6.1 - The Poisson distribution

6.1 - The Poisson distribution

A random count can have a steady average while varying considerably from one interval to the next. Learn when a Poisson model is appropriate, calculate probabilities over different intervals, and use its single parameter to describe both its centre and spread.

Events at a constant rate

If a telephone line receives an average of two calls per minute, must every minute contain two calls? No: a steady average describes the process over many intervals, not a schedule for individual events. Some minutes can have no calls and others several.

A binomial model starts with a fixed number of trials. Here we fix a length of time and count events that can occur anywhere within it. There is no prescribed maximum count.

Poisson model

A Poisson distribution models the number of events in a fixed interval of time, length or area when events occur independently, singly and at a constant average rate. Write X∼Po⁡(λ)X\sim\operatorname{Po}(\lambda), where λ>0\lambda>0 is the expected count in that interval. Its possible counts are 0,1,2,…0,1,2,\ldots.

“Constant rate” does not mean equally spaced events. It means that equal-sized intervals have the same expected count. “Independently” means that an event does not make further events more or less likely; counts in disjoint intervals are independent under this model. “Singly” rules out events that necessarily arrive in batches. In a sufficiently short interval, the chance of two or more events is negligible compared with the chance of one.

Worked example 1: a plausible random count

Question. A detector records individual particles arriving independently. During a stable experiment their average arrival rate is constant. Explain why a Poisson model could describe the number detected in a fixed minute.

Plan. Match the way events occur to the model assumptions.

The variable counts individual arrivals in a fixed interval. Arrivals are independent, occur singly and have a constant average rate during the experiment.

Answer. A Poisson model is appropriate under these stated assumptions.

Check. The model allows 00 arrivals or several arrivals in a minute. It does not require every minute to contain exactly the mean count.

Keep the same detector, but change the stability of the source. The average rate now changes within the interval being modelled.

Worked example 2: a changing arrival rate

Question. The source is changed so that its average arrival rate increases steadily during a ten-minute experiment. Arrivals remain independent and single. Is a single constant-rate Poisson process a suitable model throughout that experiment?

Plan. Check the assumption that changed rather than repeating the assumptions that still hold.

Two equal one-minute intervals, one near the start and one near the end, have different expected counts. The constant-rate assumption fails.

Answer. The constant-rate model is unsuitable throughout the experiment; the varying rate needs to be accounted for.

Check. Multiplying the initial per-minute rate by ten would underestimate the expected total. Independence alone does not justify that calculation.

For a real situation, describe the relevant assumption in context. Customers arriving together in family groups can violate single, independent arrivals. A daily rush can violate constant rate. Restricting attention to a suitably stable period may make a model more reasonable.

Watch out

An average of two events per minute does not mean an event occurs exactly every half minute. Regularly timed arrivals and random independent arrivals are different processes. Nor does a sample mean close to a sample variance, by itself, establish the modelling assumptions.

Choose the statement that actually describes this model.

Calculating probabilities

The Poisson parameter belongs to a particular interval. A rate of two calls per minute gives an expected count of three calls in ninety seconds, so the ninety-second count has parameter 33, not 22.

Poisson point probability

If X∼Po⁡(λ)X\sim\operatorname{Po}(\lambda), then

P(X=r)=e−λλrr!,r=0,1,2,….\mathrm P(X=r)=\frac{\mathrm e^{-\lambda}\lambda^r}{r!}, \qquad r=0,1,2,\ldots.

Here 0!=10!=1, so P(X=0)=e−λ\mathrm P(X=0)=\mathrm e^{-\lambda}.
In the formula booklet.

For a constant rate kk per unit interval and an interval of size tt, use λ=kt\lambda=kt. Convert units before multiplying. A cumulative probability is a sum of point probabilities; a tail can often be found more efficiently by subtraction from 11.

Worked example 3: changing the interval

Question. Calls arrive singly and independently at a constant average rate of two per minute. Find the probability of exactly three calls in ninety seconds.

Plan. Find the mean count for 1.51.5 minutes, then use the Poisson point formula.

λ=2(1.5)=3P(X=3)=e−3333!=0.2240418…\begin{aligned} \lambda&=2(1.5)=3\\ \mathrm P(X=3)&=\frac{\mathrm e^{-3}3^3}{3!}\\ &=0.2240418\ldots \end{aligned}

Answer. 0.2240.224, to 33 significant figures.

Check. Ninety seconds is 1.51.5 times the stated interval, so its expected count is 1.51.5 times 22. The point-probability recurrence gives P(X=3)=P(X=2)(3/3)\mathrm P(X=3)=\mathrm P(X=2)(3/3), the same as e−39/2\mathrm e^{-3}9/2.

Keep the same interval but change “exactly three” to “at least three”. This includes every count from three upwards, not just one count.

Worked example 4: a complementary tail

Question. For the same arrival process, find the probability of at least three calls in ninety seconds.

Plan. Subtract the probabilities of the excluded counts 0,1,20,1,2 from 11.

With X∼Po⁡(3)X\sim\operatorname{Po}(3),

P(X≥3)=1−P(X≤2)=1−e−3(1+3+322!)=0.5768099…\begin{aligned} \mathrm P(X\ge3)&=1-\mathrm P(X\le2)\\ &=1-\mathrm e^{-3}\left(1+3+\frac{3^2}{2!}\right)\\ &=0.5768099\ldots \end{aligned}

Answer. 0.5770.577.

Check. The excluded probabilities sum to 0.4231901…0.4231901\ldots; together with the answer they give 11. The result exceeds the probability of exactly three, as it should.

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