Poisson Distributions

Definition Let $\lambda > 0$. A random variable $X$ has the Poisson distribution with parameter $\lambda$, written $X \sim \operatorname{Pois}(\lambda)$, if its PMF is

for $k = 0, 1, 2, \dots$, and $p_X(x) = 0$ for every other $x$. The support is the set of all nonnegative integers, and each of them has positive probability. The mass at $k = 0$ uses $0! = 1$ and $\lambda^{0} = 1$, so $P(X = 0) = e^{-\lambda}$. For $\lambda = 0$ the distribution $\operatorname{Pois}(0)$ is defined as the point mass at $0$, with $P(X = 0) = 1$.

Definition The parameter $\lambda$ of $\operatorname{Pois}(\lambda)$ is a nonnegative real number, and it equals the expected count, $E[X] = \lambda$, as the card on the mean and variance shows. This one number determines the whole distribution. It need not be an integer, although the count always is. Negative values are not allowed, since then $p_X(1) = e^{-\lambda}\lambda$ would be negative. For $\lambda > 0$ and $k \ge 1$, consecutive masses satisfy

so the mass goes up from $k - 1$ to $k$ when $k < \lambda$ and goes down when $k > \lambda$. Hence the most likely count is $\lfloor \lambda \rfloor$, the greatest integer at most $\lambda$, when $\lambda$ is not an integer, and when $\lambda$ is a positive integer the counts $\lambda - 1$ and $\lambda$ tie for the largest mass.

Definition Exposure is the amount of time, length, area or other measure over which events are counted. Suppose a model states that events occur at a constant average rate $r$ per unit of exposure, so that the expected count over an exposure of size $t$ is $rt$, and that this count is Poisson. Then the count over exposure $t$ is $\operatorname{Pois}(\lambda)$ with

where $r$ and $t$ use the same unit, for example events per minute and minutes, so that $\lambda$ is a number of events. Using this formula for windows of different sizes is part of the assumed model. It does not follow from the distribution of the count in a single window.

Definition For every real number $\lambda$, the exponential series converges, and its sum is

The Poisson masses are the terms of this series multiplied by the constant $e^{-\lambda}$, so

Since every mass is nonnegative, this shows that the Poisson PMF is a valid distribution for every $\lambda > 0$.

Definition If $X \sim \operatorname{Pois}(\lambda)$ with $\lambda \ge 0$, then $E[X] = \lambda$, $\operatorname{Var}(X) = \lambda$ and $\operatorname{SD}(X) = \sqrt{\lambda}$. In the mean series the term $k = 0$ is zero, and for $k \ge 1$ the factor $k$ cancels against $k!$. With $j = k - 1$,

In the same way, for $k \ge 2$ the factor $k(k - 1)$ cancels against $k! = k(k - 1)(k - 2)!$, the terms $k = 0$ and $k = 1$ are zero, and with $j = k - 2$,

Since $X^{2} = X(X - 1) + X$, the second moment is $E[X^{2}] = \lambda^{2} + \lambda$, and $\operatorname{Var}(X) = \lambda^{2} + \lambda - \lambda^{2} = \lambda$. All of these series converge, so both moments exist.

Definition Let $\lambda > 0$, and for each $n \ge 1$ let $X_n \sim \operatorname{Bin}(n, p_n)$, where $n p_n \to \lambda$ as $n \to \infty$. Then for every fixed $k = 0, 1, 2, \dots$,

In practice, a $\operatorname{Bin}(n, p)$ count with many trials and a small success probability is approximated by $\operatorname{Pois}(np)$, which has the same mean. This is an approximation, not an equality. Le Cam’s inequality bounds its error. If $X \sim \operatorname{Bin}(n, p)$ and $Y \sim \operatorname{Pois}(np)$, then $\lvert P(X \in A) - P(Y \in A) \rvert \le np^{2}$ for every set $A$ of counts.