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
pn
[0,1]
npn
λ
\limn\to{n\choosek}
k | |
p | |
n |
n-k | |
(1-p | |
n) |
=e-λ
λk | |
k! |
Assume
λ>0
λ=0
\begin{align} \lim\limitsn → infty{n\choosek}
k | |
p | |
n |
n-k | |
(1-p | |
n) |
&=\limn\toinfty
n(n-1)(n-2)...(n-k+1) | \left( | |
k! |
λ | |
n |
(1+o(1))\right)k\left(1-
λ | |
n |
(1+o(1))\right)n-k\\ &=\limn\toinfty
nk+O\left(nk-1\right) | |
k! |
λk | |
nk |
\left(1-
λ | |
n |
(1+o(1))\right)n\left(1-
λ | |
n |
(1+o(1))\right)-k\\ &=\limn\toinfty
λk | \left(1- | |
k! |
λ | |
n |
(1+o(1))\right)n. \end{align}
Since
\limn\toinfty\left(1-
λ | |
n |
(1+o(1))\right)n=e-λ
this leaves
{n\choosek}pk(1-p)n-k\simeq
λke-λ | |
k! |
.
Using Stirling's approximation, it can be written:
\begin{align} {n\choosek}pk(1-p)n-k&=
n! | |
(n-k)!k! |
pk(1-p)n-k\ &\simeq
\sqrt{2\pin | |||
|
\right)n}{\sqrt{2\pi\left(n-k\right)}\left(
n-k | |
e |
\right)n-kk!}pk(1-p)n-k\ &=\sqrt{
n | |
n-k |
Letting
n\toinfty
np=λ
\begin{align} {n\choosek}pk(1-p)n-k&\simeq
nnpk(1-p)n-ke-k | |
\left(n-k\right)n-kk! |
\\&=
| |||||||||||||
|
\\&=
| ||||||||||
|
\ &\simeq
| ||||||||||
|
. \end{align}
As
n\toinfty
\left(1- | x |
n |
\right)n\toe-x
\begin{align} {n\choosek}pk(1-p)n-k&\simeq
λke-λe-k | |
e-kk! |
\\&=
λke-λ | |
k! |
\end{align}
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
N | |
\sum | |
k=0 |
\left[\binom{N}{k}pk(1-p)N-k\right]xk =[1+(x-1)p]N
by virtue of the binomial theorem. Taking the limit
N → infty
pN\equivλ
\limN → inftyG\operatorname{bin}(x;p,N) =\limN → infty\left[1+
λ(x-1) | |
N |
\right]N =eλ(x-1)=
infty | |
\sum | |
k=0 |
\left[
e-λλk | |
k! |
\right]xk
which is the OGF for the Poisson distribution. (The second equality holds due to the definition of the exponential function.)