Poisson Distribution

PMF

The support of the distribution is the set of non-negative integers: \(\{0,1,2,\cdots\}\). The PMF is given as follows:

\[ P(x)=\cfrac{\lambda^{x}}{x!}e^{-\lambda} \]

Here \(\lambda\) is a parameter governing the distribution. To verify that this is a valid PMF:

\[ \sum\limits_{x=0}^{\infty}\cfrac{\lambda^{x}}{x!}e^{-\lambda}=e^{-\lambda}\sum\limits_{x=0}^{\infty}\cfrac{\lambda^{x}}{x!}=1 \]

We have used the Taylor series expansion for \(e^{\lambda}\).

Bionomial to Poisson

Consider the Binomial distribution, \(\text{Binomial}(n,p)\):

\[ P(x)={n \choose x}p^{x}(1-p)^{n-x} \]

We allow \(n\rightarrow\infty\) and \(p\rightarrow0\) such that \(np=\lambda\). Rewriting the Binomial distribution, we get:

\[ \begin{aligned} P(x) & =\cfrac{n!}{x!(n-x)!}\,p^{x}(1-p)^{n-x}\\ \\ & =\cfrac{n!}{x!(n-x)!}\,\left(\cfrac{\lambda}{n}\right)^{x}\left(1-\cfrac{\lambda}{n}\right)^{n-x}\\ \\ & =\cfrac{\lambda^{x}}{x!}\cdot\cfrac{(n-x+1)\cdots n}{n^{x}}\cdot\left(1-\cfrac{\lambda}{n}\right)^{n}\left(1-\cfrac{\lambda}{n}\right)^{-x} \end{aligned} \]

We can now compute the limits:

\[ \begin{aligned} \lim_{n\rightarrow\infty}\frac{(n-x+1)\cdots n}{n^{x}} & =\lim_{n\rightarrow\infty}\left[1-\left(\cfrac{x-1}{n}\right)\right]\cdots\left[1-\left(\cfrac{x-x}{n}\right)\right]\\ \\ & =1\\ \\\lim_{n\rightarrow\infty}\left(1-\cfrac{\lambda}{n}\right)^{n} & =e^{-\lambda}\\ \\\lim_{n\rightarrow\infty}\left(1-\cfrac{\lambda}{n}\right)^{-x} & =1 \end{aligned} \]

Plugging these limits into the expression for \(P(x)\), we get the PMF of the Poisson distribution.