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.
Best Answer
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}