Poisson limit theorem

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

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).

Theorem

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

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

Proofs

Using Stirling's approximation, we can write:

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

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

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

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

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

Alternative Proof

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

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

Since

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

and

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

This leaves

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

Ordinary Generating Functions

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

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

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

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

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

See also

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