law of rare events

Let $X$ be distributed as $Bin(n,p)$, a binomial random variable with parameters $n$ and $p$. Suppose

$$\underset{n\to \mathrm{\infty }}{lim}np=\lambda ,$$

where $\lambda$ is a positive real constant, then $X$ is asymptotically distributed as $Poisson(\lambda )$, a Poisson distribution with parameter $\lambda$.

Basically, when the size of the population $n$ is very large and the occurrence of certain event $A$ is rare, where $p$, the probability of $A$ is very small, the binomial random variable $X$ can be approximated by a Poisson random variable.

Sketch of Proof. Let $X\sim Bin(n,p)$. So

$P(X=m)$$=$${\displaystyle \frac{n!}{m!(n-m)!}}{p}^m{(1-p)}^{n-m}$

$=$${\displaystyle \frac{n!}{n^m(n-m)!}}{\displaystyle \frac{{(np)}^m}{m!}}{(1-{\displaystyle \frac{np}{n}})}^{n-m}$

$=$${\displaystyle \frac{n!}{n^m(n-m)!}}{\displaystyle \frac{{(np)}^m}{m!}}{(1-{\displaystyle \frac{np}{n}})}^n{(1-{\displaystyle \frac{np}{n}})}^{-m}.$

As $n\to \mathrm{\infty }$,

$$\frac{n!}{n^m(n-m)!}=\frac{n}{n}\frac{n-1}{n}\mathrm{\cdots }\frac{n-m+1}{n}\approx 1,$$

$${(1-\frac{np}{n})}^{-m}\approx {(1-\frac{\lambda }{n})}^{-m}\approx 1,$$

$${(1-\frac{np}{n})}^n\approx {(1-\frac{\lambda }{n})}^n\approx e^{-\lambda },$$

and

$$\frac{{(np)}^m}{m!}\approx \frac{{\lambda }^m}{m!}.$$

Therefore,

$$P(X=m)\approx \frac{{\lambda }^m}{m!}{e}^{-\lambda }=Poisson(\lambda ).$$

Example. Suppose in a given year, the number of fatal automobile accidents has a binomial distribution for a particular insuarance company with five hundred automobile insurance policies. On average, there is one policy out of the five hundred that will be involved in a fatal crash. What is the probability that there will be no fatal accidents (out of five hundred policies) in any particular year?

Solution. If $X$ be the number of fatal accidents in a year from a population of 500 auto policies, then $X\sim Bin(n,p)$ with $n=500$ and $p=1/500$. $\lambda =500\times 1/500=1$ and so

$$P(X=0)\approx e^{-1}\approx 0.368.$$

Using the binomial distribution, we have

$$P(X=0)={(1-\frac{1}{500})}^{500}\approx 0.367.$$