Poisson Distribution

Posted by Beetle B. on Thu 18 May 2017

\(X\) is of a Poisson distribution if its pmf is \(p(x;\lambda)=\frac{e^{-\lambda}\lambda^{x}}{x!}\) where \(x\) is 0, 1, 2, etc.

\(\lambda\) is usually a rate (per unit time, per area, etc).

Given a binomial distribution with \(n\) and \(p\), if we let \(n\) approach \(\infty\) and \(p\) approach 0 in such a way that \(np\) approaches \(\lambda\), then the distribution approaches a Poisson one.

One can often use this as an approximation for the binomial distribution when \(n\ge 100,\ p\le 0.01\) and \(np\le 20\).

\(E(X)=\lambda, V(X)=\lambda\). One can see this directly from the limit of the binomial distribution.

A Poisson process is one where:

  1. Let \(\alpha>0\) such that for any short \(\Delta t\), the probability of one event occurring is \(\alpha\Delta t\).
  2. The number of events during \(\Delta t\) is independent of the past.

Let \(P_{k}(t)\) be the probability of receiving \(k\) events in an interval of length \(t\). Then

\begin{equation*} P_{k}(t)=e^{-\alpha t}\frac{(\alpha t)^{k}}{k!} \end{equation*}

(which is when \(\lambda=\alpha t\))

\(\alpha\) is known as the rate of the process.

You can combine Poisson processes. If you have two of them where 60% of the time it is with \(\lambda\) and 40% of the time it is with \(\mu\), then the overall pmf is \(0.6p_{\lambda}+0.4p_{\mu}\)