Distinguish Censoring from Truncation in Coverage Data

Why the Observation Mechanism Matters

A claim database does not always contain an exact observation of every ground-up loss. A policy limit may hide how large a loss was, while a deductible may determine which losses enter a severity sample at all. These mechanisms produce different likelihood contributions, even when both are described informally as “claims affected by coverage.”

The distinction is consequential: censoring retains an observation but reports only that the loss reached a threshold; truncation removes observations outside the sample’s inclusion rule. The likelihood must represent what was observed and how the observation was selected. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a ground-up loss with density $f_X$, distribution function $F_X$, and survival function $S_X(x)=P(X>x)$. Under right censoring at a policy limit $u$, a loss at or above $u$ is retained but its exact value is not observed; the recorded amount is capped at $u$. Under left truncation at a deductible $d$, only losses satisfying $X>d$ enter the severity sample. A retained truncated observation is therefore known to exceed $d$, and its exact ground-up value may be recorded.

The likelihood contributions below assume a continuous severity model, independent losses from the same distribution, known thresholds, and complete recording according to the stated rule. With a continuous model, $P(X=u)=P(X=d)=0$, so whether a threshold event is written with a strict or non-strict inequality does not change the likelihood. If losses are discrete, use the probability of the appropriate threshold event, including any mass at the threshold, rather than substituting a continuous density formula.

Likelihood Contributions: Censoring and Truncation

Suppose the recorded value is $z=\min(X,u)$. If $z<u$, the exact loss is observed and its likelihood contribution is $f_X(z)$. If $z=u$, the observation says only that $X\ge u$, so the contribution is $P(X\ge u)=S_X(u)$ for a continuous model. A capped payment is thus not an exact observation that the ground-up loss equals the cap.

For deductible truncation, consider a sample that contains only losses with $X>d$, with exact ground-up values recorded. Each sampled value has the conditional density of $X$ given that it passed the inclusion rule:

f_{X\mid X>d}(x) = \frac{f_X(x)}{P(X>d)} = \frac{f_X(x)}{S_X(d)}, \qquad x>d.

The denominator accounts for selection. It is not a probability that the observed loss exceeded the deductible after it has already been selected; it is the probability that a loss from the original population would be selected in the first place.

Identify what was recorded.
An exact amount below a cap contributes a density. A payment at the cap usually indicates an unknown ground-up loss at least as large as the cap.

Identify how records entered the data.
If losses below a deductible are absent, the observed severity sample is selected conditional on exceeding that deductible.

Write the likelihood for that observation scheme.
Use a survival probability for an individually right-censored loss, and a conditional density for an exact value from a truncated sample.

Why the Truncation Denominator Appears

The denominator follows directly from conditional probability. For a small interval of width $\Delta x$ around a possible observed value $x>d$, the probability that a loss falls in that interval and is selected is approximately $f_X(x)\Delta x$. Given that it was selected, divide by the probability $S_X(d)$ of selection:

P(x\le X<x+\Delta x\mid X>d) \approx \frac{f_X(x)\Delta x}{S_X(d)}.

Dividing by $\Delta x$ and taking the limit gives $f_X(x)/S_X(d)$. By contrast, an individual known only to have exceeded a censoring threshold contributes $S_X(d)$: there is no exact value whose density can be used. These contributions answer different questions.

Worked Applications

Worked Example: Likelihood for Claims Capped at a Policy Limit

A hypothetical insurer models ground-up claim severity by an exponential distribution with rate $\lambda=0.001$ per dollar, so $f_X(x)=0.001e^{-0.001x}$ and $S_X(x)=e^{-0.001x}$ for $x\ge0$. Four independent claims are recorded under a policy limit of $u=\$1{,}000$: exact amounts of $\$400$ and $\$900$, and two payments capped at $\$1{,}000$. Find the likelihood contribution of these observations. Then calculate the incorrect likelihood obtained by treating both capped amounts as exact ground-up losses.

Method. The two uncapped observations are exact, so their contributions are densities. Each capped observation means $X\ge1{,}000$, so its contribution is a survival probability.

The exact-loss contributions are $f_X(400)=0.001e^{-0.4}$ and $f_X(900)=0.001e^{-0.9}$. Each capped observation contributes $S_X(1{,}000)=e^{-1}$. Therefore,

L =(0.001e^{-0.4})(0.001e^{-0.9})(e^{-1})^2 =10^{-6}e^{-3.3} \approx 3.688\times10^{-8}.

For the incorrect calculation, replacing each survival contribution by $f_X(1{,}000)=0.001e^{-1}$ gives

L_{\text{wrong}} =(0.001e^{-0.4})(0.001e^{-0.9})(0.001e^{-1})^2 =10^{-12}e^{-3.3} \approx 3.688\times10^{-14}.

The likelihood is not a probability that must be dimensionless: exact observations contribute densities, whereas censored observations contribute probabilities. The wrong calculation treats the cap as an exact loss and inserts two extra density factors. Check: both capped observations are known to exceed the limit, and the correct likelihood uses two survival terms.

Worked Example: Estimating Severity from Deductible-Truncated Claims

A hypothetical portfolio follows an exponential ground-up severity model with unknown rate $\lambda$. Only claims exceeding a deductible $d=\$500$ enter the severity file, and their exact ground-up amounts are recorded as $\$700$, $\$1{,}200$, and $\$1{,}600$. Find the conditional likelihood and maximum likelihood estimate of $\lambda$. Compare with the estimate from incorrectly treating the observations as an unselected sample.

Method. For an exponential model, $f_X(x)=\lambda e^{-\lambda x}$ and $S_X(500)=e^{-500\lambda}$. Each selected exact loss contributes $f_X(x)/S_X(500)$. The three excesses above the deductible are $700-500=200$, $1{,}200-500=700$, and $1{,}600-500=1{,}100$, totaling $2{,}000$.

L(\lambda) = \prod_{x\in\{700,1200,1600\}} \frac{\lambda e^{-\lambda x}}{e^{-500\lambda}} = \lambda^3e^{-2000\lambda}.

The log-likelihood is $\ell(\lambda)=3\log\lambda-2{,}000\lambda$. Differentiating and setting the score to zero gives $3/\lambda-2{,}000=0$, hence $\widehat{\lambda}=3/2{,}000=0.0015$ per dollar. The second derivative is $-3/\lambda^2<0$, confirming a maximum. At this estimate, the likelihood is $(0.0015)^3e^{-3}\approx1.680\times10^{-10}$.

If selection is ignored, the incorrect likelihood is $\lambda^3e^{-3{,}500\lambda}$, because the ground-up amounts sum to $\$3{,}500$. It yields $\widehat{\lambda}_{\text{wrong}}=3/3{,}500\approx0.0008571$ per dollar. The estimate is lower, corresponding to a heavier-tailed exponential model, because the selected values are incorrectly treated as though small claims could also have appeared in the sample. Check: the correct score equation uses the sum of the three excesses above $d$, not the sum of the ground-up losses.

Worked Example: Truncated Sample Recorded as Payments Above the Deductible

A hypothetical insurer records only positive payments under a deductible of $d=\$400$. Ground-up losses follow an exponential model with unknown rate $\lambda$, and the three recorded payments above the deductible are $\$100$, $\$300$, and $\$600$. Find the appropriate conditional likelihood and estimate $\lambda$. Explain why treating the payments as exact ground-up losses from a sample truncated at $\$400$ is wrong.

Method. For a positive payment $y$, the corresponding ground-up loss is $x=y+400$. Since the record exists only when $X>400$, its conditional density as a payment is $f_X(y+400)/S_X(400)$. Substituting the exponential model gives $\lambda e^{-\lambda(y+400)}/e^{-400\lambda}=\lambda e^{-\lambda y}$.

The payment amounts sum to $100+300+600=1{,}000$, so the conditional likelihood and log-likelihood are

L(\lambda)=\lambda^3e^{-1000\lambda}, \qquad \ell(\lambda)=3\log\lambda-1000\lambda.

The score equation is $3/\lambda-1{,}000=0$, giving $\widehat{\lambda}=0.003$ per dollar. The second derivative, $-3/\lambda^2$, is negative. At the estimate, $L=(0.003)^3e^{-3}\approx1.344\times10^{-9}$.

The payments $\$100$ and $\$300$ are not ground-up losses below the deductible; they correspond to ground-up losses of $\$500$ and $\$700$. Treating them as ground-up observations from a sample truncated above $\$400$ would assign zero density to both, since they violate the stated inclusion rule. Check: the transformation from ground-up loss to payment is accounted for in the conditional density, and the exponential payment likelihood here simplifies to the displayed expression.

Common Traps and Practical Checks

A policy limit and a deductible can both produce incomplete-looking data, but the missing information points in different directions. A capped record tells you that a particular retained loss reached or exceeded the cap. A deductible-truncated sample tells you that the losses below the threshold are absent, while the retained exact values are observed.

Using a density at the policy limit. A payment equal to the cap is generally not an exact ground-up loss. Use the probability of reaching the cap, $S_X(u)$, for a continuous model.
Forgetting the selection denominator. An exact ground-up value in a deductible-truncated severity sample contributes $f_X(x)/S_X(d)$, not just $f_X(x)$, when the likelihood is conditional on inclusion.
Using a survival probability for an exact selected claim. The fact that a claim exceeded the deductible is already part of its inclusion in the truncated sample. If its exact value is recorded, use the conditional density.
Confusing payment with ground-up loss. A positive payment $y$ above deductible $d$ corresponds to ground-up loss $y+d$, not $y$.
Ignoring the study design. If the data include all exposed risks and identify which risks had no claim, the frequency or claim-occurrence information may also matter. A severity likelihood conditional on a claim being observed does not by itself represent that additional information.

Key takeaway: For a continuous ground-up severity model, an exact loss censored at a limit contributes a survival probability at that limit, whereas an exact loss retained by deductible truncation contributes a density divided by the probability of exceeding the deductible. Identify both the recorded value and the selection rule before writing the likelihood.

Check Your Understanding

For each problem, identify the observation mechanism before calculating.

A hypothetical exponential severity model has rate $0.001$ per dollar. One claim is observed exactly at $\$300$, and a second is capped at a policy limit of $\$800$. Write and evaluate their joint likelihood contribution. (Answer: $0.001e^{-0.3}e^{-0.8}=0.001e^{-1.1}\approx0.0003329$.)
Ground-up losses follow an exponential model with unknown rate. Only losses above a $\$250$ deductible are recorded, with observed values $\$400$, $\$700$, and $\$1{,}000$. Find the conditional likelihood and the maximum likelihood estimate of the rate. (Answer: $L(\lambda)=\lambda^3e^{-1350\lambda}$, so $\widehat{\lambda}=3/1350\approx0.002222$ per dollar.)
Under an exponential model with rate $0.0005$ per dollar, two claims are both capped at $\$1{,}200$. What is their likelihood contribution? (Answer: $S_X(1200)^2=e^{-1.2}\approx0.3010$.)
Only positive payments above a $\$300$ deductible are retained. Under an exponential ground-up model with unknown rate, observed payments are $\$80$, $\$220$, and $\$500$. Find the conditional likelihood and rate estimate. (Answer: $L(\lambda)=\lambda^3e^{-800\lambda}$, so $\widehat{\lambda}=3/800=0.00375$ per dollar.)

What You'll Learn