Exponential Event-Time Likelihood With Right Censoring

Why Censoring Still Provides Information

In a time-to-event trial, some participants experience the event during observation, while others leave the study or reach the end of follow-up without an observed event. The latter participants are right-censored: their event times are not known, but the data establish that each event time exceeds that participant’s observed follow-up. That information contributes to the likelihood and must not be discarded.

For an exponential event-time model, the likelihood has a particularly simple form. Each observed event contributes a hazard factor, while every participant contributes a survival factor for all the time they were followed. The result depends on two summaries: the number of observed events and the total observed follow-up time.

Definition: Suppose participant $i$ has latent event time $X_i$ and censoring time $C_i$. The observed follow-up is $Y_i=\min(X_i,C_i)$, and the event indicator is $\delta_i=1$ if an event is observed and $\delta_i=0$ if the observation is right-censored. Under an exponential model with hazard $\lambda>0$, the likelihood for $\lambda$, up to factors that do not depend on $\lambda$, is $$ L(\lambda)\propto \lambda^D\exp(-\lambda S), \qquad D=\sum_{i=1}^n\delta_i, \qquad S=\sum_{i=1}^nY_i. $$ Here $D$ is the number of observed events and $S$ is total observed follow-up, measured in participant-time units.

Building the Likelihood From Each Record

The exponential model assumes a constant event hazard over time. Its event-time density and survival function are

f(x\mid\lambda)=\lambda e^{-\lambda x},\quad x>0, \qquad P(X_i>y\mid\lambda)=e^{-\lambda y},\quad y\geq 0.

If an event occurs at time $y$, the contribution involving $\lambda$ is the density $f(y\mid\lambda)=\lambda e^{-\lambda y}$. If the participant is censored at time $y$, the observed information is that the event time exceeds $y$, so the contribution is the survival probability $P(X_i>y\mid\lambda)=e^{-\lambda y}$. Thus a single record contributes

\bigl[\lambda e^{-\lambda Y_i}\bigr]^{\delta_i} \bigl[e^{-\lambda Y_i}\bigr]^{1-\delta_i} = \lambda^{\delta_i}e^{-\lambda Y_i}.

For independent participants, multiply these contributions. The powers of $\lambda$ add to $D$, and the follow-up times add to $S$, giving $L(\lambda)\propto\lambda^D e^{-\lambda S}$. This is a likelihood as a function of the unknown hazard, not a probability distribution over $\lambda$. Event records contribute densities, so likelihood values can have units; likelihood ratios for the same records are dimensionless.

Assumptions: The event times are independent across participants and follow an exponential distribution with a common constant hazard $\lambda$. Censoring is independent of the event time, or is noninformative conditional on modeled covariates, and the censoring mechanism contributes no factor involving $\lambda$. If censoring depends on a participant’s unobserved event risk, the simple likelihood need not be valid.

Worked Likelihood Calculations

Worked Example: Combining Events and Censored Follow-Up

In a hypothetical trial, five participants have the following records, with times measured in weeks: an event at 2.0, censoring at 4.0, an event at 3.5, censoring at 1.5, and an event at 5.0. Assume independent exponential event times with a common hazard and independent censoring. The censoring times are assumed not to depend on the participants’ event risks.

There are three observed events, so $D=3$. Every participant contributes their observed time, whether or not an event occurred: $S=2.0+4.0+3.5+1.5+5.0=16.0$ participant-weeks. Therefore

L(\lambda)\propto \lambda^3e^{-16\lambda}.

For example, at the candidate hazard $\lambda=0.15$ per week,

L(0.15)\propto (0.15)^3e^{-(0.15)(16)} =0.003375e^{-2.4} \approx 0.003375(0.090718) \approx 0.0003062.

The numerical likelihood is proportional to the likelihood under the omitted, parameter-free censoring factors. Its units depend on the event-time density convention; its value is not the probability of this exact set of event times.

Worked Example: Why Censored Time Must Be Included

Consider a hypothetical record set with events at 2 and 4 weeks, and censoring at 2 and 6 weeks. Assume the same independent, common-hazard exponential model and noninformative censoring. Then $D=2$, while total follow-up is $S=2+4+2+6=14$ participant-weeks. The likelihood is

L(\lambda)\propto\lambda^2e^{-14\lambda}.

At $\lambda=0.10$ per week,

L(0.10)\propto (0.10)^2e^{-(0.10)(14)} =0.01e^{-1.4} \approx 0.01(0.246597) \approx 0.002466.

If the two censored participants were incorrectly omitted, the event times alone would give $S=6$, leading to $0.01e^{-0.6}\approx0.005488$. That is not the likelihood for the observed records: it discards the evidence that two additional participants remained event-free for 8 combined participant-weeks. The direction and size of any resulting inferential error depend on the full analysis, but the likelihood calculation itself is plainly incomplete.

Worked Example: Checking Candidate Hazards and Finding the Maximum

In another hypothetical trial, events are observed at 1.2, 2.8, and 6.5 weeks, while two participants are censored at 3.0 and 4.5 weeks. Assume independent exponential event times with a common hazard and independent, noninformative censoring. The total is $D=3$ and $S=1.2+2.8+6.5+3.0+4.5=18.0$ participant-weeks. Hence $L(\lambda)\propto\lambda^3e^{-18\lambda}$.

At $0.10$ events per week, the likelihood is proportional to $0.10^3e^{-1.8}=0.001(0.165299)\approx0.0001653$. At $0.20$ events per week, it is proportional to $0.20^3e^{-3.6}=0.008(0.027324)\approx0.0002186$. Thus the second candidate has the larger likelihood among these two, but that comparison alone does not show it is the maximum.

To find the maximum for $D>0$, take the log likelihood, ignoring constants:

\ell(\lambda)=D\log(\lambda)-\lambda S, \qquad \frac{d\ell}{d\lambda}=\frac{D}{\lambda}-S.

Setting the derivative to zero gives $\widehat{\lambda}=D/S=3/18=0.1667$ events per week. The second derivative is $-D/\lambda^2<0$ for $\lambda>0$, confirming a maximum. Its likelihood is proportional to

\left(\frac{1}{6}\right)^3e^{-3} =\frac{1}{216}(0.0497871) \approx0.0002305.

This candidate’s likelihood exceeds the values at both 0.10 and 0.20 per week, as expected for the maximum.

Diagnostics, Comparisons, and Limits

A useful check is to reconstruct $D$ and $S$ from participant-level records before evaluating the likelihood. Count only observed events in $D$, but include every participant’s observed time in $S$. Check that the hazard and follow-up units agree: if time is recorded in weeks, $\lambda$ is in inverse weeks, and $\lambda S$ is dimensionless. A negative or zero follow-up time, an event indicator inconsistent with the record, or an unexpected change in the likelihood when censored records are added signals a data or coding problem.

The form also provides a mathematical diagnostic. For $D>0$, the log likelihood is concave and has the unique maximum $D/S$. If a calculation or software output gives a different maximizer under this model, verify the event count, total time, and time units. If $D=0$ and $S>0$, then $L(\lambda)\propto e^{-\lambda S}$, which decreases as $\lambda$ increases; the likelihood’s supremum is at the boundary $\lambda\to0$. There is no interior maximum over strictly positive hazards. Zero events therefore do not justify inserting an event count of one.

With no censoring, every participant has an observed event, so $D=n$ and $S$ is the sum of all event times; the likelihood reduces to the uncensored exponential likelihood. A general right-censored survival likelihood represents each observed event by a density contribution and each censoring record by a survival contribution. The exponential model makes those contributions collapse to $D$ and $S$. Methods such as the Kaplan–Meier estimator provide a nonparametric description of survival, rather than imposing a single constant hazard. These approaches answer related but different questions; the exponential likelihood can be compact, but its constant-hazard assumption may be unsuitable if risk changes over time.

The likelihood reduction does not repair informative censoring, dependence among participants, or a failure of the exponential model. For example, if participants at higher event risk are more likely to withdraw, treating their censoring as independent can distort the event-time information. If hazards vary substantially over follow-up, the constant-hazard model may also fit poorly. Such situations require an appropriate censoring or event-time model rather than merely changing the arithmetic for $D$ and $S$.

Common mistake: “A censored participant had no event, so that person contributes nothing to the likelihood.” The correction is that the participant contributes the probability of remaining event-free through the censoring time, $e^{-\lambda Y_i}$. They add no $\lambda$ factor for an observed event, but their follow-up still adds to $S$.

Check Your Understanding

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

A hypothetical participant has an event at 5 days. What is their contribution to the exponential likelihood?
Two hypothetical participants are censored at 3 and 7 days, with no events observed. What are $D$, $S$, and the likelihood up to proportionality?
For $D=4$ events and $S=20$ participant-months, find the maximum-likelihood hazard and state its units.
Why is it incorrect to calculate total follow-up using only participants who had events?

Solutions. (1) The event contribution is $\lambda e^{-5\lambda}$, with $\lambda$ measured per day. (2) $D=0$, $S=3+7=10$ participant-days, and $L(\lambda)\propto e^{-10\lambda}$; each censoring record contributes survival through its observed time. (3) Since $D>0$, $\widehat{\lambda}=D/S=4/20=0.200$ events per participant-month. (4) Censored participants establish that they remained event-free through their observed follow-up, so their times contribute survival factors and must be included in $S$.

Exponential Event-Time Likelihood With Right Censoring

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