The Event-Time Contribution for Events and Censoring

Why Event and Censoring Records Contribute Differently

In a time-to-event study, an individual may experience the event during follow-up, or observation may end before the event is seen. A right-censored record tells us that the event time exceeds the observed follow-up time; it does not tell us that an event occurred at that time. The likelihood must preserve this distinction. The hazard gives the instantaneous event rate among individuals who have not yet experienced the event, while the cumulative hazard records the accumulated event risk up to a given time.

This tutorial derives the event-time contribution conditional on a subject’s covariates and random effects. It does not specify how a longitudinal trajectory enters the hazard. Write \(T_i\) for subject \(i\)’s event time, \(C_i\) for the censoring time, \(Y_i=\min(T_i,C_i)\) for the observed time, and \(\Delta_i=I(T_i\leq C_i)\) for the event indicator. Thus \(\Delta_i=1\) indicates an observed event, and \(\Delta_i=0\) indicates right censoring. Let \(b_i\) denote that subject’s random effects, and let \(x_i\) denote observed covariates. All survival quantities below are conditional on \(b_i\) and \(x_i\), even when that conditioning is left implicit.

Definition: The conditional hazard \(h_i(t\mid b_i,x_i)\) is the instantaneous event rate at time \(t\), given survival up to \(t\), the random effects \(b_i\), and covariates \(x_i\). The cumulative hazard is \(H_i(t\mid b_i,x_i)=\int_0^t h_i(u\mid b_i,x_i)\,du\). The conditional survival function is \(S_i(t\mid b_i,x_i)=P(T_i>t\mid b_i,x_i)\).

For a continuous event time, the hazard has units of inverse time, such as events per day, while the cumulative hazard is dimensionless. Assume the hazard is nonnegative and integrable over the time interval of interest. These conditions make the cumulative hazard well-defined and ensure that the survival function derived from it is nonincreasing.

Deriving Survival and the Event-Time Density

For a small positive interval \(dt\), the hazard definition says that, conditional on having survived to \(t\), the probability of an event in \([t,t+dt)\) is approximately \(h_i(t\mid b_i,x_i)\,dt\). Equivalently, the conditional probability of surviving that interval is approximately \(1-h_i(t\mid b_i,x_i)\,dt\). This leads to the differential equation for survival.

\frac{d}{dt}S_i(t\mid b_i,x_i) =-h_i(t\mid b_i,x_i)S_i(t\mid b_i,x_i).

The initial condition is \(S_i(0\mid b_i,x_i)=1\), assuming the event time is measured from the start of the risk period and no event has already occurred. Separating variables and integrating from 0 to \(t\) gives the cumulative-hazard representation:

\begin{aligned} \frac{dS_i(u\mid b_i,x_i)}{S_i(u\mid b_i,x_i)} &=-h_i(u\mid b_i,x_i)\,du,\\ \log S_i(t\mid b_i,x_i)-\log S_i(0\mid b_i,x_i) &=-\int_0^t h_i(u\mid b_i,x_i)\,du,\\ S_i(t\mid b_i,x_i)&=\exp\{-H_i(t\mid b_i,x_i)\}. \end{aligned}

For a continuous event time, its conditional density is the negative derivative of survival. Differentiating the expression above shows why the hazard appears at an observed event time:

f_i(t\mid b_i,x_i) =-\frac{d}{dt}S_i(t\mid b_i,x_i) =h_i(t\mid b_i,x_i)\exp\{-H_i(t\mid b_i,x_i)\}.

The density has units of inverse time. It accounts for both surviving until \(t\) and experiencing the event at \(t\). A right-censored observation at \(t\), by contrast, supplies only the information that \(T_i>t\), so its event-time contribution is the survival probability, not the event density.

Key result: Conditional on \(b_i\) and \(x_i\), the event-time contribution for an observed event at \(Y_i\) is \(h_i(Y_i\mid b_i,x_i)\exp\{-H_i(Y_i\mid b_i,x_i)\}\). For a right-censored observation at \(Y_i\), it is \(\exp\{-H_i(Y_i\mid b_i,x_i)\}\). Together these are \(h_i(Y_i\mid b_i,x_i)^{\Delta_i}\exp\{-H_i(Y_i\mid b_i,x_i)\}\).

Taking logarithms gives the convenient conditional log-likelihood contribution:

\ell_{T,i} =\Delta_i\log h_i(Y_i\mid b_i,x_i)-H_i(Y_i\mid b_i,x_i).

This compact expression is valid for \(\Delta_i=1\) as well as \(\Delta_i=0\). When \(\Delta_i=0\), the first term vanishes and only the cumulative-hazard penalty remains. The log-likelihood contribution is a density contribution for an event and a probability contribution for censoring; therefore, the individual likelihood contributions need not all have the same units.

Worked Contributions Under Specific Hazards

Worked Example: Constant Hazard With One Event and One Censored Time

Suppose, hypothetically, that a subject’s conditional hazard is constant at \(0.08\) per day throughout follow-up. The event process is continuous, and the hazard is assumed to apply conditional on the subject’s covariates and random effects. Find the event-time contributions for an event observed on day 10 and for a separate subject censored on day 10.

With \(h(t)=0.08\) per day, integration gives \(H(t)=0.08t\). At day 10, \(H(10)=0.08(10)=0.8\). For the observed event, the density contribution is

L_T= h(10)\exp\{-H(10)\} =(0.08)\exp(-0.8) =(0.08)(0.4493) =0.03595\ \text{per day}.

For the subject censored on day 10, the record establishes survival beyond day 10, so the contribution is

L_T=S(10)=\exp(-0.8)=0.4493.

The two values differ because one record says an event occurred at day 10 and the other says it had not occurred by day 10. Both use the same hazard and cumulative hazard; the event record includes the hazard at the observed event time, while the censored record does not.

Worked Example: A Time-Varying Hazard

Suppose, hypothetically, that a subject has hazard \(h(t)=0.01t\) per day, where \(t\) is measured in days. This hazard increases with follow-up time. Check that it is nonnegative for \(t\geq0\), and integrate it to obtain \(H(t)=0.005t^2\). Calculate the contribution for an event observed on day 8.

At day 8, \(h(8)=0.01(8)=0.08\) per day and \(H(8)=0.005(8^2)=0.32\). Therefore,

L_T=h(8)\exp\{-H(8)\} =(0.08)\exp(-0.32) =(0.08)(0.7261) =0.05809\ \text{per day}.

If instead the subject is censored on day 8, the contribution is \(S(8)=\exp(-0.32)=0.7261\). The calculation illustrates why using only the hazard value at the observed time is insufficient: the survival component depends on the hazard accumulated over the entire interval from 0 to 8 days.

Worked Example: Integrating a Piecewise-Constant Hazard

Consider a hypothetical hazard of \(0.04\) per day from day 0 to day 5 and \(0.10\) per day after day 5. Assume the piecewise hazard is known conditional on covariates and random effects. For an event observed on day 8, the cumulative hazard must include both intervals:

$$ H(8)=(0.04)(5)+(0.10)(8-5)=0.20+0.30=0.50. $$

The hazard at the event time is \(0.10\) per day, so the event contribution is

L_T=(0.10)\exp(-0.50) =(0.10)(0.6065) =0.06065\ \text{per day}.

For a subject censored on day 8 under the same hazard, the contribution is \(\exp(-0.50)=0.6065\). This calculation checks both the integration boundary and the event-time hazard: the earlier interval affects survival to day 8, but the hazard at day 8 comes from the later interval.

Censoring Assumptions and Conditional Versus Marginal Likelihood

The formula above is the event-process contribution conditional on the random effects and covariates. To use it as the observed-data likelihood contribution without separately modeling censoring, an appropriate noninformative-censoring condition is needed. A common condition is that event and censoring times are independent conditional on the observed covariates and random effects, with censoring parameters distinct from the event-process parameters. The study design and the target of inference also matter; administrative follow-up ending at a prespecified time is a typical setting where censoring can be treated as noninformative.

The qualification “conditional on the random effects” is important. If censoring is random, its mechanism may contribute information about the random effects even when its parameters do not overlap with the event-process parameters. For example, under conditional independence and continuous censoring, the full observed-data contribution conditional on \(b_i,x_i\) is

\left[f_i(Y_i\mid b_i,x_i)S_{C,i}(Y_i\mid b_i,x_i)\right]^{\Delta_i} \left[S_i(Y_i\mid b_i,x_i)f_{C,i}(Y_i\mid b_i,x_i)\right]^{1-\Delta_i},

where \(f_{C,i}\) and \(S_{C,i}\) are the censoring-time density and survival function. Conditional on \(b_i\), censoring factors that do not involve event-process parameters may sometimes be omitted when making inference about those parameters. But if \(b_i\) is latent, the likelihood must generally be integrated over its distribution. A censoring factor that depends on \(b_i\) remains inside that integral and can change the marginal likelihood for event-process parameters. It is therefore not generally valid to omit such a factor from a marginal likelihood merely because its parameters are distinct. Omission requires conditions that make censoring ignorable after accounting for the latent effects—for example, a censoring factor that does not depend on \(b_i\), together with distinct nuisance parameters—or an appropriate explicit model for the censoring mechanism.

Important distinction: The event-time factor \(h_i(Y_i\mid b_i,x_i)^{\Delta_i}\exp\{-H_i(Y_i\mid b_i,x_i)\}\) is conditional on \(b_i\). It is not, by itself, a general marginal likelihood contribution after the random effects have been integrated out. Whether censoring terms can be omitted must be assessed on the likelihood actually used, including its dependence on latent random effects.

Diagnostics, Alternatives, and Limitations

Before using an event-time contribution, check that the observation mechanism and the mathematical form agree. In particular:

Check the indicator. Confirm that \(\Delta_i=1\) means an observed event and \(\Delta_i=0\) means right censoring. Reversing these labels reverses which records receive the hazard factor.
Check the integration. Verify that \(H_i(Y_i)\) integrates the hazard from the defined start of the risk period to \(Y_i\). For a time-varying or piecewise hazard, missing an interval produces an incorrect survival term.
Check nonnegativity and units. A hazard must be nonnegative and have units of inverse time; its integral must be dimensionless. An event density should have inverse-time units, while a survival probability is dimensionless.
Check the censoring story. Ask whether censoring may depend on prognosis, covariates, or random effects. Conditional independence cannot simply be presumed when the follow-up process selects who remains observed.
Check the time scale. The hazard, cumulative hazard, and event time must use compatible time units. Converting days to months requires transforming the hazard consistently.

The hazard representation is equivalent to writing an event contribution as a density \(f_i(Y_i)\) and a censored contribution as a survival probability \(S_i(Y_i)\). The hazard form is especially useful when the model is specified through a hazard, because it exposes the event-time factor and the accumulated survival penalty. The density-survival form can be clearer when those functions are directly available. Neither representation repairs an incorrect censoring assumption.

The derivation assumes a continuous event-time distribution. With genuinely discrete event times, probability masses rather than continuous densities are needed; the continuous hazard-density formula should not be applied without adjustment. The formula also treats the event process as having a well-defined hazard over the observed risk interval. Competing events, delayed entry, interval censoring, and censoring mechanisms that are informative after conditioning require additional likelihood structure rather than a simple change to the event indicator.

A common mistake is to assign every censored subject an event density evaluated at the censoring time, reasoning that the subject was “observed until” that time. That reasoning confuses the end of observation with an event. The correction is direct: an event at \(Y_i\) contributes \(f_i(Y_i)=h_i(Y_i)S_i(Y_i)\), whereas right censoring at \(Y_i\) contributes only \(P(T_i>Y_i)=S_i(Y_i)\). The censored record does not assert that the event happened at the censoring time.

Key takeaway: An observed event contributes the hazard at its event time multiplied by survival to that time. A right-censored observation contributes survival to the censoring time alone. These are conditional event-process factors; latent random effects require care when forming a marginal likelihood, especially if censoring depends on those effects.

Check Your Understanding

For each question, work out the implication before reading the solution immediately below it.

If an individual is censored at time \(t\), which event-process contribution applies, and why is the hazard at \(t\) absent?
For a constant hazard of \(0.05\) per week, calculate the event contribution for an event at week 4 and the censoring contribution for censoring at week 4.
A hazard is \(h(t)=0.02t\) per day. Find \(H(5)\) and the event contribution for an event at day 5.
Why can a censoring factor with distinct parameters still affect a marginal likelihood when censoring depends on a latent random effect?
A researcher writes \(h(Y_i)\exp\{-h(Y_i)Y_i\}\) for every observation. State one problem with this expression.

Solutions. (1) The contribution is \(S_i(t)=\exp\{-H_i(t)\}\). Censoring establishes survival beyond \(t\), not an event at \(t\), so it supplies no event-time hazard factor. (2) Here \(H(4)=0.05(4)=0.20\). The event contribution is \(0.05\exp(-0.20)=0.04094\) per week; the censoring contribution is \(\exp(-0.20)=0.8187\). (3) Integrating gives \(H(5)=\int_0^5 0.02u\,du=0.01(25)=0.25\). Since \(h(5)=0.10\) per day, the event contribution is \(0.10\exp(-0.25)=0.07788\) per day. (4) When the likelihood is integrated over the latent effect, a censoring factor depending on that effect remains within the integral and can alter the weighting of the event-process likelihood across effect values. Distinct parameters alone do not remove that dependence. (5) It incorrectly assigns the event density to censored observations, and for a time-varying hazard it also generally substitutes \(h(Y_i)Y_i\) for the required integrated hazard \(H_i(Y_i)=\int_0^{Y_i}h(u)\,du\).