The Turnbull Estimator for Interval-Censored Survival

Why a Probability-Mass Representation Is Useful

With interval-censored data, an event time is not observed exactly. Instead, the information for subject $i$ may be that the event occurred in $(L_i,R_i]$. The unknown distribution must therefore assign probabilities to intervals, rather than match a list of observed event times. The Turnbull estimator expresses this problem as a likelihood for probability masses and maximizes it without specifying a parametric family for event times.

The important qualification is that the data may identify probabilities of certain sets of times without identifying where, within those sets, the probability lies. A fitted mass is consequently not always evidence that events occur at one particular time. The probability-mass representation makes both the estimation and this limitation visible.

Definition: Let $T$ be a nonnegative event time with unknown distribution function $F(t)=P(T\le t)$. Subject $i$ is observed to have $T_i\in(L_i,R_i]$, where $0\le L_i<R_i\le\infty$. Under independent subjects with a common distribution $F$, and an ignorable observation process, the interval-censored likelihood is $$ L(F)=\prod_{i=1}^n P(L_i<T_i\le R_i) =\prod_{i=1}^n \{F(R_i)-F(L_i)\}. $$ Here $F(\infty)=1$. The Turnbull estimator is a nonparametric maximum likelihood estimator (NPMLE): a distribution $\widehat F$ that maximizes this likelihood over all distributions on the event-time space.

The likelihood formula assumes the observation intervals themselves do not provide additional information about $T_i$ beyond the fact that $T_i$ lies in the reported interval. A fixed inspection schedule independent of event time is one setting in which this assumption is appropriate. More generally, the inspection or censoring mechanism must be ignorable for the event-time likelihood being used. Subjects are also assumed independent, and the common-$F$ formulation assumes they share one event-time distribution. If inspection timing depends on latent event time, the observation process may need to be modelled jointly.

From Intervals to Probability Masses

For a finite representation, choose candidate times $u_1,\ldots,u_m$, and let $p_j$ be the probability mass assigned to $u_j$. The masses satisfy $p_j\ge0$ and $\sum_{j=1}^m p_j=1$. Define $A_{ij}=1$ when $u_j\in(L_i,R_i]$, and $A_{ij}=0$ otherwise. Then the probability assigned to subject $i$'s observed interval is the sum of the masses whose candidate times fall in it:

P(L_i<T_i\le R_i)=\sum_{j=1}^m A_{ij}p_j, \qquad L(\boldsymbol p)=\prod_{i=1}^n\left(\sum_{j=1}^m A_{ij}p_j\right).

The matrix $A$ is an interval-incidence matrix: its rows describe which candidate times are compatible with each observation. The optimization is over the probability simplex, meaning all nonnegative mass vectors that sum to one. This finite formulation is not an assumption that event times can occur only at the selected candidate times. It is a way to represent the likelihood-relevant information in the observations.

Why a Finite Representation Suffices

There are only finitely many observations, and each time point $t$ has a membership pattern: the vector whose $i$th entry is 1 if $t\in(L_i,R_i]$ and 0 otherwise. Although the time axis is continuous, only finitely many distinct membership patterns can occur—at most $2^n$. For any distribution $F$, collect all its probability into groups according to these patterns. The likelihood depends on the probability in each group, not on how that probability is spread within the group.

Partition by membership.
Group event times that belong to exactly the same set of observed intervals. Each nonempty group has one incidence pattern.

Collect the probability.
For each group, add the $F$-probability of all times in that group. The interval probability for each subject is the sum of the group masses whose pattern includes that subject.

Place each mass at a representative time.
Choose any time within each nonempty group and put that group’s total probability there. Every subject’s interval probability, and hence the likelihood, is unchanged.

This gives the key finite-support justification: an NPMLE can be represented by finitely many masses, one per relevant membership pattern. The likelihood is continuous in those masses, and the probability simplex is closed and bounded, so a maximum exists in this finite representation. The maximizing mass vector or the underlying distribution need not be unique. In particular, moving probability within a group without changing its interval memberships preserves the likelihood.

Worked Example: Building and Maximizing a Small Likelihood

Suppose three independent hypothetical subjects have observed intervals $(0,1]$, $(2,3]$, and $(0,3]$. Assume their inspection process is ignorable and they share one event-time distribution. For a positive-likelihood representation, use candidate times $u_1=0.5$ and $u_2=2.5$, with masses $p_1$ and $p_2$, where $p_1+p_2=1$.

The first interval contains $u_1$ but not $u_2$; the second contains $u_2$ but not $u_1$; the broad third interval contains both. Thus the incidence rows are $(1,0)$, $(0,1)$, and $(1,1)$, respectively. The likelihood is

$$ L(p_1,p_2)=p_1p_2(p_1+p_2)=p_1p_2. $$

Set $p_2=1-p_1$, where $0\le p_1\le1$. Then $L(p_1)=p_1(1-p_1)$, with derivative $1-2p_1$. The derivative is zero at $p_1=0.5$; the likelihood is zero at either boundary and $0.25$ at this interior point. Therefore $\widehat p_1=\widehat p_2=0.5$, and the maximum likelihood is $0.25$. The broad interval contributes probability $p_1+p_2=1$, so it does not change the maximizing masses.

The fitted distribution can be represented by half its mass at 0.5 time units and half at 2.5 time units. Those particular times are convenient representatives of the two observed intervals; the likelihood alone does not establish that events actually occur at precisely those times.

Worked Mass Calculations and Interpretation

Worked Example: Repeated Interval Patterns

Consider four hypothetical independent subjects under the same assumptions. Two have intervals $(0,1]$, one has $(2,3]$, and one has $(0,3]$. Use candidate times $u_1=0.5$ and $u_2=2.5$, with masses $p_1$ and $p_2=1-p_1$. The broad interval contains both candidate times, so the likelihood is

$$ L(p_1,p_2)=p_1^2p_2(p_1+p_2)=p_1^2p_2. $$

Substitute $p_2=1-p_1$: $L(p_1)=p_1^2(1-p_1)$. For an interior maximum, differentiate:

\frac{dL}{dp_1}=2p_1(1-p_1)-p_1^2 =p_1(2-3p_1).

The interior stationary point is $p_1=2/3$, giving $p_2=1/3$. At the boundaries $p_1=0$ and $p_1=1$, the likelihood is zero; at the stationary point it is $(2/3)^2(1/3)=4/27\approx0.1481$. Thus the maximum is at masses $2/3$ and $1/3$. The two observations known to lie in $(0,1]$ require substantial probability there, while the subject known to lie in $(2,3]$ requires positive probability in the second interval. These are maximum-likelihood estimates of distribution mass, not estimates of the proportion of events at the representative times themselves.

Worked Example: Equal Likelihood Does Not Fix Event Times

Suppose two hypothetical subjects have intervals $(0,1]$ and $(1,2]$. Put mass $p_1$ at any representative $u_1\in(0,1]$, and $p_2$ at any representative $u_2\in(1,2]$. Each subject’s interval contains its corresponding candidate time only, so $L=p_1p_2$, with $p_1+p_2=1$. As in the first example, the maximum is at $p_1=p_2=0.5$, with likelihood $0.25$.

One maximizing distribution places the masses at 0.25 and 1.5; another places them at 0.9 and 1.5. Both assign probability 0.5 to each subject’s observed interval, so both have likelihood $0.25$. Yet at $t=0.75$, the first distribution has $F(t)=0.5$, while the second has $F(t)=0$. The data identify the mass assigned to the two observed interval regions in this representation, but do not identify the exact placement of the first mass within $(0,1]$. A reported pointwise distribution curve should therefore be interpreted with attention to what the observations actually locate.

Reading the Estimate and Checking the Fit

When the NPMLE is represented by point masses $\widehat p_j$ at times $u_j$, the corresponding distribution function is $\widehat F(t)=\sum_{j:u_j\le t}\widehat p_j$, and the estimated survival function is $\widehat S(t)=1-\widehat F(t)$. If the likelihood identifies only mass within a region rather than its location inside that region, this step-function display is one representative of the fitted distribution, not necessarily the only maximizing curve. The introductory material on interval-censored events motivates why such observations do not reveal exact event times; the mass representation states precisely how that coarsening enters the likelihood.

Before interpreting fitted masses, check the data and assumptions. Every interval must be valid, with $L_i<R_i$, and its endpoint convention must match the data definition. The assumed model should give each observation positive interval probability at a plausible fit. A zero fitted probability for an observed interval makes the product likelihood zero. Also check that the inspection process can reasonably be treated as ignorable and that a common event-time distribution is appropriate for the subjects being combined.

Optimization diagnostics include verifying that all masses are nonnegative and sum to one, evaluating the likelihood at the reported solution, and checking whether boundary solutions are plausible. Repeated or indistinguishable incidence patterns can also make individual mass allocations nonunique. If different maximizing mass vectors or representative times produce the same likelihood, that is an identifiability limitation in the observed data, not necessarily an optimization failure.

Alternatives and Common Pitfalls

Approach	What it does	Key distinction
Turnbull NPMLE	Maximizes the interval likelihood over distributions without choosing a parametric family.	Uses each subject’s full observed interval probability.
Midpoint substitution	Replaces each interval by its midpoint and analyzes those values as observed times.	Adds unsupported exact-time information and is not the interval likelihood above.
Parametric interval-censored model	Maximizes the same kind of interval probability likelihood within a specified distribution family.	Can be more structured, but conclusions depend on the chosen family.
Kaplan–Meier estimator	Uses exact event times with right-censoring information.	It is not a direct substitute for the general interval-censored likelihood.

Common mistake: “The interval is $(0,1]$, so the event happened at 0.5; using 0.5 as an observed event time gives the nonparametric estimate.” The correction is that 0.5 is at most a convenient representative. The observed contribution is $P(0<T\le1)=F(1)-F(0)$, and the likelihood does not generally distinguish among event times inside that interval. Midpoint substitution can change the statistical problem rather than solve the interval-censored one.

The Turnbull estimator is flexible, but it cannot recover detail that the observation intervals do not contain. It is also an unadjusted distribution estimator in the common-$F$ setup here: covariate effects, parametric structure, and more complex multi-state observation models require different modelling choices. In particular, the estimate’s nonparametric character does not remove the need to justify the coarsening assumptions or assess how much the inspection schedule leaves unidentified.

Key takeaway: The Turnbull estimator maximizes products of interval probabilities. Its finite probability-mass representation converts the problem to maximizing $\prod_i\sum_j A_{ij}p_j$ over nonnegative masses summing to one; the fitted likelihood may determine interval-level mass without determining exact event times.

Check Your Understanding

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

For one observation $(2,5]$, write its likelihood contribution in terms of $F$.
Two hypothetical subjects have intervals $(0,1]$ and $(1,2]$, represented by masses $p_1,p_2$ with $p_1+p_2=1$. Find the maximum likelihood masses and maximum likelihood.
Why can probability be moved within a membership-pattern group without changing the likelihood?
A fitted distribution puts mass at the midpoint of every observed interval. Does that fact alone establish a Turnbull NPMLE? Explain.

Solutions. (1) The contribution is $P(2<T\le5)=F(5)-F(2)$. (2) The likelihood is $p_1p_2=p_1(1-p_1)$. Its derivative is $1-2p_1$, so the maximum occurs at $p_1=p_2=0.5$, with likelihood $0.25$; the boundary likelihoods are zero. (3) All times in a membership-pattern group belong to exactly the same observed intervals, so moving mass among them leaves every interval probability, and thus the product likelihood, unchanged. (4) No. Midpoints are representative times, but placing mass there is not by itself proof that the associated mass vector maximizes the interval likelihood. The estimator must maximize that likelihood over admissible probability masses.