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?

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    $\begingroup$ It depends on the distribution in both cases (estimating $\sigma$ and $\sigma^3$). For estimating variance it doesn't (assuming the variance exists). You may find this post of some use. $\endgroup$
    – Glen_b
    Commented Feb 10, 2018 at 13:16

1 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.

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    $\begingroup$ Maybe it is a silly question, but what does $\mathcal{Ga}(n - 1 / 1 / 2)$ mean? $\endgroup$
    – abukaj
    Commented Feb 14, 2018 at 14:45
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    $\begingroup$ @abukaj, this is a statistical/probability notation, means a gamma distribution with shape = n-1, and scale=1/2. This is a somewhat standard notation. $\endgroup$ Commented Apr 17, 2018 at 19:07

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