Find the mean and variance of $\hat{θ}$ for a special case of Gamma Distribution. 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 $\dfrac{1}{(3n)^2}3n\theta^2=\frac{\theta^2}{3n}$.
