# 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 $$0.6p_{\lambda}+0.4p_{\mu}$$