linear')\n", - "p3=plt.plot(x_plot,clf3.predict(X3), label='3rd order')\n", - "p10=plt.plot(x_plot,clf10.predict(X10), label='10th order')\n", + "Instead of choosing the penalty parameter to balance model fit with\n", + "model complexity, cross-validation requires it (i.e. the penalty\n", + "parameter) to yield a model with good prediction\n", + "performance. Commonly, this performance is evaluated on novel\n", + "data. Novel data need not be easy to come by and one has to make do\n", + "with the data at hand. The setting of `original' and novel data is\n", + "then mimicked by sample splitting: the data set is divided into two\n", + "(groups of samples). One of these two data sets, called the *training\n", + "set*, plays the role of `original' data on which the model is\n", + "built. The second of these data sets, called the *test set*, plays the\n", + "role of the `novel' data and is used to evaluate the prediction\n", + "performance (often operationalized as the log-likelihood or the\n", + "prediction error or its square or the R2 score) of the model built on the training data set. This\n", + "procedure (model building and prediction evaluation on training and\n", + "test set, respectively) is done for a collection of possible penalty\n", + "parameter choices. The penalty parameter that yields the model with\n", + "the best prediction performance is to be preferred. The thus obtained\n", + "performance evaluation depends on the actual split of the data set. To\n", + "remove this dependence the data set is split many times into a\n", + "training and test set. For each split the model parameters are\n", + "estimated for all choices of $\\lambda$ using the training data and\n", + "estimated parameters are evaluated on the corresponding test set. The\n", + "penalty parameter that on average over the test sets performs best (in\n", + "some sense) is then selected.\n", "\n", "\n", - "plt.legend(loc=2)\n", - "plt.xlabel('$x$')\n", - "plt.ylabel('$y$')\n", - "plt.legend(loc='best')\n", - "plt.title(Title+\" (pred.)\")\n", - "plt.tight_layout()\n", - "plt.show()\n", + "## Computationally expensive\n", "\n", - "#Linear Filename\n", - "#filename_test=Title+\"pred-linear.pdf\"\n", - "#Tenth Order Filename\n", - "#filename_test=Title+\"pred-o10.pdf\"\n", - "#plt.savefig(filename_test)\n", - "#plt.ylim((-6,12))" + "The validation set approach is conceptually simple and is easy to implement. But it has two potential drawbacks:\n", + "\n", + "* The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.\n", + "\n", + "* In the validation approach, only a subset of the observations, those that are included in the training set rather than in the validation set are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.\n", + "\n", + "\n", + "## Various steps in cross-validation\n", + "\n", + "When the repetitive splitting of the data set is done randomly,\n", + "samples may accidently end up in a fast majority of the splits in\n", + "either training or test set. Such samples may have an unbalanced\n", + "influence on either model building or prediction evaluation. To avoid\n", + "this $k$-fold cross-validation structures the data splitting. The\n", + "samples are divided into $k$ more or less equally sized exhaustive and\n", + "mutually exclusive subsets. In turn (at each split) one of these\n", + "subsets plays the role of the test set while the union of the\n", + "remaining subsets constitutes the training set. Such a splitting\n", + "warrants a balanced representation of each sample in both training and\n", + "test set over the splits. Still the division into the $k$ subsets\n", + "involves a degree of randomness. This may be fully excluded when\n", + "choosing $k=n$. This particular case is referred to as leave-one-out\n", + "cross-validation (LOOCV). \n", + "\n", + "\n", + "## How to set up the cross-validation for Ridge and/or Lasso\n", + "\n", + "1. Define a range of interest for the penalty parameter.\n", + "\n", + "2. Divide the data set into training and test set comprising samples $\\{1, \\ldots, n\\} \\setminus i$ and $\\{ i \\}$, respectively.\n", + "\n", + "3. Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set as" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ - "## Lasso regression\n", + "$$\n", + "\\begin{align*}\n", + "\\hat{\\beta}_{-i}(\\lambda) & = ( \\hat{X}_{-i, \\ast}^{\\top}\n", + "\\hat{X}_{-i, \\ast} + \\lambda \\hat{I}_{pp})^{-1}\n", + "\\hat{X}_{-i, \\ast}^{\\top} \\hat{y}_{-i}\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "and the corresponding estimate of the error variance $\\hat{\\sigma}_{-i}^2(\\lambda)$.\n", + "\n", + "1. Evaluate the prediction performance of these models on the test set by $\\log\\{L[y_i, \\hat{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}$. Or, by the prediction error $|y_i - \\hat{X}_{i, \\ast} \\hat{\\beta}_{-i}(\\lambda)|$, the relative error, the error squared or the R2 score function.\n", + "\n", + "2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.\n", + "\n", + "3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\\frac{1}{n} \\sum_{i = 1}^n \\log\\{L[Y_i, \\mathbf{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}.\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.\n", + "\n", + "1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.\n", + "\n", + "## Predicted Residual Error Sum of Squares\n", + "Another approach in the LOOCV scheme is to the use the so-called Predicted Residual Error Sum of Squares (PRESS). \n", + "\n", + "We can define the optimal penalty parameter to minimize" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "$$\n", + "\\begin{align*}\n", + "\\lambda_{\\mbox{{\\tiny opt}}} = \\arg \\min_{\\lambda} \\frac{1}{n} \\sum_{i=1}^n [y_i - \\hat{X}_{i, \\ast} \\hat{\\beta}_{-i}(\\lambda)]^2.\n", + "\\end{align*}\n", + "$$" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "The LOOCV prediction performance can be\n", + "expressed analytically in terms of the known quantities derived from\n", + "the design matrix and the parameters $\\beta$.\n", "\n", "\n", - "## Logistic regression" + "\n", + "## Bootstrap\n", + "Bootstrapping is a nonparametric approach to statistical inference\n", + "that substitutes computation for more traditional distributional\n", + "assumptions and asymptotic results. Bootstrapping offers a number of\n", + "advantages: \n", + "1. The bootstrap is quite general, although there are some cases in which it fails. \n", + "\n", + "2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small. \n", + "\n", + "3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically. \n", + "\n", + "4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples)." ] } ], diff --git a/doc/pub/Regression/ipynb/ipynb-Regression-src.tar.gz b/doc/pub/Regression/ipynb/ipynb-Regression-src.tar.gz index 11e5676115c2089368680876bdad35a3ef51af6a..39e17ea5b7a69d4723e727ac0958db4b5e9f976c 100644 GIT binary patch literal 210 zcmb2|=3rp>I5C=m`R)0GSxklk#|qEe9kn@V^m%@BwDT^lmQ!(p&aHlq1 zJxAo1TqxS-C=dJ%!Z0B0{*>O*)-RaY-^kaF1k_u-_*r&WL zJNfiY@`^wE_@y%!pMJbq<$LXilbyVKZ*2RT`r^jEH^uLqb?>(V12XubUdQ0K LVCDh_4F(1Pob+c8 diff --git a/doc/pub/Regression/pdf/Regression-beamer-handouts2x3.pdf b/doc/pub/Regression/pdf/Regression-beamer-handouts2x3.pdf index c60178739d22d3a6650605fac1740a2ca066ae99..89eb8b91d088762b8c09f6225e98de2bd506a523 100644 GIT binary patch delta 167286 zcmY&<1yCH_(l(2`ySuwFaVUWM$u598sgmqe&C~m|)~6+wZ=_EP*^}p5cOfn4#8Cn@+9k zhAPs|Qvbt;9m1qSX%>27P&(-Vb;T>hc5IsMJoECy*u>n;7O!?nlB7^|Sf6GKt0JQ#DX7iu-p!r{Ha5o#E6r`$21y&^jAU8ko1D@f) 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The resulting $(n \times p)$-dimensional data matrix $\mathbf{X}$: -\begin{align*} -\mathbf{X} & = -\left( -X_{\ast,1} \, | \, \ldots \, | \, X_{\ast,p} -\right) -\, \, \, = \, \, \, -\left( -\begin{array}{c} -X_{1,\ast} \\ \vdots \\ X_{n,\ast} -\end{array} \right) -\, \, \, = \, \, \, -\left( -\begin{array}{ccc} -X_{1,1} & \ldots & X_{1,p} -\\ -\vdots & \ddots & \vdots -\\ -X_{n,1} & \ldots & X_{n,p} -\end{array} \right) -\end{align*} -from such a study contains a larger number of covariates than samples. When $p > n$ the data matrix $\mathbf{X}$ is said to be *high-dimensional*. - -In this chapter we adopt the traditional statistical notation of the -data matrix. An alternative notation would be $\mathbf{X}^{\top}$ -(rather than $\mathbf{X}$), which is employed in the field of -(statistical) bioinformatics. In $\mathbf{X}^{\top}$ the rows comprise -the samples rather than the covariates. The case for the -bioinformatics notation stems from practical arguments. A spreadsheet -is designed to have more rows than columns. In case $p > n$ the -traditional notation yields a spreadsheet with more columns than -rows. When $p > 10000$ the conventional display is impractical. In -this chapter we stick to the conventional statistical notation of the -data matrix as all mathematical expressions involving $\mathbf{X}$ are -then in line with those of standard textbooks on regression. - -The information contained in $\mathbf{X}$ is often used to explain a -particular property of the samples involved. In applications in -molecular biology $\mathbf{X}$ may contain microRNA expression data -from which the expression levels of a gene are to be described. When -the gene's expression levels are denoted by $\mathbf{Y} = (Y_{1}, -\ldots, Y_n)^{\top}$, the aim is to find the linear relation $Y_i = -\mathbf{X}_{i, \ast} \beta$ from the data at hand by means of -regression analysis. Regression is however frustrated by the -high-dimensionality of $\mathbf{X}$ (illustrated in Section -ref{sect.ridgeRegression} and at the end of Section -ref{sect.constrainedEstimation}). These notes discuss how regression -may be modified to accommodate the high-dimensionality of -$\mathbf{X}$. First, however, `standard' linear regression is -recaputilated. - - -======= Linear regression ======= - -Consider an experiment in which $p$ characteristics of $n$ samples are -measured. The data from this experiment are denoted $\mathbf{X}$, with -$\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design -matrix*. Additional information of the samples is available in the -form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is -generally referred to as the *response variable*. The aim of -regression analysis is to explain $\mathbf{Y}$ in terms of -$\mathbf{X}$ through a functional relationship like $Y_i = -f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of -$f(\cdot)$ is available, it is common to assume a linear relationship -between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to -the *linear regression model*: - - -!bt -\begin{align} -Y_{i} & = \mathbf{X}_{i,\ast} \, \beta + \varepsilon_i label{form.linRegressionModel} -\\ -\nonumber -& = \beta_1 \, X_{i,1} + \ldots + \beta_{p} \, X_{i, p} + \varepsilon_i. -\end{align} -!et - -In model (ref{form.linRegressionModel}) $\beta = (\beta_1, \ldots, -\beta_p)^{\top}$ is the *regression parameter*. The parameter -$\beta_j$, $j=1, \ldots, p$, represents the effect size of covariate -$j$ on the response. That is, for each unit change in covariate $j$ -(while keeping the other covariates fixed) the observed change in the -response is equal to $\beta_j$. The second summand on the right-hand -side of the model, $\varepsilon_i$, is referred to as the error. It -represents the part of the response not explained by the functional -part $\mathbf{X}_{i,\ast} \, \beta$ of the model -(ref{form.linRegressionModel}). In contrast to the functional part, -which is considered to be systematic (i.e. non-random), the error is -assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal -$Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= -\mathbf{X}_{i_2,\ast}$. To complete the formulation of model -(ref{form.linRegressionModel}) we need to specify the probability -distribution of $\varepsilon_i$. It is assumed that $\varepsilon_i -\sim \mathcal{N}(0, \sigma^2)$ and the $\varepsilon_{i}$ are -independent, i.e.: \begin{align*} \mbox{Cov}(\varepsilon_{i_1}, -\varepsilon_{i_2}) & = \left\{ \begin{array}{lcc} \sigma^2 & \mbox{if} -& i_1 = i_2, \\ 0 & \mbox{if} & i_1 \not= i_2. \end{array} \right. -\end{align*} The randomness of $\varepsilon_i$ implies that -$\mathbf{Y}_i$ is also a random variable. In particular, -$\mathbf{Y}_i$ is normally distributed, because $\varepsilon_i \sim -\mathcal{N}(0, \sigma^2)$ and $\mathbf{X}_{i,\ast} \, \beta$ is a -non-random scalar. To specify the parameters of the distribution of -$\mathbf{Y}_i$ we need to calculate its first two moments. Its -expectation equals: \begin{align*} \mathbb{E}(Y_i) & = -\mathbb{E}(\mathbf{X}_{i, \ast} \, \beta) + \mathbb{E}(\varepsilon_i) -\, \, \, = \, \, \, \mathbf{X}_{i, \ast} \, \beta, \end{align*} while -its variance is: \begin{align*} \mbox{Var}(Y_i) & = \mathbb{E} \{ [Y_i -- \mathbb{E}(Y_i)]^2 \} \, \, \, = \, \, \, \mathbb{E} ( Y_i^2 ) - -[\mathbb{E}(Y_i)]^2 # \\ # & = \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, -\beta + \varepsilon_i )^2] - ( \mathbf{X}_{i, \ast} \, \beta)^2 \\ & -= \mathbb{E} [ ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 \varepsilon_i -\mathbf{X}_{i, \ast} \, \beta + \varepsilon_i^2 ] - ( \mathbf{X}_{i, -\ast} \, \beta)^2 \\ # & = ( \mathbf{X}_{i, \ast} \, \beta)^2 + 2 -\mathbb{E}(\varepsilon_i) \mathbf{X}_{i, \ast} \, \beta + -\mathbb{E}(\varepsilon_i^2 ) - ( \mathbf{X}_{i, \ast} \, \beta)^2 # -\\ & = \mathbb{E}(\varepsilon_i^2 ) \, \, \, = \, \, \, -\mbox{Var}(\varepsilon_i) \, \, \, = \, \, \, \sigma^2. \end{align*} -Hence, $Y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \beta, -\sigma^2)$. This formulation (in terms of the normal distribution) is -equivalent to the formulation of model (ref{form.linRegressionModel}), -as both capture the assumptions involved: the linearity of the -functional part and the normality of the error. - -Model (ref{form.linRegressionModel}) is often written in a more condensed matrix form: - - -!bt -\begin{align} -\mathbf{Y} & = \mathbf{X} \, \beta + \vvarepsilon, label{form.linRegressionModelinMatrix} -\end{align} -!et -where $\vvarepsilon = (\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n)^{\top}$ and distributed as $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{p}, \sigma^2 \mathbf{I}_{nn})$. As above model (ref{form.linRegressionModelinMatrix}) can be expressed as a multivariate normal distribution: $\mathbf{Y} \sim \mathcal{N}(\mathbf{X} \, \beta, \sigma^2 \mathbf{I}_{nn})$. - -Model (ref{form.linRegressionModelinMatrix}) is a so-called -hierarchical model. This terminology emphasizes that $\mathbf{X}$ and -$\mathbf{Y}$ are not on a par, they play different roles in the -model. The former is used to explain the latter. In model -(ref{form.linRegressionModel}) $\mathbf{X}$ is referred as the -*explanatory* or *independent* variable, while the variable -$\mathbf{Y}$ is generally referred to as the *response* or *dependent* -variable. - -The covariates, the columns of $\mathbf{X}$, may themselves be -random. To apply the linear model they are temporarily assumed -fixed. The linear regression model is then to be interpreted as -$\mathbf{Y} \, | \, \mathbf{X} \sim \mathcal{N}(\mathbf{X} \, \beta, -\sigma^2 \mathbf{I}_{nn})$ - - -The linear regression model (ref{form.linRegressionModel}) involves the unknown parameters: $\beta$ and $\sigma^2$, which need to be learned from the data. The parameters of the regression model, $\beta$ and $\sigma^2$ are estimated by means of likelihood maximization. Recall that $Y_i \sim \mathcal{N}( \mathbf{X}_{i,\ast} \, \beta, \sigma^2)$ with corresponding density: $ f_{Y_i}(y_i) = (2 \, \pi \, \sigma^2)^{-1/2} \, \exp[ - (y_i - \mathbf{X}_{i\ast} \, \beta)^2 / 2 \sigma^2 ]$. The likelihood thus is: - - -!bt -\begin{align*} -L(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = -# f_{\mathbf{Y}}(\mathbf{y}) \, \, \, = \, \, \, \prod_{i=1}^n f_{Y_i}(y_i) \, \, \, = \, \, \, -\prod_{i=1}^n \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, \exp[ - (Y_i - \mathbf{X}_{i, \ast} \, \beta)^2 / 2 \sigma^2 ], -\end{align*} -!et - -in which the independence of the observations has been used. Because -of the concavity of the logarithm, the maximization of the likelihood -coincides with the maximum of the logarithm of the likelihood (called -the log-likelihood). Hence, to obtain maximum likelihood (ML) -estimates of the parameter it is equivalent to find the maximum of the -log-likelihood. The log-likelihood is: - -!bt -\begin{align*} -\mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = -\log[ L(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) ] -\, \, \, = \, \, \, % \log \Big[ \prod_{i=1}^n f_{Y_i}(y_i) \Big] -# \, \, \, = \, \, \, \sum_{i=1}^n \log[ f_{Y_i}(y_i) ] -# \\ -# & = \sum_{i=1}^n [ -\log(\sqrt{2 \, \pi} \, \sigma) - (y_i - \mathbf{X}_{i\ast} \, \beta)^2 / 2 \sigma^2 ] -# \\ -# & = --n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \sum_{i=1}^n (y_i - \mathbf{X}_{i, \ast} \, \beta)^2. -\end{align*} -After noting that $\sum_{i=1}^n (Y_i - \mathbf{X}_{i, \ast} \, \beta)^2 = \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 \, \, \, = \, \, \, (\mathbf{Y} - \mathbf{X} \, \beta)^{\top} \, (\mathbf{Y} - \mathbf{X} \, \beta)$, the log-likelihood can be written as: -\begin{align*} -\mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = -n \, \log(\sqrt{2 \, \pi} \, \sigma) - \frac{1}{ 2 \sigma^2} \, \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. -\end{align*} -!et - -In order to find the maximum of the log-likelihood, take its derivate with respect to $\beta$: -!bt -\begin{align*} -\frac{\partial }{\partial \, \beta} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = - \frac{1}{ 2 \sigma^2} \, \frac{\partial }{\partial \, \beta} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 \, \, \, = \, \, \, \frac{1}{\sigma^2} \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \, \beta). -\end{align*} -!et - -Equate this derivative to zero gives the estimating equation for $\beta$: - -!bt -\begin{align} label{form.normalEquation} -\mathbf{X}^{\top} \mathbf{X} \, \beta & = \mathbf{X}^{\top} \mathbf{Y}. -\end{align} -!et - -Equation (ref{form.normalEquation}) is called to the *normal equation*. Pre-multiplication of both sides of the normal equation by $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ now yields the ML estimator of the regression parameter: $\hat{\beta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}$, in which it is assumed that $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ is well-defined. - -Along the same lines one obtains the ML estimator of the residual variance. Take the partial derivative of the log-likelihood with respect to $\sigma^2$: - -!bt -\begin{align*} -\frac{\partial }{\partial \, \sigma} \mathcal{L}(\mathbf{Y}, \mathbf{X}; \beta, \sigma^2) & = - \frac{n}{\sigma} + \frac{1}{\sigma^3} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. -\end{align*} -!et - - -Equate the right-hand side to zero and solve for $\sigma^2$ to find -$\hat{\sigma}^2 = \tfrac{1}{n} \| \mathbf{Y} - \mathbf{X} \, \beta -\|^2_2$. In this expression $\beta$ is unknown and the ML estimate of -$\beta$ is plugged-in. \\ \\ With explicit expressions of the ML -estimators at hand, we can study their properties. The expectation of -the ML estimator of the regression parameter $\beta$ is: - -!bt -\begin{align*} -\mathbb{E}(\hat{\beta}) & = \mathbb{E}[ -(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] -\, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, -\mathbf{X}^{\top} \mathbb{E}[ \mathbf{Y}] # \\ \, \, \, = \, \, \, -(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{X} \, -\beta \, \, \, \, \, = \, \, \, \beta. -\end{align*} -!et - -Hence, the ML -estimator of the regression coefficients is unbiased. - -The variance of the ML estimator of $\beta$ is: - -!bt -\begin{align*} -\mbox{Var}(\hat{\beta}) & = \mathbb{E} \{ [\hat{\beta} - \mathbb{E}(\hat{\beta})] [\hat{\beta} - \mathbb{E}(\hat{\beta})]^{\top} \} -\\ -& = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \beta] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} - \beta]^{\top} \} -\\ -# & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}] \, [(\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y}]^{\top} \} - \beta \, \beta^{\top} -# \\ -# & = \mathbb{E} \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \, \mathbf{Y}^{\top} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \} - \beta \, \beta^{\top} -# \\ -& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{\top} \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} -\\ -& = (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \, \{ \mathbf{X} \, \beta \, \beta^{\top} \, \mathbf{X}^{\top} + \SSigma \} \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} -# \\ -# & = (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \beta \, \beta^T \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1} -# \\ -# & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \beta \beta^T -\\ -& = \beta \, \beta^{\top} + \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1} - \beta \, \beta^{\top} -\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{\top} \mathbf{X})^{-1}, -\end{align*} -!et - -in which we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{\top}) -= \mathbf{X} \, \beta \, \beta^{\top} \, \mathbf{X}^{\top} + \sigma^2 -\, \mathbf{I}_{nn}$. From $\mbox{Var}(\hat{\beta}) = \sigma^2 \, -(\mathbf{X}^{\top} \mathbf{X})^{-1}$, one obtains an estimate of the -variance of the estimate of the $j$-th regression coefficient: -$\hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ -[(\mathbf{X}^{\top} \mathbf{X})^{-1}]_{jj} }$. This may be used to -construct a confidence interval for the estimates or test the -hypothesis $H_0: \beta_j = 0$. In the latter $\hat{\sigma}^2$ should -not be the maximum likelihood estimator, as it is biased. It is then -to be replaced by the residual sum-of-squares divided by $n-p$ rather -than $n$. - - -The prediction of $Y_i$, denoted $\widehat{Y}_i$, is the expected -value of $Y_i$ according the linear regression model (with its -parameters replaced by their estimates). The prediction of $Y_i$ thus -equals $\mathbb{E}(Y_i; \hat{\beta}, \hat{\sigma}^2) = \mathbf{X}_{i, -\ast} \hat{\beta}$. In matrix notation the prediction is: - -!bt -\begin{align*} -\widehat{\mathbf{Y}} & = \mathbf{X} \, \hat{\beta} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, := \, \, \, \mathbf{H} \mathbf{Y}, -\end{align*} -!et - -where $\mathbf{H}$ is the *hat matrix*, as it `puts the hat' on -$\mathbf{Y}$. Note that the hat matrix is a projection matrix, -i.e. $\mathbf{H}^2 = \mathbf{H}$ for - -!bt -\begin{align*} -\mathbf{H}^2 & = \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \, \, \, = \, \, \, \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}. -\end{align*} -!et - -Thus, the prediction $\widehat{\mathbf{Y}}$ is an orthogonal -projection of $\mathbf{Y}$ onto the space spanned by the columns of -$\mathbf{X}$. - -With $\widehat{\beta}$ available, an estimate of the errors -$\hat{\varepsilon}_i$, dubbed the *residuals* are obtained via: - -!bt -\begin{align*} -\hat{\vvarepsilon} & = \mathbf{Y} - \widehat{\mathbf{Y}} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, \hat{\beta} \, \, \, = \, \, \, \mathbf{Y} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, [ \mathbf{I} - \mathbf{X} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} ] \, \mathbf{Y}. -\end{align*} -!et - -Thus, the residuals are a projection of $\mathbf{Y}$ onto the -orthogonal complement of the space spanned by the columns of -$\mathbf{X}$. The residuals are to be used in diagnostics, -e.g. checking of the normality assumption by means of a normal -probability plot. - - - -======= Ridge regression ======= - -When the design matrix is high-dimensional, the covariates (the -columns of $\mathbf{X}$) are super-collinear. Recall *collinearity* in -regression analysis refers to the event of two (or multiple) -covariates being highly linearly related. Consequently, the subspace -spanned by collinear covariates may not be (or close to not being) of -full rank. When the subspace (onto which $\mathbf{Y}$ is projected) -is (close to) rank deficient, it is (almost) impossible to separate -the contribution of the individual covariates. The uncertainty with -respect to the covariate responsible for the variation explained in -$\mathbf{Y}$ is often reflected in the fit of the linear regression -model to data by a large error of the estimates of the regression -parameters corresponding to the collinear covariates. - - -Consider the design matrix: - -!bt -\begin{align*} -\mathbf{X} & = \left( -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right) -\end{align*} -!et - -The columns of $\mathbf{X}$ are linearly dependent: the first column -is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of space spanned by the -column vectors. Hence, the rank of $\mathbf{X}$ is equal to the number -of linearly independent columns: $\mbox{rank}(\mathbf{X}) = 2$. - - - - -Super-collinearity of an $(n \times p)$-dimensional design matrix $\mathbf{X}$ implies[^footnote1][^footnote1]: If the (column) rank of $\mathbf{X$ is smaller than $p$, there exists a non-trivial $\mathbf{v} \in \mathbb{R}^p$ such that $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. Multiplication of this inequality by $\mathbf{X}^{\top}$ yields $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. As $\mathbf{v} \not= \mathbf{0}_{p}$, this implies that $\mathbf{X}^{\top} \mathbf{X}$ is not invertible.} that the rank of the $(p \times p)$-dimensional matrix $\mathbf{X}^{\top} \mathbf{X}$ is smaller than $p$, and, consequently, it is singular. A square matrix that does not have an inverse is called *singular*. A matrix $\mathbf{A}$ is singular if and only if its determinant is zero: $\mbox{det}(\mathbf{A}) = 0$. - - - -Consider the matrix $\mathbf{A}$ given by: -!bt -\begin{align*} -\mathbf{A} & = \left( -\begin{array}{rr} -1 & 2 -\\ -2 & 4 -\end{array} \right) -\end{align*} -!et -Clearly, $\mbox{det}(\mathbf{A}) = a_{11} a_{22} - a_{12} a_{21} = 1 \times 4 - 2 \times 2 = 0$. Hence, $\mathbf{A}$ is singular and its inverse is undefined. - -As $\mbox{det}(\mathbf{A})$ is equal to the product of the eigenvalues -$\nu_j$ of $\mathbf{A}$, the matrix $\mathbf{A}$ is singular if one -(or more) of the eigenvalues of $\mathbf{A}$ is zero. To see this, -consider the spectral decomposition of $\mathbf{A}$: - -!bt -\begin{align*} -\mathbf{A} & = \sum_{j=1}^p \nu_j \, \mathbf{v}_j \, \mathbf{v}_j^{\top}, -\end{align*} -!et -where $\mathbf{v}_j$ is the eigenvector corresponding to $\nu_j$. The inverse of $\mathbf{A}$ is then: -!bt -\begin{align*} -\mathbf{A}^{-1} & = \sum_{j=1}^p \nu_j^{-1} \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. -\end{align*} -!et - -The right-hand side is undefined if $\nu_j =0$ for any $j$. - - -Matrix $\mathbf{A}$ has eigenvalues $\nu_1 =5$ and $\nu_2=0$. According to the spectral decomposition, the inverse of $\mathbf{A}$ is: -!bt -\begin{align*} -\mathbf{A}^{-1} & = \frac{1}{5} \, \mathbf{v}_1 \, \mathbf{v}_1^{\top} + \frac{1}{0} \, \mathbf{v}_2 \, \mathbf{v}_2^{\top}. -\end{align*} -!et - -This expression is undefined as we divide by zero in the second summand on the right-hand side. - - - -In summary, the columns of a high-dimensional design matrix -$\mathbf{X}$ are linearly dependent and this super-collinearity causes -$\mathbf{X}^{\top} \mathbf{X}$ to be singular. Now recall the ML -estimator of the parameter of the linear regression model: - -!bt -\begin{align} -\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}. -\end{align} -!et - - -This estimator is only well-defined if $(\mathbf{X}^{\top} -\mathbf{X})^{-1}$ exits. Hence, when $\mathbf{X}$ is high-dimensional -the regression parameter $\beta$ cannot be estimated. - - -Above only the practical consequence of high-dimensionality is presented: the expression $( \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ cannot be evaluated numerically. But the problem arising from the high-dimensionality of the data is more fundamental. To appreciate this, consider the normal equations: - -!bt -\begin{align*} -\mathbf{X}^{\top} \mathbf{X} \beta & = \mathbf{X}^{\top} \mathbf{Y}. -\end{align*} -!et - -The matrix $\mathbf{X}^{\top} \mathbf{X}$ is of rank $n$, while $\beta$ is a vector of length $p$. Hence, while there are $p$ unknowns, the system of linear equations from which these are to be solved effectively comprises $n$ degrees of freedom. If $p > n$, the vector $\beta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = -\mathbf{X}^{\top} \mathbf{0}_{p} = \mathbf{0}_{n}$, the solution of the normal equations is: - -!bt -\begin{align*} -\hat{\beta} & = ( \mathbf{X}^{\top} \mathbf{X})^{-} \mathbf{X}^{\top} \mathbf{Y} + \mathbf{v} \qquad \mbox{for all } \mathbf{v} \in V, -\end{align*} -!et - -where $\mathbf{A}^{-}$ denotes the Moore-Penrose inverse of the matrix $\mathbf{A}$, which is defined as: - -!bt -\begin{align*} -\mathbf{A}^{-} & = \sum_{j=1}^p \nu_j^{-1} \, I_{\{ \nu_j \not= 0 \} } \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. -\end{align*} -!et - -The solution of the normal equations is thus only determined up to an -element from a non-trivial space $V$, and there is no unique estimator -of the regression parameter. - -To obtain an estimate of the regression parameter $\beta$ when -$\mathbf{X}$ is (close to) super-collinearity, cite{Hoer1970} proposed -an ad-hoc fix to resolve the (almost) singularity of -$\mathbf{X}^{\top} \mathbf{X}$. Simply replace $\mathbf{X}^{\top} -\mathbf{X}$ by $\mathbf{X}^{\top} \mathbf{X} + \lambda -\mathbf{I}_{pp}$ with $\lambda \in [0, \infty)$. The scalar $\lambda$ -is a tuning parameter, henceforth called the *penalty parameter*. - - - -Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Then, for (say) $\lambda = 1$: -!bt -\begin{align*} -\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} & = \left( -\begin{array}{rrr} -5 & 2 & 2 -\\ -2 & 7 & -4 -\\ -2 & -4 & 7 -\end{array} \right). -\end{align*} -!et -The eigenvalues of this matrix are 11, 7, and 1. Hence, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ has no zero eigenvalue and its inverse is well-defined. - -With the ad-hoc fix for the singularity of $\mathbf{X}^{\top} \mathbf{X}$, cite{Hoer1970} proceed to define the *ridge regression estimator*: - - -!bt -\begin{align} label{form.ridgeRegressionEstimator} -\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}, -\end{align} -!et - -for $\lambda \in [0, \infty)$. Clearly, this is - for $\lambda$ -strictly positive - a well-defined estimator, even if $\mathbf{X}$ is -high-dimensional. However, each choice of $\lambda$ leads to a -different ridge regression estimate. The set of all ridge regression -estimates $\{ \hat{\beta}(\lambda) \, : \, \lambda \in [0, \infty) \}$ -is called the *solution* or *regularization path* of the ridge -estimator. - - - - -Recall the super-collinear design matrix $\mathbf{X}$ of Example ref{example.supercollinearity}. Suppose that the corresponding response vector is $\mathbf{Y} = (1.3, -0.5, 2.6, 0.9)^{\top}$. -The ridge regression estimates for, e.g. $\lambda = 1, 2$, and $10$ are then: -\begin{align*} -\hat{\beta}(1) & = (0.614, 0.548, 0.066)^{\top}, -\\ -\hat{\beta}(2) & = (0.537, 0.490, 0.048)^{\top}, -\\ -\hat{\beta}(10) & = (0.269, 0.267, 0.002)^{\top}. -\end{align*} -The full solution path of the ridge estimator is plotted in Figure ref{fig.ridgeSolPathPlusVar}. - - -Having obtained an estimate of the regression parameter $\beta$, one can define the fit $\widehat{\mathbf{Y}}$. It is defined analogous to the standard case: - -!bt -\begin{align*} -\widehat{\mathbf{Y}}(\lambda) & = \mathbf{X} \hat{\beta}(\lambda) -\, \, \, = \, \, \, \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\, \, \, := \, \, \, \mathbf{H}(\lambda) \mathbf{Y}. -\end{align*} -!et - -Previously, when using the ML estimator, the fit could be understood as a projection of $\mathbf{Y}$ onto the subspace spanned by the columns of $\mathbf{X}$. The fit $\widehat{\mathbf{Y}}(\lambda)$ corresponding to the ridge estimator is not a projection of $\mathbf{Y}$ onto $\mathbf{X}$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$a). Consequently, the `ridge residuals' $\mathbf{Y} - \widehat{\mathbf{Y}}(\lambda)$ are not orthogonal to the fit $\widehat{\mathbf{Y}}(\lambda)$ (confer Exercise ref{question.ridgeResidualsProjection} $\!$b). - - -======= Eigenvalue shrinkage ======= - -The effect of the ridge penalty may also studied from the perspective of singular values. Let the singular value decomposition of the $(n \times p)$-dimensional design matrix $\mathbf{X}$ be: - -!bt -\begin{align*} -\mathbf{X} & = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}, -\end{align*} -!et - -where $\mathbf{D}_x$ an $(n \times n)$-dimensional diagonal matrix -with the singular values, $\mathbf{U}_x$ an $(n \times n)$-dimensional -matrix with columns containing the left singular vectors (denoted -$\mathbf{u}_i$), and $\mathbf{V}_x$ a $(p \times n)$-dimensional -matrix with columns containing the right singular vectors (denoted -$\mathbf{v}_i$). The columns of $\mathbf{U}_x$ and $\mathbf{V}_x$ are -orthogonal: $\mathbf{U}_x^{\top} \mathbf{U}_x = \mathbf{I}_{nn} = -\mathbf{V}_x^{\top} \mathbf{V}_x$. - -The OLS estimator can then be rewritten in terms of the SVD-matrices as: - - -\begin{align*} -\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x \mathbf{D}_x^{-2} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}, -\end{align*} -where $\mathbf{D}_x^{-2} \mathbf{D}_x$ is not simplified further to emphasize the effect of the ridge penalty. Similarly, the ridge estimator can be rewritten in terms of the SVD-matrices as: -\begin{align*} -\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{D}_x^2 \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x (\mathbf{D}_x^{2} + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. -\end{align*} -Combining the two results and writing $(\mathbf{D}_x)_{jj} = d_{x,jj}$ we have: -\begin{align*} -d_{x,jj}^{-1} & \geq & \frac{d_{x,jj}}{d_{x,jj}^2 + \lambda} \qquad \mbox{ for all } \lambda > 0. -\end{align*} -Thus, the ridge penalty shrinks the singular values. -\\ -\\ -Return to the problem of the super-collinearity of $\mathbf{X}$ in the high-dimensional setting ($p > n$). The super-collinearity implies the singularity of $\mathbf{X}^{\top} \mathbf{X}$ and prevents the calculation of the OLS estimator of the regression coefficients. However, $\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}$ is non-singular, with inverse: -\begin{align*} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} & = \sum_{j=1}^p (d_{x,jj}^2 + \lambda)^{-1} \mathbf{v}_j \mathbf{v}_j^{\top}. -\end{align*} -The right-hand side is well-defined for $\lambda > 0$. - - -===== Principal components regression ===== - -Principal component regression is a close relative to ridge regression that can also be applied in a high-dimensional context. Principal components regression explains the response not by the covariates themselves but by linear combinations of the covariates as defined by the principal components of $\mathbf{X}$. Let $\mathbf{U} \mathbf{D} \mathbf{V}^{\top}$ be the singular value decomposition of $\mathbf{X}$. The $i$-th principal component of $\mathbf{X}$ is then $\mathbf{X} \mathbf{v}_i$, henceforth denoted $\mathbf{z}_i$. Let $\mathbf{Z}_k$ be the matrix of the first $k$ principal components, i.e. $\mathbf{Z}_k = \mathbf{X} \mathbf{V}_k$ where $\mathbf{V}_k$ contains the first $k$ right singular vectors as columns. Principal components regression then amounts to regressing the response $\mathbf{Y}$ onto $\mathbf{Z}_{k}$, that is, it fits the model $\mathbf{Y} = \mathbf{Z}_k \ggamma + \vvarepsilon$. The least squares estimator of $\ggamma$ then is (with some abuse of notation): -\begin{align*} -\hat{\ggamma} & = (\mathbf{Z}_k^{\top} \mathbf{Z}_k)^{-1} \mathbf{Z}_k^{\top} \mathbf{Y} \, \, \, = \, \, \, (\mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{X} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{U} \mathbf{D} \mathbf{V}^{\top} \mathbf{V}_k)^{-1} \mathbf{V}_k^{\top} \mathbf{V} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} -\\ -& = (\mathbf{I}_{kn} \mathbf{D}^2 \mathbf{I}_{nk})^{-1} \mathbf{I}_{kn} \mathbf{D} \mathbf{U}^{\top} \mathbf{Y} -\\ -& = \mathbf{D}_k^{-2} \widetilde{\mathbf{D}}_k \mathbf{U}^{\top} \mathbf{Y} \, \, \, = \, \, \ \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}, -\end{align*} -where $\mathbf{D}_k$ and $\widetilde{\mathbf{D}}_k$ are submatrices of $\mathbf{D}$. The matrix $\mathbf{D}_k$ is obtained from $\mathbf{D}$ by removal of the last $n-p$ rows and columsn, while for $\widetilde{\mathbf{D}}_k$ only the last $n-k$ rows are dropped. Similarly, $\mathbf{I}_{kn}$ and $\mathbf{I}_{nk}$ are obtained from $\mathbf{I}_{nn}$ by removal of the last $n-k$ rows and columns, respectively. The principal component regression estimator of $\beta$ then is $\hat{\beta}_{\mbox{{\tiny pcr}}} = \mathbf{V}_k \widetilde{\mathbf{D}}_k^{-1} \mathbf{U}^{\top} \mathbf{Y}$. When $k$ is set equal to the column rank of $\mathbf{X}$, and thus to the rank of $\mathbf{X}^{\top} \mathbf{X}$, the -principal component regression estimator $\hat{\beta}_{\mbox{{\tiny pcr}}} = (\mathbf{X}^{\top} \mathbf{X})^- \mathbf{X}^{\top} \mathbf{Y}$, where $\mathbf{A}^-$ denotes the Moore-Penrose inverse of matrix $\mathbf{A}$. - -The relation between ridge and principal component regression becomes clear when their corresponding estimators are written in terms of the singular value decomposition of $\mathbf{X}$: -\begin{align*} -\hat{\beta}_{\mbox{{\tiny pcr}}} & = \mathbf{V}_x (\mathbf{I}_{nk} \mathbf{D}_x \mathbf{I}_{kn})^{-1} \mathbf{U}_x^{\top} \mathbf{Y}, -\\ -\hat{\beta} (\lambda) & = \mathbf{V}_x (\mathbf{D}_x^2 + \lambda \mathbf{I}_{nn})^{-1} \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{Y}. -\end{align*} -Both operate on the singular values of the design matrix. But where principal component regression thresholds the singular values of $\mathbf{X}$, ridge regression shrinks them (depending on their size). Hence, one applies a discrete map on the singular values while the other a continuous one. - - - - - -======= Moments ======= - -The first two moments of the ridge regression estimator are derived. Next the performance of the ridge regression estimator is studied in terms of the mean squared error, which combines the first two moments. - - -===== Expectation ===== - -The left panel of Figure ref{fig.ridgeSolPathPlusVar} shows ridge estimates of the regression parameters converging to zero as the penalty parameter tends to infinity. This behaviour of the ridge estimator does not depend on the specifics of the data set. To see this study the expectation of the ridge estimator: -\begin{align*} -\mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} \big] -\\ -& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{X}^{\top} \mathbf{Y} \big] -\\ -& = \mathbb{E} \big[ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \big] -\\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \mathbb{E} ( \hat{\beta} ) -\\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta. -\end{align*} -Clearly, $\mathbb{E} \big[ \hat{\beta}(\lambda) \big] \not= \beta$ for any $\lambda > 0$. Hence, the ridge estimator is biased. - -From the expression above it is clear that the expectation of the ridge estimator vanishes as $\lambda$ tends to infinity: -\begin{align*} -\lim_{\lambda \rightarrow \infty} \mathbb{E} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \beta \, \, \, = \, \, \, \mathbf{0}_{p}. -\end{align*} -Hence, all regression coefficients are shrunken towards zero as the penalty parameter increases. This also holds for $\mathbf{X}$ with $p > n$. Furthermore, this behaviour is not strictly monotone in $\lambda$: $\lambda_{a} > \lambda_b$ does not necessarily imply $|\hat{\beta}_j (\lambda_a) | < |\hat{\beta}_j (\lambda_b) |$. Upon close inspection this can be witnessed from the ridge solution path of $\beta_3$ in Figure ref{fig.ridgeSolPathPlusVar}. - - -\begin{example} *Orthonormal design matrix* label{example.orthoronormalDesign} -\\ -Consider an orthonormal design matrix $\mathbf{X}$, i.e.: -\begin{align*} -\mathbf{X}^{\top} \mathbf{X} & = \mathbf{I}_{pp} \, \, \, = \, \, \, (\mathbf{X}^{\top} \mathbf{X})^{-1}. -\end{align*} -An example of an orthonormal design matrix would be: -\begin{align*} -\mathbf{X} & = \frac{1}{2} \left( -\begin{array}{rr} --1 & -1 -\\ --1 & 1 -\\ -1 & -1 -\\ -1 & 1 -\end{array} \right). -\end{align*} -This design matrix is orthonormal as $\mathbf{X}^{\top} \mathbf{X} = \mathbf{I}_{22}$, which is easily verified: -\begin{align*} -\mathbf{X}^{\top} \mathbf{X} & = \frac{1}{4} -\left( -\begin{array}{rrrr} --1 & -1 & 1 & 1 -\\ --1 & 1 & -1 & 1 -\end{array} \right) -\left( -\begin{array}{rr} --1 & -1 -\\ --1 & 1 -\\ -1 & -1 -\\ -1 & 1 -\end{array} \right) \, \, \, = \, \, \, -\frac{1}{4} -\left( -\begin{array}{rr} -4 & 0 -\\ -0 & 4 -\end{array} \right) \, \, \, = \, \, \, \mathbf{I}_{22}. -\end{align*} -In case of an orthonormal design matrix the relation between the OLS and ridge estimator is: -\begin{align*} -\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\, \, \, = \, \, \, (\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (1 + \lambda)^{-1} \mathbf{I}_{pp} \mathbf{X}^{\top} \mathbf{Y} -\qquad \, \, = \, \, \, (1 + \lambda)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (1 + \lambda)^{-1} \hat{\beta}. -\end{align*} -Hence, the ridge estimator scales the OLS estimator by a factor. When taking the expectation on both sides, it is evident that the ridge estimator converges to zero as $\lambda \rightarrow \infty$. -\end{example} - - - -===== Variance ===== - -As for the ML estimate of the regression parameter $\beta$ of model (ref{form.linRegressionModelinMatrix}), we derive the second moment of the ridge estimator. Hereto define: -\begin{align*} -\mathbf{W}_{\lambda} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X}. -\end{align*} -Using $\mathbf{W}_{\lambda}$ the ridge estimator $\hat{\beta}(\lambda)$ can be expressed as $\mathbf{W}_{\lambda} \hat{\beta}$ for: -\begin{align*} -\mathbf{W}_{\lambda} \hat{\beta} & = \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = \{ (\mathbf{X}^{\top} \mathbf{X})^{-1} [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ] \}^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = \hat{\beta}(\lambda). -\end{align*} -The linear operator $\mathbf{W}_{\lambda}$ thus transforms the ML estimator of the regression parameter into the ridge estimator. - -It is now easily seen that: -\begin{align*} -\mbox{Var}[ \hat{\beta}(\lambda) ] & = \mbox{Var}[ \mathbf{W}_{\lambda} \hat{\beta} ] \qquad \qquad \, \, \, \, \, \, = \, \, \, \mathbf{W}_{\lambda} \mbox{Var}[\hat{\beta} ] \mathbf{W}_{\lambda}^{\top} -\\ -& = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}, -\end{align*} -in which we have used $\mbox{Var}(\mathbf{A} \mathbf{Y}) = \mathbf{A} \mbox{Var}( \mathbf{Y}) \mathbf{A}^{\top}$ for a non-random matrix $\mathbf{A}$, the fact that $\mathbf{W}_{\lambda}$ is non-random, and $ \mbox{Var}[\hat{\beta} ] = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1}$. - - -Like the expectation the variance of the ridge estimator vanishes as $\lambda$ tends to infinity: -\begin{align*} -\lim_{\lambda \rightarrow \infty} \mbox{Var} \big[ \hat{\beta}(\lambda) \big] & = \lim_{\lambda \rightarrow \infty} \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \, \, \, = \, \, \, \mathbf{0}_{pp}. -\end{align*} -Hence, the variance of the ridge regression coefficient estimates decreases towards zero as the penalty parameter becomes large. This is illustrated in the right panel of Figure ref{fig.ridgeSolPathPlusVar} for the data of Example ref{example.supercollinearity}. - - -With an explicit expression of the variance of the ridge estimator at hand, we can compare it to that of the OLS estimator: -\begin{align*} -\mbox{Var}[ \hat{\beta} ] - \mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 [(\mathbf{X}^{\top} \mathbf{X})^{-1} - \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] -\\ -& = \sigma^2 \mathbf{W}_{\lambda} \{ [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ] (\mathbf{X}^{\top} \mathbf{X})^{-1} [\mathbf{I} + \lambda (\mathbf{X}^{\top} \mathbf{X})^{-1} ]^{\top} - (\mathbf{X}^{\top} \mathbf{X})^{-1} \} \mathbf{W}_{\lambda}^{\top} -\\ -& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} -\\ -& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top}. -\end{align*} -The difference is non-negative definite as each component in the matrix product is non-negative definite. Hence, the variance of the ML estimator exceeds (in the positive definite ordering) that of the ridge estimator: - -!bt - -!bt -\begin{align} label{form.VarInequalityMLandRidge} -\mbox{Var}[ \hat{\beta} ] & \succeq & \mbox{Var}[ \hat{\beta}(\lambda) ], -\end{align} -!et -!et -with the inequality being strict if $\lambda > 0$. In other words, the variance of the ML estimator is larger than that of the ridge estimator (in the sense that their difference is non-negative definite). The variance inequality (ref{form.VarInequalityMLandRidge}) can be interpreted in terms of the stochastic behaviour of the estimator. This is illustrated by the next example. - -# original latex figure with scale=0.45, angle=0 - -FIGURE: [varMLellipses.eps, width=400 frac=1.0] Level sets of the distribution of the ML (left panel) and ridge (right panel) regression estimators.} label{fig.varOLSandRidge -# \afterpage{} - - -\begin{example} *Variance comparison* -\\ -Consider the design matrix: -\begin{align*} -\mathbf{X} & = \left( -\begin{array}{rr} --1 & 2 -\\ -0 & 1 -\\ -2 & -1 -\\ -1 & 0 -\end{array} \right). -\end{align*} -The variances of the ML and ridge (with $\lambda=1$) estimates of the regression coefficients then are: -\begin{align*} -\mbox{Var}(\hat{\beta}) & = \sigma^2 \left( -\begin{array}{rr} -0.3 & 0.2 -\\ -0.2 & 0.3 -\end{array} \right) -\qquad \mbox{and} \qquad -\mbox{Var}[\hat{\beta}(\lambda)] \, \, \, = \, \, \, \sigma^2 \left( -\begin{array}{rr} -0.1524 & 0.0698 -\\ -0.0698 & 0.1524 -\end{array} \right). -\end{align*} -These variances can be used to construct levels sets of the distribution of the estimates. The level sets that contain 50\%, 75\% and 95\% of the distribution of the ML and ridge estimates are plotted in Figure ref{fig.varOLSandRidge}. In line with inequality (ref{form.VarInequalityMLandRidge}) the level sets of the ridge estimate are smaller than that of the ML estimate: it thus varies less. -\end{example} - - - - -\begin{contexample}_ref{example.orthoronormalDesign_} *Orthonormal design matrix (continued)* -\\ -Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{Var}[ \hat{\beta} ] = \sigma^2 \mathbf{I}_{pp}$ and -\begin{align*} -\mbox{Var}[ \hat{\beta}(\lambda) ] & = \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} -\, \, \, = \, \, \, \sigma^2 [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{I}_{pp} \{ [\mathbf{I}_{pp} + \lambda \mathbf{I}_{pp} ]^{-1} \}^{\top} -\, \, \, = \, \, \, \sigma^2 (1 + \lambda )^{-2} \mathbf{I}_{pp} . -\end{align*} -As the penalty parameter $\lambda$ is non-negative the former exceeds the latter. In particular, this expression vanishes as $\lambda \rightarrow \infty$. -\end{contexample} - -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -The full distribution of the ridge regression estimator is now known. The estimator, $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ is a linear estimator, linear in $\mathbf{Y}$. As $\mathbf{Y}$ is normally distributed, so is $\hat{\beta}(\lambda)$. Moreover, the normal distribution is fully characterized by its first two moments, which are available. Hence: -\begin{align*} -\hat{\beta}(\lambda) & \sim & \mathcal{N} \big( (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{X} \, \beta, \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}\}^{\top} \big). -\end{align*} -Given $\lambda$ and $\beta$, the random behavior of the estimator is thus known. - - - -===== Mean squared error} label{sect:ridgeMSE ===== -Previously, we motivated the ridge estimator as an ad hoc solution to collinearity. An alternative motivation comes from studying the Mean Squared Error (MSE) of the ridge regression estimator: for a suitable choice of $\lambda$ the ridge regression estimator may outperform the ML regression estimator in terms of the MSE. Before we prove this, we first derive the MSE of the ridge estimator and quote some auxiliary results. - -Recall that (in general) for any estimator of a parameter $\theta$: -\begin{align*} -\mbox{MSE}( \hat{\theta} ) & = \mathbb{E} [ ( \hat{\theta} - \theta)^2 ] -\, \, \, = \, \, \, \mbox{Var}( \hat{ \theta} ) + [\mbox{Bias} ( \hat{\theta} )]^2. -\end{align*} -Hence, the MSE is a measure of the quality of the estimator. - -The MSE of the ridge estimator is: - -!bt - -!bt -\begin{align} -\mbox{MSE}[\hat{\beta}(\lambda)] & = \mathbb{E} [ (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta)^{\top} \, (\mathbf{W}_{\lambda} \, \hat{\beta} - \beta) ] \nonumber -\\ -& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber -\\ -& = \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) + \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) \nonumber -\\ -& & - \mathbb{E} ( \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta ) + \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \hat{\beta}) + \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta) \nonumber -\\ -& & - \mathbb{E} ( \beta^{\top} \, \mathbf{W}_{\lambda} \, \hat{\beta}) - \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta) + \mathbb{E} ( \beta^{\top} \beta) \nonumber -\\ -& = \mathbb{E} [ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) ] \nonumber -\\ -& & - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \beta + \beta^{\top} \, \mathbf{W}_{\lambda}^{\top} \mathbf{W}_{\lambda} \, \beta + \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \beta \nonumber -\\ -& & - \beta^{\top} \, \mathbf{W}_{\lambda} \, \beta - \beta^{\top} \mathbf{W}_{\lambda}^{\top} \, \beta + \beta^{\top} \beta \nonumber -\\ -& = \mathbb{E} \big\{ ( \hat{\beta} - \beta )^{\top} \mathbf{W}_{\lambda}^{\top} \, \mathbf{W}_{\lambda} \, (\hat{\beta} - \beta) \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta \nonumber -\\ -& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta. label{form.ridgeMSE} -\end{align} -!et -!et -In the last step we have used $\hat{\beta} \sim \mathcal{N}( \beta, \sigma^2 \, [\mathbf{X}^{\top} \mathbf{X}]^{-1} )$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ for some nonrandom symmetric positive definite matrix $\LLambda$ is (cf. \citealt{Math1992}): -\begin{align*} -\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) & = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}, -\end{align*} -of course replacing $\vvarepsilon$ by $\hat{\beta}$ in this expectation. The first summand in the final derived expression for $\mbox{MSE}[\hat{\beta}(\lambda)]$ is the sum of the variances of the ridge estimator, while the second summand can be thought of the ``squared bias'' of the ridge estimator. In particular, $\lim_{\lambda \rightarrow \infty} \mbox{MSE}[\hat{\beta}(\lambda)] = \beta^{\top} \beta$, which is the squared biased for an estimator that equals zero (as does the ridge estimator in the limit). - - - -\begin{example} *Orthonormal design matrix* -\\ -Assume the design matrix $\mathbf{X}$ is orthonormal. Then, $\mbox{MSE}[ \hat{\beta} ] = p \, \sigma^2$ and -\begin{align*} -\mbox{MSE}[ \hat{\beta}(\lambda) ] & = \frac{p \, \sigma^2}{ (1+ \lambda)^{2}} + \frac{\lambda^2}{ (1+ \lambda)^{2}} \beta^{\top} \beta. -\end{align*} -The latter achieves its minimum at: $\lambda = p \sigma^2 / \beta^{\top} \beta$. -\end{example} - - - -The following theorem and proposition are required for the proof of the main result. - -\begin{theorem} *(Theorem 1 of \citealp{Theo1974*}) label{theo.Theobald1} -\\ -Let $\hat{\ttheta}_1$ and $\hat{\ttheta}_2$ be (different) estimators of $\ttheta$ with second order moments: -\begin{align*} -\mathbf{M}_k & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta) (\hat{\ttheta}_k - \ttheta)^{\top} ] \qquad \mbox{for } k=1,2, -\end{align*} -and -\begin{align*} -\mbox{MSE}(\hat{\ttheta}_k) & = \mathbb{E} [ (\hat{\ttheta}_k - \ttheta)^{\top} \mathbf{A} (\hat{\ttheta}_k - \ttheta) ] \qquad \mbox{for } k=1,2, -\end{align*} -where $\mathbf{A} \succeq 0$. Then, $\mathbf{M}_1 - \mathbf{M}_2 \succeq 0$ if and only if $\mbox{MSE}(\hat{\ttheta}_1) - \mbox{MSE}(\hat{\ttheta}_2) \geq 0$ for all $\mathbf{A} \succeq 0$. -\end{theorem} - - -\begin{proposition} *(\citealp{Fare1976*}) label{prop.Farebrother} -\\ -Let $\mathbf{A}$ be a $(p \times p)$-dimensional, positive definite matrix, $\mathbf{b}$ be a nonzero $p$ dimensional vector, and $c \in \mathbb{R}_+$. Then, $c \mathbf{A} - \mathbf{b} \mathbf{b}^{\top} \succ 0$ if and only if $\mathbf{b}^{\top} \mathbf{A}^{-1} \mathbf{b} > c$. -\end{proposition} - -We are now ready to proof the main result, formalized as Theorem ref{theo.Theobald2}, that for some $\lambda$ the ridge regression estimator yields a lower MSE than the ML regression estimator. - -\begin{theorem} *(Theorem 2 of \citealp{Theo1974*}) label{theo.Theobald2} -\\ -There exists $\lambda > 0$ such that $\mbox{MSE}[\hat{\beta}(\lambda)] < \mbox{MSE}[\hat{\beta}(0)] = \mbox{MSE}[\hat{\beta}]$. -\end{theorem} - -\begin{proof} -The second order moment matrix of the ridge estimator is: -\begin{align*} -\mathbf{M} (\lambda) & := & \mathbb{E} [ (\hat{\beta}(\lambda) - \beta) (\hat{\beta} (\lambda) - \beta)^{\top} ] -\\ -& = \mathbb{E} \{ \hat{\beta}(\lambda) [\hat{\beta}(\lambda)]^{\top} \} - \mathbb{E} [ \hat{\beta}(\lambda) ] \{ \mathbb{E} [ \hat{\beta}(\lambda) ] \}^{\top} + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top} -\\ -& = \mbox{Var}[ \hat{\beta}(\lambda) ] + \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \{ \mathbb{E} [\hat{\beta} (\lambda) - \beta)] \}^{\top}. -\end{align*} -Then: -\begin{align*} -\mathbf{M} ( 0 ) - \mathbf{M}(\lambda) & = \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \sigma^2 \mathbf{W}_{\lambda} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} ] -\\ -& & - (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \beta \beta^{\top} (\mathbf{W}_{\lambda} -\mathbf{I}_{pp})^{\top} -\\ -& = \sigma^2 \mathbf{W}_{\lambda} [ 2 \, \lambda \, (\mathbf{X}^{\top} \mathbf{X})^{-2} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-3} ] \mathbf{W}_{\lambda}^{\top} -\\ -& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} -\\ -& = \sigma^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \lambda \, \mathbf{I}_{pp} + \lambda^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} -\\ -& & - \lambda^2 [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \beta \beta^{\top} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top} -\\ -& = \lambda [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} [ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} ] \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}]^{-1} \}^{\top}. -\end{align*} -This is positive definite if and only if $ 2 \, \sigma^2 \, \mathbf{I}_{pp} + \lambda \sigma^2 (\mathbf{X}^{\top} \mathbf{X})^{-1} - \lambda \beta \beta^{\top} \succ 0$. Hereto it suffices to show that $2 \, \sigma^2 \, \mathbf{I}_{pp} - \lambda \beta \beta^{\top} \succ 0$. By Proposition ref{prop.Farebrother} this holds for $\lambda$ such that $2 \sigma^2 (\beta^{\top} \beta)^{-1} > \lambda$. For these $\lambda$, we thus have $\mathbf{M} ( 0 ) - \mathbf{M}(\lambda)$. Application of Theorem ref{theo.Theobald1} now concludes the proof. -\end{proof} -This result of cite{Theo1974} is generalized by cite{Fare1976} to the class of design matrices $\mathbf{X}$ with $\mbox{rank}(\mathbf{X}) < p$. -\\ -\\ -Theorem ref{theo.Theobald2} can be used to illustrate that the ridge regression estimator strikes a balance between the bias and variance. This is illustrated in the left panel of Figure ref{fig.MSEridge}. For small $\lambda$, the variance of the ridge estimator dominates the MSE. This may be understood when realizing that in this domain of $\lambda$ the ridge estimator is close to the unbiased ML regression estimator. For large $\lambda$, the variance vanishes and the bias dominates the MSE. For small enough values of $\lambda$, the decrease in variance of the ridge regression estimator exceeds the increase in its bias. As the MSE is the sum of these two, the MSE first decreases as $\lambda$ moves away from zero. In particular, as $\lambda = 0$ corresponds to the ML regression estimator, the ridge regression estimator yields a lower MSE for these values of $\lambda$. In the right panel of Figure ref{fig.MSEridge} $\mbox{MSE}[ \hat{\beta}(\lambda)] < \mbox{MSE}[ \hat{\beta}(0)]$ for $\lambda < 7$ (roughly) and the ridge estimator outperforms the ML estimator. -# original latex figure with scale=0.45, angle=0 - -FIGURE: [MSE_ridge.eps, width=400 frac=1.0] Left panel: mean squared error, and its `bias' and `variance' parts, of the ridge regression estimator (for artificial data). Right panel: mean squared error of the ridge and ML estimator of the regression coefficient vector (for the same artificial data).} label{fig.MSEridge -# \afterpage{} -\\ -\\ -# #if FORMAT in ("latex", "pdflatex") -\noindent -# #endif -Besides another motivation behind the ridge regression estimator, the use of Theorem ref{theo.Theobald2} is limited. The optimal choice of $\lambda$ depends on the quantities $\beta$ and $\sigma^2$. These are unknown in practice. Then, the penalty parameter is chosen in a data-driven fashion by means of cross-validation (see Section ref{subsect.crossvalidation}). -\\ -\\ -\begin{remark} \mbox{ } -\\ -Theorem ref{theo.Theobald2} can also be used to conclude on the biasedness of the ridge regression estimator. The Gauss-Markov theorem \citep{Rao1973} states (under some assumptions) that the ML regression estimator is the best linear unbiased estimator (BLUE) with the smallest MSE. As the ridge regression estimator is a linear estimator and outperforms (in terms of MSE) this ML estimator, it must be biased (for it would otherwise refute the Gauss-Markov theorem). -\end{remark} - - - - -======= Constrained estimation} label{sect.constrainedEstimation ======= -The ad-hoc fix of cite{Hoer1970} to super-collinearity of the design matrix (and, consequently the singularity of the matrix $\mathbf{X}^{\top} \mathbf{X}$) has been motivated post-hoc. -The ridge estimator minimizes the *ridge loss function*, which is defined as: - -!bt - -!bt -\begin{align} -\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2 -\, \, \, = \, \, \, \sum_{i=1}^n (Y_i - \mathbf{X}_{i\ast} \, \beta)^2 + \lambda \sum_{j=1}^p \beta_j^2. label{form.ridgeLossFunction} -\end{align} -!et -!et -This loss function is the traditional sum-of-squares augmented with a *penalty*. The particular form of the penalty, $\lambda \| \beta \|^2_2$ is referred to as the *ridge penalty* and $\lambda$ as the *penalty parameter*. For $\lambda=0$, minimization of the ridge loss function yields the ML estimator. For any $\lambda > 0$, the ridge penalty contributes to the loss function, affecting its minimum and its location. The minimum of the sum-of-squares is well-known. The minimum of the ridge penalty is attained at $\beta = \mathbf{0}_{p}$ whenever $\lambda > 0$. The $\beta$ that minimizes $\mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda)$ then balances the sum-of-squares and the penalty. The effect of the penalty in this balancing act is to shrink the regression coefficients towards zero, its minimum. In particular, the larger $\lambda$, the larger the contribution of the penalty to the loss function, the stronger the tendency to shrink non-zero regression coefficients to zero (and decrease the contribution of the penalty to the loss function). This motivates the name `penalty' as non-zero elements of $\beta$ increase (or penalize) the loss function. - - -To verify that the ridge estimator indeed minimizes the ridge loss function, proceed as usual. Take the derivative with respect to $\beta$: -\begin{align*} -\frac{\partial}{\partial \beta} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = -2 \, \mathbf{X}^{\top} (\mathbf{Y} - \mathbf{X} \beta) + 2 \, \lambda \, \mathbf{I}_{pp} \, \beta \, \, \, = \, \, \, -2 \, \mathbf{X}^{\top} \mathbf{Y} + 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}) \beta. -\end{align*} -Equate the derivative to zero and solve for $\beta$. This yields the ridge regression estimator. - -The ridge estimator is thus a stationary point of the ridge loss function. A stationary point corresponds to a minimum if the Hessian matrix with second order partial derivatives is positive definite. The Hessian of the ridge loss function is -\begin{align*} -\frac{\partial^2}{\partial \beta \, \partial \beta^{\top}} \mathcal{L}_{\mbox{{\footnotesize ridge}}}(\beta; \lambda) & = 2 \, ( \mathbf{X}^{\top} \mathbf{X} + \lambda \, \mathbf{I}_{pp}). -\end{align*} -This Hessian is the sum of the semi-positive definite matrix $\mathbf{X}^{\top} \mathbf{X}$ and the positive definite matrix $\lambda \, \mathbf{I}_{pp}$. Lemma 14.2.4 of cite{Harv2008} then states that the sum of these matrices is itself a positive definite matrix. Hence, the Hessian is positive definite and the ridge loss function has a stationary point at the ridge estimator, which is a minimum. - -The ridge regression estimator minimizes the ridge loss function. It rests to verify that it is a global minimum. To this end we introduce the concept of a convex function. As a prerequisite, a set $\mathcal{S} \subset \mathbb{R}^p$ is called *convex* if for all $\beta_1, \beta_2 \in \mathcal{S}$ their weighted average $\beta_{\theta} = (1 - \theta) \beta_1 + \theta \beta_2$ for all $\theta \in [0, 1]$ is itself an element of $\mathcal{S}$, thus $\beta_{\theta} \in \mathcal{S}$. If for all $\theta \in (0, 1)$, the weighted average $\beta_{\theta}$ is inside $\mathcal{S}$ and not on its boundary, the set is called *strict convex*. Examples of (strict) convex and nonconvex sets are depicted in Figure ref{fig.ridgeAsConstrainedEst}. A function $f(\cdot)$ is *(strict) convex* if the set $\{ y \, : \, y \geq f(\beta) \mbox{ for all } \beta \in \mathcal{S} \mbox{ for any convex } \mathcal{S} \}$, called the epigraph of $f(\cdot)$, is (strict) convex. Examples of (strict) convex and nonconvex functions are depicted in Figure ref{fig.ridgeAsConstrainedEst}. The ridge loss function is the sum of two parabola's: one at least convex and the other a strict convex function in $\beta$. The sum of convex and strict convex function is itself strict convex (confer Lemma 9.4.2 of \citealt{Flet2008}). The ridge loss function is thus strict convex. Theorem 9.4.1 of \citealt{Flet2008} then warrants, by the strict convexity of the ridge loss function, that the ridge estimator is a global minimum. -\\ -\\ -From the ridge loss function the limiting behavior of the variance of the ridge regression estimator can be understood. The ridge penalty with its minimum $\beta = \mathbf{0}_{p}$ does not involve data and, consequently, the variance of its minimum equals zero. With the ridge regression being a compromise between the ML estimator and the minimum of the penalty, so is its variance a compromise of their variances. As $\lambda$ tends to infinity, the ridge estimator and its variance converge to the minimum and the variance of the minimum, respectively. Hence, in the limit (large $\lambda$) the variance of the ridge regression estimator vanishes. Understandably, as the penalty now fully dominates the loss function and, consequently, it does no longer involve data (i.e. randomness). -# original latex figure with scale=0.38, angle=0 - -FIGURE: [convexSets.eps, width=400 frac=1.0] Top panels show examples of convex (left) and nonconvex (right) sets. Middle panels show examples of convex (left) and nonconvex (right) functions. The left bottom panel illustrates the ridge estimation as a constrained estimation problem. The ellipses represent the contours of the ML loss function, with the blue dot at the center the ML estimate. The circle is the ridge parameter constraint. The red dot is the ridge estimate. It is at the intersection of the ridge constraint and the smallest contour with a non-empty intersection with the constraint. The right bottom panel shows the data corresponding to Example ref{exam.overfitting}. The grey line represents the `true' relationship, while the black line the fitted one. } label{fig.ridgeAsConstrainedEst -\afterpage{} -\\ -\\ -Above it has been shown that the ridge estimator can be defined as: - -!bt - -!bt -\begin{align} label{form.ridgeEstViaPenEst} -\hat{\beta}(\lambda) & = \arg \min_{\beta} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \lambda \| \beta \|^2_2. -\end{align} -!et -!et -This minimization problem can be reformulated into the following constrained optimization problem (illustrated in Figure ref{fig.ridgeAsConstrainedEst}): - -!bt - -!bt -\begin{align} label{form.constrEstProblemRidge} -\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2, -\end{align} -!et -!et -for some suitable $c > 0$. The constrained optimization problem (ref{form.constrEstProblemRidge}) can be solved by means of the Karush-Kuhn-Tucker (KKT) multiplier method, which minimizes a function subject to inequality constraints. The KKT multiplier method states that, under some regularity conditions (all met here), there exists a constant $\nu \geq 0$, called the *multiplier*, such that the solution $\hat{\beta}(\nu)$ of the constrained minimization problem (ref{form.constrEstProblemRidge}) satisfies the so-called KKT conditions. The first KKT condition (referred to as the stationarity condition) demands that the gradient (with respect to $\beta$) of the Lagrangian associated with the minimization problem equals zero at the solution $\hat{\beta}(\nu)$. The Lagrangian for problem (ref{form.constrEstProblemRidge}) is: -\begin{align*} -\| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2 + \nu ( \| \beta \|^2_2 - c). -\end{align*} -The second KKT condition (the complementarity condition) requires that $\nu (\| \hat{\beta}(\nu) \|_2^2 - c) = 0$. If $\nu = \lambda$ and $c = \| \hat{\beta}(\lambda) \|_2^2$, the ridge estimator $\beta (\lambda)$ satisfies both KKT conditions. Hence, both problems have the same solution when $c = \| \hat{\beta}(\lambda) \|_2^2$. -\\ -\\ -The relevance of viewing the ridge regression estimator as the solution to a constrained estimation problem becomes obvious when considering a typical threat to high-dimensional data analysis: overfitting. *Overfitting* refers to the phenomenon of modelling the noise rather than the signal. In case the true model is parsimonious (few covariates driving the response) and data on many covariates are available, it is likely that a linear combination of all covariates yields a higher likelihood than a combination of the few that are actually related to the response. As only the few covariates related to the response contain the signal, the model involving all covariates then cannot but explain more than the signal alone: it also models the error. Hence, it overfits the data. In high-dimensional settings overfitting is a real threat. The number of explanatory variables exceeds the number of observations. It is thus possible to form a linear combination of the covariates that perfectly explains the response, including the noise. - -Large estimates of regression coefficients are often an indication of overfitting. Augmentation of the estimation procedure with a constraint on the regression coefficients is a simple remedy to large parameter estimates. As a consequence it decreases the probability of overfitting. Overfitting is illustrated in the next example. - - -\begin{example} *(Overfitting)* label{exam.overfitting} -\\ -Consider an artificial data set comprising of ten observations on a response $Y_i$ and nine covariates $X_{i,j}$. All covariate data are sampled from the standard normal distribution: $X_{i,j} \sim \mathcal{N}(0, 1)$. The response is generated by $Y_i = X_{i,1} + \varepsilon_i$ with $\varepsilon_{i} \sim \mathcal{N}(0, 1/4)$. Hence, only the first covariate contributes to the response. - -The regression model $Y_i = \sum_{j=1}^9 X_{i,j} \beta_j+ \varepsilon_i$ is fitted to the artificial data using `R`. This yields the regression parameter estimates: -\begin{align*} -\hat{\beta}^{\top} & = (0.048, -2.386, -5.528, 6.243, -4.819, 0.760, -3.345, -4.748, 2.136). -\end{align*} -As $\beta^{\top} = (1, 0, \ldots, 0)$, many regression coefficient are clearly over-estimated. - -The fitted values $\widehat{Y}_i = \mathbf{X}_i \hat{\beta}$ are plotted against the values of the first covariates in the right bottom panel of Figure ref{fig.ridgeAsConstrainedEst}. As a reference the line $x=y$ is added, which represents the `true' model. The fitted model follows the `true' relationship. But it also captures the deviations from this line that represent the errors. -\end{example} - - - -======= Bayesian regression} label{sect:Bayes ======= -Ridge regression has a close connection to Bayesian linear regression. Bayesian linear regression assumes the parameters $\beta$ and $\sigma^2$ to be the random variables, while at the same time considering $\mathbf{X}$ and $\mathbf{Y}$ as fixed. Within the regression context, the conjugate priors of $\beta$ and $\sigma^2$ are: -\begin{align*} -\beta \, | \, \sigma^2 \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp}) \qquad \mbox{and} \qquad \sigma^2 \sim \mathcal{IG}(\alpha_0, \beta_0), -\end{align*} -where $\mathcal{IG}$ denotes the inverse Gamma distribution with shape parameter $\alpha_0$ and scale parameter $\beta_0$. The penalty parameter can be interpreted as the precision of the prior, determining how informative the prior should be. A smaller penalty (i.e. precision) corresponds to a wider prior, and a larger penalty to a more informative, concentrated prior (Figure ref{fig.ridgePriorOfBeta}). -# original latex figure with scale=0.45, angle=0 - -FIGURE: [ridgePriorOfBeta.eps, width=400 frac=1.0] Conjugate prior of the regression parameter $\beta$ for various choices of $\lambda$, the penalty parameters c.q. precision.} label{fig.ridgePriorOfBeta - -Under the assumption of the conjugate priors above, the joint posterior distribution of $\beta$ and $\sigma^2$ is then: -\begin{align*} -f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) -\\ -& \propto & \sigma^{-n} \exp \Big[ - \frac{1}{2\sigma^2} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) \Big] -\\ -& & \times \, \, \sigma^{-p} \exp \Big[ - \frac{1}{2\sigma^2} \lambda \beta^{\top} \beta \Big] \, \times \, \, [\sigma^2]^{-\alpha_0-1} \exp \Big[ - \frac{\beta_0}{2\sigma^2} \Big]. -\end{align*} -As -\begin{align*} -& & \hspace{-1.5cm} ( \mathbf{Y} - \mathbf{X} \beta)^{\top} ( \mathbf{Y} - \mathbf{X} \beta) + \lambda \beta^{\top} \beta -\\ -& = \mathbf{Y}^{\top} \mathbf{Y} - -\beta^{\top} \mathbf{X}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} \beta + \beta^{\top} \mathbf{X}^{\top} \mathbf{X} \beta + \lambda \beta^{\top} \beta -\\ -& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& & - \, \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta -\\ -& = \mathbf{Y}^{\top} \mathbf{Y} - \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) -\hat{\beta} (\lambda) -\\ -& & - \, [ \hat{\beta} (\lambda) ]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta + \beta^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \beta -\\ -& = \mathbf{Y}^{\top} \mathbf{Y} - \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& & + \, \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big], -\end{align*} -the posterior distribution can be rewritten to: -\begin{align*} -f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) -& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, -g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) -\end{align*} -with -\begin{align*} -g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) -& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\lambda) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) \big[ \beta - \hat{\beta}(\lambda) \big] \Big\}. -\end{align*} -Then, clearly the conditional posterior mean of $\beta$ is $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\lambda)$. Hence, the ridge regression estimator can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a Gaussian prior on the regression parameter. - -With little extra work we may also obtain the conditional posterior of $\sigma^2$ from the joint posterior distribution: -\begin{align*} -f_{\sigma^2} (\sigma^2 \, | \, \beta, \mathbf{Y}, \mathbf{X}) & \propto & (\sigma^2)^{-[(n+p)/2 + \alpha_0 + 1]} -\exp [ - \frac{1}{2\sigma^2} ( \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + -\lambda \| \beta \|_2^2 + \beta_0) ], -\end{align*} -in which one can recognize the shape of an inverse gamma distribution. -\\ -\\ -A Bayesian estimator of a parameter $\ttheta$ is the estimator that minimizes the Bayes risk over a prior distribution of the parameter $\ttheta$. The Bayes risk is defined as $\int_{\ttheta} \mathbb{E} [(\hat{\ttheta} - \ttheta)^{\top} (\hat{\ttheta} - \ttheta)] \pi_{\ttheta}(\ttheta; \aalpha) d\ttheta$, where $\pi_{\ttheta}(\ttheta; \aalpha)$ is the prior distribution of $\ttheta$ with hyperparameter $\aalpha$. It is thus a weighted average of the Mean Squared Error, with weights specified through the prior. The Bayes risk is minimized by the mean posterior $\mathbb{E}_{\ttheta}(\ttheta \, | \, \mbox{data})$ (cf., e.g., \citealp{Bijm2017}). The Bayesian estimator of $\ttheta$ thus yields the smallest possible expected MSE, under the assumption of the employed prior. - -The Bayes risk of the ridge estimator over the normal prior $\beta \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \lambda^{-1} \mathbf{I}_{pp})$ is: -\begin{align*} -\mathbb{E}_{\beta} \{\mbox{MSE}[\hat{\beta}(\lambda)] -\, | \, \sigma^2, \mathbf{Y}, \mathbf{X} \} & = \sigma^2 \, \mbox{tr}\big\{ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big\} + \mathbb{E}_{\beta} [ \beta^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp}) \, \beta] -\\ -& = \sigma^2 \, \big\{ \mbox{tr} \big[ \mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \, \mathbf{W}_{\lambda}^{\top} \big] + -\lambda^{-1} \mbox{tr} [(\mathbf{W}_{\lambda} - \mathbf{I}_{pp})^{\top} (\mathbf{W}_{\lambda} - \mathbf{I}_{pp})] \big\} -\\ -& = \sigma^2 \sum_{j=1}^p (d_{jj}^2 + \lambda)^{-1}, -\end{align*} -in which we have used *i)* the previously derived explicit expression (ref{form.ridgeMSE}) of the ridge estimator's MSE, *ii)* the expectation of the quadratic form of a multivariate random variable \citep{Math1992}, *iii)* the singular value decomposition of $\mathbf{X}$ with singular values $d_{jj}$, and *iv)* the fact that the trace of a square matrix equals the sum of its eigenvalues. As the ridge estimator coincides with the posterior mean, this is the minimal achievable MSE under a zero-centered normal prior with an uncorrelated and equivariant covariance matrix. - -Above the Bayes risk of the ridge estimator factorizes with respect to $\sigma^2$ and $\lambda$. Hence, the larger the hyperparameter $\lambda$ the lower the Bayes risk of the ridge estimator. In particular, its Bayes risk converges to zero as $\lambda \rightarrow \infty$. This can be understood as follows. The limit corresponds to an infinite precision of the prior, thus reducing the variance contribution to the MSE. Moreover, as the ridge estimator shrinks towards zero and the prior distribution of $\beta$ has a zero mean, the bias too vanishes as $\lambda \rightarrow \infty$. - -The calculation of the Bayes risk above relates the Bayesian and frequentist statements on the MSE of the ridge estimator. For the latter revisit Theorem ref{theo.Theobald2} of Section ref{sect:ridgeMSE}, which states the existence of a $\lambda$ such that the resulting ridge estimator has a superior MSE over that of the ML estimator. This result made no assumption on (the distribution of) $\beta$. In fact, it can be viewed as a statement of the MSE conditional on $\beta$. The Bayesian result integrates out the uncertainty - specified by the prior - in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. - - - -# The Bayesian result states that the ridge regression estimator minimizes the weighted MSE (over a normal prior on the parameter $\beta$). It thus integrates out the uncertainty in $\beta$ from the (frequentist's) conditional MSE to arrive at the unconditional MSE. - -======= Degrees of freedom ======= - -The degrees of freedom consumed by ridge regression is calculated. The degrees of freedom may be used in combination with an information criterion to decide on the value of the penalty parameter. Recall from ordinary regression that: -\begin{align*} -\widehat{\mathbf{Y}} & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} \, \, \, = \, \, \, \mathbf{H} \mathbf{Y}, -\end{align*} -where $\mathbf{H}$ is the hat matrix. The degrees of freedom used in the regression is then equal to $\mbox{tr}(\mathbf{H})$, the trace of $\mathbf{H}$. In particular, if $ \mathbf{X}$ is of full rank, i.e. $\mbox{rank}(\mathbf{X}) = p$, then $\mbox{tr}(\mathbf{H}) = p$. - -By analogy, the ridge-version of the hat matrix is: -\begin{align*} -\mathbf{H}(\lambda) & = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}. -\end{align*} -Continuing this analogy, the degrees of freedom of ridge regression is given by the trace of the ridge hat matrix $\mathbf{H}(\lambda)$: -\begin{align*} -\mbox{tr}[ \mathbf{H}(\lambda)] & = \mbox{tr}[ \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} ] \, \, \, = \, \, \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda}. -\end{align*} -The degrees of freedom consumed by ridge regression is monotone decreasing in $\lambda$. In particular: -\begin{align*} -\lim_{\lambda \rightarrow \infty} \mbox{tr}[ \mathbf{H}(\lambda)] & = 0. -\end{align*} -That is, in the limit no information from $\mathbf{X}$ is used. Indeed, $\beta$ is forced to equal $\mathbf{0}_{p}$ which is not derived from data. - - - -======= Efficient calculation ======= - -In the high-dimensional setting the number of covariates $p$ is large compared to the number of samples $n$. In a microarray experiment $p = 40000$ and $n= 100$ is not uncommon. To perform ridge regression in this context, the following expression needs to be evaluated numerically: -\begin{align*} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}. -\end{align*} -For $p=40000$ this requires the inversion of a $40000 \times 40000$ dimensional matrix. This is not feasible on most desktop computers. However, there is a workaround. - -Revisit the singular value decomposition of $\mathbf{X} = \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top}$ and write $\mathbf{R}_x = \mathbf{U}_x \mathbf{D}_x$. As both $\mathbf{U}_x$ and $\mathbf{D}_x$ are $(n \times n)$-dimensional matrices, so is $\mathbf{R}_x$. Consequently, $\mathbf{X}$ is now decomposed as $\mathbf{X} = \mathbf{R}_x \mathbf{V}_x^{\top}$. The ridge estimator can be rewritten in terms of $\mathbf{R}_x$ and $\mathbf{V}_x$: -\begin{align*} -\hat{\beta}(\lambda) & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{R}_x \mathbf{V}_x^{\top} + \lambda \mathbf{V}_x \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{V}_x^{\top} \mathbf{V}_x \mathbf{R}_x^{\top} \mathbf{Y} -\\ -& = \mathbf{V}_x (\mathbf{R}_x^{\top} \mathbf{R}_x + \lambda \mathbf{I}_{nn})^{-1} \mathbf{R}_x^{\top} \mathbf{Y}. -\end{align*} -Hence, the reformulated ridge estimator involves the inversion of an $(n \times n)$-dimensional matrix. With $n= 100$ this is feasible on most standard computers. - -cite{Hast2004} point out that the number of computation operations reduces from $\mathcal{O}(p^3)$ to $\mathcal{O}(p n^2)$. In addition, they point out that this computational short-cut can be used in combination with other loss functions, for instance that of standard generalized linear models. -\\ -\\ -Avoidance of the inversion of the $(p \times p)$-dimensional matrix may be achieved in an other way. Hereto one needs the Woodbury identity. Let $\mathbf{A}$, $\mathbf{U}$ and $\mathbf{V}$ be $(p \times p)$-, $(p \times n)$- and $(n \times p)$-dimensional matrices, respectively. -The (simplified form of the) Woodbury identity then is: -\begin{align*} -(\mathbf{A} + \mathbf{U} \mathbf{V})^{-1} & = \mathbf{A}^{-1} - \mathbf{A}^{-1} \mathbf{U} (\mathbf{I}_{nn} + \mathbf{V} \mathbf{A}^{-1} \mathbf{U})^{-1} \mathbf{V} \mathbf{A}^{-1}. -\end{align*} -Application of the Woodbury identity to the matrix inverse in the ridge estimator of the regression parameter gives: -\begin{align*} -(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} & = \lambda^{-1} \mathbf{I}_{pp} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X}. -\end{align*} -This gives: -\begin{align*} -(\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} & = \lambda^{-1} \mathbf{X}^{\top} \mathbf{Y} - \lambda^{-2} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = \lambda^{-1} \mathbf{X}^{\top} \left[ \mathbf{Y} - \lambda^{-1} \mathbf{X}^{\top} (\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top})^{-1} \mathbf{X} \mathbf{Y} \right]. -\end{align*} -The inversion of the $(p \times p)$-dimensional matrix $\lambda \mathbf{I}_{pp} + \mathbf{X}^{\top} \mathbf{X}$ is thus replaced by that of the $(n \times n)$-dimensional matrix $\mathbf{I}_{nn} + \lambda^{-1} \mathbf{X} \mathbf{X}^{\top}$. In addition, this expression of the ridge regression estimator avoids the singular value decomposition of $\mathbf{X}$, which may in some cases introduce additional numerical errors (e.g. at the level of machine precision). - - - -======= Choice of the penalty parameter ======= - -Throughout the introduction of ridge regression and the subsequent discussion of its properties the penalty parameter is considered known or `given'. In practice, it is unknown and the user needs to make an informed decision on its value. Several strategies to facilitate such a decision are presented. - -===== Information criterion ===== - -A popular strategy is to choose a penalty parameter that yields a good but parsimonious model. Information criteria measure the balance between model fit and model complexity. Here we present the Aikaike's information criterion (AIC), but many other criteria have been presented in the literature (e.g. \citealp{Akai1974}, \citealp{Schw1978}). The AIC measures model fit by the log-likelihood and model complexity is measured by the number of parameters used by the model. The number of model parameters in regular regression simply corresponds to the number of covariates in the model. Or, by the degrees of freedom consumed by the model, which is equivalent to the trace of the hat matrix. For ridge regression it thus seems natural to define model complexity -analogously by the trace of the ridge hat matrix. This yields the AIC for the linear regression model with ridge estimates: -\begin{align*} -\mbox{AIC}(\lambda) & = 2 \, p - 2 \log(\hat{L}) -\\ -& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\} -\\ -& = 2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda} -+ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2. -\end{align*} -The value of $\lambda$ which minimizes $\mbox{AIC}(\lambda)$ corresponds to the `optimal' balance of model complexity and overfitting. - -Information criteria guide the decision process when having to decide among various different models. Different models use different sets of explanatory variables to explain the behaviour of the response variable. In that sense, the use of information criteria for the deciding on the ridge penalty parameter may be considered inappropriate: ridge regression uses the same set of explanatory variables irrespective of the value of the penalty parameter. Moreover, often ridge regression is employed to predict a response and not to provide an insightful explanatory model. The latter need not yield the best predictions. Finally, empirically we observe that the AIC often does not show an optimum *inside* the domain of the ridge penalty parameter. Henceforth, we refrain from the use of the AIC (or any of its relatives) in determining the optimal ridge penalty parameter. - - -===== Cross-validation} label{subsect.crossvalidation ===== -Instead of choosing the penalty parameter to balance model fit with model complexity, cross-validation requires it (i.e. the penalty parameter) to yield a model with good prediction performance. Commonly, this performance is evaluated on novel data. Novel data need not be easy to come by and one has to make do with the data at hand. The setting of `original' and novel data is then mimicked by sample splitting: the data set is divided into two (groups of samples). One of these two data sets, called the *training set*, plays the role of `original' data on which the model is built. The second of these data sets, called the *test set*, plays the role of the `novel' data and is used to evaluate the prediction performance (often operationalized as the log-likelihood or the prediction error) of the model built on the training data set. This procedure (model building and prediction evaluation on training and test set, respectively) is done for a collection of possible penalty parameter choices. The penalty parameter that yields the model with the best prediction performance is to be preferred. The thus obtained performance evaluation depends on the actual split of the data set. To remove this dependence the data set is split many times into a training and test set. For each split the model parameters are estimated for all choices of $\lambda$ using the training data and estimated parameters are evaluated on the corresponding test set. The penalty parameter that on average over the test sets performs best (in some sense) is then selected. - -When the repetitive splitting of the data set is done randomly, samples may accidently end up in a fast majority of the splits in either training or test set. Such samples may have an unbalanced influence on either model building or prediction evaluation. To avoid this $k$-fold cross-validation structures the data splitting. The samples are divided into $k$ more or less equally sized exhaustive and mutually exclusive subsets. In turn (at each split) one of these subsets plays the role of the test set while the union of the remaining subsets constitutes the training set. Such a splitting warrants a balanced representation of each sample in both training and test set over the splits. Still the division into the $k$ subsets involves a degree of randomness. This may be fully excluded when choosing $k=n$. This particular case is referred to as leave-one-out cross-validation (LOOCV). For illustration purposes the LOOCV procedure is detailed fully below: -\begin{compactitem} -\item[0)] Define a range of interest for the penalty parameter. - -\item[1)] Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively. - -\item[2)] Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set. This yields: -\begin{align*} -\hat{\beta}_{-i}(\lambda) & = ( \mathbf{X}_{-i, \ast}^{\top} -\mathbf{X}_{-i, \ast} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{-i, \ast}^{\top} \mathbf{Y}_{-i} -\end{align*} -and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$. - -\item[3)] Evaluate the prediction performance of these models on the test set by $\log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)|$, possibly squared. - -\item[4)] Repeat steps 1) to 3) such that each sample plays the role of the test set once. - -\item[5)] Average the prediction performances of the test sets at each grid point of the penalty parameter: -\begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. -\end{align*} -The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. - -\item[6)] The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. -\end{compactitem} -The procedure is straightforwardly adopted to $k$-fold cross-validation, a different criterion, and different estimators. - -In the LOOCV procedure above resampling can be avoided when the prediction performance is measured by Allen's PRESS (Predicted Residual Error Sum of Squares) statistic \citep{Alle1974}. For then, the LOOCV prediction performance can be expressed analytically in terms of the known quantities derived from the design matrix and response (as pointed out but not detailed in \citealt{Golu1979}). Define the optimal penalty parameter to minimize Allen's PRESS statistic: -\begin{align*} -\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. -\end{align*} -To derive an analytic expression for the right-hand side first -rewrite $(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1}$ by means of the Woodbury identity as: -\begin{align*} -(\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp} - \mathbf{X}_{i, \ast}^{\top} \mathbf{X}_{i, \ast})^{-1} -\\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -+ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}]^{-1} -\\ -& & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -+ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\end{align*} -with $\mathbf{H}_{ii}(\lambda) = \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. -Furthermore, $\mathbf{X}_{-i}^{\top} \mathbf{Y}_{-i} = \mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i$. Substitute both in the leave-one-out ridge regression estimator and manipulate: -\begin{align*} -\hat{\beta}_{- i}(\lambda) & = (\mathbf{X}_{- i, \ast}^{\top} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{- i, \ast}^{\top} \mathbf{Y}_{- i} -\\ -& = \{(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -+ (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}\} -\\ -& & \times (\mathbf{X}^{\top} \mathbf{Y} - \mathbf{X}_{i, \ast}^{\top} Y_i ) -\\ -& = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y} - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} Y_i -\\ -& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii} (\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}^{\top} \mathbf{Y} -\\ -& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} -\mathbf{X}_{i, \ast}^{\top} Y_i -\\ -& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i -\\ -& & + (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) -\\ -& & - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{H}_{ii} (\lambda) Y_i -\\ -& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \{ [ 1 - \mathbf{H}_{ii}(\lambda)] Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii}(\lambda) Y_i \} -\\ -& = \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ]. -\end{align*} -The latter enables the reformulation of the prediction error as: -\begin{align*} -Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda) & = Y_i - \mathbf{X}_{i, \ast} \{ \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda)] \} -\\ -& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] -\\ -& = Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) + \mathbf{H}_{ii} (\lambda)[ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast} \hat{\beta}(\lambda) ] -\\ -& = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - \mathbf{X}_{i, \ast}^{\top} \hat{\beta}(\lambda) ], -\end{align*} -which in turn results in the re-expression of Allen's PRESS statistic: -\begin{align*} -\lambda_{\mbox{{\tiny opt}}} & = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [Y_i - \mathbf{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2 -\, \, \, = \, \, \, \arg \min_{\lambda} \tfrac{1}{n} \| \mathbf{B}(\lambda) [\mathbf{I}_{nn} - \mathbf{H}(\lambda)] \mathbf{Y} \|_ F^2, -\end{align*} -where $\mathbf{B}(\lambda)$ is diagonal with $[\mathbf{B}(\lambda)]_{ii} = [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1}$. Hence, the prediction performance for a given $\lambda$ can be assessed directly from the ridge hat matrix and the response vector without the recalculation of the $n$ leave-one-out ridge estimators. Computationally, this is a considerable gain. - - - -======= Simulations ======= - -Simulations are presented that illustrate properties of the ridge estimator not discussed explicitly in the previous sections of this chapter. - -===== Role of the variance of the covariates} label{ridge:covariateVariances ===== -In many applications of high-dimensional data the covariates are standardized prior to the execution of the ridge regression. Before we discuss whether this is appropriate, we first illustrate the effect of ridge penalization on covariates with distinct variances using simulated data. - -The simulation involves one response to be (ridge) regressed on fifty covariates. Data (with $n=1000$) for the covariates, denoted $\mathbf{X}$, are drawn from a multivariate normal distribution: $\mathbf{X} \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{\Sigma})$ with $\mathbf{\Sigma}$ diagonal and $(\mathbf{\Sigma})_{jj} = j / 10$. From this the response is generated through $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = \mathbf{1}_{50}$ and $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_{50}, \mathbf{I}_{50 \times 50})$. - -With the simulated data at hand the ridge regression estimates of $\beta$ are evaluated for a large grid of the penalty parameter $\lambda$. The resulting ridge regularization paths of the regression coefficients are plotted (Figure ref{fig.effectOfRidge_effectOfVariances}). All paths start ($\lambda=0$) close to one and vanish as $\lambda \rightarrow \infty$. However, ridge regularization paths of regression coefficients corresponding to covariates with a large variance dominate those with a low variance. - - -# original latex figure with scale=0.45, angle=0 - -FIGURE: [covariatesWithDifferentVariance.eps, width=400 frac=1.0] Top panel: Ridge regularization paths for coefficients of the 50 uncorrelated covariates with distinct variances. Color and line type indicated the grouping of the covariates by their variance. Bottom panels: Graphical illustration of the effect of a covariate's variance on the ridge estimator. The grey circle depicts the ridge parameter constraint. The dashed black ellipsoids are the level sets of the least squares loss function. The red dot is the ridge regression estimate. Left and right panels represent the cases with equal and unequal, respectively, variances of the covariates.} label{fig.effectOfRidge_effectOfVariances - - - - - -Ridge regression's preference of covariates with a large variance can intuitively be understood as follows. First note that the ridge regression estimator now can be written as: -\begin{align*} -\beta (\lambda) & = [ \mbox{Var}(\mathbf{X}) + \lambda \mathbf{I}_{50 \times 50}]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) -\\ -& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma [ \mbox{Var}(\mathbf{X}) ]^{-1} \mbox{Cov}(\mathbf{X}, \mathbf{Y}) -\\ -& = ( \SSigma + \lambda \mathbf{I}_{50 \times 50})^{-1} \SSigma \beta. -\end{align*} -Plug in the employed parametrization of $\mathbf{\Sigma}$, which gives: -\begin{align*} -[\beta (\lambda)]_j & = \frac{j}{j + 50 \lambda} \, (\beta)_j. -\end{align*} -Hence, the larger the covariate's variance (corresponding to the larger $j$), the larger its ridge regression coefficient estimate. Ridge regression thus prefers (among a set of covariates with comparable effect sizes) those with larger variances. - - - -The reformulation of ridge penalized estimation as a constrained estimation problem offers a geometrical interpretation of this phenomenon. Let $p=2$ and the design matrix $\mathbf{X}$ be orthogonal, while both covariates contribute equally to the response. Contrast the cases with $\mbox{Var}(X_1) \approx \mbox{Var}(X_2)$ and $\mbox{Var}(X_1) \gg \mbox{Var}(X_2)$. The level sets of the least squares loss function associated with the former case are circular, while that of the latter are strongly ellipsoidal (see Figure ref{fig.effectOfRidge_effectOfVariances}). The diameters along the principal axes (that - due to the orthogonality of $\mathbf{X}$ - are parallel to that of the $\beta_1$- and $\beta_2$-axes) of both circle and ellipsoid are reciprocals of the variance of the covariates. When the variances of both covariates are equal, the level sets of the loss function expand equally fast along both axis. With the two covariates having the same regression coefficient, the point of these level sets closest to the parameter constraint is to be found on the line $\beta_1 = \beta_2$ (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). Consequently, the ridge regression estimate satisfies $\hat{\beta}_1 (\lambda) \approx \hat{\beta}_2(\lambda)$. With unequal variances between the covariates, the ellipsoidal level sets of the loss function have diameters of rather different sizes. In particular, along the $\beta_1$-axis it is narrow (as $\mbox{Var}(X_1)$ is large), and - vice versa - wide along the $\beta_2$-axis. Consequently, the point of these level sets closest to the circular parameter constraint will be closer to the $\beta_1$- than to the $\beta_2$-axis (Figure ref{fig.effectOfRidge_effectOfVariances}, left panel). For the ridge estimates of the regression parameter this implies $0 \ll \hat{\beta}_1 (\lambda) < 1$ and $0 < \hat{\beta}_2 (\lambda) \ll 1$. Hence, the covariate with a larger variance yields the larger ridge regression estimate. - - -Should one thus standardize the covariates prior to ridge regression analysis? When dealing with gene expression data from microarrays, the data have been subjected to a series of pre-processing steps (e.g. quality control, background correction, within- and between-normalization). The purpose of these steps is to make the expression levels of genes comparable both within and between hybridizations. The preprocessing should thus be considered an inherent part of the measurement. As such it is to be done independently of whatever down-stream analysis is to follow and further tinkering with the data is preferably to be avoided (as it may mess up the `comparable-ness' of the expression levels as achieved by the preprocessing). For other data types different considerations may apply. - -Among the considerations to decide on standardization of the covariates, one should also include the fact that ridge estimates prior and posterior to scaling do not simply differ by a factor. To see this assume that the covariates have been centered. Scaling of the covariates amounts to post-multiplication of the design matrix by a $(p \times p)$-dimensional diagonal matrix $\mathbf{A}$ with the reciprocals of the covariates' scale estimates on its diagonal \citep{Sard2008}. Hence, the ridge estimator (for the rescaled data) is then given by: -\begin{align*} -\min_{\beta} \| \mathbf{Y} - \mathbf{X} \mathbf{A} \beta \|_2^2 + \lambda \| \beta \|_ 2^2. -\end{align*} -Apply the change-of-variable $\ggamma = \mathbf{A} \beta$ and obtain: -\begin{align*} -\min_{\ggamma} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \lambda \| \mathbf{A}^{-1} \ggamma \|_ 2^2 & = \min_{\beta} \| \mathbf{Y} - \mathbf{X} \gamma \|_2^2 + \sum_{j=1}^p \lambda [(\mathbf{A})_{jj}]^{-2} \gamma_j^2. -\end{align*} -Effectively, the scaling is equivalent to covariate-wise penalization. The `scaled' ridge estimator may then be derived along the same lines as before in Section ref{sect.constrainedEstimation}: -\begin{align*} -\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) & = \mathbf{A}^{-1} \hat{\ggamma} (\lambda) -\, \, \, = \, \, \, \mathbf{A}^{-1} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{A}^{-2})^{-1} \mathbf{X}^{\top} \mathbf{Y}. -\end{align*} -In general, this is unequal to the ridge estimator without the rescaling of the columns of the design matrix. Moreover, it should be clear that $\hat{\beta}^{\mbox{{\tiny (scaled)}}} (\lambda) \not= \mathbf{A} \hat{\beta}(\lambda)$. - -===== Ridge regression and collinearity} label{sect:collinearCovariates ===== -Initially, ridge regression was motivated as an ad-hoc fix of (super)-collinear covariates in order to obtain a well-defined estimator. We now study the effect of this ad-hoc fix on the regression coefficient estimates of collinear covariates. In particular, their ridge regularization paths are contrasted to those of `non-collinear' covariates. - -To this end, we consider a simulation in which one response is regressed on 50 covariates. The data of these covariates, stored in a design matrix denoted $\mathbf{X}$, are sampled from a multivariate normal distribution, with mean zero and a $5 \times 5$ blocked covariance matrix: -\begin{align*} -\mathbf{\Sigma} & = \left( -\begin{array}{ccccc} -\mathbf{\Sigma}_{11} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -\\ -\mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{22} & \mathbf{0}_{10 \times 10} -& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -\\ -\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -& \mathbf{\Sigma}_{33} & \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -\\ -\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -& \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{44} & \mathbf{0}_{10 \times 10} -\\ -\mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} -& \mathbf{0}_{10 \times 10} & \mathbf{0}_{10 \times 10} & \mathbf{\Sigma}_{55} -\end{array} -\right) -\end{align*} -with -\begin{align*} -\mathbf{\Sigma}_{kk} & = \frac{k-1}{5} \, \mathbf{1}_{10 \times 10} + \frac{6-k}{5} \, \mathbf{I}_{10 \times 10}. -\end{align*} -The data of the response variable $\mathbf{Y}$ are then obtained through: $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, with $\vvarepsilon \sim \mathcal{N}( \mathbf{0}_{n}, \mathbf{I}_{nn})$ and $\beta = \mathbf{1}_{50}$. Hence, all covariates contribute equally to the response. Would the columns of $\mathbf{X}$ be orthogonal, little difference in the ridge estimates of the regression coefficients is expected. - -The results of this simulation study with sample size $n=1000$ are presented in Figure ref{fig.ridgeEstimates_correlatedCovariates}. All 50 regularization paths start close to one as $\lambda$ is small and converge to zero as $\lambda \rightarrow \infty$. But the paths of covariates of the same block of the covariance matrix $\mathbf{\Sigma}$ quickly group, with those corresponding to a block with larger off-diagonal elements above those with smaller ones. Thus, ridge regression prefers (i.e. shrinks less) coefficient estimates of strongly positively correlated covariates. - -# original latex figure with scale=0.45, angle=0 - -FIGURE: [correlatedCovariates.eps, width=400 frac=1.0] Left panel: Ridge regularization paths for coefficients of the 50 covariates, with various degree of collinearity but equal variance. Color and line type correspond to the five blocks of the covariate matrix $\mathbf{\Sigma}$. Right panel: Graphical illustration of the effect of the collinearity among covariates on the ridge estimator. The solid and dotted grey circles depict the ridge parameter constraint for the collinear and orthogonal cases, respectively. The dashed black ellipsoids are the level sets of the sum-of-squares squares loss function. The red dot and violet diamond are the ridge regression for the positive collinear and orthogonal case, respectively.} label{fig.ridgeEstimates_correlatedCovariates -# \afterpage{} - - - -Intuitive understanding of the observed behaviour may be obtained from the $p=2$ case. Let $U$, $V$ and $\varepsilon$ be independent random variables with zero mean. Define $X_1 = U + V$, $X_2 = U - V$, and $Y = \beta_1 X_1 + \beta_2 X_2 + \varepsilon$ with $\beta_1$ and $\beta_2$ constants. Hence, $\mathbb{E}(Y) = 0$. Then: -\begin{align*} -# Y & = \beta_1 X_1 + \beta_2 X_2 + \varepsilon -# \\ -# & = \beta_1 (U + V) + \beta_2 (U - V) + \varepsilon -# \\ -Y & = (\beta_1 + \beta_2) U + (\beta_1 - \beta_2) V + \varepsilon -\\ -& = \gamma_u U + \gamma_v V + \varepsilon -\end{align*} -and $\mbox{Cor}(X_{1}, X_{2}) = [\mbox{Var}(U) - \mbox{Var}(V)] / [ \mbox{Var}(U) + \mbox{Var}(V) ]$. -# \begin{align*} -# \mbox{Cor}(X_{1}, X_{2}) -# & = \frac{ \mbox{Cov}(X_{1}, X_{2}) }{ \sqrt{\mbox{Var}(X_{1})} \sqrt{\mbox{Var}(X_{2})} } -# \\ -# & = \frac{ \mbox{Cov}(U + V, U - V) }{ \sqrt{\mbox{Var}(U + V)} \sqrt{\mbox{Var}(U - V)} } -# \\ -# & = -# \frac{ \mbox{Var}(U) - \mbox{Var}(V) }{ \mbox{Var}(U) + \mbox{Var}(V) }. -# \end{align*} -The random variables $X_1$ and $X_2$ are strongly positively correlated if $\mbox{Var}(U) \gg \mbox{Var}(V)$. - -The ridge regression estimator associated with regression of $Y$ on $U$ and $V$ is: -\begin{align*} -\ggamma(\lambda) & = \left( -\begin{array}{rr} -\mbox{Var}(U) + \lambda & 0 -\\ -0 & \mbox{Var}(V) + \lambda -\end{array} -\right)^{-1} -\left( -\begin{array}{r} -\mbox{Cov}(U, Y) -\\ -\mbox{Cov}(V, Y) -\end{array} -\right). -\end{align*} -For large enough $\lambda$ -\begin{align*} -\ggamma(\lambda) -# & & \frac{1}{\lambda} \left( \begin{array}{rr} \mbox{Var}(U) & 0 \\ 0 \mbox{Var}(V) \end{array} \right) \left( \begin{array}{rr} \mbox{Var}(U) & 0 -# \\ -# 0 & \mbox{Var}(V) -# \end{array} -# \right)^{-1} -# \left( -# \begin{array}{r} -# \mbox{Cov}(U, Y) -# \\ -# \mbox{Cov}(V, Y) -# \end{array} -# \right) -# \\ -& \approx \frac{1}{\lambda} -\left( -\begin{array}{rr} -\mbox{Var}(U) & 0 -\\ -0 & \mbox{Var}(V) -\end{array} -\right) -\left( -\begin{array}{r} -\beta_1 + \beta_2 -\\ -\beta_1 - \beta_2 -\end{array} -\right). -\end{align*} -When $\mbox{Var}(U) \gg \mbox{Var}(V)$ and $\beta_1 \approx \beta_2$, the ridge estimate of $\gamma_v$ vanishes for large $\lambda$. Hence, ridge regression prefers positively covariates with similar effect sizes. - - -This phenomenon too can be explained geometrically. For the illustration consider ridge estimation with $\lambda=1$ of the linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\beta = (3, 3)^{\top}$, $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_2, \mathbf{I}_{22})$ and the columns of $\mathbf{X}$ strongly and positively collinear. The level sets of the sum-of-squares loss, $\| \mathbf{Y} - \mathbf{X} \beta \|_2^2$, are plotted in the right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}. Recall that the ridge estimate is found by looking for the smallest loss level set that hits the ridge contraint. The sought-for estimate is then the point of intersection between this level set and the constraint, and - for the case at hand - is on the $x=y$-line. This is no different from the case with orthogonal $\mathbf{X}$ columns. Yet their estimates differ, even though the same $\lambda$ is applied. The difference is to due to fact that the radius of the ridge constraint depends on $\lambda$, $\mathbf{X}$ and $\mathbf{Y}$. This is immediate from the fact that the radius of the constraint equals $\| \hat{\beta}(\lambda) \|_2^2$ (see Section ref{sect.constrainedEstimation}). To study the effect of $\mathbf{X}$ on the radius, we remove its dependence on $\mathbf{Y}$ by considering its expectation, which is: -\begin{align*} -\mathbb{E}[ \| \hat{\beta}(\lambda) \|_2^2 ] & = \mathbb{E} \{ [(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta}]^{\top} \, (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X}) \, \hat{\beta} \} \nonumber % \, \, \, = \, \, \, \mathbb{E} ( \hat{\beta}^{\top} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, \hat{\beta} ) \nonumber -\\ -# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{W}_{\lambda}^{\top} \,\mathbf{W}_{\lambda} \, (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber -# \\ -# & = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} (\mathbf{X}^{\top} \mathbf{X}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp}) (\mathbf{X}^{\top} \mathbf{X})(\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber -# \\ -& = \mathbb{E} [ \mathbf{Y}^{\top} \mathbf{X} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \mathbf{Y} ] \nonumber -\\ -& = \sigma^2 \, \mbox{tr}\big\{ \mathbf{X} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \big\} + \beta^{\top} \mathbf{X}^{ \top} \mathbf{X} -(\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-2} \mathbf{X}^{\top} \, \mathbf{X} \beta. % label{form.expectationPenalty} -\end{align*} -In the last step we have used $\mathbf{Y} \sim \mathcal{N}( \mathbf{X} \beta, \sigma^2 \mathbf{I}_{pp})$ and the expectation of the quadratic form of a multivariate random variable $\vvarepsilon \sim \mathcal{N}(\mmu_{\varepsilon}, \SSigma_{\varepsilon})$ is $\mathbb{E} ( \vvarepsilon^{\top} \, \LLambda \, \vvarepsilon) = \mbox{tr} ( \LLambda \, \SSigma_{\varepsilon}) + \mmu_{\varepsilon}^{\top} \, \LLambda \, \mmu_{\varepsilon}$ (cf. \citealp{Math1992}). The expression for the expectation of the radius of the ridge constraint can now be evaluated for the orthogonal $\mathbf{X}$ and the strongly, positively collinear $\mathbf{X}$. It turns out that the latter is larger than the former. This results in a larger ridge constraint. For the larger ridge constraint there is a smaller level set that hits it first. The point of intersection, still on the $x=y$-line, is now thus closer to $\beta$ and further from the origin (cf. right panel of Figure ref{fig.ridgeEstimates_correlatedCovariates}). The resulting estimate is thus larger than that from the orthogonal case. - -The above needs some attenuation. Among others it depends on: *i)* the number of covariates in each block, *ii)* the size of the effects, i.e. regression coefficients of each covariate, and -*iii)* the degree of collinearity. Possibly, there are more factors influencing the behaviour of the ridge estimator presented in this subsection. - -This behaviour of ridge regression is to be understood when using (say) gene expression data to predict a certain clinical outcome. Genes work in concert to fulfil a certain function in the cell. Consequently, one expects their expression levels to be correlated. Indeed, gene expression studies exhibit many co-expressed genes, that is, genes with correlating transcript levels. - - - - -======= Illustration} label{sect.ridgeRegressionDataIllustration ======= -The application of ridge regression to actual data aims to illustrate its use in practice. - -===== MCM7 expression regulation by microRNAs ===== - -Recently, a new class of RNA was discovered, referred to as microRNA. MicroRNAs are non-coding, single stranded RNAs of approximately 22 nucleotides. Like mRNAs, microRNAs are encoded in and transcribed from the DNA. MicroRNAs play an important role in the regulatory mechanism of the cell. MicroRNAs down-regulate gene expression by either of two post-transcriptional mechanisms: mRNA cleavage or transcriptional repression. This depends on the degree of complementarity between the microRNA and the target. Perfect or nearly perfect complementarity of the mRNA to the microRNA will lead to cleavage and degradation of the target mRNA. Imperfect complementarity will repress the productive translation and reduction in protein levels without affecting the mRNA levels. -A single microRNA can bind to and regulate many different mRNA targets. Conversely, several microRNAs can bind to and cooperatively control a single mRNA target (\citealp{Bart2004}; \citealp{Esqu2006}; \citealp{Kim2006}). - -In this illustration we wish to confirm the regulation of mRNA expression by microRNAs in an independent data set. We cherry pick an arbitrary finding from literature reported in cite{Ambs2008}, which focusses on the microRNA regulation of the MCM7 gene in prostate cancer. The MCM7 gene is involved in DNA replication \citep{Tye1999}, a cellular process often derailed in cancer. Furthermore, MCM7 interacts with the tumor-suppressor gene RB1 \citep{Ster1998}. Several studies indeed confirm the involvement of MCM7 in prostate cancer \citep{Padm2004}. And recently, it has been reported that in prostate cancer MCM7 may be regulated by microRNAs \citep{Ambs2008}. - -We here assess whether the MCM7 down-regulation by microRNAs can be observed in a data set other than the one upon which the microRNA-regulation of MCM7 claim has been based. To this end we download from the Gene Expression Omnibus (GEO) a prostate cancer data set (presented by \citealp{Wang2009GeneNetworks}). This data set (with GEO identifier: GSE20161) has both mRNA and microRNA profiles for all samples available. The preprocessed (as detailed in \citealp{Wang2009GeneNetworks}) data are downloaded and require only minor further manipulations to suit our purpose. These manipulations comprise *i)* averaging of duplicated profiles of several samples, *ii)* gene- and mir-wise zero-centering of the expression data, *iii)* averaging the expression levels of the probes that interrogate MCM7. Eventually, this leaves 90 profiles each comprising of 735 microRNA expression measurements. - - - -\lstinputlisting{prostate_ridgeAnalysis.r} - - -With this prostate data set at hand we now investigate whether MCM7 is regulated by microRNAs. Hereto we fit a linear regression model regressing the expression levels of MCM7 onto those of the microRNAs. As the number of microRNAs exceeds the number of samples, ordinary least squares fails and we resort to the ridge estimator of the regression coefficients. First, an informed choice of the penalty parameter is made through maximization of the LOOCV log-likelihood, resulting in $\lambda_{\mbox{{\tiny opt}}} = 1812.826$. Having decided on the value of the to-be-employed penalty parameter, the ridge regression estimator can now readily be evaluated. The thus fitted model allows for the evaluation of microRNA-regulation of MCM7. E.g., by the proportion of variation of the MCM7 expression levels by the microRNAs as expressed in coefficient of determination: $R^2 = 0.4492$. Alternatively, but closely related, observed expression levels may be related to the linear predictor of the MCM7 expression levels: $\hat{\mathbf{Y}}(\lambda_{\mbox{{\tiny opt}}}) = \mathbf{X} \hat{\beta} (\lambda_{\mbox{{\tiny opt}}})$. The Spearman correlation of response and predictor equals 0.6295. A visual inspection is provided by the left panel of Figure ref{fig.RidgeProstateExample}. Note the difference in scale of the $x$- and $y$-axes. This is due to the fact that the regression coefficients have been estimated in penalized fashion, consequently shrinking estimates of the regression coefficients towards zero leading to small estimates and in turn compressing the range of the linear prediction. The above suggests there is indeed association between the microRNA expression levels and those of MCM7. - - - -# original latex figure with scale=0.45, angle=0 - -FIGURE: [prostate_ridgeFit.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values. Right panel: Histogram of the ridge regression coefficient estimates.} label{fig.RidgeProstateExample - - -The overall aim of this illustration was to assess whether microRNA-regulation of MCM7 could also be observed in this prostate cancer data set. In this endeavour the dogma (stating this regulation should be negative) has nowhere been used. A first simple assessment of the validity of this dogma studies the signs of the estimated regression coefficients. The ridge regression estimate has 394 out of the 735 microRNA probes with a negative coefficient. Hence, a small majority has a sign in line with the `microRNA $\downarrow$ mRNA' dogma. When, in addition, taking the size of these coefficients into account (Figure ref{fig.RidgeProstateExample}, right panel), the negative regression coefficient estimates do not substantially differ from their positive counterparts (as can be witnessed from their almost symmetrical distribution around zero). Hence, the value of the `microRNA $\downarrow$ mRNA' dogma is not confirmed by this ridge regression analysis of the MCM7-regulation by microRNAs. Nor is it refuted. - - -The implementation of ridge regression in the {\tt penalized}-package offers the possibility to fully obey the dogma on negative regulation of mRNA expression by microRNAs. This requires all regression coefficients to be negative. Incorporation of the requirement into the ridge estimation augments the constrained estimation problem with an additional constraint: -\begin{align*} % label{form.constrEstProblemRidge} -\hat{\beta}(\lambda) & = \arg \min_{\| \beta \|_2^2 \leq c (\lambda) \atop \beta_j \leq 0 \, \mbox{{\tiny for all $j$}}} \| \mathbf{Y} - \mathbf{X} \, \beta \|^2_2. -\end{align*} -With the additional non-positivity constraint on the parameters, there is no explicit solution for the estimator. The ridge estimate of the regression parameters is then found by numerical optimization using e.g. the Newton-Raphson algorithm or a gradient descent approach. The next listing gives the R-code for ridge estimation with the non-positivity constraint of the linear regression model. - -\lstinputlisting{prostate_ridgeAnalysisWithConstraint.r} - -The linear regression model linking MCM7 expression to that of the microRNAs is fitted by ridge regression while simultaneously obeying the `negative regulation of mRNA by microRNA'-dogma to the prostate cancer data. In the resulting model 401 out of 735 microRNA probes have a nonzero (and negative) coefficient. There is a large overlap in microRNAs with a negative coefficient between those from this and the previous fit. The models are also compared in terms of their fit to the data. The Spearman rank correlation coefficient between response and predictor for the model without positive regression coefficients equals 0.679 and its coefficient of determination 0.524 (confer the left panel of ref{fig.RidgeProstateExample_constrainedAnalysis} for a visualization). This is a slight improvement upon the unconstrained ridge estimated model. The improvement may be small but it should be kept in mind that the number of parameters used by both models is 401 (for the model without positive regression coefficients) vs. 735. Hence, with close to half the number of parameters the dogma-obeying model gives a somewhat better description of the data. This may suggest that there is some value in the dogma as inclusion of this prior information leads to a more parsimonious model without any loss in fit. - - - -# original latex figure with scale=0.45, angle=0 - -FIGURE: [prostate_ridgeFitConstrained.eps, width=400 frac=1.0] Left panel: Observed vs. (ridge) fitted MCM7 expression values (with the non-positive constraint on the parameters in place). Right panel: Histogram of the ridge regression coefficient estimates (from the non-positivity constrained analysis).} label{fig.RidgeProstateExample_constrainedAnalysis - - - -The dogma-obeying model selects 401 microRNAs that aid in the explanation of the variation in the gene expression levels of MCM7. There is an active field of research, called *target prediction*, trying to identify which microRNAs target the mRNA of which genes. Within {\tt R} there is a collection of packages that provide the target prediction of known microRNAs. The packages differ on the method (e.g. experimental or sequence comparison) that has been used to arrive at the prediction. These target predictions may be used to evaluate the value of the found 401 microRNAs. Ideally, there would be a substantial amount of overlap. The {\tt R}-script that loads the target predictions and does the comparison is below. - -\lstinputlisting{prostate_link2predictionTools.r} - - -|-----------------------------------------------------------------| -| # #if FORMAT in ("latex", "pdflatex") -\vspace{-7pt} -# #endif | | | -|---------r--------------------r--------------------r-------------| -| # #if FORMAT in ("latex", "pdflatex") -\vspace{3pt} -# #endif | $\hat{\beta}_j = 0$ | $\hat{\beta}_j < 0$ | -| # #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif | | | -| microRNA not target | 323 | 390 | -| microRNA target | 11 | 11 | -| # #if FORMAT in ("latex", "pdflatex") -\vspace{-9pt} -# #endif | | | -|-----------------------------------------------------------------| - - -\caption{Cross-tabulation of the microRNAs being a potential target of MCM7 vs. the value of its regression coefficient in the dogma-obeying model.} label{table.mir2mcm7prediction} - - -With knowledge available on each microRNA whether it is predicted (by at least one target prediction package) to be a potential target of MCM7, it may be cross-tabulated against its corresponding regression coefficient estimate in the dogma-obeying model being equal to zero or not. Table ref{table.mir2mcm7prediction} contains the result. Somewhat superfluous considering the data, we may test whether the targets of MCM7 are overrepresented in the group of strictly negatively estimated regression coefficients. The corresponding chi-squared test (with Yates' continuity correction) yields the test statistic $\chi^2 = 0.0478$ with a $p$-value equal to 0.827. Hence, there is no enrichment among the 401 microRNAS of those that have been predicted to target MCM7. This may seem worrisome. However, the microRNAs have been selected for their predictive power of the expression levels of MCM7. Variable selection has not been a criterion (although the sign constraint implies selection). Moreover, criticism on the value of the microRNA target prediction has been accumulating in recent years. - - -======= Conclusion ======= - -We discussed ridge regression as a modification of linear regression to overcome the empirical non-identifiability of the latter when confronted with high-dimensional data. The means to this end was the addition of a (ridge) penalty to the sum-of-squares loss function of the linear regression model, which turned out to be equivalent to constraining the parameter domain. This warranted the identification of the regression coefficients, but came at the cost of introducing bias in the estimates. Several properties of ridge regression like moments, MSE, and its Bayesian interpretation have been reviewed. Finally, its behaviour and use have been illustrated in simulation and omics data. - - -======= Exercises ======= -\begin{question}[^footnote2][^footnote2]: This exercise is freely rendered from cite{Drap1998} \mbox{ } -\\ -Find the ridge regression solution for the data below for a general value of $\lambda$ and for the straight line model $Y = \beta_0 + \beta_1 X + \varepsilon$ (only apply the ridge penalty to the slope parameter, not to the intercept). Show that when $\lambda$ is chosen as 0.4, the ridge solution fit is $\hat{Y} = 40 + 1.75 X$. Data: $\mathbf{X}^{\top} = (X_1, X_2, \ldots, X_{8})^{\top} = (-2, -1, -1, -1, 0, 1, 2, 2)^{\top}$, and $\mathbf{Y}^{\top} = (Y_1, Y_2, \ldots, Y_{8})^{\top} = (35, 40, 36, 38, 40, 43, 45, 43)^{\top}$. -\end{question} - -\begin{question}[^footnote3][^footnote3]: This exercise is freely rendered from cite{Hast2009, but can be found in many other places. The original source is unknown to the author.} label{question:ridgeAugmentation} \mbox{ } -\\ -Show that the ridge regression estimates can be obtained by ordinary least squares regression on an augmented data set. We augment the centered matrix $\mathbf{X}$ with $p$ additional row $\sqrt{\lambda} \mathbf{I}$, and augment $\mathbf{y}$ with $p$ zeros. -\end{question} - -\begin{question} label{question.ridgeResidualsProjection} \mbox{ } -\\ -The coefficients $\beta$ of a linear regression model, $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, are estimated by $\hat{\beta} = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. The associated fitted values then given by $\widehat{\mathbf{Y}} = \mathbf{X} \, \hat{\beta} = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y} = \mathbf{H} \mathbf{Y}$, where $\mathbf{H} =\mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top}$ referred to as the hat matrix. The matrix $\mathbf{P}$ is a projection matrix and satisfies $\mathbf{H} = \mathbf{H}^ 2$. Hence, linear regression projects the response $\mathbf{Y}$ onto the vector space spanned by the columns of $\mathbf{Y}$. Consequently, the residuals $\hat{\vvarepsilon}$ and $\hat{\mathbf{Y}}$ are orthogonal. Now consider the ridge estimator of the regression coefficients: $\hat{\beta}(\lambda) = (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top} \mathbf{Y}$. Let $\hat{\mathbf{Y}}(\lambda) = \mathbf{X} \hat{\beta}(\lambda)$ be the vector of associated fitted values. - -\begin{compactitem} -\item[*a)*] Show that the matrix $\mathbf{H}(\lambda) = \mathbf{X} (\mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}^{\top}$, associated with ridge regression, is not a projection matrix (for any $\lambda > 0$). - -\item[*b)*] Show that the `ridge fit' $\widehat{\mathbf{Y}}(\lambda)$ is not orthogonal to the associated `ridge residuals' $\hat{\vvarepsilon}(\lambda)$ (for any $\lambda > 0$). - -\item[*c)*] Derive the distribution of the `ridge residuals'. -\end{compactitem} -\end{question} - - -\begin{question} label{question.MSEofLinearPredictor} \mbox{ } -\\ -Recall that there exists $\lambda > 0$ such that $MSE(\hat{\beta}) > MSE[\hat{\beta}(\lambda)]$. Verify that this carries over to the linear predictor. That is, there exists a $\lambda > 0$ such that$MSE(\widehat{\mathbf{Y}}) = MSE(\mathbf{X} \hat{\beta}) > MSE[\mathbf{X}\hat{\beta}(\lambda)]$. -\end{question} - -\begin{question} \mbox{ } -\\ -Consider a 3-gene pathway. Expression levels of these three genes have been measured in an observational study involving hundred individuals. In order to assess how the expression levels of gene A are affect by that of genes B and C, a medical researcher fits the -\begin{align*} -Y_{i}^{(A)} &= \beta_b Y_{i}^{(B)} + \beta_c Y_{i}^{(C)} + -\varepsilon_{i}, -\end{align*} -with $\varepsilon_i \sim \mathcal{N}(0, \sigma^2)$. This model fitted by means of ridge regression, but with a separate penalty parameter, $\lambda_{2,b}$ and $\lambda_{2,c}$, for the two regression coefficient, $\beta_b$ and $\beta_c$, respectively. - -\begin{compactitem} -\item[*a)*] Write down the ridge penalized loss function employed by the researcher. - -\item[*b)*] Does a different choice of penalty parameter for the second regression coefficient affect the estimation of the first regression coefficient? Motivate your answer. - -\item[*c)*] The researcher decides that the second covariate $Y_i^{(C)}$ is irrelevant. Instead of removing the covariate from model, the researcher decides to set $\lambda_{2,c} = -\infty$. Show that this results in the same ridge estimate for $\beta_b$ as when fitting (again by means of ridge regression) the model without the second covariate. -\end{compactitem} -\end{question} - -\begin{question} \mbox{ } -\\ -The expression levels of the $j$-the gene are explained by a linear regression model in terms of those of all other genes. Consider the following two ridge regression estimators of the regression parameter of this model, defined as: -\begin{align*} -\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + \lambda \| \beta_j \|_ 2^2 -\quad -\mbox{and} \quad -\arg \max_\beta \sum_{i=1}^n (Y_{i,j} - \mathbf{Y}_{i,\setminus j} \beta_j)^2 + n \lambda \| \beta_j \|_ 2^2. -\end{align*} -Which do you prefer? Motivate. -\end{question} - - - - -========= Generalizing ridge regression ========= label{chap:genRidge} -# ======= Generalized ridge regression ======= - -The expos\'{e} on ridge regression may be generalized in many ways. Among others different generalized linear models may be considered (confer Section ref{sect.ridgeLogistic}). In this section we stick to the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with the usual assumptions, but fit it in weighted fashion and generalize the common, spherical penalty. The loss function corresponding to this scenario is: - -!bt - -!bt -\begin{align} label{form:generalizedRidgeLoss} -(\mathbf{Y} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Y} - \mathbf{X} \beta) + (\beta - \beta_0)^{\top} \mathbf{\Delta} -(\beta - \beta_0), -\end{align} -!et -!et -which comprises a weighted least squares criterion and a generalized ridge penalty. In this $\mathbf{W}$ is a $(n \times n)$-dimensional, diagonal matrix with $(\mathbf{W})_{ii} \in [0,1]$ representing the weight of the $i$-th observation. The penalty is now a quadratic form with penalty parameter $\mathbf{\Delta}$, a $(p \times p)$-dimensional, positive definite, symmetric matrix. When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, one regains the spherical penalty of `regular ridge regression'. This penalty shrinks each element of the regression parameter $\beta$ equally along the unit vectors $\mathbf{e}_j$. Generalizing $\mathbf{\Delta}$ to the class of symmetric, positive definite matrices $\mathcal{S}_{++}$ allows for *i)* different penalization per regression parameter, and *ii)* joint (or correlated) shrinkage among the elements of $\beta$. The penalty parameter $\mathbf{\Delta}$ determines the speed and direction of shrinkage. The $p$-dimensional column vector $\beta_0$ is a user-specified, non-random target towards which $\beta$ is shrunken as the penalty parameter increases. When recasting generalized ridge estimation as a constrained estimation problem, the implications of the penalty may be visualized (Figure ref{fig:generalizedRidgeIllustration}, left panel). The generalized ridge penalty is a quadratic form centered around $\beta_0$. In Figure ref{fig:generalizedRidgeIllustration} the parameter constraint clearly is ellipsoidal (and not spherical). Moreover, the center of this ellipsoid is not at zero. - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [generalizedRidgeIllustration_penalty.eps, width=400 frac=1.0] Left panel: the contours of the likelihood (grey solid ellipsoids) and the parameter constraint implied by the generalized penalty (black dashed ellipsoid. Right panel: generalized (fat coloured lines) and `regular' (thin coloured lines) regularization paths of four regression coefficients. The dotted grey (straight) lines indicated the targets towards the generalized ridge penalty shrinks regression coefficient estimates.} label{fig:generalizedRidgeIllustration - - -The addition of the generalized ridge penalty to the sum-of-squares ensures the existence of a unique regression estimator in the face of super-collinearity. The generalized penalty is a non-degenerated quadratic form in $\beta$ due to the positive definiteness of the matrix $\mathbf{\Delta}$. As it is non-degenerate, it is strictly convex. Consequently, the generalized ridge regression loss function (ref{form:generalizedRidgeLoss}), being the sum of a convex and strictly convex function, is also strictly convex. This warrants the existence of a unique global minimum and, thereby, a unique estimator. - - -Like for the `regular' ridge loss function (ref{form.ridgeLossFunction}), there is an explicit expression for the optimum of the generalized ridge loss function (ref{form:generalizedRidgeLoss}). To see this, obtain the estimating equation of $\beta$ through equating its derivative with respect to $\beta$ to zero: -\begin{align*} -2 \mathbf{X}^{\top} \mathbf{W} \mathbf{Y} - 2 \mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta - 2 \mathbf{\Delta} \beta + 2 \mathbf{\Delta} \beta_0 & = \mathbf{0}_{p}. -\end{align*} -This is solved by: - -!bt - -!bt -\begin{align} -\hat{\beta}(\mathbf{\Delta}) & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{Y} + \mathbf{\Delta} \beta_0). -\end{align} -!et -!et -Clearly, this reduces to the `regular' ridge estimator by setting $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$. The effects of the generalized ridge penalty on the estimates can be seen in the regularization paths of the estimates. Figure ref{fig:generalizedRidgeIllustration} (right panel) contains an example of the regularization paths for coefficients of a linear regression model with four explanatory variables. Most striking is the limiting behaviour of the estimates of $\beta_3$ and $\beta_4$ for large values of the penalty parameter $\lambda$: they convergence to a non-zero value (as was specified by a nonzero $\beta_0$). More subtle is the (temporary) convergence of the regularization paths of the estimates of $\beta_2$ and $\beta_3$. That of $\beta_2$ is pulled away from zero (its true value and approximately its unpenalized estimate) towards the estimate of $\beta_3$. In the regularization path of $\beta_3$ this can be observed in a delayed convergence to its nonzero target value (for comparison consider that of $\beta_4$). For reference the corresponding regularization paths of the `regular' ridge estimates (as thinner lines of the same colour) are included in Figure ref{fig:generalizedRidgeIllustration}. - - - -\begin{example} *Fused ridge estimation* label{example.gRidge2orthonormalDesign} -\\ -An example of a generalized ridge penalty is the *fused ridge penalty*. Consider the standard linear model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$. The fused ridge estimator of $\beta$ then minimizes: - -!bt - -!bt -\begin{align} label{form:fusedRidgeLoss} -\| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2. -\end{align} -!et -!et -The penalty in the loss function above can be written as a generalized ridge penalty: -\begin{align*} -\lambda \sum_{j=2}^p \| \beta_{j} - \beta_{j-1} \|_2^2 & = \beta^{\top} -\left( -\begin{array}{rrrrrr} -\lambda & -\lambda & 0 & \ldots & \ldots & 0 -# #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif -\\ --\lambda & 2 \lambda & -\lambda & \ddots & & \vdots -# #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif -\\ -0 & -\lambda & 2 \lambda & \ddots & \ddots & \vdots -# #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif -\\ -\vdots & \ddots & \ddots & \ddots & \ddots & 0 -# #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif -\\ -\vdots & & \ddots & \ddots & \ddots & -\lambda -# # #if FORMAT in ("latex", "pdflatex") -\vspace{-4pt} -# #endif -\\ -0 & \ldots & \ldots & 0 & -\lambda & \lambda -\end{array}\right) -\beta. -\end{align*} -The matrix $\mathbf{\Delta}$ employed above is semi-positive definite and therefore the loss function (ref{form:fusedRidgeLoss}) is not strictly convex. Hence, often a regular ridge penalty $\| \beta \|_2^2$ is added (with its own penalty parameter). - -To illustrate the effect of the fused ridge penalty on the estimation of the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$, let $\beta_j = \phi_{0,1}(z_j)$ with $z_j =-30 + \tfrac{6}{50} j$ for $j=1, \ldots, 500$. Sample the elements of the design matrix $\mathbf{X}$ and those of the error vector $\vvarepsilon$ from the standard normal distribution, then form the response $\mathbf{Y}$ from the linear model. The regression parameter is estimated through fused ridge loss minimization with $\lambda=1000$. The estimate is shown in Figure ref{fig:fusedRidgeIllustration} (red line). For reference the figure includes the true $\beta$ (black line) and the `regular ridge' estimate with $\lambda=1$ (blue line). Clearly, the fused ridge estimate yields a nice smooth vector of $\beta$ estimates -\end{example} - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [fusedRidgeIllustration_simulation.eps, width=400 frac=1.0] Left panel: illustration of the fused ridge estimator (in simulation). The true parameter $\beta$ and its ridge and fused ridge estimates against their spatial order. Right panel: Ridge vs. fused ridge estimates of the DNA copy effect on KRAS expression levels. The dashed, grey vertical bar indicates the location of the KRAS gene. } label{fig:fusedRidgeIllustration - - - - -======= Moments ======= - -The expectation and variance of $\hat{\beta}(\mathbf{\Delta})$ are obtained through application of the same matrix algebra and expectation and covariance rules used in the derivation of their counterparts of the `regular' ridge regression estimator. This leads to: -\begin{align*} -\mathbb{E}[\hat{\beta}(\mathbf{\Delta})] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \mathbf{\Delta} \beta_0), -\\ -\mbox{Var}[\hat{\beta}(\mathbf{\Delta})] & = \sigma^2 (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}. -\end{align*} -From these expressions similar limiting behaviour as for the `regular' ridge regression case can be deduced. To this end let $\mathbf{V}_{\delta} \mathbf{D}_{\delta} \mathbf{V}_{\delta}^{\top}$ be the eigendecomposition of $\mathbf{\Delta}$ and $d_{\delta,j} = (\mathbf{D}_{\delta})_{jj}$. Furthermore, define (with some abuse of notation) $\lim_{\mathbf{\Delta} \rightarrow \infty}$ as the limit of all $d_{\delta,j}$ simultaneously tending to infinity. Then, $\lim_{\mathbf{\Delta} \rightarrow \infty} \mathbb{E}[\hat{\beta}(\mathbf{\Delta})] = \beta_0$ and $\lim_{\mathbf{\Delta} \rightarrow \infty} \mbox{Var}[\hat{\beta}(\mathbf{\Delta})] = \mathbf{0}_{pp}$. - - -\begin{example} \mbox{ } label{example.gRidge2orthonormalDesign} -\\ -Let $\mathbf{X}$ be an $(n \times p)$-dimensional, orthonormal design matrix. Contrast the regular and generalized ridge regression estimator, the latter with $\mathbf{W} = \mathbf{I}_{pp}$, $\beta_0 = \mathbf{0}_p$ and $\mathbf{\Delta} = \lambda \mathbf{R}$ where $\mathbf{R} = (1-\rho) \mathbf{I}_{pp} + \rho \mathbf{1}_{pp}$ for $\rho \in (-(p-1)^{-1}, 1)$. For $\rho =0$ the two estimators coincide. The variance of the generalized ridge regression estimator then is $\mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] = (\mathbf{I}_{pp} + \mathbf{\Delta})^{-2}$. The efficiency of this estimator, measured by its generalized variance, is: -\begin{align*} -\det \{ \mbox{Var}[ \hat{\beta}(\mathbf{\Delta})] \} & = \{ [1 + \lambda + (p-1) \rho] (1 + \lambda-\rho)^{p-1} \}^{-2}. -\end{align*} -This efficiency attains its minimum at $\rho = 0$. In the present case, the regular ridge regression estimator is thus more efficient than its generalized counterpart. -\end{example} - - -\begin{example} *(MSE with perfect target)* -\\ -Set $\beta_0 = \beta$, i.e. the target is equal to the true value of the regression parameter. Then: -\begin{align*} -\mathbb{E}[\hat{\beta}(\Delta)] & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \Delta)^{-1} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} \beta + \Delta \beta) \, \, \, = \, \, \, \beta. -\end{align*} -Hence, irrespective of the choice of $\Delta$, the generalized ridge is then unbiased. Thus: -\begin{align*} -\mbox{MSE}[\hat{\beta}(\Delta)] & = \mbox{tr} \{ -\mbox{Var}[\hat{\beta}(\Delta)] \} -\\ -& = \mbox{tr}[ \sigma^{2} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-1}] -\\ -& = \sigma^2 \mbox{tr}[ \mathbf{X}^{\top} \mathbf{W}^2 \mathbf{X} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \mathbf{\Delta})^{-2}]. -\end{align*} -When $\mathbf{\Delta} = \lambda \mathbf{I}_{pp}$, this MSE is smaller than that of the ML regression estimator, irrespective of the choice of $\lambda$. -\end{example} - - -======= The Bayesian connection} label{sect:genRidgeBayes ======= -This generalized ridge estimator can, like the regular ridge estimator, be viewed as a Bayesian estimator. It requires to replace the conjugate prior on $\beta$ by a more general normal law, $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$, but retains the gamma prior on $\sigma^2$. The joint posterior distribution of $\beta$ and $\sigma^2$ is then obtained analogously (the details are left as Exercise ref{question.generalizedRidgeAndBayes}) to Section ref{sect:Bayes}: -\begin{align*} -f_{\beta, \sigma^2} (\beta, \sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) & = f_Y (\mathbf{Y} \, | \, \mathbf{X}, \beta, \sigma^2) \, f_{\beta}(\beta | \sigma^2) \, f_{\sigma}(\sigma^2) -\\ -& \propto & g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) \, -g_{\sigma^2} (\sigma^2 \, | \, \mathbf{Y}, \mathbf{X}) -\end{align*} -with -\begin{align*} -g_{\beta} (\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) -& \propto & \exp \Big\{ - \frac{1}{2\sigma^2} \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big]^{\top} (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta}) \big[ \beta - \hat{\beta}(\mathbf{\Delta}) \big] \Big\}. -\end{align*} -This implies $\mathbb{E}(\beta \, | \, \sigma^2, \mathbf{Y}, \mathbf{X}) = \hat{\beta}(\mathbf{\Delta})$. Hence, the generalized ridge regression estimator too can be viewed as the Bayesian posterior mean estimator of $\beta$ when imposing a multivariate Gaussian prior on the regression parameter. - - - - - -======= Application ======= - -An illustration involving omics data can be found in the explanation of a gene's expression levels in terms of its DNA copy number. The latter is simply the number of gene copies encoded in the DNA. For instance, for most genes on the autosomal chromosomes the DNA copy number is two, as there is a single gene copy on each chromosome and autosomal chromosomes come in pairs. Alternatively, in males the copy number is one for genes that map to the X or Y chromosome, while in females it is zero for genes on the Y chromosome. In cancer the DNA replication process has often been compromised leading to a (partially) reshuffled and aberrated DNA. Consequently, the cancer cell may exhibit gene copy numbers well over a hundred for classic oncogenes. A faulted replication process does - of course - not nicely follow the boundaries of gene encoding regions. This causes contiguous genes to commonly share aberrated copy numbers. With genes being transcribed from the DNA and a higher DNA copy number implying an enlarged availability of the gene's template, the latter is expected to lead to elevated expression levels. Intuitively, one expects this effect to be localized (a so-called *cis*-effect), but some suggest that aberrations elsewhere in the DNA may directly affect the expression levels of distant genes (referred to as a *trans*-effect). - -The *cis*- and *trans*-effects of DNA copy aberrations on the expression levels of the KRAS oncogene in colorectal cancer are investigated. Data of both molecular levels from the TCGA (The Cancer Genome Atlas) repository are downloaded \citep{TCGA2012colon}. The gene expression data are limited to that of KRAS, while for the DNA copy number data only that of chromosome 12, which harbors KRAS, is retained. This leaves genomic profiles of 195 samples comprising 927 aberrations. Both molecular data types are zero centered feature-wise. Moreover, the data are limited to ten - conveniently chosen? - samples. The KRAS expression levels are explained by the DNA copy number aberrations through the linear regression model. The model is fitted by means of ridge regression, with $\lambda \mathbf{\Delta}$ and $\mathbf{\Delta} = \mathbf{I}_{pp}$ and a single-banded $\mathbf{\Delta}$ with unit diagonal and the elements of the first off-diagonal equal to the arbitrary value of $-0.4$. The latter choice appeals to the spatial structure of the genome and encourages similar regression estimates for contiguous DNA copy numbers. The penalty parameter is chosen by means of leave-one-out cross-validation using the squared error loss. - -\lstinputlisting{fusedRidge_KRASexample.r} - -The right panel of Figure ref{fig:fusedRidgeIllustration} shows the ridge regression estimate with both choices of $\mathbf{\Delta}$ and optimal penalty parameters plotted against the chromosomal order. The location of KRAS is indicated by a vertical dashed bar. The ordinary ridge regression estimates show a minor peak at the location of KRAS but is otherwise more or less flat. In the generalized ridge estimates the peak at KRAS is emphasized. Moreover, the region close to KRAS exhibits clearly elevated estimates, suggesting co-abberated DNA. For the remainder the generalized ridge estimates portray a flat surface, with the exception of a single downward spike away from KRAS. Such negative effects are biologically nonsensible (more gene templates leading to reduced expression levels?). On the whole the generalized ridge estimates point towards the *cis*-effect as the dominant genomic regulation mechanism of KRAS expression. The isolated spike may suggest the presence of a *trans*-effect, but its sign is biological nonsensible and the spike is fully absent in the ordinary ridge estimates. This leads us to ignore the possibility of a genomic *trans*-effect on KRAS expression levels in colorectal cancer. - -The sample selection demands justification. It yields a clear illustrate-able difference between the ordinary and ridge estimates. When all samples are left in, the *cis*-effect is clearly present, discernable from both estimates that yield a virtually similar profile. - - -======= Generalized ridge regression ======= - -What is generally referred to as `generalized ridge regression' (cf. \citealp{Hoer1970, Hemm1975}) is the particular case of loss function (ref{form:generalizedRidgeLoss}) in which $\mathbf{W} = \mathbf{I}_{nn}$, $\beta_0 = \mathbf{0}_{p}$, and $\mathbf{\Delta} = \mathbf{V}_{x} \mathbf{\Lambda} \mathbf{V}_x^{\top}$, where $\mathbf{V}_x$ is obtained from the singular value decomposition of $\mathbf{X}$ (i.e., $\mathbf{X} = \mathbf{U}_{x} \mathbf{D}_x \mathbf{V}_x^{\top}$ with its constituents endowed with the usual interpretation) and $\mathbf{\Lambda}$ a positive definite diagonal matrix. This gives the estimator: -\begin{align*} -\hat{\beta}(\mathbf{\Lambda}) & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} \mathbf{X}^{\top} \mathbf{Y} -\\ -& = (\mathbf{V}_x \mathbf{D}_x \mathbf{U}_x^{\top} \mathbf{U}_x \mathbf{D}_x \mathbf{V}_x^{\top} + \mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1} \mathbf{V}_x \mathbf{D}_x \mathbf{U}_x \mathbf{Y} -\\ -& = \mathbf{V}_x (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x \mathbf{U}_x \mathbf{Y}. -\end{align*} -From this last expression it becomes clear how this estimator generalizes the `regular ridge estimator'. The latter shrinks all eigenvalues, irrespectively of their size, in the same manner through a common penalty parameter. The `generalized ridge estimator', through differing penalty parameters (i.e. the diagonal elements of $\mathbf{\Lambda}$), shrinks them individually. - -The generalized ridge estimator coincides with the Bayesian linear regression estimator with the normal prior $\mathcal{N}[\mathbf{0}_p, (\mathbf{V}_x \mathbf{\Lambda} \mathbf{V}_x^{\top})^{-1}]$ on the regression parameter $\beta$ (and preserving the inverse gamma prior on the error variance). Assume $\mathbf{X}$ to be of full column rank and choose $\mathbf{\Lambda} = g^{-1} \mathbf{D}_x^2$ with $g$ a positive scalar. The prior on $\beta$ then - assuming $(\mathbf{X}^{\top} \mathbf{X})^{-1}$ exits - reduces to Zellner's $g$-prior: $\beta \sim \mathcal{N}[\mathbf{0}_p, g (\mathbf{X}^{\top} \mathbf{X})^{-1}]$ \citep{Zell1986}. The corresponding estimator of the regression coefficient is: $\hat{\beta}(g) = g (1+g)^{-1} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$, which is proportional to the unpenalized ordinary least squares estimator of $\beta$. - -For convenience of notation in the analysis of the generalized ridge estimator the linear regression model is usually rewritten as: -\begin{align*} -\mathbf{Y} & = \mathbf{X} \beta + \vvarepsilon \, \, \, = \, \, \, -\mathbf{X} \mathbf{V}_x \mathbf{V}_x^{\top} \beta + \vvarepsilon \, \, \, = \, \, \, \tilde{\mathbf{X}} \aalpha + \vvarepsilon, -\end{align*} -with $\tilde{\mathbf{X}} = \mathbf{X} \mathbf{V}_x = \mathbf{U}_x -\mathbf{D}_x$ (and thus $\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} = \mathbf{D}_x^ 2$) and $\aalpha = \mathbf{V}_x^{\top} \beta$ with loss function $(\mathbf{Y} - \tilde{\mathbf{X}} \aalpha)^{\top} (\mathbf{Y} - \tilde{\mathbf{X}} \aalpha) + \aalpha^{\top} \mathbf{\Lambda} \aalpha$. In the notation above the generalized ridge estimator is then: -\begin{align*} -\hat{\aalpha}(\mathbf{\Lambda}) & = (\tilde{\mathbf{X}}^{\top} \tilde{\mathbf{X}} + \mathbf{\Lambda})^ {-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y} = (\mathbf{D}_x^ 2 + \mathbf{\Lambda})^{-1} \tilde{\mathbf{X}}^{\top} \mathbf{Y}, -\end{align*} -from which one obtains $\hat{\beta}(\mathbf{\Lambda}) = \mathbf{V}_x \hat{\aalpha}(\mathbf{\Lambda})$. Using $\mathbb{E}[\hat{\aalpha}(\mathbf{\Lambda})] = (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 \aalpha$ and $\mbox{Var}[\hat{\aalpha}(\mathbf{\Lambda})] = \sigma^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1} \mathbf{D}_x^2 (\mathbf{D}_x^2 + \mathbf{\Lambda})^{-1}$, the MSE for the generalized ridge estimator can be written as: -\begin{align*} -\mbox{MSE}[\hat{\aalpha}(\mathbf{\Lambda})] & = \sum_{j=1}^p ( \sigma^2 d_{x,j}^2 + \alpha_j^2 \lambda_{j}^2 ) (d_{x,j}^2 + \lambda_{j} )^{-2}, -\end{align*} -where $d_{x,j} = (\mathbf{D}_x)_{jj}$ and $\lambda_j = (\mathbf{\Lambda})_{jj}$. Having $\aalpha$ and $\sigma^ 2$ available, it is easily seen (equate the derivative w.r.t. $\lambda_j$ to zero and solve) that the MSE of $\hat{\aalpha}(\mathbf{\Lambda})$ is minimized by $\lambda_j = \sigma^2 / \alpha_j^2$ for all $j$. With $\aalpha$ and $\sigma^2$ unknown, cite{Hoer1970} suggest an iterative procedure to estimate the $\lambda_j$'s. Initiate the procedure with the OLS estimates of $\aalpha$ and $\sigma^2$, followed by sequentially updating the $\lambda_j$'s and -the estimates of $\aalpha$ and $\sigma^2$. An analytic expression of the limit of this procedure exists (\citealp{Hemm1975}). This limit, however, still depends on the observed $\mathbf{Y}$ and as such it does not necessarily yield the minimal attainable value of the MSE. This limit may nonetheless still yield a potential gain in MSE. This is investigated in cite{Lawl1981}. Under a variety of cases it seems to indeed outperform the OLS estimator, but there are exceptions. - -======= Conclusion ======= - -To conclude: a note of caution. The generalized ridge penalty is extremely flexible. It can incorporate any prior knowledge on the parameter values (through specification of $\beta_0$) and the relations among these parameters (via $\mathbf{\Delta}$). While a pilot study or literature may provide a suggestion for $\beta_0$, it is less obvious how to choose an informative $\mathbf{\Delta}$ (although a spatial structure is a nice exception). In general, exact knowledge on the parameters should not be incorporated implicitly via the penalty (read: prior) but preferably be used explicitly in the model - the likelihood - itself. Though this may be the viewpoint of a prudent frequentist and a subjective Bayesian might disagree. - - -======= Exercises ======= -\begin{question} \mbox{ } -\\ -Consider the linear regression model $Y_i = \beta_1 X_{i,1} + \beta_2 X_{i,2} + \varepsilon_i$ for $i=1, \ldots, n$. -Suppose estimates of the regression parameters $(\beta_1, \beta_2)$ of this model are obtained through the minimization of the sum-of-squares augmented with a ridge-type penalty: -\begin{align*} -\Big[ \sum_{i=1}^n (Y_i - \beta_1 X_{i,1} - \beta_2 X_{i,2})^2 \Big] + \lambda (\beta_1^2 + \beta_2^2 + 2 \nu \beta_1 \beta_2), -\end{align*} -with penalty parameters $\lambda \in \mathbb{R}_{> 0}$ and $\nu \in (-1, 1)$. - -\begin{compactitem} -\item[*a)*] Sketch (for both $\nu=0$ and $\nu=0.9$) the shape of the parameter constraint induced by the penalty above and describe in words the qualitative difference between both shapes. - -\item[*b)*] When $\nu = -1$ and $\lambda \rightarrow \infty$ the estimates of $\beta_1$ and $\beta_2$ (resulting from minimization of the penalized loss function above) converge towards each other: -$\lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1) = \lim_{\lambda \rightarrow \infty} \hat{\beta}_2(\lambda, -1)$. Motivated by this observation a data scientists incorporates the equality constraint $\beta_1 = \beta = \beta_2$ explicitly into the model, and s/he estimates the `joint regression parameter' $\beta$ through the minimization (with respect to $\beta$) of: -\begin{align*} -\Big[ \sum_{i=1}^n (Y_i - \beta X_{i,1} - \beta X_{i,2})^2 \Big] + \delta \beta^2, -\end{align*} -with penalty parameter $\delta \in \mathbb{R}_{> 0}$. The data scientist is surprised to find that resulting estimate $\hat{\beta}(\delta)$ does not have the same limiting (in the penalty parameter) behavior as the $\hat{\beta}_1(\lambda, -1)$, i.e. $\lim_{\delta \rightarrow \infty} \hat{\beta} (\delta) \not= \lim_{\lambda \rightarrow \infty} \hat{\beta}_1(\lambda, -1)$. Explain the misconception of the data scientist. - -\item[*c)*] Assume that *i)* $n \gg 2$, *ii)* the unpenalized estimates $(\hat{\beta}_1(0, 0), \hat{\beta}_2(0, 0))$ equal $(-2,2)$, and *iii)* that the two covariates $X_1$ and $X_2$ are zero-centered, have equal variance, and are strongly negatively correlated. Consider $(\hat{\beta}_1(\lambda, \nu), \hat{\beta}_2(\lambda, \nu))$ for both $\nu=-0.9$ and $\nu=0.9$. For which value of $\nu$ do you expect the sum of the absolute value of the estimates to be largest? *Hint:* Distinguish between small and large values of $\lambda$ and think geometrically! -\end{compactitem} -\end{question} - -\begin{question} label{question.generalizedRidgeAndBayes} \mbox{ } -\\ -Consider the linear regression model $\mathbf{Y} = \mathbf{X} \beta + \vvarepsilon$ with $\vvarepsilon \sim \mathcal{N}(\mathbf{0}_p, \sigma^2 \mathbf{I}_{pp})$. Assume $\beta \sim \mathcal{N}(\beta_0, \sigma^2 \mathbf{\Delta}^{-1})$ with $\beta_0 \in \mathbb{R}^p$ and $\mathbf{\Delta} \succ 0$ and a gamma prior on the error variance. Verify (i.e., work out the details of the derivation) that the posterior mean coincides with the generalized ridge estimator defined as: -\begin{align*} -\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X} + \mathbf{\Delta})^{-1} (\mathbf{X}^{\top} \mathbf{Y} + \mathbf{\Delta} \beta_0). -\end{align*} -\end{question} - - - -\begin{question} label{question.ridgeAR1penalty} \mbox{ } -\\ -The ridge penalty may be interpreted as a multivariate normal prior on the regression coefficients: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \mathbf{I}_{pp})$. Different priors may be considered. In case the covariates are spatially related in some sense (e.g. genomically), it may of interest to assume a first-order autoregressive prior: $\beta \sim \mathcal{N}(\mathbf{0}, \lambda^{-1} \SSigma_A)$, in which $\SSigma_A$ is a $p \times p$-correlation matrix with $(\SSigma_A)_{j_1, j_2} = \rho^{ | j_1 - j_2 | } $ for some correlation coefficient $\rho \in [0, 1)$. Hence, -\begin{align*} -\SSigma_A \, \, \, = \, \, \, -\left( -\begin{array}{cccc} -1 & \rho & \ldots & \rho^{p-1} -\\ -\rho & 1 & \ldots & \rho^{p-2} -\\ -\vdots & \vdots & \ddots & \vdots -\\ -\rho^{p-1} & \rho^{p-2} & \ldots & 1 -\end{array} -\right). -\end{align*} - -\begin{compactitem} -\item[*a)*] The penalized loss function associated with this AR(1) prior is: -\begin{align*} -\mathcal{L}(\beta; \lambda, \SSigma_A) & = \| \mathbf{Y} - \mathbf{X} \beta \|_2^2 + \lambda \beta^{\top} \SSigma_A^{-1} \beta. -\end{align*} -Find the minimizer of this loss function. - -\item[*b)*] What is the effect of $\rho$ on the ridge estimates? Contrast this to the effect of $\lambda$. Illustrate this on (simulated) data. - -\item[*c)*] Instead of an AR(1) prior assume a prior with a uniform correlation between the elements of $\beta$. That is, replace $\SSigma_A$ by $\SSigma_U$, given by: -\begin{align*} -\SSigma_U \, \, \, = \, \, \, \left( -\begin{array}{ccc} -1 & \rho & \rho -\\ -\rho & 1 & \rho -\\ -\rho & \rho & 1 -\end{array} -\right) -\end{align*} -\\ -Investigate (again on data) the effect of changing from the AR(1) to the uniform prior on the ridge regression estimates. -\end{compactitem} -\end{question} - - - - -========= Ridge logistic regression ========= label{sect.ridgeLogistic} -# ======= Ridge logistic regression ======= - -Ridge penalized estimation is not limited to the standard linear regression model, but may be used to estimate (virtually) any model. Here we illustrate how it may be used to fit the logistic regression model. To this end we first recap this model and the (unpenalized) maximum likelihood estimation of its parameters. After which the model is estimated by means of ridge penalized maximum likelihood, which will turn out to be a relatively straightforward modification of unpenalized estimation. - -======= Logistic regression ======= - -The logistic regression model explains a binary response variable (through some transformation) by a linear combination of a set of covariates (as in the linear regression model). Denote this response of the $i$-th sample by $Y_i$ with $Y_i \in \{ 0, 1 \}$ for $i=1, \ldots, n$. The $n$-dimensional column vector $\mathbf{Y}$ stacks these $n$ responses. For each sample information on the $p$ explanatory variables $X_{i,1}, \ldots, X_{i,p}$ is available. In row vector form this information is denoted $\mathbf{X}_{i,\ast} = (X_{i,1}, \ldots, X_{i,p})$. Or, in short, $\mathbf{X}_i$ when the context tolerates no confusion. The $(n \times p)$-dimensional matrix $\mathbf{X}$ aggregates these vectors, such that $\mathbf{X}_i$ is the $i$-th row vector. - -The binary response cannot be modelled as in the linear model like $Y_i = \mathbf{X}_i \beta + \varepsilon_i$. With each element of $\mathbf{X}_i$ and $\beta$ assuming a value in $\mathbb{R}$, the linear predictor is not restricted to the domain of the response. This is resolved by modeling $p_i = P(Y_i = 1)$ instead. Still the linear predictor may exceed the domain of the response ($p_i \in [0,1]$). Hence, a transformation is applied to map $p_i$ to $\mathbb{R}$, the range of the linear predictor. -# original latex figure with scale=0.40, angle=0 - -FIGURE: [logistic_effectOfBeta0.eps, width=400 frac=1.0] Top row, left panel: the response curve for various choices of the intercept $\beta_0$. Top row, right panel: the response curve for various choices of the regression coefficent $\beta_1$. Bottom row, left panel: the responce curve for various choices of the link function. Bottom panel, right panel: observations, fits and their deviations. } label{fig.logisticRidge_modelAndFitIllustration -The transformation associated with the logistic regression model is the logarithm of the odds, with the odds defined as: $\mbox{*odds*} = P(\mbox{succes}) / P(\mbox{failure}) = p_i/ (1-p_i)$. The logistic model is then written as $\log[ p_i / (1-p_i)] = \mathbf{X}_i \beta$ for all $i$. Or, expressed in terms of the response: -\begin{align*} -p_i & = P(Y_i = 1) \, \, \, = \, \, \, g^{-1}(\mathbf{X}_i; \beta) \, \, \, = \, \, \,\frac{\exp(\mathbf{X}_i \beta) }{1 + \exp(\mathbf{X}_i \beta) }. -\end{align*} -The function $g(\cdot; \cdot)$ is called the *link function*. It links the response to the explanatory variables. The one above is called the logistic link function. Or short, logit. The regression parameters have tangible interpretations. When the first covariate represents the intercept, i.e. $X_{i,j} = 1$ for all $i$, then $\beta_1$ determines where the link function equals a half when all other covariates fail to contribute to the linear predictor (i.e. where $P (Y_i = 1 \, | \, \mathbf{X}_{i}) = 0.5$ when $\mathbf{X}_{i} \beta = \beta_1$). This is illustrated in the top-left panel of Figure ref{fig.logisticRidge_modelAndFitIllustration} for various choices of the intercept. On the other hand, the regression parameters are directly related to the odds ratio: $\mbox{*odds ratio*} = \mbox{odds}(X_{i,j}+1) / \mbox{odds}(X_{i,j}) = \exp(\beta_j)$. Hence, the effect of a unit change in the $j$-th covariate on the odds ratio is $\exp(\beta_j)$ (see Figure ref{fig.logisticRidge_modelAndFitIllustration}, top-right panel). Other link functions (depicted in Figure ref{fig.logisticRidge_modelAndFitIllustration}, bottom-left panel) are common, e.g. the *probit*: $p_i = \Phi_{0,1}(\mathbf{X}_i \beta)$; the *cloglog*: $p_i = \frac{1}{\pi} \arctan(\mathbf{X}_i \beta) + \frac{1}{2}$; the *Cauchit*: $p_i = \exp[ - \exp(\mathbf{X}_i \beta)]$. All these link function are invertible. Irrespective of the choice of the link function, the binary data are thus modelled as $Y_i \sim \mathcal{B}[g^{-1}(\mathbf{X}_i; \beta), 1]$. That is, as a single draw from the Binomial distribution with success probability $g^{-1}(\mathbf{X}_i; \beta)$. - - -Let us now estimate the parameter of the logistic regression model by means of the maximum likelihood method. The likelihood of the experiment is then: -\begin{align*} -L(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i}. -\end{align*} -After taking the logarithm and some ready algebra, the log-likelihood is found to be: -\begin{align*} -\mathcal{L}(\mathbf{Y} \, | \, \mathbf{X}; \beta) & = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\}. -\end{align*} -Differentiate the log-likelihood with respect to $\beta$, equate it zero, and obtain the estimating equation for $\beta$: - -!bt - -!bt -\begin{align} label{form:logisticRidge_estimatingEquationOfBeta} -\frac{\partial \mathcal{L}}{\partial \beta } & = \sum_{i=1}^n \Big[ Y_i - \frac{\exp(\mathbf{X}_i \beta)}{ 1 + \exp(\mathbf{X}_i \beta)} \Big] \mathbf{X}_i^{\top} \, \, \, = \, \, \, \mathbf{0}_p. -\end{align} -!et -!et -The ML estimate of $\beta$ strikes a (weighted by the $\mathbf{X}_i$) balance between observation and model. Put differently (and illustrated in the bottom-right panel of Figure ref{fig.logisticRidge_modelAndFitIllustration}), a curve is fit through data by minimizing the distance between them: at the ML estimate of $\beta$ a weighted average of their deviations is zero. - -The maximum likelihood estimate of $\beta$ is evaluated by solving Equation (ref{form:logisticRidge_estimatingEquationOfBeta}) with respect to $\beta$ by means of the Newton-Raphson algorithm. The Newton-Raphson algorithm iteratively finds the zeros of a smooth enough function $f(\cdot)$. Let $x_0$ denote an initial guess of the zero. Then, approximate $f(\cdot)$ around $x_0$ by means of a first order Taylor series: $f(x) \approx x_0 + (x - x_0) \, (d f / d x) |_{x=x_0}$. Solve this for $x$ and obtain: $x = x_0 - [ (d f / d x) |_{x=x_0} ]^{-1} f(x_0)$. Let $x_1$ be the solution for $x$, use this as the new guess and repeat the above until convergence. When the function $f(\cdot)$ has multiple arguments, is vector-valued and denoted by $\vec{\mathbf{f}}$, and the Taylor approximation becomes: $\vec{\mathbf{f}}(\mathbf{x}) \approx \mathbf{x}_0 + J \vec{\mathbf{f}} \big|_{\mathbf{x}=\mathbf{x}_0} (\mathbf{x} - \mathbf{x}_0)$ with -\begin{align*} -J \vec{\mathbf{f}} = \left( -\begin{array}{llll} -\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \ldots & \frac{\partial f_1}{\partial x_p} -\\ -\frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \ldots & \frac{\partial f_2}{\partial x_p} -\\ -\vdots & \vdots & \ddots & \vdots -\\ -\frac{\partial f_q}{\partial x_1} & \frac{\partial f_q}{\partial x_2} & \ldots & \frac{\partial f_q}{\partial x_p} -\end{array} -\right), -\end{align*} -the Jacobi matrix. An update of $x_0$ is now readily constructed by solving (the approximation for) $\vec{\mathbf{f}}(\mathbf{x}) = \mathbf{0}$ for $\mathbf{x}$. - -When applied here to the maximum likelihood estimation of the regression parameter $\beta$ of the logistic regression model, the Newton-Raphson update is: -\begin{align*} -\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} - \Big( \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \Big)^{-1} \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } \, \, \frac{\partial \mathcal{L}}{\partial \beta } \Big|_{\beta = \hat{\beta}^{\mbox{{\tiny old}}} } -\end{align*} -where the Hessian of the log-likelihood equals: -\begin{align*} -\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} & = - \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1 + \exp(\mathbf{X}_i \beta)]^2} \mathbf{X}_i^{\top} \mathbf{X}_i. -\end{align*} -Iterative application of this updating formula converges to the ML estimate of $\beta$. - -The Newton-Raphson algorithm is often reformulated as an iteratively re-weighted least squares algorithm. Hereto, first write the gradient and Hessian in matrix notation: -\begin{align*} -\frac{\partial \mathcal{L}}{\partial \beta } \, \, \, = \, \, \, \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta)] -& \mbox{ and } & -\frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, - \mathbf{X}^{\top} \mathbf{W} \mathbf{X}, -\end{align*} -where $\vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta) = [g^{-1}( \mathbf{X}_{1, \ast}; \beta), \ldots, g^{-1}( \mathbf{X}_{n, \ast}; \beta)]^{\top}$ with $g^{-1}(\cdot; \cdot) = \exp(\cdot; \cdot) / [1 + \exp(\cdot; \cdot)]$ and $\mathbf{W}$ diagonal with $(\mathbf{W})_{ii} = \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) [ 1 + \exp(\mathbf{X}_i \hat{\beta}^{\mbox{{\scriptsize old}}} ) ]^{-2}$. The updating formula of the estimate then becomes: -\begin{align*} -\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] -\\ -& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} -\\ -& = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, -\end{align*} -where $\mathbf{Z} = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}$. The Newton-Raphson update is thus the solution to the following weighted least squares problem: -\begin{align*} -\hat{\beta}^{\mbox{{\scriptsize new}}} & = \arg \min_{\beta} (\mathbf{Z} - \mathbf{X} \beta)^{\top} \mathbf{W} (\mathbf{Z} - \mathbf{X} \beta). -\end{align*} -Effectively, at each iteration the *adjusted response* $\mathbf{Z}$ is regressed on the covariates that comprise $\mathbf{X}$. For more on logistic regression confer the monograph of cite{Hosm2013}. - - -======= Ridge estimation ======= - -High-dimensionally, the linear predictor $\mathbf{X} \beta$ may be uniquely defined, but the maximum likelihood estimate of the logistic regression parameter is not. Assume $p > n$ and an estimate $\hat{\beta}$ available. Due to the high-dimensionality, the null space of $\mathbf{X}$ is non-trivial. Hence, let $\ggamma \in \mbox{null}(\mbox{span}(\mathbf{X}))$. Then: $\mathbf{X} \hat{\beta} = \mathbf{X} \hat{\beta} + \mathbf{X} \ggamma = \mathbf{X} (\hat{\beta} + \ggamma)$. As the null space is a $p-n$-dimensional subspace, $\ggamma$ need not equal zero. Hence, an infinite number of estimates of the logistic regression parameter exists that yield the same log-likelihood. Augmentation of the loss function with a ridge penalty resolves the matter, as their sum is strictly concave in $\beta$ (not convex as a maximum rather than a minimum is sought here) and thereby has a unique maximum. - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [logisticPenLL2ridgePenalty.eps, width=400 frac=1.0] Top row, left panel: contour plot of the penalized log-likelihood of a logistic regression model with the ridge constraint (red line). Top row, right panel: the regularization paths of the ridge estimator of the logistic regression parameter. Bottom row, left panel: variance of the ridge estimator of the logistic regression parameter against the logarithm of the penalty parameter. Bottom panel, right panel: the predicted success probability versus the linear predictor for various choices of the penalty parameter. } label{fig.logisticRidge_effectOfPenalization -\mbox{ } - - -Ridge maximum likelihood estimates of the logistic model parameters are found by the maximization of the ridge penalized loglikelihood (cf. \citealt{Scha1984,LeCe1992}): -\begin{align*} -\mathcal{L}^{\mbox{{\tiny pen}}}(\mathbf{Y}, \mathbf{X}; \beta, \lambda) & = \mathcal{L} (\mathbf{Y}, \mathbf{X}; \beta) - \tfrac{1}{2} \lambda \| \beta \|_2^2 -\\ -& = \sum_{i=1}^n \big\{ Y_i \mathbf{X}_i \beta - \log [ 1 + \exp(\mathbf{X}_i \beta) ] \big\} - \tfrac{1}{2} \lambda \beta^{\top} \beta, -\end{align*} -where the second summand is the ridge penalty (the sum of the square of the elements of $\beta$) with $\lambda$ the penalty parameter. Note that as in Section ref{sect.constrainedEstimation} maximization of this penalized loss function can be reformulated as a constrained estimation problem. This is illustrated by the top left panel of Figure ref{fig.logisticRidge_effectOfPenalization}, which depicts the contours (black lines) of the log-likelihood and the spherical domain of the parameter (red line). The optimization of the above loss function proceeds, due to the differentiability of the penalty, fully analogous to the unpenalized case and uses the Newton-Raphson algorithm for solving the (penalized) estimating equation. Hence, the unpenalized ML estimation procedure is modified straightforwardly by replacing gradient and Hessian by their `penalized' counterparts: -\begin{align*} -\frac{\partial \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta } \, \, \, = \, \, \, \frac{\partial \mathcal{L}}{\partial \beta } - \lambda \beta & \mbox{ and } & \frac{\partial^2 \mathcal{L}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \, \, \, = \, \, \, \frac{\partial^2 \mathcal{L}}{\partial \beta \partial \beta^{\top}} - \lambda \mathbf{I}_{pp}. -\end{align*} -With these at hand, the Newton-Raphson algorithm is (again) reformulated as an iteratively re-weighted least squares algorithm with the updating step changes accordingly to: -\begin{align*} -\hat{\beta}^{\mbox{{\scriptsize new}}} & = \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \{ \mathbf{X}^{\top} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}}) ] - \lambda \beta^{\mbox{{\scriptsize old}}} \} -\\ -& = \mathbf{V}^{-1} \mathbf{V} \hat{\beta}^{\mbox{{\scriptsize old}}} - \lambda \mathbf{V}^{-1} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] -\\ -& = \mathbf{V}^{-1} \mathbf{X}^{\top} \mathbf{W} \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbf{Y} - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \} -\\ -& = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{Z}, -\end{align*} -where $\mathbf{V} = \mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp}$ and $\mathbf{W}$ and $\mathbf{Z}$ as before. Hence, use this to update the estimate of $\beta$ until convergence, which yields the desired ridge ML estimate. - -Obviously, the ridge estimate of the logistic regression parameter tends to zero as $\lambda \rightarrow \infty$. Now consider a linear predictor with an intercept that is left unpenalized. When $\lambda$ tends to infinity, all regression coefficients but the intercept vanish. The intercept is left to model the success probability. Hence, in this case $\lim_{\lambda \rightarrow \infty} \hat{\beta}_0 (\lambda) = \log [ \tfrac{1}{n} \sum_{i=1}^n Y_i / \tfrac{1}{n} \sum_{i=1}^n (1-Y_i)]$. - - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [logisticRidge_effectOnPrediction_lambda0.eps, width=400 frac=1.0] The realized design as scatter plot ($X_1$ vs $X_2$ overlayed by the success (\textcolor{red}{RED}) and failure regions (\textcolor{green}{GREEN}) for various choices of the penalty parameter: $\lambda = 0$ (top row, left panel), $\lambda = 10$ (top row, right panel) $\lambda = 40$ (bottom row, left panel), $\lambda = 100$ (bottom row, right panel).} label{fig.logisticRidge_effectOnPrediction - - - -The effect of the ridge penalty on parameter estimates propagates to the predictor $\hat{p}_i$. The linear predictor of the linear regression model involving the ridge estimator $\mathbf{X}_i \hat{\beta}(\lambda)$ shrinks towards a common value for each $i$, leading to a scale difference between observation and predictor (as seen before in Section ref{sect.ridgeRegressionDataIllustration}). This behaviour transfers to the ridge logistic regression predictor, as is illustrated on simulated data. The dimension and sample size of these data are $p=2$ and $n=200$, respectively. The covariate data are drawn from the standard normal, while that of the response is sampled from a Bernoulli distribution with success probability $P(Y_i=1) = \exp(2 X_{i,1} - 2 X_{i,2}) / [ 1 + \exp(2 X_{i,1} - 2 X_{i,2})]$. The logistic regression model is estimated from these data by means of ridge penalized likelihood maximization with various choices of the penalty parameter. The bottom right plot in Figure ref{fig.logisticRidge_effectOfPenalization} shows the predicted success probability versus the linear predictor for various choices of the penalty parameter. Larger values of the penalty parameter $\lambda$ flatten the slope of this curve. Consequently, for larger $\lambda$ more excessive values of the covariates are needed to achieve the same predicted success probability as those obtained with smaller $\lambda$ at more moderate covariate values. The implications for the resulting classification may become clearer when studying the effect of the penalty parameter on the `failure' and `success regions' respectively defined by: -\\ -\indent $\{(x_1, x_2) : P({\color{green}{\mathbf{Y=0}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$, -\\ -\indent $\{(x_1, x_2) : P({\color{red}{\mathbf{Y=1}}} \, | \, X_1=x_1, X_2=x_2, \hat{\beta}(\lambda)) > 0.75 \}$. -\\ -This separates the design space in a light red (`failure') and light green (`success') domain. The white bar between them is the domain where samples cannot be classified with high enough certainty. As $\lambda$ grows, so does the white area that separates the failure and success regions. Hence, as stronger penalization shrinks the logistic regression parameter estimate towards zero, it produces a predictor that is less outspoken in its class assignments. - - - - -======= Moments ======= - -The $1^{\mbox{{\tiny st}}}$ and $2^{\mbox{{\tiny nd}}}$ order moment of the ridge ML parameter of the logistic model may be approximated by the final update of the Newton-Raphson estimate. Assume the one-to-last update $\hat{\beta}^{\mbox{{\scriptsize old}}}$ to be non-random and proceed as for the ridge estimator of the linear regression model parameter to arrive at: -# \begin{align*} -# & & \hat{\beta}^{\mbox{{\scriptsize new}}} = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + 2 \lambda \mathbf{I}_{p \times p} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbf{z} -# \end{align*} -\begin{align*} -\mathbb{E} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \mathbb{E}( \mathbf{Z}), -\\ -\mbox{Var} \big( \hat{\beta}^{\mbox{{\scriptsize new}}} \big) & = [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1} \mathbf{X}^{\top} \mathbf{W} \big[ \mbox{Var} ( \mathbf{Z} ) \big] \mathbf{W} \mathbf{X} [\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp} ]^{-1}, -\end{align*} -with -\begin{align*} -\mathbb{E}(\mathbf{Z}) & = \{ \mathbf{X} \hat{\beta}^{\mbox{{\scriptsize old}}} + \mathbf{W}^{-1} [\mathbb{E}(\mathbf{Y}) - \vec{\mathbf{g}}^{-1}( \mathbf{X}; \beta^{\mbox{{\scriptsize old}}})] \}, -\\ -\mbox{Var}(\mathbf{Z}) & = \mathbf{W}^{-1} \mbox{Var}(\mathbf{Y}) \mathbf{W}^{-1} = \mathbf{W}^{-1}, -\end{align*} -where the identity $\mbox{Var}(\mathbf{Y}) = \mathbf{W}$ follows from the variance of a Binomial distributed random variable. From these expressions similar properties as for the ridge ML estimate of the regression parameter of the linear model may be deduced. For instance, the ridge ML estimate of the logistic regression parameter converges to zero as the penalty parameter tends to infinity (confer the top right panel of Figure ref{fig.logisticRidge_effectOfPenalization}). Similarly, their variances vanish as $\lambda \rightarrow \infty$ (illustrated in the bottom left panel of Figure ref{fig.logisticRidge_effectOfPenalization}). - -======= The Bayesian connection ======= - -All penalized estimators can be formulated as Bayesian estimators, including the ridge logistic estimator. In particular, ridge estimators correspond to Bayesian estimators with a multivariate normal prior on the regression coefficients. Thus, assume $\beta \sim \mathcal{N}(\mathbf{0}_p, \mathbf{\Delta}^{-1})$. The posterior distribution of $\beta$ then is: -\begin{align*} -f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X}) & \propto & \Big\{ \prod_{i=1}^n \big[ P(Y_i = 1 \, | \, \mathbf{X}_i) \big]^{Y_i} \big[ P(Y_i = 0 \, | \, \mathbf{X}_i) \big]^{1-Y_i} \Big\} \exp( - \tfrac{1}{2} \beta \mathbf{\Delta} \beta). -\end{align*} -This does not coincide with any standard distribution. But, under appropriate conditions, the posterior distribution is asymptotically normal. This invites a (multivariate) normal approximation to the posterior distribution above. The Laplace's method provides (cf. \citealp{Bish2006}). - -# original latex figure with scale=0.40, angle=0 - -FIGURE: [logisticRidge_laplace2posterior.eps, width=400 frac=1.0] Right panel: Laplace approximation to the posterior density of the Bayesian logistic regression parameter.} label{fig.logisticRidge_MSEandLaplace2posterior - -Laplace's method *i)* centers the normal approximation at the mode of the posterior, and *ii)* chooses the covariance to match the curvature of the posterior at the mode. The posterior mode is the location of the maximum of the posterior distribution. The location of this maximum coincides with that of the logarithm of the posterior. The latter is the log-likelihood augmented with a ridge penalty. Hence, the posterior mode, which is taken as the mean of the approximating Gaussian, coincides -with the ridge logistic estimator. For the covariance of the approximating Gaussian, the logarithm of the posterior is approximated by a second order Taylor series around the posterior mode and limited to second order terms: -\begin{align*} -\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] & \propto & -\left. \log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] -\right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} -\\ -& & + \tfrac{1}{2} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top} \left. \frac{\partial^2}{\partial \beta \partial \beta^{\top}} -\log[f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})] - \right|_{\beta = \hat{\beta}_{\mbox{{\tiny MAP}}}} (\beta - \hat{\beta}_{\mbox{{\tiny MAP}}})^{\top}, -\end{align*} -in which the first order term cancels as the derivative of $f_{\beta}(\beta \, | \, \mathbf{Y}, \mathbf{X})$ with respect to $\beta$ vanishes at the posterior mode - its maximum. Take the exponential of this approximation and match its arguments to that of a multivariate Gaussian $\exp[-\tfrac{1}{2} (\beta - \mmu_{\beta})^{\top} \mathbf{\Sigma}_{\beta}^{-1} (\beta - \mmu_{\beta})]$. The covariance of the sought Gaussian approximation is thus the inverse of the Hessian of the negative penalized log-likelihood. Put together the posterior is approximated by: -\begin{align*} -\beta \, | \, \mathbf{Y}, \mathbf{X} \sim \mathcal{N} \Big( \hat{\beta}_{\mbox{{\tiny MAP}}}, \Big\{ \mathbf{\Delta} + \sum_{i=1}^n \frac{\exp(\mathbf{X}_i \beta)}{ [1+ \exp(\mathbf{X}_i \beta) ]^2 } \mathbf{X}_i \mathbf{X}_i^{\top} \Big\}^{-1} \Big). -\end{align*} -The Gaussian approximation is convenient but need not be good. Fortunately, the Bernstein-Von Mises Theorem \citep{VdVa2000} tells it is very accurate when the model is regular, the prior smooth, and the sample size sufficiently large. The quality of the approximation for an artificial example data set is shown in Figure ref{fig.logisticRidge_MSEandLaplace2posterior}. - -======= Penalty parameter selection ======= - -As before the penalty parameter may be chosen through $K$-fold cross-validation. For the $K=n$ case cite{Meij2013} describe a computationally efficient approximation of the leave-one-out cross-validated loglikelihood. It is based on the exact evaluation of the LOOCV loss, discussed in Section ref{subsect.crossvalidation}, that avoided resampling. The approach of cite{Meij2013} hinges upon the first-order Taylor expansion of the left-out penalized loglikelihood of the left-out estimate $\hat{\beta}_{-i} (\lambda)$ around $\hat{\beta} (\lambda)$, which yields an approximation of the former: -\begin{align*} -\hat{\beta}_{-i} (\lambda) & \approx \hat{\beta} (\lambda) - -\left( \left. \frac{\partial^2 \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta \partial \beta^{\top}} \right|_{\beta = \hat{\beta}(\lambda)} \right)^{-1} \left. \frac{\partial \mathcal{L}_{-i}^{\mbox{{\tiny pen}}}}{\partial \beta } \right|_{\beta = \hat{\beta}(\lambda)} -\\ -& = \hat{\beta} (\lambda) + (\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} \{ \mathbf{X}_{- i, \ast}^{\top} [\mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda))] - \lambda \hat{\beta}(\lambda) \}. -\end{align*} -This approximation involves the inverse of a $p \times p$ dimensional matrix, which amounts to the evaluation of $n$ such inverses for the LOOCV loss. As in Section ref{subsect.crossvalidation} this may be avoided. Rewrite both the gradient and the Hessian of the left-out loglikelihood in the approximation of the preceding display: -\begin{align*} -\mathbf{X}_{-i, \ast}^{\top} \{ \mathbf{Y}_{-i} - \vec{\mathbf{g}}^{-1}(\mathbf{X}_{-i, \ast}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) -& = \mathbf{X}^{\top} \{ \mathbf{Y} - \vec{\mathbf{g}}^{-1}[\mathbf{X}; \hat{\beta}(\lambda)]\} - \lambda \hat{\beta}(\lambda) -- \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} -\\ -& = - \mathbf{X}_{i, \ast}^{\top} \{Y_{i} - g^{-1}[\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)]\} -\end{align*} -and -\begin{align*} -(\mathbf{X}_{- i, \ast}^{\top} \mathbf{W}_{-i, -i} \mathbf{X}_{- i, \ast} + \lambda \mathbf{I}_{pp})^{-1} & = (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} + \mathbf{W}_{ii} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} -\\ -& & \qquad \qquad \qquad \qquad \qquad \quad [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} \mathbf{X}_{i, \ast} (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1}, -\end{align*} -where the Woodbury identity has been used and now $\mathbf{H}_{ii}(\lambda) = \mathbf{W}_{ii} \mathbf{X}_{i, \ast}(\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top}$. Substitute both in the approximation of the left-out ridge logistic regression estimator and manipulate as in Section ref{subsect.crossvalidation} to obtain: -\begin{align*} -\hat{\beta}_{- i}(\lambda) & \approx \hat{\beta}(\lambda) - (\mathbf{X}^{\top} \mathbf{W} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} \mathbf{X}_{i, \ast}^{\top} [ 1 - \mathbf{H}_{ii}(\lambda)]^{-1} [ Y_i - g^{-1}(\mathbf{X}_{i, \ast}; \hat{\beta}(\lambda)) ]. -\end{align*} -Hence, the leave-one-out cross-validated loglikelihood $\sum_{i=1}^n \mathcal{L} [Y_i \, | \, \mathbf{X}_{i, \ast}, \hat{\beta}_{-i}(\lambda)]$ can now be evaluated by means of a single inverse of a $p \times p$ dimensional matrix and some matrix multiplications. For the performance of this approximation in terms of accuracy and speed confer cite{Meij2013}. - - - - - - From 4291b46bcbc11e4fa29c31a278725a853770a0de Mon Sep 17 00:00:00 2001 From: mhjensen Date: Fri, 7 Sep 2018 05:40:44 +0200 Subject: [PATCH 10/12] update on html slides for regression --- .../Regression/html/._Regression-bs046.html | 333 +++++++++++++++++ .../Regression/html/._Regression-bs047.html | 339 ++++++++++++++++++ .../Regression/html/._Regression-bs048.html | 336 +++++++++++++++++ .../Regression/html/._Regression-bs049.html | 316 ++++++++++++++++ .../Regression/html/._Regression-bs050.html | 324 +++++++++++++++++ .../Regression/html/._Regression-bs051.html | 338 +++++++++++++++++ .../Regression/html/._Regression-bs052.html | 326 +++++++++++++++++ .../Regression/html/._Regression-bs053.html | 320 +++++++++++++++++ .../Regression/html/Regression-reveal.html | 8 +- .../Regression/html/Regression-solarized.html | 8 +- doc/pub/Regression/html/Regression.html | 8 +- doc/pub/Regression/ipynb/Regression.ipynb | 10 +- .../ipynb/ipynb-Regression-src.tar.gz | Bin 210 -> 211 bytes .../pdf/Regression-beamer-handouts2x3.pdf | Bin 411149 -> 411066 bytes doc/pub/Regression/pdf/Regression-beamer.pdf | Bin 366764 -> 366730 bytes doc/pub/Regression/pdf/Regression-minted.pdf | Bin 405955 -> 405967 bytes doc/src/Regression/Regression.do.txt | 10 +- 17 files changed, 2648 insertions(+), 28 deletions(-) create mode 100644 doc/pub/Regression/html/._Regression-bs046.html create mode 100644 doc/pub/Regression/html/._Regression-bs047.html create mode 100644 doc/pub/Regression/html/._Regression-bs048.html create mode 100644 doc/pub/Regression/html/._Regression-bs049.html create mode 100644 doc/pub/Regression/html/._Regression-bs050.html create mode 100644 doc/pub/Regression/html/._Regression-bs051.html create mode 100644 doc/pub/Regression/html/._Regression-bs052.html create mode 100644 doc/pub/Regression/html/._Regression-bs053.html diff --git a/doc/pub/Regression/html/._Regression-bs046.html b/doc/pub/Regression/html/._Regression-bs046.html new file mode 100644 index 000000000..e12f107dc --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs046.html @@ -0,0 +1,333 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +

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    Resampling approaches can be computationally expensive

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    +Resampling approaches can be computationally expensive, because they +involve fitting the same statistical method multiple times using +different subsets of the training data. However, due to recent +advances in computing power, the computational requirements of +resampling methods generally are not prohibitive. In this chapter, we +discuss two of the most commonly used resampling methods, +cross-validation and the bootstrap. Both methods are important tools +in the practical application of many statistical learning +procedures. For example, cross-validation can be used to estimate the +test error associated with a given statistical learning method in +order to evaluate its performance, or to select the appropriate level +of flexibility. The process of evaluating a model’s performance is +known as model assessment, whereas the process of selecting the proper +level of flexibility for a model is known as model selection. The +bootstrap is widely used. +

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    +A popular strategy is to choose a penalty parameter that yields a good +but parsimonious model. Information criteria measure the balance +between model fit and model complexity. One possibility is Aikaike's +information criterion (AIC). +The AIC measures model fit by the log-likelihood +and model complexity is measured by the number of parameters used by +the model. The number of model parameters in regular regression simply +corresponds to the number of covariates in the model. Or, by the +degrees of freedom consumed by the model, which is equivalent to the +trace of the hat matrix. For ridge regression it thus seems natural to +define model complexity analogously by the trace of the ridge hat +matrix. This yields the AIC for the linear regression model with ridge +estimates: + +$$ +\begin{align*} +\mbox{AIC}(\lambda) & = 2 \, p - 2 \log(\hat{L}) +\\ +& = 2 \, \mbox{tr} [\mathbf{H}(\lambda)] - 2 \log\{L[\hat{\beta}(\lambda), \hat{\sigma}^2(\lambda)]\} +\\ +& = 2 \, \sum_{j=1}^p \frac{d_{jj}^2}{d_{jj}^2 + \lambda} ++ 2 n \, \log[\sqrt{2 \, \pi} \, \hat{\sigma}(\lambda)] + \frac{1}{\hat{\sigma}^2(\lambda)} \sum_{i=1}^n [y_i - \mathbf{X}_{i, \ast} \, \hat{\beta}(\lambda)]^2. +\end{align*} +$$ + +The value of \( \lambda \) which minimizes \( \mbox{AIC}(\lambda) \) corresponds to the `optimal' balance of model complexity and overfitting. + +

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    Cross-validation

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    +Instead of choosing the penalty parameter to balance model fit with +model complexity, cross-validation requires it (i.e. the penalty +parameter) to yield a model with good prediction +performance. Commonly, this performance is evaluated on novel +data. Novel data need not be easy to come by and one has to make do +with the data at hand. The setting of `original' and novel data is +then mimicked by sample splitting: the data set is divided into two +(groups of samples). One of these two data sets, called the training +set, plays the role of `original' data on which the model is +built. The second of these data sets, called the test set, plays the +role of the `novel' data and is used to evaluate the prediction +performance (often operationalized as the log-likelihood or the +prediction error or its square or the R2 score) of the model built on the training data set. This +procedure (model building and prediction evaluation on training and +test set, respectively) is done for a collection of possible penalty +parameter choices. The penalty parameter that yields the model with +the best prediction performance is to be preferred. The thus obtained +performance evaluation depends on the actual split of the data set. To +remove this dependence the data set is split many times into a +training and test set. For each split the model parameters are +estimated for all choices of \( \lambda \) using the training data and +estimated parameters are evaluated on the corresponding test set. The +penalty parameter that on average over the test sets performs best (in +some sense) is then selected. + +

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    Computationally expensive

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    +The validation set approach is conceptually simple and is easy to implement. But it has two potential drawbacks: + +

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    • The validation estimate of the test error rate can be highly variable, depending on precisely which observations are included in the training set and which observations are included in the validation set.
      • +
    • In the validation approach, only a subset of the observations, those that are included in the training set rather than in the validation set are used to fit the model. Since statistical methods tend to perform worse when trained on fewer observations, this suggests that the validation set error rate may tend to overestimate the test error rate for the model fit on the entire data set.
      • +
    + +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs050.html b/doc/pub/Regression/html/._Regression-bs050.html new file mode 100644 index 000000000..150bbb796 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs050.html @@ -0,0 +1,324 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    +
    + +
    + +

     

     

     

    + + + + +

    Various steps in cross-validation

    + +

    +When the repetitive splitting of the data set is done randomly, +samples may accidently end up in a fast majority of the splits in +either training or test set. Such samples may have an unbalanced +influence on either model building or prediction evaluation. To avoid +this \( k \)-fold cross-validation structures the data splitting. The +samples are divided into \( k \) more or less equally sized exhaustive and +mutually exclusive subsets. In turn (at each split) one of these +subsets plays the role of the test set while the union of the +remaining subsets constitutes the training set. Such a splitting +warrants a balanced representation of each sample in both training and +test set over the splits. Still the division into the \( k \) subsets +involves a degree of randomness. This may be fully excluded when +choosing \( k=n \). This particular case is referred to as leave-one-out +cross-validation (LOOCV). + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs051.html b/doc/pub/Regression/html/._Regression-bs051.html new file mode 100644 index 000000000..7542dbe4a --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs051.html @@ -0,0 +1,338 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    +
    + +
    + +

     

     

     

    + + + + +

    How to set up the cross-validation for Ridge and/or Lasso

    + +
      +
    1. Define a range of interest for the penalty parameter.
      2. +
    2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
      3. +
    3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
      4. +
    + +$$ +\begin{align*} +\hat{\beta}_{-i}(\lambda) & = ( \hat{X}_{-i, \ast}^{\top} +\hat{X}_{-i, \ast} + \lambda \hat{I}_{pp})^{-1} +\hat{X}_{-i, \ast}^{\top} \hat{y}_{-i} +\end{align*} +$$ + + +
      +
    1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
      2. +
    2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
      3. +
    3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
      4. +
    + +$$ +\begin{align*} +\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\end{align*} +$$ + + +
      +
    1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
      2. +
    + +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs052.html b/doc/pub/Regression/html/._Regression-bs052.html new file mode 100644 index 000000000..daaad1808 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs052.html @@ -0,0 +1,326 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    +
    + +
    + +

     

     

     

    + + + + +

    Predicted Residual Error Sum of Squares

    +
    +
    +

    +Another approach in the LOOCV scheme is to the use the so-called Predicted Residual Error Sum of Squares (PRESS). + +

    +We can define the optimal penalty parameter to minimize +$$ +\begin{align*} +\lambda_{\mbox{{\tiny opt}}} = \arg \min_{\lambda} \frac{1}{n} \sum_{i=1}^n [y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)]^2. +\end{align*} +$$ + +

    +The LOOCV prediction performance can be +expressed analytically in terms of the known quantities derived from +the design matrix and the parameters \( \beta \). +

    +
    + + +

    +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/._Regression-bs053.html b/doc/pub/Regression/html/._Regression-bs053.html new file mode 100644 index 000000000..ad25d53b8 --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs053.html @@ -0,0 +1,320 @@ + + + + + + + +Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis + + + + + + + + + + + + + + + + + + + + + + + +
    +
    + + Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis +
    + +
    + +
    +
    +
    + +
    + +

     

     

     

    + + + + +

    Bootstrap

    +
    +
    +

    +Bootstrapping is a nonparametric approach to statistical inference +that substitutes computation for more traditional distributional +assumptions and asymptotic results. Bootstrapping offers a number of +advantages: + +

      +
    1. The bootstrap is quite general, although there are some cases in which it fails.
      2. +
    2. Because it does not require distributional assumptions (such as normally distributed errors), the bootstrap can provide more accurate inferences when the data are not well behaved or when the sample size is small.
      3. +
    3. It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically.
      4. +
    4. It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples).
      5. +
    +
    +
    + + +

    + +

    + + +
    + + + + + + + +
    + +
    + + + + + + diff --git a/doc/pub/Regression/html/Regression-reveal.html b/doc/pub/Regression/html/Regression-reveal.html index ee2d9bcc5..63a356a89 100644 --- a/doc/pub/Regression/html/Regression-reveal.html +++ b/doc/pub/Regression/html/Regression-reveal.html @@ -1907,7 +1907,7 @@ cross-validation (LOOCV).

    1. Define a range of interest for the penalty parameter.

    2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
      3. -

    3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set as
      4. +

    4. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as

     
    $$ @@ -1919,22 +1919,20 @@ $$ $$

     
    -and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \).

    1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.

    2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
      3. -

    3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
      4. +

    4. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as

     
    $$ \begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. \end{align*} $$

     
    -The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.

    1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
      2. diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html index afaf59f99..88feb089c 100644 --- a/doc/pub/Regression/html/Regression-solarized.html +++ b/doc/pub/Regression/html/Regression-solarized.html @@ -1866,7 +1866,7 @@ cross-validation (LOOCV).
      1. Define a range of interest for the penalty parameter.
      2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
        3. -
      3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set as
        4. +
      4. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
      $$ @@ -1877,21 +1877,19 @@ $$ \end{align*} $$ -and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \).
      1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
      2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
        3. -
      3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
        4. +
      4. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
      $$ \begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. \end{align*} $$ -The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.
      1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
        2. diff --git a/doc/pub/Regression/html/Regression.html b/doc/pub/Regression/html/Regression.html index 366f94425..a53c0e587 100644 --- a/doc/pub/Regression/html/Regression.html +++ b/doc/pub/Regression/html/Regression.html @@ -1871,7 +1871,7 @@ cross-validation (LOOCV).
        1. Define a range of interest for the penalty parameter.
        2. Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
          3. -
        3. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set as
          4. +
        4. Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
        $$ @@ -1882,21 +1882,19 @@ $$ \end{align*} $$ -and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \).
        1. Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
        2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.
          3. -
        3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter
          4. +
        4. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
        $$ \begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[Y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. +\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. \end{align*} $$ -The quantity above is called the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.
        1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
          2. diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index e36087719..09033bf6f 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -2271,7 +2271,7 @@ "\n", "2. Divide the data set into training and test set comprising samples $\\{1, \\ldots, n\\} \\setminus i$ and $\\{ i \\}$, respectively.\n", "\n", - "3. Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set as" + "3. Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\\hat{\\sigma}_{-i}^2(\\lambda)$, as" ] }, { @@ -2291,13 +2291,11 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "and the corresponding estimate of the error variance $\\hat{\\sigma}_{-i}^2(\\lambda)$.\n", - "\n", "1. Evaluate the prediction performance of these models on the test set by $\\log\\{L[y_i, \\hat{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}$. Or, by the prediction error $|y_i - \\hat{X}_{i, \\ast} \\hat{\\beta}_{-i}(\\lambda)|$, the relative error, the error squared or the R2 score function.\n", "\n", "2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.\n", "\n", - "3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter" + "3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as" ] }, { @@ -2306,7 +2304,7 @@ "source": [ "$$\n", "\\begin{align*}\n", - "\\frac{1}{n} \\sum_{i = 1}^n \\log\\{L[Y_i, \\mathbf{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}.\n", + "\\frac{1}{n} \\sum_{i = 1}^n \\log\\{L[y_i, \\mathbf{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}.\n", "\\end{align*}\n", "$$" ] @@ -2315,8 +2313,6 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "The quantity above is called the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.\n", - "\n", "1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. 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z+M!j_3u29Kpt^scx*)FTb^D#vKGF-)3#CV4K`N0?@9!yX(dWFvft)g3Mwc$M~)WNyQ;9b+4!{azIi+QZdUaU*}L) zdJ1z)y2bNASD~&5E=~YR1xdweP)+8U8ooGQs~PtzrJ4%df1KA; zm_Kq;Q~q$zf~I0mY`deW#1m5+HI;he{Vz?6JlXtHQ<*1^=QNdj^64i{i#=60tf|6N z*Y9hp^wjf9nwEI#?PpC@p8E7eQ?;kQb!l4a>4_mt%RK#TPE*!1<(-;pU_0o?V+Ra? zel3Pz7 Date: Fri, 7 Sep 2018 05:45:51 +0200 Subject: [PATCH 11/12] update on html slides for regression --- .../Regression/html/._Regression-bs051.html | 14 +++++++------- .../Regression/html/Regression-reveal.html | 14 +++++++------- .../Regression/html/Regression-solarized.html | 14 +++++++------- doc/pub/Regression/html/Regression.html | 14 +++++++------- doc/pub/Regression/ipynb/Regression.ipynb | 14 +++++++------- .../ipynb/ipynb-Regression-src.tar.gz | Bin 211 -> 210 bytes .../pdf/Regression-beamer-handouts2x3.pdf | Bin 411066 -> 410824 bytes doc/pub/Regression/pdf/Regression-beamer.pdf | Bin 366730 -> 366730 bytes doc/pub/Regression/pdf/Regression-minted.pdf | Bin 405967 -> 405910 bytes doc/src/Regression/Regression.do.txt | 14 +++++++------- 10 files changed, 42 insertions(+), 42 deletions(-) diff --git a/doc/pub/Regression/html/._Regression-bs051.html b/doc/pub/Regression/html/._Regression-bs051.html index 7542dbe4a..26b39e68a 100644 --- a/doc/pub/Regression/html/._Regression-bs051.html +++ b/doc/pub/Regression/html/._Regression-bs051.html @@ -262,11 +262,11 @@ MathJax.Hub.Config({

          How to set up the cross-validation for Ridge and/or Lasso

          -
            +
            • Define a range of interest for the penalty parameter.
            • Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
            • Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
              • -
          + $$ \begin{align*} @@ -277,11 +277,11 @@ $$ $$ -
            +
            • Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
              • -
            • Repeat steps 1) to 3) such that each sample plays the role of the test set once.
              • +
            • Repeat the first three steps such that each sample plays the role of the test set once.
            • Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
              • -
          + $$ \begin{align*} @@ -290,9 +290,9 @@ $$ $$ -
            +
            • The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
              • -
          +

          diff --git a/doc/pub/Regression/html/Regression-reveal.html b/doc/pub/Regression/html/Regression-reveal.html index 63a356a89..df03e6c5f 100644 --- a/doc/pub/Regression/html/Regression-reveal.html +++ b/doc/pub/Regression/html/Regression-reveal.html @@ -1904,11 +1904,11 @@ cross-validation (LOOCV).

          How to set up the cross-validation for Ridge and/or Lasso

          -
            +

            • Define a range of interest for the penalty parameter.

            • Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.

            • Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
              • -
          +

           
          $$ \begin{align*} @@ -1920,11 +1920,11 @@ $$

           
          -

            +

            • Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
              • -

            • Repeat steps 1) to 3) such that each sample plays the role of the test set once.
              • +

            • Repeat the first three steps such that each sample plays the role of the test set once.

            • Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
              • -
          +

           
          $$ \begin{align*} @@ -1934,9 +1934,9 @@ $$

           
          -

            +

            • The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
              • -
          +
          diff --git a/doc/pub/Regression/html/Regression-solarized.html b/doc/pub/Regression/html/Regression-solarized.html index 88feb089c..575b33a94 100644 --- a/doc/pub/Regression/html/Regression-solarized.html +++ b/doc/pub/Regression/html/Regression-solarized.html @@ -1863,11 +1863,11 @@ cross-validation (LOOCV).

          How to set up the cross-validation for Ridge and/or Lasso

          -
            +
            • Define a range of interest for the penalty parameter.
            • Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
            • Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
              • -
          + $$ \begin{align*} @@ -1878,11 +1878,11 @@ $$ $$ -
            +
            • Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
              • -
            • Repeat steps 1) to 3) such that each sample plays the role of the test set once.
              • +
            • Repeat the first three steps such that each sample plays the role of the test set once.
            • Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
              • -
          + $$ \begin{align*} @@ -1891,9 +1891,9 @@ $$ $$ -
            +
            • The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
              • -
          +









          diff --git a/doc/pub/Regression/html/Regression.html b/doc/pub/Regression/html/Regression.html index a53c0e587..a6ee19e22 100644 --- a/doc/pub/Regression/html/Regression.html +++ b/doc/pub/Regression/html/Regression.html @@ -1868,11 +1868,11 @@ cross-validation (LOOCV).

          How to set up the cross-validation for Ridge and/or Lasso

          -
            +
            • Define a range of interest for the penalty parameter.
            • Divide the data set into training and test set comprising samples \( \{1, \ldots, n\} \setminus i \) and \( \{ i \} \), respectively.
            • Fit the linear regression model by means of ridge estimation for each \( \lambda \) in the grid using the training set, and the corresponding estimate of the error variance \( \hat{\sigma}_{-i}^2(\lambda) \), as
              • -
          + $$ \begin{align*} @@ -1883,11 +1883,11 @@ $$ $$ -
            +
            • Evaluate the prediction performance of these models on the test set by \( \log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\} \). Or, by the prediction error \( |y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)| \), the relative error, the error squared or the R2 score function.
              • -
            • Repeat steps 1) to 3) such that each sample plays the role of the test set once.
              • +
            • Repeat the first three steps such that each sample plays the role of the test set once.
            • Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the cross-validated log-likelihood. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as
              • -
          + $$ \begin{align*} @@ -1896,9 +1896,9 @@ $$ $$ -
            +
            • The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.
              • -
          +









          diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index 09033bf6f..f86beff59 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -2267,11 +2267,11 @@ "\n", "## How to set up the cross-validation for Ridge and/or Lasso\n", "\n", - "1. Define a range of interest for the penalty parameter.\n", + "* Define a range of interest for the penalty parameter.\n", "\n", - "2. Divide the data set into training and test set comprising samples $\\{1, \\ldots, n\\} \\setminus i$ and $\\{ i \\}$, respectively.\n", + "* Divide the data set into training and test set comprising samples $\\{1, \\ldots, n\\} \\setminus i$ and $\\{ i \\}$, respectively.\n", "\n", - "3. Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\\hat{\\sigma}_{-i}^2(\\lambda)$, as" + "* Fit the linear regression model by means of ridge estimation for each $\\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\\hat{\\sigma}_{-i}^2(\\lambda)$, as" ] }, { @@ -2291,11 +2291,11 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "1. Evaluate the prediction performance of these models on the test set by $\\log\\{L[y_i, \\hat{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}$. Or, by the prediction error $|y_i - \\hat{X}_{i, \\ast} \\hat{\\beta}_{-i}(\\lambda)|$, the relative error, the error squared or the R2 score function.\n", + "* Evaluate the prediction performance of these models on the test set by $\\log\\{L[y_i, \\hat{X}_{i, \\ast}; \\hat{\\beta}_{-i}(\\lambda), \\hat{\\sigma}_{-i}^2(\\lambda)]\\}$. Or, by the prediction error $|y_i - \\hat{X}_{i, \\ast} \\hat{\\beta}_{-i}(\\lambda)|$, the relative error, the error squared or the R2 score function.\n", "\n", - "2. Repeat steps 1) to 3) such that each sample plays the role of the test set once.\n", + "* Repeat the first three steps such that each sample plays the role of the test set once.\n", "\n", - "3. Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as" + "* Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as" ] }, { @@ -2313,7 +2313,7 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "1. The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions.\n", + "* The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. 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-1539,11 +1539,11 @@ cross-validation (LOOCV). !split ===== How to set up the cross-validation for Ridge and/or Lasso ===== -o Define a range of interest for the penalty parameter. +* Define a range of interest for the penalty parameter. -o Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively. +* Divide the data set into training and test set comprising samples $\{1, \ldots, n\} \setminus i$ and $\{ i \}$, respectively. -o Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$, as +* Fit the linear regression model by means of ridge estimation for each $\lambda$ in the grid using the training set, and the corresponding estimate of the error variance $\hat{\sigma}_{-i}^2(\lambda)$, as !bt \begin{align*} \hat{\beta}_{-i}(\lambda) & = ( \hat{X}_{-i, \ast}^{\top} @@ -1552,18 +1552,18 @@ o Fit the linear regression model by means of ridge estimation for each $\lambd \end{align*} !et -o Evaluate the prediction performance of these models on the test set by $\log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)|$, the relative error, the error squared or the R2 score function. +* Evaluate the prediction performance of these models on the test set by $\log\{L[y_i, \hat{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|y_i - \hat{X}_{i, \ast} \hat{\beta}_{-i}(\lambda)|$, the relative error, the error squared or the R2 score function. -o Repeat steps 1) to 3) such that each sample plays the role of the test set once. +* Repeat the first three steps such that each sample plays the role of the test set once. -o Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as +* Average the prediction performances of the test sets at each grid point of the penalty bias/parameter by computing the *cross-validated log-likelihood*. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data. It is defined as !bt \begin{align*} \frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \hat{\beta}_{-i}(\lambda), \hat{\sigma}_{-i}^2(\lambda)]\}. \end{align*} !et -o The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions. +* The value of the penalty parameter that maximizes the cross-validated log-likelihood is the value of choice. Or we can use the MSE or the R2 score functions. !split ===== Predicted Residual Error Sum of Squares ===== From 02d6db36970b6d35b1c77d75c50f73d7bcefe80c Mon Sep 17 00:00:00 2001 From: mhjensen Date: Tue, 11 Sep 2018 11:04:28 +0200 Subject: [PATCH 12/12] Updating project 1 --- .../gaussian/autocorrelation.eps | 4926 --------- .../gaussian/autocorrelation.png | Bin 22990 -> 0 bytes .../gaussian/blocking.eps | 1608 --- .../gaussian/blocking.png | Bin 38346 -> 0 bytes .../gaussian/data.eps | 8886 ----------------- .../gaussian/data.png | Bin 32999 -> 0 bytes .../gaussian/hist.eps | 1784 ---- .../gaussian/hist.png | Bin 26785 -> 0 bytes doc/pub/Regression/ipynb/Regression.ipynb | 3325 +++++- .../Projects/2018/Project1/Project1.do.txt | 11 +- doc/src/Regression/Regression.do.txt | 324 +- 11 files changed, 3609 insertions(+), 17255 deletions(-) delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/autocorrelation.eps delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/autocorrelation.png delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/blocking.eps delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/blocking.png delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/data.eps delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/data.png delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/hist.eps delete mode 100644 doc/Programs/ResamplingAnalysisScripts/gaussian/hist.png diff --git a/doc/Programs/ResamplingAnalysisScripts/gaussian/autocorrelation.eps b/doc/Programs/ResamplingAnalysisScripts/gaussian/autocorrelation.eps deleted file mode 100644 index d8c6a2f60..000000000 --- a/doc/Programs/ResamplingAnalysisScripts/gaussian/autocorrelation.eps +++ /dev/null @@ -1,4926 +0,0 @@ -%!PS-Adobe-3.0 EPSF-3.0 -%%Title: gaussian/autocorrelation.eps -%%Creator: matplotlib version 2.0.0, http://matplotlib.org/ -%%CreationDate: Wed Oct 4 14:08:30 2017 -%%Orientation: portrait -%%BoundingBox: 75 223 536 568 -%%EndComments -%%BeginProlog -/mpldict 9 dict def -mpldict begin -/m { moveto } bind def -/l { lineto } bind def -/r { rlineto } bind def -/c { curveto } bind def -/cl { closepath } bind def -/box { -m -1 index 0 r -0 exch r -neg 0 r -cl -} bind def -/clipbox { -box -clip -newpath -} bind def -%!PS-Adobe-3.0 Resource-Font -%%Title: DejaVu Serif -%%Copyright: Copyright (c) 2003 by Bitstream, Inc. 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matplotlib version 2.0.0, http://matplotlib.org/ -%%CreationDate: Wed Oct 4 14:08:30 2017 -%%Orientation: portrait -%%BoundingBox: 75 223 536 568 -%%EndComments -%%BeginProlog -/mpldict 9 dict def -mpldict begin -/m { moveto } bind def -/l { lineto } bind def -/r { rlineto } bind def -/c { curveto } bind def -/cl { closepath } bind def -/box { -m -1 index 0 r -0 exch r -neg 0 r -cl -} bind def -/clipbox { -box -clip -newpath -} bind def -%!PS-Adobe-3.0 Resource-Font -%%Title: DejaVu Serif -%%Copyright: Copyright (c) 2003 by Bitstream, Inc. All Rights Reserved. DejaVu changes are in public domain -%%Creator: Converted from TrueType to type 3 by PPR -25 dict begin -/_d{bind def}bind def -/_m{moveto}_d -/_l{lineto}_d -/_cl{closepath eofill}_d -/_c{curveto}_d -/_sc{7 -1 roll{setcachedevice}{pop pop pop pop pop pop}ifelse}_d -/_e{exec}_d -/FontName /DejaVuSerif def -/PaintType 0 def -/FontMatrix[.001 0 0 .001 0 0]def -/FontBBox[-770 -347 2105 1109]def -/FontType 3 def -/Encoding [ /space /B /S /V /a /c /e /g /i /k /l /n /o /r /z ] def -/FontInfo 10 dict dup begin -/FamilyName (DejaVu Serif) def -/FullName (DejaVu Serif) def -/Notice (Copyright (c) 2003 by Bitstream, Inc. All Rights Reserved. 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All Rights Reserved. Copyright (c) 2006 by Tavmjong Bah. All Rights Reserved. DejaVu changes are in public domain -%%Creator: Converted from TrueType to type 3 by PPR -25 dict begin -/_d{bind def}bind def -/_m{moveto}_d -/_l{lineto}_d -/_cl{closepath eofill}_d -/_c{curveto}_d -/_sc{7 -1 roll{setcachedevice}{pop pop pop pop pop pop}ifelse}_d -/_e{exec}_d -/FontName /DejaVuSans def -/PaintType 0 def -/FontMatrix[.001 0 0 .001 0 0]def -/FontBBox[-1021 -463 1793 1232]def -/FontType 3 def -/Encoding [ /period /zero /one /two /three /four /five /seven /e /minus ] def -/FontInfo 10 dict dup begin -/FamilyName (DejaVu Sans) def -/FullName (DejaVu Sans) def -/Notice (Copyright (c) 2003 by Bitstream, Inc. All Rights Reserved. Copyright (c) 2006 by Tavmjong Bah. All Rights Reserved. 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n3-kXIR{xKv`@g(AxAObVt)f(3!45oQh@z&XuK4Ddq3?eIcy~(0 diff --git a/doc/pub/Regression/ipynb/Regression.ipynb b/doc/pub/Regression/ipynb/Regression.ipynb index f86beff59..bdde29e8d 100644 --- a/doc/pub/Regression/ipynb/Regression.ipynb +++ b/doc/pub/Regression/ipynb/Regression.ipynb @@ -919,9 +919,7 @@ { "cell_type": "code", "execution_count": 1, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", @@ -932,9 +930,7 @@ { "cell_type": "code", "execution_count": 2, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "%matplotlib inline\n", @@ -980,9 +976,7 @@ { "cell_type": "code", "execution_count": 3, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "# Importing various packages\n", @@ -1064,9 +1058,7 @@ { "cell_type": "code", "execution_count": 4, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "# Importing various packages\n", @@ -1203,9 +1195,7 @@ { "cell_type": "code", "execution_count": 5, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", @@ -1250,9 +1240,7 @@ { "cell_type": "code", "execution_count": 6, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "import numpy as np \n", @@ -1407,9 +1395,7 @@ { "cell_type": "code", "execution_count": 7, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", @@ -1461,9 +1447,7 @@ { "cell_type": "code", "execution_count": 8, - "metadata": { - "collapsed": false - }, + "metadata": {}, "outputs": [], "source": [ "# Importing various packages\n", @@ -1767,11 +1751,802 @@ }, { "cell_type": "code", - "execution_count": 9, - "metadata": { - "collapsed": false - }, - "outputs": [], + "execution_count": 3, + "metadata": {}, + "outputs": [ + { + "data": { + "application/javascript": [ + "/* Put everything inside the global mpl namespace */\n", + "window.mpl = {};\n", + "\n", + "\n", + "mpl.get_websocket_type = function() {\n", + " if (typeof(WebSocket) !== 'undefined') {\n", + " return WebSocket;\n", + " } else if (typeof(MozWebSocket) !== 'undefined') {\n", + " return MozWebSocket;\n", + " } else {\n", + " alert('Your browser does not have WebSocket support.' +\n", + " 'Please try Chrome, Safari or Firefox ≥ 6. ' +\n", + " 'Firefox 4 and 5 are also supported but you ' +\n", + " 'have to enable WebSockets in about:config.');\n", + " };\n", + "}\n", + "\n", + "mpl.figure = function(figure_id, websocket, ondownload, parent_element) {\n", + " this.id = figure_id;\n", + "\n", + " this.ws = websocket;\n", + "\n", + " this.supports_binary = (this.ws.binaryType != undefined);\n", + "\n", + " if (!this.supports_binary) {\n", + " var warnings = document.getElementById(\"mpl-warnings\");\n", + " if (warnings) {\n", + " warnings.style.display = 'block';\n", + " warnings.textContent = (\n", + " \"This browser does not support binary websocket messages. \" +\n", + " \"Performance may be slow.\");\n", + " }\n", + " }\n", + "\n", + " this.imageObj = new Image();\n", + "\n", + " this.context = undefined;\n", + " this.message = undefined;\n", + " this.canvas = undefined;\n", + " this.rubberband_canvas = undefined;\n", + " this.rubberband_context = undefined;\n", + " this.format_dropdown = undefined;\n", + "\n", + " this.image_mode = 'full';\n", + "\n", + " this.root = $('
          ');\n", + " this._root_extra_style(this.root)\n", + " this.root.attr('style', 'display: inline-block');\n", + "\n", + " $(parent_element).append(this.root);\n", + "\n", + " this._init_header(this);\n", + " this._init_canvas(this);\n", + " this._init_toolbar(this);\n", + "\n", + " var fig = this;\n", + "\n", + " this.waiting = false;\n", + "\n", + " this.ws.onopen = function () {\n", + " fig.send_message(\"supports_binary\", {value: fig.supports_binary});\n", + " fig.send_message(\"send_image_mode\", {});\n", + " if (mpl.ratio != 1) {\n", + " fig.send_message(\"set_dpi_ratio\", {'dpi_ratio': mpl.ratio});\n", + " }\n", + " fig.send_message(\"refresh\", {});\n", + " }\n", + "\n", + " this.imageObj.onload = function() {\n", + " if (fig.image_mode == 'full') {\n", + " // Full images could contain transparency (where diff images\n", + " // almost always do), so we need to clear the canvas so that\n", + " // there is no ghosting.\n", + " fig.context.clearRect(0, 0, fig.canvas.width, fig.canvas.height);\n", + " }\n", + " fig.context.drawImage(fig.imageObj, 0, 0);\n", + " };\n", + "\n", + " this.imageObj.onunload = function() {\n", + " fig.ws.close();\n", + " }\n", + "\n", + " this.ws.onmessage = this._make_on_message_function(this);\n", + "\n", + " this.ondownload = ondownload;\n", + "}\n", + "\n", + "mpl.figure.prototype._init_header = function() {\n", + " var titlebar = $(\n", + " '
          ');\n", + " var titletext = $(\n", + " '
          ');\n", + " titlebar.append(titletext)\n", + " this.root.append(titlebar);\n", + " this.header = titletext[0];\n", + "}\n", + "\n", + "\n", + "\n", + "mpl.figure.prototype._canvas_extra_style = function(canvas_div) {\n", + "\n", + "}\n", + "\n", + "\n", + "mpl.figure.prototype._root_extra_style = function(canvas_div) {\n", + "\n", + "}\n", + "\n", + "mpl.figure.prototype._init_canvas = function() {\n", + " var fig = this;\n", + "\n", + " var canvas_div = $('
          ');\n", + "\n", + " canvas_div.attr('style', 'position: relative; clear: both; outline: 0');\n", + "\n", + " function canvas_keyboard_event(event) {\n", + " return fig.key_event(event, event['data']);\n", + " }\n", + "\n", + " canvas_div.keydown('key_press', canvas_keyboard_event);\n", + " canvas_div.keyup('key_release', canvas_keyboard_event);\n", + " this.canvas_div = canvas_div\n", + " this._canvas_extra_style(canvas_div)\n", + " this.root.append(canvas_div);\n", + "\n", + " var canvas = $('');\n", + " canvas.addClass('mpl-canvas');\n", + " canvas.attr('style', \"left: 0; top: 0; z-index: 0; outline: 0\")\n", + "\n", + " this.canvas = canvas[0];\n", + " this.context = canvas[0].getContext(\"2d\");\n", + "\n", + " var backingStore = this.context.backingStorePixelRatio ||\n", + "\tthis.context.webkitBackingStorePixelRatio ||\n", + "\tthis.context.mozBackingStorePixelRatio ||\n", + "\tthis.context.msBackingStorePixelRatio ||\n", + "\tthis.context.oBackingStorePixelRatio ||\n", + "\tthis.context.backingStorePixelRatio || 1;\n", + "\n", + " mpl.ratio = (window.devicePixelRatio || 1) / backingStore;\n", + "\n", + " var rubberband = $('');\n", + " rubberband.attr('style', \"position: absolute; left: 0; top: 0; z-index: 1;\")\n", + "\n", + " var pass_mouse_events = true;\n", + "\n", + " canvas_div.resizable({\n", + " start: function(event, ui) {\n", + " pass_mouse_events = false;\n", + " },\n", + " resize: function(event, ui) {\n", + " fig.request_resize(ui.size.width, ui.size.height);\n", + " },\n", + " stop: function(event, ui) {\n", + " pass_mouse_events = true;\n", + " fig.request_resize(ui.size.width, ui.size.height);\n", + " },\n", + " });\n", + "\n", + " function mouse_event_fn(event) {\n", + " if (pass_mouse_events)\n", + " return fig.mouse_event(event, event['data']);\n", + " }\n", + "\n", + " rubberband.mousedown('button_press', mouse_event_fn);\n", + " rubberband.mouseup('button_release', mouse_event_fn);\n", + " // Throttle sequential mouse events to 1 every 20ms.\n", + " rubberband.mousemove('motion_notify', mouse_event_fn);\n", + "\n", + " rubberband.mouseenter('figure_enter', mouse_event_fn);\n", + " rubberband.mouseleave('figure_leave', mouse_event_fn);\n", + "\n", + " canvas_div.on(\"wheel\", function (event) {\n", + " event = event.originalEvent;\n", + " event['data'] = 'scroll'\n", + " if (event.deltaY < 0) {\n", + " event.step = 1;\n", + " } else {\n", + " event.step = -1;\n", + " }\n", + " mouse_event_fn(event);\n", + " });\n", + "\n", + " canvas_div.append(canvas);\n", + " canvas_div.append(rubberband);\n", + "\n", + " this.rubberband = rubberband;\n", + " this.rubberband_canvas = rubberband[0];\n", + " this.rubberband_context = rubberband[0].getContext(\"2d\");\n", + " this.rubberband_context.strokeStyle = \"#000000\";\n", + "\n", + " this._resize_canvas = function(width, height) {\n", + " // Keep the size of the canvas, canvas container, and rubber band\n", + " // canvas in synch.\n", + " canvas_div.css('width', width)\n", + " canvas_div.css('height', height)\n", + "\n", + " canvas.attr('width', width * mpl.ratio);\n", + " canvas.attr('height', height * mpl.ratio);\n", + " canvas.attr('style', 'width: ' + width + 'px; height: ' + height + 'px;');\n", + "\n", + " rubberband.attr('width', width);\n", + " rubberband.attr('height', height);\n", + " }\n", + "\n", + " // Set the figure to an initial 600x600px, this will subsequently be updated\n", + " // upon first draw.\n", + " this._resize_canvas(600, 600);\n", + "\n", + " // Disable right mouse context menu.\n", + " $(this.rubberband_canvas).bind(\"contextmenu\",function(e){\n", + " return false;\n", + " });\n", + "\n", + " function set_focus () {\n", + " canvas.focus();\n", + " canvas_div.focus();\n", + " }\n", + "\n", + " window.setTimeout(set_focus, 100);\n", + "}\n", + "\n", + "mpl.figure.prototype._init_toolbar = function() {\n", + " var fig = this;\n", + "\n", + " var nav_element = $('
          ')\n", + " nav_element.attr('style', 'width: 100%');\n", + " this.root.append(nav_element);\n", + "\n", + " // Define a callback function for later on.\n", + " function toolbar_event(event) {\n", + " return fig.toolbar_button_onclick(event['data']);\n", + " }\n", + " function toolbar_mouse_event(event) {\n", + " return fig.toolbar_button_onmouseover(event['data']);\n", + " }\n", + "\n", + " for(var toolbar_ind in mpl.toolbar_items) {\n", + " var name = mpl.toolbar_items[toolbar_ind][0];\n", + " var tooltip = mpl.toolbar_items[toolbar_ind][1];\n", + " var image = mpl.toolbar_items[toolbar_ind][2];\n", + " var method_name = mpl.toolbar_items[toolbar_ind][3];\n", + "\n", + " if (!name) {\n", + " // put a spacer in here.\n", + " continue;\n", + " }\n", + " var button = $('');\n", + " button.click(method_name, toolbar_event);\n", + " button.mouseover(tooltip, toolbar_mouse_event);\n", + " nav_element.append(button);\n", + " }\n", + "\n", + " // Add the status bar.\n", + " var status_bar = $('');\n", + " nav_element.append(status_bar);\n", + " this.message = status_bar[0];\n", + "\n", + " // Add the close button to the window.\n", + " var buttongrp = $('
          ');\n", + " var button = $('');\n", + " button.click(function (evt) { fig.handle_close(fig, {}); } );\n", + " button.mouseover('Stop Interaction', toolbar_mouse_event);\n", + " buttongrp.append(button);\n", + " var titlebar = this.root.find($('.ui-dialog-titlebar'));\n", + " titlebar.prepend(buttongrp);\n", + "}\n", + "\n", + "mpl.figure.prototype._root_extra_style = function(el){\n", + " var fig = this\n", + " el.on(\"remove\", function(){\n", + "\tfig.close_ws(fig, {});\n", + " });\n", + "}\n", + "\n", + "mpl.figure.prototype._canvas_extra_style = function(el){\n", + " // this is important to make the div 'focusable\n", + " el.attr('tabindex', 0)\n", + " // reach out to IPython and tell the keyboard manager to turn it's self\n", + " // off when our div gets focus\n", + "\n", + " // location in version 3\n", + " if (IPython.notebook.keyboard_manager) {\n", + " IPython.notebook.keyboard_manager.register_events(el);\n", + " }\n", + " else {\n", + " // location in version 2\n", + " IPython.keyboard_manager.register_events(el);\n", + " }\n", + "\n", + "}\n", + "\n", + "mpl.figure.prototype._key_event_extra = function(event, name) {\n", + " var manager = IPython.notebook.keyboard_manager;\n", + " if (!manager)\n", + " manager = IPython.keyboard_manager;\n", + "\n", + " // Check for shift+enter\n", + " if (event.shiftKey && event.which == 13) {\n", + " this.canvas_div.blur();\n", + " event.shiftKey = false;\n", + " // Send a \"J\" for go to next cell\n", + " event.which = 74;\n", + " event.keyCode = 74;\n", + " manager.command_mode();\n", + " manager.handle_keydown(event);\n", + " }\n", + "}\n", + "\n", + "mpl.figure.prototype.handle_save = function(fig, msg) {\n", + " fig.ondownload(fig, null);\n", + "}\n", + "\n", + "\n", + "mpl.find_output_cell = function(html_output) {\n", + " // Return the cell and output element which can be found *uniquely* in the notebook.\n", + " // Note - this is a bit hacky, but it is done because the \"notebook_saving.Notebook\"\n", + " // IPython event is triggered only after the cells have been serialised, which for\n", + " // our purposes (turning an active figure into a static one), is too late.\n", + " var cells = IPython.notebook.get_cells();\n", + " var ncells = cells.length;\n", + " for (var i=0; i= 3 moved mimebundle to data attribute of output\n", + " data = data.data;\n", + " }\n", + " if (data['text/html'] == html_output) {\n", + " return [cell, data, j];\n", + " }\n", + " }\n", + " }\n", + " }\n", + "}\n", + "\n", + "// Register the function which deals with the matplotlib target/channel.\n", + "// The kernel may be null if the page has been refreshed.\n", + "if (IPython.notebook.kernel != null) {\n", + " IPython.notebook.kernel.comm_manager.register_target('matplotlib', mpl.mpl_figure_comm);\n", + "}\n" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "data": { + "text/html": [ + "" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], "source": [ "# Generate Test Data\n", "\n", @@ -1885,9 +3451,9 @@ "s_test = sigma_test*np.random.randn(N_test)\n", "\n", "#Linear\n", - "y_test=2*x_test+s_test\n", + "#y_test=2*x_test+s_test\n", "#Tenth order\n", - "#y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", + "y_test=2*x_test-10*x_test**5+15*x_test**10+s_test\n", "\n", "#Make design matrices for prediction\n", "x_plot=np.linspace(0,max_x, 200)\n", @@ -1928,11 +3494,35 @@ }, { "cell_type": "code", - "execution_count": 11, - "metadata": { - "collapsed": false - }, - "outputs": [], + "execution_count": 7, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "X_train: (37, 1)\n", + "y_train: (37,)\n", + "X_test: (13, 1)\n", + "y_test: (13,)\n", + "------------------------------------\n", + "Ordinary Least Squares\n", + "Prediction Shape: (13,)\n", + "Coefficients: \n", + " [0.48493051]\n", + "Mean squared error: 2.55\n", + "Variance score: 0.53\n", + "------------------------------------\n", + "Ridge Regression\n", + "Ridge Coefficient: [0.48378356]\n", + "Ridge Intercept: 5.38546842215786\n", + "------------------------------------\n", + "Lasso\n", + "Lasso Coefficient: [0.48405332]\n", + "Lasso Intercept: 5.380612793763472\n" + ] + } + ], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", @@ -2017,11 +3607,1630 @@ }, { "cell_type": "code", - "execution_count": 12, - "metadata": { - "collapsed": false - }, - "outputs": [], + "execution_count": 8, + "metadata": {}, + "outputs": [ + { + "data": { + "application/javascript": [ + "/* Put everything inside the global mpl namespace */\n", + "window.mpl = {};\n", + "\n", + "\n", + "mpl.get_websocket_type = function() {\n", + " if (typeof(WebSocket) !== 'undefined') {\n", + " return WebSocket;\n", + " } else if (typeof(MozWebSocket) !== 'undefined') {\n", + " return MozWebSocket;\n", + " } else {\n", + " alert('Your browser does not have WebSocket support.' +\n", + " 'Please try Chrome, Safari or Firefox ≥ 6. ' +\n", + " 'Firefox 4 and 5 are also supported but you ' +\n", + " 'have to enable WebSockets in about:config.');\n", + " };\n", + "}\n", + "\n", + "mpl.figure = function(figure_id, websocket, ondownload, parent_element) {\n", + " this.id = figure_id;\n", + "\n", + " this.ws = websocket;\n", + "\n", + " this.supports_binary = (this.ws.binaryType != undefined);\n", + "\n", + " if (!this.supports_binary) {\n", + " var warnings = document.getElementById(\"mpl-warnings\");\n", + " if (warnings) {\n", + " warnings.style.display = 'block';\n", + " warnings.textContent = (\n", + " \"This browser does not support binary websocket messages. \" +\n", + " \"Performance may be slow.\");\n", + " }\n", + " }\n", + "\n", + " this.imageObj = new Image();\n", + "\n", + " this.context = undefined;\n", + " this.message = undefined;\n", + " this.canvas = undefined;\n", + " this.rubberband_canvas = undefined;\n", + " this.rubberband_context = undefined;\n", + " this.format_dropdown = undefined;\n", + "\n", + " this.image_mode = 'full';\n", + "\n", + " this.root = $('
          ');\n", + " this._root_extra_style(this.root)\n", + " this.root.attr('style', 'display: inline-block');\n", + "\n", + " $(parent_element).append(this.root);\n", + "\n", + " this._init_header(this);\n", + " this._init_canvas(this);\n", + " this._init_toolbar(this);\n", + "\n", + " var fig = this;\n", + "\n", + " this.waiting = false;\n", + "\n", + " this.ws.onopen = function () {\n", + " fig.send_message(\"supports_binary\", {value: fig.supports_binary});\n", + " fig.send_message(\"send_image_mode\", {});\n", + " if (mpl.ratio != 1) {\n", + " fig.send_message(\"set_dpi_ratio\", {'dpi_ratio': mpl.ratio});\n", + " }\n", + " fig.send_message(\"refresh\", {});\n", + " }\n", + "\n", + " this.imageObj.onload = function() {\n", + " if (fig.image_mode == 'full') {\n", + " // Full images could contain transparency (where diff images\n", + " // almost always do), so we need to clear the canvas so that\n", + " // there is no ghosting.\n", + " fig.context.clearRect(0, 0, fig.canvas.width, fig.canvas.height);\n", + " }\n", + " fig.context.drawImage(fig.imageObj, 0, 0);\n", + " };\n", + "\n", + " this.imageObj.onunload = function() {\n", + " fig.ws.close();\n", + " }\n", + "\n", + " this.ws.onmessage = this._make_on_message_function(this);\n", + "\n", + " this.ondownload = ondownload;\n", + "}\n", + "\n", + "mpl.figure.prototype._init_header = function() {\n", + " var titlebar = $(\n", + " '
          ');\n", + " var titletext = $(\n", + " '
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          ')\n", + " nav_element.attr('style', 'width: 100%');\n", + " this.root.append(nav_element);\n", + "\n", + " // Define a callback function for later on.\n", + " function toolbar_event(event) {\n", + " return fig.toolbar_button_onclick(event['data']);\n", + " }\n", + " function toolbar_mouse_event(event) {\n", + " return fig.toolbar_button_onmouseover(event['data']);\n", + " }\n", + "\n", + " for(var toolbar_ind in mpl.toolbar_items) {\n", + " var name = mpl.toolbar_items[toolbar_ind][0];\n", + " var tooltip = mpl.toolbar_items[toolbar_ind][1];\n", + " var image = mpl.toolbar_items[toolbar_ind][2];\n", + " var method_name = mpl.toolbar_items[toolbar_ind][3];\n", + "\n", + " if (!name) {\n", + " // put a spacer in here.\n", + " continue;\n", + " }\n", + " var button = $('');\n", + " button.click(method_name, toolbar_event);\n", + " button.mouseover(tooltip, toolbar_mouse_event);\n", + " nav_element.append(button);\n", + " }\n", + "\n", + " // Add the status bar.\n", + " var status_bar = $('');\n", + " nav_element.append(status_bar);\n", + " this.message = status_bar[0];\n", + "\n", + " // Add the close button to the window.\n", + " var buttongrp = $('
          ');\n", + " var button = $('');\n", + " button.click(function (evt) { fig.handle_close(fig, {}); } );\n", + " button.mouseover('Stop Interaction', toolbar_mouse_event);\n", + " buttongrp.append(button);\n", + " var titlebar = this.root.find($('.ui-dialog-titlebar'));\n", + " titlebar.prepend(buttongrp);\n", + "}\n", + "\n", + "mpl.figure.prototype._root_extra_style = function(el){\n", + " var fig = this\n", + " el.on(\"remove\", function(){\n", + "\tfig.close_ws(fig, {});\n", + " });\n", + "}\n", + "\n", + "mpl.figure.prototype._canvas_extra_style = function(el){\n", + " // this is important to make the div 'focusable\n", + " el.attr('tabindex', 0)\n", + " // reach out to IPython and tell the keyboard manager to turn it's self\n", + " // off when our div gets focus\n", + "\n", + " // location in version 3\n", + " if (IPython.notebook.keyboard_manager) {\n", + " IPython.notebook.keyboard_manager.register_events(el);\n", + " }\n", + " else {\n", + " // location in version 2\n", + " IPython.keyboard_manager.register_events(el);\n", + " }\n", + "\n", + "}\n", + "\n", + "mpl.figure.prototype._key_event_extra = function(event, name) {\n", + " var manager = IPython.notebook.keyboard_manager;\n", + " if (!manager)\n", + " manager = IPython.keyboard_manager;\n", + "\n", + " // Check for shift+enter\n", + " if (event.shiftKey && event.which == 13) {\n", + " this.canvas_div.blur();\n", + " event.shiftKey = false;\n", + " // Send a \"J\" for go to next cell\n", + " event.which = 74;\n", + " event.keyCode = 74;\n", + " manager.command_mode();\n", + " manager.handle_keydown(event);\n", + " }\n", + "}\n", + "\n", + "mpl.figure.prototype.handle_save = function(fig, msg) {\n", + " fig.ondownload(fig, null);\n", + "}\n", + "\n", + "\n", + "mpl.find_output_cell = function(html_output) {\n", + " // Return the cell and output element which can be found *uniquely* in the notebook.\n", + " // Note - this is a bit hacky, but it is done because the \"notebook_saving.Notebook\"\n", + " // IPython event is triggered only after the cells have been serialised, which for\n", + " // our purposes (turning an active figure into a static one), is too late.\n", + " var cells = IPython.notebook.get_cells();\n", + " var ncells = cells.length;\n", + " for (var i=0; i= 3 moved mimebundle to data attribute of output\n", + " data = data.data;\n", + " }\n", + " if (data['text/html'] == html_output) {\n", + " return [cell, data, j];\n", + " }\n", + " }\n", + " }\n", + " }\n", + "}\n", + "\n", + "// Register the function which deals with the matplotlib target/channel.\n", + "// The kernel may be null if the page has been refreshed.\n", + "if (IPython.notebook.kernel != null) {\n", + " IPython.notebook.kernel.comm_manager.register_target('matplotlib', mpl.mpl_figure_comm);\n", + "}\n" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "data": { + "text/html": [ + "" + ], + "text/plain": [ + "" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "lambda = 0.0001\n", + "r2 for scikit: 0.999716\n", + "r2 for own code, not centered: 0.999716\n", + "r2 for own, centered: 0.963648\n", + "\n", + "lambda = 0.001\n", + "r2 for scikit: 0.999716\n", + "r2 for own code, not centered: 0.999716\n", + "r2 for own, centered: 0.963648\n", + "\n", + "lambda = 0.01\n", + "r2 for scikit: 0.999699\n", + "r2 for own code, not centered: 0.999699\n", + "r2 for own, centered: 0.963646\n", + "\n", + "lambda = 10\n", + "r2 for scikit: 0.875174\n", + "r2 for own code, not centered: 0.875174\n", + "r2 for own, centered: 0.695243\n", + "\n", + "lambda = 100\n", + "r2 for scikit: 0.265902\n", + "r2 for own code, not centered: 0.265902\n", + "r2 for own, centered: 0.15429\n", + "\n", + "lambda = 10000\n", + "r2 for scikit: -1.05663\n", + "r2 for own code, not centered: -1.05663\n", + "r2 for own, centered: 0.00175797\n", + "\n" + ] + } + ], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", @@ -2357,7 +5566,25 @@ ] } ], - "metadata": {}, + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.7.0" + } + }, "nbformat": 4, "nbformat_minor": 2 } diff --git a/doc/src/Projects/2018/Project1/Project1.do.txt b/doc/src/Projects/2018/Project1/Project1.do.txt index ac2a84a73..9b5e2939f 100644 --- a/doc/src/Projects/2018/Project1/Project1.do.txt +++ b/doc/src/Projects/2018/Project1/Project1.do.txt @@ -3,10 +3,19 @@ AUTHOR: "Data Analysis and Machine Learning FYS-STK3155/FYS4155":"http://www.uio DATE: May 2018 -===== Regression analysis and classification ===== +===== Regression analysis and resampling methods ===== === Introduction === +Det som var tanken, var å bruke verdier fra siden som representerer +høyde z(x,y), f.eks SRTM data som er en digital terreng modell tatt +utgangspunkt i radar data. Støy i målinger kan da unngås, samtidig som +en kan se hvordan representasjon av terreng (eller detaljerte digitale +terreng modeller) kan forenkles ved å bruke f.eks et kvadratisk +polynom i to dimensjoner. Da må studentene tilpasse dataene på formen +[1 x y x^2 x*y y^2] til høyden z(x,y) isteden, der z(x,y) blir da +målingene/estimeringene til høyden av et terrenget tatt fra f.eks +nettsiden. === Part a): The data === diff --git a/doc/src/Regression/Regression.do.txt b/doc/src/Regression/Regression.do.txt index 914eed14e..efab5bbc9 100644 --- a/doc/src/Regression/Regression.do.txt +++ b/doc/src/Regression/Regression.do.txt @@ -1582,8 +1582,14 @@ expressed analytically in terms of the known quantities derived from the design matrix and the parameters $\beta$. !eblock + !split -===== Bootstrap ===== +===== Resampling methods: Jackknife ===== + + + +!split +===== Resampling methods: Bootstrap ===== !bblock Bootstrapping is a nonparametric approach to statistical inference that substitutes computation for more traditional distributional @@ -1594,3 +1600,319 @@ o Because it does not require distributional assumptions (such as normally distr o It is possible to apply the bootstrap to statistics with sampling distributions that are difficult to derive, even asymptotically. o It is relatively simple to apply the bootstrap to complex data-collection plans (such as stratified and clustered samples). !eblock + + + + + +\textcite{efron_jackknife_1987} +explains that resampling methods +'scramble' the observations which describe the parameter +$\vec{\theta}$ in some way. The purpose of scrambling the data is to +obtain useful estimates of the probability distribution of the +estimator $\widehat{\vec{\theta}}$. This is often done if deriving the +distribution of $\widehat{\vec{\theta}}$ by analytical means is +impossible or inconvenient. The significance of this is reflected in +that Efron's original paper has more than 16 000 citations by early +spring 2018. Although these citations have come from all the sciences, +a lot of work has been done by statisticians and mathematicians. On +'Web of Science', a search for the topic \textit{bootstrap} returns +nearly 6 500 papers in journals on statistics and probability theory +alone. A similar search on 'Scopus' returns more than 7 000 papers in +the field of mathematics. In addition, there has been a renaissance in +the study of resampling methods in the 21st century, with more than 6 +000 papers in just 18 years in mathematics. Part of the reason is +that, even though the ideas which will be presented here seem innocent +and simple, the required mathematics is deep. In fact, there exists +conjectures too deep for present mathematics +\parencite{efron_jackknife_1987}. This will become apparent to us +because often we will only give intuitive explanations for why the +methods are valid. We could have done substantially more with measure +theory in place, but this is not economical in light of the present +results. However, using our introduction to real analysis, it is +possible to state and understand a few results in some detail. See for +example theorem \ref{thm:independent_strap_frechet}.\\ \\ Two famous +resampling methods are \textit{the independent bootstrap} and +\textit{the jackknife}. It would make most sense to start by +discussing the independent bootstrap, because the jackknife method +follows by making a linearization of the parameters of interest +\parencite{efron_jackknife_1987,efron_bootstrap_1979}. As such, the +jackknife is a special case of the independent bootstrap +\parencite{efron_jackknife_1987}. Still, the jackknife was made +popular prior to the independent bootstrap. And as the popularity of +the independent bootstrap soared, new variants, such as \textit{the + dependent bootstrap}\footnote{We will only consider non-parametric + bootstrap, but there exists a popular variant called parametric + bootstrap, which assumes knowledge of the probability distribution + of the observations} or stationary bootstrap were introduced, see +for example \textcite{politis_stationary_1994} or +\textcite{politis_automatic_2006}. There also exists textbooks on the +subject. The mathematical complexity of the latter variants is also +greater, and consequently it is pedagogical to introduce the methods +in this order.\\ \\ The Jackknife and independent bootstrap work for +independent, identically distributed random variables +\parencite{efron_jackknife_1987}. If these conditions are not +satisfied, the methods will fail. This is important for the results of +the thesis, because here the variables are dependent, and we will need +the dependent bootstrap. Yet, it should be said that if the data are +independent, identically distributed, and we only want to estimate the +variance of $\overline{X}$ (which often is the case), then there is no +need for bootstrapping. For if $X_1,X_2,\cdots,X_n$ are independent +identically distributed and come from an unknown distribution $F$, +then the standard error is easily computed by taking the square root +of the following expression: \eqref{eq:sigma2n} +\[ +V(\overline{X}) \stackrel{\eqref{eq:sigma2n} }{=} \frac{\sigma^2}{n} \approx \frac{\widehat{\sigma}^2}{n} \stackrel{ \eqref{eq:defn_var} }{=} \frac{1}{n^2} \sum_{i=1}^{n} (X_i - \overline{X})^2. +\] +And consequently, these methods are most useful when the data are dependent or the estimator is not the sample mean. +\subsubsection{The Jackknife} +The Jackknife works by making many replicas of the estimator $\widehat{\vec{\theta}}$. Since the jackknife is a resampling method, we explained that this happens by scrambling the data in some way. When using the jackknife, this is done by systematically leaving out one observation from the vector of observed values $\vec{X} = (X_1,X_2,\cdots,X_n)$ \parencite{tukey_abstracts_1958}. Let $\vec{X}_i$ denote the vector +\[ +\vec{X}_i = (X_1,X_2,\cdots,X_{i-1},X_{i+1},\cdots,X_n), +\] +which equals the vector $\vec{X}$ with the exception that observation number $i$ is left out. Using this notation, define $\widehat{\vec{\theta}}_i$ to be the estimator $\widehat{\vec{\theta}}$ computed using $\vec{X}_i$. According to \textcite{efron_jackknife_1987}, to get an estimate for the bias and standard error of $\widehat{\vec{\theta}}$, use the following estimators for each component of $\widehat{\vec{\theta}}$: +\[ +\widehat{\mathrm{Bias}}(\widehat \theta,\theta) = (n-1)\left( - \widehat{\theta} + \frac{1}{n}\sum_{i=1}^{n} \widehat \theta_i \right) \qquad \text{and} \qquad \widehat{\sigma}^2_{\widehat{\theta} } = \frac{n-1}{n}\sum_{i=1}^{n}( \widehat{\theta}_i - \frac{1}{n}\sum_{j=1}^{n}\widehat \theta_j )^2. +\] +Sample code is contained in figure \ref{fig:jack_code} and is available for download from the url \texttt{github.com/computative/resample} . +\begin{figure} +\begin{center} +\begin{lstlisting}[language=python] + # jack.py + + def jack(data, stat): + n = len(data);t = zeros(n); inds = arange(n); t0 = time() + # 'jackknifing' by leaving out an observation for each i + for i in range(n): + t[i] = stat(delete(data,i) ) + + return t + + # define a function which returns your chosen estimator theta-hat + def stat(data): + theta-hat = mean(data) + return theta-hat + + # boot returns the bootstrap sample + t = jack(X, stat) + +\end{lstlisting} \caption{The code follows the algorithm outlined in the text. Consider first the function \texttt{jack()}. In the \texttt{for}-loop, this function repeatedly estimates the function called \texttt{statistic()} under the resampled data by systematically leaving out one observation from the data. The function \texttt{stat()} is passed as an argument to \texttt{jack()}. The array \texttt{t} is eventually returned, which contains all the estimates $\widehat{\vec{\theta}}$, and can be plotted or analysed in other ways, such as by calling \texttt{std(t)} from \texttt{numpy} to estimate the standard error of $\widehat{\vec{\theta}}$. The function \texttt{std(t)} is just the estimator $\widehat{\sigma}^2$.}\label{fig:jack_code} +\end{center} +\end{figure} +\subsubsection{The independent bootstrap}\label{sec:ind_boot} +Many authors explain the bootstrap in an algebraic manner, similar to the way it was originally proposed by \textcite{efron_jackknife_1987}. In this thesis, I give a different view of the method; building our intuition upon \textcite{parr_bootstrap_1985}: Since $\widehat{\vec{\theta}} = \widehat{\vec{\theta}}(\vec{X})$ is a function of random variables, $\widehat{\vec{\theta}}$ itself must be a random variable. Thus it has a pdf, call this function $p(\vec{t})$. The aim of the bootstrap is to estimate $p(\vec{t})$ by the relative frequency of $\widehat{\vec{\theta}}$. You can think of this as using a histogram in the place of $p(\vec{t})$. If the relative frequency closely resembles $p(\vec{t})$, then using numerics, it is straight forward to estimate all the interesting parameters of $p(\vec{t})$ using point estimators. +%In the case that $\widehat{\vec{\theta}}$ has exactly one component, use for example the sample mean or the sample variance, as previously defined. In the case that $\widehat{\vec{\theta}}$ has more than one component, and the components are independent, use the same estimator on each component separately. +If the probability density function of $X_i$, $p(x)$, had been known, then it would have been straight forward to do this by: (1) Drawing lots of numbers from $p(x)$, suppose we call one such set of numbers $(X_1^*, X_2^*, \cdots, X_n^*)$. (2) Then using these numbers, we could compute a replica of $\widehat{\vec{\theta}}$ called $\widehat{\vec{\theta}}^*$. By repeated use of (1) and (2), many estimates of $\widehat{\vec{\theta}}$ could have been obtained. The idea is to use the relative frequency of $\widehat{\vec{\theta}}^*$ (think of a histogram again) as an estimate of $p(\vec{t})$.\\ +\\ +But unless there is enough information available about the process that generated $X_1,X_2,\cdots,X_n$, $p(x)$ is in general unknown. Therefore, Bradley \textcite{efron_bootstrap_1979} asked the natural question: What if we replace $p(x)$ by the relative frequency of the observation $X_i$; if we draw observations in accordance with the relative frequency of the observations, will we obtain the same result in some asymptotic sense? The answer is yes. The paper of \textcite{efron_bootstrap_1979} gave little in the way of general theory \parencite{efron_bootstrap_1979}. In contrast, he gave computational examples showing that in many cases, it was reasonable. It is standard to make a tweak that also speeds up computation: Instead of generating the histogram for the relative frequency of the observation $X_i$, just draw the values $(X_1^*,X_2^*,\cdots,X_n^*)$ with replacement from the vector $\vec{X}$. The end result is exactly the same, as bootstrapping is a demonstration of. \\ +\\ +See figure \ref{fig:distr_boot} for a pictorial explanation and figure \ref{fig:rithm_boot} for a concise summary and the final algorithm. Sample code for python is contained in figure \ref{fig:boot_code} and is available for download at \texttt{github.com/computative/resample} . +\begin{figure} +\begin{center} +\begin{lstlisting}[language=python] + # boot.py + + def boot(data, statistic, R): + t = zeros(R); n = len(data); inds = arange(n); t0 = time() + + # non-parametric bootstrap + for i in range(R): + t[i] = statistic(data[randint(0,n,n)]) + + return t + + # define a function which returns your chosen estimator theta-hat + def stat(data): + theta-hat = mean(data) + return theta-hat + + t = boot(X, stat, 2**9) +\end{lstlisting} \caption{The code follows the algorithm \ref{fig:rithm_boot}. Consider first the function \texttt{boot()}. In the \texttt{for}-loop, this function repeatedly estimates the function called \texttt{statistic()} under the resampled data \texttt{data[randint(0,n,n)]}. The function \texttt{statistic()} is passed as an argument to \texttt{boot()}. The array \texttt{t} is eventually returned, which contains all the estimates $\widehat{\vec{\theta}}$, and can be plotted or analysed in other ways, such as by calling \texttt{std(t)} from \texttt{numpy} to estimate the standard error of $\widehat{\vec{\theta}}$. The function \texttt{std(t)} is just the estimator $\widehat{\sigma}^2$.}\label{fig:boot_code} +\end{center} +\end{figure} +\begin{figure}[!htbp] +\center +Philosophy of the independent bootstrap \\ +\hspace{100px} +\hbox{\hspace{-1.5cm} +\input{/home/marius/Dokumenter/master/figures/bootdistr.tex} +} +\vspace{0px} +\caption{Suppose you wanted to estimate the probability distribution $p(t)$ of the estimator $\widehat{\theta}$. The obvious way to do this is to compute many replicas of some $\widehat \theta = \widehat \theta(\vec{X})$ by drawing lots of numbers from some pdf $p(x)$, such as the one of fig \ref{fig:distr_boot}(a). Then by plotting the histogram of the replicas $\widehat \theta$, you obtain the estimate of the pdf $p(t)$, namely (b). Bradley Efron asked the question: What happens if we replace the exact distribution $p(x)$ by an estimate, namely the histogram of the relative frequency (b)? It turns out that in an asymptotical sense, we still obtain the same estimate, (c). This is the philosophy of the independent bootstrap. After the estimate (c) has been obtained we can estimate any statistic thereof, for example $V(\widehat \theta)$ using $\widehat{S^2}$.}\label{fig:distr_boot} +\end{figure} +\begin{figure}[!htbp] +\center +Flow chart of the independent bootstrap \\ +\hspace{100px} +%\tikzstyle{decision} = [diamond, draw, text width=8em, text badly centered, node distance=3cm, inner sep=0pt, aspect=3] +%\tikzstyle{block} = [rectangle, draw, text width=8em, text centered, minimum height=2em] +%\tikzstyle{terminal} = [rectangle, draw, text width=8em, text centered, rounded corners, minimum height=2em] +%\tikzstyle{line} = [draw, -latex] +%\usetikzlibrary{shapes,arrows} + +\begin{tikzpicture}[node distance = 1cm, auto] + \node [terminal] (init) {\footnotesize Estimate $p(t)$}; + \node [block, below of=init, node distance=1.2cm] (index) {\footnotesize Set $i=1$}; + \node [block, below of=index, node distance=1.8cm] (pick) {\footnotesize Draw with replacement $n$ numbers from $\vec{X}$ called $\vec{X}^*$}; + \node [block, below of=pick, node distance=1.8cm] (estimator) {\footnotesize Compute $\widehat \theta^* = \widehat \theta (\vec{x}^*)$}; + \node [block, right of=estimator, node distance=4cm] (iterate) {\footnotesize Set $i = i+1$}; + \node [decision, below of=estimator, node distance=1.5cm] (decide) {\footnotesize Is $i\leq k$?}; + \node [terminal, below of=decide, node distance=1.65cm] (final) {\footnotesize Return histogram of $\widehat{\theta}^*$}; + % Draw edges + \path [line] (init) -- (index); + \path [line] (index) -- (pick); + \path [line] (pick) -- (estimator); + \path [line] (estimator) -- (decide); + \path [line] (decide) -- (final)node [near start] {no}; + \path [line] (decide) -| node [near start] {yes} (iterate); + \path [line] (iterate) |- (pick); +\end{tikzpicture} +\vspace{0px} +\caption{The independent bootstrap works like this: (1) Draw with replacement $n$ numbers for the observed variables $\vec{x} = (x_1,x_2,\cdots,x_n)$. (2) Define a vector $\vec{x}^*$ containing the values which were drawn from $\vec{x}$. (3) Using the vector $\vec{x}^*$ compute $\widehat{\theta}^*$ by evaluating $\widehat \theta$ under the observations $\vec{x}^*$. Repeat this process $k$ times. When you are done, you can draw a histogram of the relative frequency of $\widehat \theta^*$. This is your estimate of the probability distribution $p(t)$. Using this probability distribution you can estimate any statistic thereof. In principle you never draw the histogram of the relative frequency of $\widehat{\theta}^*$. Instead you use the estimators corresponding to the statistic of interest. For example, if you are interested in estimating the variance of $\widehat \theta$, apply the esimator $\widehat \sigma^2$ to the values $\widehat \theta ^*$.}\label{fig:rithm_boot} +\end{figure}\\ +\\ +As we explained, much theoretical work has gone into making the mathematics of the various types of bootstrap rigorous. However, some of the most important results were published relatively soon after. Only two years after \textcite{efron_bootstrap_1979}, proof in the case that $(\theta,\widehat{\theta}) = (\E{X}, \overline{X})$ came from \textcite{bickel_asymptotic_1981}. Despite the importance of these quantities, we will not be fully satisfied with this result. This is because bootstrapping is most useful when $\widehat{\vec{\theta}}$ is not the sample mean, as we discussed. Before formulating the theorem, some convenient notation is introduced. In ordinary real analysis, we are often interested in convergence of sequences of real numbers $\{a_n\}_{n=1}^\infty$ with limit $a \in \mathbb{R}$. In the case that the sequence $\{A_n\}_{n=1}^\infty$ is comprised of random variables, it is clearly nonsense to say that the limit is a real number. One could argue that it makes more sense to say that the limit is a random variable. However, due to amount of additional structure that comes with measure theory, there are many interesting ways of defining such limits. And \textcite{van_der_vaart_asymptotic_1998} covers some of the most useful modes of convergence: +\begin{itemize} +\item If $F_{A_n}$ is the cdf of $A_n$ and $F_A$ is the cdf of $A$, and $F_{A_n}$ converges pointwise to $F_{A}$ then we say that there is \defn{convergence in distribution}\index{convergence!{in distribution}} and denoted by $A_n \stackrel{\diff}{\to} A$.\index[s]{Ed@$\stackrel{\mathrm{d}}{\to} $ convergence in distribution} +\item If for every $\varepsilon >0$, the sequence of real numbers $a_n = P(\| A_n - A\| > \varepsilon)$ converge to zero, then we say that there is \defn{convergence in probability}\index{convergence!{in probability}}, denoted by $A_n \stackrel{P}{\to} A$\index[s]{EdP@$\stackrel{P}{\to} $ convergence in probability}. +\item If for every $\varepsilon>0$, there is an $N \in \mathbb{N}$ such that for all $n \geq N$ we have $\|A_n - A\|< \varepsilon$ with probability 1, then we say that there is \defn{almost sure convergence}\index{convergence!{almost surely}} denoted by $A_n \stackrel{\mathrm{a.s.}}{\to} A$.\index[s]{EAs@$\stackrel{\mathrm{a.s.}}{\to} $ almost sure convergence} +\end{itemize} +More generally, we will say that an event $B$ happens \defn{almost surely}\index{almost surely} if $P(B) = 1$ \parencite{oksendal_stochastic_2014,mcdonald_course_2012}. On that note we are ready for the theorem due to \textcite{bickel_asymptotic_1981}: +\begin{theorem}[Bickel-Freedman theorem] +Assume $X_1,X_2,\cdots$ are independent identically distributed random variables with variance $\sigma^2$, and assume $X \sim \mathrm{N}(0,\sigma^2)$, then given $X_1,X_2,\cdots,X_n$, +\begin{itemize} +\item $n^{1/2}(\overline{X}^* - \overline{X}) \stackrel{\diff}{\to} X $ as $n \to \infty$ almost surely. +\item $\widehat \sigma^* \stackrel{P}{\to} \sigma$ as $n \to \infty$ almost surely. +\end{itemize} +\end{theorem} +\noindent For more a general $\vec{\widehat \theta}$ the problem is harder because the assumptions are weaker. \textcite{parr_bootstrap_1985} established the result under relaxed conditions on $\widehat{\vec{\theta}}$. We will consider the case that $\vec{\widehat \theta} = {\widehat \theta}$ is one-dimensional and require the strong type of differentiability which was defined in section \ref{sec:real}; namely Fréchet differentiability. All estimators encountered thus far were expressed in terms of $\vec{X}$. However, \textcite{huber_robust_2009} explains that many estimators with practical use can also be expressed in terms of their cumulative distribution function. This area of estimation contains \defn{M-estimation}\index{estimate!estimator!M@$M$}, and is not economical to discuss in full generality. However, using our experience with real analysis, it is possible to tackle the theory relevant to maximum likelihood estimation on compact intervals $[a,b]$. Going back to the way justification was given for maximum likelihood estimators in section \ref{sec:Fisherians} it is clear that the maximum likelihood estimator is +\begin{equation} +\widehat{{\theta}} = \mathrm{argmin}_{{\theta}} \sum_{i=1}^{n} - \log(f(x_i; {\theta})) = \sum_{i=1}^{n} - \log\left( \frac{\partial F}{\partial x}(x_i; {\theta} )\right) = \widehat{{\theta}}(F). \label{eq:robust} +\end{equation} +This shows that the Maximum likelihood estimators are $M$-estimators. In fact $M$ estimation was motivated by maximum likelihood and is a generalization thereof \parencite{huber_robust_2009}. These estimators suffices for this thesis, since the maximum likelihood estimators are MVUE according to section \ref{sec:Fisherians} and is the best we can do with our present tools. We introduce the Fréchet derivative for this space of estimators. Let $H([a,b])$ be the largest open set of cdfs on $[a,b]$. It is well known that $[a,b]$ is compact \parencite{munkres_topology_2000}. Therefore, example \ref{ex:linearbounded} shows that $H([a,b])$ is contained in the Banach space $C([a,b],\mathbb{R})$ under the sup-norm, $\| \cdot \|_\infty$ from example \ref{ex:sup-norm}. Example \ref{ex:linearbounded} also shows that by letting $\psi$ be a continuous function on $X \equiv [a,b]$ such that (1) $\int_X \psi \ \diff F = 0$, (2) $\int_X \psi^2 \ \diff F < \infty$ and (3) $F \in H$ be the cdf of $X_i$ with pdf $f$, then +\[ +(A(F))({\theta}) = \Big(\int_X \psi \ \diff F\Big)({\theta}) = \int_X \psi(x;{\theta}) f(x;{\theta}) \ \diff x +\] +is a bounded linear functional on $H$. So if $T$ is any $H([a,b]) \to \mathbb{R}$ function, we say that $T$ is Fréchet differentiable at $F$ if +\[ +\lim_{C \to \textit{0}} \frac{|T(F) - T(F+C) - (A(C))(T)| }{\|C\|_\infty} = 0 \qquad\qquad \text{\parencite{parr_bootstrap_1985}}. +\] +You may wonder what the function $\psi$ is. According to \textcite{huber_robust_2009}, it defines the type of estimation equation used. In the case of maximum likelihood estimation with $F$ is twice continuously differentiable, +\[ +\psi(x;T) = - \frac{\partial \log f}{\partial {\theta}}(x;T) = - \Big(\frac{1}{\partial F / \partial x} \frac{\partial^2 F}{\partial x \partial \theta} \Big)(x;T) \quad \text{\parencite{huber_robust_2009}}. +\] +Using this definition, we are finally ready for a theorem due to \textcite{parr_bootstrap_1985} which explains consistency of the bootstrap estimator more generally. It says that if the Fréchet derivative of $\widehat{{\theta}}$ exists at $F \in H$, then the bootstrap estimator is consistent: +\begin{theorem}[Parr theorem]\label{thm:independent_strap_frechet} +Assume $X_1,X_2,\cdots,X_n$ are independent identically distributed with support $X=[a,b]$, $\widehat{{\theta}}$ is Fréchet differentiable at $F$, $\sigma^2 = V(n^{1/2}[\widehat{{\theta}}^* - \widehat{{\theta}} ])$ and $Z \sim \mathrm{N}(0,1)$, then given $X_1,X_2,\cdots,X_n$ we have $n^{1/2}(\widehat{{\theta}}^* - \widehat{{\theta}} ) \stackrel{P}{\to} \sigma Z$ as $n \to \infty$. +\end{theorem} +\subsubsection{The dependent bootstrap} +In the case that the variables $X_1,X_2,\cdots,X_n$ are dependent, the above procedure breaks down. At least one of the problems are: Independent bootstrapping assumes that $X_1,X_2,\cdots,X_n$ all come from the same marginal probability distribution, $p(x) = p(x_i)$ for all $1 \leq i \leq n$. This is clear because variables are independent, so it follows from the definitions of independence and conditional probability that $p(x_i) = p(x_i|x_j \neq x_i)$. Also the second definition given of independence says that the joint probability distribution of all the $X_i$ is $p(x_1,x_2,\cdots, x_n) = p(x)^n$ by the product rule. Since this does not carry over in the case that the variables are dependent, there are at least two problems: +\begin{enumerate} +\item Since there is dependence between the observation, observing $x_j$ reveal information about $x_i$ for some $1 \leq i \leq n$. The extra information supplied by observing $x_j$ means that $p(x_i) \neq p(x_i|x_j \neq x_i)$ as explained above. That means if we treated the variables as independent (i.e. let $p(x_i) = p(x_i|x_j \neq x_i)$), then all estimators which are sensitive to the difference between $p(x_i)$ and $p(x_i|x_j \neq x_i)$ estimate wrong systematically. Consider for example the case that $\{X_i\}$, is a time series where we wrongly assumed that the $X_i$ were independent, then the autocovariance estimator $\widehat{\gamma}(1) = \widehat{\cov}(X_i,X_{i+1}) = \widehat{0} = 0$ would estimate zero. In independent bootstrapping is induced by pick single observations with replacement as we explained. We explained above that this is equivalent to constructing the histogram for $X_i$ and drawing observations from the histogram. But by hypothesis, that is precisely the estimate of the marginal distribution $p(x)$. +\item If we do not assume that $p(x_i) = p(x_i|x_j \neq x_i)$, but instead assume that $p(x_i) \neq p(x_i|x_j \neq x_i)$, then the natural way to proceed is to treat the whole set $\{X_i\}$ as one observation $\vec{X} = (X_1,X_2,\cdots,x_n)$ and give it a multivariate probability distribution with covariance $\Sigma$ which encodes the dependence. But there is at least one problem, in this case we only have one observation, namely $\vec{X}$! This problem is however fixable in the case that the dependent data are a stationary time series, as explained by \textcite{politis_stationary_1994} and adapted here. +\end{enumerate} +That is to assume that the dependence in the data set can be related to the linear dependence, which is the covariance. In the case that the data are a stationary time series and the autocovariance $\gamma(h) \to 0$ as $h \to 0$. Since the autocovariance measures linear dependence, we assume that when the linear dependence is zero, the variables can be treated as independent. If $H \in \mathbb{N}$ a number such that $\gamma(h) \approx 0$ for all $h \geq H$, then we treat $X_{i}$ and $X_{i+h}$ as independent for all $h \geq H$ and all $1 \leq i \leq n-h$. If we split the observation and make the following definitions: +\[ +\vec{X} = (\underbrace{X_1,X_2,\cdots ,X_{H}}_{\equiv \vec{X}'_1},X_{H+1},\cdots,X_{2H},\underbrace{X_{2H+1},\cdots,X_{3H}}_{\equiv \vec{X}'_3},X_{3H+1},\cdots,X_{n} ). +\] +Then the components of $\vec{X}_1'$ and $\vec{X}_3'$ are almost independent, moreover section \ref{sec:time_series} explains that a stationary time series is identically distributed. So to these vectors can be treated as independent and identically distributed, so we can use the usual machinery of independent bootstrapping. But since the size of each vector $\vec{X}'_i$ is larger than one, we require to draw less than $n$ such vectors with replacement to compute $\widehat \theta^*$. Instead we concatenate the drawn observations $\vec{X}'_i$ into one long vector $\vec{X}^*$ of length $n$, and discard any observations left over. This procedure has the advantage that the vector $\vec{X}^*$ has the same autocovariance as $\vec{X}$, so it can be used to estimate $\gamma$ and consequently $V(\widehat \theta)$ according to chapter \ref{sec:time_series}. See figure \ref{fig:tsboot_code} for sample code which follows the description given above. The code is also available for download from \texttt{github.com/computative/resample} . +%Les mer på https://epubs.siam.org/doi/abs/10.1137/1.9781611970319.ch1 +\begin{figure} +\begin{center} +\begin{lstlisting}[language=python] + # tsboot.py + + def tsboot(data,statistic,R,l): + t = zeros(R); n = len(data); k = ceil(float(n)/l); + inds = arange(n); t0 = time() + + # time series bootstrap + for i in range(R): + # construct bootstrap sample from + # k chunks of data. The chunksize is l + _data = concatenate([data[j:j+l] for j in randint(0,n-l,k)])[0:n]; + t[i] = statistic(_data) + + return t + + # define a function which returns your chosen estimator theta-hat + def stat(data): + theta-hat = mean(data) + return theta-hat + + t = tsboot(X, stat, 2**12, 2**10) + + +\end{lstlisting} \caption{The code follows the algorithm outlined in the text. Consider first the function \texttt{tsboot()}. In the \texttt{for}-loop, this function repeatedly estimates the function called \texttt{statistic()} under the resampled data by concatenating chunks of the data which are uncorrelated by \texttt{concatenate([data[j:j+l] for j in randint(0,n-l,k)])[0:n]}. The function \texttt{statistic()} is passed as an argument to \texttt{tsboot()}. The array \texttt{t} is eventually returned, which contains all the estimates $\widehat{\vec{\theta}}$, and can be plotted or analysed in other ways, such as by calling \texttt{std(t)} from \texttt{numpy} to estimate the standard error of $\widehat{\vec{\theta}}$. The function \texttt{std(t)} is just the estimator $\widehat{\sigma}^2$.}\label{fig:tsboot_code} +\end{center} +\end{figure} +%\subsubsection{Cross validation} +\subsection{Manual blocking method} +The manual blocking method was made popular by Flyvbjerg and Pedersen (1989) and has become one of the standard ways to estimate $V(\widehat{\theta})$ for exactly one $\widehat{\theta}$, namely $\widehat{\theta} = \overline{X}$. Their paper has become a citation classic (cited more than 1000 times according to Google scholar). But the proof given by Flyvbjerg and Pedersen (1989) is not rigorous. I corresponded with Associate professor Flyvbjerg in the summer of 2017 and it became clear to me that a paper on the blocking method with rigorous modern mathematics and numerics could be useful. Thus the missing proof of the manual blocking method is one of the main results of this thesis, and contained in the results, see theorem \ref{thm:blocking}. But this begs the question: What can then be said about the blocking method in the methods of this thesis? I think it is appropriate to give the idea of the blocking method, analogous to the treatment given for bootstrapping, as well as an overview of the mathematics of \textcite{flyvbjerg_error_1989}. +\begin{figure}[!htbp] +\hbox{\hspace{-1.7cm}\input{/home/marius/Dokumenter/master/figures/block_cov.tex} } + %\vspace{-40px} +\caption{In the left panel: A typical autocovariance function $\gamma(h)$ plotted against $h$. The dashed lines illustrate how $\gamma_{k}(h) \leq \gamma_{k-1}(h)$ for all $1 \leq k \leq d-1$ in the case of applying blocking transformation. In fact, the results show that $\gamma_k$ converges to the zero-function of $\mathbb{N}$.\\ +\\ +In the right panel: It is a consequence of the behavior $\gamma_{k}(h) \leq \gamma_{k-1}(h)$ that for all $1 \leq k \leq d-1$ we have $\sigma_{k}^2/n_{k} \geq \sigma_{k-1}^2/n_{k-1}$ for all $1 \leq k \leq d-1$. The solid line is the "unestimated estimate", ${\sigma}^2_k/n_k$. That is, the estimate of $V(\overline{X})$ where the only source of error is the truncation error $e_k$. It is clear that this estimate initially is too optimistic (too small), and then as we apply blocking transformations, the estimate rises up to the correct value. By equation \eqref{eq:stationary}, it is clear that this means that ${\sigma}^2_k/n_k$ becomes constant, as we clearly see in the plot. \\ +\\ +The dashed line is the estimat\textit{or} $\widehat{\sigma}^2_k/n_k$ which contains an extra error since ${\sigma}^2_k$ itself is unknown and has to be estimated. As $k \to d$ (here $d=20$), we have $n_k\to 1$, and so the \textit{standard error} of $V(\widehat \sigma_k^2/n_k)$ becomes very large according to equation \eqref{eq:departure}. We see this in the figure, because the estimate starts to depart from the value ${\sigma}^2_k/n_k$. This is the reason it is important to stop the algorithm at the right time. This can done with a plot, like the one above, or using the automated scheme proposed here which takes care of everything for you. \\ +\\ +In this case it is relatively clear from the plot of $\widehat \sigma_k^2/n_k$ for which $k$ we have that $\sigma_k^2/n_k$ becomes constant. However if the amount of data is smaller, i.e. $n_k$ is small, this becomes more difficult to determine, because then the estimator $\widehat{\sigma}^2_k/n_k$ diverges sooner. Perhaps even before the graph of $\sigma_k^2/n_k$ becomes constant.} \label{fig:block_cov} +\end{figure}\\ +\\ +Assume $n = 2^d$ for some integer $d>1$ and $X_1,X_2,\cdots, X_n$ is a stationary time series to begin with. This guarantees that $\gamma(h)$ exists according to section \ref{sec:time_series}. Moreover, assume that the time series is asymptotically uncorrelated. We switch to vector notation by arranging $X_1,X_2,\cdots,X_n$ in an $n$-tuple. Define: +\begin{align*} +\vec{X} = (X_1,X_2,\cdots,X_n). +\end{align*} +The strength of the blocking method is evident when the number of observations, $n$ is large. For large $n$, the complexity of dependent bootstrapping scales poorly, but the blocking method does not, moreover, it becomes more accurate the larger $n$ is, as the results will show. As such the method is relatively ad-hoc. We now define blocking transformations. The idea is to take the mean of subsequent pair of elements from $\vec{X}$ and form a new vector $\vec{X}_1$. Continuing in the same way by taking the mean of subsequent pairs of elements of $\vec{X}_1$ we obtain $\vec{X}_2$, and so on. In accordance with \textcite{flyvbjerg_error_1989} define $\vec{X}_i$ recursively by: +\begin{align} +(\vec{X}_0)_k &\equiv (\vec{X})_k \nonumber \\ +(\vec{X}_{i+1})_k &\equiv \frac{1}{2}\Big( (\vec{X}_i)_{2k-1} + (\vec{X}_i)_{2k} \Big) \qquad \text{for all} \qquad 1 \leq i \leq d-1 \label{eq:blocking} +\end{align} +In this way, we say that $\vec{X}_k$ is subject to $k$ \defn{blocking transformations}\index{blocking transformation}. We now have $d$ vectors $\vec{X}_0, \vec{X}_1,\cdots,\vec X_{d-1}$ containing the subsequent averages of observations. It turns out that if the components of $\vec{X}$ is a stationary time series, then the components of $\vec{X}_i$ is a stationary time series for all $0 \leq i \leq d-1$ (see lemma \ref{lem:init} or \textcite{flyvbjerg_error_1989} ). And so we can compute the autocovariance, the variance, sample mean, and number of observations for each $i$. Let $\gamma_i, \sigma_i^2, \overline{X}_i$ denote the autocovariance, variance and average of the elements of $\vec{X}_i$ and let $n_i$ be the number of elements of $\vec{X}_i$. It follows by induction that $n_i = n/2^i$. Using the definition of the blocking transformation and the distributive property of the covariance, it is clear that since $h = |i-j|$\index[s]{gammak@$\gamma_k(h) $ autocovariance of $\vec{X}_k$}\index[s]{sigmak@$\sigma_k^2$ variance of elements in $\vec{X}_k$}\index[s]{Xk@$\overline{X}_k$ mean of elements of $\vec{X}_k$}\index[s]{Xk@$\vec{X}_k$ time series subject to $k$ blocking transformations}\index[s]{n@$n$ usually size of $\vec{X}$}\index[s]{nk@$n_k$ size of $\vec{X}_k$} +\begin{align} +\gamma_{k+1}(h) &= \cov\left( ({X}_{k+1})_{i}, ({X}_{k+1})_{j} \right) \nonumber \\ +&= \frac{1}{4}\cov\left( ({X}_{k})_{2i-1} + ({X}_{k})_{2i}, ({X}_{k})_{2j-1} + ({X}_{k})_{2j} \right) \nonumber \\ +&= +\begin{cases} +\frac{1}{2}\gamma_{k}(2h) + \frac{1}{2}\gamma_k(2h+1) \qquad\qquad\quad \ \ \text{if $h = 0$} \\ +\frac{1}{4}\gamma_k(2h-1) + \frac{1}{2}\gamma_k(2h) + \frac{1}{4}\gamma_k(2h+1) \quad \text{else} +\end{cases}. + \label{eq:gammas} +\end{align} +According to lemma \ref{lem:init}, since $\vec{X}$ is asymptotic uncorrelated by assumption, $\vec{X}_k$ is also asymptotic uncorrelated. Let's turn our attention to the variance of the sample mean $V(\overline{X})$. According to equation \eqref{eq:varepsilon} we have +\begin{align} +V(\overline{X}_k) = \frac{\sigma_k^2}{n_k} + \underbrace{\frac{2}{n_k} \sum_{h=1}^{n_k-1}\left( 1 - \frac{h}{n_k} \right)\gamma_k(h)}_{\equiv e_k} = \frac{\sigma^2_k}{n_k} + e_k \quad \text{if} \quad \gamma_k(0) = \sigma_k^2. \label{eq:varepsilon} +\end{align} +The term $e_k$ is called the \defn{truncation error}\index{truncation error}: +\begin{equation} +e_k = \frac{2}{n_k} \sum_{h=1}^{n_k-1}\left( 1 - \frac{h}{n_k} \right)\gamma_k(h). \label{eq:error} +\end{equation} +We show that $V(\overline{X}_i) = V(\overline{X}_j)$ for all $0 \leq i \leq d-1$ and $0 \leq j \leq d-1$. This follows by induction. For the induction step write: +\begin{align} +n_{j+1} \overline{X}_{j+1} &= \sum_{i=1}^{n_{j+1}} (\vec{X}_{j+1})_i \stackrel{ \eqref{eq:blocking} }{=} \frac{1}{2}\sum_{i=1}^{n_{j}/2} (\vec{X}_{j})_{2i-1} + (\vec{X}_{j})_{2i} \nonumber \\ +&= \frac{1}{2}\left[ (\vec{X}_j)_1 + (\vec{X}_j)_2 + \cdots + (\vec{X}_j)_{n_j} \right] = \underbrace{\frac{n_j}{2}}_{=n_{j+1}} \overline{X}_j = n_{j+1}\overline{X}_j. +\label{eq:means} +\end{align} +And so by repeated use of this equation we get $V(\overline{X}_i) = V(\overline{X}_0) = V(\overline{X})$ for all $0 \leq i \leq d-1$. This has the consequence that +\begin{align} +V(\overline{X}) = \frac{\sigma_k^2}{n_k} + e_k \qquad \text{for all} \qquad 0 \leq k \leq d-1. \label{eq:convergence} +\end{align} +\textcite{flyvbjerg_error_1989} claims that the sequence $\{e_k\}_{k=0}^{d-1}$ is decreasing, and conjecture that the term $e_k$ can be made as small as we would like by making $k$ (and hence $d$) sufficiently large. The sequence is decreasing because it is possible to show (as we will in proposition \ref{prop:uniform}) that $\gamma_k$ converges uniformly to the zero-function on $\mathbb{N}$. That means we can apply blocking transformations until $e_k$ is sufficiently small, and then estimate $V(\overline{X})$ by $\widehat{\sigma}^2_k/n_k$. Moreover, since $\gamma_k$ converges uniformly to the zero function on $\mathbb{N}$, see figure \ref{fig:block_cov} for an illustration.\\ +\\ +It is natural then to think the best estimate we could then make is $\widehat{\sigma}^2_{d-1}/n_{d-1}$. One could expect this because if $k = d-1$, then the truncation error $e_{d-1} \leq e_k$ for all $k \geq d-1$. But there is a problem with this, as we shall see next: If $k \to d-1$, then $V(\widehat{\sigma}^2_{d-1}/n_{d-1})$ grows to an appreciable size, and so the standard error of $\widehat{\sigma}^2_{d-1}/n_{d-1}$ can become unacceptably large. In that case, $\widehat{\sigma}^2_{d-1}/n_{d-1}$ is not very useful \parencite{flyvbjerg_error_1989}.\\ +\\ +The idea of \textcite{flyvbjerg_error_1989} is that if the conditions of the central limit theorem for dependent random variables are satisfied, see theorem \ref{thm:clt2}, then the components of $\vec{X}_k$ are asymptotically independent identically normal distributed as $k$ grows. The asymptotic independence follows by theorem \ref{thm:independent_iff_uncorrelated} because each elements of $\vec{X}_k$ is the mean of random variables which are asymptotic uncorrelated, as we explained above. In this case, it is immediate by theorem \ref{thm:S2chi} that +\begin{align} +V\left( \frac{\widehat{\sigma}_k^2}{n_k} \right) &= 2 \frac{\sigma_k^4}{n_k^4}(n-1) = \underbrace{\left(\frac{\sigma^2}{n_k} \right)^2}_{(V(\overline{X})- e_k)^2} 2\frac{n_k-1}{n_k^2} = \Big(V(\overline{X}) - e_k\Big)^2\frac{2}{n_k}\underbrace{\left( 1 - \frac{1}{n_k} \right)}_{\geq 1/n_k} \nonumber \\ +&\geq \Big(V(\overline{X}) - e_k\Big)^2\frac{2}{n_k^2}, \label{eq:departure} +\end{align} +since the truncation error $e_k$ is decreasing toward zero, this shows that the standard error of $\widehat \sigma_k^2/n_k$ grows as $n_k$ decreases. This may seem as bad news, because then the estimate of $V(\overline{X})$ has a relatively large error due to the standard error of $\widehat{\sigma}_k^2/n_k$ even though the truncation error $e_k$ is small. The question then is, how can we find the ideal $k$ such that essentially $e_k = 0$, but simultaneously ensure that the standard error of $\widehat{\sigma}^2_k/n_k$ is as small as possible? Assume $j \geq k$, and $e_k = 0$, then also $e_j$ must be zero and so it is possible to interpret from \textcite{flyvbjerg_error_1989} that +\begin{align} +0 &= |0| = |V(\overline{X}) - V(\overline{X})| = |V(\overline{X}_0) - V(\overline{X}_0)| \stackrel{ \eqref{eq:means} }{=} |V(\overline{X}_j) - V(\overline{X}_k)| \nonumber \\ +&\stackrel{ \eqref{eq:varepsilon} }{=} \Big|\frac{\sigma_j^2}{n_j} + \underbrace{e_k}_{=0} - \frac{\sigma_k^2}{n_k} - e_k \Big| = \Big|\frac{\sigma_j^2}{n_j} - \frac{\sigma_k^2}{n_k} \Big|. \label{eq:stationary} +\end{align} +That means that there could be some point on the graph of $\sigma_k^2/n_k$ where $\sigma_k^2/n_k$ become constant. See figure \ref{fig:block_cov} for a demonstration of what this looks like in practice. Manual blocking is exactly this, to plot the graph of $\sigma_k^2/n_k$ against $k$, and hope to find a stationary point where $\sigma_k^2/n_k$ becomes constant. The stationary point on this graph is where you stop increasing $k$. \\ +\\ +Despite using a few results from the present thesis to justify the claims, the above is my interpretation of \textcite{flyvbjerg_error_1989}. As you can see, it is not at all mathematically precise, but the results of this thesis will fix this and proposes an automated way to estimate $V(\overline{X})$.

    +Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. diff --git a/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz b/doc/Projects/2018/hw2/ipynb/ipynb-hw2-src.tar.gz index eddf88e9ac632bd4f11a05b825b072182e4eb6a8..dda671634a7ce1694cfc77fac4167cf1504b5782 100644 GIT binary patch literal 207 zcmV;=05Ja_iwFQo?2lUj1MSbjY63A3#&OR+#XP~1?5^v%P!Api5nmv((M?+uO(ME) z-&LVK)ti*k|L^iK!!XROW-qUcS&=V!^fXGTrmp3h(`Ir@Ro?3xtxM5WRaeud(OQcd ziisMDdfrLjPAQu>oYK8npRE6JZW!BcAMClmB47DOtUH%vzU*@sy!epqCWLqC&Sl{a zz28V1Y`XR(+T+PZE5`R?+SX+ky!jX(V%SBqPsRARKLG#$0000000000004mhHG005Z`V@Lo1 literal 207 zcmb2|=3wZ#(;v;i{Pw(W7L%huTjKS3i~J0}{Mp6#;!vx?Bb|KnAcGJ4YZ#+4>WmpQ G7#INHYheoj diff --git a/doc/Projects/2018/hw2/pdf/hw2.p.tex b/doc/Projects/2018/hw2/pdf/hw2.p.tex index 7ba297184..101bd04f8 100644 --- a/doc/Projects/2018/hw2/pdf/hw2.p.tex +++ b/doc/Projects/2018/hw2/pdf/hw2.p.tex @@ -211,11 +211,13 @@ Discuss these quantities as functions of the variable $\lambda$ in the Ridge and \subsection{Exercise 5} Using the singular value decomposition, show that the variance of the direction vector -$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie \emph{et al.}) +$\hat{z}_i=\hat{X}\hat{v}_i=\hat{u}_1d_1$ is equal to (equation (3.49) of Hastie \emph{et al.}) \[ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, \] -where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where $d_i$ are the singular values of the matrix $\hat{X}$. In Hastie \emph{et al}, the matrix elements of $X$ are centered. The consequence is that the mean values of for example $\hat{u}_i$ are zero. + +Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. % ------------------- end of main content --------------- diff --git a/doc/Projects/2018/hw2/pdf/hw2.pdf b/doc/Projects/2018/hw2/pdf/hw2.pdf index fd9334f70031d9bec14058f19f6f1f9682f17753..7fbb0fe019d42440a70eb6064227153ad0e46a69 100644 GIT binary patch delta 14814 zcmaibQ*_{M>}GA-wrx*2wQbwRuQsQ)?RKWNZF6d4YIncgv;V!_yCf$kZ!U6@oF~sa zAA@iokI>+T0M5n6n%uZa16=9oCTz7}2J9LZN+JF=_M%=UVV1b$kkI`-(;|&Z9VoEi z135dh_G{!~`bU7Tut7-@N6xyt9R|%D^)?bR^6VpjiPaNT@w%rbhrzd(|Kl=zS#vN( z*2U!;RdGqfi#U54t9cX`-QolalbVNr`e({6fk$@F`c21j7B=7RE>IKD^L4Mu9GY`$ zlmu!S%kszQHrfSqBquwaJ#N)UD7I!r1>^O(ljWwf=_-5Lc=*nKT}ZbUQr&x3C^X?y zNUwGJ+P=+K>mS)J@y4wh=l1cOyOxAkkK@}W&s__Vbe~U2aN9wVWUaJB8{z<`5k=_O 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b/doc/Projects/2018/hw2/pdf/hw2.tex index bb1c897c9..8670ad27a 100644 --- a/doc/Projects/2018/hw2/pdf/hw2.tex +++ b/doc/Projects/2018/hw2/pdf/hw2.tex @@ -181,11 +181,13 @@ Discuss these quantities as functions of the variable $\lambda$ in the Ridge and \subsection*{Exercise 5} Using the singular value decomposition, show that the variance of the direction vector -$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie \emph{et al.}) +$\hat{z}_i=\hat{X}\hat{v}_i=\hat{u}_1d_1$ is equal to (equation (3.49) of Hastie \emph{et al.}) \[ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, \] -where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where $d_i$ are the singular values of the matrix $\hat{X}$. In Hastie \emph{et al}, the matrix elements of $X$ are centered. The consequence is that the mean values of for example $\hat{u}_i$ are zero. + +Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. % ------------------- end of main content --------------- diff --git a/doc/src/Projects/2018/Exercises/hw2.do.txt b/doc/src/Projects/2018/Exercises/hw2.do.txt index 2b6e8071f..987cef3ec 100644 --- a/doc/src/Projects/2018/Exercises/hw2.do.txt +++ b/doc/src/Projects/2018/Exercises/hw2.do.txt @@ -54,10 +54,12 @@ Discuss these quantities as functions of the variable $\lambda$ in the Ridge and ===== Exercise 5 ===== Using the singular value decomposition, show that the variance of the direction vector -$\hat{z}_i=\hat{X}\hat{v}_i$ is equal to (equation (3.49) of Hastie *et al.*) +$\hat{z}_i=\hat{X}\hat{v}_i=\hat{u}_1d_1$ is equal to (equation (3.49) of Hastie *et al.*) !bt \[ \mathrm{Var}(\hat{z}_i)=\frac{d_i^2}{N}, \] !et -where $d_i$ are the singular values of the matrix $\hat{X}$. Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. +where $d_i$ are the singular values of the matrix $\hat{X}$. In Hastie *et al*, the matrix elements of $X$ are centered. The consequence is that the mean values of for example $\hat{u}_i$ are zero. + +Give an interpretation of these results, in particular in connection with the variance of the coefficients you obtained in the previous exercise. From eb9bc8a6eae0eec4f717a23ee8ee393f97ada97e Mon Sep 17 00:00:00 2001 From: mhjensen Date: Wed, 5 Sep 2018 22:28:34 +0200 Subject: [PATCH 07/12] Update on regression slides --- doc/src/Regression/Regression.do.txt | 197 ++++++++++----------------- 1 file changed, 74 insertions(+), 123 deletions(-) diff --git a/doc/src/Regression/Regression.do.txt b/doc/src/Regression/Regression.do.txt index f57d66c11..216fda678 100644 --- a/doc/src/Regression/Regression.do.txt +++ b/doc/src/Regression/Regression.do.txt @@ -18,6 +18,39 @@ A regression model aims at finding a likelihood function $p(y\vert \hat{x})$, th !eblock + +Consider an experiment in which $p$ characteristics of $n$ samples are +measured. The data from this experiment are denoted $\mathbf{X}$, with +$\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design +matrix*. Additional information of the samples is available in the +form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is +generally referred to as the *response variable*. The aim of +regression analysis is to explain $\mathbf{Y}$ in terms of +$\mathbf{X}$ through a functional relationship like $Y_i = +f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of +$f(\cdot)$ is available, it is common to assume a linear relationship +between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to +the *linear regression model*: + +In model (ref{form.linRegressionModel}) $\beta = (\beta_1, \ldots, +\beta_p)^{\top}$ is the *regression parameter*. The parameter +$\beta_j$, $j=1, \ldots, p$, represents the effect size of covariate +$j$ on the response. That is, for each unit change in covariate $j$ +(while keeping the other covariates fixed) the observed change in the +response is equal to $\beta_j$. The second summand on the right-hand +side of the model, $\varepsilon_i$, is referred to as the error. It +represents the part of the response not explained by the functional +part $\mathbf{X}_{i,\ast} \, \beta$ of the model +(ref{form.linRegressionModel}). In contrast to the functional part, +which is considered to be systematic (i.e. non-random), the error is +assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal +$Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= +\mathbf{X}_{i_2,\ast}$. To complete the formulation of model +(ref{form.linRegressionModel}) we need to specify the probability +distribution of $\varepsilon_i$. + + + !split ===== General linear models ===== !bblock @@ -1090,6 +1123,8 @@ probability plot. + + !split ===== The singular value decompostion ===== !bblock @@ -1109,9 +1144,6 @@ We have then !eblock -!split -===== Ridge regression ===== - !split ===== Lasso regression ===== @@ -1208,67 +1240,22 @@ plt.show() -Consider an experiment in which $p$ characteristics of $n$ samples are -measured. The data from this experiment are denoted $\mathbf{X}$, with -$\mathbf{X}$ as above. The matrix $\mathbf{X}$ is called the *design -matrix*. Additional information of the samples is available in the -form of $\mathbf{Y}$ (also as above). The variable $\mathbf{Y}$ is -generally referred to as the *response variable*. The aim of -regression analysis is to explain $\mathbf{Y}$ in terms of -$\mathbf{X}$ through a functional relationship like $Y_i = -f(\mathbf{X}_{i,\ast})$. When no prior knowledge on the form of -$f(\cdot)$ is available, it is common to assume a linear relationship -between $\mathbf{X}$ and $\mathbf{Y}$. This assumption gives rise to -the *linear regression model*: -!bt -\begin{align} -Y_{i} & = \mathbf{X}_{i,\ast} \, \beta + \varepsilon_i label{form.linRegressionModel} -\\ -\nonumber -& = \beta_1 \, X_{i,1} + \ldots + \beta_{p} \, X_{i, p} + \varepsilon_i. -\end{align} -!et - -In model (ref{form.linRegressionModel}) $\beta = (\beta_1, \ldots, -\beta_p)^{\top}$ is the *regression parameter*. The parameter -$\beta_j$, $j=1, \ldots, p$, represents the effect size of covariate -$j$ on the response. That is, for each unit change in covariate $j$ -(while keeping the other covariates fixed) the observed change in the -response is equal to $\beta_j$. The second summand on the right-hand -side of the model, $\varepsilon_i$, is referred to as the error. It -represents the part of the response not explained by the functional -part $\mathbf{X}_{i,\ast} \, \beta$ of the model -(ref{form.linRegressionModel}). In contrast to the functional part, -which is considered to be systematic (i.e. non-random), the error is -assumed to be random. Consequently, $Y_{i_1,\ast}$ need not equal -$Y_{i_2,\ast}$ for $i_1 \not= i_2$, even if $\mathbf{X}_{i_1,\ast}= -\mathbf{X}_{i_2,\ast}$. To complete the formulation of model -(ref{form.linRegressionModel}) we need to specify the probability -distribution of $\varepsilon_i$. - -======= Ridge regression ======= - -When the design matrix is high-dimensional, the covariates (the -columns of $\mathbf{X}$) are super-collinear. Recall *collinearity* in -regression analysis refers to the event of two (or multiple) -covariates being highly linearly related. Consequently, the subspace -spanned by collinear covariates may not be (or close to not being) of -full rank. When the subspace (onto which $\mathbf{Y}$ is projected) -is (close to) rank deficient, it is (almost) impossible to separate -the contribution of the individual covariates. The uncertainty with -respect to the covariate responsible for the variation explained in -$\mathbf{Y}$ is often reflected in the fit of the linear regression -model to data by a large error of the estimates of the regression -parameters corresponding to the collinear covariates. -Consider the design matrix: +!split +===== From standard regression to Ridge regressions ===== +One of the typical problems we encounter with linear regression, in particular +when the matrix $\hat{X}$ (our so-called design matrix) is high-dimensional, +are problems with near singular or singular matrices. The column vectors of $\hat{X}$ +may be linearly dependent, normally referred to as super-collinearity. +This means that the matrix may be rank deficient and it is basically impossible to +to model the data using linear regression. As an example, consider the matrix !bt \begin{align*} -\mathbf{X} & = \left( +\mathbf{X} & = \left[ \begin{array}{rrr} 1 & -1 & 2 \\ @@ -1277,92 +1264,56 @@ Consider the design matrix: 1 & 2 & -1 \\ 1 & 1 & 0 -\end{array} \right) +\end{array} \right] \end{align*} !et -The columns of $\mathbf{X}$ are linearly dependent: the first column -is the row-wise sum of the other two columns. The rank (more correct, -the column rank) of a matrix is the dimension of space spanned by the +The columns of $\hat{X}$ are linearly dependent. We se this easily since the +the first column is the row-wise sum of the other two columns. The rank (more correct, +the column rank) of a matrix is the dimension of the space spanned by the column vectors. Hence, the rank of $\mathbf{X}$ is equal to the number -of linearly independent columns: $\mbox{rank}(\mathbf{X}) = 2$. +of linearly independent columns. In this particular case the matrix has rank 2. - - - -Super-collinearity of an $(n \times p)$-dimensional design matrix $\mathbf{X}$ implies[^footnote1][^footnote1]: If the (column) rank of $\mathbf{X$ is smaller than $p$, there exists a non-trivial $\mathbf{v} \in \mathbb{R}^p$ such that $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. Multiplication of this inequality by $\mathbf{X}^{\top}$ yields $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{0}_{p}$. As $\mathbf{v} \not= \mathbf{0}_{p}$, this implies that $\mathbf{X}^{\top} \mathbf{X}$ is not invertible.} that the rank of the $(p \times p)$-dimensional matrix $\mathbf{X}^{\top} \mathbf{X}$ is smaller than $p$, and, consequently, it is singular. A square matrix that does not have an inverse is called *singular*. A matrix $\mathbf{A}$ is singular if and only if its determinant is zero: $\mbox{det}(\mathbf{A}) = 0$. - - - -Consider the matrix $\mathbf{A}$ given by: +Super-collinearity of an $(n \times p)$-dimensional design matrix $\mathbf{X}$ implies +that the inverse of the matrix $\hat{X}^T\hat{x}$ (the matrix we needto invert to solve the linear regression equations) is non-invertible. If we have a square matrix that does not have an inverse, we say this matrix singular. The example here demonstrates this !bt \begin{align*} -\mathbf{A} & = \left( +\hat{X} & = \left[ \begin{array}{rr} -1 & 2 +1 & -1 \\ -2 & 4 -\end{array} \right) +1 & -1 +\end{array} \right]. \end{align*} !et -Clearly, $\mbox{det}(\mathbf{A}) = a_{11} a_{22} - a_{12} a_{21} = 1 \times 4 - 2 \times 2 = 0$. Hence, $\mathbf{A}$ is singular and its inverse is undefined. +We see easily that $\mbox{det}(\hat{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0$. Hence, $\mathbf{X}$ is singular and its inverse is undefined. +This equivalent to saying that the matrix $\hat{X}$ has at least an eigenvalue which is zero. -As $\mbox{det}(\mathbf{A})$ is equal to the product of the eigenvalues -$\nu_j$ of $\mathbf{A}$, the matrix $\mathbf{A}$ is singular if one -(or more) of the eigenvalues of $\mathbf{A}$ is zero. To see this, -consider the spectral decomposition of $\mathbf{A}$: - -!bt -\begin{align*} -\mathbf{A} & = \sum_{j=1}^p \nu_j \, \mathbf{v}_j \, \mathbf{v}_j^{\top}, -\end{align*} -!et -where $\mathbf{v}_j$ is the eigenvector corresponding to $\nu_j$. The inverse of $\mathbf{A}$ is then: -!bt -\begin{align*} -\mathbf{A}^{-1} & = \sum_{j=1}^p \nu_j^{-1} \, \mathbf{v}_j \, \mathbf{v}_j^{\top}. -\end{align*} -!et - -The right-hand side is undefined if $\nu_j =0$ for any $j$. - - -Matrix $\mathbf{A}$ has eigenvalues $\nu_1 =5$ and $\nu_2=0$. According to the spectral decomposition, the inverse of $\mathbf{A}$ is: -!bt -\begin{align*} -\mathbf{A}^{-1} & = \frac{1}{5} \, \mathbf{v}_1 \, \mathbf{v}_1^{\top} + \frac{1}{0} \, \mathbf{v}_2 \, \mathbf{v}_2^{\top}. -\end{align*} -!et - -This expression is undefined as we divide by zero in the second summand on the right-hand side. - - - -In summary, the columns of a high-dimensional design matrix -$\mathbf{X}$ are linearly dependent and this super-collinearity causes -$\mathbf{X}^{\top} \mathbf{X}$ to be singular. Now recall the ML -estimator of the parameter of the linear regression model: +!split +===== Fixing the singularity ===== +If our design matrix $\hat{X}$ which enters the linear regression problem !bt \begin{align} -\hat{\beta} & = (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}. +\hat{\beta} & = (\hat{X}^{T} \hat{X})^{-1} \hat{X}^{T} \hat{y}, \end{align} !et +has linearly dependent column vectors, we will not be able to compute the inverse +of $\hat{X}^T\hat{X}$ and we cannot find the parameters (estimators) $\beta_i$. +The estimators are only well-defined if $(\hat{X}^{T}\hat{X})^{-1}$ exits. +This is more likely to happen when the matrix $\hat{X}$ is high-dimensional. In this case it is likely to encounter a situation where +the regression parameters $\beta_i$ cannot be estimated. - -This estimator is only well-defined if $(\mathbf{X}^{\top} -\mathbf{X})^{-1}$ exits. Hence, when $\mathbf{X}$ is high-dimensional -the regression parameter $\beta$ cannot be estimated. - - -Above only the practical consequence of high-dimensionality is presented: the expression $( \mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{Y}$ cannot be evaluated numerically. But the problem arising from the high-dimensionality of the data is more fundamental. To appreciate this, consider the normal equations: - +The *ad hoc* which was introduced in the 70s was simply to add a diagonal component to the matrix to invert, that is we change !bt -\begin{align*} -\mathbf{X}^{\top} \mathbf{X} \beta & = \mathbf{X}^{\top} \mathbf{Y}. -\end{align*} -!et +\[ +\hat{X}^{T} \hat{X} \rightarrow \hat{X}^{T} \hat{X}+\lambda \hat{I}, +\] +!et +where $\hat{I}$ is the identity matrix. +!split +===== The matrix $\mathbf{X}^{\top} \mathbf{X}$ is of rank $n$, while $\beta$ is a vector of length $p$. Hence, while there are $p$ unknowns, the system of linear equations from which these are to be solved effectively comprises $n$ degrees of freedom. If $p > n$, the vector $\beta$ cannot uniquely be determined from this system of equations. To make this more specific let $U$ be the $n$-dimensional space spanned by the columns of $\mathbf{X}$ and the $p-n$-dimensional space $V$ be orthogonal complement of $U$, i.e. $V = U^{\perp}$. Then, $\mathbf{X} \mathbf{v} = \mathbf{0}_{p}$ for all $\mathbf{v} \in V$. So, $V$ is the non-trivial null space of $\mathbf{X}$. Consequently, as $\mathbf{X}^{\top} \mathbf{X} \mathbf{v} = \mathbf{X}^{\top} \mathbf{0}_{p} = \mathbf{0}_{n}$, the solution of the normal equations is: From f5b9eb32a26d34cf37cd92f11df41a01da17a3df Mon Sep 17 00:00:00 2001 From: mhjensen Date: Thu, 6 Sep 2018 11:28:45 +0200 Subject: [PATCH 08/12] update on regression analysis --- .../Regression/html/._Regression-bs000.html | 155 +- .../Regression/html/._Regression-bs001.html | 153 +- .../Regression/html/._Regression-bs002.html | 179 +- .../Regression/html/._Regression-bs003.html | 172 +- .../Regression/html/._Regression-bs004.html | 192 +- .../Regression/html/._Regression-bs005.html | 195 +- .../Regression/html/._Regression-bs006.html | 180 +- .../Regression/html/._Regression-bs007.html | 181 +- .../Regression/html/._Regression-bs008.html | 177 +- .../Regression/html/._Regression-bs009.html | 182 +- .../Regression/html/._Regression-bs010.html | 178 +- .../Regression/html/._Regression-bs011.html | 167 +- .../Regression/html/._Regression-bs012.html | 176 +- .../Regression/html/._Regression-bs013.html | 172 +- .../Regression/html/._Regression-bs014.html | 167 +- .../Regression/html/._Regression-bs015.html | 170 +- .../Regression/html/._Regression-bs016.html | 172 +- .../Regression/html/._Regression-bs017.html | 187 +- .../Regression/html/._Regression-bs018.html | 224 +- .../Regression/html/._Regression-bs019.html | 178 +- .../Regression/html/._Regression-bs020.html | 206 +- .../Regression/html/._Regression-bs021.html | 221 +- .../Regression/html/._Regression-bs022.html | 176 +- .../Regression/html/._Regression-bs023.html | 176 +- .../Regression/html/._Regression-bs024.html | 184 +- .../Regression/html/._Regression-bs025.html | 202 +- .../Regression/html/._Regression-bs026.html | 204 +- .../Regression/html/._Regression-bs027.html | 199 +- .../Regression/html/._Regression-bs028.html | 178 +- .../Regression/html/._Regression-bs029.html | 187 +- .../Regression/html/._Regression-bs030.html | 212 +- .../Regression/html/._Regression-bs031.html | 226 +- .../Regression/html/._Regression-bs032.html | 185 +- .../Regression/html/._Regression-bs033.html | 204 +- .../Regression/html/._Regression-bs034.html | 258 +- .../Regression/html/._Regression-bs035.html | 180 +- .../Regression/html/._Regression-bs036.html | 236 +- .../Regression/html/._Regression-bs037.html | 207 +- .../Regression/html/._Regression-bs038.html | 307 +++ .../Regression/html/._Regression-bs039.html | 392 +++ .../Regression/html/._Regression-bs040.html | 304 +++ .../Regression/html/._Regression-bs041.html | 291 +++ .../Regression/html/._Regression-bs042.html | 365 +++ .../Regression/html/._Regression-bs043.html | 327 +++ .../Regression/html/._Regression-bs044.html | 278 +++ .../Regression/html/._Regression-bs045.html | 275 +++ doc/pub/Regression/html/Regression-bs.html | 155 +- .../Regression/html/Regression-reveal.html | 656 ++++- .../Regression/html/Regression-solarized.html | 725 ++++-- doc/pub/Regression/html/Regression.html | 725 ++++-- doc/pub/Regression/ipynb/Regression.ipynb | 916 ++++--- .../ipynb/ipynb-Regression-src.tar.gz | Bin 211 -> 211 bytes .../pdf/Regression-beamer-handouts2x3.pdf | Bin 345650 -> 389595 bytes doc/pub/Regression/pdf/Regression-beamer.pdf | Bin 324702 -> 351781 bytes doc/pub/Regression/pdf/Regression-minted.pdf | Bin 341147 -> 391306 bytes doc/src/Regression/Regression.do.txt | 2135 ++--------------- 56 files changed, 9374 insertions(+), 5675 deletions(-) create mode 100644 doc/pub/Regression/html/._Regression-bs038.html create mode 100644 doc/pub/Regression/html/._Regression-bs039.html create mode 100644 doc/pub/Regression/html/._Regression-bs040.html create mode 100644 doc/pub/Regression/html/._Regression-bs041.html create mode 100644 doc/pub/Regression/html/._Regression-bs042.html create mode 100644 doc/pub/Regression/html/._Regression-bs043.html create mode 100644 doc/pub/Regression/html/._Regression-bs044.html create mode 100644 doc/pub/Regression/html/._Regression-bs045.html diff --git a/doc/pub/Regression/html/._Regression-bs000.html b/doc/pub/Regression/html/._Regression-bs000.html index 9560a3323..c92c12f8a 100644 --- a/doc/pub/Regression/html/._Regression-bs000.html +++ b/doc/pub/Regression/html/._Regression-bs000.html @@ -41,17 +41,14 @@ Automatically generated HTML file from DocOnce source @@ -152,42 +173,50 @@ MathJax.Hub.Config({ Contents