I've completed most of the question, however i'm not sure if i'm correct so far in order to proceed with the rest of it.
Find the mean and variance of $\hat{θ}$ for a special case of Gamma Distribution. Assume the standard situation, that is, let $X_1, . . . , X_n$ be independent and identically distributed with $X_k ∼ P_θ$, where $P_θ(x) = 1/{2θ^3}.x^2.e^{-x/θ}$ , where $0<x<∞$
We are given that the mean that mean and variance of this distribution in terms of $θ$ are $ E(X) = 3θ$ and $V(X) = 3θ^2$ respectively.
For the first part of the question i demonstrated that the maximum likelihood estimator $\hat{θ}$ for $θ$ is
$\hat{θ}$ = $1/(3n) $$\sum_{i} x_i$.
Now im asked to find the mean and variance of $\hat{θ}$ . This is what i've done: $E(\bar{X}) =\int dx_1....dx_n. P_θ(x_1)....P_θ(x_n)(1/n.\sum_{k} x_k)$ $=1/n.\sum_{k} (\int dx_k. x_k.P_θ(x_k))(\int dx. P_θ(x))^{n-1}$ and $(\int dx. P_θ(x))^{n-1}=1 $ so we get $=1/n.\sum_{k} x_k(\int dx_k. x_k.1/(2θ^3). x_k^2.exp(-x_k/θ)$ $=1/n.n.3θ = 3θ$ since $E(X_k)= 3θ$ and therefore is an unbiased estimator.
For the variance part: I know that
$E(Var\bar(X))= E(\bar{X^2})- E(\bar{X})^2$. But im not sure how to calculate $E(\bar{X^2})$. Is it just the same as above by instead of having $x_k$ we have $x_k^2$?
Would really appreciate some guidance.
The mean is $\frac{1}{3n}\sum_i 3\theta=\theta$. The $x_i$ are independent, so $\sum_i x_i$ has variance $\sum_i 3\theta^2=3n\theta^2$. Hence $\hat{\theta}$ has mean $\theta$, variance $\dfrac{1}{(3n)^2}3n\theta^2=\frac{\theta^2}{3n}$.