This question is related to Poisson CDF as lower bound to binomial CDF . That is, I seek to prove the inequality

$$\Pr[X\le a] \ge \Pr[Y\le a],$$ where $X\sim \text{Binomial}(n,p)$ and $Y\sim \text{Poisson}(np)$.

The difference is that I add the condition $np\le a$, which from numerical experiments seems to be exactly what is needed, as seen in the plot below:

enter image description here

For instance, this implies the inequality $$\Pr[X\le \lceil np\rceil] \ge \Pr[Y\le \lceil np\rceil],$$ which is likewise easy to test numerically for reasonable ranges of $n$ and $p$.

The case $a=0$ is trivial, since in this instance we must have $p=0$ or $n=0$ and both sides are 1.

The $a=1$ case is more tricky, but I can prove $$(1-p)^n + np(1-p)^{n-1} \ge e^{-np} + np e^{-np}$$ when $np\le 1$ using analytical methods.

The general case seems untractable to me, however. I wonder if this coupling is well known? Perhaps there is a probabilistic argument? Maybe something using characteristic functions?

  • $\begingroup$ did u see did's answer in the question you linked? $\endgroup$ Aug 24, 2021 at 23:54
  • $\begingroup$ @mathworker21 Yes, Did's result is nice, but he increases the mean of the Poisson distribution from $np$ to $n\log\frac{1}{1-p}$, which moves probability mass out of the tail. The bound is thus weaker than the one I suggest/need. Perhaps the proof transfers in some way - that is using coupling - but I don't see how. $\endgroup$ Aug 25, 2021 at 0:03

1 Answer 1


[Not an answer -- sharing some thoughts that didn't fit in a comment. Please let me know if this is the wrong way of doing it.]

Have you thought about the problem as sum of iid RVs? Let $X_1, \ldots, X_n$ be iid RVs where $X_i \sim \text{Binomial}(1,p)$. Also, let $Y_1, \ldots, Y_n$ be iid RVs where $Y_i \sim \text{Poisson}(p)$. Then, let $X^{(n)} = \sum_{i \in [n]} X_i$ and $Y^{(n)} = \sum_{i \in [n]} Y_i$. In this case $X^{(n)}$ and $Y^{(n)}$ are identically distributed as $X$ and $Y$ in the original problem.

A few observations:

(a) $n=1$ is easy, i.e., $P[X^{(1)} \leq a] \geq P[Y^{(1)} \leq a]$ for any $a \geq p.$ Perhaps some sort of induction on $n$ could be done?

(b) Let $M_{X_i}(t) = (1-p) + pe^t$ and $M_{Y_i}(t) = e^{p(e^t-1)}$ be the MGFs of $X_i$ and $Y_i$ respectively. Then, for all $t,$ $M_{X_i}(t) \leq M_{Y_i}(t)$ (direct consequence of $e^x \geq 1+x$). This also implies that $M_{X^{(n)}}(t) \leq M_{Y^{(n)}}(t)$ for all $n$ and $t$. Does this help to prove the claim?

  • $\begingroup$ The sum approach is what @Did uses in the coupling answer I link to. I think it can be a good approach. I'm sceptical about (b) since the mgf inequality holds independent of $np\le a$, which we know is needed to prove the result. $\endgroup$ Aug 27, 2021 at 14:13
  • 1
    $\begingroup$ What I had in mind is some extension of the following: If we have a RV $X$ with MGF $M_X,$ and mean $\mu,$ and $Y = \mu,$ then $M_X(t) \geq M_Y(t) = e^{\mu t},$ and also $P[X\leq a] \leq P[Y \leq a]$ for all $a > \mu.$ I wonder if (some variation of) this holds in general for $X$ and $Y$. $\endgroup$ Aug 27, 2021 at 14:36
  • $\begingroup$ That would be an amazing result if true. $\endgroup$ Aug 28, 2021 at 1:57
  • $\begingroup$ Is this even true for $n=1$ and $p$? I think you need $a\geq \lceil np \rceil$. $\endgroup$ Aug 28, 2021 at 20:34
  • $\begingroup$ Since $X$ is an integer, I'm assuming so is $a$. $\endgroup$ Aug 29, 2021 at 13:25

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