Poisson limit theorem

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The law of rare events or Poisson limit theorem gives a Poisson approximation to the binomial distribution, under certain conditions.[1] The theorem was named after Siméon Denis Poisson (1781–1840).

Statement[edit]

If

n \rightarrow \infty, p \rightarrow 0, such that np \rightarrow \lambda

then

\frac{n!}{(n-k)!k!} p^k (1-p)^{n-k} \rightarrow e^{-\lambda}\frac{\lambda^k}{k!}.

Example[edit]

Suppose that in an interval [0, 1000], 500 points are placed randomly. Now what is the number of points that will be placed in [0, 10]?

The probabilistically precise way of describing the number of points in the sub-interval would be to describe it as a binomial distribution p_n(k).

If we look here, the probability (that a random point will be placed in the sub-interval) is p = 10/1000 = 0.01. Here n=500 so np=5.

The probability that k points lie in the sub-interval is

p_n(k)=\frac{n!}{(n-k)!k!} p^k (1-p)^{n-k}.

where: p is the probability of falling with in the interval. n!/(k! \cdot (n-k)!) gives the number of ways in which k elements can be selected. p^k gives the probability of the k elements falling in the interval. (1-p)^{n-k} counts the probability that {n-k} elements fall outside of the interval

But using the Poisson Theorem we can approximate it as

e^{-\lambda}\frac{\lambda^k}{k!} = e^{-5}\frac{5^k}{k!}.

Proofs[edit]

Accordingly to factorial's rate of growth, we replace factorials of large numbers with approximations:

\frac{n!}{(n-k)!k!} p^k (1-p)^{n-k} \rightarrow \frac{ \sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{ \sqrt{2\pi \left(n-k\right)}\left(\frac{n-k}{e}\right)^{n-k}k!} p^k (1-p)^{n-k}.

After simplifying the fraction:

\frac{ \sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{ \sqrt{2\pi \left(n-k\right)}\left(\frac{n-k}{e}\right)^{n-k}k!} p^k (1-p)^{n-k}\rightarrow \frac{ \sqrt{n}n^np^k (1-p)^{n-k}}{ \sqrt{n-k}\left(n-k\right)^{n-k}e^kk!}\rightarrow \frac{n^np^k (1-p)^{n-k}}{\left(n-k\right)^{n-k}e^kk!}.

After using the condition np \rightarrow \lambda:

 \frac{n^np^k (1-p)^{n-k}}{\left(n-k\right)^{n-k}e^kk!} \rightarrow \frac{n^k\left(\frac{\lambda}{n}\right)^k (1-\frac{\lambda}{n})^{n-k}}{\left(1-\frac{k}{n}\right)^{n-k}e^kk!}=\frac{\lambda^k \left(1-\frac{\lambda}{n}\right)^{n-k}}{\left(1-\frac{k}{n}\right)^{n-k}e^kk!}\rightarrow\frac{\lambda^k \left(1-\frac{\lambda}{n}\right)^{n}}{\left(1-\frac{k}{n}\right)^{n}e^kk!}

Apply, that due to n\rightarrow \infty we get \left(1+\frac{x}{n}\right)^n \rightarrow e^x:

\frac{\lambda^k \left(1-\frac{\lambda}{n}\right)^{n}}{\left(1-\frac{k}{n}\right)^{n}e^kk!}\rightarrow\frac{\lambda^k e^{-\lambda}}{e^{-k}e^kk!}=\frac{\lambda^k e^{-\lambda}}{k!}

Q.E.D.

Alternative Proof[edit]

If we make the stronger assumption  np=\lambda (rather than  np\rightarrow \lambda ) then a simpler proof is possible without needing approximations for the factorials. Since np=\lambda, we can rewrite p=\lambda / n. We now have:

\lim_{n\to\infty}\frac{n!}{(n-k)!k!} \left(\frac{\lambda}{n}\right)^k  \left(1- \frac{\lambda}{n}\right)^{n-k}
= \lim_{n\to\infty}\frac{n(n-1)(n-2)\dots(n-k+1)}{k!} \frac{\lambda^k}{n^k}  \left(1- \frac{\lambda}{n}\right)^{n-k}

Taking each of these terms in sequence, n(n-1)(n-2)\dots(n-k+1)=n^k+O\left(n^{k-1}\right), meaning that \lim_{n\to\infty} \frac{n(n-1)(n-2)\dots(n-k+1)}{n^k k!}= \frac{1}{k!}.

Now \left(1- \frac{\lambda}{n}\right)^{n-k}=\left(1- \frac{\lambda}{n}\right)^{n}\left(1- \frac{\lambda}{n}\right)^{-k}. The first portion of this converges to e^{-\lambda}, and the second portion goes to 1, as \lim_{n\to\infty}\left(1- \frac{\lambda}{n}\right)^{-k}=\lim_{n\to\infty}\left(1- 0\right)^{-k}=1

This leaves us with \frac{1}{k!}\lambda^k e^{-\lambda}. Q.E.D.

Ordinary Generating Functions[edit]

It is also possible to demonstrate the theorem through the use of Ordinary Generating Functions (OGF). Indeed, the OGF of the binomial distribution is


  G_\mathrm{bin}(x;p,N) 
    \equiv \sum_{k=0}^{N} \left[ \binom{N}{k} p^k (1-p)^{N-k} \right] x^k
    = \Big[ 1 + (x-1)p \Big]^{N}

by virtue of the Binomial Theorem. Taking the limit N \rightarrow \infty while keeping the product pN\equiv\lambda constant, we find

 
  \lim_{N\rightarrow\infty} G_\mathrm{bin}(x;p,N)
    = \lim_{N\rightarrow\infty} \Big[ 1 + \frac{\lambda(x-1)}{N} \Big]^{N} 
    = \mathrm{e}^{\lambda(x-1)}
    = \sum_{k=0}^{\infty} \left[ \frac{\mathrm{e}^{-\lambda}\lambda^k}{k!} \right] x^k

which is the OGF for the Poisson distribution. (The second equality holds due to the definition of the Exponential function.)

See also[edit]

References[edit]

  1. ^ Papoulis, Pillai, Probability, Random Variables, and Stochastic Processes, 4th Edition