# On the Poisson binomial distribution bounded by the binomial distribution at its center

The statement I am trying to prove is the following:

Given a Poisson binomial $$S$$ whose parameters $$p=(p_1, \dots, p_n)$$ are bounded by 0.5, and where $$n$$ is even. Proof that, $$P(S=n/2) \leq P(B=n/2),$$ where $$B$$ is the binomial distribution with parameters n and 0.5.

I think it should be true, but I could not prove it...

The pmf found on Wikipedia is hard to work with. I found that $$p_1 = p_2 = \dots = p_n = 0.5$$ is a saddle point of $$P(S=n/2)$$, but I don't know if there are other stationary points inside $$[0, 0.5]^n$$.

Alternatively, using this answer, it is possible to bound the cmf of $$S$$ using the cmf of $$B$$, i.e., $$P(S\geq k) \leq P(B\geq k)$$ for all $$k\in \{1,\dots,n\}$$. But I'm not able to transfer it to the pmf.

Let $$n=2p$$. Using the pmf of the Poisson-Binomial distribution we have $$\mathbb{P}(S=p)=\sum_{T\subset\{1,\cdots,n\}, |T|=p}\prod_{i\in T}p_i\prod_{i\in T^c}(1-p_i).$$ This expression is linear in each $$p_i$$. Therefore, if all values $$p_i$$ are fixed except for some $$p_j$$ then $$\mathbb{P}(S=p)$$ is either increasing or deacreasing wrt $$p_j$$. This shows that $$\mathbb{P}(S=p)$$ reaches its maximum value at some point $$(p_1,\cdots, p_n)$$ such that $$p_i=0$$ or $$1/2$$ for all $$i$$. Let $$E$$ be the set of $$i$$ such that $$p_i=0$$. We consider that if $$i\notin E$$ then $$p_i=1/2$$. In this case, $$\mathbb{P}(S=p)$$ can be rewritten the following way: \begin{align*} \mathbb{P}(S=p)&=\sum_{T\subset\{1,\cdots,n\}, |T|=p, T\cap E=\emptyset}\prod_{i\in T}p_i\prod_{i\in T^c}(1-p_i) + \sum_{T\subset\{1,\cdots,n\}, |T|=p, T\cap E\neq\emptyset}\prod_{i\in T}p_i\prod_{i\in T^c}(1-p_i).\\ &= \sum_{T\subset\{1,\cdots,n\}\backslash E, |T|=p}\frac{1}{2^{n-|E|}} + 0\\ &= \binom{n-|E|}{p}\frac{1}{2^{n-|E|}}. \end{align*} If $$|E|\geq 1$$ then using Pascal's triangle recursion formula we get $$\binom{n-|E|}{p} + \binom{n-|E|}{p-1} = \binom{n-(|E|-1)}{p}.$$ Now, we know that $$\binom{n}{k}$$ is increasing wrt $$k$$ when $$k\leq n/2$$ and decreasing when $$k\geq n/2$$. In our case, since $$p=n/2$$ then $$p\geq p-1 \geq \frac{n-|E|}{2}$$ hence $$\binom{n-|E|}{p-1}\geq \binom{n-|E|}{p}$$ and thus $$\binom{n-(|E|-1)}{p}\geq 2 \binom{n-|E|}{p}$$ and finally $$\binom{n-(|E|-1)}{p} \frac{1}{2^{n-(|E|-1)}} \geq \binom{n-|E|}{p}\frac{1}{2^{n-|E|}}.$$ $$\mathbb{P}(S=p)$$ is thus maximized when $$|E|=0$$ and thus $$\mathbb{P}(S=p)\leq \binom{n}{p}\frac{1}{2^n} = \mathbb{P}(B=p).$$