Why Censored-Data Likelihoods Need Careful Bookkeeping
A censored loss record does not identify a single loss value. It identifies a set of values that could have produced the record. For example, a right-censored observation at threshold \(c\) may tell us only that \(X>c\). Its likelihood contribution is therefore the probability of that whole event, not the probability of a particular value above \(c\).
For a nonparametric likelihood, we estimate the probabilities of the possible loss values or ranges directly, rather than assuming a named parametric distribution. The key step is to account for every probability-mass interval compatible with each record. A value can be compatible with several records at once, so intervals must not be left out or counted as mutually exclusive when they overlap.
This tutorial uses the usual discrete-distribution formulation of the nonparametric likelihood, in which probability masses may be placed at observed exact values and at points in censoring intervals. If a model instead assumes a continuous distribution and treats exact observations through densities, the likelihood formulation must use those densities; point masses and densities are not interchangeable.
From Records to Probability-Mass Cells
State the censoring rule before writing its likelihood contribution. Here, a left-censored record at \(c\) means \(X\leq c\), a right-censored record means \(X>c\), and an interval-censored record \((l,r]\) means \(l<X\leq r\). These endpoint conventions determine which cells enter each probability.
- An exact observation \(X=x\) contributes \(P(X=x)\).
- A left-censored observation \(X\leq c\) contributes \(P(X\leq c)\).
- A right-censored observation \(X>c\) contributes \(P(X>c)\).
- An interval-censored observation \(l<X\leq r\) contributes \(P(l<X\leq r)\).
To form a likelihood, split the loss range at the censoring thresholds and exact observations. Assign a probability to each resulting cell. The event probability for a record is the sum of the masses in all cells that satisfy its observation rule. Cells may enter more than one event probability; that is how the likelihood reflects overlap.
where \(C_i\) is the set of losses consistent with record \(i\), including a singleton set for an exact observation under the discrete formulation. The mass assigned to all cells must be nonnegative and sum to one.
Why the Likelihood Uses Event Probabilities
Under independent sampling, the probability of observing a specified collection of records is the product of the probabilities of the corresponding events. A censored record tells us that the loss lies somewhere in a set \(C_i\), not which value in \(C_i\) occurred. Its probability is therefore the sum of the probabilities of the compatible cells.
The cells must form a partition so that each possible loss is allocated exactly once in the total distribution. The event for each record can then be assembled by adding the masses of its compatible cells. Maximizing the resulting product chooses cell masses that make the observed records as probable as possible, subject to the total-mass constraint.
Write each exact or censored record as a precise set of possible loss values, respecting the endpoint convention.
Use all exact values and censoring boundaries to define cells that do not overlap and together cover the modeled loss range.
For each record, sum the masses of every cell contained in its compatible set. A cell may appear in several record probabilities.
Maximize the product over nonnegative cell masses summing to one, then state what the fitted distribution does and does not identify.
Worked Likelihoods and Distribution Estimates
Worked Example: Two Right-Censoring Thresholds and Exact Losses
A hypothetical sample of four independent losses contains exact observations \(X=4\) and \(X=7\), plus right-censored records \(X>5\) and \(X>6\). Find a nonparametric maximum likelihood estimate. In particular, include all compatible probability-mass cells.
Method. Partition the possible losses at the exact values and censoring thresholds. Define \(a=P(X=4)\), \(b=P(X=7)\), \(u=P(5<X\leq6)\), \(v=P(6<X<7)\), and \(w=P(X>7)\). Let \(z\) be all remaining probability, including values compatible with neither right-censored event. Thus \(a+b+u+v+w+z=1\).
The exact observations contribute \(a\) and \(b\). The event \(X>5\) includes the cells with masses \(u,v,b,w\); the event \(X>6\) includes \(v,b,w\). Therefore the likelihood is
Write \(q=u+v+b+w=P(X>5)\) and \(r=v+b+w=P(X>6)\). Since \(z\geq0\), \(q\leq1-a\); also \(r\leq q\), and \(b\leq r\). Hence
The second inequality uses \(b\leq1-a\). To maximize the final expression, differentiate:
The maximum on \(0\leq a\leq1\) occurs at \(a=1/4\). Equality in the likelihood bounds is attained by taking \(b=1-a=3/4\) and \(u=v=w=z=0\). The maximum likelihood is \((1/4)(3/4)^3=27/256\).
Thus one NPMLE places probability \(1/4\) at 4 and \(3/4\) at 7. Its estimated distribution function is 0 below 4, \(1/4\) from 4 up to but not including 7, and 1 at and above 7. Substitution checks that both right-censored records are satisfied. The common error is to omit the cells \((5,6]\) and \((6,7)\): they are compatible with one or both censoring records, even though the maximizing estimate happens to assign them zero mass.
Worked Example: An Exact Loss, a Right-Censored Loss, and Two Interval Records
A hypothetical sample has one exact loss \(X=2\), one right-censored record \(X>2\), and two interval-censored records \(2<X\leq4\). Find the NPMLE and its maximum likelihood.
Method. Let \(a=P(X=2)\), and let \(q=P(2<X\leq4)\). The right-censored event has probability \(1-a\). Since the interval is a subset of \(X>2\), \(q\leq1-a\). Any remaining mass outside the interval and above 2 contributes to the right-censored record but not to either interval record. The likelihood is
For fixed \(a\), the likelihood increases as \(q\) increases, so it is maximized at \(q=1-a\). This puts all mass other than the exact mass inside the interval. The likelihood becomes \(a(1-a)^3\), whose derivative is \((1-a)^2(1-4a)\). It is maximized at \(a=1/4\), giving \(q=3/4\) and maximum likelihood \(27/256\).
An estimated distribution places mass \(1/4\) at 2 and \(3/4\) at any selected point in \((2,4]\), such as 3. The likelihood check is \((1/4)(3/4)(3/4)^2=27/256\). The data identify the interval mass, not the precise location of that mass within \((2,4]\); choosing 3 is a convenient representation, not additional information from the records.
Worked Example: Left-Censoring and an Interval That Starts Above the Exact Loss
A hypothetical sample contains one exact loss \(X=5\), two left-censored records \(X\leq5\), and one interval-censored record \(5<X\leq8\). Find an NPMLE.
Method. Let \(a=P(X=5)\), \(l=P(X\leq5,\ X\ne5)\), and \(q=P(5<X\leq8)\); let \(r=P(X>8)\). These disjoint masses sum to one. The exact record contributes \(a\), each left-censored record contributes \(a+l\), and the interval record contributes \(q\). Thus
Set \(s=a+l\), the probability of the left-censored event. For fixed \(s\), \(a\leq s\) and \(q\leq1-s\), so \(L\leq s^3(1-s)\). Equality is possible by setting \(a=s\), \(l=0\), \(q=1-s\), and \(r=0\). Differentiate the bound:
The maximum occurs at \(s=3/4\). Therefore one NPMLE puts mass \(3/4\) at 5 and \(1/4\) at any point in \((5,8]\), for example 6. Its likelihood is \((3/4)^3(1/4)=27/256\). The left-censored records include the exact mass at 5; treating them as \(l\) alone would incorrectly exclude a compatible value.
Interpreting the Estimate and Avoiding Traps
A nonparametric likelihood estimate need not determine a unique loss value for every censored observation. It determines the masses that maximize the likelihood, while the allocation of mass within a cell can remain unrestricted. In the interval example, moving the estimated mass from 3 to another point in \((2,4]\) leaves every event probability and the likelihood unchanged.
When the likelihood is written in terms of cells, each cell appears once in the total-mass constraint but may appear in multiple record probabilities. This distinction is essential: overlapping observation events do not make their underlying probability masses overlap in the partition.
- Omitting a compatible cell: A loss between two thresholds can contribute to one or more censored events even if it is not an exact observed value.
- Using only one threshold: For \(X>6\), include only values strictly above 6 under the convention used here. The endpoint 6 belongs to \(X\leq6\), not \(X>6\).
- Confusing a fitted representative with identified information: Choosing a point inside an interval to display an estimate does not mean the data identified that loss value.
- Forgetting the sum constraint: The cell masses must be nonnegative and sum to one, including cells that do not contribute to any observed event.
This cell-based formulation is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus because it tests the central judgment in nonparametric likelihood with censored losses: translating incomplete observations into correctly overlapping probability constraints.
Check Your Understanding
For each problem, write the event probabilities before optimizing. Assume independent losses and use the endpoint conventions stated above.
- A hypothetical sample has one exact observation \(X=2\) and two right-censored records \(X>2\). Find the NPMLE masses and maximum likelihood. (Answer: mass \(1/3\) at 2 and \(2/3\) above 2; maximum likelihood \(4/27\).)
- A sample has one left-censored record \(X\leq4\) and two interval records \(4<X\leq7\). Find the maximizing event masses. (Answer: probability \(1/3\) on \(X\leq4\), probability \(2/3\) on \(4<X\leq7\); maximum likelihood \(4/27\).)
- A sample has one exact observation \(X=6\), one interval record \(6<X\leq9\), and one right-censored record \(X>6\). Find the NPMLE cell masses. (Answer: mass \(1/3\) at 6 and \(2/3\) in \((6,9]\); maximum likelihood \(4/27\).)
- A sample has one exact observation \(X=2\), one right-censored record \(X>3\), and one right-censored record \(X>5\). Include the cell \((3,5]\) when writing the likelihood, then find the NPMLE. (Answer: mass \(1/3\) at 2 and \(2/3\) above 5, with zero mass in \((3,5]\); maximum likelihood \(4/27\).)