Let $X$ be a random variable with pdf
$$ f\left(x;\theta\right) = \left\{ \begin{array}{lr} \frac{3\theta^3}{(x+\theta)^4} & \text{if } 0<x<\infty \text{ and } 0 < \theta <\infty\\ 0 & \text{otherwise} \end{array} \right.\\$$
Show that $\hat\theta = 2\overline X$ is an unbiased estimator of $\theta$ and determine the efficiency of $\hat\theta$.
Where I Am:
I already showed that $\hat\theta$ is an unbiased estimator, but I'm having trouble with the second part. I know that the efficiency is given by:
$$ \text{eff}(\hat\theta) = \left(\frac{1}{nI(\theta)} \right)\left(\frac{1}{Var(\hat\theta)} \right)$$
where $I(\theta)$ is the Fisher information for $\theta$.
I was able to compute the Fisher information for $\theta$, but am having trouble with the variance of $\hat\theta$. I have the following:
$$ Var(2\overline X) = 4 Var(\overline X) = 4Var\left(\frac{1}{n}\sum_{i = 1}^nX_i\right) = \frac{4}{n^2}Var\left(\sum_{i = 1}^nX_i\right)$$
Now, here, I don't think I can pull the sum out because we don't know if these random variables are uncorrelated or not. Obviously, I can break it up into
$$ \sum_{i = 1}^nVar(X_i) = E\left[\left(\sum_{i = 1}^nX_i\right)^2\right] - \left(E\left[\sum_{i = 1}^nX_i\right]\right)^2 $$
and I can easily compute the following term:
$$ \left(E\left[\sum_{i = 1}^nX_i\right]\right)^2 = \frac{(n\theta)^2}{4}$$
but, I'm not sure how to deal with this guy: $$ E\left[\left(\sum_{i = 1}^nX_i\right)^2\right] .$$
Any tips?
$$ E\left[\left(\sum_{i = 1}^nX_i\right)^2\right]=nE\left[X_1^2\right]+n(n-1)E\left[X_1]E[X_2\right]\;. $$
The second term you already know, and you can calculate the expectation in the first term by integrating by parts (much like you probably did for $E[X_1]$):
$$ \int_0^\infty x^2\frac{3\theta^3}{(x+\theta)^4}\mathrm dx=\int_0^\infty2x\frac{\theta^3}{(x+\theta)^3}=\int_0^\infty\frac{\theta^3}{(x+\theta)^2}\mathrm dx=\theta^2\;. $$