Per-Claim Excess-of-Loss Layers

How a Per-Claim Layer Allocates a Loss

A per-claim excess-of-loss treaty protects an insurer against the larger amounts of individual claims. For each claim, the reinsurer pays only the part of the loss above the attachment point, subject to a finite recovery limit. The insurer retains the part below the attachment point and any amount above the layer’s maximum recovery.

The calculation is made claim by claim. It does not depend only on the portfolio’s total loss: splitting the same total amount among different claims can change the total recovery. The loss $X$ used here is the amount the insurer owes on one claim before applying this reinsurance treaty, after any relevant underlying policy terms.

Definition: Consider a per-claim excess-of-loss layer with attachment point $d$ and maximum recovery $u$, where $d \geq 0$ and $u > 0$. For a claim amount $X \geq 0$, the reinsurer’s payment is $$ C(X)=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ The insurer’s retained payment is $$ R(X)=X-C(X). $$ Here, $u$ means the maximum amount the reinsurer pays on one claim, not the upper endpoint of the original loss. The layer therefore covers the part of a claim between $d$ and $d+u$.

This convention makes the finite limit explicit: once a claim is large enough to use the full recovery $u$, the reinsurer pays no more on that claim. The insurer is still responsible for the excess above the layer. The payment allocation applies to the insurer and reinsurer; it does not by itself change the amount owed to the policyholder.

Piecewise Payments and Their Relationship

The compact formula can be read in three ranges. If the loss does not reach the attachment point, there is no recovery. If it enters the layer, the reinsurer pays the amount above $d$. If it passes the top of the layer, the recovery is capped at $u$.

C(X)= \begin{cases} 0, & 0 \leq X \leq d,\\ X-d, & d < X < d+u,\\ u, & X \geq d+u, \end{cases} \qquad R(X)= \begin{cases} X, & 0 \leq X \leq d,\\ d, & d < X < d+u,\\ X-u, & X \geq d+u. \end{cases}

At the boundary $X=d$, the recovery is zero. At $X=d+u$, the recovery has reached its cap $u$. The adjoining expressions agree at both boundaries, so the payment functions are well-defined there.

Why the Formula Works

The reinsurer should pay none of the loss below the attachment point. The amount above the attachment point is $(X-d)_+$: this is zero when $X \leq d$, and equals $X-d$ when $X>d$. But the treaty cannot pay more than $u$. Taking the smaller of the amount above attachment and $u$ enforces both conditions at once.

The retained payment follows from the allocation of the claim. For each claim, the original insurer liability is divided between the insurer and reinsurer, with no amount lost or counted twice. Thus

X=C(X)+R(X) \quad\Longrightarrow\quad R(X)=X-C(X).

For a claim inside the layer, $d<X<d+u$, the reinsurer pays $X-d$, leaving $R(X)=X-(X-d)=d$. For a claim above the layer, $X \geq d+u$, the reinsurer pays $u$, leaving $R(X)=X-u$. This explains why the insurer’s retained amount is not always capped at $d$: a sufficiently large claim penetrates beyond the finite layer.

Identify the treaty terms.
Record the attachment point $d$ and maximum recovery $u$ per claim. Confirm that $u$ is a recovery cap under the stated convention.

Apply the layer to one claim.
Calculate the amount above attachment, $(X-d)_+$, then cap it at $u$ to obtain $C(X)$.

Find the retained amount.
Subtract the recovery from the claim: $R(X)=X-C(X)$. Check that the two payments sum to $X$.

Repeat separately for every claim.
For a collection of claims, add the individual recoveries and retained amounts only after applying the treaty to each claim.

Worked Applications

The examples below use hypothetical claim amounts. They involve no interest or mortality assumptions: the allocation is made to claim amounts at the same valuation point, with no discounting.

Worked Example: Locate a Claim Within the Layer

A hypothetical insurer has a per-claim layer with attachment $d=\$25{,}000$ and maximum recovery $u=\$40{,}000$. Calculate the reinsurer’s payment and the insurer’s retained payment for claims of \$18,000, \$47,000, and \$92,000.

Method. For each claim, calculate $C(X)=\min((X-d)_+,u)$, then calculate $R(X)=X-C(X)$. The layer’s upper endpoint is $d+u=\$25{,}000+\$40{,}000=\$65{,}000$.

Claim of \$18,000. The amount above attachment is $(\$18{,}000-\$25{,}000)_+=\$0$. Therefore $C(X)=\min(\$0,\$40{,}000)=\$0$, and $R(X)=\$18{,}000-\$0=\$18{,}000$.

Claim of \$47,000. The amount above attachment is $\$47{,}000-\$25{,}000=\$22{,}000$. This is below the recovery cap, so $C(X)=\min(\$22{,}000,\$40{,}000)=\$22{,}000$. The retained amount is $R(X)=\$47{,}000-\$22{,}000=\$25{,}000$.

Claim of \$92,000. The amount above attachment is $\$92{,}000-\$25{,}000=\$67{,}000$. The recovery cap applies: $C(X)=\min(\$67{,}000,\$40{,}000)=\$40{,}000$. Hence $R(X)=\$92{,}000-\$40{,}000=\$52{,}000$.

Check. For the three claims, recovery plus retention equals the original amount: \$0+\$18,000=\$18,000; \$22,000+\$25,000=\$47,000; and \$40,000+\$52,000=\$92,000.

Common mistake avoided. Treating the layer as though it ended at a total claim amount of \$40,000 would misread $u$. Here, \$40,000 is the maximum recovery, so the layer extends from \$25,000 to \$65,000.

Worked Example: Apply the Treaty Separately to Each Claim

A hypothetical insurer has a per-claim layer with attachment $d=\$30{,}000$ and maximum recovery $u=\$25{,}000$. Four claims are \$12,000, \$38,000, \$51,000, and \$80,000. Find the recovery and retention for each claim, and then find the portfolio totals.

Method. The treaty is applied to each amount before summing. The layer extends from \$30,000 to $d+u=\$30{,}000+\$25{,}000=\$55{,}000$.

For the \$12,000 claim, the amount above attachment is $(\$12{,}000-\$30{,}000)_+=\$0$. Recovery is \$0 and retention is $\$12{,}000-\$0=\$12{,}000$.

For the \$38,000 claim, the amount above attachment is $\$38{,}000-\$30{,}000=\$8{,}000$. Recovery is $\min(\$8{,}000,\$25{,}000)=\$8{,}000$, and retention is $\$38{,}000-\$8{,}000=\$30{,}000$.

For the \$51,000 claim, the amount above attachment is $\$51{,}000-\$30{,}000=\$21{,}000$. Recovery is $\min(\$21{,}000,\$25{,}000)=\$21{,}000$, and retention is $\$51{,}000-\$21{,}000=\$30{,}000$.

For the \$80,000 claim, the amount above attachment is $\$80{,}000-\$30{,}000=\$50{,}000$. The cap applies, so recovery is $\min(\$50{,}000,\$25{,}000)=\$25{,}000$, and retention is $\$80{,}000-\$25{,}000=\$55{,}000$.

The gross total is $\$12{,}000+\$38{,}000+\$51{,}000+\$80{,}000=\$181{,}000$. Total recovery is $\$0+\$8{,}000+\$21{,}000+\$25{,}000=\$54{,}000$. Total retention is $\$12{,}000+\$30{,}000+\$30{,}000+\$55{,}000=\$127{,}000$.

Check. Total recovery plus total retention is $\$54{,}000+\$127{,}000=\$181{,}000$, equal to the gross total.

Common mistake avoided. Applying the treaty once to the \$181,000 portfolio total would produce a different result. This is a per-claim treaty, so its attachment and cap apply four times, once to each claim.

Worked Example: Convert a Layer Width to the Recovery Cap

A hypothetical reinsurance contract covers each claim from \$15,000 up to \$35,000. Interpret the layer as an attachment of $d=\$15{,}000$ and a layer width of \$20,000. For claims of \$14,500, \$28,000, \$36,000, and \$52,500, calculate the recovery and retention.

Method. The stated top of the layer is \$35,000, so the maximum recovery is the width above the \$15,000 attachment: $u=\$35{,}000-\$15{,}000=\$20{,}000$. Use $C(X)=\min((X-\$15{,}000)_+,\$20{,}000)$.

For $X=\$14{,}500$, the amount above attachment is \$0. Thus recovery is \$0 and retention is \$14,500.

For $X=\$28{,}000$, the amount above attachment is $\$28{,}000-\$15{,}000=\$13{,}000$. Recovery is $\min(\$13{,}000,\$20{,}000)=\$13{,}000$, and retention is $\$28{,}000-\$13{,}000=\$15{,}000$.

For $X=\$36{,}000$, the amount above attachment is $\$21{,}000$, which exceeds the \$20,000 recovery cap. Recovery is \$20,000 and retention is $\$36{,}000-\$20{,}000=\$16{,}000$.

For $X=\$52{,}500$, the amount above attachment is $\$37{,}500$. Recovery remains capped at \$20,000, so retention is $\$52{,}500-\$20{,}000=\$32{,}500$.

Check. The layer begins at \$15,000 and ends at \$35,000; its width and maximum recovery are both \$20,000. The recoveries are \$0, \$13,000, \$20,000, and \$20,000, while each retained amount plus its recovery equals its claim amount.

Common mistake avoided. The layer’s top, \$35,000, is not the maximum recovery. The recovery cap is the width of the layer, \$35,000 minus \$15,000, or \$20,000.

Exam Checks and Common Traps

A reliable solution keeps the treaty terms and the claim amount distinct. First identify whether the limit is stated as a maximum recovery or as the top of a layer. Then apply the attachment and limit to each claim, and verify the allocation identity. If a claim exceeds the layer, do not stop at the attachment: the insurer also retains the amount above the layer.

A frequent incorrect calculation is $C(X)=\min(X-d,u)$ without a positive-part operation. For example, with $X=\$18{,}000$, $d=\$25{,}000$, and $u=\$40{,}000$, this gives $\min(-\$7{,}000,\$40{,}000)=-\$7{,}000$, an impossible negative recovery. The correction is

C(X)=\min\bigl((X-d)_+,u\bigr) =\min(\max(X-d,0),u).

This makes recovery nonnegative and no greater than $u$. A second check is $0 \leq C(X) \leq \min(X,u)$ for nonnegative claims and attachment $d \geq 0$. These checks catch many arithmetic and interpretation errors.

Key takeaway: For each claim, the reinsurer pays the amount above attachment, capped at the maximum recovery: $C(X)=\min((X-d)_+,u)$. The insurer retains $R(X)=X-C(X)$. Apply both calculations separately to every claim in a per-claim treaty.

Check Your Understanding

For each problem, calculate the requested per-claim recoveries and retentions before checking the final answer.

A hypothetical treaty has $d=\$10{,}000$ and maximum recovery $u=\$18{,}000$. Find recovery and retention for claims of \$7,000, \$19,000, \$31,000, and \$40,000. (Answer: Recoveries \$0, \$9,000, \$18,000, and \$18,000; retentions \$7,000, \$10,000, \$13,000, and \$22,000.)
A per-claim treaty has attachment \$24,000 and maximum recovery \$16,000. Claims are \$20,000, \$35,000, \$46,000, and \$70,000. Find total recovery and total retention. (Answer: Total recovery \$43,000; total retention \$128,000.)
Three hypothetical claims are \$12,000, \$32,000, and \$40,000. A per-claim layer attaches at \$20,000 and has maximum recovery \$15,000. Find each recovery and retention and both totals. (Answer: Recoveries \$0, \$12,000, and \$15,000, total \$27,000; retentions \$12,000, \$20,000, and \$25,000, total \$57,000.)
A claim is \$63,000 under a treaty with attachment \$25,000 and maximum recovery \$30,000. Calculate the amount above attachment, the recovery, and the insurer’s retention. (Answer: Amount above attachment \$38,000; recovery \$30,000; retention \$33,000.)
Two claims are \$28,000 and \$42,000, and each is subject to attachment \$20,000 and maximum recovery \$15,000. Compare the total recovery and retention with those for one claim of \$70,000 under the same terms. (Answer: Two claims: total recovery \$23,000 and retention \$47,000; one claim: recovery \$15,000 and retention \$55,000.)

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