With simple linear regression, there are only two regression coefficients - b0 and b1. There are only two normal equations. Finding a least-squares solution involves solving two equations with two unknowns - a task that is easily managed with ordinary algebra.
With multiple regression, things get more complicated. There are k independent variables and k + 1 regression coefficients. There are k + 1 normal equations. Finding a least-squares solution involves solving k + 1 equations with k + 1 unknowns. This can be done with ordinary algebra, but it is unwieldy.
To handle the complications of multiple regression, we will use matrix algebra.
To follow the discussion on this page, you need to understand a little matrix algebra. Specifically, you should be familiar with matrix addition, matrix subtraction, and matrix multiplication. And you should know about matrix transposes and matrix inverses.
If you are unfamiliar with these topics, check out the free matrix algebra tutorial on this site.
The Regression Equation in Matrix Form
With multiple regression, there is one dependent variable and k dependent variables. The regression equation is:
ŷ = b0 + b1x1 + b2x2 + … + bk-1xk-1 + bkxk
where ŷ is the predicted value of the dependent variable, bk are regression coefficients, and xk is the value of independent variable k.
To express the regression equation in matrix form, we need to define three matrices: Y, b, and X.
Here, the data set consists of n records. Each record includes scores for 1 dependent variable and k independent variables. Y is an n x 1 vector that holds predicted values of the dependent variable; and b is a k + 1 x 1 vector that holds estimated regression coefficients. Matrix X has a column of 1's plus k columns of values for each independent variable in the regression equation.
Given these matrices, the multiple regression equation can be expressed concisely as:
Y = Xb
It is sort of cool that this simple expression describes the regression equation for 1, 2, 3, or any number of independent variables.
Normal Equations in Matrix Form
Just as the regression equation can be expressed compactly in matrix form, so can the normal equations. The least squares normal equations can be expressed as:
X'Y = X'Xb or X'Xb = X'Y
Here, matrix X' is the transpose of matrix X. To solve for regression coefficients, simply pre-multiply by the inverse of X'X:
(X'X)-1X'Xb = (X'X)-1X'Y
b = (X'X)-1X'Y
where (X'X)-1X'X = I, the identity matrix.
In the real world, you will probably never compute regression coefficients by hand. Generally, you will use software, like SAS, SPSS, mini-tab, or excel. In the problem below, however, we will compute regression coefficients manually; so you will understand what is going on.
Test Your Understanding
Consider the table below. It shows three performance measures for five students.
|Student||Test score||IQ||Study hours|
Using least squares regression, develop a regression equation to predict test score, based on (1) IQ and (2) the number of hours that the student studied.
For this problem, we have some raw data; and we want to use this raw data to define a least-squares regression equation:
ŷ = b0 + b1x1 + b2x2
where ŷ is the predicted test score; b0, b1, and b2 are regression coefficients; x1 is an IQ score; and x2 is the number of hours that the student studied.
On the right side of the equation, the only unknowns are the regression coefficients. To define the regression coefficients, we use the following equation:
b = (X'X)-1X'Y
To solve this equation, we need to complete the following steps:
- Define X.
- Define X'.
- Compute X'X.
- Find the inverse of X'X.
- Define Y.
Let's begin with matrix X. Matrix X has a column of 1's plus two columns of values for each independent variable. So, this is matrix X and its transpose X':
Given X' and X, it is a simple matter to compute X'X.
Finding the inverse of X'X takes a little more effort. A way to find the inverse is described on this site at https://stattrek.com/matrix-algebra/how-to-find-inverse. Ultimately, we find:
Next, we define Y, the vector of dependent variable scores. For this problem, it is the vector of test scores.
With all of the essential matrices defined, we are ready to compute the least squares regression coefficients.
b = (X'X)-1X'Y
To conclude, here is our least-squares regression equation:
ŷ = 20 + 0.5x1 +0.5x2
where ŷ is the predicted test score; x1 is an IQ score; and x2 is the number of hours that the student studied. The regression coefficients are b0 = 20, b1 = 0.5, and b2 = 0.5.