We have $n$ iid random variables all taken from an $Exponential( \lambda)$ distribution.
I have found the MLE of $\lambda$ to be $\frac{n}{\Sigma _{i=1}^{n} X_i}$.
However I am struggling to prove that this is a consistent estimator. I have had two ideas as to how to approach this:
1) Through the weak law of large numbers which involves, in this case, the reciprocals of both the estimator and what I'm trying to estimate.
2) Through showing that the expected value of the MLE converges in probability to $\lambda$ and that it's variance converges to $0$ as $n$ tends to $\infty$.
However, I am struggling with 1 because I cannot manipulate the reciprocal, and I am struggling with 2 because I don't know how to calculate the expected value of the reciprocal of a sum of the RVs.
Any help?
Best Answer
(1) Note that $ \frac{1}{n} \sum_{i=1}^n X_i \xrightarrow{p} \frac{1}{\lambda}$ by WLLN. By the continuous mapping theorem, we have $$ 1 \Big/ \left( \frac{1}{n} \sum_{i=1}^n X_i \right) \xrightarrow{p} \lambda $$
(2) Alternatively, if $X_i \sim $ Exponential$(\text{rate}=\lambda)$, then $\sum_{i=1}^n X_i \sim \text{Gamma}(n,\lambda)$. Let $Z \sim \Gamma(n,\lambda)$. \begin{align*} E\left[ \frac{n}{\sum X_i} \right] &= n E[1/Z] \\ &= n \int_0^\infty z^{-1} \frac{1}{\Gamma(n)} \lambda ^n z^{n-1} e^{-\lambda z} \; dz \\ &= \frac{n \lambda \Gamma(n-1)}{\Gamma(n)} \underbrace{\int_0^\infty \frac{1}{\Gamma(n-1)} \lambda^{n-1} z^{n-2} e^{-\lambda z} \; dz}_{=1} \\ &= \frac{n \lambda (n-2)!}{(n-1)!} = \frac{n \lambda}{n-1} \end{align*} and as $n \to \infty$, this converges to $\lambda$.