Poisson as a Limit to a Binomial DistributionDate: 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/ |
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