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Statistics · Statistical distributions

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The Poisson distribution

Modelling the number of random events in a fixed interval — when the Poisson model is valid, P(X = x) = e^{−λ}λ^x/x!, mean = variance = λ, scaling λ to the length of the interval, adding independent Poissons, and using Poisson as an approximation to the binomial when n is large and p is small.

Statistics · Statistical distributions

The Poisson distribution

Modelling the number of random events in a fixed interval — when the Poisson model is valid, P(X = x) = e^{−λ}λ^x/x!, mean = variance = λ, scaling λ to the length of the interval, adding independent Poissons, and using Poisson as an approximation to the binomial when n is large and p is small.

Why it works

Some counts have no natural "number of trials": calls arriving at a switchboard, flaws in a metre of cloth, misprints on a page. There is no nn to put in a binomial — events just happen, one at a time, at some average rate. The Poisson distribution is the model for exactly this situation. If events occur
  • singly (two can't land at exactly the same instant),
  • independently (one arriving tells you nothing about the next), and
  • at a constant average rate,
then the number XX of events in a fixed interval satisfies X∼Po(λ)X \sim \mathrm{Po}(\lambda), where λ\lambda is the mean number of events in that interval, and P(X=x)=e−λλxx!,x=0,1,2,…\mathrm{P}(X = x) = \frac{\mathrm{e}^{-\lambda}\lambda^x}{x!}, \qquad x = 0, 1, 2, \ldots

Mean and variance are BOTH λ\lambda. E(X)=λ,Var(X)=λ.\mathrm{E}(X) = \lambda, \qquad \mathrm{Var}(X) = \lambda. This is the Poisson's fingerprint: real data with a sample mean close to its sample variance suggests a Poisson model; a variance far from the mean argues against one.

Scale λ\lambda to the interval. The rate is per interval, so a different interval means a different λ\lambda. Calls at 33 per hour give Po(3)\mathrm{Po}(3) for one hour, Po(6)\mathrm{Po}(6) for two hours, Po(1.5)\mathrm{Po}(1.5) for half an hour. Always restate λ\lambda for the interval in the question before computing anything.

"At least" goes through the complement. XX has no upper limit, so P(X⩾2)\mathrm{P}(X \geqslant 2) must be computed as 1−P(X=0)−P(X=1)1 - \mathrm{P}(X = 0) - \mathrm{P}(X = 1) — never by adding terms upwards.

Adding independent Poissons. If X∼Po(λ)X \sim \mathrm{Po}(\lambda) and Y∼Po(μ)Y \sim \mathrm{Po}(\mu) are independent, then X+Y∼Po(λ+μ).X + Y \sim \mathrm{Po}(\lambda + \mu). Two independent streams of random events merge into one random stream at the combined rate.

Poisson as an approximation to the binomial. When nn is large and pp is small (as a guide, n>50n > 50 and np<5np < 5), computing B(n,p)\mathrm{B}(n, p) probabilities becomes awkward and unnecessary: B(n,p)≈Po(np).\mathrm{B}(n, p) \approx \mathrm{Po}(np). Rare events in many trials behave like random arrivals. State the conditions and the value λ=np\lambda = np whenever you use it.