# $X-Y$, where $X$ and $Y$ are sums of Bernoulli random variables

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 $$\mathbb{P}(X-Y =2) \geq \mathbb{P}(X-Y=1)$$ always holds?

• Do you assume that all random variables $x_1,\ldots,x_n,y_1,\ldots,y_n$ are independent? – Dieter Kadelka Jun 29 at 17:54
• Yes, all the variables are independent. – Masood Jun 29 at 18:08
• 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. – Bjørn Kjos-Hanssen 2 days ago
• Do you have investigated the limiting case that $X \sim Po(\lambda)$ and $Y \sim Po(\lambda-c)$ with $c > 5$ and $\lambda > 5$ and $X,Y$ independent? What if you replace $c$ by some smaller value? – Dieter Kadelka yesterday
• 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 log-concave (arxiv.org/pdf/0912.0581.pdf) and 2- when one distribution is unimodal while the other one is log-concave, 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? – Masood 5 hours ago