FYS-STK4155/doc/pub/week36/ipynb/week36.ipynb

Week 36: Statistical interpretation of Linear Regression and Resampling techniques

Morten Hjorth-Jensen, Department of Physics, University of Oslo and Department of Physics and Astronomy and National Superconducting Cyclotron Laboratory, Michigan State University

Date: Sep 7, 2021

Copyright 1999-2021, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

Plans for week 36

• Thursday: Summary from last week on SVD, Statistics, probability theory and linear regression

• Friday: Linear Regression and links with Statistics, Resampling methods and presentation of first project.

Thursday September 9

Summary from last Week and Examples

Linking with the SVD

We saw earlier that

\boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T=\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T.

Since the matrices here have dimension p\times p, with p corresponding to the singular values, we defined earlier the matrix

\boldsymbol{\Sigma}^T\boldsymbol{\Sigma} = \begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \\ \boldsymbol{0}\\ \end{bmatrix},

where the tilde-matrix \tilde{\boldsymbol{\Sigma}} is a matrix of dimension p\times p containing only the singular values \sigma_i, that is

\tilde{\boldsymbol{\Sigma}}=\begin{bmatrix} \sigma_0 & 0 & 0 & \dots & 0 & 0 \\
                                    0 & \sigma_1 & 0 & \dots & 0 & 0 \\
				    0 & 0 & \sigma_2 & \dots & 0 & 0 \\
				    0 & 0 & 0 & \dots & \sigma_{p-2} & 0 \\
				    0 & 0 & 0 & \dots & 0 & \sigma_{p-1} \\
\end{bmatrix},

meaning we can write

\boldsymbol{X}^T\boldsymbol{X}=\boldsymbol{V}\tilde{\boldsymbol{\Sigma}}^2\boldsymbol{V}^T.

Multiplying from the right with \boldsymbol{V} (using the orthogonality of \boldsymbol{V}) we get

\left(\boldsymbol{X}^T\boldsymbol{X}\right)\boldsymbol{V}=\boldsymbol{V}\tilde{\boldsymbol{\Sigma}}^2.

What does it mean?

This means the vectors \boldsymbol{v}_i of the orthogonal matrix \boldsymbol{V} are the eigenvectors of the matrix \boldsymbol{X}^T\boldsymbol{X} with eigenvalues given by the singular values squared, that is

\left(\boldsymbol{X}^T\boldsymbol{X}\right)\boldsymbol{v}_i=\boldsymbol{v}_i\sigma_i^2.

In other words, each non-zero singular value of \boldsymbol{X} is a positive square root of an eigenvalue of \boldsymbol{X}^T\boldsymbol{X}. It means also that the columns of \boldsymbol{V} are the eigenvectors of \boldsymbol{X}^T\boldsymbol{X}. Since we have ordered the singular values of \boldsymbol{X} in a descending order, it means that the column vectors \boldsymbol{v}_i are hierarchically ordered by how much correlation they encode from the columns of \boldsymbol{X}.

Note that these are also the eigenvectors and eigenvalues of the Hessian matrix.

If we now recall the definition of the covariance matrix (not using Bessel's correction) we have

\boldsymbol{C}[\boldsymbol{X}]=\frac{1}{n}\boldsymbol{X}^T\boldsymbol{X},

meaning that every squared non-singular value of \boldsymbol{X} divided by n ( the number of samples) are the eigenvalues of the covariance matrix. Every singular value of \boldsymbol{X} is thus a positive square root of an eigenvalue of \boldsymbol{X}^T\boldsymbol{X}. If the matrix \boldsymbol{X} is self-adjoint, the singular values of \boldsymbol{X} are equal to the absolute value of the eigenvalues of \boldsymbol{X}.

And finally \boldsymbol{X}\boldsymbol{X}^T

For \boldsymbol{X}\boldsymbol{X}^T we found

\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T\boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{U}^T=\boldsymbol{U}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{U}^T.

Since the matrices here have dimension n\times n, we have

\boldsymbol{\Sigma}\boldsymbol{\Sigma}^T = \begin{bmatrix} \tilde{\boldsymbol{\Sigma}} \\ \boldsymbol{0}\\ \end{bmatrix}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}}  \boldsymbol{0}\\ \end{bmatrix}=\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix},

leading to

\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}\boldsymbol{U}^T.

Multiplying with \boldsymbol{U} from the right gives us the eigenvalue problem

(\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U}=\boldsymbol{U}\begin{bmatrix} \tilde{\boldsymbol{\Sigma}} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{0}\\ \end{bmatrix}.

It means that the eigenvalues of \boldsymbol{X}\boldsymbol{X}^T are again given by the non-zero singular values plus now a series of zeros. The column vectors of \boldsymbol{U} are the eigenvectors of \boldsymbol{X}\boldsymbol{X}^T and measure how much correlations are contained in the rows of \boldsymbol{X}.

Since we will mainly be interested in the correlations among the features of our data (the columns of \boldsymbol{X}, the quantity of interest for us are the non-zero singular values and the column vectors of \boldsymbol{V}.

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

{\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.

or we can state it as

{\displaystyle \min_{\boldsymbol{\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 \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2,

where we have used the definition of a norm-2 vector, that is

\vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}.

By minimizing the above equation with respect to the parameters \boldsymbol{\beta} we could then obtain an analytical expression for the parameters \boldsymbol{\beta}. We can add a regularization parameter \lambda by defining a new cost function to be optimized, that is

{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2

which leads to the Ridge regression minimization problem where we require that \vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t, where t is a finite number larger than zero. By defining

C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1,

we have a new optimization equation

{\displaystyle \min_{\boldsymbol{\beta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1

which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator.

Here we have defined the norm-1 as

\vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert.

Deriving the Ridge Regression Equations

Using the matrix-vector expression for Ridge regression and dropping the parameter 1/n in front of the standard means squared error equation, we have

C(\boldsymbol{X},\boldsymbol{\beta})=\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta},

and taking the derivatives with respect to \boldsymbol{\beta} we obtain then a slightly modified matrix inversion problem which for finite values of \lambda does not suffer from singularity problems. We obtain the optimal parameters

\hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \left(\boldsymbol{X}^T\boldsymbol{X}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y},

with \boldsymbol{I} being a p\times p identity matrix with the constraint that

\sum_{i=0}^{p-1} \beta_i^2 \leq t,

with t a finite positive number.

When we compare this with the ordinary least squares result we have

\hat{\boldsymbol{\beta}}_{\mathrm{OLS}} = \left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y},

which can lead to singular matrices. However, with the SVD, we can always compute the inverse of the matrix \boldsymbol{X}^T\boldsymbol{X}.

We see that Ridge regression is nothing but the standard OLS with a modified diagonal term added to \boldsymbol{X}^T\boldsymbol{X}. The consequences, in particular for our discussion of the bias-variance tradeoff are rather interesting. We will see that for specific values of \lambda, we may even reduce the variance of the optimal parameters \boldsymbol{\beta}. These topics and other related ones, will be discussed after the more linear algebra oriented analysis here.

Using our insights about the SVD of the design matrix \boldsymbol{X} We have already analyzed the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix \boldsymbol{U} as

\tilde{\boldsymbol{y}}_{\mathrm{OLS}}=\boldsymbol{X}\boldsymbol{\beta}  =\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}.

For Ridge regression this becomes

\tilde{\boldsymbol{y}}_{\mathrm{Ridge}}=\boldsymbol{X}\boldsymbol{\beta}_{\mathrm{Ridge}} = \boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{\Sigma}^2\boldsymbol{V}^T+\lambda\boldsymbol{I} \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\sum_{j=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\frac{\sigma_j^2}{\sigma_j^2+\lambda}\boldsymbol{y},

with the vectors \boldsymbol{u}_j being the columns of \boldsymbol{U} from the SVD of the matrix \boldsymbol{X}.

Interpreting the Ridge results

Since \lambda \geq 0, it means that compared to OLS, we have

\frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1.

Ridge regression finds the coordinates of \boldsymbol{y} with respect to the orthonormal basis \boldsymbol{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. More about this when we have covered the material on a statistical interpretation of various linear regression methods.

More interpretations

For the sake of simplicity, let us assume that the design matrix is orthonormal, that is

\boldsymbol{X}^T\boldsymbol{X}=(\boldsymbol{X}^T\boldsymbol{X})^{-1} =\boldsymbol{I}.

In this case the standard OLS results in

\boldsymbol{\beta}^{\mathrm{OLS}} = \boldsymbol{X}^T\boldsymbol{y}=\sum_{i=0}^{p-1}\boldsymbol{u}_j\boldsymbol{u}_j^T\boldsymbol{y},

and

\boldsymbol{\beta}^{\mathrm{Ridge}} = \left(\boldsymbol{I}+\lambda\boldsymbol{I}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\left(1+\lambda\right)^{-1}\boldsymbol{\beta}^{\mathrm{OLS}},

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 article is highly recommended. Similarly, Mehta et al's article is also recommended.

Deriving the Lasso Regression Equations

Using the matrix-vector expression for Lasso regression and dropping the parameter 1/n in front of the standard means squared error equation, we have the following cost function

C(\boldsymbol{X},\boldsymbol{\beta})=\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\vert\vert\boldsymbol{\beta}\vert\vert_1,

Taking the derivative with respect to \boldsymbol{\beta} and recalling that the derivative of the absolute value is (we drop the boldfaced vector symbol for simplicty)

\frac{d \vert \beta\vert}{d \boldsymbol{\beta}}=\mathrm{sgn}(\boldsymbol{\beta})=\left\{\begin{array}{cc} 1 & \beta > 0 \\ 0 & \beta =0\\-1 & \beta < 0, \end{array}\right.

we have that the derivative of the cost function is

\frac{\partial C(\boldsymbol{X},\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}=-2\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})+\lambda sgn(\boldsymbol{\beta})=0,

and reordering we have

\boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta})+\lambda sgn(\boldsymbol{\beta})=2\boldsymbol{X}^T(\boldsymbol{y}.

This equation does not lead to a nice analytical equation as in either Ridge regression or ordinary least squares. This equation can however be solved by using standard convex optimization algorithms using for example the Python package CVXOPT. We will discuss this later.

Simple example to illustrate Ordinary Least Squares, Ridge and Lasso Regression

Let us assume that our design matrix is given by unit (identity) matrix, that is a square diagonal matrix with ones only along the diagonal. In this case we have an equal number of rows and columns n=p.

Our model approximation is just \tild{\boldsymbol{y}}=\boldsymbol{\beta} and the mean squared error and thereby the cost function for ordinary least sqquares (OLS) is then (we drop the term 1/n)

C(\boldsymbol{\beta})=\sum_{i=0}^{p-1}(y_i-\beta_i)^2,

and minimizing we have that

\hat{\boldsymbol{\beta}}_{\mathrm{OLS}} = \boldsymbol{y}.

Ridge Regression

For Ridge regression our cost function is

C(\boldsymbol{\beta})=\sum_{i=0}^{p-1}(y_i-\beta_i)^2+\lambda\sum_{i=0}^{p-1}\beta_i^2,,

and minimizing we have that

\hat{\boldsymbol{\beta}}_{\mathrm{Ridge}} = \frac{\boldsymbol{y}}{1+\lambda}.

Linking the regression analysis with a statistical interpretation

The advantage of doing linear regression is that we actually end up with analytical expressions for several statistical quantities.
Standard least squares and Ridge regression allow us to derive quantities like the variance and other expectation values in a rather straightforward way.

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

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

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

Recall that \boldsymbol{X} is a matrix of dimensionality n\times p. The notation above \mathbf{X}_{i,\ast} means that we are looking at the row number i and perform a sum over all values p.

Assumptions made

The assumption we have made here can be summarized as (and this is going to be useful when we discuss the bias-variance trade off) that there exists a function f(\boldsymbol{x}) and a normal distributed error \boldsymbol{\varepsilon}\sim \mathcal{N}(0, \sigma^2) which describe our data

\boldsymbol{y} = f(\boldsymbol{x})+\boldsymbol{\varepsilon}

We approximate this function with our model from the solution of the linear regression equations, that is our function f is approximated by \boldsymbol{\tilde{y}} where we want to minimize (\boldsymbol{y}-\boldsymbol{\tilde{y}})^2, our MSE, with

\boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta}.

Expectation value and variance

We can calculate the expectation value of \boldsymbol{y} for a given element i

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

while its variance is

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

Hence, y_i \sim \mathcal{N}( \mathbf{X}_{i, \ast} \, \boldsymbol{\beta}, \sigma^2), that is \boldsymbol{y} follows a normal distribution with mean value \boldsymbol{X}\boldsymbol{\beta} and variance \sigma^2 (not be confused with the singular values of the SVD).

Expectation value and variance for \boldsymbol{\beta}

With the OLS expressions for the parameters \boldsymbol{\beta} we can evaluate the expectation value

\mathbb{E}(\boldsymbol{\beta}) = \mathbb{E}[ (\mathbf{X}^{\top} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1}\mathbf{X}^{T} \mathbb{E}[ \mathbf{Y}]=(\mathbf{X}^{T} \mathbf{X})^{-1} \mathbf{X}^{T}\mathbf{X}\boldsymbol{\beta}=\boldsymbol{\beta}.

This means that the estimator of the regression parameters is unbiased.

We can also calculate the variance

The variance of \boldsymbol{\beta} is

\begin{eqnarray*}
\mbox{Var}(\boldsymbol{\beta}) & = & \mathbb{E} \{ [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})] [\boldsymbol{\beta} - \mathbb{E}(\boldsymbol{\beta})]^{T} \}
\\
& = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} - \boldsymbol{\beta}]^{T} \}
\\
% & = & \mathbb{E} \{ [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}] \, [(\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y}]^{T} \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}
% \\
% & = & \mathbb{E} \{ (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \mathbf{Y} \, \mathbf{Y}^{T} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1}  \} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}
% \\
& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \mathbb{E} \{ \mathbf{Y} \, \mathbf{Y}^{T} \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}
\\
& = & (\mathbf{X}^{T} \mathbf{X})^{-1} \, \mathbf{X}^{T} \, \{ \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^{T} \,  \mathbf{X}^{T} + \sigma^2 \} \, \mathbf{X} \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}
% \\
% & = & (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T \, \mathbf{X} \, \boldsymbol{\beta} \, \boldsymbol{\beta}^T \,  \mathbf{X}^T \, \mathbf{X} \, (\mathbf{X}^T % \mathbf{X})^{-1}
% \\
% & & + \, \, \sigma^2 \, (\mathbf{X}^T \mathbf{X})^{-1} \, \mathbf{X}^T  \, \mathbf{X} \, (\mathbf{X}^T \mathbf{X})^{-1} - \boldsymbol{\beta} \boldsymbol{\beta}^T
\\
& = & \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}  + \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1} - \boldsymbol{\beta} \, \boldsymbol{\beta}^{T}
\, \, \, = \, \, \, \sigma^2 \, (\mathbf{X}^{T} \mathbf{X})^{-1},
\end{eqnarray*}

where we have used that $\mathbb{E} (\mathbf{Y} \mathbf{Y}^{T}) = \mathbf{X} , \boldsymbol{\beta} , \boldsymbol{\beta}^{T} , \mathbf{X}^{T} + \sigma^2 , \mathbf{I}_{nn}$. From $\mbox{Var}(\boldsymbol{\beta}) = \sigma^2 , (\mathbf{X}^{T} \mathbf{X})^{-1}$, one obtains an estimate of the variance of the estimate of the $j$-th regression coefficient: $\boldsymbol{\sigma}^2 (\boldsymbol{\beta}j ) = \boldsymbol{\sigma}^2 \sqrt{ [(\mathbf{X}^{T} \mathbf{X})^{-1}]{jj} }$. This may be used to construct a confidence interval for the estimates.

In a similar way, we can obtain analytical expressions for say the expectation values of the parameters \boldsymbol{\beta} and their variance when we employ Ridge regression, allowing us again to define a confidence interval.

It is rather straightforward to show that

\mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big]=(\mathbf{X}^{T} \mathbf{X} + \lambda \mathbf{I}_{pp})^{-1} (\mathbf{X}^{\top} \mathbf{X})\boldsymbol{\beta}^{\mathrm{OLS}}.

We see clearly that \mathbb{E} \big[ \boldsymbol{\beta}^{\mathrm{Ridge}} \big] \not= \boldsymbol{\beta}^{\mathrm{OLS}} for any \lambda > 0. We say then that the ridge estimator is biased.

We can also compute the variance as

\mbox{Var}[\boldsymbol{\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},

and it is easy to see that if the parameter \lambda goes to infinity then the variance of Ridge parameters \boldsymbol{\beta} goes to zero.

With this, we can compute the difference

\mbox{Var}[\boldsymbol{\beta}^{\mathrm{OLS}}]-\mbox{Var}(\boldsymbol{\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}.

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 \boldsymbol{\beta} obtained with the Ridge estimator. This has interesting consequences when we discuss the so-called bias-variance trade-off below.

Friday September 10

Why resampling methods

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

1. look at statistical properties, including a discussion of mean values, variance and the so-called bias-variance tradeoff

2. 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.

Two resampling methods are often used in Machine Learning analyses,

1. The bootstrap method

2. and Cross-Validation

In addition there are several other methods such as the Jackknife and the Blocking methods. We will discuss in particular cross-validation and the bootstrap method.

Resampling approaches can be computationally expensive

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

Why resampling methods ?

Statistical analysis.

• 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.

Statistical analysis

• 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.

Resampling methods

With all these analytical equations for both the OLS and Ridge regression, we will now outline how to assess a given model. This will lead us to a discussion of the so-called bias-variance tradeoff (see below) and so-called resampling methods.

One of the quantities we have discussed as a way to measure errors is the mean-squared error (MSE), mainly used for fitting of continuous functions. Another choice is the absolute error.

In the discussions below we will focus on the MSE and in particular since we will split the data into test and training data, we discuss the

1. prediction error or simply the test error \mathrm{Err_{Test}}, where we have a fixed training set and the test error is the MSE arising from the data reserved for testing. We discuss also the

2. training error \mathrm{Err_{Train}}, which is the average loss over the training data.

As our model becomes more and more complex, more of the training data tends to used. The training may thence adapt to more complicated structures in the data. This may lead to a decrease in the bias (see below for code example) and a slight increase of the variance for the test error. For a certain level of complexity the test error will reach minimum, before starting to increase again. The training error reaches a saturation.

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 \boldsymbol{x} = (x_1,x_2,\cdots,X_n). Let \boldsymbol{x}_i denote the vector

\boldsymbol{x}_i = (x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n),

which equals the vector \boldsymbol{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

In [ ]:

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)

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:

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

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

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

4. 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}(\boldsymbol{X}) is a function of random variables, \widehat{\theta} itself must be a random variable. Thus it has a pdf, call this function p(\boldsymbol{t}). The aim of the bootstrap is to estimate p(\boldsymbol{t}) by the relative frequency of \widehat{\theta}. You can think of this as using a histogram in the place of p(\boldsymbol{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(\boldsymbol{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:

1. Drawing lots of numbers from p(x), suppose we call one such set of numbers (X_1^*, X_2^*, \cdots, X_n^*).

2. Then using these numbers, we could compute a replica of \widehat{\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(\boldsymbol{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 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 \boldsymbol{X}.

Resampling methods: Bootstrap steps

The independent bootstrap works like this:

1. Draw with replacement n numbers for the observed variables \boldsymbol{x} = (x_1,x_2,\cdots,x_n).

2. Define a vector \boldsymbol{x}^* containing the values which were drawn from \boldsymbol{x}.

3. Using the vector \boldsymbol{x}^* compute \widehat{\theta}^* by evaluating \widehat \theta under the observations \boldsymbol{x}^*.

4. 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.

In [ ]:

%matplotlib inline

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()

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 \boldsymbol{\sigma}_{-i}^2(\lambda), as

\begin{align*}
\boldsymbol{\beta}_{-i}(\lambda) & =  ( \boldsymbol{X}_{-i, \ast}^{T}
\boldsymbol{X}_{-i, \ast} + \lambda \boldsymbol{I}_{pp})^{-1}
\boldsymbol{X}_{-i, \ast}^{T} \boldsymbol{y}_{-i}
\end{align*}

Warning:

Output truncated. This notebook contains too many cells to display efficiently.