Why does the correlation between decision trees have to be positive?

Question

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 $\rho\geq 0$?

Answer 1

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 $-(1-\rho)/\rho$.