The Poisson Compensator and Its Martingale

Why Subtract the Mean Count?

A Poisson counting process records the cumulative number of events. Its expected count grows linearly with time, so the process itself has a systematic upward trend. Subtracting that trend produces a process whose conditional expected future value equals its current value. The resulting martingale is the basic role of the Poisson compensator: it separates the predictable rate of event accumulation from the random fluctuation in the count.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ and a rate-$\lambda$ Poisson process $N_t$, where $\lambda\ge0$. The Brownian motion is not used below. Take the filtration to be the usual augmentation of the joint natural filtration of $B$ and $N$, and assume $B$ and $N$ are independent. In particular, for $0\le s<t$, the increment $N_t-N_s$ is independent of $\mathcal{F}_s$. A rate-$\lambda$ Poisson process starts at zero, has independent increments, and satisfies $N_t-N_s\sim\operatorname{Poisson}(\lambda(t-s))$. We use right-continuous paths, so $N_t$ counts events in $(0,t]$. No stochastic integral is used in this tutorial.

Definition: A process $M_t$ is a martingale with respect to $(\mathcal{F}_t)$ if it is adapted, $\mathbb{E}[|M_t|]<\infty$ for each $t$, and $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ almost surely whenever $0\le s\le t$. In this setting, a compensator for $N_t$ is a predictable, increasing process $A_t$ with $A_0=0$ such that $N_t-A_t$ is a local martingale. A continuous adapted process, such as $A_t=\lambda t$, is predictable. We will prove the stronger fact that the compensated process is a true martingale.

A local martingale is a process that becomes a martingale after stopping at each time in a suitable sequence of stopping times increasing to infinity. A true martingale satisfies the integrability and conditional-expectation conditions directly, without that localization. The distinction matters: identifying a local martingale alone does not establish the conditional expectation identity needed for expected event counts.

The Poisson Compensator Theorem

Theorem: Let $N_t$ be a rate-$\lambda$ Poisson process whose increments after time $s$ are independent of $\mathcal{F}_s$. Then $$ M_t=N_t-\lambda t $$ is an integrable martingale with respect to $(\mathcal{F}_t)$. Consequently, $A_t=\lambda t$ is a compensator for $N_t$.

Proof. The process $M_t$ is adapted because $N_t$ is adapted and $t\mapsto\lambda t$ is deterministic. For each fixed $t$, $N_t$ has a Poisson distribution with parameter $\lambda t$, so $\mathbb{E}[N_t]=\lambda t<\infty$. The triangle inequality gives $\mathbb{E}[|M_t|]\le\mathbb{E}[N_t]+\lambda t=2\lambda t<\infty$. Now fix $0\le s\le t$. The increment $N_t-N_s$ is independent of $\mathcal{F}_s$ and has mean $\lambda(t-s)$. Since $N_s$ is $\mathcal{F}_s$-measurable, the conditional expectation rules give $$ \mathbb{E}[M_t\mid\mathcal{F}_s] =\mathbb{E}[N_s+(N_t-N_s)-\lambda t\mid\mathcal{F}_s] =N_s+\lambda(t-s)-\lambda t =N_s-\lambda s =M_s $$ almost surely. Thus $M_t$ is a true martingale. The process $A_t=\lambda t$ starts at zero, is predictable and increasing, and $N_t-A_t=M_t$ is in particular a local martingale. Hence $A_t$ is a compensator. If $\lambda=0$, then $N_t=0$ almost surely for every $t$, and the same identities hold with $M_t=A_t=0$. $\square$

For $\lambda>0$, the compensator grows continuously even though $N_t$ changes by jumps. It records the expected accumulation rate, rather than predicting the times of individual events. The martingale $N_t-\lambda t$ is the difference between the observed count and this deterministic accumulation.

Worked Example: Expected Count in One Time Window

Suppose $\lambda=2$, and count events in the interval $(1,\tfrac52]$. Find the expected count, including conditionally on the information available at time $1$.

The count in the interval is $N_{5/2}-N_1$. From the martingale theorem, or directly from the independent-increment distribution, its conditional expectation is $$ \mathbb{E}[N_{5/2}-N_1\mid\mathcal{F}_1] =\lambda\left(\frac52-1\right) =2\cdot\frac32 =3. $$ Taking expectations gives $\mathbb{E}[N_{5/2}-N_1]=3$ as well.

Check: the interval has length $\frac52-1=\frac32$, and a rate of $2$ events per time unit gives $2\cdot\frac32=3$ expected events. This is an expectation, not a claim that the interval contains exactly three events.

Counting Events on Deterministic Sets

The martingale identity immediately gives the expected count on any deterministic interval. If $0\le a\le b$, then $N_b-N_a$ counts events in $(a,b]$, and $$ \mathbb{E}[N_b-N_a] =\mathbb{E}[N_b]-\mathbb{E}[N_a] =\lambda b-\lambda a =\lambda(b-a). $$ More strongly, conditioning on the information at the interval's left endpoint gives $$ \mathbb{E}[N_b-N_a\mid\mathcal{F}_a]=\lambda(b-a) $$ almost surely. The conditional statement follows because the increment is independent of $\mathcal{F}_a$; it is not a statement about conditioning on an event of probability zero.

Counts on disjoint deterministic intervals can be added. If $(a_j,b_j]$, $j=1,\ldots,m$, are pairwise disjoint intervals with $0\le a_j\le b_j$, then the total count is the sum of the interval counts. Linearity of expectation therefore gives the sum of their individual expected counts. This calculation does not require independence; disjointness and additivity of the count, together with linearity of expectation, suffice. (For a Poisson process the counts are also independent, but that extra fact is not needed to compute their mean.)

Proposition: For any finite collection of pairwise disjoint deterministic intervals $(a_j,b_j]$, $$ \mathbb{E}\left[\sum_{j=1}^{m}(N_{b_j}-N_{a_j})\right] =\lambda\sum_{j=1}^{m}(b_j-a_j). $$ The conditional mean on a single interval, given $\mathcal{F}_{a_j}$, is $\lambda(b_j-a_j)$.

Proof. For each $j$, the Poisson increment has mean $\lambda(b_j-a_j)$, so linearity of expectation gives $$ \mathbb{E}\left[\sum_{j=1}^{m}(N_{b_j}-N_{a_j})\right] =\sum_{j=1}^{m}\mathbb{E}[N_{b_j}-N_{a_j}] =\sum_{j=1}^{m}\lambda(b_j-a_j) =\lambda\sum_{j=1}^{m}(b_j-a_j). $$ For the conditional assertion, $N_{b_j}-N_{a_j}$ is independent of $\mathcal{F}_{a_j}$ and integrable, so its conditional expectation is its mean, $\lambda(b_j-a_j)$, almost surely. If an interval has zero length, its count is zero and both formulas give zero; thus the result includes that case. $\square$

Worked Example: Two Separate Observation Windows

Let $\lambda=\frac34$. Find the expected total number of events observed in $(0,2]$ and $(4,5]$.

The two counts are $N_2-N_0$ and $N_5-N_4$. Their interval lengths are $2-0=2$ and $5-4=1$. The expected total is therefore $$ \mathbb{E}[(N_2-N_0)+(N_5-N_4)] =\frac34\cdot2+\frac34\cdot1 =\frac64+\frac34 =\frac94. $$ The expected count in the first window is $\frac32$, and the expected count in the second is $\frac34$; adding them gives $\frac32+\frac34=\frac64+\frac34=\frac94$.

Check: the total observed time is $2+1=3$ time units. Multiplying this by the rate gives $\frac34\cdot3=\frac94$, the same answer. The unobserved gap $(2,4]$ contributes no count to this total.

Conditional Counts and a Martingale Check

Independent increments also make the compensated process useful when the past count is known. For $0\le s\le t$, $$ \mathbb{E}[N_t\mid\mathcal{F}_s] =N_s+\lambda(t-s). $$ Thus knowing the history changes the conditional expected count at time $t$ by retaining the already-observed count $N_s$; the expected additional count is still $\lambda(t-s)$. Subtracting $\lambda t$ cancels precisely that additional mean and leaves $N_s-\lambda s$.

Worked Example: A Known Past Count

Suppose $\lambda=\frac32$, and at time $1$ the observed count is $N_1=2$. Find the conditional expected count at time $3$, and the conditional expected number of new events in $(1,3]$, given $\mathcal{F}_1$.

On the event $N_1=2$, the conditional expected count at time $3$ is $$ \mathbb{E}[N_3\mid\mathcal{F}_1] =N_1+\lambda(3-1) =2+\frac32\cdot2 =2+3 =5. $$ The conditional expected number of new events is $$ \mathbb{E}[N_3-N_1\mid\mathcal{F}_1] =\lambda(3-1) =\frac32\cdot2 =3. $$ These conditional quantities are well-defined as conditional expectations; the displayed values on $N_1=2$ are understood almost surely on that event. The event has positive probability, since $N_1\sim\operatorname{Poisson}(\frac32)$ and $\mathbb{P}(N_1=2)=e^{-3/2}(\frac32)^2/2>0$.

Check: adding the known count $2$ to the expected future count $3$ gives $5$, as in the conditional expectation of $N_3$. Subtracting the compensator increment $\lambda(3-1)=3$ from the expected total count $5$ leaves $2=N_1$, exactly the martingale identity.

Common Pitfall: The Count Alone Is Not a Martingale

A common incorrect argument says that $N_t$ is a martingale because future increments are independent of the past. Independence does not mean that a future increment has conditional mean zero. In fact, for $s<t$, $$ \mathbb{E}[N_t\mid\mathcal{F}_s] =N_s+\lambda(t-s), $$ which is larger than $N_s$ when $\lambda>0$ and $t>s$. The incorrect calculation drops the positive conditional mean of $N_t-N_s$. The correction is to subtract the mean increment: $$ \mathbb{E}[N_t-\lambda t\mid\mathcal{F}_s] =N_s+\lambda(t-s)-\lambda t =N_s-\lambda s. $$ For $\lambda=0$, the count is identically zero, so $N_t$ is a martingale in that degenerate case too.

The compensator is useful because expected event counts can be found from elapsed time and rate without predicting individual event times. For a single interval, its increase is $\lambda(b-a)$; for a finite union of disjoint intervals, the increases add. This is the elementary form of compensation for Poisson counts. Expectations involving random, history-dependent weights on individual jumps require additional integration theory, which is not needed for these deterministic-window calculations.

Check Your Understanding

Use the martingale identity and the interval-count formula to answer the following.

For rate $\lambda=5$, what is the expected number of events in $(2,7]$, and what is its conditional expectation given $\mathcal{F}_2$?
For rate $\lambda=\frac12$, find the expected total count in $(0,2]$ and $(5,9]$.
Explain why independent increments alone do not make $N_t$ a martingale when $\lambda>0$.
State the integrability condition and conditional-expectation identity that verify $N_t-\lambda t$ is a true martingale.
What are the compensator and compensated process when $\lambda=0$?