The Poisson Process

The The Poisson Distribution counts how many rare events land in one fixed window. The Poisson process zooms out and watches the whole timeline: events popping up one after another — orders hitting a book, trades printing, defaults in a loan pool — arriving at random moments but at a steady long-run rate. It is the default model for "things that show up unpredictably but at a known average pace."

The whole process is controlled by a single number, the rate $\lambda$ (lambda): the average number of events per unit of time. From that one number everything else follows.

Two views of the same process

There are two ways to describe the same stream of arrivals, and knowing both is the whole trick.

Counting view. Let $N(t)$ be the number of events that have happened up to time $t$. For a Poisson process, the count in a window of length $t$ is Poisson-distributed with mean $\lambda t$:

$$P\big(N(t) = k\big) = e^{-\lambda t}\,\frac{(\lambda t)^k}{k!}.$$

Here $\lambda t$ is the expected number of events in a window of length $t$, and $k$ is the count you are asking about. Counts in non-overlapping windows are independent — the morning tells you nothing about the afternoon.

Waiting-time view. Instead of counting, ask how long you wait between events. The gaps between consecutive arrivals are independent and follow the The Exponential Distribution with rate $\lambda$:

$$P(\text{gap} > t) = e^{-\lambda t}.$$

So the average gap is $1/\lambda$, and the chance of waiting longer than $t$ decays exponentially.

Worked example: waiting on the next order

A trading desk receives large orders as a Poisson process at $\lambda = 3$ per minute. You just handled one. What is the chance the next order takes more than 40 seconds?

Convert the rate to seconds: $\lambda = 3/60 = 0.05$ per second. The waiting time is exponential, so

$$P(\text{gap} > 40) = e^{-\lambda \cdot 40} = e^{-0.05 \times 40} = e^{-2} \approx 0.135.$$

About a 14% chance of a lull longer than 40 seconds. The average gap is $1/\lambda = 1/0.05 = 20$ seconds, and the expected number of orders in a full minute is simply $\lambda \times 1 = 3$, matching the Poisson count.

The memoryless twist

Here is the property that trips people up. Suppose you have already been waiting 30 seconds with no order. How much longer should you expect to wait? The answer is still 20 seconds — exactly as if you had just started. The exponential gap is memoryless: the process does not "get due." A long dry spell does not make the next arrival any sooner.

Where it misleads

• Real order flow clusters. Poisson assumes events never trigger one another, but a large trade often provokes more trades, and news arrives in bursts. When arrivals excite further arrivals, the Hawkes process — a self-exciting cousin — fits far better, and the true tails are much heavier.
• The rate rarely stays flat. $\lambda$ drifts across the day: heavy at the open and close, quiet at lunch. A single constant rate smears over that structure.
• Merging and thinning are clean, though. Two independent Poisson streams combined form another Poisson process whose rate is just the sum, and randomly keeping a fraction $p$ of events gives a Poisson process with rate $p\lambda$. This additivity is why the model is so convenient when it does apply.

The Poisson process is the bridge between the The Poisson Distribution (the counts) and the The Exponential Distribution (the gaps), and it is the simplest continuous-time Markov arrival model underneath much of queueing theory and order-book modelling.