\(X\) 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:

- Let \(\alpha>0\) such that for any short \(\Delta t\), the
probability of
**one**event occurring is \(\alpha\Delta t\). - 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

(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 \(0.6p_{\lambda}+0.4p_{\mu}\)