Question:
Show that the average of $B$ i.d. (identically distributed, but not necessarily independent) random variables, each having a variance of $\sigma^2$, with positive pairwise correlation $\rho$ has a variance of $\rho \sigma^2+\frac{1-\rho}{B} \sigma^2$.
Answer:
The variance of the estimate of the mean is given by $\operatorname{Var}\left(\frac{1}{B} \sum x_i\right)=\frac{1}{B^2}\left(\mathrm{E}\left[\left(\sum x_i\right)^2\right]-\mathrm{E}\left[\sum x_i\right]^2\right)$. Notice that, this follows from the variance formula $\operatorname{Var}(x)=\mathrm{E}\left[x^2\right]-\mathrm{E}[x]^2$. Assume that the expected value of the identical random variables is denoted with $\mathrm{E}\left[x_i\right]=\bar{x}$ for all $i=1, \ldots, B$. Then, the second term $\mathrm{E}\left[\sum x_i\right]^2$ becomes $B^2 \bar{x}^2$. Recall that, $\operatorname{Var}\left(x_i\right)=\sigma^2$ for all $i$ and the correlation coefficient among each pair of random variables is $\rho$ which is calculated from $\mathrm{E}\left[x_i x_j\right]=\rho \sigma^2+\bar{x}^2$.
My question:
Why does this hold?
$\mathrm{E}\left[x_i x_j\right]=\rho \sigma^2+\bar{x}^2$