Bounding Poisson CDF by the Binomial: $\Pr[B(n,p)\le a] \ge \Pr[P(np)\le a]$ when $np\le a$

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:

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?

• did u see did's answer in the question you linked? Aug 24, 2021 at 23:54
• @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. Aug 25, 2021 at 0:03

[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?

• 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. Aug 27, 2021 at 14:13
• 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 $Y$. Aug 27, 2021 at 14:36
• That would be an amazing result if true. Aug 28, 2021 at 1:57
• Is this even true for $n=1$ and $p$? I think you need $a\geq \lceil np \rceil$. Aug 28, 2021 at 20:34
• Since $X$ is an integer, I'm assuming so is $a$. Aug 29, 2021 at 13:25