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.
Answer
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_iX_j)^2.
Since \mathbb{E}[(X_iX_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_iX_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_iX_j)^2\sigma^2\right] \left[{1\over2}(X_kX_\ell)^2\sigma^2\right]
depending on the size of the intersection \{i,j\}\cap\{k,\ell\}.

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

There are n(n1)(n2) terms where \{i,j\}\cap\{k,\ell\}=1 and each has an expected cross product of (\mu_4\sigma^4)/4.

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\,(n3)\over n\,(n1)}. Here \mu_4=\mathbb{E}[(X\mu)^4] is the fourth central moment of X.
Attribution
Source : Link , Question Author : MathMan , Answer Author : Community