Why Opposing Hazard Associations Need Not Cancel
A longitudinal biomarker may be associated with greater risk of one event cause and lower risk of another. For example, a higher latent biomarker level could accompany a higher instantaneous hazard of cause 1 but a lower instantaneous hazard of cause 2. Those directions alone do not tell us how the probabilities of experiencing each cause accumulate over time: each cause competes with the other, because the first event prevents a later event of the other type from being observed.
The relevant quantities are the cause-specific cumulative incidence functions. Each one combines the hazard for its particular cause with the probability of remaining event-free up to each time. In a joint model, predictions can be conditional on a subject’s latent biomarker trajectory and random effects, or averaged over their distribution. This tutorial develops the conditional calculation first, then explains the distinction.
Hazards, Survival, and Cumulative Incidence
Write \(m_i(t)\) for the latent underlying biomarker value at time \(t\) from the longitudinal model, and \(w_i\) for covariates in the event model. A two-cause current-value association model can be written as
Here \(\lambda_{0k}(t)\) is the baseline hazard for cause \(k\), \(\gamma_k\) is its covariate coefficient vector, and \(\alpha_k\) is the association parameter for the biomarker. Holding the other terms fixed, \(\alpha_1>0\) means that a one-unit increase in \(m_i(t)\) multiplies the cause-1 hazard by \(\exp(\alpha_1)\); \(\alpha_2<0\) means it multiplies the cause-2 hazard by a factor below one. Hazards have units of inverse time, while the \(\alpha_k\) coefficients are on the log-hazard scale per biomarker unit.
The two cause-specific hazards add to give the hazard of any first event. Conditional event-free survival therefore depends on their sum, not on either cause in isolation:
The cause-specific cumulative incidence is the accumulated chance of having cause \(k\) at time \(t\). At each time \(u\), the cause-\(k\) hazard contributes only for subjects still event-free at \(u\):
Why the Formula Has This Form
Over a short interval from \(u\) to \(u+du\), the conditional probability of cause \(k\) occurring is approximately \(\lambda_{ik}(u)\,du\), given event-free survival just before \(u\). The probability of reaching that interval event-free is \(S_i(u)\). Their product is thus the approximate probability of a cause-\(k\) event in that interval, and integrating these contributions from 0 to \(t\) gives \(F_{ik}(t)\).
This argument also explains why integrating the cause-specific hazard alone is not sufficient. A subject who has already had cause 1 cannot later have cause 2 as their first event. Accordingly, the survival factor in the cause-2 integral must reflect removal by both causes. Differentiating the survival function gives \(S_i'(u)=-S_i(u)\{\lambda_{i1}(u)+\lambda_{i2}(u)\}\). Summing the two cumulative-incidence derivatives and integrating therefore yields
when there are exactly two possible first-event causes and no additional event type. This identity is a useful calculation check: the two cause-specific probabilities must sum to the probability of any event by \(t\).
Worked Example: Opposing Associations With Equal Total Hazard
Consider hypothetical constant cause-specific hazards for two biomarker levels, measured in events per person-year. Let the low level be \(m=0\) and the high level be \(m=1\). Suppose the cause-1 baseline hazard is \(0.10\), the cause-2 baseline hazard is \(0.20\), and the association parameters are \(\alpha_1=\log 2\) and \(\alpha_2=-\log 2\). There are no other covariate effects.
At \(m=0\), the hazards are \(0.10\) for cause 1 and \(0.20\) for cause 2. At \(m=1\), they are \(0.10\exp(\log 2)=0.20\) and \(0.20\exp(-\log 2)=0.10\), respectively. The biomarker doubles the cause-1 hazard and halves the cause-2 hazard. The total hazard is \(0.30\) per year in both groups.
For constant hazards, with total hazard \(q=\lambda_1+\lambda_2\), integrating the cumulative-incidence formula gives \(F_k(t)=(\lambda_k/q)\{1-\exp(-qt)\}\). At three years, the probability of any event is \(1-\exp(-0.30\times3)=1-\exp(-0.9)=0.5934\). At \(m=0\), the cause-specific probabilities are
At \(m=1\), they are
Thus the higher biomarker level has a greater three-year cumulative incidence for cause 1 and a lower one for cause 2, while the overall event probability is unchanged in this particular example. The two cause-specific probabilities sum to \(0.5934\) in each group, as required.
When the Total Hazard Also Changes
Opposing associations do not generally leave overall survival unchanged. If the two hazard changes differ in size, their sum can increase or decrease. That changes how much time subjects remain at risk for either cause and therefore changes both cumulative-incidence curves.
Worked Example: Opposing Effects With Different Overall Risk
Suppose the low biomarker group has constant hazards \(0.08\) per year for cause 1 and \(0.12\) per year for cause 2. In the high group, the cause-1 hazard doubles to \(0.16\), while the cause-2 hazard halves to \(0.06\). The total hazards are \(0.20\) and \(0.22\) per year, respectively.
At four years, the low-group probability of any event is \(1-\exp(-0.20\times4)=1-\exp(-0.8)=0.5507\). Its cause-specific probabilities are
For the high group, the probability of any event is \(1-\exp(-0.22\times4)=1-\exp(-0.88)=0.5852\). The cause-specific probabilities are
At four years, the high group has both a larger cause-1 cumulative incidence and a larger overall event probability, but a smaller cause-2 cumulative incidence. The cause-2 probability falls even though the total event probability rises, because the cause-2 share of the total hazard is much lower. These are hypothetical model-based probabilities conditional on the specified hazards, not causal effects of changing a person’s biomarker.
Piecewise Hazards and Accumulation Over Time
When hazards vary with time, the constant-hazard shortcut does not apply across the full follow-up period. One can still calculate cumulative incidence by integrating over time, or by summing interval contributions when hazards are piecewise constant. On an interval of length \(\Delta\) with constant hazards \(\lambda_1,\lambda_2\), let \(q=\lambda_1+\lambda_2\). If event-free survival at the start of the interval is \(S_{\text{start}}\), the interval contributes
to the cumulative incidence of cause \(k\). This follows from integrating \(S_{\text{start}}\exp(-qs)\lambda_k\) from \(s=0\) to \(s=\Delta\).
Worked Example: Piecewise Hazards With Opposing Associations
Suppose the low biomarker level has hazards \((0.10,0.20)\) per year during year 0–1 and \((0.20,0.40)\) during year 1–2, where each pair lists causes 1 and 2. The high level doubles cause 1 and halves cause 2, giving \((0.20,0.10)\) and \((0.40,0.20)\). The total hazard is \(0.30\) per year in both groups during each interval.
In year 0–1, the event-free survival at the start is 1 and the event probability is \(1-\exp(-0.30)=0.2592\). The low-level group allocates one-third of that event probability to cause 1 and two-thirds to cause 2: the increments are \(0.08639\) and \(0.1728\). At the high level, the increments are reversed.
At year 1, event-free survival is \(\exp(-0.30)=0.7408\) in either group. During year 1–2, the total hazard is \(0.60\) per year, so the event probability conditional on reaching year 1 is \(1-\exp(-0.60)=0.4512\). The unconditional event probability contributed in that interval is
For the low level, the cause-1 share of the interval hazard is \(0.20/0.60=1/3\), so the cause-1 increment is \(0.3342/3=0.1114\); the cause-2 increment is \(0.2228\). At the high level, those increments are \(0.2228\) and \(0.1114\). Adding the interval increments gives low-level cumulative incidences \(F_1(2)=0.08639+0.1114=0.1978\) and \(F_2(2)=0.1728+0.2228=0.3956\). At the high level, they are \(F_1(2)=0.3956\) and \(F_2(2)=0.1978\). In both groups the sum is \(0.5934=1-\exp(-0.90)\); the remaining event-free probability is \(\exp(-0.90)=0.4066\).
Conditional Predictions, Population Averaging, and Checks
In a joint model, a subject’s \(m_i(t)\) is typically estimated through a longitudinal model, and may depend on subject-specific random effects. A cumulative-incidence curve calculated for one specified trajectory and random-effect value is conditional on those values. A population-averaged prediction instead averages the conditional curve over the relevant distribution of random effects and, if required, covariates. Because the cumulative-incidence formula is nonlinear, evaluating it at an average trajectory generally does not equal averaging the curves:
in general, where \(b_i\) denotes the random effects. State which prediction is being reported; otherwise, a conditional risk for a typical specified trajectory can be mistaken for a population risk.
The calculation assumes a correctly specified pair of cause-specific hazards and a well-defined event-free survival process. For a joint-model analysis, the longitudinal model and its link to the hazards must also be adequate for the prediction being made. In estimation, the event cause must be observed as modeled, and censoring must be handled under the study’s stated censoring assumptions; for example, likelihood-based inference commonly requires censoring to be independent of the event process conditional on the modeled information. The numerical examples assume exact event times and causes and no censoring in the calculation.
Several checks help identify errors. Each cause-specific cumulative incidence should be nondecreasing and between 0 and 1. The sum \(F_{i1}(t)+F_{i2}(t)\) should not exceed 1 and, with only these two first-event causes, should equal \(1-S_i(t)\). Under constant hazards, the cause-specific shares of the event probability should equal \(\lambda_k/(\lambda_1+\lambda_2)\). Large discrepancies may indicate that a cause was omitted from the total hazard, the survival factor was not updated across intervals, or a hazard ratio was applied in the wrong direction.
A common mistake is to treat the cause-1 cumulative incidence as though it were \(1-\exp\{-\int_0^t\lambda_{i1}(u)\,du\}\). That expression ignores removal by cause 2 and is not the probability of cause 1 as the first event. The correct integrand is \(S_i(u)\lambda_{i1}(u)\), where \(S_i\) accounts for all modeled causes.
How This Prediction Differs From Alternatives
Cause-specific hazards describe instantaneous rates among subjects who remain event-free. Their fitted values can be combined as above to obtain cause-specific cumulative incidence. A subdistribution-hazard model instead parameterizes a different hazard linked directly to a cause-specific cumulative-incidence function; its coefficients are not interchangeable with cause-specific hazard coefficients. A net-risk calculation that treats competing events as if they could not occur answers a different, hypothetical question and can overstate the observed probability of a particular cause. Choose the approach according to the target quantity: for the probability of each first-event cause by a time point, calculate or estimate cumulative incidence while accounting for all competing causes.
These curves also depend on the model, the covariates or trajectories being conditioned on, and the time horizon. Sparse events for one cause, misspecified baseline hazards, or poorly estimated association parameters can make a curve unstable. A smooth curve is not evidence that its cause-specific hazards are well estimated; inspect cause-specific event counts and uncertainty in the fitted parameters, and report uncertainty intervals alongside predictions when available.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- With constant hazards \(0.12\) per year for cause 1 and \(0.08\) per year for cause 2, what are the two cumulative incidences at two years?
- Why is the probability of cause 1 not obtained by integrating its hazard without a survival factor?
- For the high group in the second worked example, what is the probability of remaining event-free at four years?
- What identity should hold between the two cumulative incidences and event-free survival when there are exactly two causes?
Solutions. (1) The total hazard is \(0.20\) per year and the event probability is \(1-\exp(-0.40)=0.3297\). Thus \(F_1(2)=(0.12/0.20)(0.3297)=0.1978\) and \(F_2(2)=(0.08/0.20)(0.3297)=0.1319\). (2) Subjects who experience cause 2 are no longer at risk of experiencing cause 1 as their first event. The correct integrand weights the cause-1 hazard by survival from all causes. (3) It is \(\exp(-0.22\times4)=\exp(-0.88)=0.4148\). (4) \(F_1(t)+F_2(t)=1-S(t)\), so the two cumulative incidences sum to the probability of any event by time \(t\).