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). A generalization of this theorem is Le Cam's theorem.

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

${\displaystyle \underset{n\to \mathrm{\infty }}{\lim }(\genfrac{}{}{0ex}{}{n}{k})p_n^k(1-p_n{)}^{n-k}=e^{-\lambda }\frac{{\lambda }^k}{k!}}$

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

${\displaystyle \begin{array}{rl}\lim {}_{n\to \mathrm{\infty }}(\genfrac{}{}{0ex}{}{n}{k})p_n^k(1-p_n{)}^{n-k}& =\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n-1)(n-2)\dots (n-k+1)}{k!}{(\frac{\lambda }{n}(1+o(1)))}^k{(1-\frac{\lambda }{n}(1+o(1)))}^{n-k}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{n^k+O\left(n^{k-1}\right)}{k!}\frac{{\lambda }^k}{n^k}{(1-\frac{\lambda }{n}(1+o(1)))}^n{(1-\frac{\lambda }{n}(1+o(1)))}^{-k}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{{\lambda }^k}{k!}{(1-\frac{\lambda }{n}(1+o(1)))}^n.\end{array}}$

Since

${\displaystyle \underset{n\to \mathrm{\infty }}{\lim }{(1-\frac{\lambda }{n}(1+o(1)))}^n=e^{-\lambda }}$

this leaves

${\displaystyle (\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}\simeq \frac{{\lambda }^k{e}^{-\lambda }}{k!}.}$

Using Stirling's approximation, it can be written:

${\displaystyle \begin{array}{rl}(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}& =\frac{n!}{(n-k)!k!}{p}^k(1-p{)}^{n-k}\\ & \simeq \frac{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n}{\sqrt{2\pi (n-k)}{\left(\frac{n-k}{e}\right)}^{n-k}k!}{p}^k(1-p{)}^{n-k}\\ & =\sqrt{\frac{n}{n-k}}\frac{n^n{e}^{-k}}{{(n-k)}^{n-k}k!}{p}^k(1-p{)}^{n-k}.\end{array}}$

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

${\displaystyle \begin{array}{rl}(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}& \simeq \frac{n^n{p}^k(1-p{)}^{n-k}{e}^{-k}}{{(n-k)}^{n-k}k!}\\ & =\frac{n^n{\left(\frac{\lambda }{n}\right)}^k{(1-\frac{\lambda }{n})}^{n-k}{e}^{-k}}{n^{n-k}{(1-\frac{k}{n})}^{n-k}k!}\\ & =\frac{{\lambda }^k{(1-\frac{\lambda }{n})}^{n-k}{e}^{-k}}{{(1-\frac{k}{n})}^{n-k}k!}\\ & \simeq \frac{{\lambda }^k{(1-\frac{\lambda }{n})}^n{e}^{-k}}{{(1-\frac{k}{n})}^{n}k!}.\end{array}}$

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

${\displaystyle \begin{array}{rl}(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}& \simeq \frac{{\lambda }^k{e}^{-\lambda }{e}^{-k}}{e^{-k}k!}\\ & =\frac{{\lambda }^k{e}^{-\lambda }}{k!}\end{array}}$

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

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

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

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

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

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

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