# Variance of $\overline{X}_n^2$

Here is a problem a have got in my homework.

Given a set of $X_1, ... X_n \sim F$ i.i.d values find the variance of $T_n = \overline{X}_n^2$ where $\overline{X}_n = \frac{\sum_i{X_i}}{n}$.

I actualy have an answer for this problem: $V(T_n) = \frac{4 \mu^2 \alpha_2}{n} + \frac{4 \mu \alpha_3}{n^2} + \frac{\alpha_4}{n^3}$ where $\mu = E(X_1)$ and $\alpha_k = \int \left| x - \mu \right|^k dF(x)$. But I can't figure out how it was obtained.

Here is what I got so far: $V(\overline{X}_n^2) = E\left[\overline{X}_n^4\right] - \left[E(\overline{X}_n^2)\right]^2 = E\left[\overline{X}_n^4\right] - \left[V(\overline{X}_n) + \left(E(\overline{X}_n)\right)^2\right]^2$ where $E(\overline{X}_n) = \mu$, $V(\overline{X}_n) = \frac{\alpha_2}{n}$ so $E(\overline{X}_n^2) = \mu^2 + \frac{\alpha_2}{n}$ the question is what $E(\overline{X}_n^4)$ equals to.

If you have any thought about that problem, hint or solution, please, tell me.

-

Hint: Write $$E[\overline{X}_n^4]={1\over n^4}\sum E[X_i X_j X_k X_\ell]$$ where all four indices run from $1$ to $n$. The expectations $E[X_i X_j X_k X_\ell]$ differ depending on whether $i,j,k,\ell$ are all distinct, one pair, two pairs, three of a kind, or four of a kind.
And please get rid of the absolute value sign in the definition of $\alpha_3$. – Did Mar 22 '11 at 18:50