Why the Independence Question Arises
In competing-risks data, each subject experiences at most one first event from a set of mutually exclusive causes. It is tempting to imagine that every cause has a latent event time: the time the subject would experience that cause if the other causes could not occur first. This picture can prompt a seemingly necessary question: must these latent times be independent before we can define a cause-specific hazard?
No. A cause-specific hazard is defined from the event time and cause that are observed, among subjects who remain event-free up to the time in question. It is a conditional description of the observed process. Dependence among hypothetical times can affect who remains event-free and which cause occurs first, but it does not prevent that observed conditional quantity from being defined.
The conditioning event \(T\geq t\) matters: it selects subjects who have not yet experienced any competing event. A cause-specific hazard is therefore not the probability of cause \(k\) over an interval, and it is not automatically the hazard of a hypothetical time to cause \(k\) in a world where competing causes are removed.
From Latent Times to the Observed Hazard
To examine the independence issue precisely, suppose, for the moment, that each cause \(j\) has a well-defined hypothetical time \(T_j\). The observed first-event time is \(T=\min_j T_j\), and \(J\) is the cause attaining that minimum. Assume ties have probability zero. Then being event-free at \(t\) means that every \(T_j\) exceeds \(t\), and an observed event from cause \(k\) at time \(t\) means that \(T_k\) occurs then while the other times exceed it.
If the vector of latent times has a joint density \(f(t_1,\ldots,t_K)\), the event-free probability is the joint survival probability, not necessarily a product of marginal survival probabilities. The instantaneous cause-\(k\) event density among the whole population is obtained by placing the \(k\)th time at \(t\) and integrating over values of the other times greater than \(t\). Dividing by the probability of still being event-free gives:
Here \(d\mathbf{t}_{-k}\) means integration over the \(K-1\) time coordinates other than \(k\). This expression uses the joint distribution, whatever its dependence structure. Independence is needed only to factor the joint survival probability and the joint density into marginal pieces. It is not needed for the conditional hazard itself.
For two causes, \(A\) and \(B\), independence would give \(P(T_A>t,T_B>t)=P(T_A>t)P(T_B>t)\), and the numerator for cause \(A\) would factor into the marginal density of \(T_A\) times the survival probability for \(T_B\). The resulting \(\lambda_A(t)\) would equal the marginal hazard of \(T_A\). Under dependence, those factorizations generally fail: \(\lambda_A(t)\) remains well-defined, but need not equal the marginal hazard of the hypothetical time \(T_A\).
Worked Example: A Dependent Shared-Frailty Model
Worked Example: A Dependent Shared-Frailty Model
Hypothetical model. Suppose two causes, \(A\) and \(B\), have latent times \(T_A=E_A/Z\) and \(T_B=E_B/Z\). The random variables \(E_A\) and \(E_B\) are independent exponential random variables with rate 1. The shared variable \(Z\) equals 0.5 or 2, each with probability 0.5, and is independent of the \(E\)'s. Conditional on \(Z=z\), each cause-specific time has rate \(z\). Unconditionally, the two times are dependent because they share \(Z\).
Conditional on \(Z=z\), the probability of remaining event-free through time \(t\) is \(e^{-zt}e^{-zt}=e^{-2zt}\). Thus the probability of still being event-free at \(t=1\), before accounting for the value of \(Z\), is
Among subjects still event-free at time 1, the conditional probability that \(Z=2\) is
Given \(Z=z\), the cause-\(A\) hazard among event-free subjects is \(z\). Averaging over the values of \(Z\) among those who have survived both causes gives the observed cause-\(A\) hazard at time 1:
The calculation conditions on surviving both causes; it does not assume that their latent times are independent. Indeed, the shared \(Z\) makes them dependent. The changing mixture of \(Z\) among survivors also explains why the marginal observed hazard is not simply the unconditional mean rate \(E[Z]=1.25\) per time unit.
A Discrete Joint Distribution Shows the Risk Set Directly
The same point can be seen without a continuous-time model. The following hypothetical joint distribution gives each subject’s two latent times. There are no tied times in any row. The times are in years, and the probabilities sum to one.
| \(T_A\) | \(T_B\) | Probability |
|---|---|---|
| 1 | 4 | 0.20 |
| 2 | 3 | 0.30 |
| 3 | 1 | 0.10 |
| 4 | 2 | 0.40 |
Worked Example: Cause-Specific Event Probabilities in a Dependent Risk Set
At time 1, the subjects represented by the first and third rows have their first event: cause \(A\) for probability 0.20 and cause \(B\) for probability 0.10. The probability of remaining event-free just after time 1 is therefore \(0.30+0.40=0.70\).
Among that event-free group, the second row has an \(A\) event at time 2, with probability 0.30 in the original population; the fourth row has a \(B\) event at time 2, with probability 0.40. Conditional on having remained event-free through time 1, the cause-specific event probabilities at time 2 are
These are discrete-time conditional probabilities for the next event time, not continuous-time rates. They follow directly from the joint distribution and the event-free risk set. To check dependence, note that \(P(T_A>1)=0.80\) and \(P(T_B>1)=0.90\), whose product is \(0.72\), whereas \(P(T_A>1,T_B>1)=0.70\). The joint survival does not factor, yet the observed conditional probabilities are unambiguous.
Estimation from Observed Risk-Set Data
In practice, a cause-specific hazard is estimated using observed event times, causes, and risk sets. A latent time for a cause that did not occur is not required for the basic definition or estimation. For example, under a piecewise-constant cause-specific hazard model, let \(Y\) be the total person-time at risk during an interval and \(d_k\) the number of observed events from cause \(k\). Ignoring terms not involving the hazards, the likelihood for the interval is proportional to \(\prod_k \lambda_k^{d_k}\exp[-Y\sum_k\lambda_k]\), yielding the estimate \(d_k/Y\) for each cause-specific hazard.
Worked Example: Estimating Rates Without Latent Times
Hypothetical data. In a study interval, 20 subjects are initially event-free. Their total observed time at risk before an event, censoring, or the interval end is 9.0 person-years. Three experience cause \(A\), two experience cause \(B\), and the remaining subjects have no event in the interval. Assume event times and causes are recorded correctly, censoring is noninformative for this model, and each cause-specific hazard is constant over the interval.
The interval likelihood has hazard-dependent part \(\lambda_A^3\lambda_B^2\exp[-9(\lambda_A+\lambda_B)]\). Maximizing it gives
These estimates describe cause-specific rates in the observed event-free risk set under the stated constant-rate model. Neither the likelihood nor these calculations requires an unobserved time to the cause that did not occur. The assumptions concern the observed-data model and follow-up, not independence of hypothetical cause-specific times.
Assumptions, Ties, and Failure Modes
The definition requires an event time, an event-free status just before that time, and a cause label for the observed first event. To interpret the limit as a finite instantaneous rate, the relevant conditional event probabilities must behave regularly as the interval width shrinks. Estimation adds assumptions appropriate to the method, such as correct cause classification, adequate observation of event times, and a censoring mechanism that is handled appropriately. A piecewise-constant hazard likelihood, for example, additionally assumes constant cause-specific rates within each modeled interval.
Ties need explicit treatment. A joint density for the vector of latent times assigns probability zero to the diagonal where two times are equal. But merely saying that each latent time has a continuous marginal distribution is not sufficient: one can have \(T_A=T_B\) always, with their common time drawn from a continuous distribution. If simultaneous events can occur, the analysis must specify whether they are recorded as a separate cause, classified by a rule, or handled by another stated convention. An undefined tie rule makes \(J\) ambiguous.
- Wrong reasoning: “The latent times might be dependent, so the cause-specific hazard is undefined.” The correction is to define the observed \(T\) and \(J\), then condition on \(T\geq t\). Dependence affects the joint probabilities, not the existence of this observed-data definition.
- Wrong reasoning: “The cause-specific hazard must equal the marginal hazard of its latent time.” That equality generally requires a factorization such as latent-time independence. Without it, describe the observed conditional hazard as such; do not silently assign it a latent marginal interpretation.
- Wrong reasoning: “A continuous distribution for each latent time rules out ties.” It does not. Check whether the joint distribution puts probability on simultaneous times, or state the rule used for simultaneous events.
- Wrong reasoning: “The hazard denominator includes everyone who has not yet had cause \(k\).” The observed cause-specific risk set includes only those who have had no event yet. A prior event from another cause removes a subject from that risk set.
A practical diagnostic is to verify the event-free risk-set definition, event timing, cause coding, and any tie convention before fitting a cause-specific model. A hazard trend alone does not diagnose latent-time independence: changes in the risk set, including changes in its composition, can change an observed cause-specific hazard. The observed-data analysis and a mechanistic model for hypothetical times answer different questions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- State the conditioning event in the definition of the cause-\(k\) hazard and explain why it is needed.
- Suppose \(P(T_A>2)=0.8\), \(P(T_B>2)=0.7\), and \(P(T_A>2,T_B>2)=0.50\). Is independence needed to define the observed cause-specific hazard at time 2?
- In a discrete interval, 50 subjects are event-free at its start; 5 have cause-\(A\) events and 3 have cause-\(B\) events. What are the cause-specific event probabilities for that interval, conditional on being event-free at its start?
- Why does a joint density rule out ties between two latent times, while continuous marginal distributions alone do not?
Solutions. (1) The conditioning event is \(T\geq t\): no first event has occurred by time \(t\). It identifies the observed event-free risk set in which the instantaneous cause-specific rate is defined. (2) No. The joint survival value differs from the product \(0.8(0.7)=0.56\), so the times are not independent at that threshold, but the observed cause-specific hazard is still defined by the conditional event probability among those event-free. (3) The probabilities are \(5/50=0.10\) for cause \(A\) and \(3/50=0.06\) for cause \(B\). These are interval probabilities, not rates per unit time unless a rate model and interval width are also specified. (4) A joint density assigns probability zero to the diagonal \(T_A=T_B\), a set of area zero. Continuous marginal distributions do not prevent the joint distribution from being concentrated on that diagonal, as when \(T_A=T_B\) is a shared continuously distributed time.