Workshop: Allocate Payments Across Claims And An Aggregate Cap

From Individual Claims to an Aggregate-Capped Payment

A policy-period aggregate limit applies to covered payments over the policy period, while an occurrence deductible or occurrence limit applies to losses grouped within a particular event. To calculate claim payments correctly, keep those levels distinct: first group claims into occurrences, then calculate each occurrence’s covered amount, and finally apply the aggregate limit to those amounts in the stated order.

The examples use a hypothetical policy and amounts in thousands of dollars. Each example states an allocation convention because the way an occurrence-level payment is divided among individual claims can depend on the policy terms. Here, unless stated otherwise, payments are allocated among claims in proportion to their gross losses.

Definition: For occurrence $j$, let $X_{jk}$ be the gross loss for claim $k$ in that occurrence, and let $L_j=\sum_k X_{jk}$ be the occurrence’s total gross loss. With an occurrence deductible $d$ and an occurrence payment limit $u$, the amount eligible for payment before the policy-period aggregate is $C_j=\min\{(L_j-d)_+,u\}$, where $(z)_+=\max(z,0)$. If the policy-period aggregate limit is $A$, the remaining capacity starts at $R_0=A$. Processing occurrences in order, the insurer pays $P_j=\min(C_j,R_{j-1})$, and the remaining capacity becomes $R_j=R_{j-1}-P_j$.

The occurrence limit $u$ in this definition is a maximum insurer payment for one occurrence, after applying the deductible. It is not the policy-period aggregate $A$. The occurrence limit constrains each $C_j$; the aggregate limit constrains the combined payments $P_j$ across the policy period.

Why the Calculation Has Two Stages

The occurrence deductible is applied to the total loss from the occurrence, not automatically to each claim within it. The insurer’s payment before the occurrence limit is therefore the portion of $L_j$ above $d$, floored at zero. The occurrence limit then caps that amount. These rules produce the following cases:

C_j= \begin{cases} 0, & L_j\le d,\\ L_j-d, & d<L_j<d+u,\\ u, & L_j\ge d+u. \end{cases}

The aggregate calculation is a second cap. At step $j$, no more than $R_{j-1}$ remains available, so the insurer cannot pay more than either the occurrence’s eligible amount $C_j$ or the remaining aggregate capacity. Subtracting the actual payment gives the next remaining balance.

P_j=\min(C_j,R_{j-1}),\qquad R_j=R_{j-1}-P_j.

If the allocation convention is proportional to gross loss, the payment assigned to claim $k$ in occurrence $j$ is

Y_{jk}= \begin{cases} P_j\dfrac{X_{jk}}{L_j}, & L_j>0,\\ 0, & L_j=0. \end{cases}

This allocation reconciles to the occurrence payment: when $L_j>0$, summing over claims gives $\sum_k Y_{jk}=P_j\sum_k X_{jk}/L_j=P_j$. Proportional allocation is a convention for dividing the payment among claims; it does not change the amount calculated for the occurrence or the amount payable under the aggregate cap. If the problem specifies another allocation rule, use that rule instead.

A Reliable Calculation Sequence

Group claims into occurrences.
Use the event groupings given in the problem. Do not infer that two claims share an occurrence merely because they happened close together in time.

Calculate each occurrence’s gross loss.
Add the gross losses of the claims assigned to that occurrence to obtain $L_j$.

Apply the occurrence terms.
Subtract the occurrence deductible from $L_j$, floor the result at zero, and cap it at the occurrence limit $u$. This gives $C_j$, before the aggregate limit.

Apply the aggregate sequentially.
Compare $C_j$ with the remaining aggregate $R_{j-1}$. The smaller amount is $P_j$, and the next balance is $R_j=R_{j-1}-P_j$.

Allocate and reconcile.
Divide $P_j$ among the occurrence’s claims using the specified convention. Check that claim payments sum to $P_j$, and that all occurrence payments sum to no more than $A$.

Worked Examples

Worked Example: Two Occurrences and a Partly Used Aggregate

A hypothetical policy has a deductible of $10 thousand per occurrence, an occurrence payment limit of $30 thousand, and a policy-period aggregate limit of $40 thousand. Claims within an occurrence are paid proportionally to their gross losses. Occurrence 1 has claims of $8 thousand, $12 thousand, and $20 thousand. Occurrence 2 has claims of $15 thousand and $25 thousand. Process the occurrences in that order and find each claim payment and the remaining aggregate.

Method. Add the claims within each occurrence, calculate the occurrence amount after its deductible and limit, and then apply the aggregate to that amount. Proportional allocation uses each claim’s share of its occurrence’s gross loss.

Occurrence 1. The gross occurrence loss is $L_1=8+12+20=40$ thousand. After the $10 thousand deductible, the amount is $40-10=30$ thousand. Applying the $30 thousand occurrence limit gives $C_1=\min(30,30)=30$ thousand. The aggregate has $40 thousand available, so $P_1=\min(30,40)=30$ thousand. The remaining aggregate is $R_1=40-30=10$ thousand.

The three claims have gross-loss shares $8/40$, $12/40$, and $20/40$. Their allocated payments are $30(8/40)=6$, $30(12/40)=9$, and $30(20/40)=15$ thousand. They sum to $6+9+15=30$ thousand.

Occurrence 2. Its gross occurrence loss is $L_2=15+25=40$ thousand. After the deductible, the amount is $40-10=30$ thousand. The occurrence limit leaves $C_2=\min(30,30)=30$ thousand eligible before the aggregate. Only $10 thousand of aggregate capacity remains, so $P_2=\min(30,10)=10$ thousand. The remaining aggregate is $R_2=10-10=0$.

The claims have shares $15/40$ and $25/40$. Their allocated payments are $10(15/40)=3.75$ and $10(25/40)=6.25$ thousand. They sum to $10 thousand.

Check. Total insurer payments are $30+10=40$ thousand, exactly the aggregate limit. The five claim payments sum to $6+9+15+3.75+6.25=40$ thousand, and the final remaining aggregate is zero. The common mistake this method avoids is paying the full $30 thousand for the second occurrence without checking that only $10 thousand remains.

Worked Example: A Shared Occurrence Deductible

A hypothetical policy has a $6 thousand deductible and a $20 thousand occurrence payment limit. Its policy-period aggregate is $18 thousand. The first occurrence contains two claims, of $4 thousand and $8 thousand. The second occurrence is a separate $10 thousand claim. The third occurrence is a separate $20 thousand claim. Allocate each occurrence payment among its claims proportionally to gross loss and process the occurrences in the order given.

Method. Apply the deductible once to the combined loss for each occurrence, not once to each claim. Then apply the aggregate to each occurrence’s eligible payment.

First occurrence. Its gross loss is $4+8=12$ thousand. After one $6 thousand occurrence deductible, $12-6=6$ thousand is eligible. The occurrence limit does not reduce this amount, so $C_1=6$ thousand. The aggregate has enough capacity, giving $P_1=6$ thousand and $R_1=18-6=12$ thousand. The claims’ gross-loss shares are $4/12$ and $8/12$, so their payments are $6(4/12)=2$ and $6(8/12)=4$ thousand.

Second occurrence. Its gross loss is $10 thousand. After the deductible, $10-6=4$ thousand is eligible. Thus $C_2=4$ thousand, $P_2=\min(4,12)=4$ thousand, and $R_2=12-4=8$ thousand. The single claim receives $4 thousand.

Third occurrence. Its gross loss is $20 thousand. After the deductible, $20-6=14$ thousand is eligible, below the $20 thousand occurrence limit. The aggregate has only $8 thousand remaining, so $P_3=\min(14,8)=8$ thousand and $R_3=0$. The single claim receives $8 thousand.

Check. The allocated payments are $2 thousand, $4 thousand, $4 thousand, and $8 thousand, for a total of $2+4+4+8=18$ thousand. This equals the aggregate limit. A mistaken claim-by-claim deductible calculation for the first occurrence would produce $\max(4-6,0)+\max(8-6,0)=2$ thousand instead of the correct $6 thousand, because the contract applies one deductible to the occurrence total.

Worked Example: Aggregate Exhaustion Partway Through the Claim Sequence

A hypothetical policy has a $4 thousand deductible per occurrence, an $18 thousand occurrence payment limit, and a $25 thousand aggregate limit. Claims are allocated proportionally within each occurrence. In chronological order, Occurrence 1 has claims of $5 thousand, $10 thousand, and $15 thousand; Occurrence 2 has claims of $6 thousand, $9 thousand, and $15 thousand; Occurrence 3 has claims of $3 thousand and $5 thousand. Find every claim payment and the final aggregate balance.

Method. Calculate each occurrence’s eligible payment independently of the aggregate first. Then process the occurrences chronologically, allocating only the amount the aggregate permits.

Occurrence 1. Its gross loss is $5+10+15=30$ thousand. After the deductible, $30-4=26$ thousand is eligible, but the occurrence limit reduces this to $C_1=\min(26,18)=18$ thousand. The aggregate initially has $25 thousand available, so $P_1=18$ thousand and $R_1=25-18=7$ thousand. The claims have shares $5/30$, $10/30$, and $15/30$, so their payments are $18(5/30)=3$, $18(10/30)=6$, and $18(15/30)=9$ thousand.

Occurrence 2. Its gross loss is $6+9+15=30$ thousand. After the deductible, $30-4=26$ thousand is eligible, capped by the occurrence limit at $C_2=18$ thousand. Only $7 thousand remains in the aggregate, so $P_2=\min(18,7)=7$ thousand and $R_2=7-7=0$. The claims have shares $6/30$, $9/30$, and $15/30$. Their payments are $7(6/30)=1.4$, $7(9/30)=2.1$, and $7(15/30)=3.5$ thousand.

Occurrence 3. Its gross loss is $3+5=8$ thousand. After the deductible, $8-4=4$ thousand is eligible, below the occurrence limit. However, the aggregate is exhausted, so $P_3=\min(4,0)=0$ thousand. Both claims receive zero.

Check. The occurrence payments are $18+7+0=25$ thousand, exactly the aggregate limit. The claim payments also sum to $3+6+9+1.4+2.1+3.5=25$ thousand. Occurrence 3 has a positive amount eligible before the aggregate, but that does not create any payment once the remaining aggregate is zero.

What the Aggregate Cap Changes

The aggregate cap can reduce an occurrence payment without changing that occurrence’s underlying deductible and occurrence-limit calculation. In the examples, $C_j$ records what would be payable for the occurrence before considering the policy-period aggregate; $P_j$ records what is actually payable after considering remaining capacity. Keeping both amounts in the calculation makes it easier to identify which contract term caused a reduction.

The order of occurrences determines which occurrence is allocated the remaining capacity when the aggregate is insufficient to cover all eligible amounts. For example, if one occurrence is paid before another exhausts the aggregate, the later occurrence may receive only the remaining balance or nothing. The overall policy-period payment, when all occurrences are processed, is $\min(A,\sum_j C_j)$; changing the order does not change that total, but it can change how much is attributed to each occurrence and its claims.

Amount	What it represents	When it is calculated
$L_j$	Total gross loss in occurrence $j$	After grouping that occurrence’s claims
$C_j$	Occurrence payment before the aggregate cap	After the occurrence deductible and occurrence limit
$P_j$	Actual payment for occurrence $j$	After comparing $C_j$ with remaining aggregate capacity
$R_j$	Aggregate capacity remaining after occurrence $j$	After subtracting $P_j$ from $R_{j-1}$

Common Mistakes and Reconciliation Checks

Applying an occurrence deductible to every claim. If the deductible applies per occurrence, add the losses in that occurrence before applying it. Apply a separate deductible to each claim only if the contract specifies a per-claim deductible.
Using the aggregate as if it were an occurrence limit. The occurrence limit caps each occurrence’s eligible payment; the aggregate limit caps payments over the policy period.
Forgetting to update the remaining balance. After each payment, calculate $R_j=R_{j-1}-P_j$. Comparing every occurrence with the original aggregate can cause total payments to exceed the policy limit.
Allocating the eligible amount instead of the actual payment. When the aggregate truncates an occurrence, allocate $P_j$, not $C_j$, to its claims.
Assuming one allocation rule is universal. Proportional allocation is used here because it is stated in the examples. If a problem gives a different rule for allocating the occurrence payment, follow that rule.

Key takeaway: Group first, calculate the occurrence payment before the aggregate second, and apply the remaining aggregate third. Allocate only the actual occurrence payment to claims, and verify that claim allocations reconcile to occurrence payments and that total payments do not exceed the aggregate limit.

Check Your Understanding

For each problem, use the stated order and allocation convention. Amounts are in thousands of dollars.

A policy has a $4 thousand occurrence deductible and no binding occurrence or aggregate limit. One occurrence contains claims of $7 thousand and $9 thousand. Allocate the occurrence payment proportionally to gross loss. (Answer: occurrence payment $12 thousand; claim payments $5.25 thousand and $6.75 thousand.)
A policy has a $3 thousand occurrence deductible, an occurrence limit of $12 thousand, and an aggregate limit of $30 thousand. One occurrence contains claims of $5 thousand and $10 thousand; a second occurrence contains one $20 thousand claim. Allocate proportionally within an occurrence. Process the first occurrence before the second. (Answer: first occurrence payment $12 thousand, allocated $4 thousand and $8 thousand; second occurrence payment $12 thousand; total $24 thousand; aggregate remaining $6 thousand.)
A policy has a $2 thousand occurrence deductible, an occurrence limit of $9 thousand, and a $10 thousand aggregate. In order, the occurrences have gross losses of $6 thousand and $8 thousand. Find the payment for each occurrence and the final remaining aggregate. (Answer: payments $4 thousand and $6 thousand; remaining aggregate $0.)
A policy has a $5 thousand occurrence deductible and a $15 thousand occurrence limit, with a $16 thousand aggregate. The first occurrence has claims of $4 thousand and $11 thousand; the second has claims of $6 thousand and $9 thousand. Allocate proportionally within each occurrence, in that order. (Answer: first occurrence payment $10 thousand, allocated $2.6667 thousand and $7.3333 thousand; second occurrence payment $6 thousand, allocated $2.4 thousand and $3.6 thousand; total $16 thousand; remaining aggregate $0.)
A policy has a $3 thousand occurrence deductible, a $10 thousand occurrence limit, and a $12 thousand aggregate. There are two separate occurrences with gross losses of $12 thousand and $8 thousand. Calculate the payments first in the given order, then in reverse order. (Answer: total payment is $12 thousand in either order; in the given order the payments are $9 thousand and $3 thousand, while in reverse order they are $5 thousand and $7 thousand. The occurrence receiving each payment depends on order.)

Amount	What it represents	When it is calculated
\(L_j\)	Total gross loss in occurrence \(j\)	After grouping that occurrence’s claims
\(C_j\)	Occurrence payment before the aggregate cap	After the occurrence deductible and occurrence limit
\(P_j\)	Actual payment for occurrence \(j\)	After comparing \(C_j\) with remaining aggregate capacity
\(R_j\)	Aggregate capacity remaining after occurrence \(j\)	After subtracting \(P_j\) from \(R_{j-1}\)

What You'll Learn