Competing-Risks Models With Individual Effects

Why Competing Events Need Their Own Probabilities

In time-to-event data, a person may experience one of several mutually exclusive first events. For example, a participant may discontinue treatment because of lack of efficacy or because of toxicity. Once one event occurs, the other cannot be observed as the first event for that person. These are competing events. The probability of experiencing a particular event by a given time depends not only on its own event rate, but also on the rates of the competing events.

In an NLME model, individuals may also differ in their underlying event rates. Covariates describe measured differences, while random effects represent modeled between-individual heterogeneity. This tutorial defines the event probabilities conditional on those features, then shows how to average over random effects for a population-level probability.

Definition: For individual $i$, let $T_i$ be the time to the first event and $J_i\in\{1,\ldots,K\}$ identify its cause. Given covariates $x_i$ and random effects $b_i$, the cause-specific hazard for cause $k$ is $$ \lambda_{ik}(t\mid b_i,x_i) = \lim_{\Delta t\to 0} \frac{P(t\leq T_i<t+\Delta t,\ J_i=k\mid T_i\geq t,b_i,x_i)}{\Delta t}. $$ It is the instantaneous rate of cause $k$ at time $t$, among individuals who have not yet experienced any event.

The definition concerns an event-free individual at time $t$. A cause-specific hazard is not the probability of that cause occurring by time $t$, and a hazard ratio is not generally a ratio of cumulative probabilities. All cause-specific hazards matter because any one of the events removes the individual from the event-free state.

From Hazards to Event Probabilities

Write the total hazard as $\Lambda_i(t\mid b_i,x_i)=\sum_{k=1}^K\lambda_{ik}(t\mid b_i,x_i)$. Conditional on covariates and random effects, the event-free survival probability is

S_i(t\mid b_i,x_i) = P(T_i>t\mid b_i,x_i) = \exp\left\{-\int_0^t\Lambda_i(u\mid b_i,x_i)\,du\right\}.

The cumulative incidence function (CIF) for cause $k$ is the probability that cause $k$ is the first event and occurs by time $t$. Its conditional form is the integral of the cause-specific event density:

F_{ik}(t\mid b_i,x_i) = P(T_i\leq t,J_i=k\mid b_i,x_i) = \int_0^t S_i(u\mid b_i,x_i)\lambda_{ik}(u\mid b_i,x_i)\,du.

The factor $S_i(u\mid b_i,x_i)$ is essential: cause $k$ can occur at time $u$ only if no cause has occurred earlier. As a check, the cause-specific CIFs sum to the probability of any event by time $t$:

\sum_{k=1}^K F_{ik}(t\mid b_i,x_i)=1-S_i(t\mid b_i,x_i).

Result: If the cause-specific hazards are constant over time for a given individual, write them as $\lambda_{ik}$ and let $\Lambda_i=\sum_k\lambda_{ik}$. Then $$ F_{ik}(t\mid b_i,x_i) = \frac{\lambda_{ik}}{\Lambda_i}\left(1-e^{-\Lambda_i t}\right). $$ The expression applies when $\Lambda_i>0$; if all hazards are zero, every CIF is zero.

To justify this result, constant hazards give $S_i(u\mid b_i,x_i)=e^{-\Lambda_i u}$. Substitution into the CIF integral yields

F_{ik}(t\mid b_i,x_i) =\int_0^t e^{-\Lambda_i u}\lambda_{ik}\,du =\lambda_{ik}\left[\frac{1-e^{-\Lambda_i t}}{\Lambda_i}\right] =\frac{\lambda_{ik}}{\Lambda_i}(1-e^{-\Lambda_i t}).

Thus the event probability is not simply $1-e^{-\lambda_{ik}t}$, which would describe a setting where cause $k$ is the only possible event. The other hazards reduce the time available for cause $k$ to occur first.

Worked Example: Constant Hazards and Two Competing Events

Suppose, hypothetically, that an individual has constant cause-specific hazards of $0.12$ per year for event 1 and $0.08$ per year for event 2. Calculate the event-free probability and both CIFs at 3 years.

The total hazard is $\Lambda=0.12+0.08=0.20$ per year. Therefore, $S(3)=e^{-(0.20)(3)}=e^{-0.60}=0.5488$. The probability of any event by 3 years is $1-0.5488=0.4512$.

For event 1, $F_1(3)=(0.12/0.20)(1-e^{-0.60})=0.6(0.4512)=0.2707$. For event 2, $F_2(3)=(0.08/0.20)(1-e^{-0.60})=0.4(0.4512)=0.1805$. Both probabilities are dimensionless. They sum to $0.4512$, as required by $F_1(3)+F_2(3)=1-S(3)$.

Covariates and Individual Effects

A common conditional model specifies a cause-specific hazard for each cause using a baseline hazard, covariates, and individual random effects. For example, with a scalar covariate $x_i$, a vector of random effects $b_i$, and cause-specific coefficient vectors $\beta_k$ and $z_{ik}$, one may write

\lambda_{ik}(t\mid b_i,x_i) = \lambda_{0k}(t)\exp\left(x_i^\mathsf{T}\beta_k+z_{ik}^\mathsf{T}b_i\right), \qquad b_i\sim N(0,\Omega).

Here $\lambda_{0k}(t)$ is the baseline hazard for cause $k$, and $\Omega$ is the random-effects covariance matrix. This model assumes that the specified covariates and random effects adequately describe the modeled variation in the cause-specific hazards. Subjects are typically assumed independent conditional on model parameters, with random effects drawn from the stated distribution. If random effects influence multiple causes, their joint distribution matters: a characteristic that raises one cause-specific hazard may also raise or lower another.

A covariate coefficient describes a conditional effect on a cause-specific hazard. It does not, by itself, determine the direction or size of the covariate’s effect on that cause’s CIF, because the covariate can also alter the competing hazards and hence the event-free survival factor. Population probabilities require another step: integration over the random-effects distribution.

Worked Example: A Covariate Changes Both Causes

Consider a hypothetical two-cause model with constant hazards. Let a binary covariate $x$ be 0 or 1. At $x=0$, let the cause 1 and cause 2 hazards be $0.10$ and $0.15$ per year. Suppose $x=1$ doubles the cause 1 hazard and halves the cause 2 hazard. Compare the 2-year CIFs.

For $x=0$, the total hazard is $0.10+0.15=0.25$ per year. Since $1-e^{-(0.25)(2)}=1-e^{-0.50}=0.3935$, $F_1(2\mid x=0)=(0.10/0.25)(0.3935)=0.1574$, and $F_2(2\mid x=0)=(0.15/0.25)(0.3935)=0.2361$.

For $x=1$, the hazards are $0.20$ and $0.075$ per year, so the total is $0.275$ per year. Now $1-e^{-(0.275)(2)}=1-e^{-0.55}=0.4231$. Thus $F_1(2\mid x=1)=(0.20/0.275)(0.4231)=0.3077$, and $F_2(2\mid x=1)=(0.075/0.275)(0.4231)=0.1154$.

The cause 1 hazard ratio is 2, but the cause 1 CIF ratio is approximately $0.3077/0.1574=1.955$, not 2. The difference arises because changing $x$ also changes the competing hazard and total event-free survival.

For a population-level CIF, average the conditional CIF over the distribution of random effects:

F_k(t\mid x_i) = \int F_{ik}(t\mid b,x_i) f_B(b)\,db,

where $f_B$ is the random-effects density. In general, this is not the conditional CIF evaluated at $b=0$, even when the random-effects distribution has mean zero. The exponential hazard model and the CIF integral are nonlinear in $b$, so averaging and substituting the mean do not generally give the same result.

Worked Example: Averaging Individual CIFs

For a transparent hypothetical illustration, suppose half the population has $b=\log(2)$ and half has $b=-\log(2)$. This two-point random-effects distribution has mean zero. Let the constant conditional hazards be $\lambda_1=0.10e^b$ and $\lambda_2=0.10e^{-b}$ per year. Find the population-average cause 1 CIF at 2 years and compare it with the CIF at $b=0$.

At $b=\log(2)$, the hazards are $0.20$ and $0.05$ per year, with total $0.25$. The cause 1 CIF is $F_1(2\mid b)= (0.20/0.25)(1-e^{-0.50})=0.3148$. At $b=-\log(2)$, the hazards are $0.05$ and $0.20$ per year, also with total $0.25$. The cause 1 CIF is $F_1(2\mid b)= (0.05/0.25)(1-e^{-0.50})=0.07869$.

Averaging the two conditional probabilities gives $F_1(2)=0.5(0.3148)+0.5(0.07869)=0.1967$. At $b=0$, both hazards equal $0.10$ per year, so $F_1(2\mid b=0)=(0.10/0.20)(1-e^{-0.40})=0.1648$. The population average, $0.1967$, differs from the CIF at the mean random effect. The two-point distribution is used only to make the integration arithmetic explicit; a fitted model would average over its specified random-effects distribution.

Likelihood Contributions and Interpretation

For an exact first event of cause $k$ at time $t$, the conditional event-time density is $S_i(t\mid b_i,x_i)\lambda_{ik}(t\mid b_i,x_i)$. For right censoring at time $c$, the conditional contribution is $S_i(c\mid b_i,x_i)$, provided censoring is independent of the event process conditional on the modeled information. With random effects, the subject-level likelihood contribution is obtained by integrating the conditional contribution over the random-effects distribution. The censoring assumptions established in the earlier material on censoring and truncation remain important; adding competing causes does not make informative censoring harmless.

Worked Example: Integrating an Event Contribution

Use the two-point random-effects model from the previous example. Suppose a hypothetical individual has a cause 1 event at exactly 1 year, and their random effect was not observed. Calculate the marginal event-time density contribution using equal weights for the two possible random effects.

For $b=\log(2)$, $\lambda_1=0.20$, total hazard $\Lambda=0.25$, and the density is $S(1\mid b)\lambda_1=e^{-0.25}(0.20)=0.1558$ per year. For $b=-\log(2)$, $\lambda_1=0.05$, with the same total hazard, giving $e^{-0.25}(0.05)=0.03894$ per year.

The marginal density contribution is $0.5(0.1558)+0.5(0.03894)=0.09735$ per year. The survival part is retained because the individual must remain event-free until the observed event time. In a full likelihood, contributions from all subjects are combined, and the random effects are integrated for each subject.

Cause-specific hazard models and direct CIF models answer related but distinct questions. Cause-specific hazards describe event rates among those still event-free and combine naturally to describe the event process. A subdistribution-hazard approach instead targets regression of the CIF through a specially defined risk set. It can be useful when the target is the CIF association itself, but its hazard has a different interpretation and is not the ordinary instantaneous rate among event-free individuals. Whichever approach is used, absolute event probabilities should be obtained and interpreted on the CIF scale when that is the scientific question.

Common Pitfalls and Model Checks

Treating a competing event as ordinary censoring to estimate absolute risk. This removes the competing event from the analysis as though it could still occur later. The resulting single-cause calculation does not estimate the observed probability of that cause occurring first. For cause 1, $1-e^{-\int_0^t\lambda_1(u)\,du}$ is generally not its CIF when other causes occur.
Interpreting a cause-specific hazard ratio as a CIF ratio. The CIF depends on every cause-specific hazard and on event-free survival, so the two ratios need not match.
Using only the mean random effect for population prediction. Nonlinear averaging means $E_B[F_k(t\mid B,x)]$ generally differs from $F_k(t\mid E[B],x)$. Use integration or a suitable numerical approximation.
Ignoring cause labels or event timing. A model for total event time alone cannot distinguish which event occurred first, even though cause-specific probabilities are the target.

Useful checks include comparing predicted and observed event counts by cause over clinically relevant time intervals, checking whether predicted CIFs remain between 0 and 1 and sum with event-free survival to 1, and examining whether predictions are sensitive to the random-effects distribution. Sparse events for a cause, few observed competing events, or limited follow-up can make cause-specific and random-effects estimates imprecise. Model-based predictions also rely on the specified hazard forms, covariate effects, and random-effects structure; good numerical fit does not establish those assumptions as true.

Key takeaway: A cause-specific hazard is a conditional instantaneous rate among individuals still event-free. A CIF is an absolute probability built from that cause’s hazard and survival under all competing hazards. With individual effects, population CIFs require averaging the conditional probabilities over the random-effects distribution.

Check Your Understanding

Try each question, then compare with the brief solutions below.

With constant hazards $\lambda_1=0.06$ and $\lambda_2=0.04$ per year, find the event-free probability and both CIFs at 2 years.
Why is the cause 1 CIF generally not $1-e^{-\lambda_1t}$ when cause 2 is possible?
In a random-effects model, why might the population CIF differ from the CIF computed at the mean random effect?
If a covariate doubles the cause 1 hazard but also changes the cause 2 hazard, can the cause 1 CIF ratio be assumed to equal 2? Explain.

Solutions. (1) The total hazard is $0.10$ per year, so $S(2)=e^{-0.20}=0.8187$. The event probability is $0.1813$; $F_1(2)=(0.06/0.10)(0.1813)=0.1088$ and $F_2(2)=(0.04/0.10)(0.1813)=0.07251$. (2) The other cause can occur first, so the CIF integral includes survival under the total hazard, not just cause 1’s hazard. (3) The CIF is nonlinear in the random effect, so averaging conditional CIFs generally differs from evaluating at the mean. (4) No. The competing hazard also affects event-free survival, so the CIF ratio need not equal the cause-specific hazard ratio.