Variance of squared sum of i.i.d random variables $X_1, X_2, ..., X_n$ are i.i.d random variables with zero mean and unit variance. Let $X = \sum_{i=1}^n X_i$, I want to express $\mathrm{Var}[X^2]$ in terms of $n$, $\mathrm{E}[X_1^4]$, and constants.
I'm able to make the following progress:
$$\mathrm{Var}[X^2] = \mathrm{E}[X^4] - \mathrm{E}[X^2]^2$$
$$ = \mathrm{E}[X^4] - (\mathrm{Var}[X] + \mathrm{E}[X]^2)^2$$
$$ = \mathrm{E}[X^4] - (n\mathrm{Var}[X_1] + 0)^2$$
$$ = \mathrm{E}[X^4] - n^2$$
But then I'm lost on how to approach the $\mathrm{E}[X^4]$ term. I know it can be expanded into $\sum_i \sum_j \sum_k \sum_l \mathrm{E}[X_i X_j X_k X_l]$, but I don't know how it can be further simplified.
 A: For the fourth moment of the sum $X$, we have
\begin{align*}
\mathbb{E}[X^4]
&= \sum_{i,j,k,l} \mathbb{E}[X_i X_j X_k X_l].
\end{align*}
What complicates the computation is the overlapping of indices. But one crucial observation is that, if one index is different from all the others then the term vanishes. For instance, if $i \notin \{j, k, l\}$ then $X_i$ is independent of $X_j X_k X_l$ and thus
$$ \mathbb{E}[X_i X_j X_k X_l]
= \mathbb{E}[X_i] \mathbb{E}[X_j X_k X_l] = 0. $$
So removing all such terms from the summation, any surviving term must satisfy that for each index $i$ there is another index which has the same value as $i$. This leaves only two possible cases:


*

*Case 1. We can have $i = j = k = l$. In this case, we have $\mathbb{E}[X_i X_j X_k X_l] = \mathbb{E}[X_1^4]$.

*Case 2. We can pair indices such that indices in different pairs have different values. So we have $\mathbb{E}[X_i X_j X_k X_l] = \mathbb{E}[X_1^2]^2$.
So it follows that
$$ \mathbb{E}[X^4] = n \mathbb{E}[X_1^4] + 3n(n-1)\mathbb{E}[X_1^2]^2. $$
You can simply further by plugging $\mathbb{E}[X_1^2] = 1$, which follows from the assumption.
A: You have reached the solution, as the final result must be in terms of $n$ and $E[X_1^4]=E[X^4]$.
