Interval-Censored Event-Time Likelihoods

Why Interval Censoring Changes the Likelihood

In many PK/PD studies, an event is not observed at the instant it occurs. A biomarker may cross a threshold between scheduled visits, or a clinical event may be known to have happened since the last assessment but not precisely when. The observed data then locate the event in an interval rather than at an exact time. Treating one point in that interval as the event time discards this distinction and can change parameter estimates.

Let \(T\) denote the event time, \(S(t)=P(T>t)\) the survival function, \(h(t)\) the hazard, \(H(t)=\int_0^t h(u)\,du\) the cumulative hazard, and \(f(t)\) the event-time density. For a continuous event time, \(S(t)=\exp[-H(t)]\) and \(f(t)=h(t)S(t)\). A subject whose event is known to occur after assessment \(L\) and by assessment \(R\) contributes the probability \(P(L<T\le R)\), not a density evaluated at an imputed time.

Definition: An event time is interval-censored when the observed information establishes that \(L<T\le R\), for known assessment times \(L<R\), but does not identify the exact value of \(T\). Under a continuous event-time model, the likelihood contribution is \(S(L)-S(R)\), equivalently \(\exp[-H(L)]-\exp[-H(R)]\).

Deriving the Likelihood Contribution

The event occurs in \((L,R]\) exactly when the subject survives beyond \(L\) but does not survive beyond \(R\). These events are nested, so subtracting their probabilities gives the probability of the interval event:

P(L<T\le R)=P(T>L)-P(T>R)=S(L)-S(R).

Using \(S(t)=\exp[-H(t)]\), the contribution becomes \(\exp[-H(L)]-\exp[-H(R)]\). The same result follows by integrating the density over all possible event times in the observed interval:

\int_L^R f(t)\,dt =\int_L^R h(t)S(t)\,dt =S(L)-S(R).

For the last equality, \(S'(t)=-h(t)S(t)=-f(t)\). Thus integrating \(f(t)=-S'(t)\) from \(L\) to \(R\) gives \(S(L)-S(R)\). This derivation also shows why the interval contribution uses probability mass over a range, while an exact event time \(t\) contributes the density \(f(t)\).

For independent subjects, their contributions multiply. An exact event at time \(t_i\) contributes \(f(t_i)\); an interval-censored event in \((L_i,R_i]\) contributes \(S(L_i)-S(R_i)\); and a subject known to be event-free through time \(C_i\) contributes \(S(C_i)\). A subject whose event is known only to have occurred by \(R_i\) contributes \(1-S(R_i)\). The full likelihood is the product of the applicable contributions, evaluated under the same model for \(S\), \(H\), and any PK/PD covariate effects.

Worked Example: Constant-Hazard Interval Probability

Suppose, hypothetically, that a constant event hazard is \(\lambda=0.2000\) per month. A participant is assessed at months 2 and 5, and the event is absent at month 2 but present at month 5. Calculate the likelihood contribution.

For a constant hazard, \(H(t)=\lambda t\) and \(S(t)=\exp(-\lambda t)\). The observation says \(2<T\le5\), so

P(2<T\le5)=S(2)-S(5) =e^{-(0.2000)(2)}-e^{-(0.2000)(5)} =e^{-0.4}-e^{-1.0}.

Using unrounded exponential values, \(e^{-0.4}=0.6703200\) and \(e^{-1.0}=0.3678794\), giving a contribution of \(0.3024406\). Under this model, the probability of an event in the observed interval is about 0.3024. It is not a probability that the event occurred at a particular month within the interval.

If instead the event had been known to occur exactly at month 3.4, its contribution would be the density \(f(3.4)=\lambda e^{-\lambda(3.4)}=0.2000e^{-0.68}\approx0.1013\) per month. This has units of inverse time, unlike the dimensionless interval probability. The two contributions represent different observations and should not be interchanged.

Estimation From Interval-Censored Data

Parameter estimation proceeds by maximizing the likelihood formed from the observed intervals and any other censoring types. The interval endpoints are data; the event time inside an interval is not. The next example shows directly how the interval likelihood determines an estimate.

Worked Example: Exponential-Hazard Estimate From Two Intervals

Consider hypothetical independent participants whose event times both fall in \((1,2]\) months. Assume an exponential event-time distribution with unknown constant hazard \(\lambda>0\), measured in inverse months. Find the maximum-likelihood estimate.

For either participant, the interval contribution is \(e^{-\lambda}-e^{-2\lambda}\). Independence gives

L(\lambda)=\left(e^{-\lambda}-e^{-2\lambda}\right)^2 =\left[e^{-\lambda}(1-e^{-\lambda})\right]^2.

It is convenient to maximize the log-likelihood, which has the same maximizer because the logarithm is increasing:

\ell(\lambda)=2\left[-\lambda+\log(1-e^{-\lambda})\right], \qquad \frac{d\ell}{d\lambda} =2\left[-1+\frac{1}{e^\lambda-1}\right].

Setting the derivative to zero gives \(1/(e^\lambda-1)=1\), hence \(e^\lambda=2\) and \(\hat{\lambda}=\log(2)=0.6931\) per month. This is a maximum: as \(\lambda\) approaches zero or infinity, the interval probability approaches zero, while at \(\log(2)\) it is \(0.25\) for each participant and the likelihood is \(0.25^2=0.06250\).

For comparison, replacing both unknown event times by the interval midpoint, 1.5 months, would treat the data as two exact events at 1.5. The resulting exponential exact-time likelihood is proportional to \(\lambda^2e^{-3\lambda}\), whose maximizer is \(2/3=0.6667\) per month. This is a different analysis based on invented exact times, not the likelihood for the recorded intervals. The numerical difference in this small example is modest, but midpoint imputation need not be harmless in other designs or models.

Time-Varying Hazards and Exposure

An interval contribution remains \(S(L)-S(R)\) when the hazard varies with time, including when it depends on a changing concentration profile. The cumulative hazard must account for the hazard across the full interval. If \(h(t)\) depends on exposure \(C(t)\) and model parameters, then \(H(t)=\int_0^t h(u\mid C(u))\,du\); the interval probability uses the cumulative hazard at both assessment times. This is why substituting a concentration measured at one visit into a constant-hazard expression can be inappropriate when exposure and hazard change between visits.

Worked Example: Interval Probability With a Piecewise Hazard

Suppose, hypothetically, that the hazard is \(0.1000\) per month from time 0 through month 2 and \(0.3000\) per month after month 2. An event is absent at month 1 and present at month 3. Compute the interval contribution for \((1,3]\).

The cumulative hazard at month 1 is \(H(1)=(0.1000)(1)=0.1000\). At month 3, the first two months contribute \(0.1000(2)=0.2000\), and the next month contributes \(0.3000(1)=0.3000\), so \(H(3)=0.5000\). Therefore,

P(1<T\le3)=e^{-H(1)}-e^{-H(3)} =e^{-0.1000}-e^{-0.5000} =0.9048374-0.6065307 =0.2983068.

The contribution is approximately 0.2983. It includes the hazard accumulated between both assessment times; it does not assign the event to month 2 or to either assessment date.

Assumptions, Diagnostics, and Alternatives

The likelihood above assumes that the event-time model describes the event process and that the recorded interval is accurate: the subject is known to be event-free at \(L\) and to have experienced the event by \(R\). In a continuous-time model, the choice of open or closed endpoint does not change the probability because a single time has probability zero. If times are discrete, however, endpoint conventions and the model’s probability masses must be handled explicitly.

A further key assumption concerns the assessment process. The usual likelihood treats the inspection schedule as fixed by design or noninformative for the unobserved event time, conditional on modeled information. If visits are triggered by symptoms or suspected events, the observation schedule may carry information about event risk. Ignoring such a mechanism can bias inference; the likelihood may need to model the assessment process as well. The subjects’ event-time contributions are also multiplied as independent only when independence is appropriate, or when a specified joint model accounts for dependence.

Useful checks include verifying that \(R>L\), that every interval contribution is between zero and one, and that \(S(L)\ge S(R)\) for a valid survival curve. For a time-varying hazard, confirm that \(H(R)\) includes all hazard segments from the time origin through \(R\), not only the segment inside \((L,R]\). Plotting or tabulating observed assessment intervals can reveal that events are localized very coarsely; such data may contain little information about the detailed shape of the hazard between visits. Implausibly precise parameter estimates despite wide intervals deserve scrutiny.

The main alternatives answer different questions or impose different approximations. Exact-time likelihoods use \(f(t)\) only when event times are actually observed with adequate precision. Right-censoring uses \(S(C)\) when a participant is event-free through a known censoring time. Midpoint substitution is simple but treats an unknown time as known, can distort the likelihood, and generally fails to propagate uncertainty about within-interval event times. A discrete-time model can instead represent event probabilities over assessment periods, but its time scale and hazard interpretation differ from a continuous-time model. Direct interval-likelihood estimation preserves the information actually observed, though it cannot recover fine temporal detail that the assessment schedule did not record.

Common pitfall: Treating an event recorded in \((L,R]\) as an exact event at \(L\), \(R\), or the midpoint and then using \(f(t)\) silently changes the data. The correct continuous-time contribution is \(S(L)-S(R)\), because the observation identifies a probability over the whole interval, not a density at one selected time.

Interval censoring matters in PK/PD because assessments are often scheduled, concentrations may change between visits, and an event can occur without being observed immediately. Correctly integrating the modeled hazard over the observation interval lets the analysis use the available information without claiming temporal precision the study did not provide. It also makes clear what the data cannot identify: if intervals are broad, estimates of rapid hazard changes may depend heavily on model assumptions.

Check Your Understanding

Answer each question, then compare with the solutions below.

Under a constant hazard of \(0.1000\) per month, what is the likelihood contribution for an event observed in \((3,6]\) months?
Under a constant hazard of \(0.2500\) per month, what contribution comes from a participant known to be event-free through month 4?
Explain why an event known to occur in \((0,2]\) contributes a probability rather than the density at month 1.
A hazard is \(0.2000\) per month from time 0 to month 1 and \(0.4000\) per month from month 1 to month 2. Calculate the probability of an event in \((0.5,1.5]\).

Solutions.

1. \(S(3)-S(6)=e^{-0.3}-e^{-0.6}=0.7408182-0.5488116=0.1920066\), or about 0.1920.

2. Right-censoring at month 4 contributes \(S(4)=e^{-(0.2500)(4)}=e^{-1}=0.3679\).

3. The observation says only that the event time lies somewhere in the interval, so its contribution is \(P(0<T\le2)=S(0)-S(2)=1-S(2)\). The density \(f(1)\) would be appropriate for an exact event time at month 1, which was not observed.

4. \(H(0.5)=0.2000(0.5)=0.1000\), while \(H(1.5)=0.2000(1)+0.4000(0.5)=0.4000\). Thus \(P(0.5<T\le1.5)=e^{-0.1}-e^{-0.4}=0.9048374-0.6703200=0.2345174\), approximately 0.2345.