Estimator of variance of an exponential distribution

Question

I have a sample of $X_1,\ldots,X_n$ observations from an exponential distribution ($p(x)=\lambda \exp(-\lambda x)I(x>0)$. I'm struggling with how to prove that $(\bar{X})^2$ is an asymptotic normal estimator of $\lambda^{-2}$ and how to find the variance of this estimator. I would really appreciate some help here!

Answer 1

$\newcommand\la\lambda\newcommand\th\theta$This follows by the delta method, in this case used with $X_n=\bar X$, $\th=E\bar X=\dfrac1\la$, $\sigma^2=\dfrac1{\la^2}$, $g(\th)=\th^2$ -- which implies that $\bar X^2$ is asymptotically normal with asymptotic mean $\dfrac1{\la^2}$ and asymptotic variance $\dfrac4{n\la^4}$.

Since $\bar X$ has the gamma distribution with parameters $n$ and $\dfrac1{n\la}$, the exact mean and variance of $\bar X^2$ are $\dfrac{1+1/n}{\la^2}$ and $\dfrac{4(1+\frac1n)(1+\frac3{2n})}{n\la^4}$, respectively, which are close for large $n$ to the asymptotic mean $\dfrac1{\la^2}$ and variance $\dfrac4{n\la^4}$.