Poisson CDF as lower bound to binomial CDF

Question

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!

Thanks

Answer 1

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.

Nevertheless...

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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 $$ F_X(k)=P(X\leqslant k)\geqslant P(Y\leqslant k)=F_Y(k) $$

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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.

Answer 2

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:

$Binomial(k=0)=(1-p)^n$

$Poisson(k=0)=exp(-np)$

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

Hope that helps.