Why the Cause Matters When Event Times Are Intervals
In interval-censored data, the exact event time is not recorded. Instead, an observation may establish that the first event occurred after one time and by a later time. In a competing-risks setting, the event may also be classified by its cause. These two pieces of information determine a subject’s likelihood contribution: an event of cause \(k\) in an interval is not the same observation as simply being event-free at the end of that interval.
For example, suppose a participant is known to be free of either of two first events at a visit at time \(L\). At a later visit at time \(R\), a first event is identified and its cause is known to be cause 1. The event time is not known, but it is known to lie in \((L,R]\). The contribution must account for the chance of surviving event-free to each possible time in that interval and then experiencing cause 1. By contrast, if the participant is still event-free at \(R\), the contribution is the probability of remaining event-free through \(R\).
The focus here is the subject-level likelihood contribution given these observed statuses and intervals. It is not necessary to know the exact event time to write that contribution. The earlier material on likelihood contributions from event-time intervals supplies the basic interval-probability idea; here the probability is divided by cause.
Model and Likelihood Contributions
Assume a first-event competing-risks model with cause-specific hazard \(\lambda_k(t)\). For a subject still event-free immediately before time \(t\), \(\lambda_k(t)\) is the instantaneous rate of a cause-\(k\) event at \(t\). Write the total hazard as \(\lambda_{\cdot}(t)=\sum_{k=1}^K\lambda_k(t)\). Under continuous event times, the event-free survival function is
The cause-\(k\) cumulative incidence function accumulates the chance of having survived event-free until \(u\) and then experiencing cause \(k\) at \(u\):
These definitions imply three basic contributions. An event of known cause \(k\) in \((L,R]\) contributes the probability \(P(L<T\le R,J=k)\). An event-free observation through \(R\) contributes \(P(T>R)\). If the cause is not observed for an event known to occur in \((L,R]\), the contribution sums over all possible causes.
The last equality follows because the probability of any first event in \((L,R]\) is the probability of being event-free at \(L\) minus the probability of still being event-free at \(R\). It does not mean that an event of a particular cause has probability \(S(L)-S(R)\); that probability includes every cause.
Why the Cause-Specific Interval Formula Holds
For a small time increment \(du\), conditional on being event-free immediately before \(u\), the chance of a cause-\(k\) event in that increment is approximately \(\lambda_k(u)\,du\). The chance of reaching \(u\) event-free from time zero is \(S(u)\). Thus the unconditional probability of a cause-\(k\) event in that increment is approximately \(S(u)\lambda_k(u)\,du\). Adding these probabilities over \((L,R]\) gives the integral.
The same result follows directly by subtracting cumulative incidences. By definition, \(F_k(R)\) includes cause-\(k\) events by \(R\), while \(F_k(L)\) includes those by \(L\). Subtraction leaves precisely events of cause \(k\) after \(L\) and by \(R\). Differentiating the integral definition of \(F_k(t)\) gives \(dF_k(t)/dt=S(t)\lambda_k(t)\), so the interval probability is the integral shown above.
If the event is observed exactly at time \(t\), its cause-specific density is \(S(t)\lambda_k(t)\). An interval-censored contribution integrates that density over all event times consistent with the observation. It is therefore a probability, not a density evaluated at the interval midpoint.
Worked Example: Known Cause Under Constant Hazards
Suppose, hypothetically, that two competing causes have constant hazards \(\lambda_1=0.08\) per year and \(\lambda_2=0.04\) per year. A participant is known to be event-free at time zero, and a cause-1 event is detected in \((2,5]\) years. Find the likelihood contribution.
The total hazard is \(0.08+0.04=0.12\) per year, so \(S(t)=e^{-0.12t}\). For constant hazards, \(F_1(t)=\{\lambda_1/(\lambda_1+\lambda_2)\}\{1-e^{-(\lambda_1+\lambda_2)t}\}\). Therefore:
The contribution is \(0.1585\), the probability that this participant experiences cause 1 in years 2 through 5 as the first event. The hazards are specified for both causes, and their sum is used in survival: competing events remove a participant from the event-free state even when they are not cause 1.
Worked Example: Separating Event-Free and Cause-Specific Observations
Suppose, hypothetically, that cause A and cause B have constant hazards \(0.05\) and \(0.15\) per year. An event of known cause A occurs in \((1,4]\) for one participant; another participant is known to be event-free through year 4. Calculate both contributions, and compare with the contribution when the cause of an event in \((1,4]\) is unknown.
The total hazard is \(0.20\) per year. The cause-A fraction of the total hazard is \(0.05/0.20=0.25\), and the cause-B fraction is \(0.75\). Since \(S(t)=e^{-0.20t}\), the probability of any first event in \((1,4]\) is:
The cause-A contribution is the cause-A share of that interval probability:
An event-free observation through year 4 instead contributes:
If the event in \((1,4]\) is known to have occurred but its cause is not recorded, its contribution is \(0.3694\), the sum of the cause-A and cause-B probabilities. These three contributions describe different observed outcomes; the event-free contribution is not the remainder of the cause-A contribution.
Conditioning on Entry After Time Zero
Sometimes the likelihood is constructed conditional on a participant being event-free at a time \(L>0\), such as when follow-up for the analysis begins at that time. In that case, probabilities from time zero must be divided by the probability of having reached \(L\) event-free. For a known cause-\(k\) event in \((L,R]\), the conditional contribution is:
For event-free status through \(R\), conditional on being event-free at \(L\), the contribution is \(S(R)/S(L)\). These conditional forms are not additional event mechanisms; they are the same probabilities expressed for a subject known to have entered the risk set at \(L\). Use either a full likelihood from the chosen origin or an appropriately conditioned likelihood, not a mixture of the two.
Worked Example: A Cause-Specific Event After Conditional Entry
Suppose, hypothetically, that cause A has constant hazard \(0.05\) per year and cause B has constant hazard \(0.15\) per year. A participant is known to be event-free at year 2, enters the analysis risk set then, and is found to have experienced cause B by year 5. Find the conditional likelihood contribution.
The total hazard is \(0.20\) per year. Conditional on being event-free at year 2, the cause-B event probability over the next 3 years is the cause-B hazard fraction multiplied by the probability of any event during those 3 years:
The contribution is approximately \(0.3384\). The calculation uses the 3 years after entry, rather than the 5 years since time zero, because the likelihood is conditional on survival to year 2. The cause-B label is also essential: without it, the contribution would be \(1-e^{-0.60}=0.4512\), the chance of either cause.
Assumptions, Diagnostics, and Common Errors
The formulas above describe a first-event process and require an observation model under which the recorded intervals and causes can be used as stated. In practice, check the following conditions before multiplying subject contributions into a likelihood:
- First event and cause are well defined. Each subject has one relevant first event, and the recorded cause classifies that event. A later event after a competing first event is not treated as another first event.
- The interval is recorded consistently. For the convention \((L,R]\), the subject is known to be event-free through \(L\), with the event known to have occurred by \(R\). If endpoint events can have positive probability, the endpoint convention must be adapted to the model rather than ignored.
- Cause ascertainment is valid. A cause recorded as known should genuinely be observed. If cause is uncertain or misclassified, the simple cause-specific contribution does not represent that observation mechanism.
- The observation process is ignorable for the stated likelihood. The inspection or censoring mechanism must not carry unaccounted information about the latent event time or cause. If observation times depend on unobserved event status, the observation process may need to be modeled as well.
- The model represents the relevant hazards. Hazards must be nonnegative and defined over the time range used. Covariates, if included, are handled by evaluating the hazards conditional on the subject’s covariates.
A practical diagnostic is to compare every record’s status with its contribution. A cause-\(k\) interval record should use \(F_k(R)-F_k(L)\), while an event-free record through \(R\) should use \(S(R)\). Contributions should lie between zero and one, and the sum of cause-specific interval probabilities should equal the any-cause interval probability:
If an interval contribution is negative, exceeds one, or fails this sum check, investigate endpoint ordering, cause labels, survival calculations, and units. A likely mistake is to use only the cause-\(k\) hazard to calculate event-free survival when evaluating a cause-\(k\) contribution. That would omit removal from the risk set by other causes. Another common mistake is to treat “event-free at \(R\)” as though the event occurred sometime before \(R\); the correction is to use \(S(R)\), not an interval-event probability.
The main alternatives correspond to different observed information. An exact event time with known cause uses the cause-specific density \(S(t)\lambda_k(t)\); an interval event with unknown cause sums the cause-specific probabilities; and a right-censored, event-free record uses survival to the censoring time. These are not interchangeable analysis choices. Each follows from what is actually observed. As in the earlier material on competing risks, treating competing causes as ordinary censoring can answer a different question; for the likelihood here, all causes contribute to the total hazard governing event-free survival.
A limitation is that interval data generally do not reveal where inside the interval the event occurred. The likelihood integrates over compatible times, but this does not create exact event-time information. If inspection timing is informative, the causes are not reliably recorded, or the cause-specific hazard model is misspecified, the resulting inference may be biased or poorly identified. Constructing the contribution correctly is necessary, but it does not by itself establish that the observation process or hazard model is adequate.
Check Your Understanding
Answer each question using the stated interval convention \((L,R]\). Solutions follow the questions.
- For an event of known cause 2 in \((L,R]\), write the contribution using \(F_2\) and using the cause-specific hazard.
- With two causes, suppose \(F_1(R)-F_1(L)=0.12\) and \(F_2(R)-F_2(L)=0.08\). What is the contribution if an event in that interval is known but its cause is unknown?
- For an event-free observation through \(R\), which contribution is used, and why is it not \(F_k(R)-F_k(L)\)?
- With constant hazards \(\lambda_1=0.10\) and \(\lambda_2=0.10\) per year, calculate the probability of a cause-1 event in \((0,2]\).
- If a subject is known to be event-free at \(L\), how is the cause-\(k\) interval probability conditioned on that fact?
Solutions. (1) The contribution is \(F_2(R)-F_2(L)=\int_L^R S(u)\lambda_2(u)\,du\). (2) The unknown-cause contribution is the sum, \(0.12+0.08=0.20\). (3) Use \(S(R)\), because the observation establishes no first event by \(R\); the interval-event expression instead describes an event that occurred in \((L,R]\). (4) The total hazard is \(0.20\) per year, so the cause-1 probability is \(\{0.10/0.20\}\{1-e^{-0.20(2)}\}=0.5(1-e^{-0.40})=0.1648\). (5) Divide the unconditional interval probability by \(S(L)\), giving \(\{F_k(R)-F_k(L)\}/S(L)\), provided the subject is conditioned on being event-free at \(L\).