30
Nov 18

Application: estimating sigma squared




Application: estimating sigma squared

Consider multiple regression

(1) y=X\beta +e

where

(a) the regressors are assumed deterministic, (b) the number of regressors k is smaller than the number of observations n, (c) the regressors are linearly independent, \det (X^TX)\neq 0, and (d) the errors are homoscedastic and uncorrelated,

(2) Var(e)=\sigma^2I.

Usually students remember that \beta should be estimated and don't pay attention to estimation of \sigma^2. Partly this is because \sigma^2 does not appear in the regression and partly because the result on estimation of error variance is more complex than the result on the OLS estimator of \beta .

Definition 1. Let \hat{\beta}=(X^TX)^{-1}X^Ty be the OLS estimator of \beta. \hat{y}=X\hat{\beta} is called the fitted value and r=y-\hat{y} is called the residual.

Exercise 1. Using the projectors P=X(X^TX)^{-1}X^T and Q=I-P show that \hat{y}=Py and r=Qe.

Proof. The first equation is obvious. From the model we have r=X\beta+e-P(X\beta +e). Since PX\beta=X\beta, we have further r=e-Pe=Qe.

Definition 2. The OLS estimator of \sigma^2 is defined by s^2=\Vert r\Vert^2/(n-k).

Exercise 2. Prove that s^2 is unbiased: Es^2=\sigma^2.

Proof. Using projector properties we have

\Vert r\Vert^2=(Qe)^TQe=e^TQ^TQe=e^TQe.

Expectations of type Ee^Te and Eee^T would be easy to find from (2). However, we need to find Ee^TQe where there is an obstructing Q. See how this difficulty is overcome in the next calculation.

E\Vert r\Vert^2=Ee^TQe (e^TQe is a scalar, so its trace is equal to itself)

=Etr(e^TQe) (applying trace-commuting)

=Etr(Qee^T) (the regressors and hence Q are deterministic, so we can use linearity of E)

=tr(QEee^T) (applying (2)) =\sigma^2tr(Q).

tr(P)=k because this is the dimension of the image of P. Therefore tr(Q)=n-k. Thus, E\Vert r\Vert^2=\sigma^2(n-k) and Es^2=\sigma^2.

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