Let $Y_1, Y_2,…,Y_n$ be a random sample from a Gamma(2, $\beta$) distribution with pdf
$$f(y) = \frac{y e^{-y/\beta}}{\beta^2} \;, \quad y>0$$
If the maximum likelihood estimator for $\beta$ is:
$$\hat{\beta} = \frac{1}{2n}\sum_\limits{i=1}^{n}Y_i$$
a) Show that $\hat{\beta}$ is an unbiased estimator for $\beta$.
b) Show that $\hat{\beta}$ is a consistent estimator for $\beta$.
Does anyone know how to do this?
I know the bias of an estimator is
$$\hat{\beta} = E[\hat{\beta}] – \beta$$
Adding this after help from @FoobazJohn and reading
Show consistent by satisfying the two limits below:
$$\lim_{n \to \infty}E[\hat{\beta}] = \beta \\
\lim_{n \to \infty}Var[\hat{\beta}] = 0 \\
E[\hat{\beta}] = \frac{1}{2n}E\bigg[\sum_\limits{i=1}^{n}Y_i\bigg] = \frac{n2\beta}{2n} = \beta \\
\lim_{n \to \infty}E[\hat{\beta}] = \lim_{n \to \infty}E[\beta] = \beta \\
Var{\hat{\beta}} = \frac{1}{4n^2}Var\bigg[\sum_\limits{i=1}^{n}Y_i\bigg] = \frac{n2\beta^2}{4n^2}=\frac{\beta^2}{2n} \\
\lim_{n \to \infty}Var[\hat{\beta}] = \lim_{n \to \infty}Var[\frac{\beta^2}{2n}] = 0$$
Satisfies consistency.
Best Answer
To show that the estimate is unbiased we have to show that $E\hat{\beta}=\beta$. Since the $Y_i$ are identically distributed and $EY_1=2\beta$, it follows that $E\hat{\beta}=(2n)^{-1}\times n\times 2\beta=\beta$ as desired. To show that it is a consistent estimator one can use the strong law of large numbers to deduce that $$ \hat{\beta}=\frac{1}{2}\times\bar{Y}_n\to\frac{1}{2}EY_1=\beta $$ a.s as $n\to \infty$ as desired.