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Homelaw of rare events

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# law of rare events

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

$\lim_{{n\rightarrow\infty}}np=\lambda,$ |

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

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

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

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

As $n\rightarrow\infty$,

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

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

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

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

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

## Mathematics Subject Classification

62P05*no label found*60E99

*no label found*60F99

*no label found*

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