Poisson Models for Event Counts, statistics lesson

A help desk wants to know how likely a quiet ten-minute stretch is. Requests do not arrive through a fixed number of yes or no trials. There is only a length of time, and any number of requests can arrive in it. The Poisson distribution models counts like this, and its parameter is the expected count over the stated window. This lesson connects that parameter to a rate and an exposure, states the process assumptions behind the model, and shows how to notice counts that vary more than the model allows.

Key terms

Poisson distribution

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

P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}

for $k = 0, 1, 2, \dots$ , where $e \approx 2.71828$ is the exponential constant and $0! = 1$ . The support is the set of all nonnegative integers, with no upper endpoint, and the probabilities add to $1$ .

In plain words A Poisson count records how many events happen in a stated window of time, area or other exposure. There is no fixed number of trials and no largest possible count. A binomial count, by contrast, counts successes in $n$ fixed trials and never exceeds $n$ . The two models can give similar numbers. As $n$ grows and $p$ shrinks with $np$ held at $\lambda$ , the $\operatorname{Binomial}(n, p)$ probabilities approach the $\operatorname{Poisson}(\lambda)$ probabilities. For any particular $n$ and $p$ this is an approximation whose accuracy should be checked, not an identity.

Example For $\lambda = 3$ the probabilities, rounded to four decimals, are below.

$k$	$0$	$1$	$2$	$3$	$4$	$5$	$6$ or more
$P(X = k)$	$0.0498$	$0.1494$	$0.2240$	$0.2240$	$0.1680$	$0.1008$	$0.0839$

For instance $P(X = 0) = e^{-3} \approx 0.0498$ and $P(X = 2) = e^{-3} \cdot 9/2 \approx 0.2240$ . A binomial count with $n = 100$ and $p = 0.03$ also has mean $3$ , and its probability of no successes is $0.97^{100} \approx 0.0476$ , close to $0.0498$ but not equal.

Exposure and event rate

Definition The exposure of a count is the amount of opportunity over which events are counted, such as a length of time, an area, a volume or a number of person-years, stated with its unit. The event rate is the expected number of events per unit of exposure, with units such as events per minute. When events occur at a constant rate $r$ over an exposure $t$ , the expected count is $rt$ . An observed count divided by its exposure estimates the rate.

In plain words A count says little until you know how much opportunity produced it. Two counts become comparable as rates only after each is divided by its own exposure. A rate always carries units, and the same rate has different numbers in different units.

Example One help desk logs 12 requests in 60 minutes and another logs 9 requests in 30 minutes. The first count is larger, but the observed rates are $12/60 = 0.2$ and $9/30 = 0.3$ requests per minute, so the second desk had the higher observed rate. A rate of $0.3$ requests per minute is the same as $18$ requests per hour.

Poisson parameter

Definition The parameter $\lambda$ of a Poisson count is its expected value, the expected number of events in the stated exposure. It is a number of events, not a rate per unit of time. When events occur at a constant rate $r$ over an exposure $t$ ,

\lambda = rt.

In plain words One process gives different Poisson models for windows of different lengths. Under a constant rate a longer window has a larger parameter, and its probabilities change through the formula, not in proportion to the window length. A value of $\lambda$ quoted without its exposure does not say which count the model describes.

Example Assume that requests arrive at a constant rate of $0.3$ per minute and that the count in any window is Poisson. A 10-minute window then has

\begin{aligned} \lambda &= 0.3 \times 10 \\[0.4em] &= 3 \text{ requests}, \end{aligned}

and the probability of no requests is $e^{-3} \approx 0.0498$ . A 20-minute window under the same constant rate has $\lambda = 6$ , and its probability of no requests is $e^{-6} \approx 0.00248$ , which is neither double nor half the 10-minute value.

Independent increments

Definition A homogeneous Poisson process with rate $r > 0$ places events in time so that the count in every interval of length $t > 0$ has the $\operatorname{Poisson}(rt)$ distribution, and the counts in nonoverlapping intervals are independent. The second property is called independent increments. Homogeneous means that the rate stays constant over the period being modeled. In such a process, the probability that two events occur at exactly the same instant is $0$ .

In plain words Under this model, the count in one stretch of time carries no information about the count in a separate stretch. That rules out bursts in which one event makes more events likely soon after. The constant rate is a separate assumption, and it rules out a rate that rises and falls during the period. A single total that looks Poisson does not by itself show that the process behind it has these properties, so state the process assumptions before applying the model to other windows.

Example If requests follow a homogeneous Poisson process at $0.3$ per minute, the counts in minutes $0$ to $10$ and in minutes $10$ to $20$ are independent $\operatorname{Poisson}(3)$ counts. No requests in 20 minutes means no requests in either half, so its probability is

e^{-3} \cdot e^{-3} = e^{-6} \approx 0.00248,

which agrees with the $\operatorname{Poisson}(6)$ model for the full 20 minutes. If outages sometimes sent a stream of users to the desk, a busy first interval would suggest an outage under way and make a busy second interval more likely, and this product would no longer be justified.

Mean-variance equality

Definition If $X \sim \operatorname{Poisson}(\lambda)$ , then

\begin{gathered} E(X) = \lambda, \\[0.4em] \operatorname{Var}(X) = \lambda, \\[0.4em] \operatorname{SD}(X) = \sqrt{\lambda}. \end{gathered}

In plain words One number sets both the center and the spread. The mean and the variance are numerically equal but describe different things. The mean locates the center of the count distribution, and the variance is the expected squared distance from that center. Since the SD is $\sqrt{\lambda}$ , a larger expected count has more absolute spread but less spread relative to its mean.

Example For $\lambda = 3$ the mean and the variance are both $3$ , and the SD is $\sqrt{3} \approx 1.73$ requests. For the 20-minute window with $\lambda = 6$ , the mean and the variance are $6$ and the SD is $\sqrt{6} \approx 2.45$ requests. Doubling the window doubles the variance but multiplies the SD only by $\sqrt{2}$ .

Overdispersion

Definition Overdispersion is variability larger than a specified model predicts. A Poisson model makes the variance equal to the mean, so counts are overdispersed relative to it when their variance exceeds their mean. In data, the check compares the sample variance $s^2$ of counts recorded over equal exposures with their sample mean $\bar{x}$ .

In plain words Extra variability usually has a cause, such as a rate that differs from one window to another or events that arrive in clusters. A modest gap between $s^2$ and $\bar{x}$ is not proof, because both fluctuate from sample to sample, especially in short records. A large gap is a reason to look at how the counts were produced before relying on Poisson probabilities.

Example Suppose quiet days have a rate of $1$ report per day and busy days a rate of $5$ , each kind of day is equally likely, and given its rate the count on a day is Poisson. The mean count is $0.5 \cdot 1 + 0.5 \cdot 5 = 3$ . A Poisson count satisfies $E(X^2) = \lambda + \lambda^2$ , so the second moment is $0.5 \cdot 2 + 0.5 \cdot 30 = 16$ and the variance is

16 - 3^2 = 7.

A single Poisson model with mean $3$ would have variance $3$ , so the mixture of quiet and busy days is overdispersed.

Check yourself

Events occur according to a homogeneous Poisson process with rate $0.4$ events per minute. Find the Poisson parameter for a 5-minute window, the probability of exactly one event and the probability of at least one event.

Show the answer

The parameter is $\lambda = 0.4 \times 5 = 2$ events. Then

\begin{aligned} P(X = 1) &= \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2} \\[0.4em] &\approx 0.2707, \end{aligned}

and the probability of at least one event is $1 - e^{-2} \approx 0.8647$ .

Calls arrive according to a homogeneous Poisson process with rate $2$ calls per hour. Someone computes the probability of no calls in 30 minutes as $e^{-2} \approx 0.135$ , and the probability of no calls in 90 minutes as three times the 30-minute answer. Find the correct values.

Show the answer

The parameter belongs to the window. Thirty minutes is half an hour, so $\lambda = 2 \times 0.5 = 1$ and the probability of no calls is $e^{-1} \approx 0.3679$ . The value $e^{-2}$ is the answer for a full hour. For 90 minutes $\lambda = 3$ , and the probability of no calls is $e^{-3} \approx 0.0498$ .

Tripling even the correct 30-minute answer gives about $1.10$ , which cannot be a probability. By independent increments, no calls in 90 minutes means no calls in each of three separate 30-minute intervals, so the probabilities multiply, and $(e^{-1})^3 = e^{-3}$ .

A team records the number of reports on eight days, 0, 1, 0, 9, 0, 1, 8 and 1. Is a Poisson model with one constant daily rate reasonable for these counts?

Show the answer

The sample mean is $\bar{x} = 20/8 = 2.5$ . The squared deviations from $2.5$ add to $98$ , so the sample variance is $s^2 = 98/7 = 14$ . A constant-rate Poisson model makes the variance equal to the mean, so apart from sampling fluctuation the sample variance should be near $2.5$ . Here it is $5.6$ times as large.

The two large days point the same way. Under $\operatorname{Poisson}(2.5)$ a day with 8 or more reports has probability about $0.0042$ , and the chance that at least two of eight independent days reach that level is about $0.0005$ . The counts look overdispersed, perhaps because reports cluster or the rate changes from day to day. The data alone do not say which, so the next step is to learn what happened on the fourth and seventh days before using Poisson probabilities.

Poisson Models for Event Counts

Key terms

Poisson distribution

Exposure and event rate

Poisson parameter

Independent increments

Mean-variance equality

Overdispersion

Check yourself

How this lesson connects

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