Matrix Notation for Multiple Linear Regression

Note

The following supplemental notes were created by Dr. Maria Tackett for STA 210. They are provided for students who want to dive deeper into the mathematics behind regression and reflect some of the material covered in STA 211: Mathematics of Regression. Additional supplemental notes will be added throughout the semester.

This document provides the details for the matrix notation for multiple linear regression. We assume the reader has familiarity with some linear algebra. Please see Chapter 1 of An Introduction to Statistical Learning for a brief review of linear algebra.

Introduction

Suppose we have \(n\) observations. Let the \(i^{th}\) be \((x_{i1}, \ldots, x_{ip}, y_i)\), such that \(x_{i1}, \ldots, x_{ip}\) are the explanatory variables (predictors) and \(y_i\) is the response variable. We assume the data can be modeled using the least-squares regression model, such that the mean response for a given combination of explanatory variables follows the form in Equation 1.

\[ y = \beta_0 + \beta_1 x_1 + \dots + \beta_p x_p \qquad(1)\]

We can write the response for the \(i^{th}\) observation as shown in Equation 2

\[ y_i = \beta_0 + \beta_1 x_{i1} + \dots + \beta_p x_{ip} + \epsilon_i \qquad(2)\]

such that \(\epsilon_i\) is the amount \(y_i\) deviates from \(\mu\{y|x_{i1}, \ldots, x_{ip}\}\), the mean response for a given combination of explanatory variables. We assume each \(\epsilon_i \sim N(0,\sigma^2)\), where \(\sigma^2\) is a constant variance for the distribution of the response \(y\) for any combination of explanatory variables \(x_1, \ldots, x_p\).

Matrix Representation for the Regression Model

We can represent the Equation 1 and Equation 2 using matrix notation. Let

\[ \mathbf{Y} = \begin{bmatrix}y_1 \\ y_2 \\ \vdots \\y_n\end{bmatrix} \hspace{15mm} \mathbf{X} = \begin{bmatrix}x_{11} & x_{12} & \dots & x_{1p} \\ x_{21} & x_{22} & \dots & x_{2p} \\ \vdots & \vdots & \ddots & \vdots\\ x_{n1} & x_{n2} & \dots & x_{np} \end{bmatrix} \hspace{15mm} \boldsymbol{\beta}= \begin{bmatrix}\beta_0 \\ \beta_1 \\ \vdots \\ \beta_p \end{bmatrix} \hspace{15mm} \boldsymbol{\epsilon}= \begin{bmatrix}\epsilon_1 \\ \epsilon_2 \\ \vdots \\ \epsilon_n \end{bmatrix} \qquad(3)\]

Thus,

\[\mathbf{Y} = \mathbf{X}\boldsymbol{\beta} + \mathbf{\epsilon}\]

Therefore the estimated response for a given combination of explanatory variables and the associated residuals can be written as

$$

= = -

$$ {#eq-matrix_mean}

Estimating the Coefficients

The least-squares model is the one that minimizes the sum of the squared residuals. Therefore, we want to find the coefficients, \(\hat{\boldsymbol{\beta}}\) that minimizes

$$

{i=1}^{n} e{i}^2 = ^T = ( - )^T( - )

$$ {#eq-sum_sq_resid}

where \(\mathbf{e}^T\), the transpose of the matrix \(\mathbf{e}\).

$$

( - )^T( - ) = (^T - ^T - (^{T}T + ^{T}T )

$$ {#eq-model_equation}

Note that \((\mathbf{Y^T}\mathbf{X}\hat{\boldsymbol{\beta}})^T = \hat{\boldsymbol{\beta}}{}^{T}\mathbf{X}^T\mathbf{Y}\). Since these are both constants (i.e. \(1\times 1\) vectors), \(\mathbf{Y^T}\mathbf{X}\hat{\boldsymbol{\beta}} = \hat{\boldsymbol{\beta}}{}^{T}\mathbf{X}^T\mathbf{Y}\). Thus, ?@eq-model_equation becomes

$$

^T - 2 ^T{T} + ^{T}T

$$ {#eq-model_equation}

Since we want to find the \(\hat{\boldsymbol{\beta}}\) that minimizes ?@eq-sum_sq_resid, will find the value of \(\hat{\boldsymbol{\beta}}\) such that the derivative with respect to \(\hat{\boldsymbol{\beta}}\) is equal to 0.

$$

\[\begin{aligned} \frac{\partial \mathbf{e}^T\mathbf{e}}{\partial \hat{\boldsymbol{\beta}}} & = \frac{\partial}{\partial \hat{\boldsymbol{\beta}}}(\mathbf{Y}^T\mathbf{Y} - 2 \mathbf{X}^T\hat{\boldsymbol{\beta}}{}^T\mathbf{Y} + \hat{\boldsymbol{\beta}}{}^{T}\mathbf{X}^T\mathbf{X}\hat{\boldsymbol{\beta}}) = 0 \\ &\Rightarrow - 2 \mathbf{X}^T\mathbf{Y} + 2 \mathbf{X}^T\mathbf{X}\hat{\boldsymbol{\beta}} = 0 \\ & \Rightarrow 2 \mathbf{X}^T\mathbf{Y} = 2 \mathbf{X}^T\mathbf{X}\hat{\boldsymbol{\beta}} \\ & \Rightarrow \mathbf{X}^T\mathbf{Y} = \mathbf{X}^T\mathbf{X}\hat{\boldsymbol{\beta}} \\ & \Rightarrow (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y} = (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{X}\hat{\boldsymbol{\beta}} \\ & \Rightarrow (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y} = \mathbf{I}\hat{\boldsymbol{\beta}} \end{aligned}\]

$$ {#eq-ete}

Thus, the estimate of the model coefficients is \(\hat{\boldsymbol{\beta}} = (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}\).

Variance-covariance matrix of the coefficients

We will use two properties to derive the form of the variance-covariance matrix of the coefficients:

1. \(E[\boldsymbol{\epsilon}\boldsymbol{\epsilon}^T] = \sigma^2I\)
2. \(\hat{\boldsymbol{\beta}} = \boldsymbol{\beta} + (\mathbf{X}^T\mathbf{X})^{-1}\epsilon\)

First, we will show that \(E[\boldsymbol{\epsilon}\boldsymbol{\epsilon}^T] = \sigma^2I\)

$$

\[\begin{aligned} E[\boldsymbol{\epsilon}\boldsymbol{\epsilon}^T] &= E \begin{bmatrix}\epsilon_1 & \epsilon_2 & \dots & \epsilon_n \end{bmatrix}\begin{bmatrix}\epsilon_1 \\ \epsilon_2 \\ \vdots \\ \epsilon_n \end{bmatrix} \\ & = E \begin{bmatrix} \epsilon_1^2 & \epsilon_1 \epsilon_2 & \dots & \epsilon_1 \epsilon_n \\ \epsilon_2 \epsilon_1 & \epsilon_2^2 & \dots & \epsilon_2 \epsilon_n \\ \vdots & \vdots & \ddots & \vdots \\ \epsilon_n \epsilon_1 & \epsilon_n \epsilon_2 & \dots & \epsilon_n^2 \end{bmatrix} \\ & = \begin{bmatrix} E[\epsilon_1^2] & E[\epsilon_1 \epsilon_2] & \dots & E[\epsilon_1 \epsilon_n] \\ E[\epsilon_2 \epsilon_1] & E[\epsilon_2^2] & \dots & E[\epsilon_2 \epsilon_n] \\ \vdots & \vdots & \ddots & \vdots \\ E[\epsilon_n \epsilon_1] & E[\epsilon_n \epsilon_2] & \dots & E[\epsilon_n^2] \end{bmatrix} \end{aligned}\]

$$ {#eq-expected_error}

Recall, the regression assumption that the errors \(\epsilon_i's\) are Normally distributed with mean 0 and variance \(\sigma^2\). Thus, \(E(\epsilon_i^2) = Var(\epsilon_i) = \sigma^2\) for all \(i\). Additionally, recall the regression assumption that the errors are uncorrelated, i.e. \(E(\epsilon_i \epsilon_j) = Cov(\epsilon_i, \epsilon_j) = 0\) for all \(i,j\). Using these assumptions, we can write ?@eq-expected_error as

$$

E[^T] = \[\begin{bmatrix} \sigma^2 & 0 & \dots & 0 \\ 0 & \sigma^2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \sigma^2 \end{bmatrix}\]

= ^2

$$ {#eq-expected_error2}

where \(\mathbf{I}\) is the \(n \times n\) identity matrix.

Next, we show that \(\hat{\boldsymbol{\beta}} = \boldsymbol{\beta} + (\mathbf{X}^T\mathbf{X})^{-1}\epsilon\).

Recall that the \(\hat{\boldsymbol{\beta}} = (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}\) and \(\mathbf{Y} = \mathbf{X}\mathbf{\beta} + \mathbf{\epsilon}\). Then,

$$

\[\begin{aligned} \hat{\boldsymbol{\beta}} &= (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y} \\ &= (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T(\mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}) \\ &= \boldsymbol{\beta} + (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T \mathbf{\epsilon} \\ \end{aligned}\]

$$ {#eq-est-beta}

Using these two properties, we derive the form of the variance-covariance matrix for the coefficients. Note that the covariance matrix is \(E[(\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})(\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})^T]\)

$$

\[\begin{aligned} E[(\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})(\hat{\boldsymbol{\beta}} - \boldsymbol{\beta})^T] &= E[(\boldsymbol{\beta} + (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T \boldsymbol{\epsilon} - \boldsymbol{\beta})(\boldsymbol{\beta} + (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T \boldsymbol{\epsilon} - \boldsymbol{\beta})^T]\\ & = E[(\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T \boldsymbol{\epsilon}\boldsymbol{\epsilon}^T\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}] \\ & = (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T E[\boldsymbol{\epsilon}\boldsymbol{\epsilon}^T]\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\\ & = (\mathbf{X}^T\mathbf{X})^{-1} \mathbf{X}^T (\sigma^2\mathbf{I})\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\\ &= \sigma^2\mathbf{I}(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\\ & = \sigma^2\mathbf{I}(\mathbf{X}^T\mathbf{X})^{-1}\\ & = \sigma^2(\mathbf{X}^T\mathbf{X})^{-1} \\ \end{aligned}\]

$$ {#eq-var-cov}