Raising the bar Distribution of the estimator of the error variance

Feb 22

Distribution of the estimator of the error variance

category: Advanced Statistics ST2134, EC2020, EC2020 Elements of econometrics, Econometrics, Matrix algebra, MT2175 Further linear algebra No Comments

Distribution of the estimator of the error variance

If you are reading the book by Dougherty: this post is about the distribution of the estimator $s^2$ defined in Chapter 3.

Consider regression

(1) $y=X\beta +e$

where the deterministic matrix $X$ is of size $n\times k,$ satisfies $\det \left( X^{T}X\right) \neq 0$ (regressors are not collinear) and the error $e$ satisfies

(2) $Ee=0,Var(e)=\sigma ^{2}I$

$\beta$ is estimated by $\hat{\beta}=(X^{T}X)^{-1}X^{T}y.$ Denote $P=X(X^{T}X)^{-1}X^{T},$ $Q=I-P.$ Using (1) we see that $\hat{\beta}=\beta +(X^{T}X)^{-1}X^{T}e$ and the residual $r\equiv y-X\hat{\beta}=Qe.$ $\sigma^{2}$ is estimated by

(3) $s^{2}=\left\Vert r\right\Vert ^{2}/\left( n-k\right) =\left\Vert Qe\right\Vert ^{2}/\left( n-k\right) .$

$Q$ is a projector and has properties which are derived from those of $P$

(4) $Q^{T}=Q,$ $Q^{2}=Q.$

If $\lambda$ is an eigenvalue of $Q,$ then multiplying $Qx=\lambda x$ by $Q$ and using the fact that $x\neq 0$ we get $\lambda ^{2}=\lambda .$ Hence eigenvalues of $Q$ can be only $0$ or $1.$ The equation $tr\left( Q\right) =n-k$
tells us that the number of eigenvalues equal to 1 is $n-k$ and the remaining $k$ are zeros. Let $Q=U\Lambda U^{T}$ be the diagonal representation of $Q.$ Here $U$ is an orthogonal matrix,

(5) $U^{T}U=I,$

and $\Lambda$ is a diagonal matrix with eigenvalues of $Q$ on the main diagonal. We can assume that the first $n-k$ numbers on the diagonal of $Q$ are ones and the others are zeros.

Theorem. Let $e$ be normal. 1) $s^{2}\left( n-k\right) /\sigma ^{2}$ is distributed as $\chi _{n-k}^{2}.$ 2) The estimators $\hat{\beta}$ and $s^{2}$ are independent.

Proof. 1) We have by (4)

(6) $\left\Vert Qe\right\Vert ^{2}=\left( Qe\right) ^{T}Qe=\left( Q^{T}Qe\right) ^{T}e=\left( Qe\right) ^{T}e=\left( U\Lambda U^{T}e\right) ^{T}e=\left( \Lambda U^{T}e\right) ^{T}U^{T}e.$

Denote $S=U^{T}e.$ From (2) and (5)

$ES=0,$ $Var\left( S\right) =EU^{T}ee^{T}U=\sigma ^{2}U^{T}U=\sigma ^{2}I$

and $S$ is normal as a linear transformation of a normal vector. It follows that $S=\sigma z$ where $z$ is a standard normal vector with independent standard normal coordinates $z_{1},...,z_{n}.$ Hence, (6) implies

(7) $\left\Vert Qe\right\Vert ^{2}=\sigma ^{2}\left( \Lambda z\right) ^{T}z=\sigma ^{2}\left( z_{1}^{2}+...+z_{n-k}^{2}\right) =\sigma ^{2}\chi _{n-k}^{2}.$