TITLE: Dominance Poisson/Binomial in convex order
QUESTION [6 upvotes]: Given two r.v. $X,Y$ (with well-defined expectations) such that $\mathbb{E}[X]=\mathbb{E}[Y]$, we say that $X$ is dominated by $Y$ in the convex order, denoted $X \leq_{\rm cx} Y$, if
$$
\mathbb{E}[f(X)] \leq \mathbb{E}[f(Y)]
$$
for every convex function $f\colon\mathbb{R}\to\mathbb{R}$, provided the expectations exist.
Given $n\geq 1$, $p\in[0,1]$, let $\lambda := np$, and $X,Y$ be distributed as $\textrm{Binomial}(n,p)$ and $\textrm{Poisson}(\lambda)$, respectively.

Is it true that $X \leq_{\rm cx} Y$? If so, what would be a good reference or proof for this fact?

This seems to follow from Theorem 2.2 of [1], but I may be misreading -- and, in any case, this result is much more general, it would fill like overkill to use it.
[1] Negative dependence and stochastic orderings, Fraser Daly (2015). https://arxiv.org/abs/1504.06493

REPLY [7 votes]: This follows from the fact that ${\rm Poi}(p)$ dominates ${\rm Bin}(1,p)$, and adding $n$ independent copies of these.
To see that $Y\sim{\rm Poi}(p)$ dominates $X\sim{\rm Bin}(1,p)$, note that
$$
\mathbb P(X=0)=1-p\le e^{-p}=\mathbb P(Y=0).
$$
Hence, we can couple these so that $\{X=0\}\subseteq\{Y=0\}$ is an event of probability $1-p$.
Then, $Y=0$ whenever $X=0$ so we obtain,
$$
\mathbb E[Y\vert X]=X.
$$
This holds when $X=0$ since both sides equal $0$, and hence also holds for $X=1$ since both sides have expectation $p$.
This shows that $Y$ dominates $X$ since $f(X)\le\mathbb E[f(Y)\vert X]$ for convex $f$ by Jensen's inequality.