Consider two random variables $X \sim \operatorname{Binom}[(n,p)]$ and $Y\sim\operatorname{Poisson}[(\lambda)]$, where $\lambda = n p $. Let their respective CDFs be $F_X(k)$ and $F_Y(k)$, then I would like to show that

$$ F_Y(k) \leq F_X(k) $$

but I can't quite figure out how to. I'm not 100 % sure it's true but intuitively, it seems very much like that fatter tail of the Poisson distribution should make it true.

I thought about trying to show that $X' \sim \operatorname{Binom}[(n+1,p')]$, where $(n+1)p'=\lambda$, satisfies $F_{X'}(k) \leq F_X(k)$, but didn't get something out.

Any hints/answers appreciated!


  • 3
    $\begingroup$ In general, if you have two different distribution functions with the same mean then at some point the first cdf will be greater than the second and at another point the second cdf will be greater than the first. $\endgroup$
    – Henry
    May 4, 2011 at 21:41
  • $\begingroup$ I believe $F_Y(k) \le F_X(k)$ whenever $np \le k$. This certainly holds numerically. I'm currently trying to prove it. Let me know if you had any progress since asking the question. $\endgroup$ Aug 24, 2021 at 17:38

2 Answers 2


Assume that the distribution of $X$ is Binomial $(n,p)$ and that the distribution of $Y$ is Poisson $\lambda$. Then $F_Y(0)=\mathrm{e}^{-\lambda}$ and $F_X(0)=(1-p)^n$. Since $1-p<\mathrm{e}^{-p}$ for every $p\ne0$, $(1-p)^n<\mathrm{e}^{-np}$, hence, for $\lambda=np$, $F_X(0)<F_Y(0)$, which means the assertion is wrong.


Let us first consider the simple case $n=1$. That is, we assume that the distribution of $X$ is Binomial $(1,p)$ and that the distribution of $Y$ is Poisson $\lambda$, and we look for a value of $\lambda$ such that $F_Y(k)\leqslant F_X(k)$ for every $k$. We readily see that this inequality holds everywhere if and only if it holds at $k=0$ if and only if $\mathrm{e}^{-\lambda}\leqslant1-p$, that is for every $\lambda\geqslant\lambda_p$ with $$ \color{red}{\lambda_p=-\log(1-p)} $$ This result has an almost sure version: assume that $Y$ is Poisson $\lambda_p$ and define $$ X=\min\{Y,1\} $$ Then $X$ is Bernoulli $(1,p)$ and $X\leqslant Y$ almost surely. Hence, for every $k$, $[X\leqslant k]\supset[Y\leqslant k]$, which implies $$ F_X(k)=P(X\leqslant k)\geqslant P(Y\leqslant k)=F_Y(k) $$

We can now go back to the general case. Let us introduce $n$ i.i.d. $Y_k$ with Poisson $\lambda_p$ distributions and $n$ i.i.d. $X_k$ with Binomial $(1,p)$ distributions, such that, for every $k$, $X_k\leqslant Y_k$ almost surely (to this end, one can use the construction provided above or any other one). Then $Y=Y_1+\cdots+Y_n$ has Poisson $n\lambda_p$ distribution, $X=X_1+\cdots+X_n$ has Binomial $(n,p)$ distribution, and $X\leqslant Y$ almost surely.

This proves that for every $X$ with Binomial $(n,p)$ distribution and for every $Y$ with Poisson $n\lambda_p$ distribution, $F_X(k)\geqslant F_Y(k)$ for every $k$. Hence the result the OP is interested in holds if one replaces $\lambda=np$ by $$ \color{red}{n\lambda_p=-n\log(1-p)} $$ The comparison technique above is a simple case of the so-called coupling method. For more on this, see the book Lectures on the coupling method by Torgny Lindvall.

  • $\begingroup$ Thanks, that's a clever trick. $\endgroup$
    – okintheory
    Sep 29, 2011 at 5:11

If you can't think how to prove something, a good first step is often to try and disprove it.

In this case, see if you can find an n and p for which your inequality fails.

For example, if n=10 and p=0.5 the cumulative distributions are:

k   Binom           Poisson
1   0.009765625     0.033689735
2   0.0537109375    0.1179140725
3   0.1708984375    0.2582879683
4   0.3759765625    0.4337553381
5   0.6220703125    0.6092227078
6   0.8271484375    0.755445516
7   0.9443359375    0.8598903789
8   0.98828125      0.9251684183
9   0.998046875     0.9614339957
10  0.9990234375    0.9795667844

The Poisson CDF > Binomial CDF at low k, so your proof is impossible.

In fact, you can disprove this algebraically too, at k=0:



You can show that the Poisson distribution is always greater than the Binomial at k=0.

Hope that helps.

  • $\begingroup$ Thanks! I should have spotted the k=0 case immediately. $\endgroup$
    – okintheory
    May 4, 2011 at 20:36
  • $\begingroup$ Two cheers for your first sentence. $\endgroup$
    – user940
    May 4, 2011 at 21:45

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