The Poisson Distribution

Download Poisson.Rmd

Specification and probability calculation

The Poisson distribution uses a single parameter \(\lambda\) to find probabilities of counts. The individuals that make up these counts are assumed independent. The rate parameter \(\lambda\) can be adjusted to allow for changes in the “area of opportunity” in which the counts are observed.

The probability mass function is \(P\left( {X = x} \right) = \frac{\lambda ^x e^{- \lambda}}{x!}\) for \(\lambda > 0\)

Calculation in R

The dpois()` function finds probabilities; manual calculations could use:

Lambda=5
Counts=0:2
dpois(Counts, lambda=Lambda)

[1] 0.006737947 0.033689735 0.084224337

Lambda^Counts * exp(-Lambda) / factorial(Counts)

[1] 0.006737947 0.033689735 0.084224337

Mean and variance

The mean (expected value) and variance are both equal to \(\lambda\).

Calculation of mean from first principles

Given expected value is found using \(E(X) = \sum_x {xP(X=x)\), we find that:

\[E(X)=\sum_{x=0}^{\infty} {x \times \frac{\lambda ^x e^{- \lambda}}{x!}\]

Moving \(\lambda e^{-\lambda}\) outside the summation gives

\[E(X)= \lambda e^{-\lambda} \sum_{x=0}^{\infty} {x \times \frac{\lambda ^{x-1}}{x!}\]

The first term of the summation where x=0 is zero so we can re-start the summation from x=1 and cancel the x inside the summation to give:

\[E(X)= \lambda e^{-\lambda} \sum_{x=1}^{\infty} {\frac{\lambda ^{x-1}}{(x-1)!}\]

Rephrasing this infinite summation from zero again gives:

\[E(X)= \lambda e^{-\lambda} \sum_{x=0}^{\infty} {\frac{\lambda ^x}{x!}\]

The summation is the series for the exponential so we have

\[E(X)= \lambda e^{-\lambda} e^\lambda = \lambda\]

Calculation of variance from first principles

Given the variance of a random variable is \(Var(X) = E(X^2) - \left [ E(X)\right]^2\)$ we start with

\[Var(X) = \sum_{x=0}^{\infty} {x^2 \times \frac{\lambda^x e^{- \lambda}}{x!}} - \lambda^2\]

Keeping just the summation on the right hand side of the expression, gives:

\[Var(X) + \lambda^2 = \sum_{x=0}^{\infty} {x^2 \times \frac{\lambda ^x e^{- \lambda}}{x!}}\]

Bringing \(\lambda\) out of the summation and expanding the first term of the summation gives:

\[Var(X) + \lambda^2 = \lambda \left[ 0 + \sum_{x=1}^{\infty} {x \frac{\lambda ^({x-1)} e^{- \lambda}}{(x-1)!}}\right]\]

substituting y=x-1 gives

\[Var(X) + \lambda^2 = \lambda \left[ \sum_{y=0}^{\infty} {(y+1) \frac{\lambda ^y e^{- \lambda}}{y!}}\right]\]

Splitting the summation into two summations gives:

\[Var(X) + \lambda^2 = \lambda \left[ \sum_{y=0}^{\infty} {(y) \frac{\lambda ^y e^{- \lambda}}{y!}} +\sum_{y=0}^{\infty} {(1) \frac{\lambda ^y e^{- \lambda}}{y!}} \right]\]

The first summation is an expected value of a Poisson random variable \(E(Y)=\lambda\) and the second summation is the sum of a Poisson probability mass function equals one. This gives

\[Var(X) + \lambda^2 = \lambda \left[\lambda + 1\right]\]

Expanding brackets and cancelling gives \(Var(X) = \lambda\)

Useful properties

The sum of two independent Poisson random variables is a Poisson random variable.

Relationships to other distributions

If a Poisson random variable models the counts of events, then the gap between two subsequent events follows and exponential distribution.

Parameter estimation

The single parameter \(\lambda\) is estimated using the sample mean under both a frequentist or (uninformed) Bayesian paradigm. A separate page gives greater detail on point and interval estimation of the Poisson parameter.