Variance of sum of dependent random variables Can you guys help me prove the following:
$$
\operatorname{Var}\left[\frac{1}{m}\sum_{i=1}^my_i\right]=\frac{1}{m}(1-\rho)\sigma^2+\rho\sigma^2
$$
where the sampled predictions ($y_is$) have variance $\sigma^2$ and correlation $\rho$. 
Note: $y_is$ are NOT independent. We have a dependent sample.
 A: It's quite easy to prove this once you understand the relationship between the covariance and correlation and if you recognize that the variances for both $X_i$ and $X_j$ are identically $\sigma^2$:
\begin{eqnarray*}
V\left[\frac{1}{m}\sum_{i=1}^{m}y_{i}\right] & = & \frac{1}{m^{2}}\left[\sum_{i=1}^{m}V(y_{i})+\sum_{i=1}^{m}\sum_{i\ne j}^{m}Cov(X_{i},X_{j})\right]\\
 & = & \frac{1}{m^{2}}\left[\sum_{i=1}^{m}\sigma^{2}+\sigma^{2}\sum_{i=1}^{m}\sum_{i\ne j}^{m}\frac{Cov(X_{i},X_{j})}{\sigma^{2}}\right]\\
 & = & \frac{1}{m^{2}}\left[m\sigma^{2}+\sigma^{2}\sum_{i=1}^{m}\sum_{i\ne j}^{m}\rho\right]\\
 & = & \frac{1}{m^{2}}\left[m\sigma^{2}+\sigma^{2}(m^{2}-m)\rho\right]\\
 & = & \frac{\sigma^{2}}{m}+\frac{\sigma^{2}(m-1)\rho}{m}\\
 & = & \frac{\sigma^{2}}{m}+\frac{\sigma^{2}\rho m}{m}-\frac{\sigma^{2}\rho}{m}\\
 & = & \frac{\sigma^{2}-\sigma^{2}\rho}{m}+\rho\sigma^{2}\\
 & = & \frac{\left(1-\rho\right)\sigma^{2}}{m}+\rho\sigma^{2}\\
 & = & \frac{1}{m}\left(1-\rho\right)\sigma^{2}+\rho\sigma^{2}\,\,\,\,\,\,\,\,\,\blacksquare
\end{eqnarray*}
