Time Change by Integrated Intensity

Why Change the Time Scale?

Events measured in calendar time may arrive at changing rates. A count of one event per month means something different in a period of low risk than in a period of high risk. Integrated intensity replaces calendar time with operational time: periods with greater expected event accumulation contribute more transformed time. Under suitable conditions, this rescaling turns a counting process into a unit-rate Poisson process.

The result is useful for understanding event-time models and for checking whether a proposed intensity is compatible with observed event timing. It is not automatic: the transformed process is Poisson only when the assumptions below hold. In particular, a finite observation period or a process that stops accumulating intensity does not produce a complete Poisson process on the whole nonnegative time axis.

Definition and Conditions

Definition: Let $N(t)$ be a simple counting process adapted to a filtration $\{\mathcal{F}_t\}$, with $N(0)=0$, and let $\lambda(t)$ be its nonnegative predictable intensity. Its cumulative intensity, or integrated intensity, is $$ A(t)=\int_0^t \lambda(s)\,ds. $$ If the event times are $T_1<T_2<\cdots$, their transformed times are $U_k=A(T_k)$. When $A$ is continuous and strictly increasing, its inverse maps operational time $u$ back to calendar time. The time-changed count is $N^*(u)=N(A^{-1}(u))$.

Here “simple” means that every jump of $N$ has size one, so there are no simultaneous multiple events in this count. The intensity is predictable: at time $t$, its value can depend on information available just before $t$, but not on the event that is about to occur or on future information. As in the earlier material on compensators, $A(t)$ is the compensator when $N(t)-A(t)$ is a local martingale.

A convenient set of sufficient conditions for a full time-change result is: the process starts at zero; it is nonexplosive and simple; its intensity is nonnegative, predictable, and locally integrable; $A(t)$ is continuous and strictly increasing; and $A(t)$ tends to infinity as $t$ tends to infinity. The continuous, strictly increasing condition can be ensured, for example, by an intensity that is positive almost everywhere. The divergence condition ensures that every operational time $u\geq 0$ corresponds to a calendar time.

Theorem: Under these conditions, define $\tau(u)=A^{-1}(u)$, $N^*(u)=N(\tau(u))$, and the time-changed filtration $\mathcal{F}^*_u=\mathcal{F}_{\tau(u)}$. Then $N^*(u)$ is a unit-rate Poisson process relative to $\{\mathcal{F}^*_u\}$. In particular, its increments over disjoint operational-time intervals are independent, and $N^*(v)-N^*(u)$ has a Poisson distribution with mean $v-u$, for $0\leq u<v$.

The theorem is about the count after changing its clock, not about the original event times being equally spaced in calendar time. Nor does it assert that a fitted intensity is correct merely because its transformed event times look plausible. The model assumptions and the way the intensity was specified still matter.

Why the Transformation Gives a Poisson Process

The argument uses the compensator and the standard martingale time-change reasoning. Write $M(t)=N(t)-A(t)$. For each operational time $u$, the inverse time $\tau(u)$ is a stopping time, and continuity and strict increase give $A(\tau(u))=u$. Therefore,

M(\tau(u))=N(\tau(u))-A(\tau(u))=N^*(u)-u.

Under the usual localization and integrability conditions for optional sampling, $M(\tau(u))$ is a martingale in the time-changed filtration. Thus $N^*(u)-u$ is a martingale, so $u$ is the transformed process’s compensator. A strictly increasing continuous time map preserves the unit jumps and the absence of simultaneous jumps. The Poisson characterization for a simple counting process says that a process with unit jumps and deterministic compensator $u$ is a unit-rate Poisson process. These facts yield the result.

The important role of continuity is that the compensator does not make jumps at fixed or random times. If it did, a time transformation based only on a smooth integrated rate would not account for the event probability concentrated at those times. The assumptions also explain why the theorem is stated for a full, unbounded operational-time scale: if $A(t)$ stops at a finite value, the transformed process is observed only up to that endpoint.

Worked Example: Deterministic Intensity That Increases Over Time

Suppose, hypothetically, a recurrent-event process has deterministic intensity $\lambda(t)=0.2+0.1t$ events per month, with $t$ measured in months. Assume the process is simple, starts at zero, and can continue to accrue events indefinitely. The intensity is positive, continuous, and locally integrable, and its cumulative intensity diverges, so the time-change conditions hold.

Integrating the rate gives

A(t)=\int_0^t(0.2+0.1s)\,ds=0.2t+0.05t^2.

Suppose three hypothetical observed event times are 2, 4, and 6 months. Their transformed times are

U_1=A(2)=0.4+0.2=0.6,\qquad U_2=A(4)=0.8+0.8=1.6,\qquad U_3=A(6)=1.2+1.8=3.0.

The transformed gaps, including the gap from operational time zero to the first event, are $0.6$, $1.6-0.6=1.0$, and $3.0-1.6=1.4$ operational-time units. If the model is correct, the successive gaps have the unit exponential distribution under the theorem. They need not be equal; the claim is about their probability law, not their realized values. Since the original intensity is measured in events per month, the cumulative intensity and transformed times are on the dimensionless operational-time scale.

History-Dependent Intensities and a Stopped Process

An intensity may depend on the observed history, provided that dependence is predictable. The integral $A(t)$ is then computed along the observed history. The time-change theorem still applies to a recurrent process if its assumptions, including unbounded cumulative intensity, hold. By contrast, when a process stops after one event, its integrated intensity generally stops too. In that setting the transformed first event can have an exponential distribution, but the observed count is not a complete Poisson process over all $u\geq0$.

Worked Example: A Predictable Intensity That Changes After an Event

Consider a hypothetical recurrent-event process with intensity $0.5$ events per hour before its first event and $1.5$ events per hour after its first event. Formally, use the predictable rule $\lambda(t)=0.5$ when $N(t-)=0$ and $\lambda(t)=1.5$ when $N(t-)\geq1$. Suppose the observed event times are 1, 2.5, and 4 hours.

Before the first event, the accumulated intensity is $0.5$ per hour. After that event, it accumulates at $1.5$ per hour. Thus

A(1)=0.5(1)=0.5,\quad A(2.5)=0.5(1)+1.5(2.5-1)=2.75,\quad A(4)=0.5(1)+1.5(4-1)=5.0.

The transformed event times are $0.5$, $2.75$, and $5.0$; the transformed gaps are $0.5$, $2.25$, and $2.25$. This calculation uses the intensity in effect during each interval, not one fixed rate for the whole follow-up. The intensity is predictable because the rate after an event is determined by the history just before each time. It is positive and locally integrable, and its cumulative intensity continues to grow without bound, so the full-process theorem is applicable. These particular gaps are observed values, not evidence by themselves that the model is correct.

If instead the process could have at most one event and its intensity became zero after that event, $A(t)$ would stop increasing at $A(T_1)$. The transformed observation would stop at its first jump. It would not be a full unit-rate Poisson process. Under the standard continuous-intensity survival model, the transformed first event time is unit exponential when cumulative hazard tends to infinity; if the total cumulative hazard is finite, there can also be positive probability of no event.

When the Conditions Fail

A continuous integrated intensity is central to this version of the transformation. For example, suppose a hypothetical event process has a continuous rate of $0.4$ per day and also has a conditional probability mass at day 3. Its compensator has the form $A(t)=0.4t+0.3\,I(t\geq3)$, where the jump of size $0.3$ represents the event mass at day 3. This compensator is not continuous. Simply applying the continuous-rate formula $0.4t$ omits that event mass, while treating the compensator’s jump as ordinary elapsed operational time does not satisfy the continuous time-change theorem. A different treatment is needed for such atoms; the theorem stated here does not apply.

Worked Example: A Compensator Jump Is Not a Continuous Time Change

Continue with the hypothetical compensator $A(t)=0.4t+0.3\,I(t\geq3)$. Just before day 3, $A(3-)=0.4(3)=1.2$; at day 3 it jumps to $A(3)=1.2+0.3=1.5$. The jump is $1.5-1.2=0.3$.

If one instead used only the continuous part, an event at day 3 would be assigned operational time $0.4(3)=1.2$, ignoring the mass represented by the jump. The full compensator assigns a discontinuity from 1.2 to 1.5, so it is not a continuous, strictly increasing clock with the required inverse. Neither calculation justifies calling the transformed count a unit-rate Poisson process under the continuous-intensity theorem. The appropriate conclusion is that an assumption fails, not that the Poisson result has been approximately established.

Other warning signs include an intensity that uses future information, a cumulative intensity that is flat over an interval, a process with simultaneous jumps, or a finite total cumulative intensity when a full process on $[0,\infty)$ is being claimed. Flat sections may be handled with generalized inverses in more general results, but strict invertibility is absent and the simple theorem above cannot be invoked without further conditions. In applications, also check whether the observed follow-up ends at a finite time: calculations can be made up to the transformed endpoint, but that does not establish an untruncated Poisson process beyond it.

Diagnostics and Interpretation

If a specified intensity is appropriate and the conditions hold, transformed inter-event gaps should behave like independent unit exponential variables, and counts over operational-time intervals should behave like Poisson counts with mean equal to interval length. These are model implications and can motivate diagnostics. For instance, unusually many very small transformed gaps, systematic trends in gaps over event order, or counts that vary systematically across equal-length operational intervals can signal lack of fit.

Such checks are not proofs of the model. A short series of events can look consistent by chance, and the intensity may have been estimated from the same data being checked. Parameter estimation can alter the reference behavior of residuals, so calibration that treats fitted transformed gaps as if the intensity were known may be misleading. Compare diagnostics with the model’s data collection process, observation window, covariate history, and any known changes in recording.

Time-rescaling differs from examining raw calendar-time gaps: unequal calendar gaps may be entirely consistent with a changing rate. It also differs from using the compensator residual $N(t)-A(t)$, which tracks count minus accumulated expected count on the original time scale. Time-rescaling focuses on the locations of events on operational time. Neither diagnostic replaces checking predictability and the substantive specification of the intensity.

Key takeaway: Integrated intensity is a clock, not a guarantee. A simple process with a continuous, strictly increasing, unbounded compensator becomes a unit-rate Poisson process on that clock. Finite accumulated intensity, compensator jumps, or other assumption failures change what can be concluded.

Check Your Understanding

Answer each question, then compare with the solutions below.

A process has deterministic intensity $\lambda(t)=0.3$ events per day. What is $A(t)$, and what are the transformed times for events at days 2 and 5?
Why does a continuous-intensity model for a single-event process generally not yield a full Poisson process after the event?
A compensator is flat on a nonzero time interval. Which simple time-change condition fails?
What does a transformed gap of 2 operational-time units mean under the unit-rate Poisson model?

Solutions. (1) $A(t)=\int_0^t0.3\,ds=0.3t$. The transformed times are $A(2)=0.6$ and $A(5)=1.5$ operational-time units. (2) After the sole event, the at-risk intensity is typically zero, so cumulative intensity stops increasing at a finite value. The transformed count stops at its first jump and is not a complete Poisson process on the nonnegative axis; under infinite cumulative hazard, the transformed first-event time is unit exponential. (3) Strict increase and the existence of an ordinary inverse fail on that interval. A generalized-inverse result may be possible, but the stated strict-inverse theorem does not directly apply. (4) It is a realized waiting time on the operational clock. Under the model, each transformed inter-event gap has a unit exponential distribution, but an individual gap of 2 is not itself proof of model fit.