# Solved – Proving efficiency of OLS over GLS

I'm trying to prove the efficiency of OLS over GLS when the covariance matrix of the error $$varepsilon$$ is mistakenly assumed to be $$sigma^2Sigma$$ 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. $$sigma^2I$$)?

Contents

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'MXgeq0$$ Where $$M$$ is the residual maker matrix, and since $$M$$ is psd, the statement holds.