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In the book "Elements of statistical learning" we have that the variance of a the random forest is given by

$V(\frac{1}{n} \sum X_i)= \rho \sigma^2+ \frac{1-\rho}{n}\sigma^2$

where $\rho$ is the pairwise correlation between all $n$ trees which all have variance $\sigma^2$. It is stated that $\rho\geq 0$ otherwise the proof fails. I know that $\rho\geq -1/(n-1)$ but that does not imply that $\rho>0$? Say we have $n=3$ we have that $\rho$ can take the values $-0.5$ without violating anything. How come the proof is only for $\rho\geq 0$?

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  • $\begingroup$ Try substituting any negative number (well, between $-1$ and $0$) for $\rho$, and see what happens to the sign of the variance as $n \to \infty$. $\endgroup$
    – jbowman
    Apr 14, 2020 at 15:19
  • $\begingroup$ In the limit I do see the problem, but In general that is not a problem. That is where I see the struggle $\endgroup$
    – CutePoison
    Apr 14, 2020 at 15:53
  • $\begingroup$ It's not in the limit. Regardless of what $\rho < 0$ you choose, there is a finite $n$ for which the variance is negative, and remains negative for all greater $n$. This means the proof fails. It's not a proof if it only works for certain combinations of $\rho$ and $n$ and fails for all the rest. Note also that for any given $\rho$, the number of $n$s for which the proof fails is countably infinitely greater than the (finite) number for which it works. $\endgroup$
    – jbowman
    Apr 14, 2020 at 16:03
  • $\begingroup$ True that! Thanks $\endgroup$
    – CutePoison
    Apr 14, 2020 at 16:04
  • $\begingroup$ Feel free to add it as an answer $\endgroup$
    – CutePoison
    Apr 14, 2020 at 16:05

1 Answer 1

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The proof doesn't hold for $\rho < 0$, as for any $\rho < 0$, there exists an $n_0$ such that for all $n \geq n_0$, $\rho + \frac{1-\rho}{n} < 0$, and therefore the calculated variance $V(\frac{1}{n} \sum X_i)$ will be less than $0$ as well.

We can easily find $n_0$, as it is the smallest integer greater than or equal to $-(1-\rho)/\rho$.

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