# Consistency of MLE exponential distribution

Let $y_1, \dots,y_n$ be i.i.d. random variables from $Exp(\theta)$, where $\theta$ is scale parameter. I've found the MLE $$\hat \theta=\frac{\sum^{n}_{i=1}y_i}{n}$$

Now I have to prove that $\hat \theta$ is consistent. A sufficient condition is

$$\begin{cases} \lim_{n \to \infty}\mathbb{E}(\hat \theta)=\theta \\ \lim_{n \to \infty}\mathbb{Var}(\hat \theta)=0 \end{cases}$$

For the first condition $\mathbb{E}(\hat \theta)=\mathbb{E} \left(\frac{\sum^{n}_{i=1} y_i}{n}\right)= \theta$, so $\lim_{n \to \infty}\mathbb{E}(\hat \theta)=\theta$.

I'm stuck on second condition, any help?

Edit

$\lim_{n \to \infty}\mathbb{Var}(\hat \theta)=\lim_{n \to \infty}\mathbb{Var}(\frac{\sum^{n}_{i=1}y_i}{n})=\lim_{n \to \infty}\frac{1}{n^2}\sum^{n}_{i=1}\mathbb{Var}(y_i)=\lim_{n \to \infty}\frac{\theta^2}{n}=0$

• $$\mathrm{var}\left(\frac1n\sum_{k=1}^ny_k\right)=\frac1{n^2}\sum_{k=1}^n\mathrm{var}(y_k)=\ldots$$
– Did
Aug 20, 2016 at 17:29
• @Did maybe my brain is still on holidays, thanks!
– Paul
Aug 21, 2016 at 8:59

We have $$\operatorname{Var} \left( {\hat \theta } \right) = \operatorname{Var} \left( {\frac{1}{n}\sum\limits_{j = 1}^n {{Y_j}} } \right) = \frac{1}{{{n^2}}}\sum\limits_{j = 1}^n {\operatorname{Var} \left( {{Y_j}} \right)} = \frac{1} {{{n^2}}}\sum\limits_{j = 1}^n {\operatorname{Var} \left( {{Y_1}} \right)} = \frac{{\operatorname{Var} \left( {{Y_1}} \right)}}{n}.$$ Note that since $Y_1\sim Exp(\theta)$ with $\theta$ is scale parameter, we get $\operatorname{Var} \left( {{Y_1}} \right) = \theta^2$. Thus $$\operatorname{Var} \left( {\hat \theta } \right) = \frac{{{\theta ^2}}} {n} \to 0$$ as $n\to+\infty$.
• Hi, in this case $\mathbb{Var}(y)$ should be $\theta^2$ because I used the scale parameter of exponential distribution.