Single-Event Likelihood With Right Censoring

Why Separate an Event From a Censoring Time?

For one subject followed from a known event-free starting time, the observation ends either when the event occurs or when follow-up is censored. These outcomes are not interchangeable: an observed event supplies information about the event rate at its time, while a censoring time records that the subject remained event-free up to that time. The likelihood must represent both facts.

The key structure is a product of two parts. An observed event contributes an event-intensity factor at the event time. Both an observed event and a right-censoring time contribute an exponential term for the cumulative event intensity accumulated while the subject was under observation and at risk. This distinction is the single-subject version of the multiplicative-intensity likelihood.

Definition: Let $T$ be a subject's event time, and suppose the subject is event-free at time 0. The event hazard $\lambda(t;\theta)$ is the instantaneous event rate at time $t$, conditional on remaining event-free up to $t$, under parameter $\theta$. Its cumulative hazard through time $x$ is $$ \Lambda(x;\theta)=\int_0^x \lambda(u;\theta)\,du. $$ If the subject is observed until $X=\min(T,C)$, where $C$ is the censoring time, let $\delta=1$ indicate that the event is observed and $\delta=0$ indicate right-censoring. The observed at-risk process is $Y(t)=1$ while the subject is under observation and event-free, and $Y(t)=0$ thereafter. The event-process intensity is $Y(t)\lambda(t;\theta)$.

Throughout, assume an event can occur at most once for this subject, the event-time distribution has a continuous hazard, follow-up begins at time 0 with the subject event-free, and the event and censoring status are recorded correctly. For the simple likelihood below, censoring must be independent of the event time, either unconditionally or conditional on the information held fixed in the analysis. The censoring mechanism must not contribute a factor involving the event-model parameter $\theta$. We also assume no delayed entry; the risk interval therefore begins at 0.

Likelihood Contribution and Its Derivation

Under the continuous-hazard model, the event-free survival probability through time $x$ is $S_T(x;\theta)=\exp\{-\Lambda(x;\theta)\}$. An observed event at $x$ has density $\lambda(x;\theta)S_T(x;\theta)$. By contrast, right-censoring at $x$ tells us that the event time exceeds $x$, contributing $S_T(x;\theta)$, not an event density at $x$.

Result: Ignoring censoring-mechanism factors that do not involve $\theta$, the likelihood contribution for one observed time $x$ and event indicator $\delta$ is $$ L(\theta)=\{\lambda(x;\theta)\}^{\delta} \exp\left\{-\int_0^x \lambda(u;\theta)\,du\right\} =\{\lambda(x;\theta)\}^{\delta}\exp\{-\Lambda(x;\theta)\}. $$ For an event, $\delta=1$, so $L(\theta)=\lambda(x;\theta)\exp\{-\Lambda(x;\theta)\}$. For right-censoring, $\delta=0$, so $L(\theta)=\exp\{-\Lambda(x;\theta)\}$.

A short derivation makes the two factors transparent. Over a small interval of length $dt$, conditional on being event-free at $t$, the chance of an event is $\lambda(t;\theta)dt+o(dt)$. Therefore the conditional probability of surviving that interval is $1-\lambda(t;\theta)dt+o(dt)$. Multiplying these survival probabilities over intervals and taking the limit gives

\log S_T(x;\theta) =\int_0^x -\lambda(u;\theta)\,du, \qquad S_T(x;\theta)=\exp\{-\Lambda(x;\theta)\}.

For an observed event, the density at $x$ is the probability of surviving to just before $x$, multiplied by the event rate at $x$: $f_T(x;\theta)=\lambda(x;\theta)S_T(x;\theta)$. For censoring at $x$, there is no event at $x$; the event-model contribution is simply $P(T>x)=S_T(x;\theta)$. Thus the event-intensity factor appears only when $\delta=1$.

The same result follows from the path likelihood for the observed counting process. The log-likelihood contribution from an intensity $Y(t)\lambda(t;\theta)$ is the event term minus the integrated intensity:

\int_0^X \log\{\lambda(t;\theta)\}\,dN(t) -\int_0^X Y(t)\lambda(t;\theta)\,dt = \delta\log\{\lambda(x;\theta)\}-\Lambda(x;\theta).

Here $N$ has one jump at $x$ if $\delta=1$, and no jump by $x$ if $\delta=0$. The cumulative term is exposure to the event intensity over the observed at-risk interval. It remains present in both cases. The logarithm is understood only for a positive event intensity at an observed event time; if a model assigns zero intensity there, its event likelihood is zero.

The event factor has units of inverse time because it is a density in $x$; the exponential cumulative-hazard factor is dimensionless. A censoring contribution is a probability of surviving to a time and is dimensionless. Likelihoods for continuous observations are densities, not probabilities of observing an exact real-valued time.

Worked Contributions Under Three Hazard Models

Worked Example: Constant Hazard With an Event or Censoring

Suppose, hypothetically, that one subject has a constant event hazard $\lambda(t;\theta)=\lambda=0.08$ per year. Follow-up ends at 5 years. First suppose the event is observed at 5 years; then suppose instead the subject is right-censored at 5 years. Assume the censoring time is independent of the event time and its distribution contains no $\lambda$.

The cumulative hazard through 5 years is $$ \Lambda(5)=\int_0^5 0.08\,du=0.08(5)=0.40. $$

For an observed event, the event-intensity factor is $0.08$ per year and the cumulative-exposure factor is $\exp(-0.40)=0.6703$. The contribution is $$ L=0.08\exp(-0.40)=0.08(0.6703)=0.05363\ \text{per year}. $$

For right-censoring at 5 years, there is no event factor. The contribution is $$ L=\exp(-0.40)=0.6703. $$

The censored subject still contributes information: the model must assign the observed 5 years without an event a survival probability. The event observation adds the hazard at year 5; censoring does not.

Worked Example: Event Under a Piecewise-Constant Hazard

Consider a hypothetical hazard of $0.10$ per year from time 0 through year 2, followed by $0.30$ per year after year 2. A subject experiences the event at year 3.5. Assume the hazard model is correctly specified and that censoring is independent and does not occur before the event.

The cumulative hazard is the sum of the exposure in each interval: $$ \Lambda(3.5) =\int_0^2 0.10\,du+\int_2^{3.5}0.30\,du =0.10(2)+0.30(1.5) =0.65. $$

The event occurs in the second interval, so the event-intensity factor is $0.30$ per year. The event likelihood contribution is $$ L=0.30\exp(-0.65) =0.30(0.5220) =0.1566\ \text{per year}. $$

The hazard at the event time is not applied to the entire follow-up. The cumulative exposure accounts for the first 2 years at the lower rate and the next 1.5 years at the higher rate, while the event factor uses the rate applicable at year 3.5.

Worked Example: Weibull Hazard With Right-Censoring

Suppose, hypothetically, that event times follow a Weibull model with shape $k=2$ and scale $\eta=6$ years. Its hazard and cumulative hazard are $$ \lambda(t)=\frac{k}{\eta}\left(\frac{t}{\eta}\right)^{k-1}, \qquad \Lambda(t)=\left(\frac{t}{\eta}\right)^k. $$ A subject is right-censored at year 4. Assume independent censoring and that the subject was event-free at the start of follow-up.

At year 4, the cumulative hazard is $$ \Lambda(4)=\left(\frac{4}{6}\right)^2 =\frac{16}{36}=0.4444. $$ The event-model contribution for censoring is therefore $$ L=\exp(-0.4444)=0.6412. $$

There is no hazard factor at year 4 because the event was not observed. For comparison, if the event had instead been observed at year 4, the hazard at that time would be $$ \lambda(4)=\frac{2}{6}\left(\frac{4}{6}\right) =0.2222\ \text{per year}, $$ and the contribution would be $$ L=0.2222\exp(-0.4444) =0.2222(0.6412) =0.1425\ \text{per year}. $$

When Can the Censoring Mechanism Be Omitted?

The likelihood just derived is the event-model likelihood conditional on the censoring design, or the event-model factor from a joint likelihood when the censoring distribution has a separate parameter. For example, if $T$ and $C$ are independent and both distributions are modeled, the joint density for an observed event at $x$ includes $f_T(x;\theta)P(C\ge x)$; for a censoring observation at $x$, it includes $P(T>x)f_C(x)$. If the censoring factors involve parameters separate from $\theta$, they can be omitted when estimating $\theta$.

Independence is a substantive assumption, not a consequence of having a recorded censoring time. If subjects with higher unobserved event risk tend to leave follow-up earlier, censoring may be informative. In that case, the event-time contribution alone need not represent the observed-data likelihood. A joint model for event and censoring, or another justified approach, may be required.

For continuous event and censoring times, an exact tie has probability zero under the model. If recorded times are rounded and ties occur, the observation rule for ties must be specified; simply treating every tie as an event can change the likelihood. The formula also needs modification when follow-up begins after time 0, when the subject is not known to be event-free at entry, or when the event-time model has point masses rather than a continuous hazard.

Common Errors, Diagnostics, and Alternatives

Adding a hazard factor for a censored observation. The incorrect contribution $\lambda(x;\theta)\exp\{-\Lambda(x;\theta)\}$ treats censoring as an event. The correct right-censoring contribution is $\exp\{-\Lambda(x;\theta)\}$.
Discarding censored observations. This throws away the survival information through their censoring times. Retain the cumulative-exposure term for every observed at-risk interval.
Using the event-time hazard over the whole follow-up. For a time-varying hazard, compute $\int_0^x\lambda(u;\theta)\,du$ across all intervals. Use $\lambda(x;\theta)$ only for an observed event at $x$.
Ignoring units or time scale. A hazard measured per year must be integrated over years. The cumulative hazard is dimensionless, while an event-time density has units of inverse years.
Assuming independence because censoring is recorded. Check whether the follow-up process could be related to event risk. If censoring is informative, an event-only likelihood may be inadequate.

The main alternative is to model the censoring distribution jointly with the event-time distribution rather than condition on the censoring design. That approach is appropriate when censoring-model parameters matter or dependence must be represented, but it requires additional assumptions and data support. Nonparametric survival estimators, such as the Kaplan-Meier estimator, answer a different practical need by estimating survival without specifying a parametric hazard. They do not change the basic data distinction: an event contributes an event at its time, whereas censoring contributes survival through its time.

A useful diagnostic is to reconstruct each subject's contribution from the observed status: verify that each event has one hazard factor, each censoring record has none, and both have the integrated hazard over their observed at-risk time. For a piecewise hazard, check the interval-by-interval exposure arithmetic. Also compare the modeled survival through censoring times with the observed follow-up pattern; major incompatibility can signal model or censoring assumptions that need further investigation.

Key takeaway: The single-event likelihood is an event-intensity factor raised to the event indicator, multiplied by the exponential of negative cumulative intensity exposure. Right-censoring removes the event factor, not the survival contribution accumulated before censoring.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A subject is censored at time $x$ under hazard $\lambda(t;\theta)$. Write the event-model likelihood contribution.
With a constant hazard of $0.05$ per month, a subject has an observed event at 8 months. Calculate the likelihood contribution, including units.
With hazard $0.2$ per year from year 0 to year 1 and $0.4$ per year from year 1 to year 2, a subject is censored at year 2. Calculate the contribution.
Explain why a censored observation should not receive a hazard factor at its censoring time.
Name one condition under which censoring factors can be omitted from the event-model likelihood, and one situation in which that omission may fail.

Solutions. (1) The contribution is $L(\theta)=\exp\{-\int_0^x\lambda(u;\theta)\,du\}$, because $\delta=0$. (2) The cumulative hazard is $0.05(8)=0.40$. Thus $L=0.05\exp(-0.40)=0.05(0.6703)=0.03352$ per month. (3) The cumulative hazard is $0.2(1)+0.4(1)=0.60$, so $L=\exp(-0.60)=0.5488$. (4) Censoring establishes that the subject remained event-free through that time but does not establish an event there; only an observed event contributes the instantaneous hazard factor. (5) Censoring can be omitted when it is independent of the event time (possibly conditional on modeled information) and its distribution has parameters separate from the event parameter. The omission may fail if censoring is informative about event risk or shares parameters with the event model.