An unbiased estimator of σ³

Question

As it was suggested in the linked answer, $s_n = \sqrt{\frac{\sum_{i = 1}^n (x_i - \bar{x})^2}{n - 1}}$ is not an unbiased estimator of $\sigma$.

I suspect neither $s_n^3 = \sqrt{\frac{\sum_{i = 1}^n (x_i - \bar{x})^2}{n - 1}}^3$ is an unbiased estimator of $\sigma^3$.

Then, what is the formula for the unbiased estimator?

Answer 1

Let us consider a Normal sample. Since$$\sigma^{-2}\sum_{i=1}^n (x_i-\bar{x}_n)^2\sim\chi^2_{n-1}$$which is also a Gamma $\mathcal{Ga}(n-1,1/2)$ variate, the $3/2$ moment of this variate is $$\int_0^\infty z^{3/2}z^{n-1-1}\exp\{-z/2\}2^{-(n-1)}\Gamma(n-1)^{-1}\text{d}z=\frac{2^{(n-1)+3/2}\Gamma(n-1+3/2)}{2^{n-1}\Gamma(n-1)}$$i.e. $$\frac{2^{3/2}\Gamma(n-1+3/2)}{\Gamma(n-1)}$$This implies that $$\frac{\Gamma(n-1)}{2^{3/2}\Gamma(n-1+3/2)}\left\{\sum_{i=1}^n (x_i-\bar{x}_n)^2\right\}^{3/2}$$is an unbiased estimator of $\sigma^3$ in the Normal case. Now, as discussed in my earlier answer to an earlier question, this estimator does not provide an unbiased estimator for other scale families of distributions and there is no such an estimator for all distributions, relating to a 1968 paper by Peter Bickel and Erich Lehmann.