Likelihood Contributions From Event-Time Intervals

From an Observation Pattern to a Likelihood

When an event is known only to have occurred between two inspections, the event time is not an observed point. The data still constrain the event time: if the event was absent at one visit and present at the next, it lies in the interval between those visits. A likelihood should use that interval directly rather than pretending that its midpoint, endpoint, or some other single time was observed.

This tutorial develops likelihood contributions for common observation patterns under a specified event-time distribution. The algebra is simple once the observation pattern is stated precisely, but the assumptions matter: the distribution describes the event time, while the observation process must not selectively hide or reveal event times in a way the likelihood ignores.

Definition: Let \(T\) be a nonnegative event time with density \(f_\theta(t)\), distribution function \(F_\theta(t)=P(T\le t)\), and survival function \(S_\theta(t)=P(T>t)=1-F_\theta(t)\), where \(\theta\) is an unknown parameter. An interval-censored observation records \(a<T\le b\), for known \(0\le a<b\). Its likelihood contribution is \(F_\theta(b)-F_\theta(a)\). Left censoring records \(T\le b\), contributing \(F_\theta(b)\); right censoring records \(T>a\), contributing \(S_\theta(a)\). An exact event time \(t\) contributes the density \(f_\theta(t)\).

For continuous event-time distributions, using \(a<T\le b\) or \(a\le T\le b\) gives the same probability because \(P(T=a)=P(T=b)=0\). If the distribution has point masses, endpoint conventions can change the contribution and must match the data definition. A density contribution for an exact time is not a probability that \(T\) equals that single value; it is a likelihood density with respect to time.

Assumptions and the General Likelihood

Suppose \(n\) subjects have independent event times drawn from the same parametric distribution \(F_\theta\). For each subject, the observation pattern supplies either an exact time, a left-censoring limit, a right-censoring limit, or an interval containing the event time. If the observation mechanism is fixed by design or is noninformative for the event time under the model, the event-time likelihood is the product of the corresponding contributions. In particular, inspection times must not depend on unobserved event times in a way that changes the conditional distribution of the recorded interval.

This is a likelihood for the event-time distribution conditional on the observation scheme being ignorable. If inspection schedules or loss to follow-up are informative, their mechanisms may need to be modeled as well. Independence across subjects is also required for the simple product likelihood; clustering or dependence calls for a joint model or an appropriate dependence adjustment.

For illustration, use an exponential event-time model with rate \(\lambda>0\), measured in inverse time units. Its density and survival function are

f_\lambda(t)=\lambda e^{-\lambda t},\qquad S_\lambda(t)=e^{-\lambda t},\qquad F_\lambda(t)=1-e^{-\lambda t}.

Thus an exact event at \(t\) contributes \(\lambda e^{-\lambda t}\), a left-censored event by \(b\) contributes \(1-e^{-\lambda b}\), a right-censored event beyond \(a\) contributes \(e^{-\lambda a}\), and an interval \((a,b]\) contributes \(e^{-\lambda a}-e^{-\lambda b}\). Every contribution is a function of the unknown rate and the observed time limits.

Worked Example: Three Censoring Patterns Under an Exponential Model

Assume, hypothetically, that event times follow an exponential distribution with \(\lambda=0.25\) per month. Find the contributions for an event known to occur by month 2, an event not yet observed by month 4, and an event known to occur in \((2,5]\) months. The distribution is assumed correctly specified and the observation limits are treated as noninformative.

For left censoring by month 2, the event-time probability is \(F(2)=1-e^{-0.25(2)}=1-e^{-0.5}\approx0.3935\).

For right censoring after month 4, the contribution is the probability of surviving beyond month 4: \(S(4)=e^{-0.25(4)}=e^{-1}\approx0.3679\).

For the interval \((2,5]\), subtract the probability of surviving beyond month 5 from that of surviving beyond month 2: \(F(5)-F(2)=e^{-0.25(2)}-e^{-0.25(5)}=e^{-0.5}-e^{-1.25}\approx0.6065-0.2865=0.3200\).

These are likelihood factors, not three estimates of the event probability for a single shared subject. In a sample, the applicable factor for each independent subject is multiplied into the full likelihood.

Combining Mixed Observation Patterns

Consider five hypothetical independent subjects observed in months. Their records are: event by month 2; event-free beyond month 4; event in \((1,3]\); event in \((2,5]\); and event by month 1. Assume an exponential event-time distribution and a noninformative observation process. For a proposed rate \(\lambda\), the likelihood is the product of five contributions:

L(\lambda) =(1-e^{-2\lambda})e^{-4\lambda} (e^{-\lambda}-e^{-3\lambda}) (e^{-2\lambda}-e^{-5\lambda}) (1-e^{-\lambda}).

At \(\lambda=0.2\) per month, substitution gives

\begin{aligned} L(0.2) &=(1-e^{-0.4})e^{-0.8}(e^{-0.2}-e^{-0.6}) (e^{-0.4}-e^{-1})(1-e^{-0.2})\\ &\approx(0.329680)(0.449329)(0.269919)(0.302441)(0.181269)\\ &\approx0.00219207. \end{aligned}

The likelihood value itself is not generally a probability that the parameter is correct. It measures how compatible this observed dataset is with the proposed parameter value relative to other values in the same model.

Worked Example: Estimating the Rate From the Mixed Dataset

Use the five records above to estimate \(\lambda\) by maximum likelihood. Their observation limits are assumed fixed or ignorable, and the exponential model and independent-subject assumption are retained.

It is more convenient to maximize the log-likelihood, which has the same maximizer as the likelihood:

\ell(\lambda)= \log(1-e^{-2\lambda})-4\lambda +\log(e^{-\lambda}-e^{-3\lambda}) +\log(e^{-2\lambda}-e^{-5\lambda}) +\log(1-e^{-\lambda}).

For an interval \((a,b]\), write its contribution as \(e^{-a\lambda}(1-e^{-(b-a)\lambda})\). Differentiating its log gives \(-a+(b-a)/(e^{(b-a)\lambda}-1)\). A left-censored contribution by \(b\) has log derivative \(b/(e^{b\lambda}-1)\), and the right-censored contribution \(e^{-a\lambda}\) has log derivative \(-a\). Consequently, the score for these data is

\ell'(\lambda) =-7+\frac{4}{e^{2\lambda}-1} +\frac{3}{e^{3\lambda}-1} +\frac{1}{e^\lambda-1}.

Setting this score to zero and solving numerically gives \(\hat{\lambda}\approx0.3834\) per month. This is a numerical solution of the displayed score equation, not a closed-form estimate. The rate estimate corresponds to an exponential mean event time of \(1/\hat{\lambda}\approx2.608\) months, under the assumed model.

The solution is a maximum: each left-censored and interval log contribution has a negative second derivative, while the right-censored log contribution is linear. Here at least one contribution has strictly negative curvature, so the total log-likelihood is strictly concave for \(\lambda>0\). The score decreases through zero, establishing a unique maximizer. The estimate describes the fitted exponential rate; it does not establish that the exponential assumption is adequate.

Exact Times and Interval Records in One Likelihood

The same product rule applies when exact event times occur alongside censored records. Exact observations contribute densities; interval and censoring observations contribute probabilities. The distinction is important because replacing a density with a survival probability, or treating an interval as an exact time, changes the likelihood.

Worked Example: Exact, Interval, and Right-Censored Observations

Suppose a hypothetical sample has three independent records: an exact event at 1.5 months, an event in \((1,2.5]\) months, and a subject event-free beyond month 3. Under an exponential model with \(\lambda=0.4\) per month and noninformative observation, calculate the likelihood.

The exact-time contribution is a density: \(\lambda e^{-\lambda t}=0.4e^{-0.4(1.5)}=0.4e^{-0.6}\approx0.2195\).

The interval contribution is \(e^{-0.4(1)}-e^{-0.4(2.5)}=e^{-0.4}-e^{-1}\approx0.6703-0.3679=0.3024\). The right-censored contribution is \(e^{-0.4(3)}=e^{-1.2}\approx0.3012\).

Multiplying the three factors gives \(L(0.4)\approx(0.2195)(0.3024)(0.3012)=0.02000\). This calculation combines a density and probabilities as likelihood factors. Their different measurement dimensions do not prevent likelihood construction; the likelihood is defined relative to the observed data’s measurement structure.

Derivation, Diagnostics, and Alternatives

The interval contribution follows directly from the event represented by the record. For \(a<T\le b\), the event \(\{T\le b\}\) contains the event \(\{T\le a\}\). Removing the latter leaves precisely \(a<T\le b\), so

P(a<T\le b) =P(T\le b)-P(T\le a) =F_\theta(b)-F_\theta(a).

This derivation also shows why one must not use \(F_\theta(b)\) alone for an interval beginning after time zero: that would count events before or at \(a\) as compatible with the record. For right censoring, the observed fact is \(T>a\), hence the contribution \(1-F_\theta(a)\). For left censoring by \(b\), it is \(T\le b\), hence \(F_\theta(b)\).

Useful checks include confirming that each contribution lies between zero and one for probability records, that an interval contribution is nonnegative, and that widening an interval cannot reduce its probability. Under a continuous model, \(F_\theta(b)-F_\theta(a)\) should be positive for \(a<b\) when the model assigns positive density throughout that interval. A zero contribution makes the likelihood zero and the log-likelihood undefined. Such a result may indicate a coding error, incompatible limits, or a model that assigns no probability to an observed interval.

A common mistake is to substitute an interval midpoint and then use the exact-time density \(f_\theta((a+b)/2)\). This treats an unobserved event time as if it had been measured exactly and discards the interval’s width. The correct contribution is \(F_\theta(b)-F_\theta(a)\); for a narrow interval this is approximately \(f_\theta(t)(b-a)\) near a representative time \(t\), not simply the density value. Another error is to count a right-censored subject as an event at the censoring time. The record only says \(T>a\), so its contribution is \(S_\theta(a)\).

Compared with midpoint substitution, the interval likelihood retains all observed timing information and does not invent event times. Compared with a parametric likelihood, a nonparametric interval-censored analysis can avoid choosing a particular distributional shape, but may require more specialized estimation and can have limited identification of the event-time distribution between inspection times. A method designed only for exact and right-censored times is not automatically valid if interval observations are entered as exact events or ignored.

Parametric likelihood estimates can be sensitive to distributional misspecification, especially when intervals are wide or the observation schedule is sparse. Inspect whether fitted probabilities for observed intervals are plausible, whether estimates change materially under reasonable alternative distributions, and whether the model places appreciable mass in regions unsupported by inspections. These checks do not prove the model correct. The likelihood above also omits covariates, dependent event times, and informative inspection processes; each requires additional modeling.

Key takeaway: Translate the recorded information into the event it states, assign that event its model probability (or density for an exact time), and multiply the independent subjects’ contributions. For an interval \((a,b]\), the essential factor is \(F_\theta(b)-F_\theta(a)\).

Check Your Understanding

Answer each question, then compare with the solutions below.

Under an exponential model with rate \(\lambda\), write the contribution for an event known to have occurred in \((3,6]\).
A subject is known to be event-free through month 5. What is the contribution, and why is it not an event density at month 5?
For a continuous model, show that the interval contribution is nonnegative when \(a<b\).
Why does replacing an interval \((a,b]\) with its midpoint as an exact event generally change the likelihood?

Solutions. (1) The contribution is \(F_\lambda(6)-F_\lambda(3)=e^{-3\lambda}-e^{-6\lambda}\). (2) The record states \(T>5\), so the contribution is \(S_\lambda(5)=e^{-5\lambda}\); no exact event time was observed. (3) A distribution function is nondecreasing, so \(F_\theta(b)\ge F_\theta(a)\) for \(b>a\), giving \(F_\theta(b)-F_\theta(a)\ge0\). (4) The midpoint substitution uses the density at a single invented time and ignores interval width; the observed event is that \(T\) lies anywhere in the interval, whose probability is \(F_\theta(b)-F_\theta(a)\).