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!


You must be assuming $var(Y_i) = \sigma^2$ for all i, and $corr(Y_i,Y_j) = \rho$ for all $i \ne j$.

The expansion of $var(1/n*\Sigma_{i=1}^nY_i)$ has $n^2$ terms, of which $n$ are the variance (i = j) terms, and $n^2 - n$ are the $i \ne j$ terms. So your 2nd term becomes $1/n^2*(n^2 -n)*cov(Y_i,Y_j)$

  • $\begingroup$ Hi, Mark. Yes, that's another assumption made. This makes perfect sense. Thank you! $\endgroup$ – user122514 Apr 21 '18 at 21:01

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