Calculating variance of $\overline{X_n}^2 - \frac{1}{n}$

Question

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 $\displaystyle \frac{4\mu^2}{n}+\frac{2}{n}$

You cannot split the variance into $\text{Var}(\sum_i X_i^2)$ and $\text{Var}(\sum_{i \ne j} X_i X_j)$ because $\sum_i X_i^2$ is not independent of $\sum_{i \ne j} X_i X_j$. — angryavian, Aug 11 '20 at 03:51
Using the distribution of $\overline X_n$ yields the answer quickly, since the fourth order central moment is well-known for a normal distribution: $E[\overline X_n-\mu]^4=3\sigma^4$ with $\sigma^2=1/n$. Also see https://math.stackexchange.com/q/1945448/321264. — StubbornAtom, Aug 11 '20 at 07:06

dezdichado · Accepted Answer · 2020-08-11T04:27:52.773

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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.

edited Aug 11 '20 at 04:27

answered Aug 11 '20 at 03:36

dezdichado

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I think you and OP made the mistake of assuming the variance of the sum can be decomposed as the sum of variances. – angryavian Aug 11 '20 at 03:38
And the fourth non-central moment should be $\mu^4 + 6 \mu^2 + 3$ – angryavian Aug 11 '20 at 03:40
@angryavian yeah I assumed the independence so the variance of sum is sum of variance. Also, yeah I will correct the moment. – dezdichado Aug 11 '20 at 03:47
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No, I mean that $\text{Var}(\sum_{i\ne j} X_i X_j) \ne \sum_{i \ne j} \text{Var}(X_i X_j)$. Even if the $X_i$ are independent, it does not mean that the $X_i X_j$ are independent. – angryavian Aug 11 '20 at 03:48
ah, I see. that is rather subtle. – dezdichado Aug 11 '20 at 03:53
Your updated answer handles the case when $\mu=0$, and yields $(3n^2 - n^2)/n^4 = 2/n^2$ which agrees with my answer (when $\mu=0$). – angryavian Aug 11 '20 at 04:40
You didn do centering, so when working with the fourth power, no term will vanish. I centered and there is some work left for the OP to do. – dezdichado Aug 11 '20 at 04:46
Oh that is right, thank you for catching my mistake. – angryavian Aug 11 '20 at 04:57

Calculating variance of $\overline{X_n}^2 - \frac{1}{n}$

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