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Chapter 7 Multivariate Linear Regression Models
7.1-7.3: Least Squares Estimation
Data available:
The multiple linear regression model for the above data is
where the error terms are assumed to have the following properties:
1.
2.
3.
The above data can be represented as the matrix form. Let
.
Then,
=,
where the error terms become
1.
2.
Least squares method:
The least squares method is to find the estimate of minimizing the sum of squares of residual,
since . Expanding gives
since a real number.
Note:For two matrices A and B, and
Similar to the procedure in finding the minimum of a function in calculus, the least squares estimate b can be found by solving the equation based on the first derivative of ,
The fitted values (in vector):
The residuals (in vector):
Note: (i) where and .
(ii) where is any symmetric matrix.
Note: Since
,
is a symmetric matrix.
Also,
,
is a symmetric matrix.
Note: is called the normal equation.
Note:
.
Therefore, if there is an intercept, then the first column of Zis . Then,
Note: for the linear regression model without the intercept, might not be equal to 0.
Note:
,
whereis called “hat” matrix (or projection matrix). Thus,
.
Example 1:
Heller Company manufactures lawn mowers and related lawn equipment. The managers believe the quantity of lawn mowers sold depends on the price of the mower and the price of a competitor’s mower. We have the following data:
Competitor’s Price / Heller’s Price/ Quantity sold
120 / 100 / 102
140 / 110 / 100
190 / 90 / 120
130 / 150 / 77
155 / 210 / 46
175 / 150 / 93
125 / 250 / 26
145 / 270 / 69
180 / 300 / 65
150 / 250 / 85
Theregression model for the above data is
.
The data in matrix form are
.
The least squares estimate is
.
The fitted regression equation is
.
The fitted equation implies an increase in the competitor’s price of 1 unit is associated with an increase of 0.414 unit in expected quantity sold and an increase in its own price of 1 unit is associated with a decrease of 0.269 unit in expected quantity sold. Thus,
[89.21,94.79,120.88,79.86,74.02,98.49,50.81,53.69,60.09,61.16]
and
[12.79,5.21,-0.88,-2.86,-28.02,-5.49,-24.81,15.31,4.91,23.84]
Suppose now we want to predict the quantity sold in a city where Heller prices it mower at $160 and the competitor prices its mower at $170. The quantity sold predicted is
.
Geometry of Least Squares:
.
In linear algebra,
is the linear combination of the column vector of . That is,
.
Then,
Least squares method is to find the appropriate such that the distance between and is smaller than the one between and the other linear combination of the column vectors of , for example, . Intuitively, is the information provided by covariates to interpret the response . Thus, is the information which interprets most accurately.
Further,
If we choose the estimate of such that is orthogonal every vector in , then . Thus,
.
That is, if we choose satisfying , then
and for any other estimate of ,
.
Thus, satisfying is the least squares estimate. Therefore,
Since
,
is called the projection matrix or hat matrix. projects the response vector on the space spanned by the covariate vectors. The vector of residuals is
.
We have the following two important theorems.
Properties of the least squares estimate:
1.
2. The variance –covariance matrix of the least squares estimate b is
[Derivation:]
since
.
Also,
since
Denote
,
the mean residual sum of squares (the residual sum of squares divided by n-r-1).
the sample variance estimate,
where . can be used to estimate .
Properties of the mean residual sum of squares:
1. and .
2.
3.
4.
[proof:]
1.
and
.
2.
3.
Similarly,
4.
Thus,
Therefore,
Gauss’s Least Squares Theorem:
Let , where , and Z has full rank r+1. For any c, the estimator
of has the smallest possible variance among all linear estimators of the form
that are unbiased for .
[proof:]
Let
.
Let be any unbiased estimator of with . Then,
That is, . Thus,
since
.
Useful Splus Commands:
estate=matrix(scan("E:\\T7-1.dat"),ncol=3,byrow=T)
estatelm=lm(estate[,3]~estate[,1]+estate[,2])
estatelm
summary(estatelm)
anova(estatelm)
Useful SAS Commands:
title'Regression Analysis';
data estate;
infile'E:\T7-1.dat';
input z1 z2 y;
procregdata=estate;
model y = z1 z2;
run;
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