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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.

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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 $\dfrac{1}{(3n)^2}3n\theta^2=\frac{\theta^2}{3n}$.

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  • $\begingroup$ Thanks. Do you mind showing me how you worked this out? I would really appreciate it. $\endgroup$ – Neels Feb 14 at 18:30
  • $\begingroup$ @Neels See edit. $\endgroup$ – J.G. Feb 14 at 18:39
  • $\begingroup$ Thanks. Might be a dumb question but for the mean where did the 1/3n come from? Also, do i not need to do the whole integral process like i did above? $\endgroup$ – Neels Feb 14 at 18:47
  • $\begingroup$ @Neels It comes from the $\frac{1}{3n}$ in front of $\sum_i x_i$. Means, if existent, are always additive. $\endgroup$ – J.G. Feb 14 at 19:00

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