Interval-Censored Competing Events in Joint Models

Why Interval-Censored Event Times Need a Different Contribution

In some studies, an event is not observed at its exact occurrence time. A participant may be event-free at one assessment and found to have experienced an event at the next. If the event cause is recorded, the data say that the event occurred sometime in an interval and identify its cause. Replacing that interval by its midpoint treats an unobserved time as if it were known; using an exact-event density at either endpoint makes a different, equally unjustified assumption.

The appropriate contribution is a probability: the probability that the event occurs between the two assessment times and is of the observed cause. In a joint model, this probability depends on the shared random effects that also describe the longitudinal process. It must therefore be combined with the longitudinal likelihood before integrating over those random effects.

Definition: For subject $i$, let $T_i$ be the event time and $J_i\in\{1,\ldots,K\}$ the cause. Let $L_i<R_i$ be the observation interval, so the observed event information is $L_i<T_i\le R_i$ and $J_i=k$. Conditional on the subject's random effects $b_i$ and covariates, let $\lambda_{ik}(t\mid b_i)$ be the cause-specific hazard for cause $k$. The overall event-free survival function and the cause-$k$ event density are $$ S_i(t\mid b_i)=\exp\left\{-\int_0^t\sum_{r=1}^K\lambda_{ir}(u\mid b_i)\,du\right\}, \qquad f_{ik}(t\mid b_i)=S_i(t\mid b_i)\lambda_{ik}(t\mid b_i). $$

The survival factor contains the hazards of all causes. A subject must remain free of every competing event up to time $t$ before experiencing cause $k$ at that time. For example, a common cause-specific hazard specification is $\lambda_{ik}(t\mid b_i)=\lambda_{0k}(t)\exp\{\gamma_k^\mathsf{T}w_i+\alpha_k m_i(t;b_i)\}$, where $w_i$ is a vector of covariates and $m_i(t;b_i)$ is a longitudinal-model quantity such as a marker's current value. The interval-censored contribution follows from integrating the cause-specific density over all possible event times in the observed interval.

P(L_i<T_i\le R_i,\ J_i=k\mid b_i) =\int_{L_i}^{R_i} f_{ik}(t\mid b_i)\,dt =\int_{L_i}^{R_i}S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)\,dt.

This is not the probability of any event in the interval. That probability sums the contributions over causes. Nor is it a density evaluated at an imputed time. For continuous event times, whether an endpoint is included does not change the probability, but the recorded interval must still be defined consistently with the study's assessment scheme.

Deriving the Cause-Specific Interval Probability

The derivation uses the meaning of a cause-specific hazard. Conditional on being event-free just before time $t$, the probability of cause $k$ occurring in a short interval of length $dt$ is approximately $\lambda_{ik}(t\mid b_i)dt$. The probability of still being event-free at $t$ is $S_i(t\mid b_i)$. Thus the joint probability of surviving event-free to $t$ and then experiencing cause $k$ in the short interval is approximately $S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)dt$. Summing these probabilities across the possible event times in $(L_i,R_i]$ gives the integral.

Equivalently, conditional on $b_i$, the cause-$k$ density is the product of survival from every cause and the instantaneous hazard for cause $k$. Integrating that density gives a probability, as required for interval-observed data. As a check, summing over all causes gives the probability of any event in the interval:

\sum_{k=1}^K\int_{L_i}^{R_i}S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)\,dt =S_i(L_i\mid b_i)-S_i(R_i\mid b_i).

To verify the identity, differentiate survival: $dS_i(t\mid b_i)/dt=-S_i(t\mid b_i)\sum_{k=1}^K\lambda_{ik}(t\mid b_i)$. Integrating the negative derivative from $L_i$ to $R_i$ gives the displayed difference. This check also makes clear why a cause-specific contribution is only one part of the probability of an event in the interval.

Worked Example: Constant Hazards for Two Causes

Suppose, hypothetically, that a subject has constant cause-specific hazards $\lambda_{i1}=0.10$ per year and $\lambda_{i2}=0.05$ per year. The subject is known to have experienced cause 1 between year 1 and year 2. Conditional on the stated hazards, the total hazard is $0.15$ per year and $S_i(t)=\exp(-0.15t)$. The required contribution is

\int_1^2 0.10e^{-0.15t}\,dt =\frac{0.10}{0.15}\left(e^{-0.15}-e^{-0.30}\right) =\frac{2}{3}(0.86070798-0.74081822) =0.0799265.

Thus the probability of a cause-1 event between years 1 and 2, under this model, is about $0.07993$. The probability of any event in the interval is $e^{-0.15}-e^{-0.30}=0.1199$, which is larger because it includes cause 2 as well. Both the cause-specific survival factor and the cause-specific hazard are needed; integrating $0.10$ alone would omit the requirement that the subject remain event-free until the event time.

Combining the Event Contribution With Longitudinal Data

Let $Y_i$ denote subject $i$'s longitudinal observations and $g(b_i)$ the random-effects density. Under the usual joint-model factorization, the longitudinal observations and event process are conditionally independent given $b_i$ and the included covariates. If the interval endpoints and event cause are observed as described, the subject's joint likelihood contribution is

L_i=\int p(Y_i\mid b_i) \left\{\int_{L_i}^{R_i}S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)\,dt\right\} g(b_i)\,db_i.

The expression in braces depends on the same random effects as the longitudinal likelihood. In general, it cannot be replaced by an event contribution computed at an estimated or average random effect. The integration over $b_i$ averages the product of the longitudinal and event contributions over the population distribution of random effects.

This likelihood assumes that the event interval and recorded cause are measured as stated, and that the event and longitudinal processes are conditionally independent given the modeled random effects and covariates. It also requires an appropriate model for the longitudinal measurements and for each cause-specific hazard. The assessment schedule must be noninformative for the event time after conditioning on the modeled information, or its dependence on the event process must be modeled. If the schedule is informative, the interval probability alone does not account for how the observation interval was generated.

Worked Example: Integrating Over a Shared Random Effect

Consider a hypothetical two-cause model with a binary random effect $b_i$, taking values 0 and 1 with probability $0.5$ each. Hazards are constant over years 0 to 1. When $b_i=0$, let $\lambda_{i1}=0.10$ and $\lambda_{i2}=0.10$ per year; when $b_i=1$, let $\lambda_{i1}=0.20$ and $\lambda_{i2}=0.10$ per year. The event is known to be cause 1 in $(0,1]$. The conditional interval contribution for constant hazards is

\frac{\lambda_{i1}}{\lambda_{i1}+\lambda_{i2}} \left\{1-\exp[-(\lambda_{i1}+\lambda_{i2})]\right\}.

For $b_i=0$, substitution gives $(0.10/0.20)(1-e^{-0.20})=0.0906346$. For $b_i=1$, it gives $(0.20/0.30)(1-e^{-0.30})=0.172788$. These differ because the event risk depends on the shared random effect.

Suppose, for illustration, that the longitudinal likelihood factors evaluated at these two random-effect values are $p(Y_i\mid b_i=0)=0.8$ and $p(Y_i\mid b_i=1)=0.2$, in the chosen measurement units. The joint contribution, integrating over the two-point random-effects distribution, is

$$ L_i=0.5(0.8)(0.0906346)+0.5(0.2)(0.172788) =0.0362538+0.0172788 =0.0535326. $$

The numbers $0.8$ and $0.2$ are likelihood factors, not necessarily probabilities of the observed continuous measurements. The calculation illustrates why the longitudinal and event factors are multiplied at each value of $b_i$, then averaged using $g(b_i)$. Multiplying their separate random-effects averages would generally give a different result.

Numerical Integration and Diagnostics

The inner interval integral may not have a closed form, especially when baseline hazards are piecewise, the marker trajectory varies over time, or the hazard includes nonlinear time effects. Numerical quadrature can approximate it. For a smooth integrand $f(t)=S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)$, Simpson's rule over an interval of width $h$ uses $h\{f(L_i)+4f((L_i+R_i)/2)+f(R_i)\}/6$. For piecewise hazards, split the interval at every baseline-hazard breakpoint and apply quadrature separately on each smooth segment.

Worked Example: Simpson Approximation With a Time-Varying Hazard

Suppose, hypothetically, that there are two causes with $\lambda_{i1}(t)=0.10t$ and $\lambda_{i2}(t)=0.05$, in inverse-year units, over the interval $(1,2]$. The total cumulative hazard from 0 to $t$ is $H_i(t)=0.05t^2+0.05t$, so the cause-1 integrand is $f(t)=0.10t\exp[-(0.05t^2+0.05t)]$. Its values at the left endpoint, midpoint, and right endpoint are

f(1)=0.10e^{-0.10}=0.0904837,\qquad f(1.5)=0.15e^{-0.1875}=0.1243544,\qquad f(2)=0.20e^{-0.30}=0.1481636.

Using the unrounded values in Simpson's rule, with $h=2-1=1$, gives

\int_1^2 f(t)\,dt \approx\frac{1}{6}\{0.0904837+4(0.1243544)+0.1481636\} =0.1226775\approx0.1227.

This is an approximation to the probability of a cause-1 event in the interval under the specified hazards. It is not the probability of either cause. In computation, retaining more digits until the final rounding avoids a discrepancy between the displayed integrand values and the reported approximation.

Useful checks include the following:

Probability bounds. Each interval contribution must be between 0 and 1, and the contributions across causes must sum to $S_i(L_i\mid b_i)-S_i(R_i\mid b_i)$, up to numerical error.
Interval limits. As the interval width shrinks, the contribution is approximately $f_{ik}(t\mid b_i)$ times the width near $t$. A contribution should approach zero as the width approaches zero; the density itself need not approach zero.
Quadrature stability. Refine the integration grid or increase quadrature accuracy. Material changes in the likelihood or estimates indicate that the original integration was inadequate.
Hazard breakpoints. Check that integration segments respect changes in piecewise baseline hazards. A quadrature rule that crosses an unhandled discontinuity can be inaccurate.
Scale and units. Confirm that hazards are in inverse-time units and that interval endpoints use the same time scale. The integrated contribution is dimensionless.

Alternatives, Assumptions, and Common Pitfalls

With an exactly observed event time $t_i$ and cause $k$, the event contribution is the density $S_i(t_i\mid b_i)\lambda_{ik}(t_i\mid b_i)$. With right censoring at $C_i$, when the subject is known to be event-free through $C_i$, the event contribution is $S_i(C_i\mid b_i)$. With interval censoring and an observed cause, the contribution is the integral of that density over the interval. These are distinct observation schemes and require distinct likelihood contributions.

A common mistake is to assign the interval midpoint as the event time and use the exact-event density there. The midpoint is not observed, and the resulting density is not the probability of an event anywhere in the interval. The correction is to integrate $S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)$ from the lower to the upper endpoint, then combine that probability with the longitudinal likelihood and integrate over the random effects.

This method also has limits. Wide intervals contain less information about event timing than exact times, and cause-specific estimates may be imprecise when few events of a cause are observed. The likelihood as written assumes the cause is correctly recorded; it does not account for uncertain or misclassified causes. It also does not automatically model informative assessment times. Those features require changes to the observation model rather than a different numerical evaluation of the same integral.

If subjects enter observation after the time origin, the risk-set conditioning from delayed entry must be applied consistently with the interval contribution. In particular, one should not treat survival from the origin to entry as if it were newly observed event information. The interval integration remains over the times compatible with the observed interval, but the likelihood must reflect the sampling and entry scheme.

Key takeaway: For an interval-censored event of observed cause $k$, integrate the cause-specific density $S_i(t\mid b_i)\lambda_{ik}(t\mid b_i)$ over the observed interval. In a joint model, multiply by the longitudinal likelihood conditional on the same random effects and integrate the product over their distribution.

Check Your Understanding

Answer each question, then compare with the solutions below.

In a two-cause model, why does the cause-1 density contain both cause-specific hazards through the survival function?
For constant cause-specific hazards $\lambda_1=0.08$ and $\lambda_2=0.12$ per year, write the cause-1 contribution for an event in $(0,1]$.
What identity should the sum of cause-specific interval contributions satisfy?
Why is the joint contribution over random effects not generally the product of the separately averaged longitudinal and event contributions?

Solutions. (1) The subject must remain free of every cause until the time of the cause-1 event, so survival uses the total hazard $\lambda_1+\lambda_2$, while the instantaneous event factor is $\lambda_1$. (2) The contribution is $\int_0^1 0.08e^{-0.20t}\,dt=(0.08/0.20)(1-e^{-0.20})=0.07251$. (3) The sum over causes equals $S(0)-S(1)$, the probability of any event in the interval. (4) Both conditional factors depend on the same random effect, so the likelihood averages their product over the random-effects distribution; averaging each separately discards their dependence through that shared effect.