Self-Consistency Equations for the Turnbull Estimator

From compatible intervals to probability masses

With interval-censored data, the event time is not observed exactly. A subject’s record instead restricts the event time to an interval, such as $(L_i,R_i]$. The Turnbull estimator represents the unknown event-time distribution by probability masses on support intervals that are compatible with the observed records. The central question here is how those masses must relate to the observed compatibility patterns at a self-consistent estimate.

Assume $n$ independent subjects share a common event-time distribution, and that the inspection or coarsening mechanism is noninformative for the event time, so the interval-censored likelihood can be formed from the observed compatibility information. Let $I_1,\ldots,I_m$ denote the Turnbull support intervals, using endpoint conventions consistent with the observed intervals. Write $p_j$ for the probability mass assigned to $I_j$. In this tutorial, the support includes all relevant event-time regions, so the masses sum to one.

Definition: For subject $i$, let $A_i\subseteq\{1,\ldots,m\}$ be the set of support intervals compatible with that subject’s observation. The mass $p_j$ is nonnegative, and $\sum_{j=1}^{m}p_j=1$. Subject $i$’s likelihood contribution is the total compatible mass $q_i=\sum_{j\in A_i}p_j$.

The support intervals are not necessarily ordinary, equal-width time bins. They encode regions of event time that the observed interval patterns can distinguish. The introductory material on the Turnbull estimator and maximal intersections establishes how such support regions are constructed. Once they are specified, the self-consistency calculation depends on the sets $A_i$, not on choosing a particular time within each support interval.

The likelihood and the self-consistency equations

For a proposed mass vector $\mathbf p=(p_1,\ldots,p_m)$, the likelihood and log-likelihood, up to factors that do not depend on $\mathbf p$, are

L(\mathbf p)=\prod_{i=1}^{n}q_i =\prod_{i=1}^{n}\left(\sum_{j\in A_i}p_j\right), \qquad \ell(\mathbf p)=\sum_{i=1}^{n}\log\left(\sum_{j\in A_i}p_j\right).

For any subject compatible with support interval $I_j$, the fraction of that subject’s compatible mass assigned to $I_j$ is

w_{ij}= \begin{cases} \dfrac{p_j}{q_i},&j\in A_i,\\[6pt] 0,&j\notin A_i. \end{cases}

These fractions are conditional allocation weights: for each subject, they are nonnegative and sum to one across compatible intervals. They do not assert that the event time was observed in a particular support interval. Rather, they describe how the fitted distribution allocates that subject’s compatible probability among the intervals.

Theorem: A self-consistent mass vector satisfies $$ p_j=\frac{1}{n}\sum_{i=1}^{n}w_{ij} =\frac{1}{n}\sum_{i=1}^{n} \frac{\mathbf{1}(j\in A_i)p_j}{\sum_{k\in A_i}p_k}, \qquad j=1,\ldots,m. $$ Here $\mathbf{1}(j\in A_i)$ equals one when $I_j$ is compatible with subject $i$, and zero otherwise.

In words, each support mass equals the average fitted allocation to that interval across subjects. This is a fixed-point condition: the masses determine the allocations, and those allocations, averaged over subjects, reproduce the same masses.

Why these equations hold

One justification starts by imagining an unobserved allocation indicator $Z_{ij}$, equal to one if subject $i$’s event time falls in support interval $I_j$. Conditional on the observed interval and on a proposed mass vector, the probability of that allocation is $w_{ij}$. Thus the expected number allocated to $I_j$ is $\sum_i w_{ij}$. Dividing by $n$ gives the mass that would be estimated from those expected counts. Self-consistency requires that it equal the mass used to calculate the allocation weights.

The same equations follow from the likelihood for an interior maximum. Differentiating the log-likelihood gives

\frac{\partial\ell}{\partial p_j} =\sum_{i=1}^{n}\frac{\mathbf{1}(j\in A_i)}{q_i}.

At an interior constrained maximum, the derivatives are equal across $j$, because the constraint is $\sum_jp_j=1$. Call their common value $\lambda$. Multiplying the $j$th derivative by $p_j$ and summing over $j$ gives

\lambda =\sum_{j=1}^{m}p_j\frac{\partial\ell}{\partial p_j} =\sum_{i=1}^{n}\frac{\sum_{j\in A_i}p_j}{q_i} =\sum_{i=1}^{n}1=n.

Therefore $\partial\ell/\partial p_j=n$ for a positive-mass interval at an interior maximum. Multiplying that equation by $p_j/n$ yields the self-consistency equation. The log-likelihood is concave in $\mathbf p$, since each term is the logarithm of a positive linear function. Consequently, a feasible candidate satisfying the appropriate maximum conditions is a global maximum. At a boundary, zero masses need separate attention: the fixed-point equation for a zero mass reads $0=0$, so it alone does not rule out a direction that increases the likelihood.

Worked Example: Turning Compatibility Patterns Into Allocations

Suppose there are three support intervals and three hypothetical subjects with compatibility sets $A_1=\{1,2\}$, $A_2=\{2\}$, and $A_3=\{2,3\}$. Consider the trial mass vector $(p_1,p_2,p_3)=(0.2,0.5,0.3)$. This is a valid trial vector because its entries are nonnegative and sum to one.

The compatible masses are $q_1=0.2+0.5=0.7$, $q_2=0.5$, and $q_3=0.5+0.3=0.8$. The allocation weights are $w_{11}=0.2/0.7=0.2857$, $w_{12}=0.5/0.7=0.7143$, $w_{22}=1$, $w_{32}=0.5/0.8=0.6250$, and $w_{33}=0.3/0.8=0.3750$.

The averages of these allocations are $$ \left( \frac{0.2857}{3}, \frac{0.7143+1+0.6250}{3}, \frac{0.3750}{3} \right) =(0.0952,\,0.7798,\,0.1250). $$

These values sum to one, as they should, but they differ from the trial masses $(0.2,0.5,0.3)$. The trial vector is therefore not self-consistent. This calculation illustrates the equation’s meaning; it does not by itself identify a solution.

Solving a small example with positive masses

Consider five hypothetical subjects and three ordered support intervals. Their compatibility sets are $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, and $\{2,3\}$. The first three records restrict the event time to one support interval each; the last two are compatible with adjacent pairs. Assume the support covers all possible event times represented in these records and the interval-observation mechanism is noninformative.

Worked Example: Solving the Positive-Mass Equations

The likelihood is $$ L(\mathbf p)=p_1p_2p_3(p_1+p_2)(p_2+p_3), \qquad p_1+p_2+p_3=1. $$

By the symmetry between intervals 1 and 3, seek a solution with $p_1=p_3=a$, and write $p_2=b=1-2a$. For interval 1, the singleton record contributes allocation 1, the $\{1,2\}$ record contributes $a/(a+b)$, and the other three records contribute zero. The equation for interval 1 is therefore $a=\frac15[1+a/(a+b)]$. Since $a+b=1-a$, this becomes $$ a=\frac{1}{5}\left(1+\frac{a}{1-a}\right) =\frac{1}{5(1-a)}. $$

Multiplying by $5(1-a)$ gives $5a(1-a)=1$, or $a^2-a+0.2=0$. The roots are $(1\pm\sqrt{0.2})/2$. The larger root exceeds $0.5$, which would make $b=1-2a$ negative. Thus the feasible root is $$ a=\frac{1-\sqrt{0.2}}{2}=0.2764, \qquad b=\sqrt{0.2}=0.4472. $$

To verify the middle equation, each of the two pair records allocates $b/(a+b)$ to interval 2, and its singleton record allocates one there. Hence its right-hand side is $$ \frac15\left(1+2\frac{b}{a+b}\right) =\frac15\left(1+2\frac{0.4472}{0.7236}\right) =0.4472=b. $$

The interval-3 equation matches the interval-1 equation by symmetry. The fitted masses are therefore approximately $p_1=0.2764$, $p_2=0.4472$, and $p_3=0.2764$. They are nonnegative, sum to one, and satisfy all three self-consistency equations.

Boundary solutions and what zero mass means

A support interval can be compatible with some observations and still receive zero estimated mass. Such a result is not automatically a computational error. On a boundary, the self-consistency equations for zero-mass intervals hold trivially, because their allocation weights are zero. It is necessary to assess the likelihood or the boundary score conditions as well.

Worked Example: Checking a Zero-Mass Interval

For four hypothetical subjects, let the compatibility sets be $\{1,2\}$, $\{1\}$, $\{2,3\}$, and $\{3\}$. The likelihood is $$ L(\mathbf p)=(p_1+p_2)p_1(p_2+p_3)p_3. $$

Set $x=p_1$ and $z=p_3$, so $p_2=1-x-z$ and $x,z\geq0$ with $x+z\leq1$. Then $p_1+p_2=1-z$ and $p_2+p_3=1-x$, giving $$ L=x(1-x)z(1-z). $$

For any $u\in[0,1]$, $u(1-u)\leq1/4$, since $(u-\frac12)^2\geq0$. It follows that $L\leq1/16$. Equality occurs at $x=z=1/2$, which is feasible and gives $p_2=0$. Thus $(p_1,p_2,p_3)=(0.5,0,0.5)$ is a global maximum, even though interval 2 appears in two compatibility sets.

At these masses, the first pair record allocates entirely to interval 1, and the second pair record entirely to interval 3. The singleton records allocate to their only compatible intervals. The average allocations are $(2/4,0,2/4)=(0.5,0,0.5)$, confirming self-consistency. The likelihood calculation also verifies that this boundary solution is a maximum, not merely a fixed point.

Checks, alternatives, and limitations

When checking a proposed solution, first verify that every subject has positive compatible mass $q_i$; otherwise its likelihood contribution is zero and its allocation weights are undefined. Then check that the masses are nonnegative and sum to one, and that the allocation weights for each subject sum to one. Finally, calculate the average allocations and compare them with the proposed masses. For positive-mass intervals, also check the score condition; for zero-mass intervals, check the boundary condition that their likelihood derivative does not exceed the common derivative for positive-mass intervals.

Self-consistency expresses the fitted masses through expected allocations. Directly maximizing the Turnbull likelihood instead works with the same likelihood but optimizes it over the probability simplex. A parametric interval-censored model takes a different route by restricting the event-time distribution to a specified family. These approaches answer different modelling needs: the nonparametric Turnbull representation avoids choosing a parametric event-time shape, while a parametric model imposes additional structure. Self-consistency does not remove the need to define the support correctly or to assess what the observed intervals identify.

Approach	What it uses	Main caution
Self-consistency equations	Average conditional allocations to support intervals	Fixed-point equations alone do not establish uniqueness or exclude an unsuitable boundary point
Direct likelihood maximization	The observed-data likelihood over nonnegative masses summing to one	Results can depend on support construction, and flat likelihood regions can remain
Parametric interval-censored model	A specified distributional family and its interval probabilities	Inference depends on whether the chosen family is adequate

A common mistake is to interpret $w_{ij}$ as an observed classification of subject $i$, or to count each compatible interval as if the subject contributed one full observation to it. Neither is correct: the record identifies a set of compatible regions, and the fitted distribution divides one unit of allocation across that set using $p_j/q_i$. Another pitfall is to treat any solution of the equations as necessarily unique. Compatibility patterns can leave masses weakly determined or allow multiple maximizers; a zero-mass interval can also be a valid boundary result. These equations estimate masses on the chosen support, not a fully resolved event-time distribution within each support interval.

Key takeaway: A Turnbull mass is self-consistent when it equals the average conditional allocation to its support interval. Verify the equations together with normalization, likelihood, and boundary conditions; the equations alone do not guarantee a unique solution.

Check Your Understanding

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

For one subject compatible with intervals 1 and 2, what are the allocation weights if $p_1=0.3$ and $p_2=0.7$?
Why do the allocation weights for each subject sum to one?
Can an interval have positive compatibility with observed records but estimated mass zero? What additional check is useful?
For a positive-mass interval at an interior maximum, what value does its log-likelihood derivative have under the mass-sum constraint?

Solutions. (1) The compatible mass is $q=0.3+0.7=1$, so the weights are $0.3/1=0.3$ and $0.7/1=0.7$. (2) The weights sum to $\sum_{j\in A_i}p_j/q_i=q_i/q_i=1$. (3) Yes; a boundary maximum can assign zero mass even to a compatible interval. Check the likelihood and boundary score condition, rather than relying only on the zero-mass fixed-point equation. (4) The derivative is $n$, where $n$ is the number of subjects, as follows by multiplying the constrained score equations by the masses and summing.