The Poisson Process: Counting Random Arrivals

A Poisson process counts arrivals. The tally starts at zero, steps up by exactly one at each arrival, treats counts over stretches that do not overlap as independent, and makes the count over a stretch depend on the length of that stretch and on nothing else. Its waiting time between arrivals is memoryless, so having waited a long time says nothing at all about how much longer the wait will be.

Everything in this guide follows from those four lines, and the last sentence of that answer is the one that costs people money and sleep. The claim sounds wrong. The claim sounds like a mistake in the arithmetic. A reader who has waited two years for something to happen feels, quite reasonably, that the thing is now overdue. The Poisson process says the feeling is not merely unsupported but exactly backwards, and it says so with an identity that cancels to the sixth decimal rather than with an approximation. The identity is worked through below rather than asserted, and the price of assuming it is named at the end.

What does a Poisson process count, and what four properties fix it?

Begin with the object itself. A counting processA process that records how many arrivals have happened so far, so it never falls and only ever moves upward. is a running tally. The tally sits at zero, it waits, and at certain moments it goes up by one. Something that has happened cannot be un-counted, so the tally never comes back down. A step of a half, or of three at once, never happens. Between arrivals it is completely flat, so its picture is a staircase with treads of uneven width and risers of exactly one.

One thing is missing from that description. There is no mention of how big anything is. If arrivals are deliveries, the tally says four deliveries came and says nothing about what was in them. The separation of how often from how large is doing real work in this subject area: how often is one specification, how large is a completely different one, and that split is established under the specification of arrival size. The Poisson process is entirely about the how often.

The specification that turns a generic tally into the Poisson process is four lines long. Brownian motion is specified by four lines answering the same four questions, so the two are worth reading side by side. Where does it start. Do separate stretches talk to each other. How does a stretch of a given length behave. And what can happen in a single instant. The continuous case answers the last question with never jumps; the counting case answers it with jumps by exactly one, never by two.

One difference changes how the two objects are used, so it is worth flagging straight away. The continuous specification has nothing to choose: write the four lines and the process is pinned down completely. The counting specification has one number left over, the rate at which arrivals come. The rate is the entire freedom in a Poisson process, and every fact below is a consequence of the rate and of nothing else.

The distribution named in line three carries the name of Poisson, who worked out the counting law that follows from those conditions. The name is part of the term and the term is not shortened here. Line four is the quiet one and it is doing more work than it looks: it is what makes this a counting process rather than a tally that occasionally leaps.

How does the rate turn into the chance of a given count?

The rate is fixed. Throughout this guide the intensityThe average number of arrivals per unit of time, written as a rate per year here. Intensity is the single parameter of a Poisson process. is 0.5 arrivals a year. The figure is made up, chosen so the arithmetic reconciles by hand. Over one year the average count is therefore 0.5, over two years it is 1.0, and over four years it is 2.0. Line three of the specification is doing its job. The average count scales in a straight line with the length of the window.

The chance of any particular count comes from three ingredients multiplied together: the rate times the length raised to the count, divided by the factorialThe product of every whole number up to the given one, so three factorial is one times two times three, which is six. Nought factorial is defined as one. of the count, all multiplied by the decaying factor that keeps the chances adding to one. Work it through row by row rather than reading the formula whole. Each row is two multiplications and a division.

The four rows add to 0.998248, so four or more arrivals in the year carry the remaining 0.001752 between them. The single most useful reading of that table is the top row: at this intensity the most likely outcome by a wide margin is that nothing happens all year, and its chance of 0.606531 is larger than every other outcome added together. The chance of at least one arrival is therefore one less 0.606531, or 0.393469. Computing the nothing case and subtracting is almost always shorter than adding up the rest.

Look at the shape rather than the numbers for a moment. At a rate of 0.5 a year the distribution leans hard against the left edge and falls away sharply. Raise the rate and the shape does not merely stretch; the peak lifts off zero and moves rightward, and the whole thing starts to look symmetric. At a rate of two a year the most likely counts are one and two, tied at 0.270671, and the chance of nothing has dropped to 0.135335.

The everyday version is drops of rain on one roof tile during light drizzle. Most seconds bring nothing. A second with one drop is the next most common thing. Two drops in the same second happen occasionally and three is rare. Turn the drizzle into a downpour and the tally per second stops being mostly zero and starts being mostly some number in the middle. Nothing about the mechanism changed; only the rate did.

How long is the wait between one arrival and the next?

The counting law answers how many. Asked the other way round, how long draws a second distribution out of the same specification for free. The gap between arrivals is the exponential waiting timeThe distribution of the gap between one arrival and the next, whose average is one divided by the rate., and there is a one line argument for it that needs no calculus at all.

Here it is. The waiting time is longer than some length only if the count over that length is zero. The two statements describe the identical event, so they carry the identical chance. The counting law already gave the chance of a zero count, so the chance of waiting longer than a given length is the same decaying factor written a second time. At a rate of 0.5 a year the chance of waiting more than one year is 0.606531, more than two years is 0.367879, and more than three years is 0.223130.

The average wait is one divided by the rate, so exactly two years here. The average and the halfway point are two different numbers, and the gap between them is a permanent feature of this waiting law rather than an artefact of these particular figures. The halfway point, the length by which half of all waits are over, is the natural logarithm of two divided by the rate, or 1.386294 years. So a wait of exactly the average length, two years, is already longer than most waits: the chance of exceeding it is 0.367879, not 0.500000.

Why does the elapsed wait cancel out completely?

Now the property that makes the Poisson process worth the trouble. Suppose two years have already passed with nothing arriving. How likely is at least one more year of waiting? Nobody should take the answer on trust, so the honest way is to write it as a conditional chance and grind it out.

The event in question is waiting more than three years in total. The information in hand is that more than two years have already been waited. A conditional chance is the chance of both divided by the chance of what is known, and here the first event sits inside the second, so the numerator is simply the chance of waiting more than three years. The numerator is therefore 0.223130. The denominator, the chance of waiting more than two years, is 0.367879. Dividing gives 0.606531.

Look at what that number is. The figure 0.606531 is exactly the chance of waiting more than one year from a standing start, and exactly the chance of no arrival at all during any single year at this rate, so two years of waiting bought precisely nothing. Not approximately nothing. The cancellation is algebraic rather than numerical, so the elapsed wait counts for nothing to the sixth decimal place and beyond.

Here is where the elapsed wait disappears, and it is worth staring at. A decaying factor over three years is the same decaying factor over two years multiplied by the one over the remaining year, so the numerator 0.223130 can be written as 0.367879 multiplied by 0.606531. The numerator sits on top of a denominator of 0.367879 and the shared factor cancels. The surviving factor depends only on the extra year in question. The elapsed wait was in both the top and the bottom, in identical form, so it could never have survived the division.

Why is this the property that readers refuse to believe?

Because the intuition it contradicts is usually correct. Almost everything that waits in the physical world does age. A filament in a bulb thins each hour it burns, so a bulb that has run five thousand hours really is closer to failing than a new one. A tyre wears. A promise with a deadline gets closer to its deadline. In every one of those cases the elapsed time is genuine information and using it is correct.

So the reader is not being stupid. The reader is applying a rule that works nearly everywhere and meeting one of the few objects built to have no such rule. Being memorylessHaving waited a while says nothing at all about how much longer the wait will be, so the elapsed wait carries no information. is not a claim about the world; it is a property of a model that has been chosen, and choosing it declares that elapsed waiting carries no information whatever.

The sharpest way to feel the difference is to hold the average wait constant and change nothing else. Consider a second waiting law: an arrival that is certain to come within four years, equally likely at any moment in that window. Its average wait is also exactly two years. The two laws agree on the single number most people quote, and they disagree about everything that matters once the waiting starts.

Now compare what each says to somebody who has already been waiting. The hazard rateThe rate at which an arrival is about to happen given that it has not happened yet, which is what people mean when they say something is overdue. is the honest way to ask whether pressure builds. For the memoryless law it is the rate itself, 0.5 a year, at every elapsed wait without exception. For the four year deadline the hazard rate climbs. The remaining window shrinks while the arrival is still certain to land inside it.

The crossing point is worth naming. Before 1.458506 years the memoryless law is actually the more pessimistic of the two. The memoryless law allows an arrival immediately while the deadline law has plenty of window left. After that point they swap places and the gap widens without limit. Somebody who checks a model at one elapsed wait and concludes it looks sensible may simply have checked it near the crossing.

How does the counting process drive the jumps of a jump process?

The counting process on its own moves by one and has no units. The counting process becomes useful when something else is told to move whenever the tally moves. The tally supplies the when; a separate specification, covered under the size of an arrival, supplies the how much. The two together give a path that sits still and then breaks.

The statement here is not about the standard process itself. Write the driven path with the letter reserved for a general process. The path multiplies by one factor at the first arrival, by another at the second, and so on. The upper limit of that product is the tally, the only place the counting process enters. The staircase and the broken path are the same object drawn twice: identical arrival times, one panel showing that something happened and the other showing what it did.

The picture below uses four arrival times built by a rule rather than drawn at random, so it reproduces on every reload. The rule cuts the chance scale into four equal slices and takes the wait at the middle of each. The gaps come out at 0.267063, 0.940007, 1.961659 and 4.158883 years and the arrivals therefore at 0.267063, 1.207070, 3.168729 and 7.327612 years. The four gaps average 1.831903 years rather than the true 2.000000, and the shortfall is the honest cost of slicing at midpoints rather than sampling. The four proportional moves come from the same slicing applied to the size specification, giving factors of 0.847864, 0.921397, 0.982027 and 1.067196, so a path opening at Rs 100/- passes through Rs 84.79/-, Rs 78.12/- and Rs 76.72/- before closing at Rs 81.87/-.

What does this process rule out, and when is that the wrong assumption?

Two exclusions sit inside the four properties, and neither is a technicality. Line four rules out simultaneous arrivalsTwo arrivals landing at the very same instant, which the specification forbids by allowing steps of one only.: the tally can never step by two, so two arrivals can never share a moment. Line two rules out clusteringArrivals bunching together, so that one arrival makes another more likely soon after.: separate stretches are independent by assumption, so an arrival cannot make the next one more likely.

Both exclusions fail in situations that can be named. A single common cause that trips several counters at once produces genuine ties: one power cut stops every machine on the floor in the same second, and the tally would have to step by however many machines there were. Aftershocks are the standard case of the second failure. One event demonstrably raises the rate of the next for a while. Neither of those is exotic, and neither is ruled out by evidence. The specification that was written down rules out both. Both are therefore assumptions to be argued for rather than results to be reported.

A third exclusion is quieter and catches people more often. The rate is a constant. Nothing in the four properties allows it to depend on the calendar, so a mechanism that is busier in one season than another has already left this process. There are richer constructions that let the rate vary or let arrivals excite one another, and each of them exists precisely because one of these three exclusions was unacceptable somewhere.

How does somebody checking another person's model use this?

Very few people derive any of this at work. Plenty are handed a model that counts something and have to decide whether the counting assumption survives contact with the thing being counted. Four checks do most of the work, and none of them needs the code.

1. Does the variance of the counts match their average? The counting law fixes both at the rate times the length of the window, so their ratio, the index of dispersionThe variance of the counts divided by their average. A Poisson process fixes it at one, so a reading well above one is a signal of bunching., is one. Over four years at this rate the average count is 2.000000 and the standard deviation is 1.414214.
   A ratio well above one is the fingerprint of bunching. A ratio well below one says arrivals are more regular than this process allows.
2. Does the average count scale in a straight line with the window? Half an arrival in a year has to mean one arrival in two years and two in four. Nothing else is permitted.
   If doubling the window does not double the average count, either the rate is moving or the counts are not independent.
3. Can two records share a timestamp? Line four says no. If the underlying records routinely carry ties, the model is being fitted to data it forbids.
   Look for repeated timestamps before believing any fitted rate at all.
4. Does the elapsed wait appear anywhere in the reasoning around the model? Senior people get caught by this check more than any other. A memorandum that fits this process and then argues that an arrival is due because none has come for a while has contradicted its own model in the same document.
   The model says the elapsed wait is worth nothing. If the commentary says otherwise, one of the two has to go.

The fourth check is the most valuable because it costs nothing to run. The check needs no data, no code and no fitted rate. One reading of the words around the numbers is enough to show whether they assume something the arithmetic has already ruled out.

One last note on scope prevents a wrong inference. No jurisdiction sets the definition of a process. The four lines are the same four lines everywhere, and the cancellation that makes the waiting time memoryless is the same cancellation in every country and every decade.

The standard process, the rate of 0.5 arrivals a year, the four arrival times and the four proportional moves are invented.
Educational material. Not advice on any investment, tax, budget or market position.