The Poisson distribution

Counting events in an interval

Some counts have no number of trials. A help desk does not make 60 attempts an hour at receiving a call; calls simply arrive, and the question is how many arrive in an hour. Defects in a batch of fabric, claims in a week and visits to a web page in a minute are the same kind of count. The Poisson distribution models them when three things hold:

• A steady average rate, \(\lambda\) (lambda): the mean number of events in the interval, such as 4 calls an hour.
• Independent events: one call does not make the next more or less likely.
• One at a time: two events never happen at exactly the same instant.

The count can be 0, 1, 2 and so on with no upper limit. That is still a discrete distribution: discrete means the values can be listed one by one, not that there are finitely many of them.

The formula

A help desk receives on average 4 calls an hour, so \(X \sim \text{Poisson}(4)\). The chance of a quiet hour with no calls at all is \(e^{-4} = 0.0183\). The chance of exactly 4 calls, the average, is only \(\tfrac{e^{-4}\, 4^4}{4!} = 0.1954\): the average is the likeliest count, but most hours have some other number. At most 2 calls has probability \(P(X \le 2) = 0.2381\), and a busy hour of 7 or more calls \(P(X \ge 7) = 1 - F(6) = 0.1107\), about one hour in nine.

The mean equals the variance

With 4 calls an hour, the standard deviation is \(\sqrt{4} = 2\) calls. That the variance equals the mean is a check on the model: if the counts a desk records swing far more than their mean suggests, the calls are not arriving independently at a steady rate, and the Poisson understates the busy hours.

The rate scales with the interval. Half an hour at 4 calls an hour has \(\lambda = 2\), and the chance of no call in half an hour is \(e^{-2} = 0.1353\); two hours have \(\lambda = 8\). Always match \(\lambda\) to the interval the question asks about.

Defects per batch

A production line finds on average 1.5 defects in a batch of 500 items. With \(X \sim \text{Poisson}(1.5)\), a batch is free of defects with probability \(e^{-1.5} = 0.2231\), has at most 2 defects with probability \(P(X \le 2) = 0.8088\), and has more than 3, which sends it back for inspection, with probability \(P(X \ge 4) = 1 - F(3) = 0.0656\).

The Poisson as the limit of the binomial

The defects can also be seen as a binomial: each of the 500 items is defective with a small probability \(p = 0.003\), independently, so \(np = 1.5\). When \(n\) is large and \(p\) small, the binomial and the Poisson with \(\lambda = np\) give almost the same answers:

This is why the Poisson fits rare events among many chances: faults among many components, claims among many policies, typing errors among many words. Nobody needs to know how many chances there were, only the average count.

When the Poisson is the wrong tool

• Events that cluster. After an outage, calls come in a rush; one claim after a storm brings many. The events are not independent, and the counts vary more than the mean.
• A rate that changes. A café's lunch hour and its quiet afternoon have different rates; one λ for the whole day misstates both. Model each period with its own rate.
• A ceiling. A 12-seat minibus cannot carry 13 passengers. When the count has a fixed maximum \(n\), the binomial is the model.
• Variance far from the mean. Recorded counts whose variance is much larger (or smaller) than their mean are telling you the Poisson does not fit.

In code

As with the binomial, "at least k" is the survival function at k − 1: poisson.sf(k - 1, λ).