Calculating maximum-likelihood estimation of the exponential distribution and proving its consistency
The computation of the MLE of $\lambda$ is correct.
The consistency is the fact that, if $(X_n)_{n\geqslant1}$ is an i.i.d. sequence of random variables with exponential distribution of parameter $\lambda$, then $\Lambda_n\to\lambda$ in probability, where $\Lambda_n$ denotes the random variable $$ \Lambda_n=\frac{n}{\sum\limits_{k=1}^nX_k}. $$ Thus, one is asked to prove that, for every positive $\varepsilon$, $\mathrm P(|\Lambda_n-\lambda|\geqslant\varepsilon)\to0$ when $n\to\infty$.
In the case at hand, it might be easier to prove the stronger statement that $\frac1{\Lambda_n}\to\frac1\lambda$ almost surely when $n\to\infty$. Hint: Law of large numbers.
$\hat\lambda= \frac{n}{\sum_{i=1}^n x_i}$ to be consistent estimator of $\lambda$ it should be Asymptotically
- Unbiased,
- and it's variance goes to zero.
Using $E\left\{ x\right\}=\frac{1}{\lambda}$ and $E\left\{ x^2\right\}=\frac{2}{\lambda^2}$ and the fact that $x_i$ are iid, we have
Condition 1: $\lim_{n\rightarrow \infty} E\{\hat\lambda - \lambda\}=0$
Condition 2: $\lim_{n\rightarrow \infty}E\left\{\left(\hat\lambda - E\{\hat\lambda\}\right)^2\right\}=0 $