Consider a random variable $X$ with density $f$, and a finite set of convex and increasing functions $Y_i(x)$. I am interested in bounding the variance of the max of the $Y_i$'s.
I have the following conjecture: $$ \text{Var}\left(\max_i Y_i(X)\right)\leq \sum_i p_i \text{Var}\left(Y_i(X)\right) $$ where $p_i$ is the natural distribution over $i$ created by the max operator, i.e. $p_i$ is the probability that function $i$ is selected by the max operator when integrating over $X$ $$ p_i=\int_x\mathbb{1}\left(Y_i(x)\geq Y_j(x)\text{ for all }j\right)f(x)dx $$ (if multiple $Y_i$'s yield the maximum value for some $x$ split the mass $f(x)$ equally among them).
Any idea on how to prove/disprove this result? Any other bound that is known on the variance of the max? Please feel free to impose additional assumptions if they can help.
This inequality is false and here is a counterexample:
Let $X$ be uniform on $[0,1]$, $Y_1(x) = \delta$ for all $x$, $Y_2(x) = 0$ for $0 \leq x \leq 1 - \epsilon$, and a linear function afterwards such that $Y_2(1) = M$. (Think of $\epsilon, \delta$ as small and $M$ large). You can check that $Z = \max Y_1, Y_2$ is very similar to $Y_2$, so it has a very similar variance. However, on one side of this inequality you have the variance of $Z$, and on the other, you have $\epsilon$ Var($Y_2$). (I would do the computations explicitly, but your requirement that these are convex made me put this piecewise linear function, so I am just giving you a picture instead)
Here is a general bound (it is your bound without the $p_i$s), and why it is tight: https://stats.stackexchange.com/questions/15970/what-is-the-variance-of-the-maximum-of-a-sample
EDIT: Edited $Y_1$ to $Y_2$ as per the comment.