Why Uniqueness Matters
A compensator separates the observed event count into a predictable part and a martingale residual. That separation is useful only if the predictable part is determined by the event process and the information available over time. If two different compensators could both satisfy the defining conditions, the residual decomposition would not specify a unique expected event accumulation. The uniqueness result rules this out, subject to an important qualification: the filtration, which describes the available information, is part of the setup.
Uniqueness also explains what it means to say that an intensity determines a compensator. When the compensator is an integral of an intensity, equality of compensators identifies the intensity only up to changes on sets of time with measure zero. A compensator may also have predictable jumps, in which case an ordinary rate function integrated against time does not capture its full form.
Definitions and Assumptions
In this definition, predictability means that the process is determined by information available just before each time point. The uniqueness claim is about indistinguishability: two processes agree at every time on a single event of probability one. We assume the processes are defined on the same probability space and use the same filtration. Changing the filtration can change the compensator, because it changes what is predictable.
The theorem is a uniqueness statement, not a claim that any proposed increasing predictable process is a compensator. Each candidate must separately satisfy the martingale condition. The key fact is that the difference of two valid candidates is a predictable finite-variation local martingale, and such a process cannot accumulate nonzero variation.
Why the Compensator Is Unique
Suppose \(A\) and \(B\) both satisfy the definition. Subtract their residual martingales:
The difference of two local martingales is a local martingale. Therefore \(D=B-A\) is a local martingale. It is also predictable and has finite variation on bounded intervals, because both \(A\) and \(B\) have those properties. The remaining step is the following standard fact.
Here is a justification. First localize so that the process and its total variation are bounded; this permits the martingale and expectation arguments below. Jumps of a predictable càdlàg process occur at predictable times. At any such time \(T\), the jump \(\Delta D(T)\) is known just before \(T\), so it is \(\mathcal{F}_{T-}\)-measurable. The martingale property, by contrast, gives conditional mean zero for the jump given \(\mathcal{F}_{T-}\). Thus
After accounting for its predictable jumps, the localized process is continuous. For a continuous finite-variation martingale, take increasingly fine deterministic partitions of a fixed bounded interval \([0,t]\). The sum of squared increments has expectation \(\mathbb{E}[(D(t)-D(0))^2]\), since distinct martingale increments have expectation of their product equal to zero. Pathwise, the sum of squared increments is bounded by the largest absolute increment times the total variation. The largest increment tends to zero by continuity, while the localized total variation is bounded. Hence the sum tends to zero in expectation, which forces \(\mathbb{E}[(D(t)-D(0))^2]=0\). The process is therefore constant at each fixed \(t\); continuity and a countable dense set of times give constancy simultaneously for all \(t\). Removing the localization proves the key fact for the original local martingale.
Applying that fact to \(D=B-A\) gives \(B-A=0\) indistinguishably. Since both candidates start at zero, they agree at all times, as claimed.
From an Intensity to a Compensator
In the absolutely continuous case, a nonnegative predictable process \(\lambda(t)\) is an intensity when it is locally integrable and \(N(t)-\int_0^t\lambda(s)\,ds\) is a local martingale. Its compensator is then the cumulative integral of that intensity. If two such intensities give valid compensators for the same \(N\) and filtration, uniqueness implies equality of their cumulative integrals. Differentiating those equal, absolutely continuous paths shows that the intensities agree for almost every time, almost surely. Equivalently, they agree almost everywhere with respect to the product of time measure and probability.
The qualification “almost everywhere” matters. Changing a rate at one fixed time does not change its time integral. More generally, different versions of a rate can represent the same compensator if they differ only on a set of time measure zero along almost every path. The uniqueness theorem identifies the cumulative predictable process, not a pointwise value of its derivative at every instant.
Worked Example: A Rate Changed at One Time
Consider a hypothetical counting-process model on \(0\leq t\leq 3\), with an individual remaining under observation throughout. Suppose \(\lambda_1(t)=0.12\) events per year. Define a second candidate rate \(\lambda_2(t)\) to be \(0.12\) events per year except at \(t=2\), where it is \(5\) events per year. Assume the first rate is a valid intensity for this process and use the same filtration for both candidates.
The first cumulative candidate at time \(t\) is \(A_1(t)=0.12t\). For the second, the value assigned at the single time \(t=2\) contributes nothing to an integral with respect to time. Thus \(A_2(t)=\int_0^t\lambda_2(s)\,ds=0.12t\) for every \(t\) in the interval.
For example, at \(t=2\), both compensators equal \(0.12(2)=0.24\) expected events; at \(t=3\), both equal \(0.12(3)=0.36\) expected events. The rates differ at one point, but the compensators agree at every time. There is no conflict with uniqueness: the rate functions are equivalent as time-integrated intensities, even though their chosen point values differ.
Assumption check. The rates are nonnegative and predictable, and they are integrable over the bounded interval. The example assumes the first integrated process is a valid compensator; equality of the integrals then makes the second one the same compensator.
Worked Example: Two Different Linear Candidates Cannot Both Work
Suppose a hypothetical analysis proposes \(A(t)=0.40t\) and \(B(t)=0.30t\), in expected events, as compensators for the same counting process on \(0\leq t\leq 2\). Both are predictable, nondecreasing, finite, and start at zero. Could both be valid?
If both were valid, their difference \(D(t)=B(t)-A(t)=-0.10t\) would have to be a local martingale. It is deterministic. At \(t=1\), \(D(1)=-0.10\), while \(D(0)=0\). A deterministic process with this change cannot be a martingale: its conditional expected value at time 1, given the information at time 0, is \(-0.10\), not \(D(0)=0\). Equivalently, the proposed difference is a nonconstant predictable finite-variation process, contradicting the key fact.
Therefore, the two candidates cannot both be compensators. At least one fails the defining martingale condition \(N-A\) or \(N-B\). The calculation does not establish which one is valid; that requires checking its residual against the model and filtration.
Assumption check. The contradiction uses the same \(N\), the same filtration, and the requirement that each residual be a local martingale. Without those conditions, the uniqueness theorem does not apply.
Predictable Jumps and the Limits of a Rate Function
A compensator need not be an integral of an ordinary rate with respect to time. For example, a process may have a chance of an event at a known scheduled time. At that time, the event probability is predictable from the information just before the event, and the compensator can have a jump. The general uniqueness theorem covers this case as well: it applies to predictable finite-variation compensators, not just continuous ones.
Worked Example: A Possible Event at a Scheduled Time
Consider a hypothetical process with one possible event at time \(t=2\). Let \(B\) be a Bernoulli random variable with \(P(B=1)=0.30\), and let \(N(t)=B\,1\{t\geq 2\}\). Suppose the event outcome is not revealed before time 2, and is revealed at time 2. Take \(A(t)=0.30\,1\{t\geq 2\}\).
Before time 2, both \(N(t)\) and \(A(t)\) are zero. At and after time 2, the residual is \(N(t)-A(t)=B-0.30\), whose expectation given the information just before time 2 is \(0.30-0.30=0\). After time 2 the residual does not change. Thus \(N-A\) is a martingale for this filtration, and \(A\) is a compensator.
The compensator jumps by \(0.30\) at the scheduled time. There is no ordinary time-integrable rate \(\lambda(t)\) whose integral with respect to time produces this jump. The candidate \(A\) is nevertheless predictable, because the scheduled time and the pre-event probability are known just before the event. By uniqueness, no other predictable finite-variation compensator for this process and filtration can differ from \(A\).
Assumption check. The filtration must not reveal \(B\) before time 2; otherwise the conditional event probability immediately before time 2 would not be \(0.30\). This example shows why a rate-function description needs an absolute-continuity assumption.
Diagnostics, Comparisons, and Common Pitfalls
When checking a proposed compensator, first verify that the candidates refer to the same counting process and filtration. Then check predictability, monotonicity, finite variation, and the initial value. Most importantly, check the martingale condition for the residual. Uniqueness cannot repair a failure in any of these requirements.
Two complementary ways of reasoning are useful. The Doob–Meyer decomposition states that a process with an appropriate integrability property can be decomposed into a martingale and a predictable finite-variation part; its uniqueness argument is the one proved here. In applications, one may instead derive a compensator from a conditional intensity, if that intensity exists and the relevant integral is valid. The first route emphasizes uniqueness of the decomposition; the second emphasizes how a model-based rate accumulates over time. Neither allows a pointwise interpretation of an intensity on time-null sets, and the intensity route does not cover predictable jumps unless they are represented separately.
A common mistake is to conclude that uniqueness forces two intensity formulas to agree at every time. That is too strong: the theorem makes the compensators equal, and equal time integrals imply equality of integrands only almost everywhere. Another mistake is to assume every compensator must have a smooth or even continuous rate. The scheduled-event example has a valid predictable compensator with a jump. Finally, the theorem is filtration-specific: a process that is predictable in one information structure may not be predictable in another.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Two candidate compensators for the same process and filtration are \(A(t)=0.2t\) and \(B(t)=0.2t+0.05\). Both are nondecreasing and finite, but \(B(0)\ne 0\). Does the uniqueness theorem apply as stated?
- If two valid compensators are equal at every time, what can be concluded about their absolutely continuous intensity functions?
- Why does changing an intensity at one fixed time not necessarily change its compensator?
- A candidate compensator has a predictable jump at a known scheduled time. Does the uniqueness theorem exclude it?
Solutions. (1) No. The stated definition requires a compensator to start at zero, and \(B(0)=0.05\). One cannot apply the theorem to this pair as stated. (2) The intensities agree almost everywhere in time, almost surely; pointwise equality at every time is not implied. (3) A single time has zero time measure, so changing the integrand only there leaves its time integral unchanged. (4) No. Uniqueness applies to predictable finite-variation compensators that may have jumps; absolute continuity is needed only for the representation as an ordinary time-integrated intensity.