Distinguish Aggregate Stop-Loss from Per-Claim Reinsurance

Why the Treaty Basis Matters

An excess-of-loss treaty can respond to losses in two fundamentally different ways. A per-claim treaty applies its retention and limit separately to each claim. An aggregate stop-loss treaty first totals the covered losses over a defined period, then applies one retention to that annual total. The same set of claims can therefore produce different recoveries under the two arrangements.

The distinction also matters when claim frequency is uncertain. A group of claims concentrated in a few high-frequency years can have the same expected total loss as a group spread more evenly across years, yet produce a different expected aggregate stop-loss payment. The per-claim calculation, by contrast, depends on each claim amount and the expected number of claims.

All monetary values below are hypothetical. Unless stated otherwise, losses are measured in thousands of dollars, refer to the same annual period and valuation date, and are not discounted. Interest and mortality are not relevant to these property-loss examples.

Definition: Let $N$ be the number of covered claims in a period, let $X_j$ be the covered amount of claim $j$, and let $S_N=\sum_{j=1}^{N}X_j$ be the annual aggregate covered loss. With an annual aggregate attachment $A$ and no upper limit, the aggregate stop-loss payment is $(S_N-A)_+=\max(S_N-A,0)$. With a per-claim attachment $a$ and layer width $u$, the payment on claim $j$ is $\min((X_j-a)_+,u)$, and the total per-claim recovery is $\sum_{j=1}^{N}\min((X_j-a)_+,u)$.

An attachment is therefore applied once to the annual total in the aggregate contract, but separately to every claim in the per-claim contract. The layer width $u$ caps the per-claim payment; if the per-claim layer has no upper limit, its payment is simply $(X_j-a)_+$. An aggregate stop-loss with a cap would require an additional aggregate limit, which is not included in the definition above.

Calculating Payments for a Given Year

For a known claim list, there is no need to model frequency or severity. Calculate the aggregate stop-loss payment from the total first, and calculate the per-claim recovery by applying the layer formula to each claim before summing. These are different operations: in general, applying a positive-part function to a sum does not give the same result as summing positive-part functions applied to individual claims.

Worked Example: One Aggregate, Two Different Claim Patterns

Method: Calculate the two treaty recoveries separately, using the annual total for stop-loss and each claim amount for the per-claim layer. A hypothetical annual covered loss is either three claims of $40,000 each or one claim of $120,000. The aggregate stop-loss attaches at $100,000 with no cap. The per-claim treaty has a $50,000 attachment and no upper limit. Claim amounts are deterministic in each scenario; no interest or mortality assumption applies.

Three-claim pattern. The annual total is $40+40+40=120$ thousand dollars. The aggregate payment is $(120-100)_+=20$ thousand dollars. Each $40,000 claim is below the $50,000 per-claim attachment, so each per-claim payment is zero. Total per-claim recovery is $0+0+0=0$ thousand dollars.

One-claim pattern. The annual total is also 120 thousand dollars, so the aggregate payment remains $(120-100)_+=20$ thousand dollars. The per-claim recovery is $(120-50)_+=70$ thousand dollars.

Check and trap avoided. Both patterns have the same annual total, so their aggregate stop-loss payments agree. Their per-claim recoveries differ because splitting a total into claims changes how many times the per-claim attachment applies. The common mistake is to subtract the per-claim attachment only once from the annual total and call that the per-claim recovery; that would give $70,000 for the three-claim pattern, even though none of its individual claims exceeds $50,000.

This comparison also shows that neither recovery is automatically larger. In the three-claim pattern the aggregate treaty pays more; in the one-claim pattern the per-claim treaty pays more. A meaningful comparison must state both the attachment basis and the claim pattern.

Why Clustering Affects Expected Aggregate Cost

When claim amounts are independent and identically distributed, and are independent of $N$, the expected per-claim recovery has a simple frequency-severity decomposition. Write $g(X)=\min((X-a)_+,u)$. Conditional on $N=n$, the expected sum of $n$ per-claim payments is $nE[g(X)]$, since each claim has the same expected payment. Taking expectations over $N$ gives:

E\left[\sum_{j=1}^{N}g(X_j)\right] =E[N]E[g(X)].

This result requires a well-defined expected claim count and the stated independence and identical-distribution assumptions. It does not mean the same factorization applies to aggregate stop-loss. Aggregate stop-loss applies a nonlinear function to the total $S_N$; its expected payment depends on the distribution of that total, not only on its mean.

The positive-part function $h(s)=(s-A)_+$ is convex. Thus, when the mean aggregate loss is held fixed, greater variability in that total can increase expected stop-loss payment under the usual convex-order comparison. This observation is not a claim that every pair of loss distributions can be ranked by variance alone. The next example demonstrates the effect directly, without relying on a general ranking.

Worked Example: Same Expected Frequency, More Clustering

Method: Compare the expected payments under two frequency models. In both, every covered claim is exactly $60,000. In Model 1, $N=2$ with certainty. In Model 2, $N=0$ with probability $0.5$ and $N=4$ with probability $0.5$. Each model has expected frequency 2 and expected annual loss of $120,000. The per-claim treaty attaches at $50,000 and has a $20,000 width. The aggregate stop-loss attaches at $100,000 with no cap. The severity distribution is degenerate at $60,000; there is no interest or mortality assumption.

Per-claim payment per loss. For every claim, the payment is $\min((60-50)_+,20)=\min(10,20)=10$ thousand dollars.

Model 1 per-claim expected recovery. There are two claims, so the recovery is $2(10)=20$ thousand dollars. Because the count is fixed, this is also the expected recovery.

Model 2 per-claim expected recovery. With probability $0.5$, there are no claims and recovery is zero. With probability $0.5$, there are four claims and recovery is $4(10)=40$ thousand dollars. The expectation is $0.5(0)+0.5(40)=20$ thousand dollars. The expected per-claim recovery is unchanged.

Model 1 aggregate stop-loss. The annual total is $2(60)=120$ thousand dollars, so the payment is $(120-100)_+=20$ thousand dollars.

Model 2 aggregate stop-loss. With probability $0.5$, the total is zero and payment is zero. With probability $0.5$, the total is $4(60)=240$ thousand dollars and payment is $240-100=140$ thousand dollars. The expected payment is $0.5(0)+0.5(140)=70$ thousand dollars.

Check and trap avoided. Both models have expected frequency 2 and expected annual loss $120,000, yet expected aggregate recoveries are $20,000 and $70,000. The clustered model more often has no loss but sometimes has a much larger total that penetrates the stop-loss. Substituting expected annual loss into the stop-loss formula would give $(120-100)_+=20$ thousand dollars in both models; that is not the expected payment for Model 2.

Expected Payments with Random Claim Amounts

When severity is random, the aggregate total can vary even when the number of claims is fixed. A small discrete distribution makes it possible to calculate both treaty costs by enumerating all outcomes. The aggregate recovery depends on the sum in each outcome; the per-claim recovery depends on each claim separately.

Worked Example: Two Independent Claims with Two Possible Sizes

Method: Enumerate the joint severity outcomes and average their treaty payments. There are exactly two claims. Each claim independently equals $40,000 or $100,000, each with probability $0.5$. The per-claim layer attaches at $60,000 and has width $40,000. The aggregate stop-loss attaches at $120,000 with no cap. Thus the severity distribution has two point masses, each of probability $0.5$; no interest or mortality assumption applies.

Per-claim payments. A $40,000 claim pays $\min((40-60)_+,40)=0$ thousand dollars. A $100,000 claim pays $\min((100-60)_+,40)=40$ thousand dollars.

Joint outcomes. Independence gives probability $0.25$ to each ordered pair. For claims of $(40,40)$, the total is 80, the aggregate payment is zero, and the per-claim recovery is zero. For $(40,100)$, the total is 140, the aggregate payment is $140-120=20$, and the per-claim recovery is $0+40=40$. For $(100,40)$, the corresponding payments are again 20 and 40. For $(100,100)$, the total is 200, the aggregate payment is $200-120=80$, and the per-claim recovery is $40+40=80$. All amounts in these calculations are thousands of dollars.

Expected aggregate stop-loss payment. The expectation is $0.25(0)+0.25(20)+0.25(20)+0.25(80)=30$ thousand dollars.

Expected per-claim recovery. The expectation is $0.25(0)+0.25(40)+0.25(40)+0.25(80)=40$ thousand dollars.

Check and trap avoided. The expected aggregate stop-loss payment is $30,000, while the expected per-claim recovery is $40,000. Applying the aggregate attachment separately to each claim would be incorrect: the $120,000 attachment belongs to the annual total, not to either individual loss.

A Reliable Comparison Process

Before comparing costs, confirm that both calculations use the same covered claim amounts and period. Then keep the treaty bases separate throughout the calculation:

Specify the loss basis.
Identify which claim amounts are included in the treaty calculation and the period over which they are aggregated. Do not add a deductible or coverage modification unless the stated amounts require it.

Calculate per-claim recoveries.
For each claim, apply the per-claim attachment and layer width. Sum those individual payments to obtain the period recovery.

Calculate aggregate stop-loss recovery.
Sum the covered losses for the period, subtract the aggregate attachment once, and floor the payment at zero. Apply an aggregate cap only if the contract specifies one.

Average only after calculating outcomes.
For a stochastic model, calculate treaty payment for each outcome or condition on the claim count, then take the expectation. Do not generally apply a stop-loss formula to the expected aggregate and treat the result as an expected payment.

Common Exam Trap

The most frequent error is treating “excess of loss” as if it always described one calculation. The phrase alone does not tell you whether the attachment applies per claim or to the annual aggregate. Read the contract wording for the unit of attachment, then match the formula to that unit.

Key takeaway: Per-claim excess-of-loss applies a layer to each covered claim and then sums the recoveries. Aggregate stop-loss sums covered claims first and applies one annual attachment. Claim fragmentation can change per-claim cost, while frequency clustering can change expected aggregate stop-loss cost even when expected annual loss is unchanged.

Check Your Understanding

For each problem, calculate the requested payments using the stated attachment basis. Amounts are hypothetical and in thousands of dollars unless specified otherwise.

An annual period has claims of 30, 80, and 140. A per-claim layer attaches at 50 and has width 60; an aggregate stop-loss attaches at 180 with no cap. Find both total recoveries. (Answer: per-claim recovery 90; aggregate stop-loss recovery 70.)
Each claim is exactly 40. The annual claim count is 0 or 3, each with probability 0.5. A per-claim layer attaches at 25 and has width 20; an aggregate stop-loss attaches at 100 with no cap. Find the expected recovery under each treaty. (Answer: expected per-claim recovery 22.5; expected aggregate stop-loss recovery 10.)
There are exactly two independent claims. Each is 50 or 90 with probability 0.5 for each amount. A per-claim layer attaches at 40 and has width 30; an aggregate stop-loss attaches at 100 with no cap. Find both expected recoveries. (Answer: expected per-claim recovery 40; expected aggregate stop-loss recovery 40.)
There are exactly five claims, each of 25. A per-claim treaty attaches at 30 with no upper limit; an aggregate stop-loss attaches at 100 with no cap. Find both payments. (Answer: per-claim recovery 0; aggregate stop-loss recovery 25.)

What You'll Learn