Data Analysis and Machine Learning: Linear Regression and more Advanced Regression Analysis

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

Date: Aug 30, 2019

Copyright 1999-2019, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

Why Linear Regression (aka Ordinary Least Squares and family)

Fitting a continuous function with linear parameterization in terms of the parameters $\boldsymbol{\beta}$.

• 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 $\boldsymbol{\beta}$

• 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 article is highly recommended. Similarly, Mehta et al's article 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 $\boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T$. The first variable is called the dependent, the outcome or the response variable while the set of variables $\boldsymbol{x}$ is called the independent variable, or the predictor variable or the explanatory variable.

A regression model aims at finding a likelihood function $p(\boldsymbol{y}\vert \boldsymbol{x})$, that is the conditional distribution for $\boldsymbol{y}$ with a given $\boldsymbol{x}$. The estimation of $p(\boldsymbol{y}\vert \boldsymbol{x})$ 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 $\boldsymbol{x}_i=[x_{i0}, x_{i1}, \dots, x_{ip-1}]$ 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 $\boldsymbol{y}$ and $\boldsymbol{x}$ in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.

Regression analysis, overarching aims II

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 $\boldsymbol{y}$ (also as above). The variable $\boldsymbol{y}$ is generally referred to as the response variable. The aim of regression analysis is to explain $\boldsymbol{y}$ in terms of $\boldsymbol{X}$ 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 $\boldsymbol{X}$ and $\boldsymbol{y}$. This assumption gives rise to the linear regression model where $\boldsymbol{\beta} = [\beta_0, \ldots, \beta_{p-1}]^{T}$ 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) $\boldsymbol{y}$, 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

$$ BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1}, $$

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 $\boldsymbol{X}$.

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. 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 $24 \times 30000$. This is however a classification problem and we will come back to it when we discuss Logistic Regression.

General linear models

Before we proceed let us study a case from linear algebra where we aim at fitting a set of data $\boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}]$. 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 $\boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}]$, that is $y_i = y(x_i)$ with $i=0,1,2,\dots,n-1$. The variables $x_i$ could represent physical quantities like time, temperature, position etc. We assume that $y(x)$ is a smooth function.

Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of $y$ which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree $n-1$ with $n$ points, that is

$$ y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i, $$

where $\epsilon_i$ is the error in our approximation.

Rewriting the fitting procedure as a linear algebra problem

For every set of values $y_i,x_i$ we have thus the corresponding set of equations

$$ \begin{align*} y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\ y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\ \dots & \dots \\ y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\ \end{align*} $$

Rewriting the fitting procedure as a linear algebra problem, more details

Defining the vectors

$$ \boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$

and

$$ \boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, $$

and

$$ \boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, $$

and the design matrix

$$ \boldsymbol{X}= \begin{bmatrix} 1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ 1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ 1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ 1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ \end{bmatrix} $$

we can rewrite our equations as

$$ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. $$

The above design matrix is called a 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

$$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$

Note that we have $p=n$ here. The matrix is symmetric. This is generally not the case!

Generalizing the fitting procedure as a linear algebra problem

We redefine in turn the matrix $\boldsymbol{X}$ as

$$ \boldsymbol{X}= \begin{bmatrix} x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ \end{bmatrix} $$

and without loss of generality we rewrite again our equations as

$$ \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}. $$

The left-hand side of this equation is kwown. Our error vector $\boldsymbol{\epsilon}$ and the parameter vector $\boldsymbol{\beta}$ are our unknow quantities. How can we obtain the optimal set of $\beta_i$ values?

Optimizing our parameters

We have defined the matrix $\boldsymbol{X}$ via the equations

$$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ \dots & \dots \\ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ \dots & \dots \\ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$

As we noted above, we stayed with a system with the design matrix $\boldsymbol{X}\in {\mathbb{R}}^{n\times n}$, that is we have $p=n$. For reasons to come later (algorithmic arguments) we will hereafter define our matrix as $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$, 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 liquid drop model. 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.

With $\boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1}$, it means that we will hereafter write our equations for the approximation as

$$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$

throughout these lectures.

Optimizing our parameters, more details

With the above we use the design matrix to define the approximation $\boldsymbol{\tilde{y}}$ via the unknown quantity $\boldsymbol{\beta}$ as

$$ \boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta}, $$

and in order to find the optimal parameters $\beta_i$ instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values $y_i$ (which represent hopefully the exact values) and the parameterized values $\tilde{y}_i$, namely

$$ C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, $$

or using the matrix $\boldsymbol{X}$ and in a more compact matrix-vector notation as

$$ C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}^T\boldsymbol{\beta}\right)\right\}. $$

This function is one possible way to define the so-called cost function.

It is also common to define the function $Q$ as

$$ C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2, $$

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

$$ C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}, $$

can be linked to the variance of the quantity $y_i$ if we interpret the latter as the mean value. When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret $y_i$ as a mean value

$$ y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, $$

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 $C(\boldsymbol{\beta})$, that is we are going to solve the problem

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

In practical terms it means we will require

$$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0, $$

which results in

$$ \frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0, $$

or in a matrix-vector form as

$$ \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right). $$

Interpretations and optimizing our parameters

We can rewrite

$$ \frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right), $$

as

$$ \boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta}, $$

and if the matrix $\boldsymbol{X}^T\boldsymbol{X}$ is invertible we have the solution

$$ \boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$

We note also that since our design matrix is defined as $\boldsymbol{X}\in {\mathbb{R}}^{n\times p}$, the product $\boldsymbol{X}^T\boldsymbol{X} \in {\mathbb{R}}^{p\times p}$. 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 $\boldsymbol{X}^T\boldsymbol{X}$.

Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix $\boldsymbol{X}^T\boldsymbol{X}$? What kind of problems can we expect?

Some useful matrix and vector expressions

The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and matrices as upper case boldfaced letters.

2 6

< < < ! ! M A T H _ B L O C K

2 7

< < < ! ! M A T H _ B L O C K

2 8

< < < ! ! M A T H _ B L O C K

$$ \frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T. $$

Interpretations and optimizing our parameters

The residuals $\boldsymbol{\epsilon}$ are in turn given by

$$ \boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}, $$

and with

$$ \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, $$

we have

$$ \boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0, $$

meaning that the solution for $\boldsymbol{\beta}$ 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.

Own code for Ordinary Least Squares

It is rather straightforward to implement the matrix inversion and obtain the parameters $\boldsymbol{\beta}$. After having defined the matrix $\boldsymbol{X}$ we simply need to write

Alternatively, you can use the least squares functionality in Numpy as

And finally we plot our fit with and compare with data

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 _ScikitLearn in the introductory slides. Since we are not using _Scikit-Learn here we can define our own $R2$ function as

and we would be using it as

We can easily add our MSE score as

and finally the relative error as

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

$$ \chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\}, $$

where the matrix $\boldsymbol{\Sigma}$ is a diagonal matrix with $\sigma_i$ as matrix elements.

The $\chi^2$ function

In order to find the parameters $\beta_i$ we will then minimize the spread of $\chi^2(\boldsymbol{\beta})$ by requiring

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, $$

which results in

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, $$

or in a matrix-vector form as

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right). $$

where we have defined the matrix $\boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma}$ with matrix elements $a_{ij} = x_{ij}/\sigma_i$ and the vector $\boldsymbol{b}$ with elements $b_i = y_i/\sigma_i$.

The $\chi^2$ function

We can rewrite

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right), $$

as

$$ \boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta}, $$

and if the matrix $\boldsymbol{A}^T\boldsymbol{A}$ is invertible we have the solution

$$ \boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}. $$

The $\chi^2$ function

If we then introduce the matrix

$$ \boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}, $$

we have then the following expression for the parameters $\beta_j$ (the matrix elements of $\boldsymbol{H}$ are $h_{ij}$)

$$ \beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik} $$

We state without proof the expression for the uncertainty in the parameters $\beta_j$ as (we leave this as an exercise)

$$ \sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2, $$

resulting in

$$ \sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}! $$

The $\chi^2$ function

The first step here is to approximate the function $y$ with a first-order polynomial, that is we write

$$ y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i. $$

By computing the derivatives of $\chi^2$ with respect to $\beta_0$ and $\beta_1$ show that these are given by

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0, $$

and

$$ \frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0. $$

The $\chi^2$ function

For a linear fit (a first-order polynomial) we don't need to invert a matrix!!
Defining

$$ \gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2}, $$

$$ \gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2}, $$

$$ \gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right), $$

$$ \gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2}, $$

$$ \gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2}, $$

we obtain

$$ \beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}, $$

$$ \beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}. $$

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. 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.

The code

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.

The Boston housing data example

The Boston housing
data set was originally a part of UCI Machine Learning Repository and has been removed now. The data set is now included in Scikit-Learn's library. There are 506 samples and 13 feature (predictor) variables in this data set. The objective is to predict the value of prices of the house using the features (predictors) listed here.

The features/predictors are

1. CRIM: Per capita crime rate by town

2. ZN: Proportion of residential land zoned for lots over 25000 square feet

3. INDUS: Proportion of non-retail business acres per town

4. CHAS: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise)

5. NOX: Nitric oxide concentration (parts per 10 million)

6. RM: Average number of rooms per dwelling

7. AGE: Proportion of owner-occupied units built prior to 1940

8. DIS: Weighted distances to five Boston employment centers

9. RAD: Index of accessibility to radial highways

10. TAX: Full-value property tax rate per USD10000

11. B: $1000(Bk - 0.63)^2$, where $Bk$ is the proportion of [people of African American descent] by town

12. LSTAT: Percentage of lower status of the population

13. MEDV: Median value of owner-occupied homes in USD 1000s

Housing data, the code

We start by importing the libraries

and load the Boston Housing DataSet from Scikit-Learn

Then we invoke Pandas

and preprocess the data

We can then visualize the data

It is now useful to look at the correlation matrix

From the above coorelation plot we can see that MEDV is strongly correlated to LSTAT and RM. We see also that RAD and TAX are stronly correlated, but we don't include this in our features together to avoid multi-colinearity

Now we start training our model

We split the data into training and test sets

Then we use the linear regression functionality from Scikit-Learn

The singular value decomposition

The examples we have looked at so far are cases where we normally can invert the matrix $\boldsymbol{X}^T\boldsymbol{X}$. 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. Obtaining the inverse of the design matrix is then often done via a so-called LU, QR or Cholesky decomposition.

This may however not the be case in general and a standard matrix inversion algorithm based on say LU, QR or Cholesky decomposition may lead to singularities. We will see examples of this below.

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.

Linear Regression Problems

One of the typical problems we encounter with linear regression, in particular when the matrix $\boldsymbol{X}$ (our so-called design matrix) is high-dimensional, are problems with near singular or singular matrices. The column vectors of $\boldsymbol{X}$ may be linearly dependent, normally referred to as super-collinearity.
This means that the matrix may be rank deficient and it is basically impossible to to model the data using linear regression. As an example, consider the matrix

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

The columns of $\boldsymbol{X}$ 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 $\boldsymbol{X}^T\boldsymbol{X}$ (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

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

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

Fixing the singularity

If our design matrix $\boldsymbol{X}$ which enters the linear regression problem

$$ \begin{equation} \boldsymbol{\beta} = (\boldsymbol{X}^{T} \boldsymbol{X})^{-1} \boldsymbol{X}^{T} \boldsymbol{y}, \label{_auto1} \tag{1} \end{equation} $$

has linearly dependent column vectors, we will not be able to compute the inverse of $\boldsymbol{X}^T\boldsymbol{X}$ and we cannot find the parameters (estimators) $\beta_i$. The estimators are only well-defined if $(\boldsymbol{X}^{T}\boldsymbol{X})^{-1}$ exits. This is more likely to happen when the matrix $\boldsymbol{X}$ 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

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

where $\boldsymbol{I}$ 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 $\boldsymbol{X}$ can be diagonalized if and only it is a so-called normal matrix, that is if $\boldsymbol{X}\in {\mathbb{R}}^{n\times n}$ we have $\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X}$ or if $\boldsymbol{X}\in {\mathbb{C}}^{n\times n}$ we have $\boldsymbol{X}\boldsymbol{X}^{\dagger}=\boldsymbol{X}^{\dagger}\boldsymbol{X}$. The matrix has then a set of eigenpairs

$$ (\lambda_1,\boldsymbol{u}_1),\dots, (\lambda_n,\boldsymbol{u}_n), $$

and the eigenvalues are given by the diagonal matrix

$$ \boldsymbol{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n). $$

The matrix $\boldsymbol{X}$ can be written in terms of an orthogonal/unitary transformation $\boldsymbol{U}$

$$ \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, $$

with $\boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I}$ or $\boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I}$.

Not all square matrices are diagonalizable. A matrix like the one discussed above

$$ \boldsymbol{X} = \begin{bmatrix} 1& -1 \\ 1& -1\\ \end{bmatrix} $$

is not diagonalizable, it is a so-called defective matrix. It is easy to see that the condition $\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X}$ is not fulfilled.

The SVD, a Fantastic Algorithm

However, and this is the strength of the SVD algorithm, any general matrix $\boldsymbol{X}$ can be decomposed in terms of a diagonal matrix and two orthogonal/unitary matrices. The Singular Value Decompostion (SVD) theorem states that a general $m\times n$ matrix $\boldsymbol{X}$ can be written in terms of a diagonal matrix $\boldsymbol{\Sigma}$ of dimensionality $n\times n$ and two orthognal matrices $\boldsymbol{U}$ and $\boldsymbol{V}$, where the first has dimensionality $m \times m$ and the last dimensionality $n\times n$. We have then

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

As an example, the above defective matrix can be decomposed as

$$ \boldsymbol{X} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1& 1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 0\\ \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -1 \\ 1& 1\\ \end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T, $$

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

$$ \boldsymbol{X} = \frac{1}{15}\begin{bmatrix} 14 & 2\\ 4 & 22\\ 16 & 13\end{bmatrix}=\frac{1}{3}\begin{bmatrix} 1& 2 & 2 \\ 2& -1 & 1\\ 2 & 1& -2\end{bmatrix} \begin{bmatrix} 2& 0 \\ 0& 1\\ 0 & 0\end{bmatrix}\frac{1}{5}\begin{bmatrix} 3& 4 \\ 4& -3\end{bmatrix}=\boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T. $$

This is a $3\times 2$ matrix which is decomposed in terms of a $3\times 3$ matrix $\boldsymbol{U}$, and a $2\times 2$ matrix $\boldsymbol{V}$. It is easy to see that $\boldsymbol{U}$ and $\boldsymbol{V}$ 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 $\boldsymbol{X}$ has dimension $n\times p$, the matrix is thus decomposed into an $n\times n$ orthogonal matrix $\boldsymbol{U}$, a $p\times p$ orthogonal matrix $\boldsymbol{V}$ and a diagonal matrix $\boldsymbol{\Sigma}$ with $r=\mathrm{min}(n,p)$ singular values $\sigma_i\geq 0$ 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 $\boldsymbol{U}$ are called the left singular vectors while the columns of $\boldsymbol{V}$ are the right singular vectors.

Economy-size SVD

If we assume that $n > p$, then our matrix $\boldsymbol{U}$ has dimension $n \times n$. The last $n-p$ columns of $\boldsymbol{U}$ become however irrelevant in our calculations since they are multiplied with the zeros in $\boldsymbol{\Sigma}$.

The economy-size decomposition removes extra rows or columns of zeros from the diagonal matrix of singular values, $\boldsymbol{\Sigma}$, along with the columns in either $\boldsymbol{U}$ or $\boldsymbol{V}$ 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 $\boldsymbol{U}$ and $\boldsymbol{\Sigma}$ has dimension $p\times p$. If $p > n$, then only the first $n$ columns of $\boldsymbol{V}$ are computed and $\boldsymbol{\Sigma}$ 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

$$ \boldsymbol{\tilde{y}} = \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}. $$

The matrix to invert can be rewritten in terms of our SVD decomposition as

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

Using the orthogonality properties of $\boldsymbol{U}$ we have

$$ \boldsymbol{X}^T\boldsymbol{X} = \boldsymbol{V}\boldsymbol{\Sigma}^T\boldsymbol{\Sigma}\boldsymbol{V}^T = \boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T, $$

with $\boldsymbol{D}$ being a diagonal matrix with values along the diagonal given by the singular values squared.

This means that

$$ (\boldsymbol{X}^T\boldsymbol{X})\boldsymbol{V} = \boldsymbol{V}\boldsymbol{D}, $$

that is the eigenvectors of $(\boldsymbol{X}^T\boldsymbol{X})$ are given by the columns of the right singular matrix of $\boldsymbol{X}$ and the eigenvalues are the squared singular values. It is easy to show (show this) that

$$ (\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}, $$

that is, the eigenvectors of $(\boldsymbol{X}\boldsymbol{X})^T$ are the columns of the left singular matrix and the eigenvalues are the same.

Going back to our OLS equation we have

$$ \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y}. $$

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

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

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

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

More on Ridge Regression

Using the matrix-vector expression for Ridge regression,

$$ C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\left\{(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})^T(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta})\right\}+\lambda\boldsymbol{\beta}^T\boldsymbol{\beta}, $$

by 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

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

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.

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 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

$$ (\boldsymbol{X}\boldsymbol{X}^T)\boldsymbol{U} = \boldsymbol{U}\boldsymbol{D}. $$

We can analyse the OLS solutions in terms of the eigenvectors (the columns) of the right singular value matrix $\boldsymbol{U}$ as

$$ \boldsymbol{X}\boldsymbol{\beta} = \boldsymbol{X}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}\boldsymbol{X}^T\boldsymbol{y}=\boldsymbol{U\Sigma V^T}\left(\boldsymbol{V}\boldsymbol{D}\boldsymbol{V}^T \right)^{-1}(\boldsymbol{U\Sigma V^T})^T\boldsymbol{y}=\boldsymbol{U}\boldsymbol{U}^T\boldsymbol{y} $$

For Ridge regression this becomes

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

with the vectors $\boldsymbol{u}_j$ being the columns of $\boldsymbol{U}$.

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. Actually, calculating the variance of $\boldsymbol{X}\boldsymbol{v}_j$ shows that this quantity is equal to $\sigma_j^2/n$. With a parameter $\lambda$ we can thus shrink the role of specific parameters.

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.

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

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.

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.

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:

$$ p(x) = \mathrm{prob}(X=x) $$

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:

$$ \mathrm{prob}(a\leq X\leq b) = \int_a^b p(x)dx $$

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.

Statistics, moments

A particularly useful class of special expectation values are the moments. The $n$-th moment of the PDF $p$ is defined as follows:

$$ \langle x^n\rangle \equiv \int\! x^n p(x)\,dx $$

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 $\mu$:

$$ \langle x\rangle = \mu \equiv \int\! x p(x)\,dx $$

Statistics, central moments

A special version of the moments is the set of central moments, the n-th central moment defined as:

$$ \langle (x-\langle x \rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx $$

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 $\mathrm{var}(X)$:

$$ \begin{equation} \sigma^2_X\ \ =\ \ \mathrm{var}(X) = \langle (x-\langle x\rangle)^2\rangle = \int\! (x-\langle x\rangle)^2 p(x)\,dx \label{_auto2} \tag{2} \end{equation} $$

$$ \begin{equation} = \int\! \left(x^2 - 2 x \langle x\rangle^{2} + \langle x\rangle^2\right)p(x)\,dx \label{_auto3} \tag{3} \end{equation} $$

$$ \begin{equation} = \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2 \label{_auto4} \tag{4} \end{equation} $$

$$ \begin{equation} = \langle x^2\rangle - \langle x\rangle^2 \label{_auto5} \tag{5} \end{equation} $$

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:

$$ \mathrm{cov}(X_i,\,X_j) \equiv \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \nonumber $$

$$ \begin{equation} = \int\!\cdots\!\int\!(x_i-\langle x_i \rangle)(x_j-\langle x_j \rangle)\, P(x_1,\dots,x_n)\,dx_1\dots dx_n \label{eq:def_covariance} \tag{6} \end{equation} $$

with

$$ \langle x_i\rangle = \int\!\cdots\!\int\!x_i\,P(x_1,\dots,x_n)\,dx_1\dots dx_n $$

Statistics, more covariance

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 $X_j$, ($i\neq j$):

$$ \begin{equation} \mathrm{cov}(X_i,\,X_j) = \langle(x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \label{_auto6} \tag{7} \end{equation} $$

$$ \begin{equation} =\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle \label{_auto7} \tag{8} \end{equation} $$

$$ \begin{equation} =\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j\rangle + \langle \langle x_i\rangle\langle x_j\rangle\rangle \label{_auto8} \tag{9} \end{equation} $$

$$ \begin{equation} =\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle + \langle x_i\rangle\langle x_j\rangle \label{_auto9} \tag{10} \end{equation} $$

$$ \begin{equation} =\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle \label{_auto10} \tag{11} \end{equation} $$

Covariance example

Suppose we have defined three vectors $\hat{x}, \hat{y}, \hat{z}$ with $n$ elements each. The covariance matrix is defined as

$$ \hat{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix}, $$

where for example

$$ \sigma_{xy} =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). $$

The Numpy function np.cov calculates the covariance elements using the factor $1/(n-1)$ instead of $1/n$ since it assumes we do not have the exact mean valu\ es.

The following simple function uses the np.vstack function which takes each vector of dimension $1\times n$ and produces a $3\times n$ matrix $\hat{W}$

$$ \hat{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ \dots & \dots & \dots \\ x_{n-2} & y_{n-2} & z_{n-2} \\ x_{n-1} & y_{n-1} & z_{n-1} \end{bmatrix}, $$

which in turn is converted into into the $3\times 3$ covariance matrix $\hat{\Sigma}$ via the Numpy function np.cov(). We note that we can also calculate the mean value of each set of samples $\hat{x}$ etc using the Numpy function np.mean(x). We can also extract the eigenvalues of the covariance matrix through the np.linalg.eig() function.

Covariance in numpy

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:

$$ U = \sum_i a_i X_i \qquad V = \sum_j b_j Y_j $$

By the linearity of the expectation value

$$ \mathrm{cov}(U, V) = \sum_{i,j}a_i b_j \mathrm{cov}(X_i, Y_j) $$

Statistics, more variance

Now, since the variance is just $\mathrm{var}(X_i) = \mathrm{cov}(X_i, X_i)$, we get the variance of the linear combination $U = \sum_i a_i X_i$:

$$ \begin{equation} \mathrm{var}(U) = \sum_{i,j}a_i a_j \mathrm{cov}(X_i, X_j) \label{eq:variance_linear_combination} \tag{12} \end{equation} $$

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:

1 1 1

< < < ! ! M A T H _ B L O C K

$$ \mathrm{var}(\sum_i a_i X_i) = \sum_i a_i^2 \mathrm{var}(X_i) $$

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:

$$ \{x_1, x_2,\dots\,x_k,\dots\}. $$

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 $\sigma_X^{\phantom X}$.

Statistics and sample variables

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:

$$ \bar{x}_n \equiv \frac{1}{n}\sum_{k=1}^n x_k $$

The sample variance is:

$$ \mathrm{var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_k - \bar{x}_n)^2 $$

its square root being the standard deviation of the sample. The sample covariance is:

$$ \mathrm{cov}(x)\equiv\frac{1}{n}\sum_{kl}(x_k - \bar{x}_n)(x_l - \bar{x}_n) $$

Statistics, sample variance and covariance

Note that the sample variance is the sample covariance without the cross terms. In a similar manner as the covariance in Eq. (6) 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 $\mathrm{cov}(X,Y)$.

Statistics, law of large numbers

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:

$$ \lim_{n\to\infty}\bar{x}_n = \mu_X^{\phantom X} $$

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:

$$ \overline X_n = \frac{1}{n}\sum_{i=1}^n X_i $$

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.

Statistics

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 $\bar x_n$:

$$ p_{\overline X_n}(x) = \int p_X^{\phantom X}(x_1)\cdots \int p_X^{\phantom X}(x_n)\ \delta\!\left(x - \frac{x_1+x_2+\dots+x_n}{n}\right)dx_n \cdots dx_1 $$

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:

$$ \begin{equation} \lim_{n\to\infty} p_{\overline X_n}(x) = \left(\frac{n}{2\pi\mathrm{var}(X)}\right)^{1/2} e^{-\frac{n(x-\bar x_n)^2}{2\mathrm{var}(X)}} \label{eq:central_limit_gaussian} \tag{13} \end{equation} $$

Statistics, more technicalities

The desired variance $\mathrm{var}(\overline X_n)$, i.e. the sample error squared $\mathrm{err}_X^2$, is given by:

$$ \begin{equation} \mathrm{err}_X^2 = \mathrm{var}(\overline X_n) = \frac{1}{n^2} \sum_{ij} \mathrm{cov}(X_i, X_j) \label{eq:error_exact} \tag{14} \end{equation} $$

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.

Statistics

Our estimate of $\mu_{X_i}^{\phantom X}$ is then the sample mean $\bar x$ itself, in accordance with the the central limit theorem:

$$ \mu_{X_i}^{\phantom X} = \langle x_i\rangle \approx \frac{1}{n}\sum_{k=1}^n x_k = \bar x $$

Using $\bar x$ in place of $\mu_{X_i}^{\phantom X}$ we can give an estimate of the covariance in Eq. (14)

$$ \mathrm{cov}(X_i, X_j) = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle \approx\langle (x_i - \bar x)(x_j - \bar{x})\rangle, $$

resulting in

$$ \frac{1}{n} \sum_{l}^n \left(\frac{1}{n}\sum_{k}^n (x_k -\bar x_n)(x_l - \bar x_n)\right)=\frac{1}{n}\frac{1}{n} \sum_{kl} (x_k -\bar x_n)(x_l - \bar x_n)=\frac{1}{n}\mathrm{cov}(x) $$

Statistics and sample variance

By the same procedure we can use the sample variance as an estimate of the variance of any of the stochastic variables $X_i$

$$ \mathrm{var}(X_i)=\langle x_i - \langle x_i\rangle\rangle \approx \langle x_i - \bar x_n\rangle\nonumber, $$

which is approximated as

$$ \begin{equation} \mathrm{var}(X_i)\approx \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)=\mathrm{var}(x) \label{eq:var_estimate_i_think} \tag{15} \end{equation} $$

Now we can calculate an estimate of the error $\mathrm{err}_X^{\phantom X}$ of the sample mean $\bar x_n$:

$$ \mathrm{err}_X^2 =\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) \nonumber $$

$$ \approx\frac{1}{n^2}\sum_{ij}\frac{1}{n}\mathrm{cov}(x) =\frac{1}{n^2}n^2\frac{1}{n}\mathrm{cov}(x)\nonumber $$

$$ \begin{equation} =\frac{1}{n}\mathrm{cov}(x) \label{eq:error_estimate} \tag{16} \end{equation} $$

which is nothing but the sample covariance divided by the number of measurements in the sample.

Statistics, uncorrelated results

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:

$$ \mathrm{err}_X^2=\frac{1}{n^2}\sum_{ij} \mathrm{cov}(X_i, X_j) = \frac{1}{n^2} \sum_i \mathrm{var}(X_i), $$

resulting in

$$ \begin{equation} \mathrm{err}_X^2\approx \frac{1}{n^2} \sum_i \mathrm{var}(x)= \frac{1}{n}\mathrm{var}(x) \label{eq:error_estimate_uncorrel} \tag{17} \end{equation} $$

where in the second step we have used Eq. (15). 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.

Statistics, computations

For computational purposes one usually splits up the estimate of $\mathrm{err}_X^2$, given by Eq. (16), into two parts

$$ \mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x) + \frac{1}{n}(\mathrm{cov}(x)-\mathrm{var}(x)), $$

which equals

$$ \begin{equation} \frac{1}{n^2}\sum_{k=1}^n (x_k - \bar x_n)^2 +\frac{2}{n^2}\sum_{k<l} (x_k - \bar x_n)(x_l - \bar x_n) \label{eq:error_estimate_split_up} \tag{18} \end{equation} $$

The first term is the same as the error in the uncorrelated case, Eq. (17). This means that the second term accounts for the error correction due to correlation between the measurements. For uncorrelated measurements this second term is zero.

Statistics, more on computations of errors

Computationally the uncorrelated first term is much easier to treat efficiently than the second.

$$ \mathrm{var}(x) = \frac{1}{n}\sum_{k=1}^n (x_k - \bar x_n)^2 = \left(\frac{1}{n}\sum_{k=1}^n x_k^2\right) - \bar x_n^2 $$

We just accumulate separately the values $x^2$ and $x$ for every measurement $x$ we receive. The correlation term, though, has to be calculated at the end of the experiment since we need all the measurements to calculate the cross terms. Therefore, all measurements have to be stored throughout the experiment.

Statistics, wrapping up 1

Let us analyze the problem by splitting up the correlation term into partial sums of the form:

$$ f_d = \frac{1}{n-d}\sum_{k=1}^{n-d}(x_k - \bar x_n)(x_{k+d} - \bar x_n) $$

The correlation term of the error can now be rewritten in terms of $f_d$

$$ \frac{2}{n}\sum_{k<l} (x_k - \bar x_n)(x_l - \bar x_n) = 2\sum_{d=1}^{n-1} f_d $$

The value of $f_d$ reflects the correlation between measurements separated by the distance $d$ in the sample samples. Notice that for $d=0$, $f$ is just the sample variance, $\mathrm{var}(x)$. If we divide $f_d$ by $\mathrm{var}(x)$, we arrive at the so called autocorrelation function

$$ \kappa_d = \frac{f_d}{\mathrm{var}(x)} $$

which gives us a useful measure of pairwise correlations starting always at $1$ for $d=0$.

Statistics, final expression

The sample error (see eq. (18)) can now be written in terms of the autocorrelation function:

$$ \mathrm{err}_X^2 = \frac{1}{n}\mathrm{var}(x)+\frac{2}{n}\cdot\mathrm{var}(x)\sum_{d=1}^{n-1} \frac{f_d}{\mathrm{var}(x)}\nonumber $$

$$ = \left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\mathrm{var}(x)\nonumber $$

$$ \begin{equation} =\frac{\tau}{n}\cdot\mathrm{var}(x) \label{eq:error_estimate_corr_time} \tag{19} \end{equation} $$

and we see that $\mathrm{err}_X$ can be expressed in terms the uncorrelated sample variance times a correction factor $\tau$ which accounts for the correlation between measurements. We call this correction factor the autocorrelation time:

$$ \begin{equation} \tau = 1+2\sum_{d=1}^{n-1}\kappa_d \label{eq:autocorrelation_time} \tag{20} \end{equation} $$

Statistics, effective number of correlations

For a correlation free experiment, $\tau$ equals 1. From the point of view of eq. (19) we can interpret a sequential correlation as an effective reduction of the number of measurements by a factor $\tau$. The effective number of measurements becomes:

$$ n_\mathrm{eff} = \frac{n}{\tau} $$

To neglect the autocorrelation time $\tau$ will always cause our simple uncorrelated estimate of $\mathrm{err}_X^2\approx \mathrm{var}(x)/n$ to be less than the true sample error. The estimate of the error will be too good. On the other hand, the calculation of the full autocorrelation time poses an efficiency problem if the set of measurements is very large.

Linking the regression analysis with a statistical interpretation

Finally, we are going to discuss several statistical properties which can be obtained in terms of analytical expressions. 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 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 describes 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: $\hat{\sigma}^2 (\hat{\beta}_j ) = \hat{\sigma}^2 \sqrt{ [(\mathbf{X}^{T} \mathbf{X})^{-1}]_{jj} }$. This may be used to construct a confidence interval for the estimates.

In a similar way, we cna obtain analytical expressions for say the expectation values of the parameters $\boldsymbol{\beta}$ and their variance when we employ Ridge regression, and thereby 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.

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

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

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

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

• 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 $\boldsymbol{x} = (x_1,x_2,\cdots,X_n)$. 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

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.

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.

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

$$ \boldsymbol{y}=f(\boldsymbol{x}) + \boldsymbol{\epsilon} $$

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 $\boldsymbol{\beta}$ and the design matrix $\boldsymbol{X}$ which embody our model, that is $\boldsymbol{\tilde{y}}=\boldsymbol{X}\boldsymbol{\beta}$.

Thereafter we found the parameters $\boldsymbol{\beta}$ by optimizing the means squared error via the so-called cost function

$$ C(\boldsymbol{X},\boldsymbol{\beta}) =\frac{1}{n}\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2=\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]. $$

We can rewrite this as

$$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\frac{1}{n}\sum_i(f_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2+\sigma^2. $$

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 $\boldsymbol{\epsilon}$.

To derive this equation, we need to recall that the variance of $\boldsymbol{y}$ and $\boldsymbol{\epsilon}$ are both equal to $\sigma^2$. The mean value of $\boldsymbol{\epsilon}$ is by definition equal to zero. Furthermore, the function $f$ is not a stochastics variable, idem for $\boldsymbol{\tilde{y}}$. We use a more compact notation in terms of the expectation value

$$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}})^2\right], $$

and adding and subtracting $\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]$ we get

$$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{f}+\boldsymbol{\epsilon}-\boldsymbol{\tilde{y}}+\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right], $$

which, using the abovementioned expectation values can be rewritten as

$$ \mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathbb{E}\left[(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2\right]+\mathrm{Var}\left[\boldsymbol{\tilde{y}}\right]+\sigma^2, $$

that is the rewriting in terms of the so-called bias, the variance of the model $\boldsymbol{\tilde{y}}$ and the variance of $\boldsymbol{\epsilon}$.

Example code for Bias-Variance tradeoff

Understanding what happens

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

The Ising model

The one-dimensional Ising model with nearest neighbor interaction, no external field and a constant coupling constant $J$ is given by

$$ \begin{equation} H = -J \sum_{k}^L s_k s_{k + 1}, \label{_auto11} \tag{21} \end{equation} $$

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.

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

$$ \begin{equation} H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. \label{_auto12} \tag{22} \end{equation} $$

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

$$ \begin{equation} \boldsymbol{H} = \boldsymbol{X} J, \label{_auto13} \tag{23} \end{equation} $$

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

$$ \begin{equation} \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, \label{_auto14} \tag{24} \end{equation} $$

We split the data in training and test data as discussed in the previous example

Linear regression

In the ordinary least squares method we choose the cost function

$$ \begin{equation} C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}. \label{_auto15} \tag{25} \end{equation} $$

We then find the extremal point of $C$ by taking the derivative with respect to $\boldsymbol{\beta}$ as discussed above. This yields the expression for $\boldsymbol{\beta}$ to be

$$ \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}}, $$

which immediately imposes some requirements on $\boldsymbol{X}$ as there must exist an inverse of $\boldsymbol{X}^T \boldsymbol{X}$. If the expression we are modeling contains an intercept, i.e., a constant term, we must make sure that the first column of $\boldsymbol{X}$ consists of $1$. We do this here

Singular Value decomposition

Doing the inversion directly turns out to be a bad idea since the matrix $\boldsymbol{X}^T\boldsymbol{X}$ 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 $\boldsymbol{\beta}$ as

$$ \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y}, $$

where the pseudoinverse of $\boldsymbol{X}$ is given by

$$ \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}. $$

Using singular value decomposition we can decompose the matrix $\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T$, where $\boldsymbol{U}$ and $\boldsymbol{V}$ are orthogonal(unitary) matrices and $\boldsymbol{\Sigma}$ contains the singular values (more details below). where $X^{+} = V\Sigma^{+} U^T$. This reduces the equation for $\omega$ to

$$ \begin{equation} \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}. \label{_auto16} \tag{26} \end{equation} $$

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.

When extracting the $J$-matrix we need to make sure that we remove the intercept, as is done here

A way of looking at the coefficients in $J$ is to plot the matrices as images.

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?

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

$$ \begin{equation} H = -J \sum_{k}^L s_k s_{k + 1}, \label{_auto17} \tag{27} \end{equation} $$

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.

A more general form for the one-dimensional Ising model is

$$ \begin{equation} H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. \label{_auto18} \tag{28} \end{equation} $$

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

$$ \begin{equation} H = X J, \label{_auto19} \tag{29} \end{equation} $$

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.

$$ \begin{equation} \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}. \label{_auto20} \tag{30} \end{equation} $$