Per-Claim Excess-of-Loss Across a Portfolio

Why the Treaty Must Be Applied Claim by Claim

A per-claim excess-of-loss treaty responds separately to each covered claim. For a claim to generate a recovery, its loss must exceed the treaty’s attachment point. The reinsurer then pays the portion above that attachment, subject to the treaty’s per-claim recovery limit. When the claim count is specified, the portfolio calculation is to apply those terms to every claim and add the resulting payments.

The order matters. The attachment is applied to each loss, not to the portfolio’s sum, and the limit is applied separately to each claim’s recovery. Thus two portfolios with the same total loss can produce different per-claim recoveries if their losses are distributed differently among claims. The calculations below focus on the treaty’s payments for a specified set of claims; they do not require a frequency model or an assumption that claim amounts are independent.

Definition: Suppose a portfolio has $n$ claims, with nonnegative claim amounts $X_1,\ldots,X_n$. A per-claim excess-of-loss treaty has attachment $d$ and per-claim recovery limit $u$. The recovery on claim $i$ is $$ R_i=\min\bigl((X_i-d)_+,u\bigr), $$ where $(z)_+=\max(z,0)$. The insurer’s retained payment on that claim is $Z_i=X_i-R_i$. Total recovery and total retained payment are $R_{\mathrm{tot}}=\sum_{i=1}^n R_i$ and $Z_{\mathrm{tot}}=\sum_{i=1}^n Z_i$.

Here, $d$ is the amount the insurer bears before the treaty responds to a claim, and $u$ is the maximum recovery for that claim. The recovery limit is not an upper bound on the claim amount. If $u$ is finite, the treaty’s exhaustion point for a claim is $d+u$: once $X_i$ reaches or exceeds $d+u$, the recovery is $u$. An unlimited per-claim recovery is represented by allowing $u$ to be unbounded.

Some contract descriptions instead give a maximum covered loss amount $L$, measured from zero, rather than a maximum recovery. Under that wording, the covered layer from attachment $d$ to upper endpoint $L$ has recovery $\min((X_i-d)_+,L-d)$, assuming $L\ge d$. Be sure to identify which limit the problem specifies before calculating.

Calculating Portfolio Payments

For each claim, the recovery can be found by comparing the loss to two thresholds. A loss at or below $d$ produces no recovery. A loss above $d$ but below $d+u$ produces a recovery of $X_i-d$. A loss at or above $d+u$ produces the full recovery limit $u$. The retained payment is then the claim amount less that recovery.

Claim amount	Recovery $R_i$	Retained payment $Z_i$
$X_i\le d$	0	$X_i$
$d<X_i<d+u$	$X_i-d$	$d$
$X_i\ge d+u$	$u$	$X_i-u$

This table also shows why the retained payment is not always capped at $d$. Once a claim exceeds the exhaustion point $d+u$, the recovery cannot grow further, so the insurer retains the amount above the limit as well as the attachment.

Record the terms and claim amounts.
Confirm that $d$ is the per-claim attachment and $u$ is the per-claim recovery limit. Keep all amounts in the same units.

Calculate each recovery.
For every claim $i$, find $(X_i-d)_+$, then cap that amount at $u$.

Find each retained payment.
Subtract the claim’s recovery from its original amount: $Z_i=X_i-R_i$.

Sum and reconcile.
Add the claim-level recoveries and retentions separately. Check that $R_{\mathrm{tot}}+Z_{\mathrm{tot}}=\sum_i X_i$.

Worked Example: Five Claims With a Recovery Limit

A hypothetical insurer has five covered claims of $0$, $8$, $25$, $60$, and $110$ thousand dollars. A per-claim excess-of-loss treaty has an attachment of $d=20$ thousand dollars and a per-claim recovery limit of $u=50$ thousand dollars. Calculate the recovery and retained payment for each claim, and the portfolio totals.

Method. Apply $R_i=\min((X_i-d)_+,u)$ to each claim individually. The exhaustion point is $d+u=20+50=70$ thousand dollars.

For the $0$-thousand-dollar claim, the amount above the attachment is $\max(0-20,0)=0$, so the recovery is $0$ and the retained payment is $0-0=0$.

For the $8$-thousand-dollar claim, the amount above the attachment is $\max(8-20,0)=0$, so the recovery is $0$ and the retained payment is $8-0=8$.

For the $25$-thousand-dollar claim, the amount above the attachment is $25-20=5$. This is below the $50$-thousand-dollar recovery limit, so the recovery is $5$ and the retained payment is $25-5=20$.

For the $60$-thousand-dollar claim, the amount above the attachment is $60-20=40$, also below the limit. The recovery is $40$, and the retained payment is $60-40=20$.

For the $110$-thousand-dollar claim, the amount above the attachment is $110-20=90$, but the per-claim recovery is capped at $50$. The recovery is $50$, and the retained payment is $110-50=60$.

The total recovery is $0+0+5+40+50=95$ thousand dollars. The total retained payment is $0+8+20+20+60=108$ thousand dollars. The original claims total $0+8+25+60+110=203$ thousand dollars, and $95+108=203$, as required.

Common mistake avoided: Applying the attachment and limit once to the portfolio total would give $\min((203-20)_+,50)=50$ thousand dollars. That is not the recovery under a per-claim treaty: the treaty responds separately to each of the five claims.

Why the Portfolio Reconciliation Holds

The claim-level allocation has a simple accounting justification. For each claim, the insurer’s payment and the reinsurer’s payment together equal that claim’s original amount. Adding this identity over the portfolio proves the corresponding total identity. No assumption about independence or about how the claim amounts were generated is needed.

Z_i=X_i-R_i \quad\Longrightarrow\quad R_i+Z_i=X_i.

Summing both sides over the fixed claim count gives

\sum_{i=1}^{n}R_i+\sum_{i=1}^{n}Z_i = \sum_{i=1}^{n}(R_i+Z_i) = \sum_{i=1}^{n}X_i.

This is a useful check on arithmetic, but it does not by itself confirm that the treaty was applied correctly. An incorrect allocation can still add to the original loss. Check both the claim-level formula and the portfolio reconciliation.

Worked Example: Keep the Attachment and Limit Claim-Specific

A hypothetical equipment insurer has four covered claims of $12$, $35$, $58$, and $91$ thousand dollars. The treaty has an attachment of $25$ thousand dollars and a per-claim recovery limit of $40$ thousand dollars. Find the total recovery and total retained payment.

Method. The exhaustion point is $25+40=65$ thousand dollars. For each claim, calculate the loss above $25$, floored at zero, and then cap the recovery at $40$.

The $12$-thousand-dollar claim is below the attachment: $R_1=\min(0,40)=0$, so $Z_1=12-0=12$.

The $35$-thousand-dollar claim is $35-25=10$ thousand dollars above the attachment: $R_2=\min(10,40)=10$, so $Z_2=35-10=25$.

The $58$-thousand-dollar claim is $58-25=33$ thousand dollars above the attachment: $R_3=\min(33,40)=33$, so $Z_3=58-33=25$.

The $91$-thousand-dollar claim is $91-25=66$ thousand dollars above the attachment, which exceeds the recovery limit: $R_4=\min(66,40)=40$, so $Z_4=91-40=51$.

Total recovery is $0+10+33+40=83$ thousand dollars. Total retained payment is $12+25+25+51=113$ thousand dollars. The original claims total $12+35+58+91=196$ thousand dollars, and $83+113=196$.

Common mistake avoided: The $40$-thousand-dollar limit applies to each claim’s recovery. It is not one $40$-thousand-dollar cap shared across the four claims.

Worked Example: Many Claims at and Around the Exhaustion Point

A hypothetical portfolio has seven covered claims, in thousands of dollars: $4$, $10$, $17$, $25$, $33$, $48$, and $70$. A per-claim treaty has attachment $d=10$ thousand dollars and recovery limit $u=15$ thousand dollars. Calculate both portfolio totals, paying particular attention to claims at the attachment and exhaustion points.

Method. The exhaustion point is $10+15=25$ thousand dollars. A claim of exactly $10$ has no amount above the attachment; a claim of exactly $25$ has reached the full recovery limit.

The recoveries, in the same order as the claims, are $0$, $0$, $\min(17-10,15)=7$, $\min(25-10,15)=15$, $\min(33-10,15)=15$, $\min(48-10,15)=15$, and $\min(70-10,15)=15$. Their total is $0+0+7+15+15+15+15=67$ thousand dollars.

Subtracting claim by claim gives retained payments of $4$, $10$, $10$, $10$, $18$, $33$, and $55$ thousand dollars. Their total is $4+10+10+10+18+33+55=140$ thousand dollars.

The claims total $4+10+17+25+33+48+70=207$ thousand dollars, and $67+140=207$. The claim at the attachment generates no recovery, while the claim at exhaustion generates the full $15$-thousand-dollar recovery.

Common mistake avoided: Treating a claim exactly at $d$ as having a positive recovery. The amount above the attachment is $X_i-d=0$ at equality, so the recovery is zero.

Common Traps in Portfolio Calculations

The main exam challenge is usually not the formula itself, but applying the right terms at the right level. A reliable calculation keeps each claim’s amount, recovery, and retention together until the claim-level work is complete.

Applying the treaty to the sum of losses. A per-claim treaty is not an aggregate treaty. Calculate $R_i$ for each claim before summing.
Using the recovery limit as a cap on the claim amount. The limit caps $R_i$, not $X_i$. A claim can exceed $d+u$; the insurer then retains the amount beyond the capped recovery.
Forgetting the positive-part operation. If $X_i\le d$, the recovery is zero, not a negative amount.
Applying one limit to the whole portfolio. A per-claim limit resets for each covered claim unless the contract specifies an additional aggregate cap.
Mixing units. If losses are stated in thousands of dollars, convert the attachment and limit to thousands before comparing or adding.
Assuming a fixed claim count means equal claim amounts. The count tells you how many claim-level calculations to perform; it does not change the individual losses or treaty terms.

For questions involving random claim amounts, the same payment functions define the treaty allocation for each realized portfolio. But when the task is simply to allocate a specified set of claims, first complete the deterministic claim-by-claim calculation. Keep that task distinct from calculating an expectation or studying a random aggregate distribution.

Key takeaway: For a per-claim excess-of-loss treaty, calculate $R_i=\min((X_i-d)_+,u)$ separately for each claim, set $Z_i=X_i-R_i$, and only then sum. Confirm that total recovery plus total retention equals total original loss.

Check Your Understanding

For each problem, apply the treaty separately to each listed claim. Amounts are in thousands of dollars unless stated otherwise.

A hypothetical repair portfolio has claims of $9$, $20$, $34$, and $47$. The attachment is $15$ and the per-claim recovery limit is $20$. Find the total recovery and total retained payment. (Answer: total recovery $44$; total retained payment $66$.)
A hypothetical insurer has three claims of $18$, $42$, and $83$, with attachment $30$ and per-claim recovery limit $40$. Find each recovery and the portfolio total recovery. (Answer: recoveries $0,12,40$; total recovery $52$.)
A hypothetical equipment portfolio has claims of $0$, $7$, $12$, $19$, $31$, and $55$. The treaty has attachment $10$ and per-claim recovery limit $25$. Find total recovery and total retention. (Answer: total recovery $57$; total retention $67$.)
A hypothetical portfolio has five claims of $1{,}250$, $4{,}800$, $9{,}200$, $14{,}700$, and $23{,}000$ dollars. The attachment is $5{,}000$ dollars and the per-claim recovery limit is $8{,}000$ dollars. Find each recovery and the total recovery. (Answer: recoveries $\$0,\ \$0,\ \$4{,}200,\ \$8{,}000,\ \$8{,}000$; total recovery $\$20{,}200$.)
A hypothetical portfolio has claims of $14$, $28$, $51$, and $76$, with attachment $20$ and per-claim recovery limit $35$. Find the total recovery, total retention, and verify the original portfolio total. (Answer: total recovery $74$; total retention $95$; original total $169$.)

Claim amount	Recovery \(R_i\)	Retained payment \(Z_i\)
\(X_i\le d\)	0	\(X_i\)
\(d<X_i<d+u\)	\(X_i-d\)	\(d\)
\(X_i\ge d+u\)	\(u\)	\(X_i-u\)

What You'll Learn