Why Both Contract Features Matter
A deductible determines whether a loss produces a payment, while a policy limit caps the payment on a sufficiently large loss. When both apply, the observed payment alone does not always reveal the underlying loss. A zero payment may represent a recorded loss below the deductible; an interior payment may identify the loss exactly; and a payment at the cap usually tells us only that the loss reached or exceeded a threshold.
The likelihood must reflect both the insurance contract and the way observations entered the data set. In particular, a database containing every loss event is not equivalent to a sample containing only claims with positive payments. The first can include evidence about the probability of falling below the deductible; the second is selected on exceeding it.
This convention makes \(u\) the maximum payment, not the maximum ground-up loss. The ground-up loss threshold for reaching the cap is therefore \(d+u\). If a contract instead specifies a maximum covered ground-up loss \(L\), its maximum payment is generally \(L-d\) when \(L>d\), and the threshold and payment cap must be translated before writing a likelihood.
Three Observation Types and Their Likelihood Contributions
Assume first that every loss event is recorded, including events that produce no payment. For a zero-payment record, the observation tells us that the loss lies in the interval \(0\le X\le d\). Its likelihood contribution is the probability of that interval, \(F_X(d)\). It is a probability, not a density evaluated at zero.
For an interior payment \(y\), where \(0<y<u\), the ground-up loss is \(x=d+y\). Since payment changes one-for-one with loss in this range, the density contribution is \(f_X(d+y)\). At the cap, the actual loss is not known; it is at least \(d+u\). The contribution is the tail probability \(S_X(d+u)\). Under a continuous severity model, an exact loss of \(d+u\) has probability zero, so a cap record is treated as right-censored at that threshold.
Here \(Z\), \(I\), and \(C\) index zero-payment, interior-payment, and capped records. Each interior density is expressed per unit of the loss scale. The full likelihood therefore has one factor with inverse monetary units for every interior observation; probability and tail factors are dimensionless.
Now suppose the data set contains only claims with positive payment. This is selection on \(X>d\). The likelihood for each sampled claim must be conditional on that event: divide its unconditioned contribution by \(P(X>d)=S_X(d)\). An interior payment therefore contributes \(f_X(d+y)/S_X(d)\), and a capped claim contributes \(S_X(d+u)/S_X(d)\). There are no zero-payment records in this selected sample.
These expressions assume that the selection rule is exactly positive payment, that the deductible and cap are known, and that the severity distribution applies to the ground-up losses in the population being modeled. If claims are sampled by another rule, or if the data distinguish reporting from payment in some other way, the observation mechanism must be represented separately.
Why the Contributions Have This Form
The interior contribution follows from a change of variable. For an interior payment \(Y=y\), the corresponding ground-up loss is \(X=d+y\). For a small interval of payment values from \(y\) to \(y+\Delta y\), the underlying loss lies between \(d+y\) and \(d+y+\Delta y\). Therefore, \(P(y<Y<y+\Delta y)=\int_{d+y}^{d+y+\Delta y}f_X(x)\,dx\). Dividing by \(\Delta y\) and taking the limit as \(\Delta y\) approaches zero gives density \(f_X(d+y)\), since the derivative of \(d+y\) with respect to \(y\) is 1.
For positive-payment sampling, conditional probability gives the denominator. For example, the conditional probability of a capped observation is \(P(X\ge d+u\mid X>d)=P(X\ge d+u)/P(X>d)\), because \(d+u>d\). Under a continuous model this is \(S_X(d+u)/S_X(d)\). The same conditioning argument converts the interior density to \(f_X(d+y)/S_X(d)\).
Determine whether the records include every loss event or only claims with positive payment.
Use \(X\le d\) for zero payment, \(X=d+y\) for an interior payment, and \(X\ge d+u\) for a capped payment.
Use a probability for an interval, a density for an exact interior value, and a survival probability for a censored cap. Condition on selection if only positive claims are sampled.
Worked Example: All Loss Events Are Recorded
A hypothetical portfolio records every loss event under a deductible of \(d=1\) thousand dollars and a maximum payment of \(u=3\) thousand dollars. Assume ground-up loss \(X\) is exponential with rate \(0.1\) per thousand dollars, so \(f_X(x)=0.1e^{-0.1x}\), \(F_X(x)=1-e^{-0.1x}\), and \(S_X(x)=e^{-0.1x}\), for \(x\ge0\). The three recorded payments are \(0\), \(2\), and \(3\) thousand dollars. Find their joint likelihood.
Method. Because all loss events are recorded, use the unconditional contributions. The zero payment contributes \(F_X(1)=1-e^{-0.1}=0.09516258\).
The interior payment of \(2\) thousand dollars identifies a ground-up loss of \(d+y=1+2=3\) thousand dollars. Its density contribution is \(f_X(3)=0.1e^{-0.3}=0.07408182\) per thousand dollars.
The payment of \(3\) thousand dollars is at the cap. It means \(X\ge d+u=1+3=4\) thousand dollars, so its contribution is \(S_X(4)=e^{-0.4}=0.67032005\).
Multiplying the three factors gives \((1-e^{-0.1})(0.1e^{-0.3})(e^{-0.4})=0.004725633\) per thousand dollars. Rounded to six decimal places, the likelihood is 0.004726 per thousand dollars. The single density factor accounts for the inverse monetary unit.
Check. There is one zero event, one exact interior observation, and one tail event, so the likelihood contains exactly one distribution probability, one density, and one survival probability. The cap threshold is \(4\), not \(3\), because the \(3\)-thousand-dollar cap is the payment after the deductible.
Worked Example: Only Positive-Payment Claims Are Sampled
A hypothetical insurer samples claims only when they produce positive payment. The deductible is \(d=2\) thousand dollars and the payment cap is \(u=3\) thousand dollars. Let \(X\) be exponential with rate \(0.2\) per thousand dollars. The sample contains an interior payment of \(1\) thousand dollars and one capped payment. Find the conditional likelihood of these observations.
Method. Positive payment requires \(X>2\), so each observation must be conditioned on survival beyond the deductible. For the interior payment \(y=1\), the ground-up loss is \(x=2+1=3\). Its conditional density is \(f_X(3)/S_X(2)=(0.2e^{-0.6})/e^{-0.4}=0.2e^{-0.2}=0.163746\) per thousand dollars.
For the capped payment, the ground-up threshold is \(d+u=2+3=5\). Its conditional contribution is \(S_X(5)/S_X(2)=e^{-1}/e^{-0.4}=e^{-0.6}=0.548812\).
The joint conditional likelihood is \((0.2e^{-0.2})(e^{-0.6})=0.2e^{-0.8}=0.0898658\) per thousand dollars. Rounded to six decimal places, this is 0.089866 per thousand dollars.
Check. The cap factor is a conditional tail probability, while the interior factor is a conditional density. The denominator \(S_X(2)\) appears once for each sampled claim; it is not a single denominator for the entire two-claim likelihood.
Worked Example: Comparing a Capped Claim Under Two Sampling Rules
Suppose \(X\) is exponential with rate \(0.25\) per thousand dollars. A hypothetical contract has deductible \(d=2\) thousand dollars and payment cap \(u=4\) thousand dollars. Find the likelihood contribution of one capped claim when (a) all loss events are recorded and (b) only positive-payment claims are sampled. Explain the effect of incorrectly treating the cap as a ground-up loss limit of \(4\) thousand dollars.
Method. A capped payment of \(4\) thousand dollars means \(X\ge d+u=6\) thousand dollars. With all events recorded, the contribution is \(S_X(6)=e^{-0.25(6)}=e^{-1.5}=0.223130\).
If only positive-payment claims are sampled, divide by the probability of exceeding the deductible: \(S_X(6)/S_X(2)=e^{-1.5}/e^{-0.5}=e^{-1}=0.367879\). This is the conditional probability of reaching the cap, given that the claim generated a positive payment.
A mistaken calculation would use \(S_X(4)=e^{-1}=0.367879\) as the unconditional contribution, treating \(4\) as the ground-up threshold. That omits the deductible from the threshold and answers a different question. The corrected all-events contribution is \(0.223130\); \(0.367879\) is correct only for the positive-claim conditional calculation in this example.
Check. The conditional answer happens to equal the incorrectly chosen unconditional tail probability because the exponential survival ratio simplifies to \(e^{-0.25(6-2)}=e^{-1}\). That numerical coincidence does not make the threshold error valid.
Common Errors and Practical Checks
The most consequential mistake is to treat every observed payment as an exact ground-up loss. That is true only for an uncapped interior payment, for which \(X=d+y\). A zero is an interval observation, and a capped payment is a tail observation. Applying a density to either of those records assigns the wrong likelihood contribution.
- Confusing zero payment with no record. In an all-events data set, zero payment contributes \(F_X(d)\). In a positive-payment sample, such records are excluded and the remaining likelihood is conditional on \(X>d\).
- Forgetting the deductible when locating the cap. With a maximum payment \(u\), the cap threshold is \(d+u\). It is not \(u\).
- Using a density for a capped observation. A cap indicates \(X\ge d+u\), not an exact loss at that threshold. Use a survival probability.
- Dropping the selection normalizer. If only positive claims are sampled, divide each observation contribution by \(S_X(d)\). Otherwise the likelihood describes the unselected population.
- Ignoring units. A density has inverse-loss units; probabilities do not. The number of exact interior observations determines the inverse-unit power in the likelihood.
These distinctions matter whenever a severity model is estimated from claim records subject to coverage terms. A likelihood that mishandles zeros, truncation, or censoring can shift estimates even when the chosen severity family is otherwise appropriate. Before fitting, document the sampling unit, what qualifies as a record, and whether the displayed amount is exact or only a bound on ground-up loss.
Check Your Understanding
For each problem, state whether the likelihood contribution is unconditional or conditional and keep track of the loss units.
- All loss events are recorded. \(X\) is exponential with rate \(0.5\) per thousand dollars, \(d=1\) thousand dollars, and \(u=2\) thousand dollars. What is the contribution of a zero-payment event? (Answer: \(1-e^{-0.5}=0.3935\).)
- Only positive-payment claims are sampled. \(X\) is exponential with rate \(0.25\) per thousand dollars, and \(d=2\) thousand dollars. Find the conditional density contribution for an interior payment of \(2\) thousand dollars. (Answer: \(0.25e^{-0.5}=0.1516\) per thousand dollars.)
- All events are recorded under \(d=1\) thousand dollars and \(u=3\) thousand dollars. If \(X\) is exponential with rate \(0.2\) per thousand dollars, what is the contribution of a capped payment? (Answer: \(S_X(4)=e^{-0.8}=0.4493\).)
- Only positive claims are sampled. \(X\) is exponential with rate \(0.1\) per thousand dollars, \(d=1\) thousand dollars, and \(u=2\) thousand dollars. Find the joint likelihood for an interior payment of \(1\) thousand dollars and a capped payment. (Answer: \((0.1e^{-0.1})(e^{-0.2})=0.1e^{-0.3}=0.0741\) per thousand dollars.)
- All events are recorded. \(X\) is exponential with rate \(0.1\) per thousand dollars, \(d=1\) thousand dollars, and \(u=3\) thousand dollars. The payments are \(0\), \(2\), and \(3\) thousand dollars. Write and evaluate the joint likelihood. (Answer: \((1-e^{-0.1})(0.1e^{-0.3})(e^{-0.4})=0.004726\) per thousand dollars, to six decimal places.)