# Prove that $Cov(\hat{\beta}_{LS},\hat{\beta}_{GLS})=Var(\hat{\beta}_{GLS})$

I have a problem which is as follows:

Given the linar model $y=X\beta+\epsilon$ and $Var(y)=\Sigma$, show that: $$Cov(\hat{\beta}_{LS},\hat{\beta}_{GLS})=Var(\hat{\beta}_{GLS})$$ and compute also $Cov(\hat{\beta}_{GLS},\hat{\beta}_{GLS}-\hat{\beta}_{LS})$.

I'm so lost I do not know where to begin.

Any suggestions or ideas?

EDIT: I have started with the definitions of Cov and Var. I got the $\text{Var}(\hat{\beta}_{GLS})$, but when computing $\text{Cov}(\hat{\beta}_{LS},\hat{\beta}_{GLS})=E[\hat{\beta}_{LS}\cdot\hat{\beta}_{GLS}]-E[\hat{\beta}_{LS}]E[\hat{\beta}_{GLS}]$ I got stuck in the first expected value, I got the right-hand side of the equality. I do not manage to get the expectance in a "good" way for applying formulas like $E(Ay)=A\mu$ or $E(y'Ay)=tr(AV)+\mu'A\mu$.

Any further ideas?

• What relationship btw $\hat \beta_{LS}$ and $\hat \beta_{GLS}$? – BruceET Mar 24 '17 at 0:41
• Are you asking for a relationship like $\hat{\beta}_{LS}=c\hat{\beta}_{GLS}$ roughly speaking? – plr Mar 24 '17 at 1:01

$Cov(X + Y, X) = Var(X) + Cov(Y,X).$ Then, provided that $X$ and $Y$ are independent (or at least uncorrelated), you have $Cov(X + Y, X) = Var(X).$