Time Change Construction for Nonhomogeneous Poisson Processes

The Cumulative Intensity as a Clock

A homogeneous Poisson process runs at a constant rate, so its natural clock is elapsed time. For a time-varying rate, the amount of Poisson activity accumulated by calendar time \(t\) is instead measured by the integral of the rate up to \(t\). This cumulative intensity provides a new clock: run a unit-rate Poisson process at that clock, and its increments have the desired means.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_0(u))_{u\ge0}\) be a unit-rate Poisson process, with natural filtration \(\mathcal{G}_u=\sigma(N_0(v):0\le v\le u)\). Let \(\lambda:[0,\infty)\to[0,\infty)\) be a deterministic, measurable, locally integrable function, and define its cumulative intensity by

Thus \(\Lambda(0)=0\), and \(\Lambda\) is continuous and nondecreasing. It need not be strictly increasing: intervals on which \(\lambda=0\) produce flat portions of \(\Lambda\).

For a calendar-time interval \((s,t]\), the operational-time interval is \((\Lambda(s),\Lambda(t)]\). The number of events in the first interval is exactly the number of unit-rate Poisson events in the second. This is why the cumulative intensity, rather than the instantaneous rate itself, is the clock.

The Time-Change Theorem

Proof. Fix \(0\le s\le t\). Since \(\Lambda\) is nondecreasing, \(\Lambda(s)\le\Lambda(t)\), and the definition of \(N\) gives \(N(t)-N(s)=N_0(\Lambda(t))-N_0(\Lambda(s))\). A unit-rate Poisson process has independent increments, and its increment over a deterministic interval of length \(v-u\) has a Poisson distribution with mean \(v-u\). Therefore this difference is Poisson with mean \(\Lambda(t)-\Lambda(s)=\int_s^t\lambda(r)\,dr\). If \(0\le s_1\le t_1\le s_2\le t_2\), the operational intervals \((\Lambda(s_1),\Lambda(t_1)]\) and \((\Lambda(s_2),\Lambda(t_2)]\) are disjoint, so the corresponding increments of \(N_0\), and hence of \(N\), are independent. Also \(N(0)=N_0(0)=0\). On every bounded calendar-time interval \([0,T]\), \(\Lambda(T)\) is finite by local integrability, and \(N_0(\Lambda(T))\) is finite almost surely; thus \(N\) has finitely many events on \([0,T]\). The composition is nondecreasing and right-continuous because \(N_0\) is right-continuous and \(\Lambda\) is continuous and nondecreasing. These facts establish the defining properties. \(\square\)

The filtration statement is compatible with the construction: \(N(t)\) is \(\mathcal{F}_t\)-measurable, and a future increment uses only the unit-rate process after operational time \(\Lambda(t)\). The theorem concerns increments, not stationarity: when \(\lambda\) varies, the distribution of an increment generally depends on the calendar-time interval, not only on its length.

Event Times and the Generalized Inverse

Let \(S_k=\inf\{u\ge0:N_0(u)\ge k\}\) be the \(k\)-th operational arrival time. For \(s\ge0\), define the generalized inverse of the cumulative intensity by

The generalized inverse is necessary because \(\Lambda\) may have flat intervals and may have a finite limiting value. For every finite \(t\), continuity and monotonicity give the exact equivalence \(\Lambda^{-1}(s)\le t\) if and only if \(s\le\Lambda(t)\). Indeed, if \(s\le\Lambda(t)\), then \(t\) belongs to the set in the infimum, so the infimum is at most \(t\). Conversely, if the infimum is at most \(t\), the set is nonempty and its first point occurs no later than \(t\); continuity of \(\Lambda\) ensures that the level \(s\) is attained by that time. This also shows why the non-strict inequality matters at an attained endpoint.

Proof. For each operational arrival level \(S_k\), the preceding equivalence gives \(T_k\le t\) exactly when \(S_k\le\Lambda(t)\). The number of operational arrivals at or before \(\Lambda(t)\) is \(N_0(\Lambda(t))\), which is \(N(t)\) by definition. Counting the equivalent events over \(k\) proves the identity. \(\square\)

Write \(L=\sup_{t\ge0}\Lambda(t)\), which may be infinite. If \(L<\infty\), every level \(s<L\) is reached at a finite time: by the definition of supremum there is a finite \(t\) with \(\Lambda(t)>s\), and continuity from \(\Lambda(0)=0\) implies that level \(s\) is attained. A level \(s>L\) is never reached. The endpoint \(s=L\) has a finite inverse if and only if \(\Lambda\) attains \(L\) at a finite time. Equivalently, an operational arrival has a finite calendar-time image exactly when its level is at most \(\Lambda(t)\) for some finite \(t\). The endpoint distinction is part of the exact inverse definition, even though a unit-rate Poisson arrival equals any fixed level with probability zero.

Worked Constructions

Why the Clock Must Be Cumulative

A common mistake is to substitute the instantaneous rate \(\lambda(t)\) where the cumulative intensity \(\Lambda(t)\) belongs. For the power-law rate \(\lambda(t)=3t^2\), at \(t=1/2\) the instantaneous rate is \(3(1/2)^2=3/4\), but the cumulative intensity is \((1/2)^3=1/8\). The correct construction gives \(N(1/2)=N_0(1/8)\), so \(\mathbb{P}(N(1/2)=0)=e^{-1/8}\). Using the incorrect expression \(N_0(\lambda(1/2))=N_0(3/4)\) would give \(e^{-3/4}\), which has the wrong mean and the wrong zero-count probability. A rate describes activity per unit calendar time; its integral describes the total activity accumulated by a given time.

The construction is useful both conceptually and computationally. It turns a process with varying calendar-time rate into a homogeneous process on operational time, while the inverse cumulative intensity maps event levels back to calendar time. Flat portions of \(\Lambda\) mean there is no event accumulation during those calendar intervals; unbounded \(\Lambda\) ensures every finite operational arrival level is eventually reached. When \(\Lambda\) is bounded, the finite-inverse endpoint must be treated according to whether the supremum is attained.