Per-Claim Excess Layers Versus Aggregate Stop-Loss

Why the Attachment Level Matters

Reinsurance can respond to losses at two different levels. A per-claim excess-of-loss contract applies its attachment separately to each claim. An aggregate stop-loss contract applies its attachment once to the total loss in a period. Even with the same claims, these rules can produce different recoveries because the timing of aggregation changes which losses reach the attachment.

The calculations in this tutorial use a hypothetical portfolio. Unless a problem states otherwise, claim severities are nonnegative and independent of the claim count. A recovery is never negative: if losses do not exceed an attachment, the payment is zero. The examples compare recoveries only; they do not calculate reinsurance premiums or expenses.

Definition: Let $N$ be the number of claims and let $X_1,\ldots,X_N$ be their ground-up severities. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ if $N=0$. For attachment $d$, an unlimited per-claim excess-of-loss layer pays $(X_j-d)_+=\max(X_j-d,0)$ on claim $j$. Its total recovery is $R_{\mathrm{pc}}=\sum_{j=1}^{N}(X_j-d)_+$. An aggregate stop-loss contract with attachment $D$ pays $R_{\mathrm{agg}}=(S-D)_+=\max(S-D,0)$ once on the period's aggregate loss.

The letters $d$ and $D$ distinguish the per-claim attachment from the aggregate attachment. They need not be equal. An attachment is the amount retained before the layer begins to pay; it is not itself the maximum payment. A per-claim layer with a payment limit $u$ instead pays $\min((X_j-d)_+,u)$ on claim $j$. The aggregate stop-loss definition above has no upper payment limit.

Two Different Calculation Chains

For per-claim cover, first determine the recovery on each claim, then add those recoveries. Under the standard compound model in which $N$ is independent of identically distributed severities $X_j$, linearity of expectation gives

E[R_{\mathrm{pc}}]=E[N]\,E[(X-d)_+].

This formula does not require the claim severities to be independent of one another, but it does require the stated common severity distribution and independence of $N$ from the severities. If claim amounts differ by claim, or frequency and severity are dependent, calculate the expected sum using the actual model instead.

For aggregate stop-loss, form the total $S$ first and apply the attachment once. In a discrete model, a direct calculation is

E[R_{\mathrm{agg}}]=\sum_s (s-D)_+\,P(S=s).

This calculation needs the distribution of the aggregate loss, not just the expected per-claim recovery. If the distribution of $S$ has many possible values, an aggregate-loss recursion or another appropriate method may be used; the attachment still applies to $S$ after the aggregate distribution has been determined.

Key distinction: Per-claim cover applies an attachment to every $X_j$, then sums payments. Aggregate stop-loss sums the $X_j$ first, then applies one attachment to $S$. The two operations generally do not commute.

Worked Comparisons

Worked Example: Two Claims With Different Attachments

A hypothetical portfolio has exactly two independent claims. Each claim is \$3 or \$9, each with probability $0.5$. Compare (a) unlimited per-claim excess cover with attachment $d=\$4$ and (b) aggregate stop-loss with attachment $D=\$10$. Find the expected total recovery under each contract.

Method. For the per-claim contract, calculate each claim's payment and use linearity of expectation. For aggregate stop-loss, enumerate the possible sums of the two claims. The contracts have different attachments, so the comparison is between the stated structures, not equal attachment amounts.

Per-claim recovery. A \$3 claim produces $(3-4)_+=0$. A \$9 claim produces $(9-4)_+=5$. Thus the expected recovery per claim is

$$ 0.5(0)+0.5(5)=2.5. $$

There are exactly two claims, so the expected total per-claim recovery is $2(2.5)=\$5.00$.

Aggregate recovery. The total loss is \$6 with probability $0.5(0.5)=0.25$, \$12 with probability $2(0.5)(0.5)=0.50$, and \$18 with probability $0.25$. Their aggregate recoveries are $(6-10)_+=0$, $(12-10)_+=2$, and $(18-10)_+=8$. Therefore

E[R_{\mathrm{agg}}]=0.25(0)+0.50(2)+0.25(8)=0+1+2=\$3.00.

Check and trap avoided. The per-claim expectation is \$5.00 and the aggregate expectation is \$3.00. It would be wrong to apply the aggregate attachment separately to both claims: that would change the contract being valued.

Worked Example: Same Attachment Applied at Two Levels

A hypothetical portfolio again has exactly two independent claims, but each claim is now \$2 or \$8 with probability $0.5$ each. Find the expected total recovery for unlimited per-claim cover with attachment $d=\$5$, and for aggregate stop-loss with attachment $D=\$5$.

Per-claim method. The payment on a \$2 claim is $(2-5)_+=0$; on a \$8 claim it is $(8-5)_+=3$. The expected payment per claim is $0.5(0)+0.5(3)=\$1.50$. For two claims, the expected total is $2(1.50)=\$3.00$.

Aggregate method. The possible total losses and probabilities are \$4 with probability $0.25$, \$10 with probability $0.50$, and \$16 with probability $0.25$. The corresponding stop-loss payments are $0$, $5$, and $11$. Thus

E[R_{\mathrm{agg}}]=0.25(0)+0.50(5)+0.25(11)=0+2.50+2.75=\$5.25.

Check and trap avoided. With the same \$5 attachment, aggregate recovery has expectation \$5.25, greater than the \$3.00 per-claim expectation in this example. A claim below the attachment can still contribute to an aggregate recovery by adding to other claims; it cannot generate a per-claim recovery on its own.

Worked Example: Random Claim Count and Aggregate Attachment

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$. Given a claim occurs, its severity is exactly \$6, independently of the claim count. Find expected total recovery for per-claim excess cover with $d=\$4$ and for aggregate stop-loss with $D=\$8$.

Per-claim method. Each claim produces $(6-4)_+=\$2$. The expected claim count is $E[N]=2(0.5)=1$. Therefore

E[R_{\mathrm{pc}}]=E[N](2)=1(2)=\$2.00.

Aggregate method. The claim count is $0,1,$ or $2$, with probabilities $0.25,0.50,$ and $0.25$, respectively. Aggregate losses are \$0, \$6, and \$12. Applying the \$8 attachment once gives recoveries $(0-8)_+=0$, $(6-8)_+=0$, and $(12-8)_+=4$. Hence

E[R_{\mathrm{agg}}]=0.25(0)+0.50(0)+0.25(4)=\$1.00.

Check and trap avoided. The aggregate expected recovery is \$1.00, not \$2.00. The \$2.00 result belongs to the per-claim contract, which collects \$2 for each claim that occurs. The aggregate contract does not attach until total loss exceeds \$8.

Why Aggregation Changes Recovery

For a fixed set of nonnegative claim amounts and a nonnegative attachment $d$, compare an unlimited per-claim layer and an aggregate stop-loss with that same attachment. The per-claim recovery is at most the aggregate recovery:

\sum_{j=1}^{n}(x_j-d)_+\ \leq\ \left(\sum_{j=1}^{n}x_j-d\right)_+.

To see why, let $I$ be the claims with $x_j>d$, and let $k$ be the number of such claims. If $k=0$, the left side is zero, so the inequality holds. If $k\geq1$, the left side is $\sum_{j\in I}x_j-kd$. The right side is at least $\sum_{j\in I}x_j-d$, because the omitted claim amounts are nonnegative and the positive-part operation cannot reduce a nonnegative value. Since $kd\geq d$, we have

\sum_{j\in I}x_j-kd\ \leq\ \sum_{j\in I}x_j-d \ \leq\ \left(\sum_{j=1}^{n}x_j-d\right)_+.

Thus, at the same attachment, applying the attachment once to total loss cannot yield less than summing unlimited per-claim recoveries. This is a payment-by-payment statement for the same portfolio, so it also implies the corresponding expectation inequality when the two contracts are evaluated on the same random claims. It does not rank contracts with different attachment amounts, or contracts where one layer has a payment limit.

Limits, Negative Differences, and Exam Traps

A per-claim limit caps each individual payment. For attachment $d$ and payment limit $u$, the payment is $\min((X_j-d)_+,u)$. For example, if $d=\$6$, $u=\$3$, and a claim is \$10, the uncapped excess is $10-6=\$4$, but the recovery is capped at $\min(4,3)=\$3$. Do not confuse the payment limit $u$ with an upper bound on the claim amount.

A second frequent error is to treat a negative difference as a negative insurance payment. If two claims total \$4 and the aggregate attachment is \$10, then the signed difference is $4-10=-\$6$, but the recovery is

(4-10)_+=\max(-6,0)=0,

not $-\$6$. Likewise, a calculation such as $(4+4-10)_+$ gives $\max(-2,0)=0$. The positive-part operation is essential: the insurer or reinsurer does not pay a negative amount when losses fall below the attachment.

A reliable calculation sequence is to identify the contract level, write the payment formula, and only then calculate expectations. For per-claim cover, transform each claim before summing. For aggregate stop-loss, find or enumerate aggregate losses before applying the attachment. Keep the attachment and any payment limit in the units specified by the problem, and state whether the reported result is per claim or total for the portfolio.

Key takeaway: A per-claim excess layer and an aggregate stop-loss contract are different functions of the same claim portfolio. Their expected recoveries cannot generally be equated from the attachment amount alone; the claim count, severity distribution, and order of aggregation all matter.

Check Your Understanding

For each hypothetical portfolio, calculate the requested recoveries using the stated contract structure. Unless stated otherwise, claim amounts are independent.

There are exactly two claims, each \$1 or \$7 with probability $0.5$. Find the expected total recovery for unlimited per-claim cover with attachment $d=\$3$, and for aggregate stop-loss with attachment $D=\$8$. (Answer: per-claim \$4.00; aggregate \$1.50)
There are exactly two claims, each \$2 or \$8 with probability $0.5$. Find the expected total recovery for unlimited per-claim cover and aggregate stop-loss, both with attachment \$5. (Answer: per-claim \$3.00; aggregate \$5.25)
The count is $N\sim\operatorname{Binomial}(3,1/3)$, and each claim is exactly \$9. Find expected total recovery for per-claim cover with attachment \$6 and aggregate stop-loss with attachment \$15. (Answer: per-claim \$3.00; aggregate \$10/9, or approximately \$1.11)
There are exactly three claims, each \$4 or \$10 with probability $0.5$. Per-claim cover attaches at \$6 and has a \$3 payment limit per claim. Aggregate stop-loss attaches at \$15. Find each expected total recovery. (Answer: per-claim \$4.50; aggregate \$6.38, rounded to cents)
The count is $N\sim\operatorname{Binomial}(2,0.5)$, and each claim is exactly \$5. Find expected total recovery for per-claim cover with attachment \$2 and payment limit \$2, and aggregate stop-loss with attachment \$7. (Answer: per-claim \$2.00; aggregate \$0.75)

What You'll Learn