Variance of sample variance?

What is the variance of the sample variance? In other words I am looking for Var(S2).

I have started by expanding out Var(S2) into E(S4)[E(S2)]2

I know that [E(S2)]2 is σ to the power of 4. And that is as far as I got.


Here’s a general derivation that does not assume normality.

Let’s rewrite the sample variance S2 as an average over all pairs of indices:
S^2={1\over{n\choose 2}}\sum_{\{i,j\}} {1\over2}(X_i-X_j)^2.
Since \mathbb{E}[(X_i-X_j)^2/2]=\sigma^2, we see that S^2 is an unbiased estimator for \sigma^2.

The variance of S^2 is the expected value of
\left({1\over{n\choose 2}}\sum_{\{i,j\}} \left[{1\over2}(X_i-X_j)^2-\sigma^2\right]\right)^2.

When you expand the outer square, there are 3 types of cross product terms
\left[{1\over2}(X_i-X_j)^2-\sigma^2\right] \left[{1\over2}(X_k-X_\ell)^2-\sigma^2\right]
depending on the size of the intersection \{i,j\}\cap\{k,\ell\}.

  1. When this intersection is empty, the factors are independent and the expected cross product is zero.

  2. There are n(n-1)(n-2) terms where |\{i,j\}\cap\{k,\ell\}|=1 and each has an expected cross product of (\mu_4-\sigma^4)/4.

  3. There are {n\choose 2} terms where |\{i,j\}\cap\{k,\ell\}|=2 and each has an expected cross product of (\mu_4+\sigma^4)/2.

Putting it all together shows that \mbox{Var}(S^2)={\mu_4\over n}-{\sigma^4\,(n-3)\over n\,(n-1)}. Here \mu_4=\mathbb{E}[(X-\mu)^4] is the fourth central moment of X.

Source : Link , Question Author : MathMan , Answer Author : Community

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