Suppose we have a regression model $$ y=X\beta+u,\quad E(u)=0,\quad E(uu')=\Sigma. $$ Let $\hat\beta$ and $\bar\beta$ respectively denote the OLS and GLS estimator. Then, when $X$ is fixed (or when $X$ and $u$ are independent), one can show the efficiency result $$ \text{Var}(\hat\beta)=(X'X)^{-1}X'\Sigma X(X'X)^{-1}\succeq(X'\Sigma^{-1}X)^{-1}=\text{Var}(\bar\beta). $$ Is there a similar asymptotic result that holds when $X$ is stochastic? If so, could you please provide a statement and a (sketch of) proof or point me to some references?

  • $\begingroup$ What does the inequality-like symbol mean? It must be related to matrices because I've always seen it in that contest, but $\mathbf {R}^{n,n}$ is not ordered. $\endgroup$
    – DeltaIV
    Commented Sep 17, 2016 at 16:15
  • 1
    $\begingroup$ @DeltaIV $A\succeq B$ usually means $A-B$ is positive semidefinite. $\endgroup$
    – yurnero
    Commented Sep 17, 2016 at 16:37

1 Answer 1


I think you have to look at the asymptotic variances of OLS and GLS. That is, at the variances in the multivariate normal distributions in the distribution limits of $\sqrt{n}(\hat{\beta}-\beta)$ and $\sqrt{n}(\tilde{\beta}-\beta)$. Under quite general conditions, the asymptotic results state that $$\sqrt{n}(\hat{\beta}-\beta) \stackrel{d} \rightarrow \mathcal{N}\left(0, \text{Avar}(\hat{\beta})\right)$$ $$\sqrt{n}(\tilde{\beta}-\beta) \stackrel{d} \rightarrow \mathcal{N}\left(0, \text{Avar}(\tilde{\beta})\right),$$ where $$\text{Avar}(\hat{\beta}) = \left(\text{plim} \frac{1}{n}X′X\right)^{−1} \, \text{plim} \frac{1}{n}X′\Sigma X \, \left(\text{plim} \frac{1}{n}X′X\right)^{−1}$$ $$\text{Avar}(\tilde{\beta}) = \left(\text{plim} \frac{1}{n}X′\Sigma^{−1}X\right)^{−1}.$$

From here you can see that you can draw similar conclusion asymptotically: $$\text{Avar}(\hat{\beta}) \succeq \text{Avar}(\tilde{\beta}).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.