Asymptotic distribution of sample variance

Question

Asymptotic distribution of sample variance $S^2$ for an exponential distribution $\mathcal{E}(\frac{1}{\theta})$

$S^2= \frac{1}{n}\sum (X_i-\bar{X})^2$

$\mathbb{E}(S^2)= \frac{n-1}{n}\theta^2$

$\sqrt{n}(S^2- \frac{n-1}{n}\theta^2) \rightarrow_d N(0,\frac{(n-1)(8n-6)\theta^2}{n^3})$

is it correct?

Answer 1

If $X_1,...,X_n$ is a sample of $n$ i.i.d. random variables with mean $\mu$ and variance $\sigma^2$, then we know from this post that $$\sqrt{n}\left(S^2-\sigma^2\right)\longrightarrow_d\mathcal{N}(0,\mu_4-\sigma^4)$$

Here, $S^2=\frac{1}{n-1}\sum_{k=1}^{n}(X_k-\bar{X})^2$ is the sample variance and $\mu_4=\mathbb{E}\left((X_1-\mu)^4\right)$.

In your problem $\mu=\theta,\sigma^2=\theta^2,$ and $\mu_4=9\theta^4$ which means $$\sqrt{n}\left(S^2-\theta^2\right)\longrightarrow_d \mathcal{N}\left(0,8\theta^4\right)$$

So, we can say $S^2$ is approximately $\mathcal{N}(\theta^2,\frac{8\theta^4}{n})$ for large $n$.