Poisson process

You will learn: Connect arrivals, gaps, and independent marking under a homogeneous process.

Start with: Poisson distribution · Exponential distribution

How can one sequence of arrivals answer both “how many by time T?” and “how long until arrival k?” A homogeneous Poisson process connects these questions. Follow a single path, divide its time window, and independently retain some arrivals to see which properties survive.

Start with Poisson counts, exponential waiting times, and conditioning and independence. Here the rate λ means arrivals per time unit; λT is the expected count in a window of duration T.

Filled blue circles are retained arrivals; open red circles are removed arrivals. The dashed line splits the observation window. Increasing time preserves the existing path. Changing the retention probability reuses the same independent marks.

Count in this realization	Observed	Model mean and variance
All arrivals in (0, T]	9	10.0000
Retained arrivals	2	4.00000
Removed arrivals	7	6.00000
Original arrivals in (0, split]	1	5.00000
Original arrivals in (split, T]	8	5.00000

One event, expressed as a count or a wait

N(T) ≥ 3 is true on this path. Arrival 3 occurs at time 3.50388, within T. Both event descriptions have model probability 0.9972306.

An individual gap has mean 0.50000 time units. The uncompleted final gap is right-censored: no arrival occurred during the last 0.00051 time units. It is not a completed exponential observation.

Independent streams, conditional dependence

Before observing the total, the retained and removed counts are independent Poisson variables: covariance 0. Conditional on this total of 9, they sum to a fixed number and have covariance −n p(1−p) = -2.16000. Independence of the marks is essential; keeping every second arrival gives a different process.

Inspect all observed arrivals and gaps

Arrival	Time	Completed gap	Retained?
1	0.23335	0.23335	Yes
2	3.18403	2.95068	Yes
3	3.50388	0.31985	No
4	3.51344	0.00957	No
5	3.52951	0.01607	No
6	4.16403	0.63452	No
7	4.40885	0.24483	No
8	4.57259	0.16374	No
9	4.99949	0.42690	No

Rate per time unit 2.00 Observation time T 5.00 Retention probability 0.40 Split as fraction of T 0.50 Arrival number k 3.00

One synthetic homogeneous Poisson process. Exact model quantities describe repeated realizations; the single path illustrates their events.

Specify a process, not just a histogram

A homogeneous Poisson process starts with N(0)=0, has independent counts in disjoint intervals, and assigns a Poisson(λt) count to every interval of length t. “Homogeneous” says that moving a window without changing its duration does not change its count distribution. The event times are continuous, and simultaneous arrivals have probability zero under this model.

These assumptions give more information than a Poisson-shaped count histogram. Counts could have Poisson marginals yet be dependent across windows. An observed arrival stream with a daily cycle does not have a constant rate merely because its average count is stable.

Read the same outcome two ways

Write Xᵢ for the gap before arrival i and Sₖ for the time of arrival k. Independent Exponential(λ) gaps give:

S_k=\sum_{i=1}^kX_i,\qquad \{N(T)\ge k\}=\{S_k\le T\}.

The event identity holds path by path: the kth event has happened by T exactly when at least k events have happened. Thus a gamma waiting-time probability can be evaluated through a Poisson count:

P(S_k\le T)=1-e^{-\lambda T}\sum_{j=0}^{k-1}\frac{(\lambda T)^j}{j!}.

For rate 2 per hour and T=1 hour, the chance of at least three arrivals is 1−5e⁻² ≈ 0.323324. It is also the chance that the third arrival occurs within one hour. The first gap has mean 1/2 hour; the third arrival time has mean 3/2 hours. “Mean waiting time” needs an arrival number.

Make a prediction

If only two events have arrived by T, is the third waiting time zero, unknown, or greater than T?

Explore the answer

It is greater than T. The demo knows that inequality without treating the observation boundary as another arrival. The uncompleted final gap is right-censored, so averaging only completed gaps in a short window can favor shorter gaps.

Split time or split arrivals

At rate 2, a five-hour window has mean and variance 10. Splitting it at two hours gives independent Poisson counts with means 4 and 6. Their sum recovers the original count. The split location must be a fixed time for this simple independent-increment statement; choosing a boundary by inspecting the path changes the conditioning question.

Now classify each arrival independently as retained with probability p. For p=0.4, the retained process has rate 0.8 and the removed process rate 1.2. In five hours, their means and variances are 4 and 6. The streams are independent before conditioning on their total.

Why? If retained count R=r and removed count D=d, first draw the total r+d and then its independent classifications. Their joint probability factors:

P(R=r,D=d)=\frac{e^{-\lambda T}(\lambda T)^{r+d}}{(r+d)!}\binom{r+d}{r}p^r(1-p)^d.

Splitting the exponential and powers turns this into the product of Poisson(λTp) and Poisson(λT(1−p)) probabilities. Conversely, superposing independent Poisson streams adds their rates.

Given a fixed total n, however, R is Binomial(n,p) and D=n−R. Their conditional covariance is −np(1−p). At n=10 and p=0.4 it is −2.4. A retained event then uses one of a fixed number of available events. This conditional dependence is compatible with unconditional independence.

Make a prediction

Does keeping every second arrival give a Poisson process with half the rate?

Explore the answer

No. Its long-run rate is halved, but a retained-to-retained gap sums two original exponential gaps. That gap is Gamma(shape 2, rate λ), not exponential. Independent random classification is what makes Poisson thinning work.

Where to go next

The gamma distribution describes the kth arrival time. Total variance explains why mixing different underlying rates can make count variance exceed its mean. Conjugate priors treats the rate as uncertain after observing events and exposure. These are different extensions: changing a physical rate over time and being uncertain about a common fixed rate are different models.

Sources

Random Services: introduction to the Poisson process defines its gap, arrival-time, and count descriptions. Its thinning and superposition chapter develops independent classification. The numerical examples and shared-path experiment above instantiate those assumptions; they are synthetic, not evidence that an observed arrival stream satisfies them.

Try the historical earthquake-arrival case study to compare observed counts and gaps with fitted models. Return to learning paths and foundations.

The inspection paradox compares time-based observations with equally weighted intervals, and explains why a repeating timetable has a different remaining-wait law.

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