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Prove that FGLS is asymptotically efficient. Does one have to use Cramer Rao to do this?

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    $\begingroup$ Homework tag too? $\endgroup$
    – jbowman
    Feb 26, 2012 at 21:46
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    $\begingroup$ I presume that FGLS stands for "Feasible GLS". Is this question related to your preceding question, stats.stackexchange.com/q/23700/930? $\endgroup$
    – chl
    Feb 26, 2012 at 21:49
  • $\begingroup$ yes feasible GLS. I'm unsure at which stage the proof should end, in class our lecturer ends with "it can be shown that" _ is asymp efficient. And I'm uncertain whether in proving it it is necessary to calculate the Cramer Rao variance as part of the proof that it is the min var estimator of Beta. $\endgroup$
    – user9448
    Feb 26, 2012 at 22:52

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The short answer is "No." Some hints for an approach: FGLS is closely related to another estimator, which can be shown to be asymptotically efficient (w/o using Cramer-Rao.) I'm sure you've covered it already in your class. Under what conditions does the asymptotic dist'n of FGLS converge to that of the other estimator? (They aren't terribly restrictive.)

Note that FGLS isn't guaranteed to be asymptotically efficient; if I use an estimate of the covariance matrix that isn't consistent, then FGLS won't even be consistent. So some conditions (relating to what?) are required for asymptotic efficiency.

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  • $\begingroup$ It has the same asymptotic properties as GLS if the weighted sum of squares matrix based on the Omega converges to a positive definite matrix (then estimated gamma also converges to the same matrix,) and if it has a limiting normal distribution. $\endgroup$
    – user9448
    Feb 27, 2012 at 0:38
  • $\begingroup$ What are "Omega" and "gamma"? Don't forget, not everybody uses the same notation... Regardless, the "limiting normal distribution" isn't a necessary condition, it's more of a result. $\endgroup$
    – jbowman
    Feb 27, 2012 at 0:44
  • $\begingroup$ where Omega is the variable in the covariance matrix (heteroskedastic) $\endgroup$
    – user9448
    Feb 27, 2012 at 1:05
  • $\begingroup$ You can construct cases where $X^T\widehat{\Omega}^{-1}X/N$ converging to a pds matrix isn't enough; it has to converge to a particular pds matrix. Also, there's another condition involving the errors that is analogous to one you see in proofs relating to OLS. $\endgroup$
    – jbowman
    Feb 27, 2012 at 16:29
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    $\begingroup$ @bob - $X^T\Omega^{-1}X/N$. $\endgroup$
    – jbowman
    Oct 13, 2021 at 23:25

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