Empirical Payments Under Deductibles and Limits

Why Transform Losses Before Summarizing Payments?

A ground-up loss is not necessarily the amount an insurer pays. A deductible can remove the first part of a loss, a limit can restrict the covered amount, and coinsurance can assign only part of the covered amount to the insurer. To estimate the distribution of insurer payments from observed losses, first apply the coverage rule to each loss. Then summarize the resulting payments.

This order matters. The mean of the ground-up losses is generally not the mean insurer payment, and the proportion of losses above the deductible is not necessarily the proportion with positive payments if the coverage also has a zero cap or zero coinsurance. The calculations below assume that each observation is a complete ground-up loss and that observations have equal weight unless weights are stated.

Definition: For a complete ground-up loss $x$, the insurer payment per loss under a deductible $d$, a covered-layer cap $u$, and an insurer share $c$ is $$ y^L(x)=c\min\bigl((x-d)_+,u\bigr), \qquad (x-d)_+=\max(x-d,0). $$ Here $d\geq 0$, $u\geq 0$, and $0\leq c\leq 1$. The cap $u$ applies to the covered loss after the deductible but before coinsurance. Thus, $u$ is not necessarily a cap on the insurer’s final payment.

The formula has a direct interpretation. A loss at or below the deductible produces no covered amount. Above the deductible, the covered amount is the excess loss, but no more than $u$. The insurer pays share $c$ of that amount. A policy may instead impose a cap $L$ on the insurer’s final payment. Under that different convention, the formula is $\min\bigl(c(x-d)_+,L\bigr)$. Read the coverage wording carefully: a cap on the covered layer and a cap on the insurer’s payment are not interchangeable when $c<1$.

Building the Empirical Payment Distribution

Suppose the observed complete ground-up losses are $x_1,\ldots,x_n$. Transform each observation using the same coverage rule to obtain $y_j^L=y^L(x_j)$. The empirical distribution assigns probability $1/n$ to each transformed observation, including payments of zero. For any payment threshold $y$, its empirical cumulative distribution function is

\widehat{F}_{Y^L}(y)=\frac{1}{n}\sum_{j=1}^{n} I(y_j^L\leq y),

where $I(\cdot)$ equals 1 when the condition in parentheses is true and 0 otherwise. If the data are given as distinct loss values with counts, use each count as its weight; do not treat each distinct value as though it occurred just once.

Key formulas: For $n$ equally weighted complete losses, the empirical mean payment and empirical probability of a positive payment are $$ \overline{y^L}=\frac{1}{n}\sum_{j=1}^{n}y_j^L, \qquad \widehat{P}(Y^L>0)=\frac{1}{n}\sum_{j=1}^{n}I(y_j^L>0). $$ The conditional mean among positive payments is different: $$ \overline{y^L\mid y^L>0} =\frac{\sum_{j=1}^{n}y_j^L}{\sum_{j=1}^{n}I(y_j^L>0)}, $$ provided at least one payment is positive.

For $c>0$ and $u>0$, the payment is positive exactly when $x_j>d$. If $c=0$ or $u=0$, every payment is zero, regardless of the loss. This qualification is important: comparing losses with the deductible alone is not enough when another coverage parameter can eliminate all payments.

Worked Examples

Worked Example: Deductible, Covered-Layer Cap, and Coinsurance

A hypothetical insurer reviews six complete ground-up losses: $\$80,\ \$100,\ \$250,\ \$500,\ \$700,\ \$1{,}000$. The contract has a $\$100$ deductible, a $\$400$ cap on the covered layer, and a 75% insurer share. Find the empirical payment distribution, mean payment, and probability of a positive payment.

Method. Apply the deductible first, cap the resulting covered amount at $\$400$, and then multiply by $0.75$. The payment formula is $y^L(x)=0.75\min((x-100)_+,400)$.

For $x=\$80$, the excess over the deductible is $\max(80-100,0)=0$. The payment is $0.75\min(0,400)=\$0$.

For $x=\$100$, the excess is $\max(100-100,0)=0$. The payment is $0.75\min(0,400)=\$0$.

For $x=\$250$, the excess is $250-100=\$150$, below the cap. The payment is $0.75(150)=\$112.50$.

For $x=\$500$, the excess is $\$400$. The payment is $0.75(400)=\$300$.

For $x=\$700$, the excess is $\$600$, capped at $\$400$. The payment is $0.75(400)=\$300$.

For $x=\$1{,}000$, the excess is $\$900$, also capped at $\$400$. The payment is $0.75(400)=\$300$.

The six payments are $\$0,\ \$0,\ \$112.50,\ \$300,\ \$300,\ \$300$. Their total is $0+0+112.50+300+300+300=\$1{,}012.50$, so the empirical mean is $\$1{,}012.50/6=\$168.75$. Four of the six payments are positive, so the empirical probability of a positive payment is $4/6=0.6667$, to four decimal places.

The empirical distribution places probability $2/6$ at $\$0$, $1/6$ at $\$112.50$, and $3/6$ at $\$300$. As a check, the probabilities sum to $2/6+1/6+3/6=1$, and the weighted payment total reproduces the mean. The common mistake avoided here is treating the $\$400$ covered-layer cap as a $\$400$ cap on the insurer’s payment; under this contract the maximum payment is $0.75(400)=\$300$.

Worked Example: A Cap on the Insurer’s Final Payment

A hypothetical contract has a $\$100$ deductible, an 80% insurer share, and a maximum insurer payment of $\$500$ per loss. The complete ground-up losses are $\$50,\ \$200,\ \$700,\ \$900$. Calculate the empirical mean payment and the probability of a positive payment.

Method. The limit caps the insurer’s payment after coinsurance, so use $y^L(x)=\min(0.80(x-100)_+,500)$, not a cap on the covered loss before applying the 80% share.

At $\$50$, the excess after the deductible is $\max(50-100,0)=0$, and the payment is $\min(0.80(0),500)=\$0$.

At $\$200$, the excess is $\$100$. The uncapped insurer share is $0.80(100)=\$80$, so the payment is $\min(80,500)=\$80$.

At $\$700$, the excess is $\$600$. The uncapped insurer share is $0.80(600)=\$480$, below the final-payment cap, so the payment is $\$480$.

At $\$900$, the excess is $\$800$. The uncapped insurer share is $0.80(800)=\$640$, so the payment is capped at $\$500$.

The payments are $\$0,\ \$80,\ \$480,\ \$500$, with total $\$1{,}060$. The empirical mean is $\$1{,}060/4=\$265.00$. Three of four payments are positive, giving probability $3/4=0.7500$. As a check, only the largest uncapped payment exceeds the final-payment limit. The common mistake avoided is applying the $\$500$ cap to the covered loss before multiplying by 80%; the contract states that $\$500$ is the insurer’s maximum payment.

Worked Example: Grouped Losses and a Conditional Mean

A hypothetical portfolio summary reports complete ground-up losses by value: two losses of $\$120$, three of $\$180$, two of $\$400$, and one of $\$950$. A contract has a $\$150$ deductible, a $\$500$ cap on the covered layer, and a 60% insurer share. Find the empirical mean payment, the probability of a positive payment, and the mean payment conditional on a positive payment.

Method. Calculate one payment per distinct loss value, then multiply by its count. The payment formula is $y^L(x)=0.60\min((x-150)_+,500)$.

For each $\$120$ loss, the amount above the deductible is zero, so each payment is $\$0$. These two losses contribute $2(0)=\$0$.

For each $\$180$ loss, the covered amount is $180-150=\$30$, below the cap. Each payment is $0.60(30)=\$18$. The three losses contribute $3(18)=\$54$.

For each $\$400$ loss, the covered amount is $400-150=\$250$. Each payment is $0.60(250)=\$150$. The two losses contribute $2(150)=\$300$.

For the $\$950$ loss, the amount above the deductible is $\$800$, capped at $\$500$. The payment is $0.60(500)=\$300$. The total payment across all eight losses is $0+54+300+300=\$654$.

The empirical mean payment is $\$654/8=\$81.75$. Six of the eight payments are positive, so the probability of a positive payment is $6/8=0.7500$. The conditional mean among positive payments is $\$654/6=\$109.00$. As a check, the conditional mean multiplied by the positive-payment proportion gives $109(0.75)=\$81.75$, the unconditional empirical mean. The common mistake avoided is averaging the four distinct payment values without their counts.

Why the Empirical Formulas Hold

The empirical distribution treats the observed portfolio as a finite collection of equally weighted outcomes. For each loss $x_j$, the coverage rule produces one payment $y_j^L$. The indicator $I(y_j^L\leq y)$ contributes one exactly when that observed payment is at most $y$; summing the indicators counts how many of the $n$ payments satisfy the condition. Dividing by $n$ converts that count into the empirical proportion, which gives $\widehat{F}_{Y^L}(y)$.

The mean formula follows the same finite-sample logic: add the $n$ transformed payments and divide by $n$. For the conditional mean, only observations with $y_j^L>0$ belong in the average. The numerator is the sum of their payments, since zero payments do not change that sum; the denominator counts the positive payments. Thus the conditional mean is not obtained by dividing the unconditional mean by the positive-payment count.

Confirm what the data represent.
Check that each observation is a complete ground-up loss, not an already-modified payment or an amount observed only because it exceeded a reporting threshold.

Translate the contract into a formula.
Identify whether the limit applies to the covered loss layer or to the insurer’s final payment, and put the terms in the correct order.

Transform every observation.
Apply the same coverage rule to each loss, retaining zero payments in the sample.

Summarize the transformed payments.
Use the total number of losses for the unconditional mean and positive-payment probability; use the number of positive payments for the conditional mean.

Data and Interpretation Traps

The calculations require ground-up losses, or enough information to reconstruct them. A recorded insurer payment alone generally does not reveal the original loss: several different losses may map to the same capped payment, and a zero payment may represent a loss below the deductible or another circumstance. Likewise, applying a new coverage rule directly to amounts that are already subject to a deductible or limit can apply the adjustment twice. The earlier material on censoring and truncation is relevant when deciding whether observed records contain the complete losses required here.

Dropping zero payments. Zero payments remain part of the empirical distribution and belong in the denominator for the unconditional mean and positive-payment probability.
Using the wrong cap. State whether a limit caps the covered layer or the insurer’s final payment before calculating.
Ignoring coinsurance or a zero cap. For the stated layer formula, positive payment requires $c>0$, $u>0$, and $x>d$. If either $c=0$ or $u=0$, all payments are zero.
Confusing conditional and unconditional means. The unconditional mean averages over all losses; the conditional mean averages only over losses with positive payments.
Ignoring counts in grouped data. A value occurring many times must contribute many times to the empirical distribution and total payment.

These empirical summaries describe the transformed observations under the specified hypothetical coverage. They do not, by themselves, adjust for sampling uncertainty or establish how payments would behave in a different portfolio. Correctly defining the observation and applying the contract rule consistently are the essential first steps.

The formulas also give a useful arithmetic check. If there is at least one positive payment, the unconditional mean must equal the probability of a positive payment multiplied by the conditional mean among positive payments:

\overline{y^L} = \widehat{P}(Y^L>0)\,\overline{y^L\mid y^L>0}.

This identity holds because zero payments contribute nothing to the total, while the positive-payment count is the fraction $\widehat{P}(Y^L>0)$ of the full sample. It is a useful check, but it does not replace verifying each transformed payment.

Check Your Understanding

For each problem, transform the losses first and include zero payments in the empirical distribution and unconditional mean.

A hypothetical sample has complete losses of $\$50,\ \$140,\ \$260,\ \$500$. Full coverage applies above a $\$100$ deductible, with no limit or coinsurance. Find the four payments, mean payment, and probability of a positive payment. (Answer: payments $\$0,\ \$40,\ \$160,\ \$400$; mean $\$150.00$; probability $0.7500$.)
Complete losses are $\$100,\ \$250,\ \$700$. The insurer pays 50% of the amount above a $\$100$ deductible, subject to a $\$300$ cap on the covered layer. Find the payments, mean, and positive-payment probability. (Answer: payments $\$0,\ \$75,\ \$150$; mean $\$75.00$; probability $0.6667$.)
A hypothetical contract has a $\$100$ deductible, 80% coinsurance, and a $\$250$ cap on the insurer’s final payment. Complete losses are $\$200,\ \$400,\ \$900$. Find the mean payment and probability of a positive payment. (Answer: payments $\$80,\ \$240,\ \$250$; mean $\$190.00$; probability $1.0000$.)
A grouped data set has two complete losses of $\$100$, three of $\$200$, and one of $\$600$. The insurer pays 50% of the amount above a $\$150$ deductible, with a $\$300$ cap on the covered layer. Find the unconditional mean payment and the conditional mean among positive payments. (Answer: payments by group $\$0,\ \$25,\ \$225$; total $\$300$; unconditional mean $\$50.00$; conditional mean $\$75.00$.)
Four complete losses are $\$200,\ \$300,\ \$800,\ \$1{,}000$. The contract has a $\$100$ deductible, a 70% insurer share, and a zero covered-layer cap $u=0$. Find the empirical payment distribution and the probability of a positive payment. (Answer: all four payments are $\$0$; the empirical distribution assigns probability $1$ to $\$0$; probability of a positive payment $0.0000$.)

What You'll Learn