Deriving Poisson distribution
May 21, 2017

Poisson distribution is defined as, $$ P(x; \lambda) = \frac{e^{-\lambda} \lambda^x}{x!} $$ Where,

  • $P(x;\lambda)$ is the probability that an event occurs $x$ times in the given interval,
  • $\lambda$ is the expected rate/probability of an event occuring.

In Binomial distribution, probability that one of the two events($p$ and $q$) occurs $x$ times out of $n$ trials is, $$P(X=x) = \binom {n}{x}p^xq^{n-x} \tag{1}$$ where $q = 1-p$. Now, if $\lambda$ is the expected number of successes then, $p=\lambda/n$.

Substituting value of $p$ in $(1)$, we obtain $$ P(X=x) = \binom{n}{x} \frac{\lambda^x}{n^x} (1-\frac{\lambda}{n})^{n-x}$$ Expanding $\binom{n}{x}$, $$P(X=x) = \frac{n.(n-1)...(n-x+1)}{x!}\frac{\lambda^x}{n^x}(1-\frac{\lambda}{n})^n (1-\frac{\lambda}{n})^{-x}$$ $$=\frac{n.(n-1)...(n-x+1)}{n^x}\frac{\lambda^x}{x!}(1-\frac{\lambda}{n})^n (1-\frac{\lambda}{n})^{-x}$$ As $n\to\infty$, it's obvious that $(1-\lambda/n)\to1$ and by definition, $(1-\lambda/n)^n=e^{-\lambda}$. So, $$P(X=x) = \frac{n.(n-1)...(n-x+1)}{n^x}\frac{\lambda^x}{x!}e^\lambda$$ $$= \frac{n}{n}.\frac{n-1}{n}...\frac{n-x+1}{n}\frac{\lambda^x}{x!}e^{-\lambda}$$ Thus, as $n\to\infty$ we obtain, $$ P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$$ And hence, our Poisson equation.