# Suppose $X_1\cdots X_n$ are correlated normal random variables… [closed]

Suppose $X_1\cdots X_n$ are correlated normal random variables... such that $E(X_i) = 0$, $Cov(X_i, X_j ) = ρσ^2$ for $i$ not equal to $j$, and $Var(X_i) = σ^2$.

In the case where $ρ = 0.2, σ^2 = 2$ and $n = 100$, find the covariance matrix of the multivariate Normal distribution for $X = (X_1, . . . , X_n)'$ and hence find the variance of $\sum_{i=1}^n X_i$.

All I know is that I apparently need to consider $Y = BX =\sum_{i=1}^n X_i$.

This is the only hint our lecturer gave us. But I am still unsure how to go about doing this.

Thank you

## closed as off-topic by Did, Nick Peterson, JonMark Perry, user91500, Behrouz MalekiJan 5 '17 at 7:12

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• The covariance matrix just has $\sigma^2$ on the diagonal and $\rho\sigma^2$ on the off-diagonals. It's unclear how to answer the question in RStudio without more information. Maybe just doing the matrix multiplications $B\Sigma B'$ you used to compute the variance from the correlation matrix? – spaceisdarkgreen Jan 5 '17 at 1:04
• I'm a little confused what you're asking cause you imply that you've found the variance – spaceisdarkgreen Jan 5 '17 at 1:05

The covariance matrix $Cov(X)$ has $\sigma^2$ on the diagonals and $\rho\sigma^2$ on the off diagonals.
You can express the sum $\sum_{i=1}^n X_i$ as $BX$ where $B$ is the column matrix of all ones. Then using the formula for covariance of $BX$ we get $$Var(\sum X) = Cov(BX)= B'Cov(X)B.$$