# Variance of sample Variance

In page 292 of Introduction to Mathematical Statistics by Hogg and Craig it is stated that in order for the variance of the sample variance, i.e. $\text{var}(S^2)$ to exist we need to assume that $E[X^4]<\infty$, that is the fourth (uncentered) moment of our RV is finite. My question is, how do these two connect?

I tried to derive a formula for the variance of the sample variance that does not reveal the connection as far as I can see. Since we know that $S^2=\frac{1}{n-1} \sum_{i=1}^n \left( X_i-\bar{X} \right)^2$ is an unbiased estimator of $\sigma^2$, i.e. the population variance, what we need to find is $$E\left[ \left( S^2-\sigma^2 \right) \right]^2$$ which upon simplifying becomes $$E\left(S^4\right)-\sigma^4$$.

From now on the algebra will get horrific though and I am not even sure whether we can proceed without stating the distribution of $X$. Is there perhaps another way to make sense of this statement? Thank you in advance.

• – wolfies Dec 16 '13 at 14:54
• @wolfies Yeah I've seen that. One of the answers is irrelevant while the other one is too complicated for an intermediate book like the one I am using-and consequently for me. Do you think there is a simpler derivation somewhere? – JohnK Dec 16 '13 at 14:59
• Well, the point is that, since the solution (see link above) is expressed in terms of the first 4 moments of the population, the solution also assumes the existence of those moments. – wolfies Dec 16 '13 at 15:58

## 1 Answer

Note that $S^2$ has terms involving $X_i^4$, and so $E[S^2]$ is the sum of terms involving $E[X_i^4]$. Thus, if the fourth moment is not finite, neither is $E[S^2]$ finite, nor is var$(S^2)$ finite. Some people say that various quantities such as expectations, variances, etc must be said to be undefined when the corresponding integrals/sums diverge. Others reserve the term "undefined" for cases when the integrals/sums lead to indeterminate forms such as $\infty - \infty$. The latter group would say that for a Cauchy random variable $X$, $E[X]$ is undefined, $E[X^2]$ is defined (but unbounded), and var$(X)$ is undefined (since $E[X]$ is undefined and so var$(X) = E[X^2] - (E[X])^2$ makes no sense). The former group would say that $E[X]$, $E[X^2]$, and var$(X)$ are all undefined for a Cauchy random variable.

• Thanks. I think I side with the latter group as there is a little theorem that states that if an expectation for $X_k$ exists then the expectation for all values of $n\leq k$ exist as well. – JohnK Dec 16 '13 at 15:37