Why Track Two Kinds of Variation?
A compensated counting process subtracts its predictable accumulated event rate from the observed event count. The result is a martingale: its future increments have conditional mean zero given the observed history. But a martingale can still fluctuate, and inference often requires measuring the size of those fluctuations. There are two complementary quantities: quadratic variation, which records the jumps that actually occurred, and predictable variation, which records the variation anticipated from the event intensity.
The distinction matters in event-history data. The observed jump total is available after following a path; the predictable variation is built from the intensity and can vary across paths when the risk set or covariates vary. Confusing the two can lead to incorrect variance calculations.
Here, \(\Delta M(u)=M(u)-M(u-)\) is the jump at time \(u\). Simplicity means that \(N\) has jumps of size one and no simultaneous events within this process. The assumption that \(A\) is continuous is standard when the model specifies an ordinary intensity with respect to time. It ensures that compensating for the event rate does not itself introduce jumps.
From Event Jumps to the Two Variation Processes
Between events, \(N\) is constant and \(M\) changes continuously as \(A\) accumulates. At an event time, \(\Delta N=1\); because \(A\) is continuous, \(\Delta M=1\). At all other times \(\Delta M=0\). Therefore the squared jumps sum to the event count itself:
This is a pathwise identity: it holds for each realized path satisfying the assumptions. To obtain predictable variation, use the integration-by-parts identity for a process with jumps:
Substituting \([M]=N\), and using \(N=M+A\), gives
A stochastic integral of a predictable integrand against a martingale is a local martingale; it is a true martingale when the relevant integrability conditions hold. Thus \(M^2-A\) is a martingale in the square-integrable setting, and the defining characterization gives \(\langle M\rangle=A\). For merely local conditions, the corresponding statements are understood locally, with stopping or localization as needed.
The predictable variation is closely connected to variance. For a square-integrable martingale and \(s\le t\),
This says that the conditional second moment of a future martingale increment is the conditional expectation of the future increase in predictable variation. If the increase in \(A\) is already known at time \(s\), it is the conditional variance itself. In general, \(A(t)-A(s)\) can still be random after \(s\), so it is the conditional expectation of that increase—not necessarily its unconditioned realized value—that equals the conditional second moment.
Worked Example: Unweighted Events at a Constant Rate
Consider a hypothetical simple counting process observed for 4 days, with specified intensity \(\lambda(t)=0.6\) events per day. Suppose the observed path contains three events by day 4. Assume the intensity is the valid predictable intensity for this process, the compensator is continuous, and the required moments are finite.
The compensator accumulated by day 4 is
Each event contributes a jump of size 1 to \(M=N-A\); the continuous compensator contributes no jump. Thus the observed quadratic variation is
The predictable variation is \( \langle M\rangle(4)=A(4)=2.4\) events. The observed compensated value is \(M(4)=3-2.4=0.6\) events, so \(M(4)^2=0.36\) squared events. This is not the same as either variation at this particular endpoint: the martingale identity concerns \(M^2-A\), not an identity \(M^2=A\) on every path. Across repeated realizations under the model, \(E[M(4)^2]=E[A(4)]=2.4\) squared events.
Predictably Weighted Event Processes
A predictable weight allows different events to contribute different amounts. Let \(H(t)\) be predictable and sufficiently integrable, and define the martingale transform \(M_H(t)=\int_0^t H(u)\,dM(u)\). Predictability means the weight at an event time is determined from information available just before that time; it cannot use the event's future or its unobserved jump to choose the weight. At an event, the jump of \(M_H\) is \(H(u)\Delta M(u)\). Consequently, its quadratic variation sums squared weighted jumps. Its predictable variation replaces the event count measure with the compensator:
The second formula follows because the compensator of the weighted jump-square process \(\int H^2\,dN\) is \(\int H^2\,dA\), when \(H^2\) is predictable and the integrability conditions hold. Equivalently, the stochastic-integral variation rule gives \(\langle\int H\,dM\rangle=\int H^2\,d\langle M\rangle\). These formulas separate realized weighted jump variation from the predictable accumulation implied by the intensity.
Worked Example: A Time-Varying Predictable Weight
In a hypothetical process, let the specified intensity be \(0.5\) events per unit time on \(0\le t<2\) and \(0.2\) events per unit time on \(2\le t\le5\). Let the deterministic weight be \(H(t)=1\) before time 2 and \(H(t)=3\) from time 2 onward. Suppose the observed event times are \(0.7, 1.6, 2.4,\) and \(4.2\). The weight is deterministic and hence predictable; assume the stated intensity is valid and the integrals are finite.
The predictable variation is
The observed weighted quadratic variation uses the weights at the four event times:
For comparison, the weighted martingale itself at time 5 equals the sum of weighted observed events minus the weighted compensator:
The realized quadratic variation is 20 squared weight units, while the predictable variation is 6.4 squared weight units. Their difference is not evidence of an error: one is determined by the four realized event jumps, and the other is determined by the intensity integrated with the squared weight.
Worked Example: Variation with a Changing Risk Set
Suppose, hypothetically, that two individuals are initially at risk and each has event intensity \(0.3\) per unit time while at risk. One event occurs at time \(0.5\), leaving one individual at risk; the second event occurs at time \(2.0\), leaving nobody at risk thereafter. Take the observed risk-set path as predictable, with no other changes. The individual-level intensity model is assumed valid, with continuous compensator and finite moments.
The total intensity is \(0.3Y(t)\), where \(Y(t)\) is the number at risk. The compensator accumulated through time 2 is therefore
There are two unit jumps, so \([M](2)=2\), whereas \(\langle M\rangle(2)=0.75\). The endpoint martingale value is \(M(2)=2-0.75=1.25\). The risk set makes the compensator path-dependent in general; it must be calculated from the predictable risk process over each interval, not by multiplying the terminal risk-set size by total follow-up.
Comparison, Diagnostics, and Limits
| Quantity | What it records | For \(M=N-A\) |
|---|---|---|
| Quadratic variation \([M]\) | Squared jumps that actually occurred on the observed path | \(N\) |
| Predictable variation \(\langle M\rangle\) | Predictable accumulation corresponding to squared martingale variation | \(A=\int\lambda\,du\) |
| Weighted quadratic variation \([M_H]\) | Squared weighted jumps actually observed | \(\int H^2\,dN\) |
| Weighted predictable variation \(\langle M_H\rangle\) | Intensity-based accumulation for the weighted process | \(\int H^2\,dA\) |
The observed quadratic variation is useful for describing realized path fluctuation, while predictable variation is the natural intensity-based variance quantity. They are not competing estimates that must match on one path. Under square-integrability, the expected quadratic variation agrees with expected predictable variation; for example, \(E([M](t))=E(A(t))\) when these expectations are finite.
Several checks help detect mistakes. First, verify that every jump has the size assumed: if events can have non-unit marks, the squared jump sum is not simply \(N\). Second, check that \(A\) is continuous before using \([N-A]=N\). Third, ensure a weight is predictable and its squared integral against \(A\) is finite for the desired square-integrable result. A weight selected after seeing whether an event occurred generally does not define the martingale transform used here. Finally, verify that the intensity and risk-set process are correctly specified; a misspecified compensator need not make \(N-A\) a martingale, invalidating the predictable-variation interpretation.
If a compensator has jumps, then \(M\) may have jumps at times when \(N\) does not, and \([M]\) must be calculated from all jumps of \(M\), not just the event count. If events have marks or jump sizes other than one, the pathwise quadratic variation sums their squared jump sizes. The simple formulas above rely on the stated simple-process and continuous-compensator assumptions; they should not be transferred unchanged to those settings.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A simple counting process has a continuous compensator \(A(3)=1.8\) and has two observed events by time 3. Find \([M](3)\) and \(\langle M\rangle(3)\).
- For a predictable constant weight \(H=2\), the unweighted process has \(N(4)=3\) and \(A(4)=2.5\). Find the weighted quadratic and predictable variations.
- Why does predictability matter when defining a weighted event martingale?
- State the conditional second-moment relationship involving the increase in predictable variation.
Solutions. (1) Since \([M]=N\), \([M](3)=2\); since \(\langle M\rangle=A\), \(\langle M\rangle(3)=1.8\). (2) The weight is squared, so \([M_H](4)=2^2N(4)=4(3)=12\), and \(\langle M_H\rangle(4)=2^2A(4)=4(2.5)=10\). (3) Predictability ensures the weight is determined from information available just before an increment, as required for the stochastic integral against a martingale to retain its martingale property under suitable integrability. (4) For a square-integrable martingale, \(E[(M(t)-M(s))^2\mid\mathcal{F}_s]=E[\langle M\rangle(t)-\langle M\rangle(s)\mid\mathcal{F}_s]\) for \(s\le t\).