Calculate the MSE of T = $\frac{1}{4}\overline{X}$ with respect to $$ where X~$Gamma(4,)$

Question

Suppose that $X_1,...,X_n$ is an iid sample that is distributed as $Gamma(4,)$ and suppose that $T = \frac{1}{4}\overline{X}$ is an estimator of $$. How do you calculate $MSE(T; )$?

I don't know how to start this problem because I know $MSE(T; )$ = $Var(T) - Bias(T; )^2$ but I don't know how to calculate $Var(T)$. I also calculated $Bias(T; )$ as follows:

$Bias(T; )$ = $E(T)$ - $$ = $E(\frac{1}{4}\overline{X})$ - $$ = .... = $\frac{1}{}$ - $$ = $\frac{1-}{}$ But I'm not sure if that calculation is even correct. Can someone confirm if it's correct as well?

Answer 1

First, when you say that the distribution is a $Gamma(4;\theta)$ it is not well determinated because Gamma distribution has different parametrizations.

Second when you write

Suppose that $X_1,...,X_n$ is an iid sample that is distributed as $Gamma(4,\theta)$

I hope you mean that the random sample is "from a population with distribution"... or "where every single observation is distributed as..."

Third,

$MSE=VAR + (BIAS)^2$ and not $VAR - (BIAS)^2$ as you wrote: this looks like the great is the Bias, the best is the Estimator...

Fourth,

if the problem is only in calculating

$$V(T)=V\Bigg[\frac{1}{4}\overline{X}_n\Bigg]$$

it is simply

$$V(T)=V\Bigg[\frac{1}{4}\frac{\Sigma_i X_i}{n}\Bigg]=\frac{1}{(4n)^2}\cdot n V(X_1)$$

But I'm not sure if that calculation is even correct

this depends which is the parametrization you used for the Gamma distribution. In the link you can find the two more common parametrizations

IMHO, if you use $T$ to estimate the parameter $\theta$ the parametrization that makes more sense is the one in the first column showed in the link, so that $\mathbb{E}[T]=\theta$ and you estimator is unbiased...but this is only my opinion. Only you can know the exact parametrization you have to use.