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Given the RVs $X_1, X_2, \ldots, X_N$, with each $X_i\sim\mathcal{N}(0,1)$ and $\operatorname{Cov}(X_i,X_j) = 1$.

I want to compute the variance of the sample mean $\overline{X}$ using the general definition of the variance of the sum of RVs:

\begin{align} \operatorname{Var}(\overline{X}) &= \operatorname{Var}\left(\frac1N \sum_{i=1}^{N} X_i\right) \\&= \frac1N \left( \sum_{i=1}^{N} \operatorname{Var}(X_i) + \sum_{i\neq j} \operatorname{Cov}(X_i,X_j)\right) \\&= \frac1N \left(N + \sum_{i\neq j} \operatorname{Cov}(X_i,X_j)\right) \end{align}

Is my approach correct? What is the value of the sum of the covariances?

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The factor outside is $\frac 1 { N^{2}}$ and not $\frac 1 N$. The last sum is nothing but $N(N-1)$, the number of choices for $(i,j)$ with $ i \neq j$.

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  • $\begingroup$ Did you mean that the sum evaluates to N(N-1) ? $\endgroup$
    – Morcus
    Mar 19, 2020 at 7:35
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    $\begingroup$ The answer is $\frac 1 {N^{2}} (N+(N^{2}-N))=1$. @Morcus $\endgroup$ Mar 19, 2020 at 7:38

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