Martingale Central Limit Theorems for Event Data

Why Event-Count Fluctuations Can Be Approximately Normal

An event count is often decomposed into its predictable accumulation and a residual fluctuation. If $N(t)$ has intensity $\lambda(t)$, its compensated count is $M(t)=N(t)-\int_0^t\lambda(s)\,ds$. Under suitable conditions, many small event contributions combine to give a nearly Gaussian fluctuation. The martingale central limit theorem makes “suitable conditions” precise: the predictable variation must stabilize on the normalization scale, and no small number of unusually large jumps may dominate.

The result is useful for event-history inference because score processes and weighted event-count residuals are often martingales. This tutorial focuses on when a normalized martingale has a normal limit, not on constructing a particular likelihood or variance estimator. The conditions must be checked for the martingale sequence itself, not inferred from the fact that the data contain many events.

Definition: For a square-integrable martingale $M_n(t)$, its predictable variation $\langle M_n\rangle_t$ is the predictable increasing process for which $M_n(t)^2-\langle M_n\rangle_t$ is a martingale, when $M_n(0)=0$. For a pure-jump martingale, its jump measure $\mu_n(ds,dx)$ places a unit of mass at each jump time $s$, with jump size $x=\Delta M_n(s)$; $\nu_n(ds,dx)$ denotes the predictable compensator of that jump measure.

For a compensated count with continuous compensator, jumps of the martingale are the event contributions. If an event at time $t$ contributes a predictable weight $H(t)$, then that jump has size $H(t)$. In a simple counting-process setting, the predictable variation accumulates the squared weights against the intensity. With multiple event types or possible simultaneous jumps, it is important to use the predictable variation and the actual net jump sizes of the martingale being analyzed.

The Martingale Central Limit Theorem

Consider a sequence of square-integrable, pure-jump martingales $M_n(t)$ on a fixed interval $[0,\tau]$, where $\tau$ is finite. Let $b_n$ be deterministic positive normalizing constants with $b_n\to\infty$. The theorem below is stated for the terminal value. It applies, for example, to a sum of compensated event counts or a predictable weighted sum, provided the stated conditions hold.

Theorem: Suppose $M_n(0)=0$ and, for a constant $\sigma^2>0$, $$ \frac{\langle M_n\rangle_\tau}{b_n}\ \xrightarrow{p}\ \sigma^2. $$ Suppose also that for every $\varepsilon>0$, $$ \frac{1}{b_n}\int_0^\tau\int x^2\,\mathbf{1}\{|x|>\varepsilon\sqrt{b_n}\}\,\nu_n(ds,dx)\ \xrightarrow{p}\ 0. $$ Then $$ \frac{M_n(\tau)}{\sqrt{b_n}}\ \xrightarrow{d}\ N(0,\sigma^2). $$ More generally, $M_n(0)=0$ may be replaced by $M_n(0)/\sqrt{b_n}\xrightarrow{p}0$.

The first condition says that the total predictable fluctuation, divided by $b_n$, approaches the finite positive limit $\sigma^2$. The second is the jump Lindeberg condition: on the $\sqrt{b_n}$ scale, the accumulated squared contribution from jumps larger than any fixed fraction $\varepsilon$ becomes negligible. It is a condition on the compensator of the jump measure, not simply a statement that the observed maximum jump happened to be small.

Why These Conditions Lead to a Normal Limit

Write a normalized jump as $y=x/\sqrt{b_n}$. For fixed real $u$, the characteristic-function expansion for a small jump is $$ e^{iuy}-1-iuy=-\frac{u^2y^2}{2}+r_u(y), $$ where, for small $y$, the remainder satisfies $|r_u(y)|\leq C_u|y|^3$ for a constant $C_u$ depending on $u$. Thus, on jumps with $|y|\leq\delta$, the remainder is bounded by $C_u\delta y^2$. Their total squared contribution is controlled by predictable variation. On jumps with $|y|>\delta$, the Lindeberg condition makes the accumulated squared contribution negligible. Taking $n$ large and then $\delta$ small leaves the quadratic term as the leading contribution.

That quadratic term is determined by predictable variation, which converges after normalization to $\sigma^2$. The martingale characteristic-function argument, with localization where needed, therefore has the same limiting quadratic exponent as a centered normal variable: $-u^2\sigma^2/2$. This is the characteristic exponent of $N(0,\sigma^2)$, yielding the stated limit. The argument also shows why neither condition can be casually dropped: unpredictable accumulation or non-negligible large jumps can change the limiting distribution.

Worked Example: A Sum of Independent Poisson Event Counts

Suppose, hypothetically, that $n$ independent event processes are Poisson processes with rate $0.4$ events per year. Observe each for $\tau=2$ years and define $M_n(t)$ as the total number of events by time $t$, minus its compensator. The total rate is $0.4n$ events per year, so $M_n(0)=0$.

Choose $b_n=n$. The predictable variation at two years is $$ \langle M_n\rangle_2=0.4n(2)=0.8n, \qquad \frac{\langle M_n\rangle_2}{b_n}=0.8. $$ Thus the predictable-variation condition holds with $\sigma^2=0.8$, in squared event-count units for this normalization.

Each jump has size $1$. For any fixed $\varepsilon>0$, once $n>\varepsilon^{-2}$, we have $1\leq\varepsilon\sqrt{n}$, so there are no jumps larger than the Lindeberg threshold. The Lindeberg integral is then zero. The theorem gives $$ \frac{M_n(2)}{\sqrt{n}}\ \xrightarrow{d}\ N(0,0.8). $$ For large $n$, the aggregate compensated count, measured in units of $\sqrt n$, is approximately centered normal with variance $0.8$.

Worked Example: Predictably Weighted Event Contributions

Consider a hypothetical collection of $n$ independent Poisson event processes observed for four years, each with rate $0.25$ events per year. Let half the processes contribute weight $1$ at each event and half contribute weight $2$. These weights are deterministic, hence predictable. Let $M_n$ be the sum of weighted counts minus their weighted compensators, and take $n$ even.

Each process has expected one event over four years. The squared weights sum to $n/2+4(n/2)=2.5n$, so the predictable variation is $$ \langle M_n\rangle_4=2.5n, \qquad \frac{\langle M_n\rangle_4}{n}=2.5. $$ Set $b_n=n$; the limiting variance is $\sigma^2=2.5$. Every jump has size at most $2$. For each fixed $\varepsilon>0$, when $n>(2/\varepsilon)^2$, no jump exceeds $\varepsilon\sqrt n$, so the Lindeberg integral is zero. The initial value is zero. Therefore $M_n(4)/\sqrt n$ converges in distribution to $N(0,2.5)$. The weights change the variance through their squares, but their bounded size ensures that no individual weighted event dominates asymptotically.

What Goes Wrong When a Condition Fails

Worked Example: Stable Variation but Non-Negligible Large Jumps

For a hypothetical event-count array, let $N(t)$ be a rate-one Poisson process and define $M_n(t)=\sqrt n\{N(t)-t\}$ on $[0,1]$. This is a compensated weighted event count, with predictable weight $\sqrt n$, and $M_n(0)=0$. Take $b_n=n$.

Its predictable variation is $\langle M_n\rangle_1=n$, so $\langle M_n\rangle_1/b_n=1$. But each jump has size $\sqrt n$. For any $0<\varepsilon<1$, the Lindeberg integral divided by $b_n$ is $$ \frac{1}{n}\int_0^1 n\,dt=1, $$ not a quantity tending to zero. Indeed, $M_n(1)/\sqrt n=N(1)-1$ has a centered Poisson distribution, not a normal distribution. Stable predictable variation alone does not ensure a Gaussian limit; the large-jump condition rules out this failure.

Worked Example: Why the Initial Value Must Be Controlled

Let $N_n(t)$ be a Poisson process of rate $n$, and set $M_n(t)=\sqrt n+N_n(t)-nt$ on $[0,1]$, with $b_n=n$. The process is a square-integrable martingale, and its predictable variation satisfies $\langle M_n\rangle_1/n=1$. Its jumps have size $1$, so the Lindeberg integral is zero for all sufficiently large $n$, for each fixed $\varepsilon>0$.

Nevertheless, $M_n(0)/\sqrt n=1$, not zero or negligible. Moreover, $$ \frac{M_n(1)}{\sqrt n}=1+\frac{N_n(1)-n}{\sqrt n}\ \xrightarrow{d}\ N(1,1). $$ The limit is shifted, not $N(0,1)$. The theorem’s centered conclusion requires the initial-value condition as well as the variation and jump conditions. In the common compensated-count formulation, the martingale is an integral starting at zero, so $M_n(0)=0$ holds by construction.

Checking Applicability and Choosing an Approach

A practical check starts by writing down the exact martingale whose limit is sought. Identify its initial value, normalization $b_n$, predictable variation, and jump sizes. Then verify each condition at the same terminal time $\tau$. In a weighted event process, weights should be predictable; if distinct event types can jump together, calculate the net jump of the combined martingale rather than treating simultaneous contributions as separate jumps without justification.

Center the process.
Confirm $M_n(0)=0$, or show $M_n(0)/\sqrt{b_n}\xrightarrow{p}0$. A non-negligible starting value can shift the limit.

Check predictable variation.
Establish convergence in probability of $\langle M_n\rangle_\tau/b_n$ to a finite, positive constant.

Check the jump tail.
For each fixed $\varepsilon>0$, evaluate the normalized compensator integral over jumps larger than $\varepsilon\sqrt{b_n}$.

State the conclusion on its scale.
Only after all conditions hold, report $M_n(\tau)/\sqrt{b_n}\xrightarrow{d}N(0,\sigma^2)$.

A frequent diagnostic is whether a few large weights or event contributions carry a meaningful part of the total variation. If so, the Lindeberg condition may fail even when the normalized predictable variation appears stable. Also check that the bracket is computed for the martingale of interest: a variance calculation for separate event types may omit cross-contributions when jumps can be simultaneous.

The ordinary independent-observation central limit theorem is a natural alternative when the target is a sum of independent, identically distributed variables with suitable finite variance. The martingale theorem is useful when event contributions are conditionally structured over time and their increments need not be independent. It accommodates predictable, history-dependent intensities and weights, but replaces independence assumptions with martingale and variation conditions. Neither method automatically handles a non-negligible initial offset or dominant large jumps. This result is also a scalar terminal-time statement; process-level Gaussian limits require additional conditions.

Common Mistakes and Limitations

Checking variation only. Convergence of $\langle M_n\rangle_\tau/b_n$ does not rule out a small number of jumps carrying substantial variation. Verify the Lindeberg condition separately.
Forgetting the initial value. The bracket describes subsequent martingale fluctuation; it does not remove a starting offset. Include $M_n(0)/\sqrt{b_n}\xrightarrow{p}0$, commonly by defining a compensated integral that starts at zero.
Using the wrong jump size. For a sum of event types, the relevant jump is the change in the entire martingale at that time. Simultaneous contributions may combine.
Claiming an exact finite-sample normal law. The theorem gives convergence in distribution as $n$ grows. It does not say that a finite-sample event count is exactly normal.
Ignoring the scope of the conclusion. The theorem concerns a terminal value on a fixed time interval, under its stated assumptions. It does not by itself provide a process-wide approximation or validate a misspecified intensity model.

The central lesson is that approximate normality reflects both accumulation and balance: predictable variation must settle at the chosen scale, and no jump may remain influential on that scale. A large sample size or a large total event count, by itself, is not a proof of either condition.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If $\langle M_n\rangle_\tau/n\to_p 3$, what additional conditions are needed for $M_n(\tau)/\sqrt n$ to converge to $N(0,3)$?
In a sequence with jumps all bounded by $5$ and $b_n=n$, why is the Lindeberg condition eventually satisfied?
Suppose the normalized predictable variation converges to $2$, but the Lindeberg integral converges to $0.4$ for some positive threshold. Can the stated theorem be used to conclude a normal limit?
What limit does a non-negligible initial value create in the Poisson example with $M_n(0)=\sqrt n$?

Solutions. (1) Require the initial value to be negligible on the $\sqrt n$ scale (in particular, $M_n(0)=0$) and require the Lindeberg integral to converge in probability to zero for every $\varepsilon>0$. (2) For any fixed $\varepsilon>0$, once $n>(5/\varepsilon)^2$, the threshold $\varepsilon\sqrt n$ exceeds every possible jump, so the integral is zero. (3) No. The Lindeberg condition fails at that threshold, so this theorem does not establish a normal limit. (4) The normalized terminal value converges to $N(1,1)$, a shifted normal distribution rather than $N(0,1)$.