I'm trying to prove the following property:
$$ var \Big( \frac{1}{n}\sum^n_{i=1} Y_i \Big) = \frac{\sigma^2}{n} + \frac{n-1}{n}\rho\sigma^2 $$
So far, I've gotten up to this point:
$$ \textrm{var}\Big( \frac{1}{n}\sum^n_{i=1}Y_i \Big) = \frac{1}{n^2}\textrm{var}\Big( \sum^n_{i=1}Y_i \Big) \\ = \frac{1}{n^2}\Bigg( \sum^n_{i=1}\textrm{var}(Y_i) + 2\sum^{n-1}_{i=1}\sum^n_{j={i+1}}\textrm{cov}(Y_i, Y_j)\Bigg) \\ $$
where $\rho$ is the correlation coefficient between $Y_i$ and $Y_j$, since
$$ \textrm{var}\Big( \sum^n_{i=1}Y_i \Big) = \sum^n_{i=1}\sum^n_{j=1}\textrm{cov}(Y_i, Y_j) \\ = \sum^n_{i=1}\textrm{var}(Y_i) + \sum_{i \ne j}\textrm{cov}(Y_i, Y_j) \\ = \sum^n_{i=1}\textrm{var}(Y_i) + 2\sum^{n-1}_{i=1}\sum^n_{j={i+1}}\textrm{cov}(Y_i, Y_j) $$
I know that$\sum^n_{i=1}var(Y_i) = n\sigma^2$, but I cannot see how the second term becomes $\frac{n-1}{n}\rho\sigma^2$. Any hints would be greatly appreciated. Thank you!