0

In this answer is shown that the variance of the sample variance is

$$ \text{Var}(S^2) = \frac{1}{n} \left(\mu_4 - \frac{n-3}{n-1}\sigma^4\right) $$

where $\mu_4$ is the fourth central moment, ie $E[(X-\mu)^4]$.

My question is, what prevents the variance from being negative? As far as I know, it can happen that $\mu_4 < \frac{n-3}{n-1}\sigma^4$, and then the variance would be negative, which doesn't make sense.

Am I missing something?

alexmolas
  • 288
  • 2
    Because variances cannot be negative (they are defined as expectations of squares, as shown explicitly in the answer you reference, and squares are never negative), you must conclude either that the answer is incorrect or "as far as I know" is wrong. I believe the answer. Moreover, basic inequalities establish that $\mu_4\ge \sigma^4:$ see https://stats.stackexchange.com/questions/512568 for one such inequality and its proof. The Power-Mean Inequality states, among other things, that $\mu_4^{1/4}\ge (\sigma^2)^{1/2}$ so taking fourth powers also gets you there. – whuber Feb 18 '23 at 23:26
  • "As far as I know, it can happen that $\mu_4 < \frac{n-3}{n-1}\sigma^4$" would have been more convincing with a numerical example, but unfortunately none exists – Henry Feb 19 '23 at 00:28

0 Answers0