Statistical Sampling

Poisson distribution

Assume that we count the number of times that a random event occurs over some time interval. Denote this count by N, and say that we know the average number \lambda of times that the event occurs. Then N is a stochastic variable following a Poisson distribution, with probability distribution function P(N = n) = \frac{\lambda^n}{n!} e^{-\lambda} that the event occurs n times, where n! denotes the factorial function 1 \cdot 2 \cdots (n - 1) \cdot n, and e^{-\lambda} is the exponential function.

The same distribution describes anything that is counted over some interval, area, volume. For example the number of visible stars in a given section of the night sky, the number of quartz grains in a scoop of beach sand, the number of sodium atoms in a glass of sea water, and so on. The difference between these cases lies the parameter \lambda, which equals both the mean of the distribution and the variance. On the last note, the standard deviation thus equals the square root of \lambda. From this, if the the expected number \langle N \rangle = \lambda (the mean) has a value to some order 10^{M}, the standard deviation is of the order 10^{M/2}, which for higher values of M becomes smaller and smaller compared to the mean. Thus, for the examples of sodium atoms or quartz grains, all samples (scoops, glasses) can not at all be expected to vary much in number compared to the numbers themselves. Hence the apparent homogeneity of beach sand, seawater, et.al..

Basic properties:
Notation:
N \sim \mathrm{Pois}(\lambda)
Type:
Discrete
Parameters:
\lambda \in \mathbb{R}_{> 0} = (0, \infty)
Variables:
N
Support:
\mathbb{Z}_{\ge 0} = \{ 0, 1, 2, \dots \}
PDF:
f(n) = P(N = n) = e^{-\lambda} \frac{\lambda^n}{n!}
CDF:
F(n) = P(N \le n) = e^{-\lambda} \sum\limits_{k = 0}^n \frac{\lambda^k}{k!} = \frac{1}{n!} \int\limits_\lambda^\infty e^{-u} u^n \: dx
Mean:
\lambda
Variance:
\lambda
Generating function:
e^{\lambda (t - 1)}