I'm looking for the sampling standard deviation of $\hat\sigma^\gamma$, where $\hat\sigma$ is a sample standard deviation. For simplicity, lets do the sample variance of the sample variance and take the roots later.

Before doing any analytical math, I did the following simulation:

N = 10
gam = 2.5
p = rnorm(1000,mu,sig)
SDgam = c()
sdsd = c()
for (i in 1:4000){
    s = sample(p,N)
    SDgam[i] = sd(s)^gam
    abline(v = mean(SDgam),col='red')
    sdsd[i] = sd(SDgam)
    if (i>1){

enter image description here

The distribution is quite skewed when $\gamma=2.5$, which seems natural enough, and it looks like convergence is a bit slow, which also seems natural enough.

I start with the definition of the variance $$ var(\hat\sigma^{2\gamma}) = E(\hat\sigma^{4\gamma}) - E(\hat\sigma^{2\gamma})^2 $$ then expand it out $$ var(\hat\sigma^{2\gamma}) = \int\hat\sigma^{4\gamma}f(\hat\sigma^2)d\hat\sigma^2 - \left(\int\hat\sigma^{2\gamma}f(\hat\sigma^2)d\hat\sigma^2\right)^2 $$

I can use that fact that $\frac{(n-1)\hat\sigma^2}{\sigma^2}\sim\chi^2_{n-1}$ to rewrite the $f(\hat\sigma^2)$ from above as

$$ f(\hat\sigma^2) = \frac{\hat\sigma^{2(n/2-1)}e^{-\hat\sigma^2/2}\sigma^2}{2^{n/2}\Gamma(n/2)(n-1)} $$

But it isn't clear to me what that gets me or whether it is helpful:

$$ var(\hat\sigma^{2\gamma}) = \int\hat\sigma^{4\gamma}\frac{\hat\sigma^{2(n/2-1)}e^{-\hat\sigma^2/2}\sigma^2}{2^{n/2}\Gamma(n/2)(n-1)}d\hat\sigma^2 - \left(\int\hat\sigma^{2\gamma}\frac{\hat\sigma^{2(n/2-1)}e^{-\hat\sigma^2/2}\sigma^2}{2^{n/2}\Gamma(n/2)(n-1)}d\hat\sigma^2\right)^2 $$

How would I go about getting a closed form here? I want the variance of the sample variance as a function of the sample variance, $N$, and $\gamma$. Apologies if I'm missing something obvious -- I haven't had to do analytical math in a while.

  • $\begingroup$ Could you please specify what language or program your code is written in. Otherwise, it is just gobbledegook. I'm not sure what relevance the code has to the question anyway!? $\endgroup$
    – wolfies
    Apr 11, 2015 at 5:11
  • $\begingroup$ 2. Are you assuming a Normal parent? Your question does not state this. Or is your question posed in general terms, wrt to the variance of the sample variance of any arbitrary distribution? $\endgroup$
    – wolfies
    Apr 11, 2015 at 5:15
  • $\begingroup$ 3. You need to specify what formula you are using to derive the sample variance. There are two common versions in usage: $\frac1n$ and $\frac{1}{n-1}$. If you wish to work with an estimator of sample standard deviation, you need to specify how the latter is derived: e.g. the square root of your sample variance, or something else? $\endgroup$
    – wolfies
    Apr 11, 2015 at 5:33

2 Answers 2


When faced with a non-linear function of a random variable, we usually apply the Delta Method (sometimes referred to as "error propagation" when the purpose is to obtain the variance) to obtain an approximate estimation.

If we have a random variable $X$ and a non-linear function of it $h(X)$, then we apply a first-order Taylor expansion around a fixed point $x_0$:

$$h(X) \approx h(x_0) + h'(x_0)\cdot (X-x_0)$$

Keeping in mind that $x_0$ is treated as a fixed number, consider

$${\rm Var}[h(X)] \approx {\rm Var}\big[h(x_0) + h'(x_0)\cdot (X-x_0)\big]$$

$$={\rm Var}\big[h'(x_0)\cdot (X-x_0)\big] = \left[h'(x_0)\right]^2{\rm Var}\big(X\big)$$

In your case, $h(X) = X^{\gamma} \implies [h'(x_0)]^2 = [\gamma x_0^{\gamma-1}]^2$, and your random variable is the sample standard deviation. Use the obtained estimate of it as your $x_0$ and also the estimated variance of the sample standard deviation for ${\rm Var}\big(X\big)$ (except if you know a priori the true value of this variance of course).

  • $\begingroup$ Thanks! But what do you pick for $x_0$? Is it completely arbitrary? Say I pick "1", and $\gamma$ is 2. I get 2^2 = 4 as a multiplier of the variance of the sample standard deviation (which has a closed form). But if I pick $x_0 = 2$, I get a different answer. I'm probably missing something basic... $\endgroup$ Apr 11, 2015 at 3:03
  • $\begingroup$ I am confused. Is is written explicitly in the third-to-last line of my answer. $\endgroup$ Apr 11, 2015 at 3:23
  • $\begingroup$ ah OK. so you pick $x_0 = \hat\sigma$ $\endgroup$ Apr 11, 2015 at 4:30
  • $\begingroup$ for highly skewed distributions, this variance of variance will itself have a pretty high variance. $\endgroup$ Apr 11, 2015 at 4:31

The Question

  • Let $p$ denote the sample variance. Find $Var(p^\gamma)$

Exact symbolic solutions can be attained for positive integer-valued $\gamma$. The answer will, of course, depend on what formula you are using for sample variance $p$. Such problems are known as moments of moments problems and can be solved more easily by working with power sum $s_r$ notation.


I will assume that by sample variance $p$, you are referring to the unbiased estimator of population variance $\mu_2$, i.e. the $2^{\text{nd}}$ h-statistic, namely:

$$ p \quad=\quad \frac{1}{n-1}\sum _{i=1}^n \left(X_i-\bar{X}\right){}^2 \quad = \quad \frac{n s_2-s_1^2}{n (n-1)} \quad \text{where} \quad s_r=\sum _{i=1}^n X_i^r$$

You seek $Var(p^\gamma)$. Note that the variance operator is the $2^{nd}$ Central Moment of $p^\gamma$.

  • In the case of $\gamma = 2$, the solution $\text{Var}(p^2)$ is:

enter image description here

where I am using the CentralMomentToCentral function from the mathStatica package for Mathematica to perform the calculation. The solution is expressed in terms of the central moments $\mu_i$ of the population of $X$.

  • In the case of $\gamma = 3$, the solution $\text{Var}(p^3)$ is the $2^{\text{nd}}$ central moment of $p^3$:

enter image description here

enter image description here

The Normal case

The above results are completely general and hold for any distributions whose moments exist. In the special case of a Normal parent, i.e. $X \sim N(\mu, \sigma^2)$, the first 12 central moments $\mu_i$ of the population are:

$\text{mulis}=\left\{\mu _2\to \sigma ^2,\mu _3\to 0,\mu _4\to 3 \sigma ^4, \mu _5\to 0, \mu _6\to 15 \sigma ^6, \mu _7\to 0, \mu _8\to 105 \sigma ^8, \mu _9\to 0,\mu _{10}\to 945 \sigma ^{10}, \mu _{11}\to 0, \mu _{12}\to 10395 \sigma ^{12}\right\}$

and the above solutions simplify to:

enter image description here


enter image description here

If your interest is in non-integer $\gamma$ (e.g.$\gamma = 2.5$) and need recourse to approximate solutions, the above exact solutions will provide a helpful benchmark to check how well any approximate method is working.

  • $\begingroup$ Thanks for this -- I apparently don't have enough reputation to upvote. I am indeed interested in non-normal "parent" distributions, and real-valued $\gamma$. It looks like the closed form isn't all that convenient! $\endgroup$ Apr 11, 2015 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.