# Poisson limit theorem

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. [1] The theorem was named after Siméon Denis Poisson (1781–1840).

## TheoremEdit

As ${\displaystyle n\rightarrow \infty }$  and ${\displaystyle p\rightarrow 0}$  such that the mean value ${\displaystyle np=\lambda }$  remains constant, we can approximate

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

## ProofsEdit

Using Stirling's approximation, we can write:

{\displaystyle {\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 ${\displaystyle n\to \infty }$  and ${\displaystyle np=\lambda }$ :

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

{\displaystyle {\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}}}

### Alternative ProofEdit

A simpler proof is possible without using Stirling's approximation:

{\displaystyle {\begin{aligned}{n \choose k}p^{k}(1-p)^{n-k}&\simeq \lim _{n\to \infty }{\frac {n(n-1)(n-2)\dots (n-k+1)}{k!}}\left({\frac {\lambda }{n}}\right)^{k}\left(1-{\frac {\lambda }{n}}\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}}\right)^{n-k}\\&=\lim _{n\to \infty }{\frac {\lambda ^{k}}{k!}}\left(1-{\frac {\lambda }{n}}\right)^{n-k}\end{aligned}}} .

Since

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

and

${\displaystyle \lim _{n\to \infty }\left(1-{\frac {\lambda }{n}}\right)^{-k}=1}$

This leaves

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

### Ordinary Generating FunctionsEdit

It is also possible to demonstrate the theorem through the use of Ordinary Generating Functions of the binomial distribution:

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

by virtue of the Binomial Theorem. Taking the limit ${\displaystyle N\rightarrow \infty }$  while keeping the product ${\displaystyle pN\equiv \lambda }$  constant, we find

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

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