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. The theorem was named after Siméon Denis Poisson (1781–1840). A generalization of this theorem is Le Cam's theorem.

Theorem
Let $$p_n $$ be a sequence of real numbers in $$[0,1] $$ such that the sequence $$n p_n $$ 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{align} \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{align} $$

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{align} {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{align} $$

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

\begin{align} {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{align} $$

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

{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{align}$$

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.)