Poisson limit theorem

"Poisson theorem" redirects here. For the "Poisson's theorem" in Hamiltonian mechanics, see Poisson bracket § Constants of motion.

Comparison of the Poisson distribution (black lines) and the binomial distribution with n = 10 (red circles), n = 20 (blue circles), n = 1000 (green circles). All distributions have a mean of 5. The horizontal axis shows the number of events k. As n gets larger, the Poisson distribution becomes an increasingly better approximation for the binomial distribution with the same mean.

In probability theory, the law of rare events or Poisson limit theorem states that the Poisson distribution may be used as an approximation to the binomial distribution, under certain conditions. The theorem was named after Siméon Denis Poisson (1781–1840). A generalization of this theorem is Le Cam's theorem.

For broader coverage of this topic, see Poisson distribution § Law of rare events.

Theorem

Let $p_{n}$ be a sequence of real numbers in $[0,1]$ such that the sequence $np_{n}$ converges to a finite limit $\lambda$. Then:

$\lim _{n\to \infty }{n \choose k}p_{n}^{k}(1-p_{n})^{n-k}=e^{-\lambda }{\frac {\lambda ^{k}}{k!}}$

First proof

Assume $\lambda >0$ (the case $\lambda =0$ is easier). Then

${\begin{aligned}\lim \limits _{n\rightarrow \infty }{n \choose k}p_{n}^{k}(1-p_{n})^{n-k}&=\lim _{n\to \infty }{\frac {n(n-1)(n-2)\dots (n-k+1)}{k!}}\left({\frac {\lambda }{n}}(1+o(1))\right)^{k}\left(1-{\frac {\lambda }{n}}(1+o(1))\right)^{n-k}\\&=\lim _{n\to \infty }{\frac {n^{k}+O\left(n^{k-1}\right)}{k!}}{\frac {\lambda ^{k}}{n^{k}}}\left(1-{\frac {\lambda }{n}}(1+o(1))\right)^{n}\left(1-{\frac {\lambda }{n}}(1+o(1))\right)^{-k}\\&=\lim _{n\to \infty }{\frac {\lambda ^{k}}{k!}}\left(1-{\frac {\lambda }{n}}(1+o(1))\right)^{n}.\end{aligned}}$

Since

$\lim _{n\to \infty }\left(1-{\frac {\lambda }{n}}(1+o(1))\right)^{n}=e^{-\lambda }$

this leaves

${n \choose k}p^{k}(1-p)^{n-k}\simeq {\frac {\lambda ^{k}e^{-\lambda }}{k!}}.$

Alternative proof

Using Stirling's approximation, it can be written:

${\begin{aligned}{n \choose k}p^{k}(1-p)^{n-k}&={\frac {n!}{(n-k)!k!}}p^{k}(1-p)^{n-k}\\&\simeq {\frac {{\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}{{\sqrt {2\pi \left(n-k\right)}}\left({\frac {n-k}{e}}\right)^{n-k}k!}}p^{k}(1-p)^{n-k}\\&={\sqrt {\frac {n}{n-k}}}{\frac {n^{n}e^{-k}}{\left(n-k\right)^{n-k}k!}}p^{k}(1-p)^{n-k}.\end{aligned}}$

Letting $n\to \infty$ and $np=\lambda$:

${\begin{aligned}{n \choose k}p^{k}(1-p)^{n-k}&\simeq {\frac {n^{n}\,p^{k}(1-p)^{n-k}e^{-k}}{\left(n-k\right)^{n-k}k!}}\\&={\frac {n^{n}\left({\frac {\lambda }{n}}\right)^{k}\left(1-{\frac {\lambda }{n}}\right)^{n-k}e^{-k}}{n^{n-k}\left(1-{\frac {k}{n}}\right)^{n-k}k!}}\\&={\frac {\lambda ^{k}\left(1-{\frac {\lambda }{n}}\right)^{n-k}e^{-k}}{\left(1-{\frac {k}{n}}\right)^{n-k}k!}}\\&\simeq {\frac {\lambda ^{k}\left(1-{\frac {\lambda }{n}}\right)^{n}e^{-k}}{\left(1-{\frac {k}{n}}\right)^{n}k!}}.\end{aligned}}$

As $n\to \infty$, $\left(1-{\frac {x}{n}}\right)^{n}\to e^{-x}$ so:

${\begin{aligned}{n \choose k}p^{k}(1-p)^{n-k}&\simeq {\frac {\lambda ^{k}e^{-\lambda }e^{-k}}{e^{-k}k!}}\\&={\frac {\lambda ^{k}e^{-\lambda }}{k!}}\end{aligned}}$

Ordinary generating functions

It is also possible to demonstrate the theorem through the use of ordinary generating functions of the binomial distribution:

$G_{\operatorname {bin} }(x;p,N)\equiv \sum _{k=0}^{N}\left[{\binom {N}{k}}p^{k}(1-p)^{N-k}\right]x^{k}={\Big [}1+(x-1)p{\Big ]}^{N}$

by virtue of the binomial theorem. Taking the limit $N\rightarrow \infty$ while keeping the product $pN\equiv \lambda$ constant, it can be seen:

$\lim _{N\rightarrow \infty }G_{\operatorname {bin} }(x;p,N)=\lim _{N\rightarrow \infty }\left[1+{\frac {\lambda (x-1)}{N}}\right]^{N}=\mathrm {e} ^{\lambda (x-1)}=\sum _{k=0}^{\infty }\left[{\frac {\mathrm {e} ^{-\lambda }\lambda ^{k}}{k!}}\right]x^{k}$

which is the OGF for the Poisson distribution. (The second equality holds due to the definition of the exponential function.)

See also

• De Moivre–Laplace theorem
• Le Cam's theorem