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.


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*}

| cite | improve this answer | |
  • 1
    $\begingroup$ This is a very rigorous and detailed derivation, thank you!! Just what I wanted :) $\endgroup$ – Stats Pupil Jan 23 '19 at 1:49
  • $\begingroup$ @StatsPupil, the way to indicate that this answers fulfills your needs is by clicking the accept check mark button next to the answer itself. This incentive will encourage others to answer your questions in the future. Thank you. $\endgroup$ – StatsStudent Jan 23 '19 at 1:56
  • $\begingroup$ @StatsPupil. Absolutely. That's what some of us more experienced users are here for. We try to help out some of the newcomers! Best of luck to you! $\endgroup$ – StatsStudent Jan 24 '19 at 0:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.