Let $P=${$p_1, ... , p_n$} with $0<p_i<1$ for all $i$, and let $f:P \rightarrow P$ be a bijective function, what is the $\arg max_f \sum_i p_i \frac{f(p_i)}{1-f(p_i)}$?
Using brute force I can show that for n = 2 and n = 3 that the solution is the identity mapping, i.e. $f(p_i) = p_i$ for all $i$. I suspect this is true in general but I'm stuck.
Any help greatly appreciated!
You can prove the given statement using the Rearrangement Inequality. Assume WLOG that $p_i \leqslant p_j$ when $i < j$ by arranging in ascending order. Let $x_i = p_i$ and $y_i = \frac{p_i}{1-p_i}$. Observe that $y_i \leqslant y_j$ when $i < j$ since the function $f(x) = \frac{x}{1-x} = \frac{1}{1-x}-1$ is increasing in $[0, 1)$. Now, you want to find the permutation $\sigma$ for which $\sum x_iy_{\sigma(i)}$ is maximized, and by the Rearrangement Inequality, this occurs precisely when $\sigma$ is the identity permutation.
If you are unfamiliar with the Rearrangement Inequality, here's a reference. You can try to prove this fact yourself!