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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.

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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 $Y_i$ and $Y_j$ 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(y_{i},y_{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(y_{i},y_{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*}

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    $\begingroup$ This is a very rigorous and detailed derivation, thank you!! Just what I wanted :) $\endgroup$ Jan 23, 2019 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. $\endgroup$ Jan 23, 2019 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! $\endgroup$ Jan 24, 2019 at 0:53
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    $\begingroup$ What is X for this question? $\endgroup$
    – Kuo
    Jul 21, 2023 at 11:36
  • $\begingroup$ @Kuo, I think I got sloppy with the notation and substituted $X_i$, and $X_j$ for $Y$. I'll correct this. Thanks for catching this. $\endgroup$ Jul 26, 2023 at 15:51

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