# Solved – Proving efficiency of OLS over GLS

generalized-least-squaresinequalitylinear algebraregressionself-study

I'm trying to prove the efficiency of OLS over GLS when the covariance matrix of the error $$\varepsilon$$ is mistakenly assumed to be $$\sigma^2\Sigma$$ instead of $$\sigma^2 I$$. After deriving the variances, what I have so far is: $$Var(\beta^{ols}) = \sigma^2(X'X)^{-1}$$ and
$$Var(\beta^{gls}) = \sigma^2(X'\Sigma^{-1} X)^{-1}$$.

Want to show $$Var(\beta^{gls}) – Var(\beta^{ols})$$ is psd.
$$\implies \sigma^2(X'\Sigma^{-1} X)^{-1} – \sigma^2(X'X)^{-1} \geq 0\\ \implies (X'\Sigma^{-1} X)^{-1} – (X'X)^{-1} \geq 0 \\ \iff X'X – X'\Sigma^{-1} X \geq 0$$
But this is where I don't know how to proceed. First I tried $$X'(I-\Sigma^{-1})X \geq 0$$ but got stuck. Any hints on how to continue?

Update:
After one of the comments, I want to double check the variance of $$\beta^{gls}$$:
$$Var(\beta^{gls}) = Var[(X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1}\varepsilon]\\ =(X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1} Var(\varepsilon) \Sigma^{-1} X(X'\Sigma^{-1}X)^{-1}$$
Here's is where I think I made the mistake, I replaced the variance of $$\varepsilon$$ as the one that we think is "right" (i.e. $$\sigma^{2}\Sigma)$$. Should I replace it with the real one (i.e. $$\sigma^2I$$)?

This is what I ended up doing: $$Var(\beta^{gls}) = Var[(X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1}\varepsilon]\\ =(X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1} Var(\varepsilon) \Sigma^{-1} X(X'\Sigma^{-1}X)^{-1}\\ = \sigma^2 (X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1} \Sigma^{-1} X(X'\Sigma^{-1}X)^{-1}\\$$
Want to show $$Var(\beta^{gls}) - Var(\beta^{ols})$$ is psd. $$\implies \sigma^2 (X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1} \Sigma^{-1} X(X'\Sigma^{-1}X)^{-1} - \sigma^2(X'X)^{-1} \geq 0\\ \iff X'X - X'\Sigma^{-1}X( X'\Sigma^{-1} \Sigma^{-1} X)^{-1}X'\Sigma^{-1}X \geq 0\\ \implies X'(I-\Sigma^{-1}X( X'\Sigma^{-1} \Sigma^{-1} X)^{-1}X'\Sigma^{-1})X \geq 0\\ \implies X'MX\geq0$$ Where $$M$$ is the residual maker matrix, and since $$M$$ is psd, the statement holds.