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Exam ASTAM · Empirical Loss and Survival · Tutorial 220 of 979

Interval-Censored Loss Observations

Translate interval-censored losses into defensible bounds on threshold exceedances without treating an unknown loss as exact.

Beginner 10 min read

What You'll Learn

  • Define interval-censored loss observations and state their endpoint convention
  • Classify threshold exceedances as certain, impossible, or unresolved
  • Bound the observed sample proportion exceeding a threshold
  • Handle threshold equality correctly for intervals of the form (L, U]
  • Explain why midpoint substitution does not follow from an interval record
  • Distinguish bounds on a recorded sample from a population probability claim

What an Interval-Censored Loss Tells You

A loss record does not always give the exact value of \(X\). Sometimes a claim file, audit, or data system only establishes that the loss lies between two amounts. Such a record is informative, but it does not justify replacing the unknown loss by an exact number. For threshold questions, the useful information is often whether the entire recorded interval lies above or below the threshold.

For this tutorial, an interval record is written \((L,U]\), where \(L\) is the lower bound and \(U\) is the upper bound. It means \(L<X\leq U\). This endpoint convention matters when the threshold equals a bound. If another data system uses different endpoint rules, those rules must be used instead.

Definition: An observation is interval-censored when the exact loss \(X\) is unknown but the record establishes that \(L<X\leq U\). For a threshold \(t\), the record may establish that \(X>t\), establish that \(X\leq t\), or leave both possibilities open.

For a loss distribution, \(F_X(t)=P(X\leq t)\) is the distribution function and \(S_X(t)=P(X>t)=1-F_X(t)\) is the survival function. These quantities refer to a random loss under a specified population model. A single interval record does not give an exact value of either quantity. It only restricts what the recorded loss could be; several records can, in turn, restrict the observed sample’s exceedance proportion.

Classifying a Threshold Claim

Suppose a record says \(L<X\leq U\), and the question is whether the loss exceeds \(t\). If \(L\geq t\), then \(X>L\geq t\), so exceedance is certain. If \(U\leq t\), then \(X\leq U\leq t\), so exceedance is impossible. If \(L<t<U\), the record allows values on both sides of \(t\), so the exceedance status is unresolved.

$$ \begin{array}{ll} L\geq t &: X>t\text{ is certain},\\ U\leq t &: X>t\text{ is impossible},\\ L<t<U &: X>t\text{ is unresolved}. \end{array} $$

The equalities in these rules follow from the \((L,U]\) convention. At \(t=L\), every possible loss is strictly greater than \(L\), so exceedance is certain. At \(t=U\), every possible loss is at most \(U\), so exceedance is impossible. If the record instead included its lower endpoint, \(L\leq X\leq U\), the first equality case would no longer guarantee \(X> L\).

Worked Example: One Record at Several Thresholds

A hypothetical claim audit establishes that a particular loss satisfies \(12{,}000<X\leq19{,}000\). For each threshold \(t=10{,}000,\ 12{,}000,\ 15{,}000,\ 19{,}000,\) and \(20{,}000\), determine whether the record proves \(X>t\), rules it out, or leaves it unresolved.

Method. Compare \(L=12{,}000\) and \(U=19{,}000\) with each threshold, applying the endpoint rules above.

  • At \(t=10{,}000\), \(L\geq t\). Thus \(X>10{,}000\) is certain.
  • At \(t=12{,}000\), \(L=t\). Since the interval excludes its lower endpoint, \(X>12{,}000\) is still certain.
  • At \(t=15{,}000\), \(L<t<U\). Values such as \(13{,}000\) and \(17{,}000\) are both consistent with the record, so the answer is unresolved.
  • At \(t=19{,}000\), \(U=t\). The loss is at most \(19{,}000\), so exceedance is impossible.
  • At \(t=20{,}000\), \(U<t\). Exceedance is also impossible.

Check. The record permits losses below, at, and near the upper bound, but none above \(19{,}000\). It therefore supports certainty at and below the lower bound, and rules out exceedance at and above the upper bound. The common mistake avoided here is treating equality at the lower and upper bounds identically: \((L,U]\) excludes \(L\) but includes \(U\).

Bounds for a Set of Interval Records

For \(n\) records \((L_i,U_i]\), define the observed sample exceedance proportion at threshold \(t\) as the number of the \(n\) actual losses exceeding \(t\), divided by \(n\). When some statuses are unresolved, this proportion is not known exactly. A lower bound counts only exceedances that are certain; an upper bound counts every record for which exceedance remains possible.

$$ \frac{1}{n}\sum_{i=1}^{n}\mathbf{1}(L_i\geq t) \ \leq\ \frac{1}{n}\sum_{i=1}^{n}\mathbf{1}(X_i>t) \ \leq\ \frac{1}{n}\sum_{i=1}^{n}\mathbf{1}(U_i>t). $$

Here \(\mathbf{1}(A)\) equals 1 when statement \(A\) is true and 0 otherwise. The left side counts records whose losses must exceed \(t\). The right side counts records whose losses could exceed \(t\); a record with \(U_i=t\) is not counted because its loss cannot exceed the inclusive upper bound. Every actual exceedance is included in the right count, and every certain exceedance is included in the actual count. These facts give the two inequalities.

The bounds are sharp if each loss can take any value consistent with its interval: unresolved records can be assigned values on either side of the threshold as needed. This does not mean the bounds estimate a population probability without further assumptions. They bound the proportion in the recorded sample. A claim about \(S_X(t)\) requires a suitable relationship between that sample and the population, as well as appropriate treatment of how the intervals were obtained.

Worked Example: Bounding a Sample Exceedance Proportion

A hypothetical insurer has five audited losses, recorded in thousands of dollars as \((0,4]\), \((3,8]\), \((7,11]\), \((10,14]\), and \((14,18]\). Find the sharp bounds on the proportion of these five losses exceeding \(9\) thousand dollars.

Method. Use \(t=9\). A record is a certain exceedance when \(L_i\geq9\); it is a possible exceedance when \(U_i>9\).

For the lower bound, the lower endpoints are \(0,3,7,10,14\). Exactly two are at least 9: \(10\) and \(14\). Thus at least \(2\) of the \(5\) losses exceed 9, giving \(2/5=0.4\).

For the upper bound, the upper endpoints are \(4,8,11,14,18\). Exactly three are greater than 9: \(11,14,18\). Thus at most \(3\) of the \(5\) losses exceed 9, giving \(3/5=0.6\). The observed proportion is therefore between \(0.4\) and \(0.6\), inclusive.

Check. The first two records cannot exceed 9; the fourth and fifth must exceed it; the third, \((7,11]\), could lie on either side. Assigning its loss to 8 gives two exceedances, while assigning it to 10 gives three. Both assignments obey the interval, so the bounds are attainable. The common mistake avoided is counting every interval whose upper endpoint exceeds 9 as a definite exceedance; that produces the upper bound, not the actual count.

Why Midpoints and Population Claims Need Caution

An interval record does not say that losses are uniformly distributed within the interval, concentrated near its midpoint, or equally likely to be anywhere in it. Substituting \((L+U)/2\) creates a fabricated exact observation. It may be a useful modeling choice in a separately justified analysis, but it is not a conclusion supported by the interval alone. In particular, if a midpoint lies above \(t\), that does not prove the actual loss exceeded \(t\).

Likewise, the sample bounds are not automatically bounds on a population tail probability \(S_X(t)\). They describe the unknown exceedance proportion among the recorded losses. To interpret that proportion as information about a broader population requires a sampling design or model that connects the observed records to that population. The method here makes no assumption about a distribution within an interval and does not produce an exact population estimate.

Key takeaway: For records \((L_i,U_i]\), count \(L_i\geq t\) for certain exceedances and \(U_i>t\) for possible exceedances. Divide those counts by the number of records to obtain bounds on the observed sample proportion, not automatically on a population probability.

Common Exam Traps

  • Ignoring endpoint convention. With \((L,U]\), equality \(L=t\) guarantees \(X>t\), while equality \(U=t\) rules it out. Check which endpoints the record includes.
  • Confusing possible with certain. An upper endpoint above \(t\) makes exceedance possible, not certain. A lower endpoint at least \(t\) makes it certain.
  • Replacing an interval by its midpoint. The midpoint is not observed and need not be on the same side of the threshold as the true loss.
  • Calling a sample bound a population estimate. Interval comparisons alone bound the sample’s unknown exceedance fraction; a population interpretation needs additional assumptions.

Worked Example: Two Thresholds and a Boundary Case

A hypothetical review records six losses, in thousands of dollars, as \((0,10]\), \((8,20]\), \((18,30]\), \((24,40]\), \((35,50]\), and \((45,60]\). Find the bounds on the observed proportion exceeding 25 and the observed proportion exceeding 35.

Method. At each threshold, count certain exceedances using \(L_i\geq t\), and possible exceedances using \(U_i>t\).

At \(t=25\), the lower endpoints at least 25 are \(35\) and \(45\), so there are \(2\) certain exceedances. The upper endpoints greater than 25 are \(30,40,50,60\), so there are \(4\) possible exceedances. The proportion is bounded by \(2/6=1/3\) and \(4/6=2/3\).

At \(t=35\), the lower endpoints at least 35 are \(35\) and \(45\), again giving \(2\) certain exceedances. The upper endpoints greater than 35 are \(40,50,60\), giving \(3\) possible exceedances. The proportion is bounded by \(2/6=1/3\) and \(3/6=1/2\).

Check. The record \((24,40]\) is unresolved at 35, while \((35,50]\) is a certain exceedance because its lower endpoint is excluded. The results are therefore \([1/3,2/3]\) at 25 and \([1/3,1/2]\) at 35. The common mistake avoided is treating the interval beginning at 35 as unresolved at threshold 35; under the stated convention its loss must be strictly greater than 35.

Check Your Understanding

For each problem, use the stated \((L,U]\) convention and distinguish certain, impossible, and unresolved exceedances.

  1. A loss is recorded as \((18,27]\). Classify the claims \(X>18\) and \(X>27\). (Answer: \(X>18\) is certain; \(X>27\) is impossible.)
  2. Four records are \((2,5]\), \((4,9]\), \((9,12]\), and \((11,15]\). Find the bounds on the proportion exceeding 9. (Answer: \(2/4\) to \(2/4\), so the proportion is exactly \(0.5\).)
  3. Five records are \((0,6]\), \((5,10]\), \((8,13]\), \((12,16]\), and \((15,22]\). Find the bounds on the proportion exceeding 11. (Answer: \(2/5\) to \(3/5\), or \(0.4\) to \(0.6\).)
  4. Three records are \((10,20]\), \((18,25]\), and \((25,30]\). Find the bounds on the proportion exceeding 25. (Answer: \(1/3\) to \(1/3\); only \((25,30]\) is a certain exceedance.)
  5. An analyst replaces each record \((10,30]\) by the value 20 and reports that the loss certainly exceeds 18. Is that conclusion justified by the interval alone? (Answer: No. The interval permits values at or below 18 as well as values above 18, so exceedance is unresolved.)

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