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, JMP, user91500, Behrouz MalekiJan 5 '17 at 7:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nick Peterson, JMP, user91500, Behrouz Maleki
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $$Var(\sum X) = Cov(BX)= B'Cov(X)B.$$