consistency of maximum likelihood estimator

Question

For population with n size and following density function

$$f(y, a)= (1/6a^4)y^3e^{-y/a}$$

For that, I have found the maximum likelihood estimator of a which is $\hat{a}= \bar{y}/4$

I have also shon that this is unbiased estimator of a.

But I cannot show whether this MLE estimator is consistent.

Please help me to do that.

I know that if $lim(|\hat{a}-a|>c)=0$ as n goes to infinity, then MLE is consistent estimator.

I do not know how I can show this limit.

Answer 1

Easiest is to use the Strong Law of Large Numbers to get the almost everywhere convergence:

$\hat{a}=\frac{\bar{y}}{4}\to \frac{E[Y]}{4}=\frac{4a}{4}=a$

And the consistency (convergence in probability) follows immediately.

You can also use the Week Law of Large Numbers with continuous mapping theorem, or even directly Chebyshev's inequality.

Edit: $Y$ follows a Gamma distribution with shape $k=4$ and scale $\theta=a$, so: $E[Y]=k\theta=4a$, $\operatorname{Var}(Y)=k\theta^2=4a^2$, and:

$$E[\hat{a}]=\frac{1}{4}E[\bar{y}]=\frac{1}{4}\frac{E[Y_1]+\ldots+E[Y_n]}{n}=\frac{1}{4}\frac{4an}{n}=a\text{ (unbiased)}$$

$$\operatorname{Var}(\hat{a})=\frac{1}{4^2}\operatorname{Var}(\bar{y})=\frac{1}{4^2}\frac{\operatorname{Var}(Y_1)+\ldots+\operatorname{Var}(Y_n)}{n^2}=\frac{1}{4^2}\frac{4a^2n}{n^2}=\frac{a^2}{4n}\xrightarrow{n\to\infty}0$$