How to find the variance of the estimator, $\frac{1}{\bar{X}}$, for $X_i \sim \operatorname{Expo}(\lambda)$? So I used the method of moments to find an estimator for $\lambda$: $\frac{1}{\bar{X}}$.
However, I'm having trouble deriving the variance of it:
$$\operatorname{Var}\left(\frac{1}{\bar{X}}\right) = \operatorname{Var} \left(\frac{n}{\sum_{i=1}^n X_i}\right)$$
$$= n^2 \operatorname{Var}\left(\frac{1}{\sum_{i=1}^n X_i}\right)$$
I'm not sure how to proceed from here given the denominator is a sum of $X_i$'s.
 A: The distribution of $\sum_{i=1}^n X_i$ is a gamma distribution:
$$
\frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \quad \text{for } x>0.
$$
Consequently
\begin{align}
\operatorname E\left( \frac 1 {\sum_{i=1}^n X_i} \right) & = \int_0^\infty \frac 1 x\cdot \frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \\[8pt]
& = \lambda \int_0^\infty \frac 1 {\lambda x} \cdot \frac 1 {\Gamma(n)} (\lambda x)^{n-1} e^{-\lambda x} (\lambda \, dx) \\[8pt]
& = \lambda \int_0^\infty \frac 1 u \cdot \frac 1 {\Gamma(n)} u^{n-1} e^{-u} \, du \\[8pt]
& = \frac \lambda {\Gamma(n)} \int_0^\infty u^{n-2} e^{-u} \, du \\[8pt]
& = \frac{\lambda \Gamma(n-1)}{\Gamma(n)} = \frac \lambda {n-1}.
\end{align}
Therefore
$$
\operatorname E\left( \frac 1 {\overline X} \right) = \frac{n\lambda}{n-1}.
$$
Similarly
\begin{align}
\operatorname E\left( \frac 1 {\left( \sum_{i=1}^n X_i \right)^2} \right) & = \frac {\lambda^2} {\Gamma(n)} \int_0^\infty \frac 1 {u^2} \cdot u^{n-1} e^{-u} \, du \\[8pt]
& = \frac{\lambda^2\Gamma(n-2)}{\Gamma(n)} = \frac{\lambda^2}{(n-1)(n-2)}
\end{align}
so
$$
\operatorname E\left( \frac 1 {\overline X^2} \right) = \frac{n^2\lambda^2}{(n-1)(n-2)}.
$$
Finally
\begin{align}
\operatorname{var}\left(\frac 1 {\overline X}\right) & = \operatorname E\left(\frac 1 {\overline X^2} \right) - \left( \operatorname E\left(\frac 1 {\overline X} \right) \right)^2 \\[8pt]
& = \lambda^2 \left( \frac{n^2}{(n-1)(n-2)} - \frac{n^2}{(n-1)^2} \right) \\[8pt]
& = \frac{n^2\lambda^2}{(n-1)^2(n-2)}.
\end{align}
A: A faster  way to get the same result:
$Y=\Sigma_i X_i\sim Gamma(n;\lambda)$
Thus
$\frac{1}{Y}\sim\text{Inverse Gamma}$ and thus $E\Big[\frac{1}{Y}\Big]=\frac{\lambda}{n-1}$
Concluding,
$$E\Bigg[\frac{1}{\overline{X}}_n\Bigg]=\frac{n}{n-1}\lambda$$
Similar reasoning to get  immediately variance, without any calculation
