# Sample Variance of Power of Sample Mean

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.

• I do not think there is general solution. For normal, $\bar x_m$ is still normal and you already have the answer. Jul 24 '19 at 15:54
• 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. Jul 24 '19 at 16:14
• @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.
– Hiro
Jul 24 '19 at 19:41
• @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.
– Hiro
Jul 24 '19 at 19:42
• 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. Jul 24 '19 at 20:27