TITLE: Stochastic dominance between (products of) binomials
QUESTION [8 upvotes]: Suppose $p \leq q \leq 1/2$, and $n,m\geq 1$ two integers. Let $X\sim \mathrm{Bin}(n,p)$, $Y\sim \mathrm{Bin}(m,p)$ and $X'\sim \mathrm{Bin}(n,q)$, $Y'\sim \mathrm{Bin}(m,q)$ be independent.

Is it true that
$$
\forall t\in \mathbb{R}, \Pr[ XY + (n-X)(m-Y) > t] \geq \Pr[ X'Y' + (n-X')(m-Y') > t]
$$ ?

Note/Update: I think the $>t$ above should both be $\geq t$, for the usual definition of stochastic dominance (shouldn't make a difference...)
Put differently:

Does the r.v. $Z = XY + (n-X)(m-Y)$ stochastically dominate the r.v. $Z'=X'Y' + (n-X')(m-Y')$?

It seems so to me, as $X,Y$ are "less balanced" than their counterparts $X',Y'$, but I am not sure how to formally show it.$^{(\dagger)}$ I tried to come up with a coupling of $Z,Z'$ such that $Z\geq Z'$ w.p. one, but failed. More generally, I don't know of any  way to show stochastic dominance, besides (1) coming up with a coupling, or (2) finding or comparing the CDFs explicitly.
${(\dagger)}$ Based on some experimental plots, this looks like it does hold (e.g., see below).

REPLY [2 votes]: The conjecture is false, as shown by mschauer. (At least if we allow $m\neq n$; the case $m=n$ is still unclear to me.) Below is the simplest counterexample I could find in hindsight.

Following mschauer's answer, here is the smallest counterexample I could find: $n=1,m=4$.
For $n=1$, we have
$$
\Pr[Z=0] = p(1-p)^m + p^m(1-p)
$$
and, for $m\geq 4$, this is not a monotone function of $p$ on $[0,1/2]$. So the stochastic dominance, which would require $\Pr_p[Z\geq t]$ to be a non-increasing function of $p$ for every fixed $t$, cannot hold.

In particular, one can explicitly compute the cdf as a function of p. I am plotting it below to show that the CDFs are not monotone as a function of $p$ (this should be a discrete plot, piecewise linear; I am joining the points only for visualization's sake, to show that the relative order between the CDFs changes between 0 and 1).