Grouped, Censored, and Truncated Loss Observations

Why the Observation Process Matters

A loss data set may not contain each loss as an exact dollar amount. A report may put losses into ranges, record only that a loss exceeded a threshold, or omit losses that fall below a reporting threshold. These are different observation processes. Treating them as though they all produced exact observations can distort a mean, hide uncertainty, or change which population the summary describes.

The key question is not only “What values are in the data?” but also “What information does each record provide, and which losses could not enter the data?” The distinction matters when calculating summaries and when interpreting the population represented by an observed sample.

Definition: Grouped data report how many observations fall in each interval, rather than reporting each exact value. Censored data retain an observation in the sample but give only a bound or range for its value. Truncated data omit observations outside a specified inclusion rule, so the observed sample contains only losses meeting that rule.

Grouped Losses: Intervals, Not Exact Values

Suppose a report gives a count $n_j$ for each interval $[a_j,b_j)$. It establishes that the losses in that group are at least $a_j$ and less than $b_j$; it does not reveal their locations within the interval. A class midpoint $(a_j+b_j)/2$ is often used as a representative value to estimate a mean. This is an approximation, not a recovery of the actual losses.

For nonnegative losses, the data can also give bounds on the sample mean. The lower bound assigns each grouped loss its interval’s lower endpoint, while the upper bound assigns it the upper endpoint. With half-open intervals $[a_j,b_j)$, the upper-endpoint result is a strict bound: no loss in that interval equals $b_j$.

\widehat{\bar{X}}_{\text{mid}} = \frac{\sum_j n_j(a_j+b_j)/2}{\sum_j n_j}, \qquad \frac{\sum_j n_j a_j}{\sum_j n_j} \leq \bar{X} < \frac{\sum_j n_j b_j}{\sum_j n_j}.

The midpoint estimate is most informative when losses are reasonably spread within each interval. If losses cluster near one end of a wide interval, its midpoint can be a poor representative. Narrower intervals generally provide tighter bounds and a more detailed approximation.

Worked Example: Estimating a Mean From Grouped Losses

A hypothetical insurer summarizes five losses in two intervals: two losses are in $[0,40)$ dollars and three are in $[40,100)$ dollars. Calculate the midpoint estimate of the mean and the bounds on the actual sample mean.

Method: midpoint approximation and endpoint bounds. The interval midpoints are $(0+40)/2=20$ dollars and $(40+100)/2=70$ dollars. Weight each midpoint by its count:

\widehat{\bar{X}}_{\text{mid}} = \frac{2(20)+3(70)}{5} = \frac{40+210}{5} = \frac{250}{5} = \$50.00.

For the lower bound, use the lower endpoint of each interval: \[ \bar{X}\geq \frac{2(0)+3(40)}{5}=\frac{120}{5}=\$24. \] For the upper bound, use the excluded upper endpoints: \[ \bar{X}<\frac{2(40)+3(100)}{5}=\frac{380}{5}=\$76. \] Thus the actual mean is at least $\$24$ and less than $\$76$. The midpoint estimate, $\$50.00$, lies within those bounds.

Check and common mistake. The weighted midpoint numerator is $2(20)+3(70)=40+210=250$, not 260. Using 260 would produce the incorrect estimate $\$52.00$. The interval bounds do not depend on midpoint assumptions.

Censored Losses: The Observation Is Present, Its Exact Value Is Not

With right censoring at a threshold $c$, an observation may be recorded as “at least $c$” rather than as its exact amount. It remains one of the observations in the sample. With left censoring, the record may state that a loss is at most $c$. With interval censoring, the record identifies a range, such as $[a,b)$. In each case, the exact value is unknown within the reported limit or interval.

A censored record should not automatically be replaced by its threshold. For a right-censored loss, using $c$ as if it were the exact amount gives a lower-bound contribution, not necessarily the true contribution. If losses are nonnegative, known exact values and lower bounds for censored values can produce a lower bound on the sample mean. A finite upper bound is available only when the records supply finite upper limits.

\text{If }X_k\geq c\text{ for each of }r\text{ right-censored observations, then } \bar{X}\geq \frac{\sum_{\text{exact }i}x_i+rc}{n}.

This expression is a bound, not an estimate of the exact mean. Without upper limits on the censored values, the records alone give no finite upper bound. They also do not determine the exact variance, because the unknown losses affect both the mean and the squared deviations.

Worked Example: A Sample With Right-Censored Losses

A hypothetical data file contains five claims. Three exact losses are $\$18$, $\$32$, and $\$50$; two additional claims are each reported as at least $\$90$, with no upper amounts available. Find what the records establish about the sample mean.

Method: retain every claim and use the censoring threshold as a lower bound. There are $n=5$ claims. The three exact amounts sum to $18+32+50=\$100$. The two censored claims contribute at least $2(90)=\$180$. Therefore the total loss is at least $100+180=\$280$, and:

\bar{X}\geq \frac{\$280}{5}=\$56.

The records establish a sample mean of at least $\$56$, but give no finite upper bound: either censored loss could exceed $\$90$ by an unknown amount. The mean is not known to equal $\$56$.

Check and common mistake. Replacing both censored claims by exactly $\$90$ produces $\$56$, but that calculation assumes values the records do not provide. It is valid as the lower-bound calculation, not as an exact mean. The denominator remains five because the censored claims are in the sample.

Truncated Losses: Some Observations Never Enter the Sample

Truncation differs from censoring because excluded observations do not appear as records. For example, a database that includes only losses of at least $c$ is left-truncated at $c$: losses below the threshold are absent. The observed losses are not a complete list with some values marked unknown; they are a selected subset.

When $X$ is a loss and inclusion requires $X\geq c$, the observed distribution is the distribution conditional on meeting that rule. For a continuous loss with density $f_X$, the conditional density above the threshold is $f_X(x)/P(X\geq c)$ for $x\geq c$, provided $P(X\geq c)>0$. Thus summaries of the observed sample describe losses meeting the inclusion rule. They are not automatically summaries of all losses.

f_{X\mid X\geq c}(x) = \frac{f_X(x)}{P(X\geq c)}, \qquad x\geq c.

The denominator is the probability of inclusion; dividing by it rescales the density over the part of the loss distribution that is observed. This follows because, for a small interval around $x\geq c$, the conditional probability is the probability of being in that interval and included, divided by the probability of being included. Since the interval already lies above $c$, the numerator is governed by $f_X(x)$, giving the displayed conditional density.

If the number of omitted losses and their possible range are known, bounds on a full-cohort mean may still be possible. If neither is known, the mean of the unobserved full population generally cannot be recovered from the truncated observations alone. A model or additional information is needed.

Worked Example: Separating the Observed Mean From the Cohort Mean

A hypothetical cohort has ten claims, but a report includes only claims of at least $\$50$. The four observed claims are $\$60$, $\$80$, $\$120$, and $\$140$. The other six claims are known to be nonnegative and less than $\$50$. Calculate the observed mean and bound the mean for all ten claims.

Method: calculate the selected-sample summary, then use the known omitted-claim limits. The observed total is $60+80+120+140=\$400$. There are four observed claims, so their mean is:

\bar{X}_{\text{observed}}=\frac{\$400}{4}=\$100.

That is the mean among claims meeting the reporting rule. For the full cohort, the six omitted losses contribute at least zero and less than $6(50)=\$300$. The total for all ten claims is therefore at least $\$400$ and less than $\$700$. Dividing by ten gives:

\$40\leq \bar{X}_{\text{cohort}}<\$70.

Check and common mistake. The observed mean is $\$100$, while the full-cohort mean is bounded between $\$40$ and $\$70$. Reporting $\$100$ as the cohort mean would incorrectly treat a threshold-selected sample as though it included all ten claims.

Compare the Information Each Record Provides

Observation process	What is recorded?	What a basic summary describes
Grouped	Counts in intervals	An approximation or bounds based on those intervals
Censored	A record for the loss, with an exact value replaced by a bound or range	A sample summary may be bounded, but exact values remain unknown
Truncated	Only observations meeting an inclusion rule	The selected sample, unless additional information supports a full-population summary

A useful diagnostic is to ask whether a missing exact amount still has a record. If so, the issue is censoring; if the report gives a range and count, it is grouping; if a loss is absent because it failed the inclusion rule, it is truncation. These distinctions determine both the denominator and the interpretation of any calculated mean.

Identify the observation rule.
Determine whether the data give intervals, bounds for retained observations, or only a selected subset.

State what is known about each loss.
Use exact amounts as given and represent grouped or censored amounts only by the intervals or bounds provided.

Choose the appropriate summary.
Use midpoints as an explicitly labelled approximation, derive bounds where possible, and identify when a summary applies only to the selected observations.

The most common exam trap is to treat a threshold or interval endpoint as an exact loss. A second is to use the number of records as the denominator for a full cohort when truncation has excluded claims. Always state the population represented by the calculation and whether the answer is exact, approximate, or a bound.

Key takeaway: Grouping hides within-interval values, censoring retains observations but hides exact amounts, and truncation excludes observations. A correct summary must preserve those distinctions.

Check Your Understanding

For each problem, distinguish the information actually observed from any estimate or bound that can be calculated.

A report contains three losses in $[0,20)$ and two in $[20,60)$. Find the midpoint estimate and bounds for the mean. (Answer: midpoint estimate $\$22.00$; $ \$8\leq\bar{X}<\$36$.)
Four claims have exact losses of $\$15$ and $\$35$, plus two claims reported as at least $\$80$. Find the lower bound for the mean and say whether these records give a finite upper bound. (Answer: mean at least $\$52.50$; no finite upper bound.)
Four nonnegative losses are recorded as $\$10$, at most $\$25$, $\$40$, and $\$55$. Bound their mean. (Answer: $\$26.25\leq\bar{X}\leq\$32.50$.)
A cohort has eight claims. Only claims of at least $\$30$ are reported; the observed amounts are $\$40$, $\$50$, and $\$70$. The five omitted claims are nonnegative and less than $\$30$. Find the observed mean and bounds for the cohort mean. (Answer: observed mean $\$53.33$; cohort mean $\$20\leq\bar{X}<\$38.75$.)
Among all losses, $P(X\geq100)=0.20$ and $P(X\geq200)=0.08$. In a sample truncated to include only losses at least $\$100$, what fraction of included losses are at least $\$200$? (Answer: $0.4000$, or 40.00%.)

What You'll Learn