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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?

Golden_Ratio
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alexmolas
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  • I think if you work out $\mu_4$ you can bound the integral and see that this can't happen. – Randall Feb 18 '23 at 22:44
  • @Randall But this is a general formula, regardless of the underlying distribution. Do you mean that $\mu_4 > \frac{n-3}{n-1}\sigma^4$ for any $n$ and any distribution function? – alexmolas Feb 18 '23 at 23:06
  • The answer posted below is the type of thing I was getting at. – Randall Feb 19 '23 at 00:36

1 Answers1

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By Cauchy-Schwarz,

$$\frac{n-3}{n-1}\sigma^4\leq \sigma^4=(E[(X-\mu)^2])^2\leq E[(X-\mu)^4]=\mu_4. $$

Golden_Ratio
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