Poisson as a Limit to a Binomial Distribution

Date: 05/29/2000 at 15:46:18
From: Dawn Sykes
Subject: Poisson as a limit to a binomial distribution

How do I prove that the limit of a binomial distribution with large n 
and small p is a Poisson distribution?

Date: 05/29/2000 at 16:21:45
From: Doctor Anthony
Subject: Re: Poisson as a limit to a binomial distribution

Using the binomial distribution the probability of r successes in n 
trials in which p = probability of success at each trial is:

We let np = a, so p = a/n

     P(r) = C(n,r).p^r.(1-p)^(n-r)

            n(n-1)(n-2)...(n-r+1)
          = --------------------- (a/n)^r (1 -a/n)^(n-r) 
                     r!

Take the n^r in the denominator and put one n into each term of C(n,r)

            1(1-1/n)(1-2/n)...(1 - (r-1)/n)
     P(r) = ------------------------------- x a^r x (1 - a/n)^(n-r)
                          r!

Now let n -> infinity and p -> 0 in such a way that np = a is finite.

The term (1- a/n)^(n-r) = (1-a/n)^n/(1-a/n)^r, which as n -> infinity 
becomes:

     (1 - a/n)^n
     -----------  ->  e^(-a)
          1

The terms representing C(n,r) become:

     1 x 1 x 1 x 1 ... to r terms
     ----------------------------  =  1/r!
                  r!

and so we get:

     P(r) = a^r/r! x e^(-a)

which is the required expression for Poisson probabilities for a 
distribution with mean = a

The sum of all the probabilities is given by

       e^(-a)[1 + a + a^2/2! + a^3/3! + ... to infinity]

     = e^(-a)[e^a]

     = 1

as required. We know already that the mean is a.

The variance is npq = np(1-p) = a x 1 = a  since 1-p = 1 as p -> 0. So 
the mean and variance of the distribution are both equal to a.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/