Nonhomogeneous Poisson process

Section of Poisson processes, part of Stochastic processes

Type	Section
Taught in	Poisson processes
Part	Stochastic processes

A non-homogeneous Poisson process has a deterministic rate \lambda(u)\ge 0, locally integrable. The cumulative intensity is \Lambda(t)=\int_0^t \lambda(u)\,du. Increments over disjoint intervals stay independent. Their means are differences of \Lambda, and stationarity of increments fails as soon as \lambda is not constant:

N(t)-N(s)\sim\mathrm{Poisson}\Bigl(\int_s^t \lambda(u)\,du\Bigr).

The waiting time W after time s has survival probability

\mathbb{P}(W>w)=\exp\Bigl(-\int_s^{s+w}\lambda(u)\,du\Bigr).

This is exponential in w only when \lambda is constant on the range of the integral. Successive waits, measured in calendar time, need not share a distribution.

One simulation draws a unit-rate Poisson process and reads it through the inverse of \Lambda, when that inverse exists. Another draws a homogeneous process of rate high enough to dominate \lambda(u), and keeps each point independently with probability \lambda(u) over that dominating rate. The second construction is thinning.

Example 1 (A rate that grows with the hour) Suppose \lambda(t)=2+t events per hour for t\ge 0.

Question. Find the law of N(2), the expected count from hour 1 to hour 2, and a draw of the first arrival time.

Solution. The cumulative intensity is \Lambda(t)=2t+t^2/2. Then \Lambda(2)=4+2=6 and \Lambda(1)=2.5, so N(2)\sim\mathrm{Poisson}(6) and \mathbb{P}\{N(2)=0\}=e^{-6}. The increment from 1 to 2 is \mathrm{Poisson}(6-2.5)=\mathrm{Poisson}(3.5). A homogeneous process that kept the initial rate 2 would predict mean 2 on that hour, which is the right calculation for a different rate function.

To draw the first arrival, take E\sim\mathrm{Exp}(1) and solve \Lambda(T_1)=E:

2T_1+\frac{T_1^2}{2}=E \qquad\Longrightarrow\qquad T_1=-2+\sqrt{4+2E},

the positive root. Later arrivals add independent unit-rate exponential amounts on the \Lambda scale and invert again. The clock that is homogeneous is \Lambda(t), not calendar time.

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Taught in Poisson processes
All 3 questions on Poisson processes

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