Law of Large Numbers for Compensated Event Counts

Why Compensated Counts Become Small on the Law-of-Large-Numbers Scale

An event count records realized jumps, while its compensator records the predictable accumulation implied by the event-process intensity. Their difference is a martingale: it captures the stochastic fluctuation of the count around that compensator. In many large-sample arguments, the count and compensator are each of order $n$, but their difference is smaller than $n$. This tutorial gives a condition that makes that statement precise.

The central requirement is not merely that each individual event rate is bounded. It is that the predictable variation of the compensated quantity being scaled is small relative to the square of the scale. This distinction matters when event processes can jump together: aggregate jump sizes, and therefore aggregate variation, can be much larger than the sum of the individual compensators.

Definition: Let $N_n(t)$ be a counting process adapted to a filtration, and let $A_n(t)$ be its compensator. The compensated count is $M_n(t)=N_n(t)-A_n(t)$. Its predictable variation, denoted $\langle M_n\rangle(t)$, is the predictable increasing process that measures the martingale's accumulated variability. The notation $\langle M_n\rangle$ refers to the predictable variation of the martingale being analyzed, not automatically to the compensator $A_n$.

For a simple counting process with unit jumps and a continuous compensator, the predictable variation of its compensated count equals its compensator. For an aggregate process, that equality requires the aggregate jumps to be unit-sized. If several component processes jump simultaneously, an aggregate jump can exceed one, and its predictable variation must account for the square of that larger jump.

A Martingale Law of Large Numbers

Consider a fixed time interval $[0,\tau]$ and a positive scaling sequence $a_n$ tending to infinity. Suppose $M_n(0)=0$, each $M_n$ is a square-integrable martingale, and $E\{\langle M_n\rangle(\tau)\}$ is finite. A useful sufficient condition for the compensated count to vanish uniformly after scaling is:

Theorem: If $$ \frac{E\{\langle M_n\rangle(\tau)\}}{a_n^2}\longrightarrow 0, $$ then $$ \sup_{0\leq t\leq \tau}\frac{|M_n(t)|}{a_n}\ \xrightarrow{P}\ 0. $$

This is a uniform-in-time convergence statement in probability. It says that the largest absolute discrepancy between the count and its compensator, over the whole interval, is negligible relative to $a_n$ with probability tending to one. It does not assert almost-sure convergence, nor does it say that the unscaled martingale becomes small.

Justification by a Maximal Inequality

The argument combines the martingale maximal inequality with Markov's inequality. For a square-integrable martingale starting at zero, the maximal inequality bounds the expected squared maximum by four times the expected predictable variation at the endpoint. Applying Markov's inequality at the threshold $\epsilon a_n$ gives the following bound for every $\epsilon>0$:

P\left\{\sup_{0\leq t\leq \tau}|M_n(t)|>\epsilon a_n\right\} \leq \frac{E\left\{\sup_{0\leq t\leq \tau}|M_n(t)|^2\right\}}{\epsilon^2a_n^2} \leq \frac{4E\{\langle M_n\rangle(\tau)\}}{\epsilon^2a_n^2}.

Under the theorem's condition, the final bound tends to zero for each fixed $\epsilon>0$, which proves the result. The predictable-variation condition is sufficient, not necessary: a particular martingale may be small for other reasons even when this bound does not establish it.

Independent Unit-Jump Counts

A common setting consists of $n$ individuals with counting processes $N_i(t)$, compensators $A_i(t)$, and compensated processes $M_i(t)=N_i(t)-A_i(t)$. The aggregate count and compensator are $N_n(t)=\sum_{i=1}^nN_i(t)$ and $A_n(t)=\sum_{i=1}^nA_i(t)$; hence $M_n(t)=\sum_{i=1}^nM_i(t)$, provided the component compensators are defined relative to the same filtration.

If the component event processes have no simultaneous jumps, the aggregate has unit jumps. With continuous compensators, its predictable variation is then $A_n$. Thus a bound $E\{A_n(\tau)\}\leq Cn$, for a constant $C$ not depending on $n$, implies $E\{\langle M_n\rangle(\tau)\}/n^2\leq C/n\to0$. The theorem gives $M_n/n\to0$ uniformly in probability. Independence is one way to support this type of argument, but the direct variation bound is the condition that matters; dependence is not automatically disqualifying if the aggregate variation is still controlled.

Worked Example: Equal-Rate Independent Event Processes

Suppose, hypothetically, that $n$ independent subjects each have a Poisson event process with rate 2 events per year, observed for $\tau=3$ years. Each subject's compensator is $A_i(t)=2t$. Independent Poisson processes have no simultaneous jumps with probability one, so the aggregate has unit jumps and $\langle M_n\rangle(3)=A_n(3)=6n$.

Take $a_n=n$. For any $\epsilon>0$, the maximal-inequality bound is $$ P\left\{\sup_{0\leq t\leq3}|M_n(t)|>\epsilon n\right\} \leq \frac{4(6n)}{\epsilon^2n^2} =\frac{24}{\epsilon^2n}. $$

For $n=10{,}000$ and $\epsilon=0.2$, this upper bound is $24/(0.04\times10{,}000)=0.06$. It is an upper bound, not the exact probability. As $n$ increases, the bound tends to zero for every fixed $\epsilon$, so the aggregate count minus its compensator, divided by $n$, converges uniformly to zero in probability over the three-year interval. Time units are years, and the counts and compensators are measured in events.

Worked Example: Heterogeneous Bounded Intensities

Consider a hypothetical set of 500 individuals followed for four years. Assume the individual processes have no simultaneous jumps, have continuous compensators, and have predictable intensities bounded by 0.2 events per year for 300 individuals and 0.6 events per year for the other 200. These deterministic bounds imply $$ A_n(4)\leq \{300(0.2)+200(0.6)\}(4)=720 $$ events. The no-simultaneous-jumps assumption ensures that the aggregate predictable variation equals this aggregate compensator.

For scaling $a_n=n=500$ and threshold $\epsilon=0.15$, the event in the theorem is that the maximum discrepancy exceeds $0.15(500)=75$ events. The probability bound is $$ P\left\{\sup_{0\leq t\leq4}|M_n(t)|>75\right\} \leq \frac{4(720)}{75^2} =\frac{2880}{5625}\approx0.512. $$

This bound is valid but not especially informative at this sample size: it permits a probability as high as about 0.512. Its value is that it establishes the required order. With the same intensity bounds and the same proportional threshold, the bound is $4(720n/500)/(0.15n)^2$, which decreases in proportion to $1/n$. The conclusion is asymptotic, not a promise of a tight finite-sample bound.

Why Aggregate Variation Must Be Checked

Adding individual compensators always gives the compensator of the summed counts under the shared-filtration setup above. It does not, by itself, show that the predictable variation of the aggregate martingale equals that sum. When jumps occur together, the aggregate jump is larger, and its squared size contributes to its variation. The next hypothetical example shows that this can defeat the law of large numbers at scale $n$.

Worked Example: A Shared Event Process Does Not Average Out

Suppose $n$ individuals all experience exactly the same events, represented by one Poisson process $P(t)$ of rate 0.5 events per year, over two years. For every individual, $N_i(t)=P(t)$, so each individual has compensator $0.5t$, with expected count one event over the follow-up period. But the aggregate count is $N_n(t)=nP(t)$, its compensator is $A_n(t)=0.5nt$, and its compensated count is $M_n(t)=n\{P(t)-0.5t\}$.

Each aggregate jump has size $n$. Its predictable variation at two years is $$ \langle M_n\rangle(2)=n^2(0.5)(2)=n^2, \qquad \frac{\langle M_n\rangle(2)}{n^2}=1. $$ The scaled martingale at the endpoint is $M_n(2)/n=P(2)-1$, which has the same nondegenerate distribution for every $n$. In particular, since $P(2)\sim\operatorname{Poisson}(1)$, $$ P\{|M_n(2)/n|>0.5\}=P\{P(2)\ne1\}=1-e^{-1}\approx0.6321. $$ This probability does not tend to zero. Although each individual's compensator is bounded, the aggregate does not satisfy the required small-variation condition at scale $n$. The shared jumps prevent averaging.

Key condition: For an aggregate compensated count, establish a bound on $E\{\langle M_n\rangle(\tau)\}$ for the aggregate martingale itself. An individual-intensity bound is not enough unless additional assumptions, such as no simultaneous jumps, justify the step from individual compensators to aggregate predictable variation.

Relation to Other Large-Sample Approaches

The scaling distinguishes a law of large numbers from a central limit approximation. If predictable variation is of order $n$, the theorem supports scaling by $n$, because $n/n^2\to0$. It does not generally show that $M_n/\sqrt n$ vanishes: at that scale, variation of order $n$ can leave nontrivial fluctuations. Martingale central limit arguments address such fluctuation-scale behavior under additional conditions; they answer a different question.

An ordinary law of large numbers for independent observations is another possible route when the quantity can be represented as a sum of independent subject-level terms and its assumptions hold. The compensated-count result is useful when the event data are naturally described through intensities and martingales, and when a uniform-in-time statement is needed. Directly analyzing the compensator can also be useful, but it does not replace checking whether the realized count is close to it.

A typical application is to write a normalized event count as $(N_n-A_n)/n+A_n/n$. Once the first term is negligible, the large-sample behavior of the count can be studied through the normalized compensator. The theorem alone does not guarantee that $A_n/n$ converges to a particular limit; that requires a separate argument about the compensators.

Diagnostics, Assumptions, and Limitations

Before using the result, verify that the martingale is defined relative to the intended filtration, starts at zero (or that its initial value is negligible at the chosen scale), is square-integrable on the interval, and has finite expected predictable variation. Then determine the aggregate jump structure and derive a variation bound for the actual martingale. In applications, useful checks include inspecting whether events can be shared across individuals, whether simultaneous events are possible by design or measurement, and whether the assumed intensity bounds are plausible over the full follow-up period.

Failure to obtain a small variation ratio does not prove that convergence fails; it means this sufficient condition has not established convergence. Conversely, a small average individual intensity does not establish the result if dependence creates large aggregate jumps. The fixed horizon $\tau$ also matters: a bound on one finite interval does not automatically justify a claim over an interval that grows with $n$. Finally, the theorem gives convergence in probability, not a finite-sample confidence statement or almost-sure convergence.

Common Mistakes

Equating aggregate compensator and aggregate predictable variation without checking jumps. The compensators add, but simultaneous jumps can make aggregate jump sizes exceed one. Establish no simultaneous jumps under suitable unit-jump conditions, or bound the aggregate predictable variation directly.
Scaling by $n$ without checking the order of variation. The relevant ratio is $E\{\langle M_n\rangle(\tau)\}/n^2$, not simply an individual event rate.
Reading a probability bound as an exact probability. The maximal-inequality result supplies an upper bound, which can be loose, especially at moderate sample sizes.
Claiming almost-sure convergence from this theorem. The stated conclusion is uniform convergence in probability; stronger modes of convergence need further conditions and arguments.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $E\{\langle M_n\rangle(\tau)\}\leq 8n$ and $a_n=n$, what does the theorem imply?
Why does summing individual compensators not generally determine the predictable variation of the aggregate compensated count?
For $n$ independent unit-jump processes with continuous compensators, suppose $E\{A_n(\tau)\}\leq 3n$. What is the maximal-inequality bound for $P\{\sup_{t\leq\tau}|M_n(t)|>0.1n\}$?
In the shared-process example, what is the aggregate predictable variation at time 2, and why does normalization by $n$ not give convergence to zero?

Solutions. (1) The ratio is at most $8n/n^2=8/n$, which tends to zero. Thus $\sup_{t\leq\tau}|M_n(t)|/n\to0$ in probability. (2) Simultaneous component jumps can produce an aggregate jump larger than one; predictable variation accounts for squared aggregate jump sizes, so it can exceed the sum of component compensators. (3) The bound is $4(3n)/\{(0.1n)^2\}=1200/n$. It tends to zero, although it may exceed one for smaller $n$. (4) It is $n^2$; after division by $n^2$, it remains 1, and $M_n(2)/n=P(2)-1$ has a nondegenerate distribution independent of $n$, so it does not converge to zero in probability.