From 5a2f0191bd91955efae41c055220f8acefffebc0 Mon Sep 17 00:00:00 2001 From: mhjensen Date: Sun, 22 Dec 2019 19:52:58 +0100 Subject: [PATCH] updating book on ML --- doc/LectureNotes/book.dlog | 7 + doc/LectureNotes/book.do.txt | 5858 +--------------------------------- 2 files changed, 49 insertions(+), 5816 deletions(-) diff --git a/doc/LectureNotes/book.dlog b/doc/LectureNotes/book.dlog index cfdaa7fef..fed599580 100644 --- a/doc/LectureNotes/book.dlog +++ b/doc/LectureNotes/book.dlog @@ -2984,3 +2984,10 @@ collected all required additional files in ipynb-book-src.tar.gz which must be d environments in math environments. output in book.ipynb +translating doconce text in book.do.txt to latex +figure fig/pandas.jpg must have extension(s) .eps, .ps +*** warning: need to convert from fig/pandas.jpg to fig/pandas.eps +using ImageMagick's convert program, but the result will +be loss of quality. Generate a proper fig/pandas.eps file (if possible). +....image conversion: convert fig/pandas.jpg fig/pandas.eps +output in book.p.tex diff --git a/doc/LectureNotes/book.do.txt b/doc/LectureNotes/book.do.txt index 1b8928d45..ebfd8b443 100644 --- a/doc/LectureNotes/book.do.txt +++ b/doc/LectureNotes/book.do.txt @@ -7,6 +7,7 @@ TOC: on ======= Introduction ======= + During the last two decades there has been a swift and amazing development of Machine Learning techniques and algorithms that impact many areas in not only Science and Technology but also the Humanities, @@ -81,7 +82,6 @@ Carlo methods are central elements in a proper understanding of many of algorithms and methods we will discuss. - ===== Learning outcomes ===== These sets of lectures aim at giving you an overview of central aspects of @@ -296,19 +296,21 @@ opt for the hitting the small folks instead of the monstertruck, since the likelihood of surving a collision with our future citizens, is much higher. -This leads to serious ethical aspects. Why should we -opt for such an option? Who decides and who is entitled to make such -choices? Keep in mind that many of the algorithms you will encounter in -this series of lectures or hear about later, are indeed based on -simple programming instructions. And you are very likely to be one of -the people who may end up writing such a code. Thus, developing a -sound ethical attitude to what we do, an approach well beyond the -simple mechanistic one of making our science available and -reproducible, is much needed. The example of the self-driving cars is -just one of infinitely many cases where we have to make choices. When -you analyze data on economic inequalities, who guarantees that you are -not weighting some data in a particular way, perhaps because you dearly want a -specific conclusion which may support your political views? +This leads to serious ethical aspects. Why should we opt for such an +option? Who decides and who is entitled to make such choices? Keep in +mind that many of the algorithms you will encounter in this series of +lectures or hear about later, are indeed based on simple programming +instructions. And you are very likely to be one of the people who may +end up writing such a code. Thus, developing a sound ethical attitude +to what we do, an approach well beyond the simple mechanistic one of +making our science available and reproducible, is much needed. The +example of the self-driving cars is just one of infinitely many cases +where we have to make choices. When you analyze data on economic +inequalities, who guarantees that you are not weighting some data in a +particular way, perhaps because you dearly want a specific conclusion +which may support your political views? Or what about the recent +claims that a famous IT company like Apple has a sexist bias on the +their recently "launched credit card":"https://qz.com/1748321/the-role-of-goldman-sachs-algorithms-in-the-apple-credit-card-scandal/"? We do not have the answers here, nor will we venture into a deeper discussions of these aspects, but we want you think over these topics @@ -322,9 +324,10 @@ large. Use these insights. Be a critical citizen. You owe it to our society. +======= Machine Learning Overview with Selectec Examples ======= -======= Getting started with Machine Learning ======= +===== Introduction ===== Our emphasis throughout this series of lectures is on understanding the mathematical aspects of @@ -607,7 +610,7 @@ developed in the 1970s, namely EISPACK and LINPACK. We describe them shortly he ===== Basic Matrix Features ===== - +!bblock Matrix properties reminder !bt \[ \mathbf{A} = @@ -646,7 +649,7 @@ The inverse of a matrix is defined by | $A = \left (A^{\dagger} \right )^{-1}$ | unitary | $\sum_k a_{ik} a_{jk}^{ * } = \sum_k a_{ki}^{ * } a_{kj} = \delta_{ij}$ | |----------------------------------------------------------------------| - +!eblock === Some famous Matrices === @@ -664,7 +667,7 @@ The inverse of a matrix is defined by === More Basic Matrix Features === -Some Equivalent Statements +!bblock Some Equivalent Statements For an $N\times N$ matrix $\mathbf{A}$ the following properties are all equivalent * If the inverse of $\mathbf{A}$ exists, $\mathbf{A}$ is nonsingular. @@ -673,7 +676,7 @@ For an $N\times N$ matrix $\mathbf{A}$ the following properties are all equival * The columns of $\mathbf{A}$ form a basis of $R^N$. * $\mathbf{A}$ is a product of elementary matrices. * $0$ is not eigenvalue of $\mathbf{A}$. - +!eblock ===== Numpy and arrays ===== @@ -1133,6 +1136,14 @@ the relative error (why would we prefer the MSE instead of the relative error?) \epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. \] !et + +The squared cost function results in an arithmetic mean-unbiased +estimator, and the absolute-value cost function results in a +median-unbiased estimator (in the one-dimensional case, and a +geometric median-unbiased estimator for the multi-dimensional +case). The squared cost function has the disadvantage that it has the tendency +to be dominated by outliers. + We can modify easily the above Python code and plot the relative error instead !bc pycod import numpy as np @@ -1234,7 +1245,7 @@ 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|. \] !et -Finally we present the +We present the squared logarithmic (quadratic) error !bt \[ @@ -1247,6 +1258,16 @@ 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. + +Finally, another cost function is the Huber cost function used in robust regression. +It is less sensitive to outliers in data than the squared error cost function. +A variant for classification is also sometimes used, a quantity we will meet later. +!bt +\[ +L_{\delta }(a)={\begin{cases}{\frac {1}{2}}{a^{2}}&{\text{for }}|a|\leq \delta ,\\\delta (|a|-{\frac {1}{2}}\delta ),&{\text{otherwise.}}\end{cases}}}L_{\delta }(a)={\begin{cases}{\frac {1}{2}}{a^{2}}&{\text{for }}|a|\leq \delta ,\\\delta (|a|-{\frac {1}{2}}\delta ),&{\text{otherwise.}}\end{cases}} +\] +!et + 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. @@ -1620,10 +1641,6 @@ plt.show() !ec - - - - ===== A first summary ===== The aim behind these introductory words was to present to you various @@ -1637,5799 +1654,8 @@ Machine Learning algorithms for supervised learning. Later we will meet _Tensorf Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails. +======= Review of Basic Statistics ======= - - -======= Why Linear Regression (aka Ordinary Least Squares and family) ======= - -Fitting a continuous function with linear parameterization in terms of the parameters $\bm{\beta}$. -* Method of choice for fitting a continuous function! -* Gives an excellent introduction to central Machine Learning features with _understandable pedagogical_ links to other methods like _Neural Networks_, _Support Vector Machines_ etc -* Analytical expression for the fitting parameters $\bm{\beta}$ -* Analytical expressions for statistical propertiers like mean values, variances, confidence intervals and more -* Analytical relation with probabilistic interpretations -* Easy to introduce basic concepts like bias-variance tradeoff, cross-validation, resampling and regularization techniques and many other ML topics -* Easy to code! And links well with classification problems and logistic regression and neural networks -* Allows for _easy_ hands-on understanding of gradient descent methods -* and many more features - -For more discussions of Ridge and Lasso regression, "Wessel van Wieringen's":"https://arxiv.org/abs/1509.09169" article is highly recommended. -Similarly, "Mehta et al's article":"https://arxiv.org/abs/1803.08823" is also recommended. - - -=== Regression analysis, overarching aims === - -Regression modeling deals with the description of the sampling distribution of a given random variable $y$ and how it varies as function of another variable or a set of such variables $\bm{x} =[x_0, x_1,\dots, x_{n-1}]^T$. -The first variable is called the _dependent_, the _outcome_ or the _response_ variable while the set of variables $\bm{x}$ is called the independent variable, or the predictor variable or the explanatory variable. - -A regression model aims at finding a likelihood function $p(\bm{y}\vert \bm{x})$, that is the conditional distribution for $\bm{y}$ with a given $\bm{x}$. The estimation of $p(\bm{y}\vert \bm{x})$ is made using a data set with -* $n$ cases $i = 0, 1, 2, \dots, n-1$ -* Response (target, dependent or outcome) variable $y_i$ with $i = 0, 1, 2, \dots, n-1$ -* $p$ so-called explanatory (independent or predictor) variables $\bm{x}_i=[x_{i0}, x_{i1}, \dots, x_{ip-1}]$ with $i = 0, 1, 2, \dots, n-1$ and explanatory variables running from $0$ to $p-1$. See below for more explicit examples. - The goal of the regression analysis is to extract/exploit relationship between $\bm{y}$ and $\bm{X}$ in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things. - - -Consider an experiment in which $p$ characteristics of $n$ samples are -measured. The data from this experiment, for various explanatory variables $p$ are normally represented by a matrix -$\mathbf{X}$. - -The matrix $\mathbf{X}$ is called the *design -matrix*. Additional information of the samples is available in the -form of $\bm{y}$ (also as above). The variable $\bm{y}$ is -generally referred to as the *response variable*. The aim of -regression analysis is to explain $\bm{y}$ in terms of -$\bm{X}$ 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 $\bm{X}$ and $\bm{y}$. This assumption gives rise to -the *linear regression model* where $\bm{\beta} = [\beta_0, \ldots, -\beta_{p-1}]^{T}$ are the *regression parameters*. - -Linear regression gives us a set of analytical equations for the parameters $\beta_j$. - - -=== Examples === - -In order to understand the relation among the predictors $p$, the set of data $n$ and the target (outcome, output etc) $\bm{y}$, -consider the model we discussed for describing nuclear binding energies. - -There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model. -Assuming -!bt -\[ -BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, -\] -!et -we have five predictors, that is the intercept, the $A$ dependent term, the $A^{2/3}$ term and the $A^{-1/3}$ and $A^{-1}$ terms. -This gives $p=0,1,2,3,4$. Furthermore we have $n$ entries for each predictor. It means that our design matrix is a -$p\times n$ matrix $\bm{X}$. - -Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the -so-called "credit card default data from Taiwan":"https://www.sciencedirect.com/science/article/pii/S0957417407006719?via%3Dihub". The data set contains data on $n=30000$ credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are $24$ such predictors or attributes leading to a design matrix of dimensionality $24 \times 30000$ - - -===== General linear models ===== - -Before we proceed let us study a case from linear algebra where we aim at fitting a set of data $\bm{y}=[y_0,y_1,\dots,y_{n-1}]$. 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 $\bm{x}=[x_0,x_1,\dots,x_{n-1}]$, 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 -!bt -\[ -y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, -\] -!et -where $\epsilon_i$ is the error in our approximation. - - -For every set of values $y_i,x_i$ we have thus the corresponding set of equations -!bt -\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_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ -\end{align*} -!et - - -Defining the vectors -!bt -\[ -\bm{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, -\] -!et -and -!bt -\[ -\bm{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, -\] -!et -and -!bt -\[ -\bm{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, -\] -!et -and the design matrix -!bt -\[ -\bm{X}= -\begin{bmatrix} -1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ -1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ -1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ -\end{bmatrix} -\] -!et -we can rewrite our equations as -!bt -\[ -\bm{y} = \bm{X}\bm{\beta}+\bm{\epsilon}. -\] -!et -The above design matrix is called a "Vandermonde matrix":"https://en.wikipedia.org/wiki/Vandermonde_matrix". - - - - -===== Generalizing the fitting procedure as a linear algebra problem ===== - -We are obviously not limited to the above polynomial expansions. We -could replace the various powers of $x$ with elements of Fourier -series or 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 - -!bt -\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_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} -!et - -_Note that we have $p=n$ here. The matrix is symmetric. This is generally not the case!_ - -We redefine in turn the matrix $\bm{X}$ as -!bt -\[ -\bm{X}= -\begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ -\dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ -\end{bmatrix} -\] -!et -and without loss of generality we rewrite again our equations as -!bt -\[ -\bm{y} = \bm{X}\bm{\beta}+\bm{\epsilon}. -\] -!et -The left-hand side of this equation is kwown. Our error vector $\bm{\epsilon}$ and the parameter vector $\bm{\beta}$ are our unknow quantities. How can we obtain the optimal set of $\beta_i$ values? - -We have defined the matrix $\bm{X}$ via the equations -!bt -\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_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} -!et - -As we noted above, we stayed with a system with the design matrix - $\bm{X}\in {\mathbb{R}}^{n\times n}$, that is we have $p=n$. For reasons to come later (algorithmic arguments) we will hereafter define -our matrix as $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors refering to the column numbers and the entries $n$ being the row elements. - - -===== Our model for the nuclear binding energies ===== - -In our introductory notes we looked at the so-called "liguid drop model":"https://en.wikipedia.org/wiki/Semi-empirical_mass_formula". Let us remind ourselves about what we did by looking at the code. - -We restate the parts of the code we are most interested in. -!bc pycod -# Common imports -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from IPython.display import display -import os - -# Where to save the figures and data files -PROJECT_ROOT_DIR = "Results" -FIGURE_ID = "Results/FigureFiles" -DATA_ID = "DataFiles/" - -if not os.path.exists(PROJECT_ROOT_DIR): - os.mkdir(PROJECT_ROOT_DIR) - -if not os.path.exists(FIGURE_ID): - os.makedirs(FIGURE_ID) - -if not os.path.exists(DATA_ID): - os.makedirs(DATA_ID) - -def image_path(fig_id): - return os.path.join(FIGURE_ID, fig_id) - -def data_path(dat_id): - return os.path.join(DATA_ID, dat_id) - -def save_fig(fig_id): - plt.savefig(image_path(fig_id) + ".png", format='png') - -infile = open(data_path("MassEval2016.dat"),'r') - - -# Read the experimental data with Pandas -Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11), - names=('N', 'Z', 'A', 'Element', 'Ebinding'), - widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1), - header=39, - index_col=False) - -# Extrapolated values are indicated by '#' in place of the decimal place, so -# the Ebinding column won't be numeric. Coerce to float and drop these entries. -Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce') -Masses = Masses.dropna() -# Convert from keV to MeV. -Masses['Ebinding'] /= 1000 - -# Group the DataFrame by nucleon number, A. -Masses = Masses.groupby('A') -# Find the rows of the grouped DataFrame with the maximum binding energy. -Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()]) -A = Masses['A'] -Z = Masses['Z'] -N = Masses['N'] -Element = Masses['Element'] -Energies = Masses['Ebinding'] - -# Now we set up the design matrix X -X = np.zeros((len(A),5)) -X[:,0] = 1 -X[:,1] = A -X[:,2] = A**(2.0/3.0) -X[:,3] = A**(-1.0/3.0) -X[:,4] = A**(-1.0) -# Then nice printout using pandas -DesignMatrix = pd.DataFrame(X) -DesignMatrix.index = A -DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A'] -display(DesignMatrix) -!ec - -With $\bm{\beta}\in {\mathbb{R}}^{p\times 1}$, it means that we will hereafter write our equations for the approximation as -!bt -\[ -\bm{\tilde{y}}= \bm{X}\bm{\beta}, -\] -!et -throughout these lectures. - - - -With the above we use the design matrix to define the approximation $\bm{\tilde{y}}$ via the unknown quantity $\bm{\beta}$ as -!bt -\[ -\bm{\tilde{y}}= \bm{X}\bm{\beta}, -\] -!et -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 parameterized values $\tilde{y}_i$, namely -!bt -\[ -C(\bm{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\bm{y}-\bm{\tilde{y}}\right)^T\left(\bm{y}-\bm{\tilde{y}}\right)\right\}, -\] -!et -or using the matrix $\bm{X}$ and in a more compact matrix-vector notation as -!bt -\[ -C(\bm{\beta})=\frac{1}{n}\left\{\left(\bm{y}-\bm{X}^T\bm{\beta}\right)^T\left(\bm{y}-\bm{X}^T\bm{\beta}\right)\right\}. -\] -!et -This function is one possible way to define the so-called cost function. - - - -It is also common to define -the function $Q$ as - -!bt -\[ -C(\bm{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, -\] -!et -since when taking the first derivative with respect to the unknown parameters $\beta$, the factor of $2$ cancels out. - - - - -===== Interpretations and optimizing our parameters ===== - - -The function -!bt -\[ -C(\bm{\beta})=\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\beta}\right)^T\left(\bm{y}-\bm{X}\bm{\beta}\right)\right\}, -\] -!et -can be linked to the variance of the quantity $y_i$ if we interpret the latter as the mean value. -When linking below with the maximum likelihood approach below, we will indeed interpret $y_i$ as a mean value (see exercises) -!bt -\[ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, -\] -!et - -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 $C(\bm{\beta})$, that is we are going to solve the problem -!bt -\[ -{\displaystyle \min_{\bm{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\beta}\right)^T\left(\bm{y}-\bm{X}\bm{\beta}\right)\right\}. -\] -!et -In practical terms it means we will require -!bt -\[ -\frac{\partial C(\bm{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, -\] -!et -which results in -!bt -\[ -\frac{\partial C(\bm{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, -\] -!et -or in a matrix-vector form as -!bt -\[ -\frac{\partial C(\bm{\beta})}{\partial \bm{\beta}} = 0 = \bm{X}^T\left( \bm{y}-\bm{X}\bm{\beta}\right). -\] -!et - - - -We can rewrite -!bt -\[ -\frac{\partial C(\bm{\beta})}{\partial \bm{\beta}} = 0 = \bm{X}^T\left( \bm{y}-\bm{X}\bm{\beta}\right), -\] -!et -as -!bt -\[ -\bm{X}^T\bm{y} = \bm{X}^T\bm{X}\bm{\beta}, -\] -!et -and if the matrix $\bm{X}^T\bm{X}$ is invertible we have the solution -!bt -\[ -\bm{\beta} =\left(\bm{X}^T\bm{X}\right)^{-1}\bm{X}^T\bm{y}. -\] -!et - -We note also that since our design matrix is defined as $\bm{X}\in -{\mathbb{R}}^{n\times p}$, the product $\bm{X}^T\bm{X} \in -{\mathbb{R}}^{p\times p}$. In the above case we have that $p \ll n$, -in our case $p=5$ meaning that we end up with inverting a small -$5\times 5$ matrix. This is a rather common situation, in many cases we end up with low-dimensional -matrices to invert. The methods discussed here and for many other -supervised learning algorithms like classification with logistic -regression or support vector machines, exhibit dimensionalities which -allow for the usage of direct linear algebra methods such as _LU_ decomposition or _Singular Value Decomposition_ (SVD) for finding the inverse of the matrix -$\bm{X}^T\bm{X}$. - - - -The residuals $\bm{\epsilon}$ are in turn given by -!bt -\[ -\bm{\epsilon} = \bm{y}-\bm{\tilde{y}} = \bm{y}-\bm{X}\bm{\beta}, -\] -!et -and with -!bt -\[ -\bm{X}^T\left( \bm{y}-\bm{X}\bm{\beta}\right)= 0, -\] -!et -we have -!bt -\[ -\bm{X}^T\bm{\epsilon}=\bm{X}^T\left( \bm{y}-\bm{X}\bm{\beta}\right)= 0, -\] -!et -meaning that the solution for $\bm{\beta}$ is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. - - -Let us now return to our nuclear binding energies and simply code the above equations. - -It is rather straightforward to implement the matrix inversion and obtain the parameters $\bm{\beta}$. After having defined the matrix $\bm{X}$ we simply need to -write -!bc pycod -# matrix inversion to find beta -beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies) -# and then make the prediction -ytilde = X @ beta -!ec -Alternatively, you can use the least squares functionality in _Numpy_ as -!bc pycod -fit = np.linalg.lstsq(X, Energies, rcond =None)[0] -ytildenp = np.dot(fit,X.T) -!ec - -And finally we plot our fit with and compare with data -!bc pycod -Masses['Eapprox'] = ytilde -# Generate a plot comparing the experimental with the fitted values values. -fig, ax = plt.subplots() -ax.set_xlabel(r'$A = N + Z$') -ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$') -ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2, - label='Ame2016') -ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m', - label='Fit') -ax.legend() -save_fig("Masses2016OLS") -plt.show() -!ec - -===== Adding error analysis and training set up ===== - -We can easily test our fit by computing the $R2$ score that we discussed in connection with the functionality of _Scikit_Learn_ in the introductory slides. -Since we are not using _Scikit-Learn here we can define our own $R2$ function as -!bc pycod -def R2(y_data, y_model): - return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_model)) ** 2) -!ec -and we would be using it as -!bc pycod -print(R2(Energies,ytilde)) -!ec - -We can easily add our _MSE_ score as -!bc pycod -def MSE(y_data,y_model): - n = np.size(y_model) - return np.sum((y_data-y_model)**2)/n - -print(MSE(Energies,ytilde)) -!ec -and finally the relative error as -!bc pycod -def RelativeError(y_data,y_model): - return abs((y_data-y_model)/y_data) -print(RelativeError(Energies, ytilde)) -!ec - - - -===== The $\chi^2$ function ===== - -Normally, the response (dependent or outcome) variable $y_i$ is 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 (omitting the $1/n$ term) -as - -!bt -\[ -\chi^2(\bm{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\bm{y}-\bm{\tilde{y}}\right)^T\frac{1}{\bm{\Sigma^2}}\left(\bm{y}-\bm{\tilde{y}}\right)\right\}, -\] -!et -where the matrix $\bm{\Sigma}$ is a diagonal matrix with $\sigma_i$ as matrix elements. - - -In order to find the parameters $\beta_i$ we will then minimize the spread of $\chi^2(\bm{\beta})$ by requiring -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, -\] -!et -which results in -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, -\] -!et -or in a matrix-vector form as -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \bm{\beta}} = 0 = \bm{A}^T\left( \bm{b}-\bm{A}\bm{\beta}\right). -\] -!et -where we have defined the matrix $\bm{A} =\bm{X}/\bm{\Sigma}$ with matrix elements $a_{ij} = x_{ij}/\sigma_i$ and the vector $\bm{b}$ with elements $b_i = y_i/\sigma_i$. - -We can rewrite -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \bm{\beta}} = 0 = \bm{A}^T\left( \bm{b}-\bm{A}\bm{\beta}\right), -\] -!et -as -!bt -\[ -\bm{A}^T\bm{b} = \bm{A}^T\bm{A}\bm{\beta}, -\] -!et -and if the matrix $\bm{A}^T\bm{A}$ is invertible we have the solution -!bt -\[ -\bm{\beta} =\left(\bm{A}^T\bm{A}\right)^{-1}\bm{A}^T\bm{b}. -\] -!et - -If we then introduce the matrix -!bt -\[ -\bm{H} = \left(\bm{A}^T\bm{A}\right)^{-1}, -\] -!et -we have then the following expression for the parameters $\beta_j$ (the matrix elements of $\bm{H}$ are $h_{ij}$) -!bt -\[ -\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} -\] -!et -We state without proof the expression for the uncertainty in the parameters $\beta_j$ as (we leave this as an exercise) -!bt -\[ -\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, -\] -!et -resulting in -!bt -\[ -\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! -\] -!et - -The first step here is to approximate the function $y$ with a first-order polynomial, that is we write -!bt -\[ -y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. -\] -!et -By computing the derivatives of $\chi^2$ with respect to $\beta_0$ and $\beta_1$ show that these are given by -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, -\] -!et -and -!bt -\[ -\frac{\partial \chi^2(\bm{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. -\] -!et - -For a linear fit (a first-order polynomial) we don't need to invert a matrix!! -Defining -!bt -\[ -\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, -\] -!et - -!bt -\[ -\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, -\] -!et - -!bt -\[ -\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), -\] -!et - -!bt -\[ -\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, -\] -!et - -!bt -\[ -\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, -\] -!et - -we obtain - -!bt -\[ -\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, -\] -!et - -!bt -\[ -\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. -\] -!et - -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. - - -===== Fitting an Equation of State for Dense Nuclear Matter ===== - -Before we continue, let us introduce yet another example. We are going to fit the -nuclear equation of state using results from many-body calculations. -The equation of state we have made available here, as function of -density, has been derived using modern nucleon-nucleon potentials with -"the addition of three-body -forces":"https://www.sciencedirect.com/science/article/pii/S0370157399001106". This -time the file is presented as a standard _csv_ file. - -The beginning of the Python code here is similar to what you have seen before, -with the same initializations and declarations. We use also _pandas_ -again, rather extensively in order to organize our data. - -The difference now is that we use _Scikit-Learn's_ regression tools -instead of our own matrix inversion implementation. Furthermore, we -sneak in _Ridge_ regression (to be discussed below) which includes a -hyperparameter $\lambda$, also to be explained below. - -!split -===== The code ===== - -!bc pycod -# Common imports -import os -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -import matplotlib.pyplot as plt -import sklearn.linear_model as skl -from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error - -# Where to save the figures and data files -PROJECT_ROOT_DIR = "Results" -FIGURE_ID = "Results/FigureFiles" -DATA_ID = "DataFiles/" - -if not os.path.exists(PROJECT_ROOT_DIR): - os.mkdir(PROJECT_ROOT_DIR) - -if not os.path.exists(FIGURE_ID): - os.makedirs(FIGURE_ID) - -if not os.path.exists(DATA_ID): - os.makedirs(DATA_ID) - -def image_path(fig_id): - return os.path.join(FIGURE_ID, fig_id) - -def data_path(dat_id): - return os.path.join(DATA_ID, dat_id) - -def save_fig(fig_id): - plt.savefig(image_path(fig_id) + ".png", format='png') - -infile = open(data_path("EoS.csv"),'r') - -# Read the EoS data as csv file and organize the data into two arrays with density and energies -EoS = pd.read_csv(infile, names=('Density', 'Energy')) -EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce') -EoS = EoS.dropna() -Energies = EoS['Energy'] -Density = EoS['Density'] -# The design matrix now as function of various polytrops -X = np.zeros((len(Density),4)) -X[:,3] = Density**(4.0/3.0) -X[:,2] = Density -X[:,1] = Density**(2.0/3.0) -X[:,0] = 1 - -# We use now Scikit-Learn's linear regressor and ridge regressor -# OLS part -clf = skl.LinearRegression().fit(X, Energies) -ytilde = clf.predict(X) -EoS['Eols'] = ytilde -# The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde)) -# Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(Energies, ytilde)) -# Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde)) -print(clf.coef_, clf.intercept_) - -# The Ridge regression with a hyperparameter lambda = 0.1 -_lambda = 0.1 -clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies) -yridge = clf_ridge.predict(X) -EoS['Eridge'] = yridge -# The mean squared error -print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge)) -# Explained variance score: 1 is perfect prediction -print('Variance score: %.2f' % r2_score(Energies, yridge)) -# Mean absolute error -print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge)) -print(clf_ridge.coef_, clf_ridge.intercept_) - -fig, ax = plt.subplots() -ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$') -ax.set_ylabel(r'Energy per particle') -ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2, - label='Theoretical data') -ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m', - label='OLS') -ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g', - label='Ridge $\lambda = 0.1$') -ax.legend() -save_fig("EoSfitting") -plt.show() -!ec - -The above simple polynomial in density $\rho$ gives an excellent fit -to the data. -We note also that there is a small deviation between the -standard OLS and the Ridge regression at higher densities. We discuss this in more detail -below. - - -===== Splitting our Data in Training and Test data ===== - -It is normal in essentially all Machine Learning studies to split the -data in a training set and a test set (sometimes also an additional -validation set). _Scikit-Learn_ has an own function for this. There -is no explicit recipe for how much data should be included as training -data and say test data. An accepted rule of thumb is to use -approximately $2/3$ to $4/5$ of the data as training data. We will -postpone a discussion of this splitting to the end of these notes and -our discussion of the so-called _bias-variance_ tradeoff. Here we -limit ourselves to repeat the above equation of state fitting example -but now splitting the data into a training set and a test set. - -!bc pycod -import os -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -from sklearn.model_selection import train_test_split -# Where to save the figures and data files -PROJECT_ROOT_DIR = "Results" -FIGURE_ID = "Results/FigureFiles" -DATA_ID = "DataFiles/" - -if not os.path.exists(PROJECT_ROOT_DIR): - os.mkdir(PROJECT_ROOT_DIR) - -if not os.path.exists(FIGURE_ID): - os.makedirs(FIGURE_ID) - -if not os.path.exists(DATA_ID): - os.makedirs(DATA_ID) - -def image_path(fig_id): - return os.path.join(FIGURE_ID, fig_id) - -def data_path(dat_id): - return os.path.join(DATA_ID, dat_id) - -def save_fig(fig_id): - plt.savefig(image_path(fig_id) + ".png", format='png') - -def R2(y_data, y_model): - return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_model)) ** 2) -def MSE(y_data,y_model): - n = np.size(y_model) - return np.sum((y_data-y_model)**2)/n - -infile = open(data_path("EoS.csv"),'r') - -# Read the EoS data as csv file and organized into two arrays with density and energies -EoS = pd.read_csv(infile, names=('Density', 'Energy')) -EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce') -EoS = EoS.dropna() -Energies = EoS['Energy'] -Density = EoS['Density'] -# The design matrix now as function of various polytrops -X = np.zeros((len(Density),5)) -X[:,0] = 1 -X[:,1] = Density**(2.0/3.0) -X[:,2] = Density -X[:,3] = Density**(4.0/3.0) -X[:,4] = Density**(5.0/3.0) -# We split the data in test and training data -X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2) -# matrix inversion to find beta -beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train) -# and then make the prediction -ytilde = X_train @ beta -print("Training R2") -print(R2(y_train,ytilde)) -print("Training MSE") -print(MSE(y_train,ytilde)) -ypredict = X_test @ beta -print("Test R2") -print(R2(y_test,ypredict)) -print("Test MSE") -print(MSE(y_test,ypredict)) -!ec - - - -===== The singular value decomposition ===== - -The examples we have looked at so far are cases where we normally can -invert the matrix $\bm{X}^T\bm{X}$. Using a polynomial expansion as we -did both for the masses and the fitting of the equation of state, -leads to row vectors of the design matrix which are essentially -orthogonal due to the polynomial character of our model. This may -however not the be case in general and a standard matrix inversion -algorithm based on say LU decomposition may lead to singularities. We will see an example of this below when we try to fit -the coupling constant of the widely used Ising model. -There is however a way to partially circumvent this problem and also gain some insight about the ordinary least squares approach. - -This is given by the _Singular Value Decomposition_ algorithm, perhaps -the most powerful linear algebra algorithm. Let us look at a -different example where we may have problems with the standard matrix -inversion algorithm. Thereafter we dive into the math of the SVD. - - -===== The Ising model ===== - -The one-dimensional Ising model with nearest neighbor interaction, no -external field and a constant coupling constant $J$ is given by - -!bt -\begin{align} - H = -J \sum_{k}^L s_k s_{k + 1}, -\end{align} -!et - -where $s_i \in \{-1, 1\}$ and $s_{N + 1} = s_1$. The number of spins -in the system is determined by $L$. For the one-dimensional system -there is no phase transition. - -We will look at a system of $L = 40$ spins with a coupling constant of -$J = 1$. To get enough training data we will generate 10000 states -with their respective energies. - - -!bc pycod -import numpy as np -import matplotlib.pyplot as plt -from mpl_toolkits.axes_grid1 import make_axes_locatable -import seaborn as sns -import scipy.linalg as scl -from sklearn.model_selection import train_test_split -import tqdm -sns.set(color_codes=True) -cmap_args=dict(vmin=-1., vmax=1., cmap='seismic') - -L = 40 -n = int(1e4) - -spins = np.random.choice([-1, 1], size=(n, L)) -J = 1.0 - -energies = np.zeros(n) - -for i in range(n): - energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1)) -!ec - -Here we use ordinary least squares -regression to predict the energy for the nearest neighbor -one-dimensional Ising model on a ring, i.e., the endpoints wrap -around. We will use linear regression to fit a value for -the coupling constant to achieve this. - -===== Reformulating the problem to suit regression ===== - -A more general form for the one-dimensional Ising model is - -!bt -\begin{align} - H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. -\end{align} -!et - -Here we allow for interactions beyond the nearest neighbors and a state dependent -coupling constant. This latter expression can be formulated as -a matrix-product -!bt -\begin{align} - \bm{H} = \bm{X} J, -\end{align} -!et - -where $X_{jk} = s_j s_k$ and $J$ is a matrix which consists of the -elements $-J_{jk}$. This form of writing the energy fits perfectly -with the form utilized in linear regression, that is - -!bt -\begin{align} - \bm{y} = \bm{X}\bm{\beta} + \bm{\epsilon}, -\end{align} -!et - -We split the data in training and test data as discussed in the previous example - -!bc pycod -X = np.zeros((n, L ** 2)) -for i in range(n): - X[i] = np.outer(spins[i], spins[i]).ravel() -y = energies -X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2) -!ec - - -In the ordinary least squares method we choose the cost function - -!bt -\begin{align} - C(\bm{X}, \bm{\beta})= \frac{1}{n}\left\{(\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y})\right\}. -\end{align} -!et - -We then find the extremal point of $C$ by taking the derivative with respect to $\bm{\beta}$ as discussed above. -This yields the expression for $\bm{\beta}$ to be - -!bt -\[ - \bm{\beta} = \frac{\bm{X}^T \bm{y}}{\bm{X}^T \bm{X}}, -\] -!et - -which immediately imposes some requirements on $\bm{X}$ as there must exist -an inverse of $\bm{X}^T \bm{X}$. If the expression we are modeling contains an -intercept, i.e., a constant term, we must make sure that the -first column of $\bm{X}$ consists of $1$. We do this here - -!bc pycod -X_train_own = np.concatenate( - (np.ones(len(X_train))[:, np.newaxis], X_train), - axis=1 -) -X_test_own = np.concatenate( - (np.ones(len(X_test))[:, np.newaxis], X_test), - axis=1 -) -!ec - -!bc pycod -def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray: - return scl.inv(x.T @ x) @ (x.T @ y) -beta = ols_inv(X_train_own, y_train) -!ec - - - -===== Singular Value decomposition ===== - -Doing the inversion directly turns out to be a bad idea since the matrix -$\bm{X}^T\bm{X}$ is singular. An alternative approach is to use the _singular -value decomposition_. Using the definition of the Moore-Penrose -pseudoinverse we can write the equation for $\bm{\beta}$ as - -!bt -\[ - \bm{\beta} = \bm{X}^{+}\bm{y}, -\] -!et - -where the pseudoinverse of $\bm{X}$ is given by - -!bt -\[ - \bm{X}^{+} = \frac{\bm{X}^T}{\bm{X}^T\bm{X}}. -\] -!et - -Using singular value decomposition we can decompose the matrix $\bm{X} = \bm{U}\bm{\Sigma} \bm{V}^T$, -where $\bm{U}$ and $\bm{V}$ are orthogonal(unitary) matrices and $\bm{\Sigma}$ contains the singular values (more details below). -where $X^{+} = V\Sigma^{+} U^T$. This reduces the equation for -$\omega$ to -!bt -\begin{align} - \bm{\beta} = \bm{V}\bm{\Sigma}^{+} \bm{U}^T \bm{y}. -\end{align} -!et - -Note that solving this equation by actually doing the pseudoinverse -(which is what we will do) is not a good idea as this operation scales -as $\mathcal{O}(n^3)$, where $n$ is the number of elements in a -general matrix. Instead, doing $QR$-factorization and solving the -linear system as an equation would reduce this down to -$\mathcal{O}(n^2)$ operations. - - -!bc pycod -def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray: - u, s, v = scl.svd(x) - return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y -!ec - -!bc pycod -beta = ols_svd(X_train_own,y_train) -!ec - -When extracting the $J$-matrix we need to make sure that we remove the intercept, as is done here - -!bc pycod -J = beta[1:].reshape(L, L) -!ec - -A way of looking at the coefficients in $J$ is to plot the matrices as images. - - -!bc pycod -fig = plt.figure(figsize=(20, 14)) -im = plt.imshow(J, **cmap_args) -plt.title("OLS", fontsize=18) -plt.xticks(fontsize=18) -plt.yticks(fontsize=18) -cb = fig.colorbar(im) -cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18) -plt.show() -!ec -It is interesting to note that OLS -considers both $J_{j, j + 1} = -0.5$ and $J_{j, j - 1} = -0.5$ as -valid matrix elements for $J$. -In our discussion below on hyperparameters and Ridge and Lasso regression we will see that -this problem can be removed, partly and only with Lasso regression. - -In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD? - - - -===== Linear Regression Problems ===== - -One of the typical problems we encounter with linear regression, in particular -when the matrix $\bm{X}$ (our so-called design matrix) is high-dimensional, -are problems with near singular or singular matrices. The column vectors of $\bm{X}$ -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[ -\begin{array}{rrr} -1 & -1 & 2 -\\ -1 & 0 & 1 -\\ -1 & 2 & -1 -\\ -1 & 1 & 0 -\end{array} \right] -\end{align*} -!et - -The columns of $\bm{X}$ are linearly dependent. We see 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 $\bm{X}^T\bm{x}$ (the matrix we need to 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*} -\bm{X} & = \left[ -\begin{array}{rr} -1 & -1 -\\ -1 & -1 -\end{array} \right]. -\end{align*} -!et -We see easily that $\mbox{det}(\bm{X}) = x_{11} x_{22} - x_{12} x_{21} = 1 \times (-1) - 1 \times (-1) = 0$. Hence, $\mathbf{X}$ is singular and its inverse is undefined. -This is equivalent to saying that the matrix $\bm{X}$ has at least an eigenvalue which is zero. - - - -===== Fixing the singularity ===== - -If our design matrix $\bm{X}$ which enters the linear regression problem -!bt -\begin{align} -\bm{\beta} & = (\bm{X}^{T} \bm{X})^{-1} \bm{X}^{T} \bm{y}, -\end{align} -!et -has linearly dependent column vectors, we will not be able to compute the inverse -of $\bm{X}^T\bm{X}$ and we cannot find the parameters (estimators) $\beta_i$. -The estimators are only well-defined if $(\bm{X}^{T}\bm{X})^{-1}$ exits. -This is more likely to happen when the matrix $\bm{X}$ is high-dimensional. In this case it is likely to encounter a situation where -the regression parameters $\beta_i$ cannot be estimated. - -A cheap *ad hoc* approach is simply to add a small diagonal component to the matrix to invert, that is we change -!bt -\[ -\bm{X}^{T} \bm{X} \rightarrow \bm{X}^{T} \bm{X}+\lambda \bm{I}, -\] -!et -where $\bm{I}$ is the identity matrix. When we discuss _Ridge_ regression this is actually what we end up evaluating. The parameter $\lambda$ is called a hyperparameter. More about this later. - - - - -===== Basic math of the SVD ===== - - -From standard linear algebra we know that a square matrix $\bm{X}$ can be diagonalized if and only it is -a so-called "normal matrix":"https://en.wikipedia.org/wiki/Normal_matrix", that is if $\bm{X}\in {\mathbb{R}}^{n\times n}$ -we have $\bm{X}\bm{X}^T=\bm{X}^T\bm{X}$ or if $\bm{X}\in {\mathbb{C}}^{n\times n}$ we have $\bm{X}\bm{X}^{\dagger}=\bm{X}^{\dagger}\bm{X}$. -The matrix has then a set of eigenpairs - -!bt -\[ -(\lambda_1,\bm{u}_1),\dots, (\lambda_n,\bm{u}_n), -!et -and the eigenvalues are given by the diagonal matrix -!bt -\[ -\bm{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n). -\] -!et -The matrix $\bm{X}$ can be written in terms of an orthogonal/unitary transformation $\bm{U}$ -!bt -\[ -\bm{X} = \bm{U}\bm{\Sigma}\bm{V}^T, -\] -!et -with $\bm{U}\bm{U}^T=\bm{I}$ or $\bm{U}\bm{U}^{\dagger}=\bm{I}$. - -Not all square matrices are diagonalizable. A matrix like the one discussed above -!bt -\[ -\bm{X} = \begin{bmatrix} -1& -1 \\ -1& -1\\ -\end{bmatrix} -\] -!et -is not diagonalizable, it is a so-called "defective matrix":"https://en.wikipedia.org/wiki/Defective_matrix". It is easy to see that the condition -$\bm{X}\bm{X}^T=\bm{X}^T\bm{X}$ is not fulfilled. - - - -===== The SVD, a Fantastic Algorithm ===== - - -However, and this is the strength of the SVD algorithm, any general -matrix $\bm{X}$ can be decomposed in terms of a diagonal matrix and -two orthogonal/unitary matrices. The "Singular Value Decompostion -(SVD) theorem":"https://en.wikipedia.org/wiki/Singular_value_decomposition" -states that a general $m\times n$ matrix $\bm{X}$ can be written in -terms of a diagonal matrix $\bm{\Sigma}$ of dimensionality $n\times n$ -and two orthognal matrices $\bm{U}$ and $\bm{V}$, where the first has -dimensionality $m \times m$ and the last dimensionality $n\times n$. -We have then - -!bt -\[ -\bm{X} = \bm{U}\bm{\Sigma}\bm{V}^T -\] -!et - -As an example, the above defective matrix can be decomposed as - -!bt -\[ -\bm{X} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1& 1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 0\\ \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -1 \\ 1& 1\\ \end{bmatrix}=\bm{U}\bm{\Sigma}\bm{V}^T, -\] -!et - -with eigenvalues $\sigma_1=2$ and $\sigma_2=0$. -The SVD exits always! - - - -===== Another Example ===== - -Consider the following matrix which can be SVD decomposed as - -!bt -\[ -\bm{X} = \frac{1}{15}\begin{bmatrix} 14 & 2\\ 4 & 22\\ 16 & 13\end{matrix}=\frac{1}{3}\begin{bmatrix} 1& 2 & 2 \\ 2& -1 & 1\\ 2 & 1& -2\end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 1\\ 0 & 0\end{bmatrix}\frac{1}{5}\begin{bmatrix} 3& 4 \\ 4& -3\end{bmatrix}=\bm{U}\bm{\Sigma}\bm{V}^T. -\] -!et - -This is a $3\times 2$ matrix which is decomposed in terms of a -$3\times 3$ matrix $\bm{U}$, and a $2\times 2$ matrix $\bm{V}$. It is easy to see -that $\bm{U}$ and $\bm{V}$ are orthogonal (how?). - -And the SVD -decomposition (singular values) gives eigenvalues -$\sigma_i\geq\sigma_{i+1}$ for all $i$ and for dimensions larger than $i=2$, the -eigenvalues (singular values) are zero. - -In the general case, where our design matrix $\bm{X}$ has dimension -$n\times p$, the matrix is thus decomposed into an $n\times n$ -orthogonal matrix $\bm{U}$, a $p\times p$ orthogonal matrix $\bm{V}$ -and a diagonal matrix $\bm{\Sigma}$ with $r=\mathrm{min}(n,p)$ -singular values $\sigma_i\lg 0$ on the main diagonal and zeros filling -the rest of the matrix. There are at most $p$ singular values -assuming that $n > p$. In our regression examples for the nuclear -masses and the equation of state this is indeed the case, while for -the Ising model we have $p > n$. These are often cases that lead to -near singular or singular matrices. - -The columns of $\bm{U}$ are called the left singular vectors while the columns of $\bm{V}$ are the right singular vectors. - - -===== Economy-size SVD ===== - -If we assume that $n > p$, then our matrix $\bm{U}$ has dimension $n -\times n$. The last $n-p$ columns of $\bm{U}$ become however -irrelevant in our calculations since they are multiplied with the -zeros in $\bm{\Sigma}$. - -The economy-size decomposition removes extra rows or columns of zeros -from the diagonal matrix of singular values, $\bm{\Sigma}$, along with the columns -in either $\bm{U}$ or $\bm{V}$ that multiply those zeros in the expression. -Removing these zeros and columns can improve execution time -and reduce storage requirements without compromising the accuracy of -the decomposition. - -If $n > p$, we keep only the first $p$ columns of $\bm{U}$ and $\bm{\Sigma}$ has dimension $p\times p$. -If $p > n$, then only the first $n$ columns of $\bm{V}$ are computed and $\bm{\Sigma}$ has dimension $n\times n$. -The $n=p$ case is obvious, we retain the full SVD. -In general the economy-size SVD leads to less FLOPS and still conserving the desired accuracy. - - -===== Mathematical Properties ===== - -There are several interesting mathematical properties which will be -relevant when we are going to discuss the differences between say -ordinary least squares (OLS) and _Ridge_ regression. - -We have from OLS that the parameters of the linear approximation are given by -!bt -\[ -\bm{\tilde{y}} = \bm{X}\bm{\beta} = \bm{X}\left(\bm{X}^T\bm{X}\right)^{-1}\bm{X}^T\bm{y}. -\] -!et - -The matrix to invert can be rewritten in terms of our SVD decomposition as - -!bt -\[ -\bm{X}^T\bm{X} = \bm{V}\bm{\Sigma}^T\bm{U}^T\bm{U}\bm{\Sigma}\bm{V}^T. -\] -!et -Using the orthogonality properties of $\bm{U}$ we have - -!bt -\[ -\bm{X}^T\bm{X} = \bm{V}\bm{\Sigma}^T\bm{\Sigma}\bm{V}^T = \bm{V}\bm{D}\bm{V}^T, -\] -!et -with $\bm{D}$ being a diagonal matrix with values along the diagonal given by the singular values squared. - -This means that -!bt -\[ -(\bm{X}^T\bm{X})\bm{V} = \bm{V}\bm{D}, -\] -!et -that is the eigenvectors of $(\bm{X}^T\bm{X})$ are given by the columns of the right singular matrix of $\bm{X}$ and the eigenvalues are the squared singular values. It is easy to show (show this) that -!bt -\[ -(\bm{X}\bm{X}^T)\bm{U} = \bm{U}\bm{D}, -\] -!et -that is, the eigenvectors of $(\bm{X}\bm{X})^T$ are the columns of the left singular matrix and the eigenvalues are the same. - -Going back to our OLS equation we have -!bt -\[ -\bm{X}\bm{\beta} = \bm{X}\left(\bm{V}\bm{D}\bm{V}^T \right)^{-1}\bm{X}^T\bm{y}=\bm{U\Sigma V^T}\left(\bm{V}\bm{D}\bm{V}^T \right)^{-1}(\bm{U\Sigma V^T})^T\bm{y}=\bm{U}\bm{U}^T\bm{y}. -\] -!et -We will come back to this expression when we discuss Ridge regression. - - - -===== Ridge and LASSO Regression ===== - -Let us remind ourselves about the expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is -our optimization problem is -!bt -\[ -{\displaystyle \min_{\bm{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\beta}\right)^T\left(\bm{y}-\bm{X}\bm{\beta}\right)\right\}. -\] -!et -or we can state it as -!bt -\[ -{\displaystyle \min_{\bm{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2, -\] -!et -where we have used the definition of a norm-2 vector, that is -!bt -\[ -\vert\vert \bm{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. -\] -!et - -By minimizing the above equation with respect to the parameters -$\bm{\beta}$ we could then obtain an analytical expression for the -parameters $\bm{\beta}$. We can add a regularization parameter $\lambda$ by -defining a new cost function to be optimized, that is - -!bt -\[ -{\displaystyle \min_{\bm{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_2^2 -\] -!et - -which leads to the Ridge regression minimization problem where we -require that $\vert\vert \bm{\beta}\vert\vert_2^2\le t$, where $t$ is -a finite number larger than zero. By defining - -!bt -\[ -C(\bm{X},\bm{\beta})=\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_1, -\] -!et - -we have a new optimization equation -!bt -\[ -{\displaystyle \min_{\bm{\beta}\in -{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \bm{y}-\bm{X}\bm{\beta}\vert\vert_2^2+\lambda\vert\vert \bm{\beta}\vert\vert_1 -\] -!et -which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator. - -Here we have defined the norm-1 as -!bt -\[ -\vert\vert \bm{x}\vert\vert_1 = \sum_i \vert x_i\vert. -\] -!et - -Using the matrix-vector expression for Ridge regression, - -!bt -\[ -C(\bm{X},\bm{\beta})=\frac{1}{n}\left\{(\bm{y}-\bm{X}\bm{\beta})^T(\bm{y}-\bm{X}\bm{\beta})\right\}+\lambda\bm{\beta}^T\bm{\beta}, -\] -!et - -by taking the derivatives with respect to $\bm{\beta}$ we obtain then -a slightly modified matrix inversion problem which for finite values -of $\lambda$ does not suffer from singularity problems. We obtain - -!bt -\[ -\bm{\beta}^{\mathrm{Ridge}} = \left(\bm{X}^T\bm{X}+\lambda\bm{I}\right)^{-1}\bm{X}^T\bm{y}, -\] -!et - -with $\bm{I}$ being a $p\times p$ identity matrix with the constraint that - -!bt -\[ -\sum_{i=0}^{p-1} \beta_i^2 \leq t, -\] -!et - -with $t$ a finite positive number. - -We see that Ridge regression is nothing but the standard -OLS with a modified diagonal term added to $\bm{X}^T\bm{X}$. The -consequences, in particular for our discussion of the bias-variance -are rather interesting. - -Furthermore, if we use the result above in terms of the SVD decomposition (our analysis was done for the OLS method), we had -!bt -\[ -(\bm{X}\bm{X}^T)\bm{U} = \bm{U}\bm{D}. -\] -!et - -We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix $\bm{U}$ as -!bt -\[ -\bm{X}\bm{\beta} = \bm{X}\left(\bm{V}\bm{D}\bm{V}^T \right)^{-1}\bm{X}^T\bm{y}=\bm{U\Sigma V^T}\left(\bm{V}\bm{D}\bm{V}^T \right)^{-1}(\bm{U\Sigma V^T})^T\bm{y}=\bm{U}\bm{U}^T\bm{y} -\] -!et - - -For Ridge regression this becomes - -!bt -\[ -\bm{X}\bm{\beta}^{\mathrm{Ridge}} = \bm{U\Sigma V^T}\left(\bm{V}\bm{D}\bm{V}^T+\lambda\bm{I} \right)^{-1}(\bm{U\Sigma V^T})^T\bm{y}=\sum_{j=0}^{p-1}\bm{u}_j\bm{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\bm{y}, -\] -!et - -with the vectors $\bm{u}_j$ being the columns of $\bm{U}$. - -===== Interpreting the Ridge results ===== - -Since $\lambda \geq 0$, it means that compared to OLS, we have - -!bt -\[ -\frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1. -\] -!et - -Ridge regression finds the coordinates of $\bm{y}$ with respect to the -orthonormal basis $\bm{U}$, it then shrinks the coordinates by -$\frac{\sigma_j^2}{\sigma_j^2+\lambda}$. Recall that the SVD has -eigenvalues ordered in a descending way, that is $\sigma_i \geq -\sigma_{i+1}$. - -For small eigenvalues $\sigma_i$ it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. -Actually, calculating the variance of $\bm{X}\bm{v}_j$ shows that this quantity is equal to $\sigma_j^2/n$. -With a parameter $\lambda$ we can thus shrink the role of specific parameters. - - -For the sake of simplicity, let us assume that the design matrix is orthonormal, that is - -!bt -\[ -\bm{X}^T\bm{X}=(\bm{X}^T\bm{X})^{-1} =\bm{I}. -\] -!et - -In this case the standard OLS results in -!bt -\[ -\bm{\beta}^{\mathrm{OLS}} = \bm{X}^T\bm{y}=\sum_{i=0}^{p-1}\bm{u}_j\bm{u}_j^T\bm{y}, -\] -!et - -and - -!bt -\[ -\bm{\beta}^{\mathrm{Ridge}} = \left(\bm{I}+\lambda\bm{I}\right)^{-1}\bm{X}^T\bm{y}=\left(1+\lambda\right)^{-1}\bm{\beta}^{\mathrm{OLS}}, -\] -!et - -that is the Ridge estimator scales the OLS estimator by the inverse of a factor $1+\lambda$, and -the Ridge estimator converges to zero when the hyperparameter goes to -infinity. - -We will come back to more interpreations after we have gone through some of the statistical analysis part. - -For more discussions of Ridge and Lasso regression, "Wessel van Wieringen's":"https://arxiv.org/abs/1509.09169" article is highly recommended. -Similarly, "Mehta et al's article":"https://arxiv.org/abs/1803.08823" is also recommended. - -===== Where are we going? ===== - -Before we proceed, we need to rethink what we have been doing. In our -eager to fit the data, we have omitted several important elements in -our regression analysis. In what follows we will -o look at statistical properties, including a discussion of mean values, variance and the so-called bias-variance tradeoff -o introduce resampling techniques like cross-validation, bootstrapping and jackknife and more - -This will allow us to link the standard linear algebra methods we have discussed above to a statistical interpretation of the methods. - - - -===== 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, 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. - -===== Why resampling methods ? ===== - -* Our simulations can be treated as *computer experiments*. This is particularly the case for Monte Carlo methods -* The results can be analysed with the same statistical tools as we would use analysing experimental data. -* As in all experiments, we are looking for expectation values and an estimate of how accurate they are, i.e., possible sources for errors. - - -* As in other experiments, many numerical experiments have two classes of errors: - * Statistical errors - * Systematical errors -* Statistical errors can be estimated using standard tools from statistics -* Systematical errors are method specific and must be treated differently from case to case. - -===== Statistics ===== - -The *probability distribution function (PDF)* is a function -$p(x)$ on the domain which, in the discrete case, gives us the -probability or relative frequency with which these values of $X$ occur: -!bt -\[ -p(x) = \mathrm{prob}(X=x) -\] -!et -In the continuous case, the PDF does not directly depict the -actual probability. Instead we define the probability for the -stochastic variable to assume any value on an infinitesimal interval -around $x$ to be $p(x)dx$. The continuous function $p(x)$ then gives us -the *density* of the probability rather than the probability -itself. The probability for a stochastic variable to assume any value -on a non-infinitesimal interval $[a,\,b]$ is then just the integral: -!bt -\[ -\mathrm{prob}(a\leq X\leq b) = \int_a^b p(x)dx -\] -!et -Qualitatively speaking, a stochastic variable represents the values of -numbers chosen as if by chance from some specified PDF so that the -selection of a large set of these numbers reproduces this PDF. - -A particularly useful class of special expectation values are the -*moments*. The $n$-th moment of the PDF $p$ is defined as -follows: -!bt -\[ -\langle x^n\rangle \equiv \int\! x^n p(x)\,dx -\] -!et -The zero-th moment $\langle 1\rangle$ is just the normalization condition of -$p$. The first moment, $\langle x\rangle$, is called the *mean* of $p$ -and often denoted by the letter $\mu$: -!bt -\[ -\langle x\rangle = \mu \equiv \int\! x p(x)\,dx -\] -!et - -A special version of the moments is the set of *central moments*, -the n-th central moment defined as: -!bt -\[ -\langle (x-\langle x \rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx -\] -!et -The zero-th and first central moments are both trivial, equal $1$ and -$0$, respectively. But the second central moment, known as the -*variance* of $p$, is of particular interest. For the stochastic -variable $X$, the variance is denoted as $\sigma^2_X$ or $\mathrm{var}(X)$: -!bt -\begin{align} -\sigma^2_X\ \ =\ \ \mathrm{var}(X) & = \langle (x-\langle x\rangle)^2\rangle = -\int\! (x-\langle x\rangle)^2 p(x)\,dx\\ -& = \int\! \left(x^2 - 2 x \langle x\rangle^{2} + - \langle x\rangle^2\right)p(x)\,dx\\ -& = \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2\\ -& = \langle x^2\rangle - \langle x\rangle^2 -\end{align} -!et -The square root of the variance, $\sigma =\sqrt{\langle (x-\langle x\rangle)^2\rangle}$ is called the *standard deviation* of $p$. It is clearly just the RMS (root-mean-square) -value of the deviation of the PDF from its mean value, interpreted -qualitatively as the *spread* of $p$ around its mean. - - - -===== Statistics, covariance ===== - -Another important quantity is the so called covariance, a variant of -the above defined variance. Consider again the set $\{X_i\}$ of $n$ -stochastic variables (not necessarily uncorrelated) with the -multivariate PDF $P(x_1,\dots,x_n)$. The *covariance* of two -of the stochastic variables, $X_i$ and $X_j$, is defined as follows: -!bt -\begin{align} -\mathrm{cov}(X_i,\,X_j) &\equiv \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle -\nonumber\\ -&= -\int\!\cdots\!\int\!(x_i-\langle x_i \rangle)(x_j-\langle x_j \rangle)\, -P(x_1,\dots,x_n)\,dx_1\dots dx_n -label{eq:def_covariance} -\end{align} -!et -with -!bt -\[ -\langle x_i\rangle = -\int\!\cdots\!\int\!x_i\,P(x_1,\dots,x_n)\,dx_1\dots dx_n -\] -!et - -If we consider the above covariance as a matrix $C_{ij}=\mathrm{cov}(X_i,\,X_j)$, then the diagonal elements are just the familiar -variances, $C_{ii} = \mathrm{cov}(X_i,\,X_i) = \mathrm{var}(X_i)$. It turns out that -all the off-diagonal elements are zero if the stochastic variables are -uncorrelated. This is easy to show, keeping in mind the linearity of -the expectation value. Consider the stochastic variables $X_i$ and -$X_j$, ($i\neq j$): -!bt -\begin{align} -\mathrm{cov}(X_i,\,X_j) &= \langle(x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle\\ -&=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle \\ -&=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j\rangle + -\langle \langle x_i\rangle\langle x_j\rangle\rangle\\ -&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + -\langle x_i\rangle\langle x_j\rangle\\ -&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle -\end{align} -!et - -===== Statistics, independent variables ===== - -If $X_i$ and $X_j$ are independent, we get -$\langle x_i x_j\rangle =\langle x_i\rangle\langle x_j\rangle$, resulting in $\mathrm{cov}(X_i, X_j) = 0\ \ (i\neq j)$. - -Also useful for us is the covariance of linear combinations of -stochastic variables. Let $\{X_i\}$ and $\{Y_i\}$ be two sets of -stochastic variables. Let also $\{a_i\}$ and $\{b_i\}$ be two sets of -scalars. Consider the linear combination: -!bt -\[ -U = \sum_i a_i X_i \qquad V = \sum_j b_j Y_j -\] -!et -By the linearity of the expectation value -!bt -\[ -\mathrm{cov}(U, V) = \sum_{i,j}a_i b_j \mathrm{cov}(X_i, Y_j) -\] -!et - -Now, since the variance is just $\mathrm{var}(X_i) = \mathrm{cov}(X_i, X_i)$, we get -the variance of the linear combination $U = \sum_i a_i X_i$: -!bt -\begin{equation} -\mathrm{var}(U) = \sum_{i,j}a_i a_j \mathrm{cov}(X_i, X_j) -label{eq:variance_linear_combination} -\end{equation} -!et -And in the special case when the stochastic variables are -uncorrelated, the off-diagonal elements of the covariance are as we -know zero, resulting in: -!bt -\[ -\mathrm{var}(U) = \sum_i a_i^2 \mathrm{cov}(X_i, X_i) = \sum_i a_i^2 \mathrm{var}(X_i) -\] -!et -!bt -\[ -\mathrm{var}(\sum_i a_i X_i) = \sum_i a_i^2 \mathrm{var}(X_i) -\] -!et -which will become very useful in our study of the error in the mean -value of a set of measurements. - -===== Statistics and stochastic processes ===== - -A *stochastic process* is a process that produces sequentially a -chain of values: -!bt -\[ -\{x_1, x_2,\dots\,x_k,\dots\}. -\] -!et -We will call these -values our *measurements* and the entire set as our measured -*sample*. The action of measuring all the elements of a sample -we will call a stochastic *experiment* since, operationally, -they are often associated with results of empirical observation of -some physical or mathematical phenomena; precisely an experiment. We -assume that these values are distributed according to some -PDF $p_X^{\phantom X}(x)$, where $X$ is just the formal symbol for the -stochastic variable whose PDF is $p_X^{\phantom X}(x)$. Instead of -trying to determine the full distribution $p$ we are often only -interested in finding the few lowest moments, like the mean -$\mu_X^{\phantom X}$ and the variance $\sigma_X^{\phantom X}$. - -In practical situations a sample is always of finite size. Let that -size be $n$. The expectation value of a sample, the *sample mean*, is then defined as follows: -!bt -\[ -\bar{x}_n \equiv \frac{1}{n}\sum_{k=1}^n x_k -\] -!et -The *sample variance* is: -!bt -\[ -\mathrm{var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_k - \bar{x}_n)^2 -\] -!et -its square root being the *standard deviation of the sample*. The -*sample covariance* is: -!bt -\[ -\mathrm{cov}(x)\equiv\frac{1}{n}\sum_{kl}(x_k - \bar{x}_n)(x_l - \bar{x}_n) -\] -!et - -Note that the sample variance is the sample covariance without the -cross terms. In a similar manner as the covariance in Eq.~(ref{eq:def_covariance}) is a measure of the correlation between -two stochastic variables, the above defined sample covariance is a -measure of the sequential correlation between succeeding measurements -of a sample. - -These quantities, being known experimental values, differ -significantly from and must not be confused with the similarly named -quantities for stochastic variables, mean $\mu_X$, variance $\mathrm{var}(X)$ -and covariance $\mathrm{cov}(X,Y)$. - -The law of large numbers -states that as the size of our sample grows to infinity, the sample -mean approaches the true mean $\mu_X^{\phantom X}$ of the chosen PDF: -!bt -\[ -\lim_{n\to\infty}\bar{x}_n = \mu_X^{\phantom X} -\] -!et -The sample mean $\bar{x}_n$ works therefore as an estimate of the true -mean $\mu_X^{\phantom X}$. - -What we need to find out is how good an approximation $\bar{x}_n$ is to -$\mu_X^{\phantom X}$. In any stochastic measurement, an estimated -mean is of no use to us without a measure of its error. A quantity -that tells us how well we can reproduce it in another experiment. We -are therefore interested in the PDF of the sample mean itself. Its -standard deviation will be a measure of the spread of sample means, -and we will simply call it the *error* of the sample mean, or -just sample error, and denote it by $\mathrm{err}_X^{\phantom X}$. In -practice, we will only be able to produce an *estimate* of the -sample error since the exact value would require the knowledge of the -true PDFs behind, which we usually do not have. - -===== Statistics, more on sample error ===== - -Let us first take a look at what happens to the sample error as the -size of the sample grows. In a sample, each of the measurements $x_i$ -can be associated with its own stochastic variable $X_i$. The -stochastic variable $\overline X_n$ for the sample mean $\bar{x}_n$ is -then just a linear combination, already familiar to us: -!bt -\[ -\overline X_n = \frac{1}{n}\sum_{i=1}^n X_i -\] -!et -All the coefficients are just equal $1/n$. The PDF of $\overline X_n$, -denoted by $p_{\overline X_n}(x)$ is the desired PDF of the sample -means. - -The probability density of obtaining a sample mean $\bar x_n$ -is the product of probabilities of obtaining arbitrary values $x_1, -x_2,\dots,x_n$ with the constraint that the mean of the set $\{x_i\}$ -is $\bar x_n$: -!bt -\[ -p_{\overline X_n}(x) = \int p_X^{\phantom X}(x_1)\cdots -\int p_X^{\phantom X}(x_n)\ -\delta\!\left(x - \frac{x_1+x_2+\dots+x_n}{n}\right)dx_n \cdots dx_1 -\] -!et -And in particular we are interested in its variance $\mathrm{var}(\overline X_n)$. - -===== Statistics, central limit theorem ===== - -It is generally not possible to express $p_{\overline X_n}(x)$ in a -closed form given an arbitrary PDF $p_X^{\phantom X}$ and a number -$n$. But for the limit $n\to\infty$ it is possible to make an -approximation. The very important result is called *the central limit theorem*. It tells us that as $n$ goes to infinity, -$p_{\overline X_n}(x)$ approaches a Gaussian distribution whose mean -and variance equal the true mean and variance, $\mu_{X}^{\phantom X}$ -and $\sigma_{X}^{2}$, respectively: -!bt -\begin{equation} -\lim_{n\to\infty} p_{\overline X_n}(x) = -\left(\frac{n}{2\pi\mathrm{var}(X)}\right)^{1/2} -e^{-\frac{n(x-\bar x_n)^2}{2\mathrm{var}(X)}} -label{eq:central_limit_gaussian} -\end{equation} -!et - - -The desired variance -$\mathrm{var}(\overline X_n)$, i.e. the sample error squared -$\mathrm{err}_X^2$, is given by: -!bt -\begin{equation} -\mathrm{err}_X^2 = \mathrm{var}(\overline X_n) = \frac{1}{n^2} -\sum_{ij} \mathrm{cov}(X_i, X_j) -label{eq:error_exact} -\end{equation} -!et -We see now that in order to calculate the exact error of the sample -with the above expression, we would need the true means -$\mu_{X_i}^{\phantom X}$ of the stochastic variables $X_i$. To -calculate these requires that we know the true multivariate PDF of all -the $X_i$. But this PDF is unknown to us, we have only got the measurements of -one sample. The best we can do is to let the sample itself be an -estimate of the PDF of each of the $X_i$, estimating all properties of -$X_i$ through the measurements of the sample. - -Our estimate of $\mu_{X_i}^{\phantom X}$ is then the sample mean $\bar x$ -itself, in accordance with the the central limit theorem: -!bt -\[ -\mu_{X_i}^{\phantom X} = \langle x_i\rangle \approx \frac{1}{n}\sum_{k=1}^n x_k = \bar x -\] -!et -Using $\bar x$ in place of $\mu_{X_i}^{\phantom X}$ we can give an -*estimate* of the covariance in Eq.~(ref{eq:error_exact}) -!bt -\[ -\mathrm{cov}(X_i, X_j) = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle -\approx\langle (x_i - \bar x)(x_j - \bar{x})\rangle, -\] -!et -resulting in -!bt -\[ -\frac{1}{n} \sum_{l}^n \left(\frac{1}{n}\sum_{k}^n (x_k -\bar x_n)(x_l - \bar x_n)\right)=\frac{1}{n}\frac{1}{n} \sum_{kl} (x_k -\bar x_n)(x_l - \bar x_n)=\frac{1}{n}\mathrm{cov}(x) -\] -!et - -By the same procedure we can use the sample variance as an -estimate of the variance of any of the stochastic variables $X_i$ -!bt -\[ -\mathrm{var}(X_i)=\langle x_i - \langle x_i\rangle\rangle \approx \langle x_i - \bar x_n\rangle\nonumber, -\] -!et -which is approximated as -!bt -\begin{equation} -\mathrm{var}(X_i)\approx \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)=\mathrm{var}(x) -label{eq:var_estimate_i_think} -\end{equation} -!et - -Now we can calculate an estimate of the error -$\mathrm{err}_X^{\phantom X}$ of the sample mean $\bar x_n$: -!bt -\begin{align} -\mathrm{err}_X^2 -&=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) \nonumber \\ -&\approx&\frac{1}{n^2}\sum_{ij}\frac{1}{n}\mathrm{cov}(x) =\frac{1}{n^2}n^2\frac{1}{n}\mathrm{cov}(x)\nonumber\\ -&=\frac{1}{n}\mathrm{cov}(x) -label{eq:error_estimate} -\end{align} -!et -which is nothing but the sample covariance divided by the number of -measurements in the sample. - -In the special case that the measurements of the sample are -uncorrelated (equivalently the stochastic variables $X_i$ are -uncorrelated) we have that the off-diagonal elements of the covariance -are zero. This gives the following estimate of the sample error: -!bt -\[ -\mathrm{err}_X^2=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) = -\frac{1}{n^2} \sum_i \mathrm{var}(X_i), -\] -!et -resulting in -!bt -\begin{equation} -\mathrm{err}_X^2\approx \frac{1}{n^2} \sum_i \mathrm{var}(x)= \frac{1}{n}\mathrm{var}(x) -label{eq:error_estimate_uncorrel} -\end{equation} -!et -where in the second step we have used Eq.~(ref{eq:var_estimate_i_think}). -The error of the sample is then just its standard deviation divided by -the square root of the number of measurements the sample contains. -This is a very useful formula which is easy to compute. It acts as a -first approximation to the error, but in numerical experiments, we -cannot overlook the always present correlations. - -For computational purposes one usually splits up the estimate of -$\mathrm{err}_X^2$, given by Eq.~(ref{eq:error_estimate}), into two -parts -!bt -\[ -\mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x) + \frac{1}{n}(\mathrm{cov}(x)-\mathrm{var}(x)), -\] -!et -which equals -!bt -\begin{equation} -\frac{1}{n^2}\sum_{k=1}^n (x_k - \bar x_n)^2 +\frac{2}{n^2}\sum_{k 0$. We say then that the ridge estimator is biased. - -We can also compute the variance as - -!bt -\[ -\mbox{Var}[\bm{\beta}^{\mathrm{Ridge}}]=\sigma^2[ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1} \mathbf{X}^{T} \mathbf{X} \{ [ \mathbf{X}^{\top} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}, -\] -!et -and it is easy to see that if the parameter $\lambda$ goes to infinity then the variance of Ridge parameters $\bm{\beta}$ goes to zero. - -With this, we can compute the difference - -!bt -\[ -\mbox{Var}[\bm{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\bm{\beta}^{\mathrm{Ridge}})=\sigma^2 [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}[ 2\lambda\mathbf{I} + \lambda^2 (\mathbf{X}^{T} \mathbf{X})^{-1} ] \{ [ \mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I} ]^{-1}\}^{T}. -\] -!et -The difference is non-negative definite since each component of the -matrix product is non-negative definite. -This means the variance we obtain with the standard OLS will always for $\lambda > 0$ be larger than the variance of $\bm{\beta}$ obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below. - - -===== Cross-validation ===== - -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 ===== - -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 ===== - -* 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 $\bm{\sigma}_{-i}^2(\lambda)$, as -!bt -\begin{align*} -\bm{\beta}_{-i}(\lambda) & = ( \bm{X}_{-i, \ast}^{T} -\bm{X}_{-i, \ast} + \lambda \bm{I}_{pp})^{-1} -\bm{X}_{-i, \ast}^{T} \bm{y}_{-i} -\end{align*} -!et - -* Evaluate the prediction performance of these models on the test set by $\log\{L[y_i, \bm{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}$. Or, by the prediction error $|y_i - \bm{X}_{i, \ast} \bm{\beta}_{-i}(\lambda)|$, the relative error, the error squared or the R2 score function. - -* 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 -!bt -\begin{align*} -\frac{1}{n} \sum_{i = 1}^n \log\{L[y_i, \mathbf{X}_{i, \ast}; \bm{\beta}_{-i}(\lambda), \bm{\sigma}_{-i}^2(\lambda)]\}. -\end{align*} -!et - -* 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. - - - - -===== Resampling methods: Jackknife and Bootstrap ===== - -Two famous -resampling methods are the _independent bootstrap_ and _the jackknife_. - -The jackknife is a special case of the independent bootstrap. Still, the jackknife was made -popular prior to the independent bootstrap. And as the popularity of -the independent bootstrap soared, new variants, such as _the dependent bootstrap_. - -The Jackknife and independent bootstrap work for -independent, identically distributed random variables. -If these conditions are not -satisfied, the methods will fail. 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. - - -===== Resampling methods: Jackknife ===== - -The Jackknife works by making many replicas of the estimator $\widehat{\theta}$. -The jackknife is a resampling method where we systematically leave out one observation from the vector of observed values $\bm{x} = (x_1,x_2,\cdots,X_n)$. -Let $\bm{x}_i$ denote the vector -!bt -\[ -\bm{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n), -\] -!et - -which equals the vector $\bm{x}$ with the exception that observation -number $i$ is left out. Using this notation, define -$\widehat{\theta}_i$ to be the estimator -$\widehat{\theta}$ computed using $\vec{X}_i$. - - - -===== Jackknife code example ===== -!bc pycod -from numpy import * -from numpy.random import randint, randn -from time import time - -def jackknife(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) ) - - # analysis - print("Runtime: %g sec" % (time()-t0)); print("Jackknife Statistics :") - print("original bias std. error") - print("%8g %14g %15g" % (stat(data),(n-1)*mean(t)/n, (n*var(t))**.5)) - - return t - - -# Returns mean of data samples -def stat(data): - return mean(data) - - -mu, sigma = 100, 15 -datapoints = 10000 -x = mu + sigma*random.randn(datapoints) -# jackknife returns the data sample -t = jackknife(x, stat) - -!ec - - - -===== Resampling methods: 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: -o The bootstrap is quite general, although there are some cases in which it fails. -o 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. -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). - - - - -===== Resampling methods: Bootstrap background ===== - -Since $\widehat{\theta} = \widehat{\theta}(\bm{X})$ is a function of random variables, -$\widehat{\theta}$ itself must be a random variable. Thus it has -a pdf, call this function $p(\bm{t})$. The aim of the bootstrap is to -estimate $p(\bm{t})$ by the relative frequency of -$\widehat{\theta}$. You can think of this as using a histogram -in the place of $p(\bm{t})$. If the relative frequency closely -resembles $p(\vec{t})$, then using numerics, it is straight forward to -estimate all the interesting parameters of $p(\bm{t})$ using point -estimators. - - - -===== Resampling methods: More Bootstrap background ===== - -In the case that $\widehat{\theta}$ has -more than one component, and the components are independent, we 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: -o Drawing lots of numbers from $p(x)$, suppose we call one such set of numbers $(X_1^*, X_2^*, \cdots, X_n^*)$. -o Then using these numbers, we could compute a replica of $\widehat{\theta}$ called $\widehat{\theta}^*$. - -By repeated use of (1) and (2), many -estimates of $\widehat{\theta}$ could have been obtained. The -idea is to use the relative frequency of $\widehat{\theta}^*$ -(think of a histogram) as an estimate of $p(\bm{t})$. - - -===== Resampling methods: Bootstrap approach ===== - -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, "Efron in 1979":"https://projecteuclid.org/euclid.aos/1176344552" asked the -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. - - -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 -$\bm{X}$. - - -===== Resampling methods: Bootstrap steps ===== - -The independent bootstrap works like this: - -o Draw with replacement $n$ numbers for the observed variables $\bm{x} = (x_1,x_2,\cdots,x_n)$. -o Define a vector $\bm{x}^*$ containing the values which were drawn from $\bm{x}$. -o Using the vector $\bm{x}^*$ compute $\widehat{\theta}^*$ by evaluating $\widehat \theta$ under the observations $\bm{x}^*$. -o 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 statistics 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 etsimator $\widehat \sigma^2$ to the values -$\widehat \theta ^*$. - - - -===== Code example for the Bootstrap method ===== - -The following code starts with a Gaussian distribution with mean value -$\mu =100$ and variance $\sigma=15$. We use this to generate the data -used in the bootstrap analysis. The bootstrap analysis returns a data -set after a given number of bootstrap operations (as many as we have -data points). This data set consists of estimated mean values for each -bootstrap operation. The histogram generated by the bootstrap method -shows that the distribution for these mean values is also a Gaussian, -centered around the mean value $\mu=100$ but with standard deviation -$\sigma/\sqrt{n}$, where $n$ is the number of bootstrap samples (in -this case the same as the number of original data points). The value -of the standard deviation is what we expect from the central limit -theorem. - - -!bc pycod -from numpy import * -from numpy.random import randint, randn -from time import time -import matplotlib.mlab as mlab -import matplotlib.pyplot as plt - -# Returns mean of bootstrap samples -def stat(data): - return mean(data) - -# Bootstrap algorithm -def bootstrap(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)]) - - # analysis - print("Runtime: %g sec" % (time()-t0)); print("Bootstrap Statistics :") - print("original bias std. error") - print("%8g %8g %14g %15g" % (statistic(data), std(data),mean(t),std(t))) - return t - - -mu, sigma = 100, 15 -datapoints = 10000 -x = mu + sigma*random.randn(datapoints) -# bootstrap returns the data sample -t = bootstrap(x, stat, datapoints) -# the histogram of the bootstrapped data -n, binsboot, patches = plt.hist(t, 50, normed=1, facecolor='red', alpha=0.75) - -# add a 'best fit' line -y = mlab.normpdf( binsboot, mean(t), std(t)) -lt = plt.plot(binsboot, y, 'r--', linewidth=1) -plt.xlabel('Smarts') -plt.ylabel('Probability') -plt.axis([99.5, 100.6, 0, 3.0]) -plt.grid(True) - -plt.show() - -!ec - - - -===== Code Example for Cross-validation and $k$-fold Cross-validation ===== - -The code here uses Ridge regression with cross-validation (CV) resampling and $k$-fold CV in order to fit a specific polynomial. -!bc pycod -import numpy as np -import matplotlib.pyplot as plt -from sklearn.model_selection import KFold -from sklearn.linear_model import Ridge -from sklearn.model_selection import cross_val_score -from sklearn.preprocessing import PolynomialFeatures - -# A seed just to ensure that the random numbers are the same for every run. -# Useful for eventual debugging. -np.random.seed(3155) - -# Generate the data. -nsamples = 100 -x = np.random.randn(nsamples) -y = 3*x**2 + np.random.randn(nsamples) - -## Cross-validation on Ridge regression using KFold only - -# Decide degree on polynomial to fit -poly = PolynomialFeatures(degree = 6) - -# Decide which values of lambda to use -nlambdas = 500 -lambdas = np.logspace(-3, 5, nlambdas) - -# Initialize a KFold instance -k = 5 -kfold = KFold(n_splits = k) - -# Perform the cross-validation to estimate MSE -scores_KFold = np.zeros((nlambdas, k)) - -i = 0 -for lmb in lambdas: - ridge = Ridge(alpha = lmb) - j = 0 - for train_inds, test_inds in kfold.split(x): - xtrain = x[train_inds] - ytrain = y[train_inds] - - xtest = x[test_inds] - ytest = y[test_inds] - - Xtrain = poly.fit_transform(xtrain[:, np.newaxis]) - ridge.fit(Xtrain, ytrain[:, np.newaxis]) - - Xtest = poly.fit_transform(xtest[:, np.newaxis]) - ypred = ridge.predict(Xtest) - - scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred) - - j += 1 - i += 1 - - -estimated_mse_KFold = np.mean(scores_KFold, axis = 1) - -## Cross-validation using cross_val_score from sklearn along with KFold - -# kfold is an instance initialized above as: -# kfold = KFold(n_splits = k) - -estimated_mse_sklearn = np.zeros(nlambdas) -i = 0 -for lmb in lambdas: - ridge = Ridge(alpha = lmb) - - X = poly.fit_transform(x[:, np.newaxis]) - estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold) - - # cross_val_score return an array containing the estimated negative mse for every fold. - # we have to the the mean of every array in order to get an estimate of the mse of the model - estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds) - - i += 1 - -## Plot and compare the slightly different ways to perform cross-validation - -plt.figure() - -plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score') -plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold') - -plt.xlabel('log10(lambda)') -plt.ylabel('mse') - -plt.legend() - -plt.show() - -!ec - - - -===== The bias-variance tradeoff ===== - - -We will discuss the bias-variance tradeoff in the context of -continuous predictions such as regression. However, many of the -intuitions and ideas discussed here also carry over to classification -tasks. Consider a dataset $\mathcal{L}$ consisting of the data -$\mathbf{X}_\mathcal{L}=\{(y_j, \boldsymbol{x}_j), j=0\ldots n-1\}$. - -Let us assume that the true data is generated from a noisy model - -!bt -\[ -\bm{y}=f(\boldsymbol{x}) + \bm{\epsilon} -\] -!et - -where $\epsilon$ is normally distributed with mean zero and standard deviation $\sigma^2$. - -In our derivation of the ordinary least squares method we defined then -an approximation to the function $f$ in terms of the parameters -$\bm{\beta}$ and the design matrix $\bm{X}$ which embody our model, -that is $\bm{\tilde{y}}=\bm{X}\bm{\beta}$. - -Thereafter we found the parameters $\bm{\beta}$ by optimizing the means squared error via the so-called cost function -!bt -\[ -C(\bm{X},\bm{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]. -\] -!et - -We can rewrite this as -!bt -\[ -\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\bm{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\bm{\tilde{y}}\right])^2+\sigma^2. -\] -!et - -The three terms represent the square of the bias of the learning -method, which can be thought of as the error caused by the simplifying -assumptions built into the method. The second term represents the -variance of the chosen model and finally the last terms is variance of -the error $\bm{\epsilon}$. - -To derive this equation, we need to recall that the variance of $\bm{y}$ and $\bm{\epsilon}$ are both equal to $\sigma^2$. The mean value of $\bm{\epsilon}$ is by definition equal to zero. Furthermore, the function $f$ is not a stochastics variable, idem for $\bm{\tilde{y}}$. -We use a more compact notation in terms of the expectation value -!bt -\[ -\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{f}+\bm{\epsilon}-\bm{\tilde{y}})^2\right], -\] -!et -and adding and subtracting $\mathbb{E}\left[\bm{\tilde{y}}\right]$ we get -!bt -\[ -\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{f}+\bm{\epsilon}-\bm{\tilde{y}}+\mathbb{E}\left[\bm{\tilde{y}}\right]-\mathbb{E}\left[\bm{\tilde{y}}\right])^2\right], -\] -!et -which, using the abovementioned expectation values can be rewritten as -!bt -\[ -\mathbb{E}\left[(\bm{y}-\bm{\tilde{y}})^2\right]=\mathbb{E}\left[(\bm{y}-\mathbb{E}\left[\bm{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\bm{\tilde{y}}\right]+\sigma^2, -\] -!et -that is the rewriting in terms of the so-called bias, the variance of the model $\bm{\tilde{y}}$ and the variance of $\bm{\epsilon}$. - - - - - -===== Example code for Bias-Variance tradeoff ===== -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.linear_model import LinearRegression, Ridge, Lasso -from sklearn.preprocessing import PolynomialFeatures -from sklearn.model_selection import train_test_split -from sklearn.pipeline import make_pipeline -from sklearn.utils import resample - -np.random.seed(2018) - -n = 500 -n_boostraps = 100 -degree = 18 # A quite high value, just to show. -noise = 0.1 - -# Make data set. -x = np.linspace(-1, 3, n).reshape(-1, 1) -y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2) + np.random.normal(0, 0.1, x.shape) - -# Hold out some test data that is never used in training. -x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2) - -# Combine x transformation and model into one operation. -# Not neccesary, but convenient. -model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False)) - -# The following (m x n_bootstraps) matrix holds the column vectors y_pred -# for each bootstrap iteration. -y_pred = np.empty((y_test.shape[0], n_boostraps)) -for i in range(n_boostraps): - x_, y_ = resample(x_train, y_train) - - # Evaluate the new model on the same test data each time. - y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel() - -# Note: Expectations and variances taken w.r.t. different training -# data sets, hence the axis=1. Subsequent means are taken across the test data -# set in order to obtain a total value, but before this we have error/bias/variance -# calculated per data point in the test set. -# Note 2: The use of keepdims=True is important in the calculation of bias as this -# maintains the column vector form. Dropping this yields very unexpected results. -error = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) ) -bias = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 ) -variance = np.mean( np.var(y_pred, axis=1, keepdims=True) ) -print('Error:', error) -print('Bias^2:', bias) -print('Var:', variance) -print('{} >= {} + {} = {}'.format(error, bias, variance, bias+variance)) - -plt.plot(x[::5, :], y[::5, :], label='f(x)') -plt.scatter(x_test, y_test, label='Data points') -plt.scatter(x_test, np.mean(y_pred, axis=1), label='Pred') -plt.legend() -plt.show() - -!ec - - - -===== Understanding what happens ===== -!bc pycod -import matplotlib.pyplot as plt -import numpy as np -from sklearn.linear_model import LinearRegression, Ridge, Lasso -from sklearn.preprocessing import PolynomialFeatures -from sklearn.model_selection import train_test_split -from sklearn.pipeline import make_pipeline -from sklearn.utils import resample - -np.random.seed(2018) - -n = 40 -n_boostraps = 100 -maxdegree = 14 - - -# Make data set. -x = np.linspace(-3, 3, n).reshape(-1, 1) -y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape) -error = np.zeros(maxdegree) -bias = np.zeros(maxdegree) -variance = np.zeros(maxdegree) -polydegree = np.zeros(maxdegree) -x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2) - -for degree in range(maxdegree): - model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False)) - y_pred = np.empty((y_test.shape[0], n_boostraps)) - for i in range(n_boostraps): - x_, y_ = resample(x_train, y_train) - y_pred[:, i] = model.fit(x_, y_).predict(x_test).ravel() - - polydegree[degree] = degree - error[degree] = np.mean( np.mean((y_test - y_pred)**2, axis=1, keepdims=True) ) - bias[degree] = np.mean( (y_test - np.mean(y_pred, axis=1, keepdims=True))**2 ) - variance[degree] = np.mean( np.var(y_pred, axis=1, keepdims=True) ) - print('Polynomial degree:', degree) - print('Error:', error[degree]) - print('Bias^2:', bias[degree]) - print('Var:', variance[degree]) - print('{} >= {} + {} = {}'.format(error[degree], bias[degree], variance[degree], bias[degree]+variance[degree])) - -plt.plot(polydegree, np.log10(error), label='Error') -plt.plot(polydegree, bias, label='bias') -plt.plot(polydegree, variance, label='Variance') -plt.legend() -plt.show() - - - - -!ec - - -===== Summing up ===== - - - - -The bias-variance tradeoff summarizes the fundamental tension in -machine learning, particularly supervised learning, between the -complexity of a model and the amount of training data needed to train -it. Since data is often limited, in practice it is often useful to -use a less-complex model with higher bias, that is a model whose asymptotic -performance is worse than another model because it is easier to -train and less sensitive to sampling noise arising from having a -finite-sized training dataset (smaller variance). - - - -The above equations tell us that in -order to minimize the expected test error, we need to select a -statistical learning method that simultaneously achieves low variance -and low bias. Note that variance is inherently a nonnegative quantity, -and squared bias is also nonnegative. Hence, we see that the expected -test MSE can never lie below $Var(\epsilon)$, the irreducible error. - - -What do we mean by the variance and bias of a statistical learning -method? The variance refers to the amount by which our model would change if we -estimated it using a different training data set. Since the training -data are used to fit the statistical learning method, different -training data sets will result in a different estimate. But ideally the -estimate for our model should not vary too much between training -sets. However, if a method has high variance then small changes in -the training data can result in large changes in the model. In general, more -flexible statistical methods have higher variance. - - - -===== Another Example rom Scikit-Learn's Repository ===== -!bc pycod -""" -============================ -Underfitting vs. Overfitting -============================ - -This example demonstrates the problems of underfitting and overfitting and -how we can use linear regression with polynomial features to approximate -nonlinear functions. The plot shows the function that we want to approximate, -which is a part of the cosine function. In addition, the samples from the -real function and the approximations of different models are displayed. The -models have polynomial features of different degrees. We can see that a -linear function (polynomial with degree 1) is not sufficient to fit the -training samples. This is called **underfitting**. A polynomial of degree 4 -approximates the true function almost perfectly. However, for higher degrees -the model will **overfit** the training data, i.e. it learns the noise of the -training data. -We evaluate quantitatively **overfitting** / **underfitting** by using -cross-validation. We calculate the mean squared error (MSE) on the validation -set, the higher, the less likely the model generalizes correctly from the -training data. -""" - -print(__doc__) - -import numpy as np -import matplotlib.pyplot as plt -from sklearn.pipeline import Pipeline -from sklearn.preprocessing import PolynomialFeatures -from sklearn.linear_model import LinearRegression -from sklearn.model_selection import cross_val_score - - -def true_fun(X): - return np.cos(1.5 * np.pi * X) - -np.random.seed(0) - -n_samples = 30 -degrees = [1, 4, 15] - -X = np.sort(np.random.rand(n_samples)) -y = true_fun(X) + np.random.randn(n_samples) * 0.1 - -plt.figure(figsize=(14, 5)) -for i in range(len(degrees)): - ax = plt.subplot(1, len(degrees), i + 1) - plt.setp(ax, xticks=(), yticks=()) - - polynomial_features = PolynomialFeatures(degree=degrees[i], - include_bias=False) - linear_regression = LinearRegression() - pipeline = Pipeline([("polynomial_features", polynomial_features), - ("linear_regression", linear_regression)]) - pipeline.fit(X[:, np.newaxis], y) - - # Evaluate the models using crossvalidation - scores = cross_val_score(pipeline, X[:, np.newaxis], y, - scoring="neg_mean_squared_error", cv=10) - - X_test = np.linspace(0, 1, 100) - plt.plot(X_test, pipeline.predict(X_test[:, np.newaxis]), label="Model") - plt.plot(X_test, true_fun(X_test), label="True function") - plt.scatter(X, y, edgecolor='b', s=20, label="Samples") - plt.xlabel("x") - plt.ylabel("y") - plt.xlim((0, 1)) - plt.ylim((-2, 2)) - plt.legend(loc="best") - plt.title("Degree {}\nMSE = {:.2e}(+/- {:.2e})".format( - degrees[i], -scores.mean(), scores.std())) -plt.show() -!ec - - - - -===== The one-dimensional Ising model ===== - -Let us bring back the Ising model again, but now with an additional -focus on Ridge and Lasso regression as well. We repeat some of the -basic parts of the Ising model and the setup of the training and test -data. The one-dimensional Ising model with nearest neighbor -interaction, no external field and a constant coupling constant $J$ is -given by - -!bt -\begin{align} - H = -J \sum_{k}^L s_k s_{k + 1}, -\end{align} -!et -where $s_i \in \{-1, 1\}$ and $s_{N + 1} = s_1$. The number of spins in the system is determined by $L$. For the one-dimensional system there is no phase transition. - -We will look at a system of $L = 40$ spins with a coupling constant of $J = 1$. To get enough training data we will generate 10000 states with their respective energies. - - -!bc pycod -import numpy as np -import matplotlib.pyplot as plt -from mpl_toolkits.axes_grid1 import make_axes_locatable -import seaborn as sns -import scipy.linalg as scl -from sklearn.model_selection import train_test_split -import sklearn.linear_model as skl -import tqdm -sns.set(color_codes=True) -cmap_args=dict(vmin=-1., vmax=1., cmap='seismic') - -L = 40 -n = int(1e4) - -spins = np.random.choice([-1, 1], size=(n, L)) -J = 1.0 - -energies = np.zeros(n) - -for i in range(n): - energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1)) -!ec - -A more general form for the one-dimensional Ising model is - -!bt -\begin{align} - H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. -\end{align} -!et - -Here we allow for interactions beyond the nearest neighbors and a more -adaptive coupling matrix. This latter expression can be formulated as -a matrix-product on the form -!bt -\begin{align} - H = X J, -\end{align} -!et - -where $X_{jk} = s_j s_k$ and $J$ is the matrix consisting of the -elements $-J_{jk}$. This form of writing the energy fits perfectly -with the form utilized in linear regression, viz. -!bt -\begin{align} - \bm{y} = \bm{X}\bm{\beta} + \bm{\epsilon}. -\end{align} -!et -We organize the data as we did above -!bc pycod -X = np.zeros((n, L ** 2)) -for i in range(n): - X[i] = np.outer(spins[i], spins[i]).ravel() -y = energies -X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96) - -X_train_own = np.concatenate( - (np.ones(len(X_train))[:, np.newaxis], X_train), - axis=1 -) - -X_test_own = np.concatenate( - (np.ones(len(X_test))[:, np.newaxis], X_test), - axis=1 -) -!ec - -We will do all fitting with _Scikit-Learn_, - -!bc pycod -clf = skl.LinearRegression().fit(X_train, y_train) -!ec -When extracting the $J$-matrix we make sure to remove the intercept -!bc pycod -J_sk = clf.coef_.reshape(L, L) -!ec -And then we plot the results -!bc pycod -fig = plt.figure(figsize=(20, 14)) -im = plt.imshow(J_sk, **cmap_args) -plt.title("LinearRegression from Scikit-learn", fontsize=18) -plt.xticks(fontsize=18) -plt.yticks(fontsize=18) -cb = fig.colorbar(im) -cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18) -plt.show() -!ec -The results perfectly with our previous discussion where we used our own code. - - -===== Ridge regression ===== - -Having explored the ordinary least squares we move on to ridge -regression. In ridge regression we include a _regularizer_. This -involves a new cost function which leads to a new estimate for the -weights $\bm{\beta}$. This results in a penalized regression problem. The -cost function is given by - -!bt -\begin{align} - C(\bm{X}, \bm{\beta}; \lambda) = (\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y}) + \lambda \bm{\beta}^T\bm{\beta}. -\end{align} -!et -!bc pycod -_lambda = 0.1 -clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train) -J_ridge_sk = clf_ridge.coef_.reshape(L, L) -fig = plt.figure(figsize=(20, 14)) -im = plt.imshow(J_ridge_sk, **cmap_args) -plt.title("Ridge from Scikit-learn", fontsize=18) -plt.xticks(fontsize=18) -plt.yticks(fontsize=18) -cb = fig.colorbar(im) -cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18) - -plt.show() -!ec - - -===== LASSO regression ===== - -In the _Least Absolute Shrinkage and Selection Operator_ (LASSO)-method we get a third cost function. - -!bt -\begin{align} - C(\bm{X}, \bm{\beta}; \lambda) = (\bm{X}\bm{\beta} - \bm{y})^T(\bm{X}\bm{\beta} - \bm{y}) + \lambda \sqrt{\bm{\beta}^T\bm{\beta}}. -\end{align} -!et - -Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from _Scikit-Learn_. - -!bc pycod -clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train) -J_lasso_sk = clf_lasso.coef_.reshape(L, L) -fig = plt.figure(figsize=(20, 14)) -im = plt.imshow(J_lasso_sk, **cmap_args) -plt.title("Lasso from Scikit-learn", fontsize=18) -plt.xticks(fontsize=18) -plt.yticks(fontsize=18) -cb = fig.colorbar(im) -cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18) - -plt.show() -!ec - -It is quite striking how LASSO breaks the symmetry of the coupling -constant as opposed to ridge and OLS. We get a sparse solution with -$J_{j, j + 1} = -1$. - - - - -===== Performance as function of the regularization parameter ===== - -We see how the different models perform for a different set of values for $\lambda$. - - -!bc pycod -lambdas = np.logspace(-4, 5, 10) - -train_errors = { - "ols_sk": np.zeros(lambdas.size), - "ridge_sk": np.zeros(lambdas.size), - "lasso_sk": np.zeros(lambdas.size) -} - -test_errors = { - "ols_sk": np.zeros(lambdas.size), - "ridge_sk": np.zeros(lambdas.size), - "lasso_sk": np.zeros(lambdas.size) -} - -plot_counter = 1 - -fig = plt.figure(figsize=(32, 54)) - -for i, _lambda in enumerate(tqdm.tqdm(lambdas)): - for key, method in zip( - ["ols_sk", "ridge_sk", "lasso_sk"], - [skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)] - ): - method = method.fit(X_train, y_train) - - train_errors[key][i] = method.score(X_train, y_train) - test_errors[key][i] = method.score(X_test, y_test) - - omega = method.coef_.reshape(L, L) - - plt.subplot(10, 5, plot_counter) - plt.imshow(omega, **cmap_args) - plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda)) - plot_counter += 1 - -plt.show() -!ec - -We see that LASSO reaches a good solution for low -values of $\lambda$, but will "wither" when we increase $\lambda$ too -much. Ridge is more stable over a larger range of values for -$\lambda$, but eventually also fades away. - - -===== Finding the optimal value of $\lambda$ ===== - -To determine which value of $\lambda$ is best we plot the accuracy of -the models when predicting the training and the testing set. We expect -the accuracy of the training set to be quite good, but if the accuracy -of the testing set is much lower this tells us that we might be -subject to an overfit model. The ideal scenario is an accuracy on the -testing set that is close to the accuracy of the training set. - - -!bc pycod -fig = plt.figure(figsize=(20, 14)) - -colors = { - "ols_sk": "r", - "ridge_sk": "y", - "lasso_sk": "c" -} - -for key in train_errors: - plt.semilogx( - lambdas, - train_errors[key], - colors[key], - label="Train {0}".format(key), - linewidth=4.0 - ) - -for key in test_errors: - plt.semilogx( - lambdas, - test_errors[key], - colors[key] + "--", - label="Test {0}".format(key), - linewidth=4.0 - ) -plt.legend(loc="best", fontsize=18) -plt.xlabel(r"$\lambda$", fontsize=18) -plt.ylabel(r"$R^2$", fontsize=18) -plt.tick_params(labelsize=18) -plt.show() -!ec - -From the above figure we can see that LASSO with $\lambda = 10^{-2}$ -achieves a very good accuracy on the test set. This by far surpasses the -other models for all values of $\lambda$. - - - - -===== Further Exercises ===== - -=== Exercise 1 === - -We will 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). -!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 (following the examples above) for computing the parametrization of the data set fitting a second-order polynomial. -o Use thereafter _scikit-learn_ (see again the examples in the regression slides) and compare with your own code. -o Using scikit-learn, 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 - -You can use the functionality included in scikit-learn. If you feel -for it, you can use your own program and define functions which -compute the above two functions. Discuss the meaning of these -results. Try also to vary the coefficient in front of the added -stochastic noise term and discuss the quality of the fits. - - - - -=== Exercise 2, variance of the parameters $\beta$ in linear regression === - -Show that the variance of the parameters $\beta$ in the linear regression method (chapter 3, equation (3.8) of "Trevor Hastie, Robert Tibshirani, Jerome H. Friedman, The Elements of Statistical Learning, Springer":"https://www.springer.com/gp/book/9780387848570") is given as - -!bt -\[ -\mathrm{Var}(\hat{\beta}) = \left(\hat{X}^T\hat{X}\right)^{-1}\sigma^2, -\] -!et -with -!bt -\[ -\sigma^2 = \frac{1}{N-p-1}\sum_{i=1}^{N} (y_i-\tilde{y}_i)^2, -\] -!et -where we have assumed that we fit a function of degree $p-1$ (for example a polynomial in $x$). - - - -=== Exercise 3 === - -This exercise is a continuation of exercise 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. - -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). -!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 and compute the parametrization for different values of $\lambda$. Compare and analyze your results with those from exercise 1. 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 $(\hat{X}^T\hat{X})+\lambda\hat{I})^{-1}$) or use the functionality of _scikit-learn_ and compute their variances. Discuss the results of these variances as functions - -o Repeat the previous step but add now the Lasso method. Discuss your results and compare with standard regression and the Ridge regression results. - -o Try to implement the cross-validation as well. - -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 4 === - -We will study how -to fit polynomials to a specific two-dimensional function called -"Franke's -function":"http://www.dtic.mil/dtic/tr/fulltext/u2/a081688.pdf". This -is a function which has been widely used when testing various interpolation and fitting -algorithms. Furthermore, after having established the model and the -method, we will employ resamling techniques such as the cross-validation and/or -the bootstrap methods, in order to perform a proper assessment of our models. - - -The Franke function, which is a weighted sum of four exponentials reads as follows -!bt -\begin{align*} -f(x,y) &= \frac{3}{4}\exp{\left(-\frac{(9x-2)^2}{4} - \frac{(9y-2)^2}{4}\right)}+\frac{3}{4}\exp{\left(-\frac{(9x+1)^2}{49}- \frac{(9y+1)}{10}\right)} \\ -&+\frac{1}{2}\exp{\left(-\frac{(9x-7)^2}{4} - \frac{(9y-3)^2}{4}\right)} -\frac{1}{5}\exp{\left(-(9x-4)^2 - (9y-7)^2\right) }. -\end{align*} -!et - -The function will be defined for $x,y\in [0,1]$. Our first step will -be to perform an OLS regression analysis of this function, trying out -a polynomial fit with an $x$ and $y$ dependence of the form $[x, y, -x^2, y^2, xy, \dots]$. We will also include cross-validation and -bootstrap as resampling techniques. As in homeworks 1 and 2, we -can use a uniform distribution to set up the arrays of values for $x$ -and $y$, or as in the example below just a fix values for $x$ and $y$ with a given step size. -In this case we will have two predictors and need to fit a -function (for example a polynomial) of $x$ and $y$. Thereafter we will -repeat much of the same procedure using the the Ridge and -Lasso regression methods, introducing thus a dependence on the bias -(penalty) $\lambda$. - - -The Python function for the Franke function is included here (it performs also a three-dimensional plot of it) -!bc pycod -from mpl_toolkits.mplot3d import Axes3D -import matplotlib.pyplot as plt -from matplotlib import cm -from matplotlib.ticker import LinearLocator, FormatStrFormatter -import numpy as np -from random import random, seed - -fig = plt.figure() -ax = fig.gca(projection='3d') - -# Make data. -x = np.arange(0, 1, 0.05) -y = np.arange(0, 1, 0.05) -x, y = np.meshgrid(x,y) - - -def FrankeFunction(x,y): - term1 = 0.75*np.exp(-(0.25*(9*x-2)**2) - 0.25*((9*y-2)**2)) - term2 = 0.75*np.exp(-((9*x+1)**2)/49.0 - 0.1*(9*y+1)) - term3 = 0.5*np.exp(-(9*x-7)**2/4.0 - 0.25*((9*y-3)**2)) - term4 = -0.2*np.exp(-(9*x-4)**2 - (9*y-7)**2) - return term1 + term2 + term3 + term4 - - -z = FrankeFunction(x, y) - -# Plot the surface. -surf = ax.plot_surface(x, y, z, cmap=cm.coolwarm, - linewidth=0, antialiased=False) - -# Customize the z axis. -ax.set_zlim(-0.10, 1.40) -ax.zaxis.set_major_locator(LinearLocator(10)) -ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f')) - -# Add a color bar which maps values to colors. -fig.colorbar(surf, shrink=0.5, aspect=5) - -plt.show() - -!ec - - -We will thus again generate our own dataset for a function $\mathrm{FrankeFunction}(x,y)$ where -$x,y \in [0,1]$ could be defined by random numbers computed with the uniform -distribution. The function $f(x,y)$ is the Franke function. You should explore also the addition -an added stochastic noise to this function using the normal distribution $\cal{N}(0,1)$. - -Write your own code (using either a matrix inversion or a singular value decomposition from e.g., _numpy_ ) or use your code from exercises 1 and 3 -and perform a standard least square regression analysis using polynomials in $x$ and $y$ up to fifth order. Find the confidence intervals of the parameters $\beta$ by computing their variances, evaluate the Mean Squared error (MSE) -!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 - -Perform a resampling of the data where you split the data in training data and test data. Implement the $k$-fold cross-validation algorithm and/or the bootstrap algorithm -and evaluate again the MSE and the $R^2$ functions resulting from the test data. Evaluate also the bias and variance of the final models. - - -Write then your own code for the Ridge method, either using matrix -inversion or the singular value decomposition as done for standard OLS. Perform the same analysis as in the -previous exercise (for the same polynomials and include resampling -techniques) but now for different values of $\lambda$. Compare and -analyze your results with those obtained with standard OLS. Study the -dependence on $\lambda$ while also varying eventually the strength of -the noise in your expression for $\mathrm{FrankeFunction}(x,y)$. - -Then perform the same studies but now with Lasso regression. Use the functionalities of -_scikit-learn_. Give a critical discussion of the three methods and a -judgement of which model fits the data best. - - - - - -======= Logistic Regression ======= - -In linear regression our main interest was centered on learning the -coefficients of a functional fit (say a polynomial) in order to be -able to predict the response of a continuous variable on some unseen -data. The fit to the continuous variable $y_i$ is based on some -independent variables $\hat{x}_i$. Linear regression resulted in -analytical expressions (in terms of matrices to invert) for several -quantities, ranging from the variance and thereby the confidence -intervals of the parameters $\hat{\beta}$ to the mean squared -error. If we can invert the product of the design matrices, linear -regression gives then a simple recipe for fitting our data. - - -Classification problems, however, are concerned with outcomes taking -the form of discrete variables (i.e. categories). We may for example, -on the basis of DNA sequencing for a number of patients, like to find -out which mutations are important for a certain disease; or based on -scans of various patients' brains, figure out if there is a tumor or -not; or given a specific physical system, we'd like to identify its -state, say whether it is an ordered or disordered system (typical -situation in solid state physics); or classify the status of a -patient, whether she/he has a stroke or not and many other similar -situations. - -The most common situation we encounter when we apply logistic -regression is that of two possible outcomes, normally denoted as a -binary outcome, true or false, positive or negative, success or -failure etc. - - -===== Optimization and Deep learning ===== - -Logistic regression will also serve as our stepping stone towards neural -network algorithms and supervised deep learning. For logistic -learning, the minimization of the cost function leads to a non-linear -equation in the parameters $\hat{\beta}$. The optmization of the problem calls therefore for minimization algorithms. This forms the bottle neck of all machine learning algorithms, namely how to find reliable minima of a multi-variable function. This leads us to the family of gradient descent methods. The latter are the working horses of basically all modern machine learning algorithms. - -We note also that many of the topics discussed here -regression are also commonly used in modern supervised Deep Learning -models, as we will see later. - - - -===== Basics ===== - -We consider the case where the dependent variables, also called the -responses or the outcomes, $y_i$ are discrete and only take values -from $k=0,\dots,K-1$ (i.e. $K$ classes). - -The goal is to predict the -output classes from the design matrix $\hat{X}\in\mathbb{R}^{n\times p}$ -made of $n$ samples, each of which carries $p$ features or predictors. The -primary goal is to identify the classes to which new unseen samples -belong. - -Let us specialize to the case of two classes only, with outputs $y_i=0$ and $y_i=1$. Our outcomes could represent the status of a credit card user who could default or not on her/his credit card debt. That is -!bt -\[ -y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}. -\] -!et - - - -Before moving to the logistic model, let us try to use our linear regression model to classify these two outcomes. We could for example fit a linear model to the default case if $y_i > 0.5$ and the no default case $y_i \leq 0.5$. - -We would then have our -weighted linear combination, namely -!bt -\begin{equation} -\hat{y} = \hat{X}^T\hat{\beta} + \hat{\epsilon}, -\end{equation} -!et -where $\hat{y}$ is a vector representing the possible outcomes, $\hat{X}$ is our -$n\times p$ design matrix and $\hat{\beta}$ represents our estimators/predictors. - - -The main problem with our function is that it -takes values on the entire real axis. In the case of -logistic regression, however, the labels $y_i$ are discrete -variables. - -One simple way to get a discrete output is to have sign -functions that map the output of a linear regressor to values $\{0,1\}$, -$f(s_i)=sign(s_i)=1$ if $s_i\ge 0$ and 0 if otherwise. -We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning -literature. This model is extremely simple. However, in many cases it is more -favorable to use a ``soft" classifier that outputs -the probability of a given category. This leads us to the logistic function. - -The code for plotting the perceptron can be seen here. This si nothing but the standard "Heaviside step function":"https://en.wikipedia.org/wiki/Heaviside_step_function". -!bc pycod - -!ec - - - -The perceptron is an example of a ``hard classification'' model. We -will encounter this model when we discuss neural networks as -well. Each datapoint is deterministically assigned to a category (i.e -$y_i=0$ or $y_i=1$). In many cases, it is favorable to have a ``soft'' -classifier that outputs the probability of a given category rather -than a single value. For example, given $x_i$, the classifier -outputs the probability of being in a category $k$. Logistic regression -is the most common example of a so-called soft classifier. In logistic -regression, the probability that a data point $x_i$ -belongs to a category $y_i=\{0,1\}$ is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event, -!bt -\[ -p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}. -\] -!et -Note that $1-p(t)= p(-t)$. -The following code plots the logistic function. -!bc pycod - -!ec - - - - -We assume now that we have two classes with $y_i$ either $0$ or $1$. Furthermore we assume also that we have only two parameters $\beta$ in our fitting of the Sigmoid function, that is we define probabilities -!bt -\begin{align*} -p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\ -p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}), -\end{align*} -!et -where $\hat{\beta}$ are the weights we wish to extract from data, in our case $\beta_0$ and $\beta_1$. - -Note that we used -!bt -\[ -p(y_i=0\vert x_i, \hat{\beta}) = 1-p(y_i=1\vert x_i, \hat{\beta}). -\] -!et - - -===== Maximum likelihood ===== - -In order to define the total likelihood for all possible outcomes from a -dataset $\mathcal{D}=\{(y_i,x_i)\}$, with the binary labels -$y_i\in\{0,1\}$ and where the data points are drawn independently, we use the so-called "Maximum Likelihood Estimation":"https://en.wikipedia.org/wiki/Maximum_likelihood_estimation" (MLE) principle. -We aim thus at maximizing -the probability of seeing the observed data. We can then approximate the -likelihood in terms of the product of the individual probabilities of a specific outcome $y_i$, that is -!bt -\begin{align*} -P(\mathcal{D}|\hat{\beta})& = \prod_{i=1}^n \left[p(y_i=1|x_i,\hat{\beta})\right]^{y_i}\left[1-p(y_i=1|x_i,\hat{\beta}))\right]^{1-y_i}\nonumber \\ -\end{align*} -!et -from which we obtain the log-likelihood and our _cost/loss_ function -!bt -\[ -\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\hat{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\hat{\beta}))\right]\right). -\] -!et - - -===== The cost function rewritten ===== - -Reordering the logarithms, we can rewrite the _cost/loss_ function as -!bt -\[ -\mathcal{C}(\hat{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right). -\] -!et - -The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to $\beta$. -Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that -!bt -\[ -\mathcal{C}(\hat{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right). -\] -!et -This equation is known in statistics as the _cross entropy_. Finally, we note that just as in linear regression, -in practice we often supplement the cross-entropy with additional regularization terms, usually $L_1$ and $L_2$ regularization as we did for Ridge and Lasso regression. - - -The cross entropy is a convex function of the weights $\hat{\beta}$ and, -therefore, any local minimizer is a global minimizer. - - -Minimizing this -cost function with respect to the two parameters $\beta_0$ and $\beta_1$ we obtain - -!bt -\[ -\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right), -\] -!et -and -!bt -\[ -\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right). -\] -!et - - -Let us now define a vector $\hat{y}$ with $n$ elements $y_i$, an -$n\times p$ matrix $\hat{X}$ which contains the $x_i$ values and a -vector $\hat{p}$ of fitted probabilities $p(y_i\vert x_i,\hat{\beta})$. We can rewrite in a more compact form the first -derivative of cost function as - -!bt -\[ -\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right). -\] -!et - -If we in addition define a diagonal matrix $\hat{W}$ with elements -$p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta})$, we can obtain a compact expression of the second derivative as - -!bt -\[ -\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}. -\] -!et - -===== Extending to more predictors ===== - -Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with $p$ predictors -!bt -\[ -\log{ \frac{p(\hat{\beta}\hat{x})}{1-p(\hat{\beta}\hat{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p. -\] -!et -Here we defined $\hat{x}=[1,x_1,x_2,\dots,x_p]$ and $\hat{\beta}=[\beta_0, \beta_1, \dots, \beta_p]$ leading to -!bt -\[ -p(\hat{\beta}\hat{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}. -\] -!et - - -Till now we have mainly focused on two classes, the so-called binary system. Suppose we wish to extend to $K$ classes. -Let us for the sake of simplicity assume we have only two predictors. We have then following model -!bt -\[ -\log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1, -\] -!et -!bt -\[ -\log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1, -\] -!et -and so on till the class $C=K-1$ class -!bt -\[ -\log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1, -\] -!et -and the model is specified in term of $K-1$ so-called log-odds or _logit_ transformations. - - - -===== The Softmax function ===== - -In our discussion of neural networks we will encounter the above again in terms of the so-called _Softmax_ function. - -The softmax function is used in various multiclass classification -methods, such as multinomial logistic regression (also known as -softmax regression), multiclass linear discriminant -analysis, naive Bayes classifiers, and artificial neural networks. -Specifically, in multinomial logistic regression and linear -discriminant analysis, the input to the function is the result of $K$ -distinct linear functions, and the predicted probability for the $k$-th -class given a sample vector $\hat{x}$ and a weighting vector $\hat{\beta}$ is (with two predictors): - -!bt -\[ -p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}. -\] -!et -It is easy to extend to more predictors. The final class is -!bt -\[ -p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}, -\] -!et -and they sum to one. Our earlier discussions were all specialized to the case with two classes only. It is easy to see from the above that what we derived earlier is compatible with these equations. - -To find the optimal parameters we would typically use a gradient descent method. -Newton's method and gradient descent methods are discussed in the material on "optimization methods":"https://compphysics.github.io/MachineLearning/doc/pub/Splines/html/Splines-bs.html". - - - - -===== A _scikit-learn_ example ===== - -!bc pycod -import numpy as np -import matplotlib.pyplot as plt -from sklearn import datasets -iris = datasets.load_iris() -list(iris.keys()) -['data', 'target_names', 'feature_names', 'target', 'DESCR'] -X = iris["data"][:, 3:] # petal width -y = (iris["target"] == 2).astype(np.int) # 1 if Iris-Virginica, else 0 - -from sklearn.linear_model import LogisticRegression -log_reg = LogisticRegression() -log_reg.fit(X, y) - -X_new = np.linspace(0, 3, 1000).reshape(-1, 1) -y_proba = log_reg.predict_proba(X_new) -plt.plot(X_new, y_proba[:, 1], "g-", label="Iris-Virginica") -plt.plot(X_new, y_proba[:, 0], "b--", label="Not Iris-Virginica") -plt.show() - -!ec - - - -===== A simple classification problem ===== -!bc pycod -import numpy as np -from sklearn import datasets, linear_model -import matplotlib.pyplot as plt - - -def generate_data(): - np.random.seed(0) - X, y = datasets.make_moons(200, noise=0.20) - return X, y - - -def visualize(X, y, clf): - # plt.scatter(X[:, 0], X[:, 1], s=40, c=y, cmap=plt.cm.Spectral) - # plt.show() - plot_decision_boundary(lambda x: clf.predict(x), X, y) - plt.title("Logistic Regression") - - -def plot_decision_boundary(pred_func, X, y): - # Set min and max values and give it some padding - x_min, x_max = X[:, 0].min() - .5, X[:, 0].max() + .5 - y_min, y_max = X[:, 1].min() - .5, X[:, 1].max() + .5 - h = 0.01 - # Generate a grid of points with distance h between them - xx, yy = np.meshgrid(np.arange(x_min, x_max, h), np.arange(y_min, y_max, h)) - # Predict the function value for the whole gid - Z = pred_func(np.c_[xx.ravel(), yy.ravel()]) - Z = Z.reshape(xx.shape) - # Plot the contour and training examples - plt.contourf(xx, yy, Z, cmap=plt.cm.Spectral) - plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Spectral) - plt.show() - - -def classify(X, y): - clf = linear_model.LogisticRegressionCV() - clf.fit(X, y) - return clf - - -def main(): - X, y = generate_data() - # visualize(X, y) - clf = classify(X, y) - visualize(X, y, clf) - - -if __name__ == "__main__": - main() -!ec - - -===== The two-dimensional Ising model, Predicting phase transition of the two-dimensional Ising model ===== - -The Hamiltonian of the two-dimensional Ising model without an external field for a constant coupling constant $J$ is given by -!bt -\begin{align} - H = -J \sum_{\langle ij\rangle} S_i S_j, -\end{align} -!et -where $S_i \in \{-1, 1\}$ and $\langle ij \rangle$ signifies that we only iterate over the nearest neighbors in the lattice. We will be looking at a system of $L = 40$ spins in each dimension, i.e., $L^2 = 1600$ spins in total. Opposed to the one-dimensional Ising model we will get a phase transition from an _ordered_ phase to a _disordered_ phase at the critical temperature - -!bt -\begin{align} - \frac{T_c}{J} = \frac{2}{\log\left(1 + \sqrt{2}\right)} \approx 2.26, -\end{align} -!et -as shown by Lars Onsager. - - -Here we use _logistic regression_ to predict when a phase transition -occurs. The data we will look at is a set of spin configurations, -i.e., individual lattices with spins, labeled _ordered_ `1` or -_disordered_ `0`. Our job is to build a model which will take in a -spin configuration and predict whether or not the spin configuration -constitutes an ordered or a disordered phase. To achieve this we will -represent the lattices as flattened arrays with $1600$ elements -instead of a matrix of $40 \times 40$ elements. As an extra test of -the performance of the algorithms we will divide the dataset into -three pieces. We will do a conventional train-test-split on a -combination of totally ordered and totally disordered phases. The -remaining "critical-like" states will be used as test data which we -hope the model will be able to make good extrapolated predictions on. - - -!bc pycod -import pickle -import os -import glob -import numpy as np -import pandas as pd -import matplotlib.pyplot as plt -import seaborn as sns -import sklearn.model_selection as skms -import sklearn.linear_model as skl -import sklearn.metrics as skm -import tqdm -import copy -import time -from IPython.display import display - -%matplotlib inline - -sns.set(color_codes=True) -!ec - - -Using the data from "Mehta et al.":"https://physics.bu.edu/~pankajm/ML-Review-Datasets/isingMC/" (specifically the two datasets named `Ising2DFM_reSample_L40_T=All.pkl` and `Ising2DFM_reSample_L40_T=All_labels.pkl`) we have to unpack the data into numpy arrays. - - -!bc pycod -filenames = glob.glob(os.path.join("..", "dat", "*")) -label_filename = list(filter(lambda x: "label" in x, filenames))[0] -dat_filename = list(filter(lambda x: "label" not in x, filenames))[0] - -# Read in the labels -with open(label_filename, "rb") as f: - labels = pickle.load(f) - -# Read in the corresponding configurations -with open(dat_filename, "rb") as f: - data = np.unpackbits(pickle.load(f)).reshape(-1, 1600).astype("int") - -# Set spin-down to -1 -data[data == 0] = -1 -!ec - -This dataset consists of $10000$ samples, i.e., $10000$ spin -configurations with $40 \times 40$ spins each, for $16$ temperatures -between $0.25$ to $4.0$. Next we create a train/test-split and keep -the data in the critical phase as a separate dataset for -extrapolation-testing. - - -!bc pycod -# Set up slices of the dataset -ordered = slice(0, 70000) -critical = slice(70000, 100000) -disordered = slice(100000, 160000) - -X_train, X_test, y_train, y_test = skms.train_test_split( - np.concatenate((data[ordered], data[disordered])), - np.concatenate((labels[ordered], labels[disordered])), - test_size=0.95 -) -!ec - - - -===== Logistic regression ===== - -Logistic regression is a linear model for classification. Recalling -the cost function for ordinary least squares with both L2 (ridge) and -L1 (LASSO) penalties we will see that the logistic cost function is -very similar. In OLS we wish to predict a continuous variable -$\hat{y}$ using -!bt -\begin{align} - \hat{y} = X\omega, -\end{align} -!et - -where $X \in \mathbb{R}^{n \times p}$ is the input data and $\omega^{p -\times d}$ are the weights of the regression. In a classification -setting (binary classification in our situation) we are interested in -a positive or negative answer. We can thus define either answer to be -above or below some threshold. But, in order to limit the size of the -answer and also to get a probability interpretation on how sure we are -for either answer we can compute the sigmoid function of OLS. That is, - -!bt -\begin{align} - f(X\omega) = \frac{1}{1 + \exp(-X\omega)}. -\end{align} -!et -We are thus interested in minizming the following cost function -!bt -\begin{align} - C(X, \omega) = \sum_{i = 1}^n \left\{ - - y_i\log\left( f(x_i^T\omega) \right) - - (1 - y_i)\log\left[1 - f(x_i^T\omega)\right] - \right\}, -\end{align} -!et - -where we will restrict ourselves to a value for $f(z)$ as the sigmoid -described above. We can also tack on a L2 (Ridge) or L1 (LASSO) -penalization to this cost function in the same manner we did for -linear regression. - - -The penalization factor $\lambda$ is inverted in the case of the -logistic regression model we use. We will explore several values of -$\lambda$ using both L1 and L2 penalization. We do this using a grid -search over different parameters and run a 3-fold cross validation for -each configuration. In other words, we fit a model 3 times for each -configuration of the hyper parameters. - - -!bc pycod -lambdas = np.logspace(-7, -1, 7) - -param_grid = { - "C": list(1.0/lambdas), - "penalty": ["l1", "l2"] -} -clf = skms.GridSearchCV( - skl.LogisticRegression(), - param_grid=param_grid, - n_jobs=-1, - return_train_score=True -) -t0 = time.time() -clf.fit(X_train, y_train) -t1 = time.time() - -print ( - "Time spent fitting GridSearchCV(LogisticRegression): {0:.3f} sec".format( - t1 - t0 - ) -) -!ec - -We can see that logistic regression is quite slow and using the grid -search and cross validation results in quite a heavy -computation. Below we show the results of the different -configurations. - - -!bc pycod -logreg_df = pd.DataFrame(clf.cv_results_) - -display(logreg_df) -!ec - - -===== Accuracy of a classification model ===== - -To determine how well a classification model is performing we count -the number of correctly labeled classes and divide by the number of -classes in total. The accuracy is thus given by - -!bt -\begin{align} - a(y, \hat{y}) = \frac{1}{n}\sum_{i = 1}^{n} I(y_i = \hat{y}_i), -\end{align} -!et - -where $I(y_i = \hat{y}_i)$ is the indicator function given by - -!bt -\begin{align} - I(x = y) = \begin{array}{cc} - 1 & x = y, \\ - 0 & x \neq y. - \end{array} -\end{align} -!et - -This is the accuracy provided by Scikit-learn when using _sklearn.metrics.accuracyscore_. - -Below we compute the accuracy of the best fit model on the training data (which should give a good accuracy), the test data (which has not been shown to the model) and the critical data (completely new data that needs to be extrapolated). - - -!bc pycod -train_accuracy = skm.accuracy_score(y_train, clf.predict(X_train)) -test_accuracy = skm.accuracy_score(y_test, clf.predict(X_test)) -critical_accuracy = skm.accuracy_score(labels[critical], clf.predict(data[critical])) - -print ("Accuracy on train data: {0}".format(train_accuracy)) -print ("Accuracy on test data: {0}".format(test_accuracy)) -print ("Accuracy on critical data: {0}".format(critical_accuracy)) -!ec - -We can see that we get quite good accuracy on the training data, but gradually worsening accuracy on the test and critical data. - - -===== Analyzing the results ===== - -Below we show a different metric for determining the quality of our -model, namely the _reciever operating characteristic_ (ROC). The ROC -curve tells us how well the model correctly classifies the different -labels. We plot the _true positive rate_ (the rate of predicted -positive classes that are positive) versus the _false positive rate_ -(the rate of predicted positive classes that are negative). The ROC -curve is built by computing the true positive rate and the false -positive rate for varying _thresholds_, i.e, which probability we -should acredit a certain class. - -By computing the _area under the curve_ (AUC) of the ROC curve we get an estimate of how well our model is performing. Pure guessing will get an AUC of $0.5$. A perfect score will get an AUC of $1.0$. - - -!bc pycod -fig = plt.figure(figsize=(20, 14)) - -for (_X, _y), label in zip( - [ - (X_train, y_train), - (X_test, y_test), - (data[critical], labels[critical]) - ], - ["Train", "Test", "Critical"] -): - proba = clf.predict_proba(_X) - fpr, tpr, _ = skm.roc_curve(_y, proba[:, 1]) - roc_auc = skm.auc(fpr, tpr) - - print ("LogisticRegression AUC ({0}): {1}".format(label, roc_auc)) - - plt.plot(fpr, tpr, label="{0} (AUC = {1})".format(label, roc_auc), linewidth=4.0) - -plt.plot([0, 1], [0, 1], "--", label="Guessing (AUC = 0.5)", linewidth=4.0) - -plt.title(r"The ROC curve for LogisticRegression", fontsize=18) -plt.xlabel(r"False positive rate", fontsize=18) -plt.ylabel(r"True positive rate", fontsize=18) -plt.axis([-0.01, 1.01, -0.01, 1.01]) -plt.xticks(fontsize=18) -plt.yticks(fontsize=18) -plt.legend(loc="best", fontsize=18) -plt.show() -!ec - -We can see that this plot of the ROC looks very strange. This tells us -that logistic regression is quite inept at predicting the Ising model -transition and is therefore highly non-linear. The ROC curve for the -training data looks quite good, but as the testing data is so far off -we see that we are dealing with an overfit model. - - - - - -======= Optimization and Gradient Methods ======= - - -===== Optimization, the central part of any Machine Learning algortithm ===== - -Almost every problem in machine learning and data science starts with -a dataset $X$, a model $g(\beta)$, which is a function of the -parameters $\beta$ and a cost function $C(X, g(\beta))$ that allows -us to judge how well the model $g(\beta)$ explains the observations -$X$. The model is fit by finding the values of $\beta$ that minimize -the cost function. Ideally we would be able to solve for $\beta$ -analytically, however this is not possible in general and we must use -some approximative/numerical method to compute the minimum. - - - -===== Revisiting our Logistic Regression case ===== - -In our discussion on Logistic Regression we studied the -case of -two classes, with $y_i$ either -$0$ or $1$. Furthermore we assumed also that we have only two -parameters $\beta$ in our fitting, that is we -defined probabilities - -!bt -\begin{align*} -p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\ -p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}), -\end{align*} -!et -where $\hat{\beta}$ are the weights we wish to extract from data, in our case $\beta_0$ and $\beta_1$. - - -Our compact equations used a definition of a vector $\hat{y}$ with $n$ -elements $y_i$, an $n\times p$ matrix $\hat{X}$ which contains the -$x_i$ values and a vector $\hat{p}$ of fitted probabilities -$p(y_i\vert x_i,\hat{\beta})$. We rewrote in a more compact form -the first derivative of the cost function as - -!bt -\[ -\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right). -\] -!et - -If we in addition define a diagonal matrix $\hat{W}$ with elements -$p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta})$, we can obtain a compact expression of the second derivative as - -!bt -\[ -\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}. -\] -!et -This defines what is called the Hessian matrix. - - -===== Solving using Newton-Raphson's method ===== - -If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives. - -Our iterative scheme is then given by - -!bt -\[ -\hat{\beta}^{\mathrm{new}} = \hat{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T}\right)^{-1}_{\hat{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}}\right)_{\hat{\beta}^{\mathrm{old}}}, -\] -!et -or in matrix form as - -!bt -\[ -\hat{\beta}^{\mathrm{new}} = \hat{\beta}^{\mathrm{old}}-\left(\hat{X}^T\hat{W}\hat{X} \right)^{-1}\times \left(-\hat{X}^T(\hat{y}-\hat{p}) \right)_{\hat{\beta}^{\mathrm{old}}}. -\] -!et -The right-hand side is computed with the old values of $\beta$. - -If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement. - - - -Let us quickly remind ourselves how we derive the above method. - -Perhaps the most celebrated of all one-dimensional root-finding -routines is Newton's method, also called the Newton-Raphson -method. This method requires the evaluation of both the -function $f$ and its derivative $f'$ at arbitrary points. -If you can only calculate the derivative -numerically and/or your function is not of the smooth type, we -normally discourage the use of this method. - - - -The Newton-Raphson formula consists geometrically of extending the -tangent line at a current point until it crosses zero, then setting -the next guess to the abscissa of that zero-crossing. The mathematics -behind this method is rather simple. Employing a Taylor expansion for -$x$ sufficiently close to the solution $s$, we have - - -!bt -\[ - f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots. - \label{eq:taylornr} -\] -!et - -For small enough values of the function and for well-behaved -functions, the terms beyond linear are unimportant, hence we obtain - - -!bt -\[ - f(x)+(s-x)f'(x)\approx 0, -\] -!et -yielding -!bt -\[ - s\approx x-\frac{f(x)}{f'(x)}. -\] -!et - -Having in mind an iterative procedure, it is natural to start iterating with -!bt -\[ - x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. -\] -!et - - -The above is Newton-Raphson's method. It has a simple geometric -interpretation, namely $x_{n+1}$ is the point where the tangent from -$(x_n,f(x_n))$ crosses the $x$-axis. Close to the solution, -Newton-Raphson converges fast to the desired result. However, if we -are far from a root, where the higher-order terms in the series are -important, the Newton-Raphson formula can give grossly inaccurate -results. For instance, the initial guess for the root might be so far -from the true root as to let the search interval include a local -maximum or minimum of the function. If an iteration places a trial -guess near such a local extremum, so that the first derivative nearly -vanishes, then Newton-Raphson may fail totally - - - -Newton's method can be generalized to systems of several non-linear equations -and variables. Consider the case with two equations -!bt -\[ - \begin{array}{cc} f_1(x_1,x_2) &=0\\ - f_2(x_1,x_2) &=0,\end{array} -\] -!et -which we Taylor expand to obtain - -!bt -\[ - \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1 - \partial f_1/\partial x_1+h_2 - \partial f_1/\partial x_2+\dots\\ - 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1 - \partial f_2/\partial x_1+h_2 - \partial f_2/\partial x_2+\dots - \end{array}. -\] -!et -Defining the Jacobian matrix $\bm{J}$ we have -!bt -\[ - \bm{J}=\left( \begin{array}{cc} - \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\ - \partial f_2/\partial x_1 &\partial f_2/\partial x_2 - \end{array} \right), -\] -!et -we can rephrase Newton's method as -!bt -\[ -\left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)= -\left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+ -\left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right), -\] -!et -where we have defined -!bt -\[ - \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)= - -{\bm{J}}^{-1} - \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right). -\] -!et -We need thus to compute the inverse of the Jacobian matrix and it -is to understand that difficulties may -arise in case $\bm{J}$ is nearly singular. - -It is rather straightforward to extend the above scheme to systems of -more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function. - - - - -===== Steepest descent ===== - -The basic idea of gradient descent is -that a function $F(\mathbf{x})$, -$\mathbf{x} \equiv (x_1,\cdots,x_n)$, decreases fastest if one goes from $\bf {x}$ in the -direction of the negative gradient $-\nabla F(\mathbf{x})$. - -It can be shown that if -!bt -\[ -\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), -\] -!et -with $\gamma_k > 0$. - -For $\gamma_k$ small enough, then $F(\mathbf{x}_{k+1}) \leq -F(\mathbf{x}_k)$. This means that for a sufficiently small $\gamma_k$ -we are always moving towards smaller function values, i.e a minimum. - - -The previous observation is the basis of the method of steepest -descent, which is also referred to as just gradient descent (GD). One -starts with an initial guess $\mathbf{x}_0$ for a minimum of $F$ and -computes new approximations according to - -!bt -\[ -\mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0. -\] -!et - -The parameter $\gamma_k$ is often referred to as the step length or -the learning rate within the context of Machine Learning. - - - -Ideally the sequence $\{\mathbf{x}_k \}_{k=0}$ converges to a global -minimum of the function $F$. In general we do not know if we are in a -global or local minimum. In the special case when $F$ is a convex -function, all local minima are also global minima, so in this case -gradient descent can converge to the global solution. The advantage of -this scheme is that it is conceptually simple and straightforward to -implement. However the method in this form has some severe -limitations: - -In machine learing we are often faced with non-convex high dimensional -cost functions with many local minima. Since GD is deterministic we -will get stuck in a local minimum, if the method converges, unless we -have a very good intial guess. This also implies that the scheme is -sensitive to the chosen initial condition. - -Note that the gradient is a function of $\mathbf{x} = -(x_1,\cdots,x_n)$ which makes it expensive to compute numerically. - - - -The gradient descent method -is sensitive to the choice of learning rate $\gamma_k$. This is due -to the fact that we are only guaranteed that $F(\mathbf{x}_{k+1}) \leq -F(\mathbf{x}_k)$ for sufficiently small $\gamma_k$. The problem is to -determine an optimal learning rate. If the learning rate is chosen too -small the method will take a long time to converge and if it is too -large we can experience erratic behavior. - -Many of these shortcomings can be alleviated by introducing -randomness. One such method is that of Stochastic Gradient Descent -(SGD), see below. - - - -Ideally we want our cost/loss function to be convex(concave). - -First we give the definition of a convex set: A set $C$ in -$\mathbb{R}^n$ is said to be convex if, for all $x$ and $y$ in $C$ and -all $t \in (0,1)$ , the point $(1 − t)x + ty$ also belongs to -C. Geometrically this means that every point on the line segment -connecting $x$ and $y$ is in $C$ as discussed below. - -The convex subsets of $\mathbb{R}$ are the intervals of -$\mathbb{R}$. Examples of convex sets of $\mathbb{R}^2$ are the -regular polygons (triangles, rectangles, pentagons, etc...). - - -===== Convex function ===== - -_Convex function_: Let $X \subset \mathbb{R}^n$ be a convex set. Assume that the function $f: X \rightarrow \mathbb{R}$ is continuous, then $f$ is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all $x_1, x_2 \in X$ and for all $t \in [0,1]$. If $\leq$ is replaced with a strict inequaltiy in the definition, we demand $x_1 \neq x_2$ and $t\in(0,1)$ then $f$ is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting $f(x_1)$ and $f(x_2)$, the value of the function on the interval $[x_1,x_2]$ is always below the line as illustrated below. - - -In the following we state first and second-order conditions which -ensures convexity of a function $f$. We write $D_f$ to denote the -domain of $f$, i.e the subset of $R^n$ where $f$ is defined. For more -details and proofs we refer to: "S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press":"http://stanford.edu/boyd/cvxbook/, 2004". - -!bblock First order condition -Suppose $f$ is differentiable (i.e $\nabla f(x)$ is well defined for -all $x$ in the domain of $f$). Then $f$ is convex if and only if $D_f$ -is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds -for all $x,y \in D_f$. This condition means that for a convex function -the first order Taylor expansion (right hand side above) at any point -a global under estimator of the function. To convince yourself you can -make a drawing of $f(x) = x^2+1$ and draw the tangent line to $f(x)$ and -note that it is always below the graph. -!eblock - -!bblock Second order condition -Assume that $f$ is twice -differentiable, i.e the Hessian matrix exists at each point in -$D_f$. Then $f$ is convex if and only if $D_f$ is a convex set and its -Hessian is positive semi-definite for all $x\in D_f$. For a -single-variable function this reduces to $f''(x) \geq 0$. Geometrically this means that $f$ has nonnegative curvature -everywhere. -!eblock - -This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition. - - -The next result is of great importance to us and the reason why we are -going on about convex functions. In machine learning we frequently -have to minimize a loss/cost function in order to find the best -parameters for the model we are considering. - -Ideally we want the -global minimum (for high-dimensional models it is hard to know -if we have local or global minimum). However, if the cost/loss function -is convex the following result provides invaluable information: - -!bblock Any minimum is global for convex functions -Consider the problem of finding $x \in \mathbb{R}^n$ such that $f(x)$ -is minimal, where $f$ is convex and differentiable. Then, any point -$x^*$ that satisfies $\nabla f(x^*) = 0$ is a global minimum. -!eblock - -This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum. - - -===== Some simple problems ===== - -o Show that $f(x)=x^2$ is convex for $x \in \mathbb{R}$ using the definition of convexity. Hint: If you re-write the definition, $f$ is convex if the following holds for all $x,y \in D_f$ and any $\lambda \in [0,1]$ $\lambda f(x)+(1-\lambda)f(y)-f(\lambda x + (1-\lambda) y ) \geq 0$. - -o Using the second order condition show that the following functions are convex on the specified domain. - * $f(x) = e^x$ is convex for $x \in \mathbb{R}$. - * $g(x) = -\ln(x)$ is convex for $x \in (0,\infty)$. -o Let $f(x) = x^2$ and $g(x) = e^x$. Show that $f(g(x))$ and $g(f(x))$ is convex for $x \in \mathbb{R}$. Also show that if $f(x)$ is any convex function than $h(x) = e^{f(x)}$ is convex. - -o A norm is any function that satisfy the following properties - * $f(\alpha x) = |\alpha| f(x)$ for all $\alpha \in \mathbb{R}$. - * $f(x+y) \leq f(x) + f(y)$ - * $f(x) \leq 0$ for all $x \in \mathbb{R}^n$ with equality if and only if $x = 0$ - -Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this). - - - -===== Standard steepest descent ===== - - -Before we proceed, we would like to discuss the approach called the -_standard Steepest descent_, which again leads to us having to be able -to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG). - -"The success of the CG method":"https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf" -for finding solutions of non-linear problems is based on the theory -of conjugate gradients for linear systems of equations. It belongs to -the class of iterative methods for solving problems from linear -algebra of the type -!bt -\begin{equation*} -\hat{A}\hat{x} = \hat{b}. -\end{equation*} -!et - -In the iterative process we end up with a problem like - -!bt -\begin{equation*} - \hat{r}= \hat{b}-\hat{A}\hat{x}, -\end{equation*} -!et -where $\hat{r}$ is the so-called residual or error in the iterative process. - -When we have found the exact solution, $\hat{r}=0$. - - -The residual is zero when we reach the minimum of the quadratic equation -!bt -\begin{equation*} - P(\hat{x})=\frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T\hat{b}, -\end{equation*} -!et - -with the constraint that the matrix $\hat{A}$ is positive definite and -symmetric. This defines also the Hessian and we want it to be positive definite. - - - -We denote the initial guess for $\hat{x}$ as $\hat{x}_0$. -We can assume without loss of generality that -!bt -\begin{equation*} -\hat{x}_0=0, -\end{equation*} -!et -or consider the system -!bt -\begin{equation*} -\hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, -\end{equation*} -!et -instead. - - - -One can show that the solution $\hat{x}$ is also the unique minimizer of the quadratic form -!bt -\begin{equation*} - f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. -\end{equation*} -!et -This suggests taking the first basis vector $\hat{r}_1$ (see below for definition) -to be the gradient of $f$ at $\hat{x}=\hat{x}_0$, -which equals -!bt -\begin{equation*} -\hat{A}\hat{x}_0-\hat{b}, -\end{equation*} -!et -and -$\hat{x}_0=0$ it is equal $-\hat{b}$. - -We can compute the residual iteratively as -!bt -\begin{equation*} -\hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, - \end{equation*} -!et -which equals -!bt -\begin{equation*} -\hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{r}_k), - \end{equation*} -!et -or -!bt -\begin{equation*} -(\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{r}_k, - \end{equation*} -!et -which gives - -!bt -\[ -\alpha_k = \frac{\hat{r}_k^T\hat{r}_k}{\hat{r}_k^T\hat{A}\hat{r}_k} -\] -!et -leading to the iterative scheme -!bt -\begin{equation*} -\hat{x}_{k+1}=\hat{x}_k-\alpha_k\hat{r}_{k}, - \end{equation*} -!et - -===== Simple codes for steepest descent and conjugate gradient using a $2\times 2$ matrix, in c++, Python code to come ===== - -!bc cppcod -#include -#include -#include -#include -#include "vectormatrixclass.h" -using namespace std; -// Main function begins here -int main(int argc, char * argv[]){ - int dim = 2; - Vector x(dim),xsd(dim), b(dim),x0(dim); - Matrix A(dim,dim); - - // Set our initial guess - x0(0) = x0(1) = 0; - // Set the matrix - A(0,0) = 3; A(1,0) = 2; A(0,1) = 2; A(1,1) = 6; - b(0) = 2; b(1) = -8; - cout << "The Matrix A that we are using: " << endl; - A.Print(); - cout << endl; - xsd = SteepestDescent(A,b,x0); - cout << "The approximate solution using Steepest Descent is: " << endl; - xsd.Print(); - cout << endl; -} -!ec - - - -!bc cppcod -Vector SteepestDescent(Matrix A, Vector b, Vector x0){ - int IterMax, i; - int dim = x0.Dimension(); - const double tolerance = 1.0e-14; - Vector x(dim),f(dim),z(dim); - double c,alpha,d; - IterMax = 30; - x = x0; - r = A*x-b; - i = 0; - while (i <= IterMax){ - z = A*r; - c = dot(r,r); - alpha = c/dot(r,z); - x = x - alpha*r; - r = A*x-b; - if(sqrt(dot(r,r)) < tolerance) break; - i++; - } - return x; -} -!ec - -===== Steepest descent example ===== - -!bc pycod -import numpy as np -import numpy.linalg as la - -import scipy.optimize as sopt - -import matplotlib.pyplot as pt -from mpl_toolkits.mplot3d import axes3d - -def f(x): - return 0.5*x[0]**2 + 2.5*x[1]**2 - -def df(x): - return np.array([x[0], 5*x[1]]) - -fig = pt.figure() -ax = fig.gca(projection="3d") - -xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j] -fmesh = f(np.array([xmesh, ymesh])) -ax.plot_surface(xmesh, ymesh, fmesh) -!ec -And then as countor plot -!bc pycod -pt.axis("equal") -pt.contour(xmesh, ymesh, fmesh) -guesses = [np.array([2, 2./5])] -!ec -Find guesses -!bc pycod -x = guesses[-1] -s = -df(x) -!ec -Run it! -!bc pycod -def f1d(alpha): - return f(x + alpha*s) - -alpha_opt = sopt.golden(f1d) -next_guess = x + alpha_opt * s -guesses.append(next_guess) -print(next_guess) -!ec -What happened? -!bc pycod -pt.axis("equal") -pt.contour(xmesh, ymesh, fmesh, 50) -it_array = np.array(guesses) -pt.plot(it_array.T[0], it_array.T[1], "x-") -!ec - - -===== Conjugate gradient method ===== - -In the CG method we define so-called conjugate directions and two vectors -$\hat{s}$ and $\hat{t}$ -are said to be -conjugate if -!bt -\begin{equation*} -\hat{s}^T\hat{A}\hat{t}= 0. -\end{equation*} -!et -The philosophy of the CG method is to perform searches in various conjugate directions -of our vectors $\hat{x}_i$ obeying the above criterion, namely -!bt -\begin{equation*} -\hat{x}_i^T\hat{A}\hat{x}_j= 0. -\end{equation*} -!et -Two vectors are conjugate if they are orthogonal with respect to -this inner product. Being conjugate is a symmetric relation: if $\hat{s}$ is conjugate to $\hat{t}$, then $\hat{t}$ is conjugate to $\hat{s}$. - -An example is given by the eigenvectors of the matrix -!bt -\begin{equation*} -\hat{v}_i^T\hat{A}\hat{v}_j= \lambda\hat{v}_i^T\hat{v}_j, -\end{equation*} -!et -which is zero unless $i=j$. - -Assume now that we have a symmetric positive-definite matrix $\hat{A}$ of size -$n\times n$. At each iteration $i+1$ we obtain the conjugate direction of a vector -!bt -\begin{equation*} -\hat{x}_{i+1}=\hat{x}_{i}+\alpha_i\hat{p}_{i}. -\end{equation*} -!et -We assume that $\hat{p}_{i}$ is a sequence of $n$ mutually conjugate directions. -Then the $\hat{p}_{i}$ form a basis of $R^n$ and we can expand the solution -$ \hat{A}\hat{x} = \hat{b}$ in this basis, namely - -!bt -\begin{equation*} - \hat{x} = \sum^{n}_{i=1} \alpha_i \hat{p}_i. -\end{equation*} -!et - -The coefficients are given by -!bt -\begin{equation*} - \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}. -\end{equation*} -!et -Multiplying with $\hat{p}_k^T$ from the left gives - -!bt -\begin{equation*} - \hat{p}_k^T \hat{A}\hat{x} = \sum^{n}_{i=1} \alpha_i\hat{p}_k^T \hat{A}\hat{p}_i= \hat{p}_k^T \hat{b}, -\end{equation*} -!et -and we can define the coefficients $\alpha_k$ as - -!bt -\begin{equation*} - \alpha_k = \frac{\hat{p}_k^T \hat{b}}{\hat{p}_k^T \hat{A} \hat{p}_k} -\end{equation*} -!et - -If we choose the conjugate vectors $\hat{p}_k$ carefully, -then we may not need all of them to obtain a good approximation to the solution -$\hat{x}$. -We want to regard the conjugate gradient method as an iterative method. -This will us to solve systems where $n$ is so large that the direct -method would take too much time. - -We denote the initial guess for $\hat{x}$ as $\hat{x}_0$. -We can assume without loss of generality that -!bt -\begin{equation*} -\hat{x}_0=0, -\end{equation*} -!et -or consider the system -!bt -\begin{equation*} -\hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, -\end{equation*} -!et -instead. - -One can show that the solution $\hat{x}$ is also the unique minimizer of the quadratic form -!bt -\begin{equation*} - f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. -\end{equation*} -!et -This suggests taking the first basis vector $\hat{p}_1$ -to be the gradient of $f$ at $\hat{x}=\hat{x}_0$, -which equals -!bt -\begin{equation*} -\hat{A}\hat{x}_0-\hat{b}, -\end{equation*} -!et -and -$\hat{x}_0=0$ it is equal $-\hat{b}$. -The other vectors in the basis will be conjugate to the gradient, -hence the name conjugate gradient method. - -Let $\hat{r}_k$ be the residual at the $k$-th step: -!bt -\begin{equation*} -\hat{r}_k=\hat{b}-\hat{A}\hat{x}_k. -\end{equation*} -!et -Note that $\hat{r}_k$ is the negative gradient of $f$ at -$\hat{x}=\hat{x}_k$, -so the gradient descent method would be to move in the direction $\hat{r}_k$. -Here, we insist that the directions $\hat{p}_k$ are conjugate to each other, -so we take the direction closest to the gradient $\hat{r}_k$ -under the conjugacy constraint. -This gives the following expression -!bt -\begin{equation*} -\hat{p}_{k+1}=\hat{r}_k-\frac{\hat{p}_k^T \hat{A}\hat{r}_k}{\hat{p}_k^T\hat{A}\hat{p}_k} \hat{p}_k. -\end{equation*} -!et - -We can also compute the residual iteratively as -!bt -\begin{equation*} -\hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, - \end{equation*} -!et -which equals -!bt -\begin{equation*} -\hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{p}_k), - \end{equation*} -!et -or -!bt -\begin{equation*} -(\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{p}_k, - \end{equation*} -!et -which gives - -!bt -\begin{equation*} -\hat{r}_{k+1}=\hat{r}_k-\hat{A}\hat{p}_{k}, - \end{equation*} -!et - -===== Simple implementation of the Conjugate gradient algorithm ===== - -!bc cppcod - Vector ConjugateGradient(Matrix A, Vector b, Vector x0){ - int dim = x0.Dimension(); - const double tolerance = 1.0e-14; - Vector x(dim),r(dim),v(dim),z(dim); - double c,t,d; - - x = x0; - r = b - A*x; - v = r; - c = dot(r,r); - int i = 0; IterMax = dim; - while(i <= IterMax){ - z = A*v; - t = c/dot(v,z); - x = x + t*v; - r = r - t*z; - d = dot(r,r); - if(sqrt(d) < tolerance) - break; - v = r + (d/c)*v; - c = d; i++; - } - return x; -} -!ec - -===== Broyden–Fletcher–Goldfarb–Shanno algorithm ===== - -The optimization problem is to minimize $f(\mathbf {x} )$ where $\mathbf {x}$ is a vector in $R^{n}$, and $f$ is a differentiable scalar function. There are no constraints on the values that $\mathbf {x}$ can take. - -The algorithm begins at an initial estimate for the optimal value $\mathbf {x}_{0}$ and proceeds iteratively to get a better estimate at each stage. - -The search direction $p_k$ at stage $k$ is given by the solution of the analogue of the Newton equation -!bt -\[ -B_{k}\mathbf {p} _{k}=-\nabla f(\mathbf {x}_{k}), -\] -!et - -where $B_{k}$ is an approximation to the Hessian matrix, which is -updated iteratively at each stage, and $\nabla f(\mathbf {x} _{k})$ -is the gradient of the function -evaluated at $x_k$. -A line search in the direction $p_k$ is then used to -find the next point $x_{k+1}$ by minimising -!bt -\[ -f(\mathbf {x}_{k}+\alpha \mathbf {p}_{k}), -\] -!et -over the scalar $\alpha > 0$. - - -We will use linear regression as a case study for the gradient descent -methods. Linear regression is a great test case for the gradient -descent methods discussed in the lectures since it has several -desirable properties such as: - -o An analytical solution. -o The gradient can be computed analytically. -o The cost function is convex which guarantees that gradient descent converges for small enough learning rates - -We revisit the example from homework set 1 where we had -!bt -\[ -y_i = 5x_i^2 + 0.1\xi_i, \ i=1,\cdots,100 -\] -!et -with $x_i \in [0,1] $ chosen randomly with a uniform distribution. Additionally $\xi_i$ represents stochastic noise chosen according to a normal distribution $\cal {N}(0,1)$. -The linear regression model is given by -!bt -\[ -h_\beta(x) = \hat{y} = \beta_0 + \beta_1 x, -\] -!et -such that -!bt -\[ -\hat{y}_i = \beta_0 + \beta_1 x_i. -\] -!et - - -===== Gradient descent example ===== - -Let $\mathbf{y} = (y_1,\cdots,y_n)^T$, $\mathbf{\hat{y}} = (\hat{y}_1,\cdots,\hat{y}_n)^T$ and $\beta = (\beta_0, \beta_1)^T$ - -It is convenient to write $\mathbf{\hat{y}} = X\beta$ where $X \in \mathbb{R}^{100 \times 2} $ is the design matrix given by -!bt -\[ -X \equiv \begin{bmatrix} -1 & x_1 \\ -\vdots & \vdots \\ -1 & x_{100} & \\ -\end{bmatrix}. -\] -!et -The loss function is given by -!bt -\[ -C(\beta) = ||X\beta-\mathbf{y}||^2 = ||X\beta||^2 - 2 \mathbf{y}^T X\beta + ||\mathbf{y}||^2 = \sum_{i=1}^{100} (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2 -\] -!et -and we want to find $\beta$ such that $C(\beta)$ is minimized. - - -Computing $\partial C(\beta) / \partial \beta_0$ and $\partial C(\beta) / \partial \beta_1$ we can show that the gradient can be written as -!bt -\[ -\nabla_{\beta} C(\beta) = (\partial C(\beta) / \partial \beta_0, \partial C(\beta) / \partial \beta_1)^T = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ -\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ -\end{bmatrix} = 2X^T(X\beta - \mathbf{y}), -\] -!et -where $X$ is the design matrix defined above. - - -The Hessian matrix of $C(\beta)$ is given by -!bt -\[ -\hat{H} \equiv \begin{bmatrix} -\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\ -\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\ -\end{bmatrix} = 2X^T X. -\] -!et -This result implies that $C(\beta)$ is a convex function since the matrix $X^T X$ always is positive semi-definite. - - - -===== Simple program ===== - -We can now write a program that minimizes $C(\beta)$ using the gradient descent method with a constant learning rate $\gamma$ according to -!bt -\[ -\beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots -\] -!et - -We can use the expression we computed for the gradient and let use a -$\beta_0$ be chosen randomly and let $\gamma = 0.001$. Stop iterating -when $||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8}$. - -And finally we can compare our solution for $\beta$ with the analytic result given by -$\beta= (X^TX)^{-1} X^T \mathbf{y}$. -!bc pycod -import numpy as np - -""" -The following setup is just a suggestion, feel free to write it the way you like. -""" - -#Setup problem described in the exercise -N = 100 #Nr of datapoints -M = 2 #Nr of features -x = np.random.rand(N) #Uniformly generated x-values in [0,1] -y = 5*x**2 + 0.1*np.random.randn(N) -X = np.c_[np.ones(N),x] #Construct design matrix - -#Compute beta according to normal equations to compare with GD solution -Xt_X_inv = np.linalg.inv(np.dot(X.T,X)) -Xt_y = np.dot(X.transpose(),y) -beta_NE = np.dot(Xt_X_inv,Xt_y) -print(beta_NE) -!ec - - -Another simple example is here -!bc pycod - -# Importing various packages -from random import random, seed -import numpy as np -import matplotlib.pyplot as plt -from mpl_toolkits.mplot3d import Axes3D -from matplotlib import cm -from matplotlib.ticker import LinearLocator, FormatStrFormatter -import sys - -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_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y) -print(beta_linreg) -beta = 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(beta)-y) - beta -= eta*gradients - -print(beta) -xnew = np.array([[0],[2]]) -xbnew = np.c_[np.ones((2,1)), xnew] -ypredict = xbnew.dot(beta) -ypredict2 = xbnew.dot(beta_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'Gradient descent example') -plt.show() - -!ec - - -===== And a corresponding example using _scikit-learn_ ===== - -!bc pycod -# Importing various packages -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] -beta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y) -print(beta_linreg) -sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1) -sgdreg.fit(x,y.ravel()) -print(sgdreg.intercept_, sgdreg.coef_) - -!ec - - - - -===== Gradient descent and Ridge ===== - -We have also discussed Ridge regression where the loss function contains a regularized given by the $L_2$ norm of $\beta$, -!bt -\[ -C_{\text{ridge}}(\beta) = ||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0. -\] -!et - -In order to minimize $C_{\text{ridge}}(\beta)$ using GD we only have adjust the gradient as follows -!bt -\[ -\nabla_\beta C_{\text{ridge}}(\beta) = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ -\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ -\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta). -\] -!et - -We can now extend our program to minimize $C_{\text{ridge}}(\beta)$ using gradient descent and compare with the analytical solution given by -!bt -\[ -\beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}, -\] -!et -for $\lambda = {0,1,10,50,100}$ ($\lambda = 0$ corresponds to ordinary least squares). -We can then compute $||\beta_{\text{ridge}}||$ for each $\lambda$. - -!bc pycod -import numpy as np - -""" -The following setup is just a suggestion, feel free to write it the way you like. -""" - -#Setup problem described in the exercise -N = 100 #Nr of datapoints -M = 2 #Nr of features -x = np.random.rand(N) -y = 5*x**2 + 0.1*np.random.randn(N) - - -#Compute analytic beta for Ridge regression -X = np.c_[np.ones(N),x] -XT_X = np.dot(X.T,X) - -l = 0.1 #Ridge parameter lambda -Id = np.eye(XT_X.shape[0]) - -Z = np.linalg.inv(XT_X+l*Id) -beta_ridge = np.dot(Z,np.dot(X.T,y)) - -print(beta_ridge) -print(np.linalg.norm(beta_ridge)) #||beta|| -!ec - - -===== Automatic differentiation ===== -Python has tools for so-called _automatic differentiation_. -Consider the following example -!bt -\[ -f(x) = \sin\left(2\pi x + x^2\right) -\] -!et -which has the following derivative -!bt -\[ -f'(x) = \cos\left(2\pi x + x^2\right)\left(2\pi + 2x\right) -\] -!et -Using _autograd_ we have - -!bc pycod -import autograd.numpy as np - -# To do elementwise differentiation: -from autograd import elementwise_grad as egrad - -# To plot: -import matplotlib.pyplot as plt - - -def f(x): - return np.sin(2*np.pi*x + x**2) - -def f_grad_analytic(x): - return np.cos(2*np.pi*x + x**2)*(2*np.pi + 2*x) - -# Do the comparison: -x = np.linspace(0,1,1000) - -f_grad = egrad(f) - -computed = f_grad(x) -analytic = f_grad_analytic(x) - -plt.title('Derivative computed from Autograd compared with the analytical derivative') -plt.plot(x,computed,label='autograd') -plt.plot(x,analytic,label='analytic') - -plt.xlabel('x') -plt.ylabel('y') -plt.legend() - -plt.show() - -print("The max absolute difference is: %g"%(np.max(np.abs(computed - analytic)))) -!ec - - -===== Using autograd ===== - -Here we -experiment with what kind of functions Autograd is capable -of finding the gradient of. The following Python functions are just -meant to illustrate what Autograd can do, but please feel free to -experiment with other, possibly more complicated, functions as well. - -!bc pycod -import autograd.numpy as np -from autograd import grad - -def f1(x): - return x**3 + 1 - -f1_grad = grad(f1) - -# Remember to send in float as argument to the computed gradient from Autograd! -a = 1.0 - -# See the evaluated gradient at a using autograd: -print("The gradient of f1 evaluated at a = %g using autograd is: %g"%(a,f1_grad(a))) - -# Compare with the analytical derivative, that is f1'(x) = 3*x**2 -grad_analytical = 3*a**2 -print("The gradient of f1 evaluated at a = %g by finding the analytic expression is: %g"%(a,grad_analytical)) -!ec - - - -===== Autograd with more complicated functions ===== - -To differentiate with respect to two (or more) arguments of a Python -function, Autograd need to know at which variable the function if -being differentiated with respect to. - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f2(x1,x2): - return 3*x1**3 + x2*(x1 - 5) + 1 - -# By sending the argument 0, Autograd will compute the derivative w.r.t the first variable, in this case x1 -f2_grad_x1 = grad(f2,0) - -# ... and differentiate w.r.t x2 by sending 1 as an additional arugment to grad -f2_grad_x2 = grad(f2,1) - -x1 = 1.0 -x2 = 3.0 - -print("Evaluating at x1 = %g, x2 = %g"%(x1,x2)) -print("-"*30) - -# Compare with the analytical derivatives: - -# Derivative of f2 w.r.t x1 is: 9*x1**2 + x2: -f2_grad_x1_analytical = 9*x1**2 + x2 - -# Derivative of f2 w.r.t x2 is: x1 - 5: -f2_grad_x2_analytical = x1 - 5 - -# See the evaluated derivations: -print("The derivative of f2 w.r.t x1: %g"%( f2_grad_x1(x1,x2) )) -print("The analytical derivative of f2 w.r.t x1: %g"%( f2_grad_x1(x1,x2) )) - -print() - -print("The derivative of f2 w.r.t x2: %g"%( f2_grad_x2(x1,x2) )) -print("The analytical derivative of f2 w.r.t x2: %g"%( f2_grad_x2(x1,x2) )) -!ec - -Note that the grad function will not produce the true gradient of the function. The true gradient of a function with two or more variables will produce a vector, where each element is the function differentiated w.r.t a variable. - - - -===== More complicated functions using the elements of their arguments directly ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f3(x): # Assumes x is an array of length 5 or higher - return 2*x[0] + 3*x[1] + 5*x[2] + 7*x[3] + 11*x[4]**2 - -f3_grad = grad(f3) - -x = np.linspace(0,4,5) - -# Print the computed gradient: -print("The computed gradient of f3 is: ", f3_grad(x)) - -# The analytical gradient is: (2, 3, 5, 7, 22*x[4]) -f3_grad_analytical = np.array([2, 3, 5, 7, 22*x[4]]) - -# Print the analytical gradient: -print("The analytical gradient of f3 is: ", f3_grad_analytical) -!ec - -Note that in this case, when sending an array as input argument, the -output from Autograd is another array. This is the true gradient of -the function, as opposed to the function in the previous example. By -using arrays to represent the variables, the output from Autograd -might be easier to work with, as the output is closer to what one -could expect form a gradient-evaluting function. - - -===== Functions using mathematical functions from Numpy ===== - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f4(x): - return np.sqrt(1+x**2) + np.exp(x) + np.sin(2*np.pi*x) - -f4_grad = grad(f4) - -x = 2.7 - -# Print the computed derivative: -print("The computed derivative of f4 at x = %g is: %g"%(x,f4_grad(x))) - -# The analytical derivative is: x/sqrt(1 + x**2) + exp(x) + cos(2*pi*x)*2*pi -f4_grad_analytical = x/np.sqrt(1 + x**2) + np.exp(x) + np.cos(2*np.pi*x)*2*np.pi - -# Print the analytical gradient: -print("The analytical gradient of f4 at x = %g is: %g"%(x,f4_grad_analytical)) -!ec - - - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f5(x): - if x >= 0: - return x**2 - else: - return -3*x + 1 - -f5_grad = grad(f5) - -x = 2.7 - -# Print the computed derivative: -print("The computed derivative of f5 at x = %g is: %g"%(x,f5_grad(x))) -!ec - - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f6_for(x): - val = 0 - for i in range(10): - val = val + x**i - return val - -def f6_while(x): - val = 0 - i = 0 - while i < 10: - val = val + x**i - i = i + 1 - return val - -f6_for_grad = grad(f6_for) -f6_while_grad = grad(f6_while) - -x = 0.5 - -# Print the computed derivaties of f6_for and f6_while -print("The computed derivative of f6_for at x = %g is: %g"%(x,f6_for_grad(x))) -print("The computed derivative of f6_while at x = %g is: %g"%(x,f6_while_grad(x))) -!ec -!bc pycod -import autograd.numpy as np -from autograd import grad -# Both of the functions are implementation of the sum: sum(x**i) for i = 0, ..., 9 -# The analytical derivative is: sum(i*x**(i-1)) -f6_grad_analytical = 0 -for i in range(10): - f6_grad_analytical += i*x**(i-1) - -print("The analytical derivative of f6 at x = %g is: %g"%(x,f6_grad_analytical)) -!ec - - -===== Using recursion ===== -!bc pycod -import autograd.numpy as np -from autograd import grad - -def f7(n): # Assume that n is an integer - if n == 1 or n == 0: - return 1 - else: - return n*f7(n-1) - -f7_grad = grad(f7) - -n = 2.0 - -print("The computed derivative of f7 at n = %d is: %g"%(n,f7_grad(n))) - -# The function f7 is an implementation of the factorial of n. -# By using the product rule, one can find that the derivative is: - -f7_grad_analytical = 0 -for i in range(int(n)-1): - tmp = 1 - for k in range(int(n)-1): - if k != i: - tmp *= (n - k) - f7_grad_analytical += tmp - -print("The analytical derivative of f7 at n = %d is: %g"%(n,f7_grad_analytical)) - -!ec -Note that if n is equal to zero or one, Autograd will give an error message. This message appears when the output is independent on input. - - -===== Unsupported functions ===== -Autograd supports many features. However, there are some functions that is not supported (yet) by Autograd. - -Assigning a value to the variable being differentiated with respect to -!bc pycod -import autograd.numpy as np -from autograd import grad -def f8(x): # Assume x is an array - x[2] = 3 - return x*2 - -f8_grad = grad(f8) - -x = 8.4 - -print("The derivative of f8 is:",f8_grad(x)) -!ec -Here, Autograd tells us that an 'ArrayBox' does not support item assignment. The item assignment is done when the program tries to assign x[2] to the value 3. However, Autograd has implemented the computation of the derivative such that this assignment is not possible. - - -===== The syntax a.dot(b) when finding the dot product ===== -!bc pycod -import autograd.numpy as np -from autograd import grad -def f9(a): # Assume a is an array with 2 elements - b = np.array([1.0,2.0]) - return a.dot(b) - -f9_grad = grad(f9) - -x = np.array([1.0,0.0]) - -print("The derivative of f9 is:",f9_grad(x)) -!ec - -Here we are told that the 'dot' function does not belong to Autograd's -version of a Numpy array. To overcome this, an alternative syntax -which also computed the dot product can be used: - -!bc pycod -import autograd.numpy as np -from autograd import grad -def f9_alternative(x): # Assume a is an array with 2 elements - b = np.array([1.0,2.0]) - return np.dot(x,b) # The same as x_1*b_1 + x_2*b_2 - -f9_alternative_grad = grad(f9_alternative) - -x = np.array([3.0,0.0]) - -print("The gradient of f9 is:",f9_alternative_grad(x)) - -# The analytical gradient of the dot product of vectors x and b with two elements (x_1,x_2) and (b_1, b_2) respectively -# w.r.t x is (b_1, b_2). -!ec - - -===== Recommended to avoid ===== -The documentation recommends to avoid inplace operations such as -!bc pycod -a += b -a -= b -a*= b -a /=b -!ec - - -===== Stochastic Gradient Descent ===== - -Stochastic gradient descent (SGD) and variants thereof address some of -the shortcomings of the Gradient descent method discussed above. - -The underlying idea of SGD comes from the observation that the cost -function, which we want to minimize, can almost always be written as a -sum over $n$ data points $\{\mathbf{x}_i\}_{i=1}^n$, -!bt -\[ -C(\mathbf{\beta}) = \sum_{i=1}^n c_i(\mathbf{x}_i, -\mathbf{\beta}). -\] -!et - - -This in turn means that the gradient can be -computed as a sum over $i$-gradients -!bt -\[ -\nabla_\beta C(\mathbf{\beta}) = \sum_i^n \nabla_\beta c_i(\mathbf{x}_i, -\mathbf{\beta}). -\] -!et - -Stochasticity/randomness is introduced by only taking the -gradient on a subset of the data called minibatches. If there are $n$ -data points and the size of each minibatch is $M$, there will be $n/M$ -minibatches. We denote these minibatches by $B_k$ where -$k=1,\cdots,n/M$. - - -As an example, suppose we have $10$ data points $(\mathbf{x}_1,\cdots, \mathbf{x}_{10})$ -and we choose to have $M=5$ minibathces, -then each minibatch contains two data points. In particular we have -$B_1 = (\mathbf{x}_1,\mathbf{x}_2), \cdots, B_5 = -(\mathbf{x}_9,\mathbf{x}_{10})$. Note that if you choose $M=1$ you -have only a single batch with all data points and on the other extreme, -you may choose $M=n$ resulting in a minibatch for each datapoint, i.e -$B_k = \mathbf{x}_k$. - -The idea is now to approximate the gradient by replacing the sum over -all data points with a sum over the data points in one the minibatches -picked at random in each gradient descent step -!bt -\[ -\nabla_{\beta} -C(\mathbf{\beta}) = \sum_{i=1}^n \nabla_\beta c_i(\mathbf{x}_i, -\mathbf{\beta}) \rightarrow \sum_{i \in B_k}^n \nabla_\beta -c_i(\mathbf{x}_i, \mathbf{\beta}). -\] -!et - - - -Thus a gradient descent step now looks like -!bt -\[ -\beta_{j+1} = \beta_j - \gamma_j \sum_{i \in B_k}^n \nabla_\beta c_i(\mathbf{x}_i, -\mathbf{\beta}) -\] -!et - -where $k$ is picked at random with equal -probability from $[1,n/M]$. An iteration over the number of -minibathces (n/M) is commonly referred to as an epoch. Thus it is -typical to choose a number of epochs and for each epoch iterate over -the number of minibatches, as exemplified in the code below. - - -!bc pycod -import numpy as np - -n = 100 #100 datapoints -M = 5 #size of each minibatch -m = int(n/M) #number of minibatches -n_epochs = 10 #number of epochs - -j = 0 -for epoch in range(1,n_epochs+1): - for i in range(m): - k = np.random.randint(m) #Pick the k-th minibatch at random - #Compute the gradient using the data in minibatch Bk - #Compute new suggestion for - j += 1 -!ec - -Taking the gradient only on a subset of the data has two important -benefits. First, it introduces randomness which decreases the chance -that our opmization scheme gets stuck in a local minima. Second, if -the size of the minibatches are small relative to the number of -datapoints ($M < n$), the computation of the gradient is much -cheaper since we sum over the datapoints in the $k-th$ minibatch and not -all $n$ datapoints. - - -A natural question is when do we stop the search for a new minimum? -One possibility is to compute the full gradient after a given number -of epochs and check if the norm of the gradient is smaller than some -threshold and stop if true. However, the condition that the gradient -is zero is valid also for local minima, so this would only tell us -that we are close to a local/global minimum. However, we could also -evaluate the cost function at this point, store the result and -continue the search. If the test kicks in at a later stage we can -compare the values of the cost function and keep the $\beta$ that -gave the lowest value. - - -Another approach is to let the step length $\gamma_j$ depend on the -number of epochs in such a way that it becomes very small after a -reasonable time such that we do not move at all. - -As an example, let $e = 0,1,2,3,\cdots$ denote the current epoch and let $t_0, t_1 > 0$ be two fixed numbers. Furthermore, let $t = e \cdot m + i$ where $m$ is the number of minibatches and $i=0,\cdots,m-1$. Then the function $$\gamma_j(t; t_0, t_1) = \frac{t_0}{t+t_1} $$ goes to zero as the number of epochs gets large. I.e. we start with a step length $\gamma_j (0; t_0, t_1) = t_0/t_1$ which decays in *time* $t$. - -In this way we can fix the number of epochs, compute $\beta$ and -evaluate the cost function at the end. Repeating the computation will -give a different result since the scheme is random by design. Then we -pick the final $\beta$ that gives the lowest value of the cost -function. - -!bc pycod -import numpy as np - -def step_length(t,t0,t1): - return t0/(t+t1) - -n = 100 #100 datapoints -M = 5 #size of each minibatch -m = int(n/M) #number of minibatches -n_epochs = 500 #number of epochs -t0 = 1.0 -t1 = 10 - -gamma_j = t0/t1 -j = 0 -for epoch in range(1,n_epochs+1): - for i in range(m): - k = np.random.randint(m) #Pick the k-th minibatch at random - #Compute the gradient using the data in minibatch Bk - #Compute new suggestion for beta - t = epoch*m+i - gamma_j = step_length(t,t0,t1) - j += 1 - -print("gamma_j after %d epochs: %g" % (n_epochs,gamma_j)) -!ec - - - -!bc pycod -# 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("Own inversion") -print(theta_linreg) -sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1) -sgdreg.fit(x,y.ravel()) -print("sgdreg from scikit") -print(sgdreg.intercept_, sgdreg.coef_) - - -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 frm own gd") -print(theta) - -xnew = np.array([[0],[2]]) -xbnew = np.c_[np.ones((2,1)), xnew] -ypredict = xbnew.dot(theta) -ypredict2 = xbnew.dot(theta_linreg) - - -n_epochs = 50 -t0, t1 = 5, 50 -m = 100 -def learning_schedule(t): - return t0/(t+t1) - -theta = np.random.randn(2,1) - -for epoch in range(n_epochs): - for i in range(m): - random_index = np.random.randint(m) - xi = xb[random_index:random_index+1] - yi = y[random_index:random_index+1] - gradients = 2 * xi.T.dot(xi.dot(theta)-yi) - eta = learning_schedule(epoch*m+i) - theta = theta - eta*gradients -print("theta from own sdg") -print(theta) - - - - - - -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() - -!ec - - -===== Using gradient descent methods, limitations ===== - -* _Gradient descent (GD) finds local minima of our function_. Since the GD algorithm is deterministic, if it converges, it will converge to a local minimum of our energy function. Because in ML we are often dealing with extremely rugged landscapes with many local minima, this can lead to poor performance. - -* _GD is sensitive to initial conditions_. One consequence of the local nature of GD is that initial conditions matter. Depending on where one starts, one will end up at a different local minima. Therefore, it is very important to think about how one initializes the training process. This is true for GD as well as more complicated variants of GD. - -* _Gradients are computationally expensive to calculate for large datasets_. In many cases in statistics and ML, the energy function is a sum of terms, with one term for each data point. For example, in linear regression, $E \propto \sum_{i=1}^n (y_i - \mathbf{w}^T\cdot\mathbf{x}_i)^2$; for logistic regression, the square error is replaced by the cross entropy. To calculate the gradient we have to sum over *all* $n$ data points. Doing this at every GD step becomes extremely computationally expensive. An ingenious solution to this, is to calculate the gradients using small subsets of the data called ``mini batches''. This has the added benefit of introducing stochasticity into our algorithm. - -* _GD is very sensitive to choices of learning rates_. GD is extremely sensitive to the choice of learning rates. If the learning rate is very small, the training process take an extremely long time. For larger learning rates, GD can diverge and give poor results. Furthermore, depending on what the local landscape looks like, we have to modify the learning rates to ensure convergence. Ideally, we would *adaptively* choose the learning rates to match the landscape. - -* _GD treats all directions in parameter space uniformly._ Another major drawback of GD is that unlike Newton's method, the learning rate for GD is the same in all directions in parameter space. For this reason, the maximum learning rate is set by the behavior of the steepest direction and this can significantly slow down training. Ideally, we would like to take large steps in flat directions and small steps in steep directions. Since we are exploring rugged landscapes where curvatures change, this requires us to keep track of not only the gradient but second derivatives. The ideal scenario would be to calculate the Hessian but this proves to be too computationally expensive. - -* GD can take exponential time to escape saddle points, even with random initialization. As we mentioned, GD is extremely sensitive to initial condition since it determines the particular local minimum GD would eventually reach. However, even with a good initialization scheme, through the introduction of randomness, GD can still take exponential time to escape saddle points. - - - -===== Momentum based GD ===== - -The stochastic gradient descent (SGD) is almost always used with a *momentum* or inertia term that serves as a memory of the direction we are moving in parameter space. This is typically -implemented as follows -!bt -\begin{align} -\mathbf{v}_{t}&=\gamma \mathbf{v}_{t-1}+\eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t) \nonumber \\ -\boldsymbol{\theta}_{t+1}&= \boldsymbol{\theta}_t -\mathbf{v}_{t}, -\end{align} -!et -where we have introduced a momentum parameter $\gamma$, with $0\le\gamma\le 1$, and for brevity we dropped the explicit notation to indicate the gradient is to be taken over a different mini-batch at each step. We call this algorithm gradient descent with momentum (GDM). From these equations, it is clear that $\mathbf{v}_t$ is a running average of recently encountered gradients and $(1-\gamma)^{-1}$ sets the characteristic time scale for the memory used in the averaging procedure. Consistent with this, when $\gamma=0$, this just reduces down to ordinary SGD as discussed earlier. An equivalent way of writing the updates is -!bt -\[ -\Delta \boldsymbol{\theta}_{t+1} = \gamma \Delta \boldsymbol{\theta}_t -\ \eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t), -\] -!et -where we have defined $\Delta \boldsymbol{\theta}_{t}= \boldsymbol{\theta}_t-\boldsymbol{\theta}_{t-1}$. - - -===== More on momentum based approaches ===== - -Let us try to get more intuition from these equations. It is helpful to consider a simple physical analogy with a particle of mass $m$ moving in a viscous medium with drag coefficient $\mu$ and potential -$E(\mathbf{w})$. If we denote the particle's position by $\mathbf{w}$, then its motion is described by -!bt -\[ -m {d^2 \mathbf{w} \over dt^2} + \mu {d \mathbf{w} \over dt }= -\nabla_w E(\mathbf{w}). -\] -!et -We can discretize this equation in the usual way to get -!bt -\[ -m { \mathbf{w}_{t+\Delta t}-2 \mathbf{w}_{t} +\mathbf{w}_{t-\Delta t} \over (\Delta t)^2}+\mu {\mathbf{w}_{t+\Delta t}- \mathbf{w}_{t} \over \Delta t} = -\nabla_w E(\mathbf{w}). -\] -!et -Rearranging this equation, we can rewrite this as -!bt -\[ -\Delta \mathbf{w}_{t +\Delta t}= - { (\Delta t)^2 \over m +\mu \Delta t} \nabla_w E(\mathbf{w})+ {m \over m +\mu \Delta t} \Delta \mathbf{w}_t. -\] -!et - - -===== Momentum parameter ===== -Notice that this equation is identical to previous one if we identify the position of the particle, $\mathbf{w}$, with the parameters $\boldsymbol{\theta}$. This allows -us to identify the momentum parameter and learning rate with the mass of the particle and the viscous drag as: -!bt -\[ -\gamma= {m \over m +\mu \Delta t }, \qquad \eta = {(\Delta t)^2 \over m +\mu \Delta t}. -\] -!et -Thus, as the name suggests, the momentum parameter is proportional to the mass of the particle and effectively provides inertia. Furthermore, in the large viscosity/small learning rate limit, our memory time scales as $(1-\gamma)^{-1} \approx m/(\mu \Delta t)$. - -Why is momentum useful? SGD momentum helps the gradient descent algorithm gain speed in directions with persistent but small gradients even in the presence of stochasticity, while suppressing oscillations in high-curvature directions. This becomes especially important in situations where the landscape is shallow and flat in some directions and narrow and steep in others. It has been argued that first-order methods (with appropriate initial conditions) can perform comparable to more expensive second order methods, especially in the context of complex deep learning models. - -These beneficial properties of momentum can sometimes become even more pronounced by using a slight modification of the classical momentum algorithm called Nesterov Accelerated Gradient (NAG). - -In the NAG algorithm, rather than calculating the gradient at the current parameters, $\nabla_\theta E(\boldsymbol{\theta}_t)$, one calculates the gradient at the expected value of the parameters given our current momentum, $\nabla_\theta E(\boldsymbol{\theta}_t +\gamma \mathbf{v}_{t-1})$. This yields the NAG update rule -!bt -\begin{align} -\mathbf{v}_{t}&=\gamma \mathbf{v}_{t-1}+\eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t +\gamma \mathbf{v}_{t-1}) \nonumber \\ -\boldsymbol{\theta}_{t+1}&= \boldsymbol{\theta}_t -\mathbf{v}_{t}. -\end{align} -!et -One of the major advantages of NAG is that it allows for the use of a larger learning rate than GDM for the same choice of $\gamma$. - - - -In stochastic gradient descent, with and without momentum, we still -have to specify a schedule for tuning the learning rates $\eta_t$ -as a function of time. As discussed in the context of Newton's -method, this presents a number of dilemmas. The learning rate is -limited by the steepest direction which can change depending on the -current position in the landscape. To circumvent this problem, ideally -our algorithm would keep track of curvature and take large steps in -shallow, flat directions and small steps in steep, narrow directions. -Second-order methods accomplish this by calculating or approximating -the Hessian and normalizing the learning rate by the -curvature. However, this is very computationally expensive for -extremely large models. Ideally, we would like to be able to -adaptively change the step size to match the landscape without paying -the steep computational price of calculating or approximating -Hessians. - -Recently, a number of methods have been introduced that accomplish this by tracking not only the gradient, but also the second moment of the gradient. These methods include AdaGrad, AdaDelta, RMS-Prop, and ADAM. - - -===== RMS prop ===== - -In RMS prop, in addition to keeping a running average of the first moment of the gradient, we also keep track of the second moment denoted by $\mathbf{s}_t=\mathbb{E}[\mathbf{g}_t^2]$. The update rule for RMS prop is given by -!bt -\begin{align} -\mathbf{g}_t &= \nabla_\theta E(\boldsymbol{\theta}) \\ -\mathbf{s}_t &=\beta \mathbf{s}_{t-1} +(1-\beta)\mathbf{g}_t^2 \nonumber \\ -\boldsymbol{\theta}_{t+1}&=&\boldsymbol{\theta}_t - \eta_t { \mathbf{g}_t \over \sqrt{\mathbf{s}_t +\epsilon}}, \nonumber -\end{align} -!et -where $\beta$ controls the averaging time of the second moment and is typically taken to be about $\beta=0.9$, $\eta_t$ is a learning rate typically chosen to be $10^{-3}$, and $\epsilon\sim 10^{-8} $ is a small regularization constant to prevent divergences. Multiplication and division by vectors is understood as an element-wise operation. It is clear from this formula that the learning rate is reduced in directions where the norm of the gradient is consistently large. This greatly speeds up the convergence by allowing us to use a larger learning rate for flat directions. - - - -===== ADAM optimizer ===== - -A related algorithm is the ADAM optimizer. In ADAM, we keep a running average of both the first and second moment of the gradient and use this information to adaptively change the learning rate for different parameters. In addition to keeping a running average of the first and second moments of the gradient (i.e. $\mathbf{m}_t=\mathbb{E}[\mathbf{g}_t]$ and $\mathbf{s}_t=\mathbb{E}[\mathbf{g}^2_t]$, respectively), ADAM performs an additional bias correction to account for the fact that we are estimating the first two moments of the gradient using a running average (denoted by the hats in the update rule below). The update rule for ADAM is given by (where multiplication and division are once again understood to be element-wise operations below) -!bt -\begin{align} -\mathbf{g}_t &= \nabla_\theta E(\boldsymbol{\theta}) \\ -\mathbf{m}_t &= \beta_1 \mathbf{m}_{t-1} + (1-\beta_1) \mathbf{g}_t \nonumber \\ -\mathbf{s}_t &=\beta_2 \mathbf{s}_{t-1} +(1-\beta_2)\mathbf{g}_t^2 \nonumber \\ -\hat{\mathbf{m}}_t&={\mathbf{m}_t \over 1-\beta_1^t} \nonumber \\ -\hat{\mathbf{s}}_t &={\mathbf{s}_t \over1-\beta_2^t} \nonumber \\ -\boldsymbol{\theta}_{t+1}&=\boldsymbol{\theta}_t - \eta_t { \hat{\mathbf{m}}_t \over \sqrt{\hat{\mathbf{s}}_t} +\epsilon}, \nonumber \\ -\end{align} -!et -where $\beta_1$ and $\beta_2$ set the memory lifetime of the first and second moment and are typically taken to be $0.9$ and $0.99$ respectively, and $\eta$ and $\epsilon$ are identical to RMSprop. - -Like in RMSprop, the effective step size of a parameter depends on the magnitude of its gradient squared. To understand this better, let us rewrite this expression in terms of the variance $\boldsymbol{\sigma}_t^2 = \hat{\mathbf{s}}_t - (\hat{\mathbf{m}}_t)^2$. Consider a single parameter $\theta_t$. The update rule for this parameter is given by -!bt -\[ -\Delta \theta_{t+1}= -\eta_t { \hat{m}_t \over \sqrt{\sigma_t^2 + m_t^2 }+\epsilon}. -\] -!et - - - - - -===== Practical tips ===== - -* _Randomize the data when making mini-batches_. It is always important to randomly shuffle the data when forming mini-batches. Otherwise, the gradient descent method can fit spurious correlations resulting from the order in which data is presented. - -* _Transform your inputs_. Learning becomes difficult when our landscape has a mixture of steep and flat directions. One simple trick for minimizing these situations is to standardize the data by subtracting the mean and normalizing the variance of input variables. Whenever possible, also decorrelate the inputs. To understand why this is helpful, consider the case of linear regression. It is easy to show that for the squared error cost function, the Hessian of the energy matrix is just the correlation matrix between the inputs. Thus, by standardizing the inputs, we are ensuring that the landscape looks homogeneous in all directions in parameter space. Since most deep networks can be viewed as linear transformations followed by a non-linearity at each layer, we expect this intuition to hold beyond the linear case. - -* _Monitor the out-of-sample performance._ Always monitor the performance of your model on a validation set (a small portion of the training data that is held out of the training process to serve as a proxy for the test set. If the validation error starts increasing, then the model is beginning to overfit. Terminate the learning process. This *early stopping* significantly improves performance in many settings. - -* _Adaptive optimization methods don't always have good generalization._ Recent studies have shown that adaptive methods such as ADAM, RMSPorp, and AdaGrad tend to have poor generalization compared to SGD or SGD with momentum, particularly in the high-dimensional limit (i.e. the number of parameters exceeds the number of data points). Although it is not clear at this stage why these methods perform so well in training deep neural networks, simpler procedures like properly-tuned SGD may work as well or better in these applications. - -