Combining Truncation and Censoring by Record

Why Each Record Needs Two Sets

When a severity is both truncated and censored, two distinct questions determine its likelihood contribution. Could this loss have entered the data? describes the truncation or inclusion rule. What does the recorded information say about the loss? describes the censoring. For a particular record, the likelihood must honor both conditions at once.

For example, a hypothetical study might include only losses above a reporting threshold, then record a loss as “greater than” a separate censoring threshold. The event behind that record is not simply “greater than the censoring threshold”: it is that event restricted to losses that could have been included. The general method is to represent these conditions as sets, intersect them, and condition on inclusion.

Definition: Let \(X\) be a loss with distribution indexed by parameter \(\theta\). For record \(i\), let \(A_i\) be its inclusion region, the set of loss values that could be observed under the sampling or reporting rule. Let \(B_i\) be its observed set, the set of loss values consistent with the recorded information. The record conveys that \(X_i\in A_i\cap B_i\), conditional on \(X_i\in A_i\). For an interval-censored record, \(B_i\) is an interval; for a right-censored record at \(c_i\), it is typically \((c_i,\infty)\).

This treatment assumes that the inclusion region and censoring information are correctly specified, and that the censoring mechanism is noninformative for \(X\), or otherwise has a known contribution that is included in the likelihood. Conditional on their respective inclusion rules, records are treated as independent when forming a product likelihood. If censoring depends on the unobserved loss in a way not captured by the model, the severity likelihood below is not sufficient.

The Record-Level Likelihood

For a censored record, the information is an event, so its conditional contribution is a probability. For an exact record \(X_i=x_i\) from a continuous distribution, the contribution is a conditional density. In either case, the numerator reflects the information that is both observed and eligible for inclusion.

L_i(\theta)= \frac{P_\theta(X_i\in A_i\cap B_i)} {P_\theta(X_i\in A_i)} \quad\text{for event-valued observations,} \qquad L_i(\theta)= \frac{f_\theta(x_i)\mathbf{1}_{\{x_i\in A_i\}}} {P_\theta(X_i\in A_i)} \quad\text{for an exact continuous observation.}

Here \(f_\theta\) is the density of \(X\), and \(\mathbf{1}_{\{x_i\in A_i\}}\) is 1 if the recorded exact value is in the inclusion region and 0 otherwise. An exact value outside the stated inclusion region is incompatible with the record design; it does not have a positive likelihood under that design. If the distribution is discrete, an exact record uses a probability mass in place of a density.

The justification is conditional probability. Given that record \(i\) was included, the probability of also observing information consistent with \(B_i\) is \(P_\theta(X_i\in A_i\cap B_i)/P_\theta(X_i\in A_i)\). For an exact continuous value, consider a small interval of width \(h\) around \(x_i\). If \(x_i\) is eligible, the probability of falling in that interval and being included is approximately \(f_\theta(x_i)h\); dividing by the inclusion probability and then by \(h\), as \(h\) tends to zero, gives the conditional density \(f_\theta(x_i)/P_\theta(X_i\in A_i)\).

After determining each contribution, multiply them across independent records. Inclusion probabilities generally belong in the likelihood; they can depend on \(\theta\), so they cannot be discarded merely because they are normalizing constants in a conditional distribution.

Write the inclusion region.
Translate the record’s truncation or selection rule into a set \(A_i\).

Write the observed set.
Translate the exact, interval-censored, or one-sided-censored information into a set \(B_i\), if the record is not exact.

Intersect before calculating.
Use \(A_i\cap B_i\) in the numerator, and use \(A_i\) alone in the denominator.

Form the likelihood.
Use a conditional probability for a censored record or a conditional density for an exact continuous record, then multiply record contributions.

Worked Examples

Worked Example: Left Truncation with Different Censoring Types

A hypothetical loss study has independent records from an exponential severity model with rate \(\theta\), so \(f_\theta(x)=\theta e^{-\theta x}\) and \(P_\theta(X>x)=e^{-\theta x}\) for \(x\geq0\). A record is included only when its loss is above its stated lower threshold. The study records: (1) an exact loss of 5, included when \(X>2\); (2) a right-censored loss reported as \(X>4\), included when \(X>1\); and (3) a left-censored loss reported as \(X\leq2\), included when \(0<X<6\). Find the likelihood at \(\theta=0.5\).

Method. For each record, form the observed set and intersect it with the inclusion region. Since the records are independent conditional on inclusion, multiply their contributions.

Record 1, exact. Its inclusion region is \(A_1=(2,\infty)\), and the observed value is \(x_1=5\). The conditional density is \(f_{0.5}(5)/P_{0.5}(X>2)\). The numerator is \(0.5e^{-0.5(5)}=0.5e^{-2.5}\), and the denominator is \(e^{-0.5(2)}=e^{-1}\). Thus the contribution is \(0.5e^{-1.5}=0.1116\) to four decimal places.

Record 2, right-censored. Here \(A_2=(1,\infty)\), \(B_2=(4,\infty)\), and \(A_2\cap B_2=(4,\infty)\). Its contribution is \(e^{-0.5(4)}/e^{-0.5(1)}=e^{-1.5}=0.2231\).

Record 3, left-censored. Here \(A_3=(0,6)\), \(B_3=(-\infty,2]\), and their intersection is \((0,2]\). The numerator is \(P(0<X\leq2)=1-e^{-0.5(2)}=1-e^{-1}=0.6321\). The denominator is \(P(0<X<6)=1-e^{-0.5(6)}=1-e^{-3}=0.9502\). The contribution is \(0.6321/0.9502=0.6652\).

The full likelihood at the specified rate is the product \(0.1115651\times0.2231302\times0.6652410=0.0166\), rounded to four decimal places. This is an evaluation of the likelihood, not a claim that \(\theta=0.5\) maximizes it.

Common mistake avoided: Using \(P(X\leq2)\) as the third numerator ignores that the study only includes losses above zero. The numerator must be \(P(0<X\leq2)\), the probability of the intersection.

Worked Example: Interval Censoring Within a Truncated Sample

A hypothetical portfolio has exponential losses with rate \(\theta=0.4\). A first record is included only if \(X>1.5\), and the observed information places its loss in \([3,5)\). A second record is included only if \(X>2\) and is reported as right-censored at 4.5, meaning \(X>4.5\). Find the likelihood of these two records conditional on their inclusion.

First record. Its inclusion region is \(A_1=(1.5,\infty)\), and its observed set is \(B_1=[3,5)\). Their intersection is \([3,5)\). For an exponential loss, \(P(3\leq X<5)=e^{-0.4(3)}-e^{-0.4(5)}=e^{-1.2}-e^{-2}\). The inclusion probability is \(P(X>1.5)=e^{-0.6}\). Therefore the conditional contribution is \((e^{-1.2}-e^{-2})/e^{-0.6}=e^{-0.6}-e^{-1.4}=0.3022\).

Second record. Its inclusion region is \(A_2=(2,\infty)\), and its observed set is \(B_2=(4.5,\infty)\). The intersection is \((4.5,\infty)\). The contribution is \(e^{-0.4(4.5)}/e^{-0.4(2)}=e^{-1}=0.3679\).

The joint conditional likelihood is \(0.3022146\times0.3678794=0.1112\), rounded to four decimal places.

Common mistake and correction: For the first record, using only the unconditioned interval probability \(e^{-1.2}-e^{-2}=0.1659\) omits the inclusion condition. Divide that probability by \(e^{-0.6}\), the probability of meeting the truncation rule, to obtain the correct conditional contribution \(0.3022\).

Worked Example: Double-Bounded Inclusion and Three Record Types

A hypothetical study uses exponential losses with rate \(\theta=0.2\). It includes records only when their losses lie within record-specific bounds. The records are: (1) included when \(1<X<6\), then left-censored as \(X\leq3\); (2) included when \(2<X<7\), with exact loss \(x=4.5\); and (3) included when \(1<X<8\), then right-censored as \(X>5\). Find the product likelihood conditional on inclusion.

Record 1, left-censored. The intersection of \((1,6)\) and \((-\infty,3]\) is \((1,3]\). Its numerator is \(P(1<X\leq3)=e^{-0.2}-e^{-0.6}=0.2699\). Its inclusion probability is \(P(1<X<6)=e^{-0.2}-e^{-1.2}=0.5175\). Thus its contribution is \(0.2699/0.5175=0.5215\).

Record 2, exact. The density at 4.5 is \(0.2e^{-0.2(4.5)}=0.2e^{-0.9}=0.0813\). The inclusion probability is \(P(2<X<7)=e^{-0.4}-e^{-1.4}=0.4237\). The conditional density contribution is \(0.0813/0.4237=0.1919\).

Record 3, right-censored. The intersection of \((1,8)\) and \((5,\infty)\) is \((5,8)\), not \((5,\infty)\). Its numerator is \(P(5<X<8)=e^{-1}-e^{-1.6}=0.1660\). The inclusion probability is \(P(1<X<8)=e^{-0.2}-e^{-1.6}=0.6168\). Its contribution is \(0.1660/0.6168=0.2691\).

Multiplying the contributions gives \(0.5215\times0.1919\times0.2691=0.0269\), rounded to four decimal places. The finite upper inclusion bound in the third record must remain in the numerator intersection.

Common mistake avoided: Treating the third record as merely \(X>5\) would use \(e^{-1}\) as its numerator and include losses above 8, even though those losses could not enter the data. The intersection correctly excludes them.

Checks Before Using the Likelihood

The set method is useful because it keeps truncation and censoring logically separate until the numerator is formed. Before simplifying any contribution, verify that the inclusion event is in the denominator, the intersection is in the numerator, and all intervals match the record’s stated bounds. For continuous distributions, changing an endpoint from open to closed does not change a probability; for discrete distributions or distributions with atoms, endpoints can matter.

A likelihood assembled this way is conditional on the stated inclusion rules. If the inclusion process itself is part of the inferential target, or if censoring depends informatively on \(X\), additional model terms may be required. Likewise, when an inclusion probability depends on \(\theta\), it remains part of the likelihood; treating it as constant can change the resulting estimate.

Key takeaway: For each record, first describe the losses eligible for observation and the losses consistent with the reported information. Use their intersection in the numerator and the inclusion probability in the denominator. Exact continuous observations use a density in the numerator; censored observations use a probability.

Check Your Understanding

For each item, assume the stated exponential model and conditional independence. Form the record likelihood contribution or product as requested.

A loss has exponential rate \(\theta=0.4\), is included only if \(X>2\), and is right-censored at 5, so the record says \(X>5\). Find its conditional contribution. (Answer: \(e^{-1.2}=0.3010\).)
For exponential rate \(\theta=0.25\), a loss is included if \(1<X<5\) and is observed in \([2,4)\). Find the conditional contribution. (Answer: \((e^{-0.5}-e^{-1})/(e^{-0.25}-e^{-1.25})=0.4848\).)
For exponential rate \(\theta=0.3\), an exact loss \(x=4\) is observed in a sample that includes only \(X>1\). Find its conditional density contribution. (Answer: \(0.3e^{-0.9}=0.1220\).)
For exponential rate \(\theta=0.5\), a record is included if \(2<X<6\) and is left-censored as \(X\leq4\). Find its conditional contribution. (Answer: \((e^{-1}-e^{-2})/(e^{-1}-e^{-3})=0.7310\).)
For exponential rate \(\theta=0.2\), an exact loss \(x=3\) is included if \(1<X<5\), while an independent record is included if \(X>2\) and right-censored as \(X>4\). Find the product of the two conditional contributions. (Answer: \(\frac{0.2e^{-0.6}}{e^{-0.2}-e^{-1}}\times e^{-0.4}=0.1632\).)

What You'll Learn