Show that the distribution maximizing $\Pr[\sum_{i = 1}^n {X_i} \geq \mathbb{E}[\sum_{i = 1}^n {X_i}] + 1]$, for $X_i$ being Bernoulli is the uniform.

Question

Suppose we have r.v.s $X_1, \dots X_n$ where each $X_i$ is a Bernoulli random variable with probability $p_i$, and the $p_i$'s are not necessarily equal. Let $Y = \sum_{i = 1}^n {X_i}$, and notice that $Y$ follows a Poisson binomial distribution, and that $ \mathbb{E}[Y] = \sum_{i = 1}^n {p_i}$. Let $m = \mathbb{E}[Y]$.

I am pretty sure that the distribution on the $p_i$'s which maximizes $\Pr[Y \geq m+1]$ is setting $p_i = \frac{m}{n}$ for all $i$, but I am not sure how to show this. I tried induction but is seems messy.

Is there a simple way to show this (or perhaps a counterexample)?

Answer 1

Turns out I found the answer. For anyone interested in the future, there is this wonderful paper by Hoeffding himself. The proof of the fact I wanted can be found in Theorem 4.