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Counting rare events over a stretch, the one parameter that scales with it, and the four conditions the model rests on.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will scale a rate to the stretch a question asks about, use the Poisson mass function and the ratio between consecutive terms, and check the four conditions the model rests on before using it. You will also estimate the rate from a record, which is the direction the model is used in outside a textbook.
A binomial count needs a number of trials. Many real counts have no such number: there is no list of occasions on which a typing error could have been made. The Poisson model is what the binomial becomes when the trials are numerous beyond counting and each is vanishingly unlikely.
A rate is an average count per unit of time, length or area; the stretch is how much of it a question asks about. $\lambda$ is the parameter of the model — the average count over that stretch — and it scales with the stretch, so it belongs to the question rather than to the apparatus. Arrivals is the general word for the events being counted.
A Poisson variable counts events over a stretch of time, length or area:
$$P(X = k) = \frac{e^{-\lambda} \lambda^{k}}{k!}, \qquad k = 0, 1, 2, \ldots$$
There is one parameter, $\lambda$, and it is the average number of events over the stretch you are asking about. Double the stretch and it doubles; that is what makes it a rate, and quoting it without naming the stretch says nothing.
The model earns its place by arriving as a limit. Split an hour into $n$ tiny intervals, each holding an event with probability $\lambda / n$ independently. That count is $\mathrm{Bin}(n, \lambda/n)$, and as $n \to \infty$ its mass function converges to the formula above. Four conditions come with it: the rate is steady, events arrive independently, no two arrive at the same instant, and the quantity is a count of whole events.
The formula's shape is easiest to see through the ratio of consecutive terms, $$\frac{P(X = k)}{P(X = k-1)} = \frac{\lambda}{k},$$ which is above one while $k < \lambda$ and below it afterwards. So the mass function climbs to a peak near $\lambda$ and then falls away, faster and faster.
Another way: story
A book has a hundred thousand characters and a typesetter errs on roughly one in twenty thousand. Nobody can list the occasions for an error, and nobody needs to: the count of errors on a page behaves like a Poisson variable whose parameter is the average number of errors per page, and that one number is the whole model.
Another way: steps
The ratio $\lambda / k$ means one value of the mass function gives all the others. With $\lambda = 3$:
| $k$ | Ratio to the term before | $P(X = k)$ |
|---|---|---|
| 0 | — | $e^{-3} \approx 0.0498$ |
| 1 | $3/1 = 3$ | $0.1494$ |
| 2 | $3/2 = 1.5$ | $0.2240$ |
| 3 | $3/3 = 1$ | $0.2240$ |
| 4 | $3/4 = 0.75$ | $0.1680$ |
The ratio passing through $1$ at $k = \lambda$ is why the peak sits there, and why $P(X = 2)$ and $P(X = 3)$ are equal whenever $\lambda = 3$. None of that needed the exponential, which cancels between consecutive terms.
Forgetting to scale the rate. A rate of $3$ an hour gives $\lambda = 12$ over four hours, not $3$. This is the commonest error in the lesson by a wide margin.
Using the model where the rate moves. Traffic across a morning that includes a rush is not Poisson; two Poisson models over two stretches might be.
Using it where events arrive in clumps. A coachload of passengers is one arrival of many people, and the model wants many arrivals of one.
Counting the wrong thing. A waiting time is not a count. The gaps between Poisson arrivals have a distribution of their own — the exponential — and it is a different object.
A helpline takes $2$ calls an hour on average. What is the parameter for a three-hour shift?
Ask what stretch the question is about.
$\lambda = 2 \times 3 = 6$ calls.
The rate multiplied by the stretch.
So $P(\text{no calls all shift}) = e^{-6} \approx 0.0025$, not $e^{-2}$.
Scaling first, formula second.
With $\lambda = 4$, suppose $P(X = 3)$ is known.
The ratio gives the neighbours.
$P(X = 4) = P(X = 3) \times 4/4 = P(X = 3)$: equal, because $\lambda$ is a whole number.
The ratio is exactly one there.
$P(X = 5) = P(X = 4) \times 4/5$, smaller.
Past the peak the terms fall.
$\lambda = 0.5 \times 6 = 3$ over the six metres.
Scale the rate to the stretch.
The ratio of consecutive terms is $\lambda / k$.
With $k = 2$ here.
So $P(X = 2) / P(X = 1) = 3/2 = 1.5$.
A Poisson model has parameter $\lambda = 4$ over one hour. Give the ratio of each probability to the one below it.
| Ratio | |
|---|---|
| $P(X = 1) \div P(X = 0)$ | |
| $P(X = 2) \div P(X = 1)$ | |
| $P(X = 4) \div P(X = 3)$ |
Calls reach a switchboard at an average of $8$ an hour, and the arrivals are modelled as Poisson. What is the parameter of the count over $5$ hours?
Answer:
Which of these counts is best modelled as Poisson with a rate of about $4$ per hour?
Faults appear in a cable at an average of $6$ per kilometre, modelled as Poisson. Consider a stretch of $5$ kilometres.
Parameter over the stretch: p. Ratio $P(X = 2) \div P(X = 1)$: q.
A machine logged $27$ stoppages over $3$ shifts, and the stoppages are being modelled as Poisson. Give the rate parameter per shift.
Stoppages logged shift by shift
Rate per shift:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The Poisson model rests on four conditions. Match each to the situation it rules out.
| A morning that includes a rush and a lull | Passengers who arrive together off one coach | Two machines that always fail in the same instant | The temperature of a room, measured each hour | |
|---|---|---|---|---|
| The rate is the same throughout the stretch | ||||
| Events arrive independently of one another | ||||
| No two events arrive at the same instant | ||||
| The quantity is a count of whole events |
You can scale a rate to a stretch, step between consecutive Poisson probabilities with the ratio, and name which condition fails when the model does not fit. Say in your own words why a coachload of arrivals breaks the model even though it is still a count of people.
9. Your turn: faults occur at $0.5$ per metre of cable; find the parameter for a six-metre length, and the ratio of two faults to one, step 3