Why Simultaneous Jumps Matter
When several event types are followed over time, their compensated counts form a vector of martingales. Whether two components are orthogonal depends on how their jumps line up, not simply on whether their event labels differ. If they never jump together, their cross-variation is zero. If they share jumps, those jumps contribute directly to predictable covariation, which in turn affects covariance calculations for event-process estimators and weighted martingale integrals.
The crucial condition is process-level absence of simultaneous jumps. For example, two event categories may be mutually exclusive for one individual, but counts aggregated across individuals can still jump at the same recorded time if separate individuals experience different events together. Continuous-time models with continuous event-time distributions often make exact ties have probability zero, but the relevant assumption is about the counting processes being analyzed, not just the names of the categories.
Compensated Processes and Covariation
Let \(N_j(t)\) and \(N_k(t)\) be adapted, locally finite, unit-jump counting processes for two event types. Suppose their compensators are continuous and absolutely continuous in time: \(A_j(t)=\int_0^t \lambda_j(u)\,du\) and \(A_k(t)=\int_0^t \lambda_k(u)\,du\), where each intensity is nonnegative and predictable. The compensated processes are \(M_j(t)=N_j(t)-A_j(t)\) and \(M_k(t)=N_k(t)-A_k(t)\). Under the usual integrability conditions, these are local martingales; finite expected compensators on the time interval give square-integrable martingales.
Since \(A_j\) and \(A_k\) are continuous, \(\Delta M_j(t)=\Delta N_j(t)\) and \(\Delta M_k(t)=\Delta N_k(t)\). Thus, for these processes, \([M_j,M_k](t)=\sum_{u\leq t}\Delta N_j(u)\Delta N_k(u)\). For unit jumps this is exactly the number of times both processes jump at once. This identity is specific to the setting just stated; for instance, it would not describe the full quadratic covariation of continuous martingales such as Brownian motion.
Two local martingales starting at zero are called orthogonal when their product is a local martingale; for square-integrable martingales, zero predictable covariation is the standard condition used to express this property. In the present counting-process setting, if there are no simultaneous jumps, the jump-product sum is identically zero, and therefore both optional and predictable covariation are zero. Orthogonality does not, in general, mean that the processes are independent.
Why No Simultaneous Jumps Gives Orthogonality
The product formula for semimartingales provides the key justification. For martingales starting at zero, it gives \(M_j(t)M_k(t)=\int_0^t M_j(u-)\,dM_k(u)+\int_0^t M_k(u-)\,dM_j(u)+[M_j,M_k](t)\). Under square-integrability conditions that make the stochastic integrals true martingales, they have mean zero. Also, the predictable covariation compensates the optional covariation, so their expectations agree when integrable. Consequently, \(E\{M_j(t)M_k(t)\}=E\{\langle M_j,M_k\rangle(t)\}\).
If there are no simultaneous jumps, \([M_j,M_k](t)=0\) for every \(t\). Its predictable compensator is then also zero, giving \(\langle M_j,M_k\rangle(t)=0\). The product formula shows why this is more than a visual description of two paths: under the stated integrability conditions, it also yields zero covariance between the centered martingale values at a fixed time. The converse interpretation needs care: a zero covariance at one time does not establish that simultaneous jumps never occur.
Worked Example: Mutually Exclusive First Events
Consider one hypothetical subject followed until the first of two failure types. Let \(Y(t)\) indicate that the subject is still at risk, and suppose the type-specific intensities are \(\lambda_1(t)=0.08Y(t)\) and \(\lambda_2(t)=0.12Y(t)\) per year. Each event ends follow-up, and the event-time model has continuous compensators. These assumptions mean that each type count can jump at most once and that, for this subject, the two type counts cannot jump at the same time.
The compensators are \(A_1(t)=\int_0^t 0.08Y(u)\,du\) and \(A_2(t)=\int_0^t 0.12Y(u)\,du\), and the compensated processes are \(M_1=N_1-A_1\) and \(M_2=N_2-A_2\). At every time \(u\), at most one of \(\Delta N_1(u)\) and \(\Delta N_2(u)\) is 1. Therefore, \[ \Delta N_1(u)\Delta N_2(u)=0 \quad\text{for every }u,\qquad [M_1,M_2](t)=\sum_{u\leq t}0=0. \]
It follows that \(\langle M_1,M_2\rangle(t)=0\): the compensated type-specific counts are orthogonal. This conclusion does not require the intensities to be constant or deterministic; \(Y(t)\) is a predictable risk indicator and both intensities may depend on the observed history. The conclusion does depend on analyzing one subject's mutually exclusive first-event counts. For aggregated counts across subjects, the same reasoning applies only if the aggregate processes themselves have no simultaneous jumps.
Shared Jumps and Predictable Covariation
When the two event-type processes can jump together, their shared jumps contribute positively to their optional covariation. Define \(N_{jk}(t)\) as the count of times up to \(t\) at which both unit-jump processes jump. Suppose \(N_{jk}\) has a continuous compensator \(A_{jk}(t)=\int_0^t\lambda_{jk}(u)\,du\), with predictable shared-jump intensity \(\lambda_{jk}\). Then \([M_j,M_k](t)=N_{jk}(t)\) and \(\langle M_j,M_k\rangle(t)=A_{jk}(t)=\int_0^t\lambda_{jk}(u)\,du\).
If each process may jump either alone or together, the marginal intensities include the shared events. In that case \(\lambda_j-\lambda_{jk}\) and \(\lambda_k-\lambda_{jk}\) are the respective intensities of jumps occurring in only one process, and must be nonnegative; hence \(\lambda_{jk}\leq\min(\lambda_j,\lambda_k)\) almost everywhere. Shared-jump intensity is not an additional intensity to add to each marginal after it has already been included there.
Worked Example: Covariation From a Shared Event Component
Suppose two hypothetical event counts have jumps from three independent Poisson components: events affecting type 1 only occur at rate \(0.07\) per year, events affecting type 2 only at rate \(0.04\) per year, and shared events that increment both counts occur at rate \(0.03\) per year. Let \(U_1,U_2,S\) denote these counts. Set \(N_1=U_1+S\) and \(N_2=U_2+S\). The marginal intensities are therefore \(\lambda_1=0.07+0.03=0.10\) and \(\lambda_2=0.04+0.03=0.07\) per year; the shared-jump intensity is \(\lambda_{12}=0.03\) per year. These component rates are constant and predictable, and the model explicitly permits simultaneous jumps through \(S\).
For a five-year interval, the predictable covariation is \[ \langle M_1,M_2\rangle(5)=\int_0^5 0.03\,du =0.03(5)=0.15. \] The optional covariation is the observed shared-event count \(S(5)\), whose expectation is also \(0.15\). The product identity therefore gives \(E\{M_1(5)M_2(5)\}=0.15\), or a covariance of \(0.15\) counts squared, since both martingales have mean zero. The nonzero covariance is due to the shared component, not to an error in the compensation.
Predictable Weights and Cross-Variation
Orthogonality is also useful after predictable weighting. For predictable integrands \(H_j\) and \(H_k\) for which the integrals are square-integrable, define \(Z_j(t)=\int_0^t H_j(u)\,dM_j(u)\) and \(Z_k(t)=\int_0^t H_k(u)\,dM_k(u)\). Their predictable covariation is \(\langle Z_j,Z_k\rangle(t)=\int_0^t H_j(u)H_k(u)\,d\langle M_j,M_k\rangle(u)\). Thus predictable weights preserve orthogonality when the original predictable covariation is zero. With shared jumps, the product of the weights determines whether each shared jump contributes positively or negatively to cross-covariation.
Worked Example: Oppositely Weighted Shared Jumps
Use two hypothetical event processes whose shared-jump intensity is constant at \(0.025\) per year over four years. Weight the first compensated process by \(H_1=2\) and the second by \(H_2=-1\). These are constant, hence predictable, weights, and assume finite expected counts so the integrals are square-integrable.
The predictable covariation of the weighted integrals at year 4 is \[ \langle Z_1,Z_2\rangle(4) =\int_0^4 (2)(-1)(0.025)\,du =(-0.05)(4)=-0.20. \] Because the weighted integrals are mean-zero martingales, their covariance is the expectation of this predictable covariation, also \(-0.20\) in this constant-rate model. The negative sign comes from the opposite signs of the weights; the underlying shared-jump count itself is nonnegative.
Diagnostics, Alternatives, and Limitations
A practical check begins with the path structure and the time scale used to define a jump. Verify that the compensators are continuous, that the counts have unit jumps, and that recorded event times can be compared at sufficient resolution. Then inspect the number and mechanism of simultaneous jumps. If the model treats two event types as mutually exclusive for each individual, determine whether the analysis concerns individual processes or sums across individuals. Rounding event times to days, for example, can create apparent ties that were not simultaneous in continuous time; conversely, a shared shock may create genuine common jumps.
When simultaneous jumps occur, the appropriate alternative to setting cross-covariation to zero is to model or estimate their predictable rate and use \(\int_0^t\lambda_{jk}(u)\,du\). If the common-jump rate is unknown, it cannot be inferred from the marginal intensities alone: two models can have the same \(\lambda_j\) and \(\lambda_k\) but different overlap. More generally, zero cross-covariation does not establish independence, and positive cross-covariation describes shared martingale variation rather than causal dependence.
| Situation | Predictable covariation | Interpretation |
|---|---|---|
| No simultaneous jumps of the two processes | \(\langle M_j,M_k\rangle=0\) | The compensated processes are orthogonal under the stated conditions. |
| Shared jumps with predictable rate \(\lambda_{jk}\) | \(\int_0^t\lambda_{jk}(u)\,du\) | Shared events contribute to cross-variation. |
| Marginal rates known but overlap unspecified | Not determined by marginal rates alone | The joint jump mechanism must also be specified. |
A common mistake is to argue, “These are different event types, so their martingales are orthogonal.” Different labels do not rule out simultaneous jumps, particularly for aggregated data or overlapping event definitions. The correction is to check the joint jump structure: calculate \(\Delta N_j(u)\Delta N_k(u)\) at each time, or specify its predictable rate. Another mistake is to call zero covariation independence; orthogonality is a martingale property and is generally weaker than independence.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- Two unit-jump counting processes have continuous compensators and never jump at the same time. What are their optional and predictable covariations?
- Two processes share jumps at a constant predictable rate of \(0.02\) per month. Find their predictable covariation over 18 months.
- Why does the fact that two event types are mutually exclusive for each individual not automatically prove that their aggregated counts have no simultaneous jumps?
- For predictable constants \(H_j=3\) and \(H_k=2\), the underlying shared-jump intensity is \(0.01\) per year. Find the weighted predictable covariation over five years.
- Does zero predictable covariation prove that two event-process martingales are independent? Explain.
Solutions. (1) Both covariations are zero: every jump product is zero, so the optional covariation and its predictable compensator vanish. (2) \(\int_0^{18}0.02\,du=0.36\), in expected shared jumps over the 18-month interval. (3) Different individuals can experience different event types at the same recorded time, so aggregate processes can jump together even if each individual can experience only one type at once. (4) \(\int_0^5(3)(2)(0.01)\,du=0.06(5)=0.30\). (5) No. It establishes orthogonality under the stated martingale conditions, but orthogonality alone does not imply independence.