Is there any advantage of a maximum-likelihood estimator over another estimator (for example least squares) assuming both methods of estimation are efficient, unbiased, and consistent?

I have been taught that maximum-likelihood methods are generally superior but I don't know why.

  • $\begingroup$ Because the competing estimators might not be efficient and asymptotically unbiased. $\endgroup$ – JohnK Jul 17 '17 at 15:22

The MLE is more efficient when the distributional assumptions are correctly specified. For instance, if the linear model satisfies, $Y = b X + \epsilon$ where $\epsilon$ comes from, say, a shifted 0 meanWeibull distribution, the MLE will account for the skewness of the errors and the LS will not. It happens to be the case that when $\epsilon$ is normally distributed, you arrive at the least squares estimator anyway.

  • $\begingroup$ Indeed if the MLE assumptions are true (i.e. you have the right variables, model and distribution) there exists no other estimator why is more efficient than MLE. Of course other estimators could be equally efficient, and biased estimators could be more efficient. $\endgroup$ – Repmat Jul 17 '17 at 18:04
  • $\begingroup$ @Repmat there're also superefficient estimators that beat MLEs! They're unbiased too :) $\endgroup$ – AdamO Jul 17 '17 at 18:08
  • $\begingroup$ In this example, if Y does not have a Weibull distribution but we use a Weibull MLE, would the LS estimator be more efficient and would both estimators still be unbiased and consistent? $\endgroup$ – Great38 Jul 18 '17 at 14:15
  • $\begingroup$ @Great38 you would need to use simulations or arithmetic to derive the relative efficiency on a case-by-case basis. In general, I think whichever distribution provides a better "fit" to the residuals will provide better inference, but I bet there are counter examples even to that. $\endgroup$ – AdamO Jul 18 '17 at 16:26

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