Why Censored Values Cannot Be Treated As Exact

Introduction

An exact loss tells us the amount of a claim. A right-censored record tells us something different: the loss exceeded a specified threshold, but its final amount was not observed. Treating that threshold as though it were the exact loss can therefore change counts at a threshold, distort an empirical distribution, and understate the average loss.

The key discipline is to separate what the record proves from what remains unknown. The calculations here use only the information in the records. They do not assume a particular severity distribution or attempt to adjust for how censoring occurred.

Definition: For loss $X_i$, let $Y_i$ be the recorded amount and let $\delta_i$ indicate whether the record is exact. If $\delta_i=1$, then $Y_i=X_i$. If $\delta_i=0$, the record is right-censored at threshold $C_i=Y_i$, meaning $X_i>C_i$. Thus a censored value is a lower bound on the loss, not an exact loss.

This tutorial uses the strict right-censoring convention $X_i>C_i$. An exact loss equal to a query threshold $t$ is at or below $t$; a loss known to exceed $t$ is an exceedance of $t$. Keeping strict and inclusive comparisons straight is essential.

What the Records Establish at a Threshold

For a sample of $n$ underlying losses, the empirical distribution function and strict empirical tail proportion are

F_n(t)=\frac{1}{n}\sum_{i=1}^{n} I(X_i\le t), \qquad \widehat{S}_n(t)=\frac{1}{n}\sum_{i=1}^{n} I(X_i>t),

where $I(A)=1$ when statement $A$ is true and $0$ otherwise. With censored records, some indicators cannot be determined. An exact record supplies its indicator directly. A censored record at $C_i$ has $X_i>C_i$: if $C_i\ge t$, it is definitely an exceedance of $t$; if $C_i<t$, it may or may not exceed $t$.

Key result: Let $N_E(\le t)$ be the number of exact losses at or below $t$, and let $N_C(C_i<t)$ be the number of censored records whose thresholds are strictly below $t$. Then the underlying empirical distribution function must satisfy $$ \frac{N_E(\le t)}{n} \le F_n(t)\le \frac{N_E(\le t)+N_C(C_i<t)}{n}. $$ The lower bound counts only exact losses known to be at or below $t$. The upper bound allows every censored loss with $C_i<t$ to be at or below $t$. Censored thresholds equal to or greater than $t$ cannot contribute to $F_n(t)$, because their losses are strictly greater than $t$.

The corresponding strict-tail bounds follow by counting definite exceedances. If $N_E(>t)$ is the number of exact losses strictly greater than $t$, then

\frac{N_E(>t)+N_C(C_i\ge t)}{n} \le \widehat{S}_n(t)\le \frac{N_E(>t)+N_C(C_i<t)+N_C(C_i\ge t)}{n}.

Equivalently, the upper bound is one minus the lower bound for $F_n(t)$. The lower tail bound counts exact exceedances and all censored records with $C_i\ge t$; those records are known to satisfy $X_i>t$. Each censored record with $C_i<t$ is unresolved at $t$, so it may add to the tail, or it may not.

Why Substituting a Threshold Can Misclassify a Loss

A tempting shortcut is to calculate as if every recorded value $Y_i$ were an exact loss. The resulting empirical distribution is

\widetilde{F}_n(t)=\frac{1}{n}\sum_{i=1}^{n} I(Y_i\le t).

For an exact record, $Y_i=X_i$, so its contribution is correct. For a censored record, the shortcut substitutes $C_i$ for $X_i$. When $C_i<t$, it assigns the loss to the at-or-below-$t$ group even though the actual loss could lie on either side of $t$. When $C_i=t$, it assigns the record to the at-or-below group, although $X_i>C_i=t$ proves that the loss is strictly above $t$. When $C_i>t$, both the actual loss and substituted value are above $t$, so this particular comparison is classified correctly.

The same issue appears in the strict tail. Substitution counts a censored record as an exceedance only when $C_i>t$. But a record with $C_i=t$ is also known to be an exceedance, because $X_i>t$. Records censored below $t$ are unresolved: substitution excludes them, although their actual losses could exceed $t$.

Separate exact from censored records.
Use exact losses as observed. For a censored record, retain the fact $X_i>C_i$ rather than treating $C_i$ as an observed loss.

Compare each record with the query threshold.
An exact loss is classified directly. A censored record with $C_i\ge t$ is definitely above $t$; one with $C_i<t$ is unresolved at $t$.

Report what the data identify.
Give a known count or proportion when all indicators are determined; otherwise give bounds or state that the exact empirical quantity cannot be calculated from these records.

Worked Examples

Worked Example: Exceedance Counts at a Censoring Threshold

A hypothetical portfolio has six claims. Exact losses are $40$, $80$, $130$, and $210$ dollars. Two other claims are right-censored at $100$ and $180$ dollars. At $t=100$, find the strict exceedance count supported by the records, and compare it with the count obtained by treating both thresholds as exact losses.

Method. Classify each record using whether it proves $X_i>100$. The four exact losses give indicators $0,0,1,1$: $40$ and $80$ are not strictly above $100$, while $130$ and $210$ are. The record censored at $100$ proves $X_i>100$, including when the threshold equals $t$. The record censored at $180$ also proves $X_i>100$.

The records therefore establish $1+1+1+1=4$ exceedances out of $6$, so the strict empirical tail proportion is $4/6=0.6667$ (to four decimals).

Under threshold substitution, the recorded values are $40,80,130,210,100,180$. Only $130,210,$ and $180$ are strictly greater than $100$, giving $3/6=0.5000$. The $100$-dollar threshold is counted at, not above, $100$, even though its actual loss is known to exceed $100$.

Check. The substituted count is one less than the count known from the records. The error is the censored observation with $C_i=t$. The common mistake is to call a threshold equal to $t$ a strict exceedance after substituting it as an exact value.

Worked Example: Bounding a Distribution Value and a Tail Probability

A hypothetical set of eight losses contains exact values $25,55,90,160,$ and $240$ dollars, plus records censored at $70,150,$ and $230$ dollars. At $t=150$, determine the possible range for $F_n(150)$ and $\widehat{S}_n(150)$. Also calculate the empirical mean that results from threshold substitution and state what the records imply about the actual mean.

Distribution function. Among the exact losses, $25,55,$ and $90$ are at or below $150$, so $N_E(\le150)=3$. Of the censored thresholds, only $70$ is strictly below $150$; that loss could be at or below $150$, or it could exceed $150$. The thresholds $150$ and $230$ both imply losses strictly above $150$. Therefore

\frac{3}{8}\le F_n(150)\le\frac{3+1}{8}, \qquad 0.3750\le F_n(150)\le0.5000.

Strict tail. The exact exceedances are $160$ and $240$. The records censored at $150$ and $230$ are also definite exceedances, so at least $2+2=4$ of the eight losses exceed $150$. The record censored at $70$ might also exceed $150$. Thus

\frac{4}{8}\le\widehat{S}_n(150)\le\frac{5}{8}, \qquad 0.5000\le\widehat{S}_n(150)\le0.6250.

Threshold-substitution mean. The exact losses sum to $25+55+90+160+240=570$. The thresholds sum to $70+150+230=450$. The substituted total is $570+450=1020$, so the substituted mean is $1020/8=\$127.50$. Each of the three actual censored losses is strictly greater than its threshold. Consequently, if the losses are finite, their actual sample mean is strictly greater than $\$127.50$; its exact value is not determined.

Check. Substitution gives $F$-proportion $5/8$: it counts the exact losses $25,55,90$ and the thresholds $70,150$ as at or below $150$. This exceeds the valid upper bound $4/8$, because the threshold equal to $150$ is wrongly assigned to the at-or-below group. The common mistake is to include $C_i=t$ among unresolved thresholds below $t$; under $X_i>C_i$, it is already known to be above $t$.

Worked Example: Equality Affects Both the CDF and the Mean

A hypothetical small-business cover records three exact losses of $\$30,\$75,$ and $\$140$, and two claims censored at $\$75$ and $\$120$. At $t=\$75$, find the known strict-tail proportion, compare the substituted empirical CDF at $75$ with the true value, and calculate the substituted sample mean.

Strict tail. The exact loss of $\$140$ is the only exact exceedance. Both censored records are definite exceedances: the first satisfies $X_i>75$, and the second satisfies $X_i>120>75$. The strict-tail count is $1+2=3$ of $5$, or $3/5=0.6000$.

Distribution function. The exact losses $\$30$ and $\$75$ are at or below $75$; $\$140$ is above. Neither censored loss can be at or below $75$, because their thresholds are $75$ and $120$, respectively, and each actual loss strictly exceeds its threshold. Thus $F_n(75)=2/5=0.4000$. Treating the thresholds as exact instead counts $30,75,$ and the substituted $75$ as at or below $75$, giving $\widetilde F_n(75)=3/5=0.6000$.

Mean. The exact losses sum to $30+75+140=245$; the thresholds sum to $75+120=195$. Their substituted total is $245+195=440$, and the substituted mean is $440/5=\$88.00$. Because both censored losses exceed their thresholds, the actual sample mean is strictly greater than $\$88.00$.

Check. The censored observation at $\$75$ contributes to the true strict tail, but its substituted value does not: equality is not a strict exceedance. The common mistake is to use the same comparison rule for $X_i>t$ and $X_i\ge t$, or to forget that a censored loss at $t$ is known to exceed $t$.

Limits of Empirical Summaries

Censoring does not turn an unknown loss into a known one. If some losses are censored, the ordinary empirical distribution of the underlying losses may not be fully determined by the recorded values. A count at a particular threshold may still be known—for example, if every censored threshold is at or above that point—but another count may only be bounded. The correct answer depends on the comparison being asked and on each record’s threshold.

The same caution applies to averages. Replacing every censored loss with its threshold produces a lower-bound calculation for the sample total, and therefore for the sample mean when all losses are finite. It does not recover the actual mean. A distributional model or an appropriate method for incomplete observations would require additional assumptions and methods; the threshold-substitution calculation alone supplies neither.

Exam takeaway: A right-censored record at $C_i$ establishes $X_i>C_i$, not $X_i=C_i$. At query threshold $t$, a record with $C_i=t$ is definitely a strict exceedance and definitely not at or below $t$. A record with $C_i<t$ is unresolved at $t$, while one with $C_i>t$ is definitely above it. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, distinguish exact observations from censoring thresholds before making the requested comparison.

A sample has exact losses $\$20$ and $\$60$, and right-censored losses at $\$50$ and $\$80$. At $t=\$50$, what is the true empirical proportion at or below $t$, and what proportion results from treating thresholds as exact? (Answer: True proportion $1/4=0.2500$; substituted proportion $2/4=0.5000$.)
Exact losses are $\$40,\$110,\$170$, with censoring thresholds $\$100$ and $\$150$. Give the bounds for $F_n(\$120)$. (Answer: $2/5\le F_n(120)\le3/5$, or $0.4000$ to $0.6000$.)
Exact losses are $\$35,\$95,\$130$, with censoring thresholds $\$95$ and $\$140$. What strict-tail proportion is established at $t=\$95$, and what does threshold substitution give? (Answer: Established tail proportion $3/5=0.6000$; substituted tail proportion $2/5=0.4000$.)
Exact losses are $\$50,\$100,\$200$, with censoring thresholds $\$60$ and $\$160$. Give the bounds for $F_n(\$150)$, then calculate the substituted proportion. (Answer: $2/5\le F_n(150)\le3/5$; substituted proportion $3/5=0.6000$.)
Three exact losses are $\$45,\$85,\$145$, and two losses are censored at $\$100$ and $\$200$. Calculate the mean after threshold substitution and give the implication for the actual mean, assuming finite losses. (Answer: Substituted mean $\$575/5=\$115.00$; actual mean is strictly greater than $\$115.00$.)

What You'll Learn