Discrete Stochastic Processes

Lesson 10 of 10

The Poisson Process

Arrivals at random: Poisson counts, exponential gaps that forget the past, and streams you merge by adding rates and split by scaling them.

Counting arrivals

A Poisson process with rate λ\lambda counts arrivals over time. N(t)N(t) is the number of arrivals in [0,t][0, t], and it follows two rules: counts in disjoint windows are independent, and the count in any window of length tt is Poisson with mean λt\lambda t:

P(N(t)=k)=e−λt(λt)kk!.P(N(t) = k) = \frac{e^{-\lambda t}(\lambda t)^k}{k!}.

You can build it from coin flips. Cut [0,t][0, t] into nn tiny slots and let each slot hold an arrival with probability λt/n\lambda t/n, independently. The count is Binomial(n,λt/n)\text{Binomial}(n, \lambda t/n), and as n→∞n \to \infty it becomes Poisson(λt)\text{Poisson}(\lambda t), the limit from 3.0.6. The Poisson process is that same coin-flip grid run in continuous time.

Worked example: market orders hit a book at λ=4\lambda = 4 per minute. In one minute, P(N=4)=e−444/4!≈0.195P(N = 4) = e^{-4}4^4/4! \approx 0.195. The counts 3 and 4 tie as the most likely values, and each still happens less than a fifth of the time. In a 30-second window the mean is 22, and P(no orders)=e−2≈0.135P(\text{no orders}) = e^{-2} \approx 0.135. Mean and variance are both λt\lambda t, so over an hour you expect 240 orders, give or take about 240≈15.5\sqrt{240} \approx 15.5.