I asked this over on the Cross-Validated stack exchange, but it’s probably more suited for here since it's a linear algebra/matrix analysis problem. For those familiar with GMM estimation (this is not important for this question), the asymptotic variance of the GMM estimator has the form,
$$(G^TWG)^{-1}G^TWSWG(G^TWG)^{-1} \;\;\; (*)$$
where $W$ is a symmetric positive definite $n \times n$ weight matrix, $S$ is symmetric positive definite $n \times n$ and $G$ is full rank $n \times k$.
Now suppose we have two different positive definite weight matrices $W_1$ and $W_2$ and that $W_1 - W_2$ is positive semidefinite. Then does this imply that the corresponding difference in $(*)$, $$(G^TW_1G)^{-1}G^TW_1SW_1G(G^TW_1G)^{-1} - (G^TW_2G)^{-1}G^TW_2SW_2G(G^TW_2G)^{-1} \;\;\;\;\;\;\;(**)$$ is also positive semidefinite?
I thought I made some progress by looking at the Woodbury inverse matrix identity:
$$ A^{-1}B^T(C-BA^{-1}B^{T})^{-1}BA^{-1} = (A - B^TC^{-1}B)^{-1} - A^{-1}$$
My intention was to apply this formula to both terms in $(**)$ and solve for $C$ in each term. Hopefully this would lead to a sum of positive semidefinite matrices. But unfortunately, the matrix algebra got very messy so I don't think this is the right approach.
Any other suggestions would be greatly appreciated. I did generate a few matrices in R and so far it looks like it's true.
You can try to factorize $W$ using Cholesky decomposition, so $W=LL^T$. Then with $H=G^TL$, we have: $$(HH^T)^{-1}(HL^TSLH^T)(HH^T)^{-1}$$
Here $H$ is also full-rank but it depends now on $W$. We can also define $U=L^TSL$, which makes the equation look simpler: $$(HH^T)^{-1}(HUH^T)(HH^T)^{-1}$$
I'm not sure if that helps, because $H$ and $U$ depend on $W$, but maybe you can prove something in relation to the eigenspace of these matrices, it might coincide by some choice you have made. If it does then it would be simple to prove. If it doesn't, I'm not sure if there is an easy proof.