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It is known that

  1. MLE is consistent and asymptotically efficient.
  2. OLS under certain assumptions is asymptotically normal.
  3. If the errors are gaussian, then OLS is equivalent to MLE.
  4. If the errors are gaussian with mean 0 and constant variance, then OLS is UMVU.
  5. If the errors are not gaussian and under certain assumptions, OLS is BLUE.

Given these previous facts, I would like to know if the following statements are correct.

  • By (2) and (4), even if the errors are not Gaussians, OLS is asymptotically UMVU, and as such, the best possible estimator.

  • By (2) and (3) OLS is asymptotically the same as MLE and so by (1) is asymptotically efficient and asymtotically consistent.

I tried to look in books but couldn't find these statements, I would like to know if they are correct. Any help please?

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  • $\begingroup$ The fact you have mentioned is clear to me. My question basically is this: "Is OLS asymptotically UMVU even if we don't assume gaussian errors?" I tried to come up with two possible arguments in favour of this as you can read in my post. Are my arguments correct? $\endgroup$
    – user405777
    Commented May 5 at 15:20
  • $\begingroup$ 5. is not strictly correct. You do not have to rule out the Gaussian distribution for OLS to be BLUE. $\endgroup$ Commented May 6 at 10:31

1 Answer 1

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The OLS estimate and the MLE will asymptotically approach normal distributions. But, no, they will not have the same variance, and the OLS estimate does not need to become (asymptotically) the UMVU estimator.

Take the Laplace distribution as a counter example and say we want to estimate the location parameter. Then the sample mean is the OLS estimate and the sample median is the maximum likelihood estimate.

The sample variance of the mean will relate to the variance of the variable:

$$Var[mean(x_n)] \approx \frac{1}{n} Var[x]$$

The sample variance of the median will relate to the slope of the cumulative distribution (and the height of the distribution density near the median)

$$Var[median(x_n)] \approx \frac{1}{n} \frac{1}{4f(median(x))^2}$$

If we fill in the values for a Laplace distribution with scale $b$ then we have

$$Var[mean(x_n)] \approx \frac{2b^2}{n}$$

and

$$Var[median(x_n)] \approx \frac{b^2}{n}$$


Extension based on jbowman's comment. The error in the reasoning is that the 3-th and 4-th point use as condition "If the errors are gaussian". The OLS estimate approaching a normal distribution, does not mean that those conditions in the 3-th and 4-th point, about the errors are fulfilled.

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    $\begingroup$ To extend your excellent answer a little - the fundamental mistake the OP makes is in conflating the asymptotic normality of the OLS estimator with the finite-sample normality of the errors. $\endgroup$
    – jbowman
    Commented May 5 at 16:04
  • $\begingroup$ asymptomatically? $\endgroup$ Commented May 6 at 10:32

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