Variance of sample variance in terms of kurtosis

Question

I want to prove the formula of the variance of sample variance in terms of kurtosis for general distribution (I think it is?), which can be found here on Wikipedia. Suppose $X_1, \dotsc, X_N$ IID, with variance $\sigma^2$, $$ \bar{X} = \frac{1}{N} (X_1 + \dotsb + X_N) \\ S = \sqrt {\frac {1} {N-1} \sum _{n=1} ^N (X_n -\bar{X})^2} $$ Then we claim that $$ \mathrm{Var} \left( S^2 \right) =\frac {\sigma^4} {n} \left( \kappa - \frac{n-3} {n-1} \right) $$ where $$ \mu_n = \mathbb{E} (X_1^n) \\ \kappa = \sigma^{-4} \mathbb{E} \left[ (X_1 - \mu_1)^4 \right] \\ $$ Observe $$ \sigma^4 \kappa = \mu_4 - 4 \mu_3 \mu_1 + 6 \mu_2 \mu_1^2 - 3 \mu_1^4 $$ Now, introduce a shorthand $$ \sum _{\mathbf{x}} {\mathbf{x}}^{(a_1, \dotsc, a_M)} =\sum _{i_1=1}^N \dotsb \sum _{i_M=1}^N x_{i_1}^{a_1} \dotsb x_{i_M}^{a_M} $$ Start with $$ n (n-1) S^2 = (n-1) \sum _{\mathbf{x}} {\mathbf{x}}^{(2)} - \sum _{\mathbf{x}} {\mathbf{x}}^{(1,1)} $$ I find that \begin{align} &n^2 (n-1)^2 S^4 \\ =&(n-1)^2 \left( \sum _{\mathbf{x}} {\mathbf{x}}^{(4)} +2 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,2)} \right) -2 (n-1) \left( 2 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,1,1)} +2 \sum _{\mathbf{x}} {\mathbf{x}}^{(3,1)} \right) \\ & \quad + \left( 24 \sum _{\mathbf{x}} {\mathbf{x}}^{(1,1,1,1)} +8 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,2)} +8 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,1,1)} \right) \end{align} giving \begin{align} &n^2 (n-1)^2 \mathbb{E} \left( S^4 \right) \\ =&(n-1)^2 \left[ n \mu_4 + 2 n (n-1) \mu_2^2 \right] -2 (n-1) \left[ 2 n (n-1) (n-2) \mu_2 \mu_1^2 + 2 n (n-1) \mu_3 \mu_1 \right] \\ &\quad +\left[ 24 n (n-1) (n-2) (n-3) \mu_1^4 + 8 n (n-1) \mu_2^2 + 8 n (n-1) (n-2) \mu_2 \mu_1^2 \right] \end{align} This turns out to be wrong. I am pretty sure I screw up some permutation factors before every $\Sigma$.

Answer 1

I think I get it. \begin{align} &n^2 (n-1)^2 S^4 \\ =&(n-1)^2 \left( \sum _{\mathbf{x}} {\mathbf{x}}^{(4)} +\sum _{\mathbf{x}} {\mathbf{x}}^{(2,2)} \right) -2 (n-1) \left( \sum _{\mathbf{x}} {\mathbf{x}}^{(2,1,1)} +2 \sum _{\mathbf{x}} {\mathbf{x}}^{(3,1)} \right) \\ & \quad + \left( \sum _{\mathbf{x}} {\mathbf{x}}^{(1,1,1,1)} +2 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,2)} +4 \sum _{\mathbf{x}} {\mathbf{x}}^{(2,1,1)} \right) \end{align}