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I would like to know whether it is possible to derive a closed-form expression for the sample variance of the $n$-th power of the sample mean $\overline{x}_m$ that was calculated using $m$ sample values $x_i$:

$\mathrm{Var}[\overline{x}_m^n] =\;?$,

where

$\overline{x}_m = \frac{1}{m} \sum_{i=1}^m x_i$.

It is clear that in case of $m=1$, $\overline{x}_1 = x_1$, for which this solution can be used. However, I am interested in a general solution for $m > 1$, preferably for arbitrarily distributed samples, but I expect that to be challenging or even impossible. So, I would also be happy with a solution for normally distributed samples.

Update #1: I think that I have used the term closed-form expression incorrectly and might have been too unprecise in general. Let me clarify: I am basically looking for a formula to calculate a good (preferably unbiased) estimate for $\mathrm{Var}[\overline{x}_m^n]$ based sample values $x_i$.

A comment indicated that for normally distributed $x_i$, $\overline{x}_m$ is still normally distributed and as such, the solution should be trivial. However, I still would like to have a formula to estimate the variance $\mathrm{Var}[\overline{x}_m^n]$based on the given set of samples values $x_i$.

Apologies for the misleading initial question.

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  • $\begingroup$ I do not think there is general solution. For normal, $\bar x_m$ is still normal and you already have the answer. $\endgroup$
    – user158565
    Jul 24 '19 at 15:54
  • $\begingroup$ I think you really want the population variance of blah i.e. Var(blah)--- not the sample variance of blah. There can be no closed-form expression for the sample variance other than the sample variance of blah. $\endgroup$
    – wolfies
    Jul 24 '19 at 16:14
  • $\begingroup$ @user158565 Thanks for the hint! You're correct, however, I have updated the original question in order to clarify what I'm really interested in. $\endgroup$
    – Hiro
    Jul 24 '19 at 19:41
  • $\begingroup$ @wolfies I'm afraid that I actually wanted something entirely different. I have updated the question accordingly. Thank you for contributing to the quality of the question. $\endgroup$
    – Hiro
    Jul 24 '19 at 19:42
  • $\begingroup$ From your link, it seems you are working though mgf. Then see this math.stackexchange.com/questions/912869/…, maybe you can get answer from there. $\endgroup$
    – user158565
    Jul 24 '19 at 20:27

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