Poisson Derivation 1

From Qwiki

A conceptually simple method for deriving the Poisson Distribution is based on the continuous limit of the Binomial Distribution. Let's consider very small but not quite infinitesimal time intervals Δ t, and suppose we have N of these intervals partitioning a large interval T = N Δt. In each interval, the probability of an event is (approximately) λ Δt, with the approximation getting better and better as Δt goes to 0. Then for this discrete case, the probability of observing n events over the whole interval is given by the Binomial Distribution:

$P_\lambda(n,T) \approx {N \choose n} \left(1 - \lambda\,\Delta t\right)^{N-n} (\lambda\, \Delta t)^n$

$= {N\choose n} \left(1 - \frac{\lambda T}{N}\right)^{N-n} \left( \frac{\lambda T}{N}\right)^n.$

Using Stirling's formula

The Poisson distribution can be found by taking the limit $N \rightarrow\infty$ and using Stirling's approximation $N!\approx (N/e)^N$ (which is exact in the large $N$ limit):

$P_\lambda(n,T) = \lim_{N \rightarrow \infty}\frac{N!}{n!(N-n)!} \left(1-\frac{\lambda T}{N}\right)^{N-n} \left(\frac{\lambda T}{N}\right)^n$

$= \frac{(\lambda T)^n}{n!}\lim_{N \rightarrow \infty} \frac{(N/e)^N}{\left(\frac{N-n}{e}\right)^{N-n}N^n}\left(1-\frac{\lambda T}{N}\right)^{N-n}$

$= \frac{e^{-n}(\lambda T)^n}{n!}\lim_{N \rightarrow \infty} \frac{N^{N-n}}{\left(N-n\right)^{N-n}}\left(1-\frac{\lambda T}{N}\right)^{N-n}$

$= \frac{e^{-n}(\lambda T)^n}{n!}\lim_{N \rightarrow \infty} \frac{\left(1-\frac{\lambda T}{N}\right)^{N-n}}{\left(1-\frac{n}{N}\right)^{N-n}}$

$= \frac{e^{-n}(\lambda T)^n}{n!}\frac{e^{-\lambda T}}{e^{-n}}$

$= \frac{e^{-\lambda T} (\lambda T)^n}{n!}.$

A variation on this theme that avoids the need for Stirling's formula

The following variation on this theme avoids Stirling's formula:

$P_\lambda(n,T) = \lim_{N \rightarrow \infty}\frac{N!}{n!(N-n)!} \left(1-\frac{\lambda T}{N}\right)^{N-n} \left(\frac{\lambda T}{N}\right)^n$

$= \frac{(\lambda T)^n}{n!} \lim_{N\to\infty} N(N-1)(N-2)\cdots(N-n+1) \left(1-\frac{\lambda T}{N}\right)^{N-n} \left(\frac{1}{N}\right)^n$

(it is because n and λT do not change as N → ∞ that the factor (λT)ⁿ/n! can be taken out from within the "lim")

$= \frac{(\lambda T)^n}{n!} \lim_{N\to\infty} \underbrace{\frac{N}{N}\cdot \frac{N-1}{N}\cdot \frac{N-2}{N} \cdots \frac{N-n+1}{N}}\ \underbrace{\left(1-\frac{\lambda T}{N}\right)^N}\ \underbrace{\left(1-\frac{\lambda T}{N}\right)^{-n}}.$

Now observe that the limits of the expressions over the first and third underbraces are equal to 1, and the limit of the expression over the second underbrace is e^−λt. Therefore the limit we seek is just

$\frac{(\lambda t)^n}{n!}e^{-\lambda t}.$