Calculating variance of $\overline{X_n}^2 - \frac{1}{n}$
Let $X_1,...,X_n \sim$ $N(\mu, 1)$ (independent) and $\overline{X_n} = \frac{1}{n}\sum_{i=1}^nX_i$
Calculate the variance of $\overline{X_n}^2 - \frac{1}{n}$
This seems that is should be not be difficult to calculate but for whatever reason I am having difficulty.
Var$\overline{X_n} =$ Var$( \bigg(\frac{1}{n}\sum_{i=1}^n X_i \bigg)^2 - \frac{1}{n}) =$ Var$(\frac{1}{n^2} \sum_{i=1}^n X_i^2 + 2\sum_{i \neq j} X_iX_j -\frac{1}{n}) $
$=$ $\frac{1}{n^4}\bigg [$Var$(\sum_{i=1}^n X_i^2) + $ $4$Var$(\sum_{i\neq j} X_iX_j)\bigg]$ From here I find I am either miscalculating or this is itself incorrect.
The answer I am supposed to have is $\displaystyle \frac{4\mu^2}{n}+\frac{2}{n}$
I will drop the bar notation and simply write $X = \sum X_i$ and also do the centering $X_i\leftarrow X_i-\mu$ and now $X_i$ are standard normals. $$\textrm{Var}(X^2) = E(X^4) - E(X^2)^2,$$ so it suffices to calculate each terms separately. First, $$E(X^2) = E\left(\sum_{i,j=1}^nX_iX_j\right) = nE(X_1^2) + \sum_{i\neq j}E(X_i)E(X_j) = n.$$ Similarly, $$E(X^4) = E\left(\sum_{i,j,k,s=1}^nX_iX_jX_kX_s\right) = \sum_{i,j,k,s=1}^nE(X_iX_jX_kX_s).$$ Now, when $i = j = k = s$ we get $nE(X_1)^4 = 3n.$ If one of the indices, say $i$, is different from the others, then that summand is zero because you can pull out the expectation of that using independence. Therefore, the only remaining non-vanishing terms are from configuration of indices such that: $$i = j, k = s, i\neq j.$$ The number of such arrangement is $n\times(n-1)\times 3 = 3n(n-1).$ Thus, $$E(X^4) = 3n + 3n(n-1) = 3n^2.$$ After this, you can finish the calculation.