Poisson distribution

A Poisson distribution counts independent events with an average of $\lambda$ events occurring per time $T$ .

To derive the PDF, imagine breaking up the time period into $N$ pieces (where $N$ is large).

The probability of an event occurring in one of these time ranges is $\lambda T/N$ , where $N$ is large enough that this probability is $\ll 1$ .

The probability of no event occurring is

p(0) = \lim_{N \to \infty} \left(1 - \frac{\lambda T}{N}\right)^{N} = e^{-\lambda T}.

The probability of one event occurring is

p(1) = \lim_{N \to \infty} N \left(1 - \frac{\lambda T}{N}\right)^{N-1} \frac{\lambda T}{N} = e^{-\lambda T} \lambda T.

Since there are $N$ ways to choose which period the event will occur in, and there are $N-1$ events that don’t occur.

It follows that the probability of $n$ events occurring is

p(n) = \lim_{N \to \infty} {N \choose n} \left(1 - \frac{\lambda T}{N}\right)^{N-n} \left(\frac{\lambda T}{N}\right)^n = \frac{(\lambda T)^n}{n!} e^{-\lambda T}.

The mean of a Poisson distribution is $\mu = \lambda T$ . An interesting property is that the variance equals the mean: $\Var = \mu = \lambda T$ .

For large $z = \lambda T$ , the Poisson distribution approaches a normal distribution with mean and variance $z$

\begin{align*} p_z(n) &= \frac{z^n}{n!} e^{-z} \\ \ln p_z(n) &= n \ln z - \underbrace{\ln n!}_\text{apply Stirling approx.} - z \\ &= n \ln z - n \ln n + n - z - \ln \sqrt{2\pi n} \\ \frac{\partial}{\partial n} \ln p_z(n) &= \ln z - \ln n -\frac{1}{2n} \\ \frac{\partial^2}{\partial n^2} \ln p_z(n) &\approx -\frac 1n \\ \ln p_z(n) &\approx - \frac12 \ln \sqrt{2\pi z} + \frac{(n-z)^2}{2} \left(\frac{-1}{z}\right) \\ p_z(n) &\approx \frac{1}{\sqrt{2\pi z}} \exp\left(\frac{(n-z)^2}{2z}\right). \end{align*}