I am having some trouble finding the $\text{Var}(S^2)$.
I am using this formula: $$\text{Var}(S^2)=\frac{1}{n}\left[\mu_4-\frac{n-3}{n-1}\sigma^4\right]$$
With, $$\mu_4=E\left[(x-\mu)^4\right]$$
To calculate $\mu4$, I have used the central moment generating function: $$M_{x}^{c}(t)=E\left[e^{t(x-\mu)}\right]=\exp{\{-\theta t\}}\exp{\{\theta(e^t-1)\}}=\exp{\{\theta(e^t-1)-\theta t\}}$$
So, according to Wolfram Alpha, $$\mu_4=\frac{\partial^4 M_{x}^{c}(t)}{\partial t^4}\Bigr|_{t=0}=\theta(3\theta+1)$$
Therefore, $$\text{Var}(S^2)=\frac{1}{n}\left[3\theta^2+\theta-\frac{n-3}{n-1}\theta^2\right]=\frac{1}{n}\left[\frac{2n\theta^2+n\theta-\theta}{n-1}\right]$$
But I know that the variance should be: $$\text{Var}(S^2)=\frac{\theta}{n}+\frac{2\theta^2}{n-1}$$
Am I missing something in the process?
Thank you very much!
\begin{align*} \text{var}(S^2) &= \frac{1}{n}\left( \frac{2n\theta^2 + n\theta - \theta}{n -1} \right)\\ &= \frac{2n}{n(n - 1)}\theta^2 + \frac{n - 1}{n(n - 1)}\theta \\ &= \frac{2}{n - 1}\theta^2 + \frac{1}{n}\theta. \end{align*} I have no doubt you can take it from there, comparing this with your desired solution.