Doob-Meyer Decomposition for Counting Processes

Why Separate a Count Into Two Processes?

A counting process records accumulated events. Its values rise at event times and remain constant between events, so the observed path is not itself a sequence of mean-zero increments. The Doob–Meyer decomposition separates that path into a predictable component that represents its accumulated event rate and a martingale component that represents the remaining fluctuation.

This separation is useful because the compensator accounts for the event-generating structure given the information available over time. The martingale part then has zero conditional mean increments. In event-history settings, the compensator is often constructed by integrating a conditional intensity over time.

Definition: Let $N(t)$, $0\leq t\leq \tau$, be an adapted, right-continuous counting process with left limits, $N(0)=0$, and nondecreasing paths. A Doob–Meyer decomposition is a representation $$ N(t)=M(t)+A(t), $$ where $A(0)=0$ is predictable and nondecreasing and $M(0)=0$ is a martingale with respect to the filtration $\{\mathcal{F}_t\}$. The process $A$ is the compensator of $N$, and $M=N-A$ is its martingale component.

Predictability means that the value of $A$ at a time is determined by information available just before that time, in the formal sense of the predictable sigma-field. A compensator is therefore not simply a fitted curve that follows the observed count. It is a nondecreasing process whose increments are predictable from the evolving information structure. The martingale condition is

E\{M(t)\mid\mathcal{F}_s\}=M(s),\qquad 0\leq s\leq t\leq\tau.

Equivalently, after subtracting the compensator, each future increment has conditional mean zero:

E\{N(t)-N(s)-[A(t)-A(s)]\mid\mathcal{F}_s\}=0.

Conditions for the Decomposition

A standard finite-horizon version applies when the filtration satisfies the usual regularity conditions, $N$ is adapted and right-continuous with left limits, and $E\{N(\tau)\}<\infty$. Since $N$ is nondecreasing, it is a submartingale: $N(t)\geq N(s)$ for $t\geq s$, so $E\{N(t)\mid\mathcal{F}_s\}\geq N(s)$. Also, $0\leq N(t)\leq N(\tau)$; the integrable upper bound makes the family of values $N(t)$ uniformly integrable on this interval. These conditions allow the Doob–Meyer theorem for class-D submartingales to be applied.

The result gives a predictable, nondecreasing $A$ and a martingale $M=N-A$. For a simple counting process, $N$ has unit jumps, but the decomposition does not require the compensator to have unit jumps or to be continuous. If the process is considered over an unbounded time range, the result is usually applied on finite intervals, with the required integrability checked on each interval.

When a predictable conditional intensity $\lambda(t)$ is available, a common form of the compensator is

A(t)=\int_0^t \lambda(u)\,du.

Here $\lambda(t)$ must be nonnegative and predictable, and the integral must be finite in expectation on the time interval: $E\{\int_0^\tau\lambda(u)\,du\}<\infty$. In this case the intensity describes the event rate given the history immediately before time $t$. The compensator accumulates that rate along the observed history.

Why an Intensity Produces a Martingale

The key link can be stated in terms of predictable weights. Suppose $A(t)=\int_0^t\lambda(u)\,du$ and, for every bounded predictable process $H$,

E\left\{\int_0^\tau H(u)\,dN(u)\right\} = E\left\{\int_0^\tau H(u)\lambda(u)\,du\right\}.

This identity says that, on average, events accumulated with predictable weights agree with the integrated conditional rate using the same weights. To verify that $M=N-A$ is a martingale, choose any $B\in\mathcal{F}_s$ and let $H(u)=\mathbf{1}_B\mathbf{1}_{(s,t]}(u)$. This is predictable: membership in $B$ is known at time $s$, and the interval begins just after $s$. The identity gives

E\{\mathbf{1}_B[N(t)-N(s)]\} = E\left\{\mathbf{1}_B\int_s^t\lambda(u)\,du\right\} = E\{\mathbf{1}_B[A(t)-A(s)]\}.

Subtracting the right-hand side from the left-hand side shows that $E\{\mathbf{1}_B[M(t)-M(s)]\}=0$ for every $B\in\mathcal{F}_s$. This is the defining conditional-expectation property $E\{M(t)-M(s)\mid\mathcal{F}_s\}=0$. Integrability follows from the finite expected count and finite expected compensator. Thus $M$ is a martingale.

Worked Examples

Worked Example: A Homogeneous Poisson Process

Suppose a hypothetical event process is a homogeneous Poisson process with rate $\lambda=1.2$ events per hour on $0\leq t\leq 3$ hours. Its increments are independent of the past, with expected count $1.2$ times the length of an interval. The rate is constant, predictable, and has finite integral on this interval, so the compensator is $A(t)=1.2t$.

Suppose the observed path has events at $0.4$, $1.1$, and $2.7$ hours. At $t=2$, two events have occurred, giving

A(2)=1.2(2)=2.4\ \text{events},\qquad M(2)=N(2)-A(2)=2-2.4=-0.4\ \text{events}.

At $t=3$, all three events have occurred:

A(3)=1.2(3)=3.6\ \text{events},\qquad M(3)=3-3.6=-0.6\ \text{events}.

A negative value of $M(t)$ is not an error: the observed count can be below its compensator at a particular time. The martingale property concerns conditional means of future increments, not a requirement that $M(t)$ be nonnegative or equal zero along each path.

Worked Example: A Single Event With an At-Risk Indicator

Consider a hypothetical event time $T$ with exponential distribution and constant hazard $0.4$ per day. Follow each individual until the event or a fixed administrative end at day 3. Let $N(t)=\mathbf{1}(T\leq t)$, and let $Y(t)=\mathbf{1}(T\geq t)$ indicate that the individual is still at risk just before $t$. The conditional intensity is $\lambda(t)=0.4Y(t)$: it is $0.4$ before the event and zero afterward. This intensity is predictable, and its integral is bounded by $0.4(3)$, so the finite-interval integrability condition holds.

For a hypothetical individual whose event occurs at day $1.5$,

A(t)=\int_0^t 0.4Y(u)\,du =0.4\min(t,1.5),\qquad 0\leq t\leq 3.

At day 1, the event has not occurred, so $N(1)=0$ and $A(1)=0.4(1)=0.4$. Hence $M(1)=-0.4$ events. At day 2, the event has occurred and the compensator stops accumulating at day 1.5:

N(2)=1,\qquad A(2)=0.4(1.5)=0.6,\qquad M(2)=1-0.6=0.4.

The compensator is based on time at risk, not on the full follow-up duration regardless of event. Continuing to add $0.4$ per day after the event would contradict the specified intensity.

Worked Example: An Intensity That Changes After an Event

Suppose a hypothetical recurrent-event process is followed for two hours. Its conditional intensity is $0.5$ events per hour before the first event and $1.0$ event per hour after the first event. In process notation,

\lambda(t)=0.5\,\mathbf{1}\{N(t-)=0\} +1.0\,\mathbf{1}\{N(t-)\geq 1\}.

The dependence on $N(t-)$, the count immediately before $t$, makes the intensity predictable. It is bounded by $1.0$, so its integrated value over two hours is finite. Suppose the observed path has events at $0.6$ and $1.4$ hours. Before the first event, the compensator accumulates at rate $0.5$; after it, the rate is $1.0$. Thus

A(2)=0.5(0.6)+1.0(2-0.6)=0.3+1.4=1.7\ \text{events}.

Two events have occurred by the end of follow-up, so $M(2)=2-1.7=0.3$ events. At $t=1$, one event has occurred and the compensator is $0.5(0.6)+1.0(1-0.6)=0.7$, giving $M(1)=1-0.7=0.3$. Between the second event at 1.4 hours and the end of follow-up, $N(t)$ stays at 2 while $A(t)$ increases at rate 1.0. The martingale component consequently decreases over that interval.

Compensator, Cumulative Mean, and Diagnostics

A common point of confusion is the difference between the compensator $A(t)$ and the cumulative mean function $E\{N(t)\}$. Since $M(t)$ is an integrable martingale starting at zero, $E\{M(t)\}=0$, and therefore

E\{N(t)\}=E\{A(t)\}.

This equality is between expectations; it does not say that $A(t)=E\{N(t)\}$ on each path. When the intensity depends on a changing risk set or event history, $A(t)$ is generally random. For example, in the single-event case, the compensator stops growing at the event time, whereas the cumulative mean $E\{N(t)\}=1-e^{-0.4t}$ is a deterministic number.

Quantity	What it represents	Can it depend on the observed history?
$N(t)$	Observed accumulated event count by time $t$	Yes
$A(t)$	Accumulated predictable event rate along the history	Usually, if the intensity depends on the history
$E\{N(t)\}$	Mean count at time $t$ across repeated realizations	No; it is deterministic

The decomposition also has practical checks. A proposed $A$ should start at zero, be nondecreasing and predictable, and satisfy the required integrability condition. With an ordinary intensity, it should accumulate at the stated rate and stop or change when the risk set or event history changes. A compensator that decreases, uses future information, or continues accumulating after an absorbing event when the intensity is zero is inconsistent with the model. A pathwise value of $M$ that is positive or negative is not, by itself, a diagnostic failure.

The Doob–Meyer decomposition is more general than the intensity-integral construction: it provides a compensator for an integrable counting-process submartingale even when no convenient ordinary rate function has been specified. The intensity form is especially useful when the event rate can be written explicitly. It is not appropriate to assume an intensity is constant, or that a random compensator is a deterministic mean curve, without checking the process assumptions.

Common mistake: “The expected count by time $t$ is the compensator, so I can subtract that fixed mean curve from every observed path.” The equality $E\{A(t)\}=E\{N(t)\}$ does not imply $A(t)=E\{N(t)\}$ path by path. Use the predictable compensator specified by the event history and intensity; subtracting only a marginal mean curve need not produce a martingale.

Check Your Understanding

Try the questions first, then compare with the brief solutions below.

For a Poisson process with rate $0.8$ events per day, what are $A(4)$ and $M(4)$ on a path with three events by day 4?
Why must the intensity $0.4Y(t)$ for a single event stop accumulating after the event?
Does $E\{N(t)\}=E\{A(t)\}$ imply equality of the two quantities on every sample path? Explain.
Name two conditions that a proposed compensator should satisfy, apart from being nondecreasing.

Solutions. (1) $A(4)=0.8(4)=3.2$ events and $M(4)=3-3.2=-0.2$ events. (2) After the event, $Y(t)=0$, so the specified intensity is zero and its integral no longer increases. (3) No. The equality is between expectations; the compensator may be random and differ from the deterministic mean on an individual path. (4) It should start at zero and be predictable; on a finite interval it must also satisfy the relevant integrability condition.

Quantity	What it represents	Can it depend on the observed history?
\(N(t)\)	Observed accumulated event count by time \(t\)	Yes
\(A(t)\)	Accumulated predictable event rate along the history	Usually, if the intensity depends on the history
\(E\{N(t)\}\)	Mean count at time \(t\) across repeated realizations	No; it is deterministic