Let $X = x_1 + x_2 + \ldots + x_n$ and $Y = y_1 + y_2 + \ldots + y_n$, where each $x_i$ is an independent Bernoulli random variable with success probability $p_i$ and each $y_i$ is a Bernoulli random variable with success probability $q_i$. In addition, suppose that $\sum_i p_i = c +\sum_j q_j$ for some constant $c>5$. My question is, whether or not the inequality $$ \mathbb{P}(XY =2) \geq \mathbb{P}(XY=1) $$ always holds?

1$\begingroup$ Do you assume that all random variables $x_1,\ldots,x_n,y_1,\ldots,y_n$ are independent? $\endgroup$ – Dieter Kadelka Jun 29 at 17:54

$\begingroup$ Yes, all the variables are independent. $\endgroup$ – Masood Jun 29 at 18:08

1$\begingroup$ I guess $X$ and $Y$ are what's known as "Poisson binomial" random variables, and the question is basically whether a difference of two independent Poisson binomials is unimodal. $\endgroup$ – Bjørn KjosHanssen 2 days ago

1$\begingroup$ Do you have investigated the limiting case that $X \sim Po(\lambda)$ and $Y \sim Po(\lambdac)$ with $c > 5$ and $\lambda > 5$ and $X,Y$ independent? What if you replace $c$ by some smaller value? $\endgroup$ – Dieter Kadelka yesterday

$\begingroup$ It seems that it is valid for the limiting cases. While I was searching for the correct answer, I found the following series of results: 1 both $X$ and $Y$ are unimodal and logconcave (arxiv.org/pdf/0912.0581.pdf) and 2 when one distribution is unimodal while the other one is logconcave, their difference is also unimodal (statement in the proof of proposition 7 in paper "Precision May Harm: The Comparative Statics of Imprecise Judgement."). Is it correct? $\endgroup$ – Masood 5 hours ago