# 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

## 1 Answer

I'm not totally sure about your notation. (That's one reason we ask you to show what you have tried, because that helps us get into the context of your course.)

I am pretty sure you are supposed to use the following idea:

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

If you need more, please give us some formal definitions and some of your own thoughts or doubts.

• I have edited my question with my advances. – plr Mar 24 '17 at 0:38