Multiple Excess-of-Loss Layers

Introduction

A loss that exceeds the attachment point of one excess-of-loss layer may continue into a second layer, then a third. Each layer covers a specified band of loss, so calculating the total recovery requires allocating the loss across those bands without paying the same dollars twice. The insurer may retain losses below the first attachment point and amounts above the highest layer.

The essential question is what amount is being run through the layers? For per-claim cover, apply the layers to each claim separately. For cover written on an aggregate basis, apply them to the aggregate amount specified by the treaty. These two bases can produce different recoveries even when the underlying claims are the same.

Definition: A layer with attachment point $a$ and limit $u$ covers the portion of a loss above $a$, up to a maximum payment of $u$. Its recovery on a loss amount $x$ is $\min((x-a)_+,u)$, where $(z)_+=\max(z,0)$. Its recovery cannot be negative or exceed the layer limit.

Allocating a Loss Across Consecutive Layers

Suppose the layers are ordered from lowest to highest attachment point. Write layer $j$'s attachment point as $a_j$ and its limit as $u_j$. This layer covers the interval from $a_j$ up to $a_j+u_j$. For consecutive layers, the next layer attaches exactly where the preceding layer exhausts: $a_{j+1}=a_j+u_j$. Thus, the covered bands meet but do not overlap.

For a loss amount $x$, calculate the recovery in each layer separately, then add the recoveries. If the layers do not overlap, the insurer’s amount remaining is the original loss less the sum of those recoveries. This remaining amount includes any loss below the lowest attachment point and any loss above the top of the highest layer.

R_j(x)=\min\bigl((x-a_j)_+,u_j\bigr), \qquad R_{\text{total}}(x)=\sum_{j=1}^{m}R_j(x), \qquad I(x)=x-R_{\text{total}}(x).

Here $R_j(x)$ is the recovery from layer $j$, $R_{\text{total}}(x)$ is the total recovery, and $I(x)$ is the insurer’s remaining amount. These formulas assume that each stated layer applies to the same loss amount $x$, that its payment is capped at its limit, and that recoveries from distinct layers are not counted more than once. For a layer with a gap between it and the next layer, the same individual-layer formula applies, but the gap remains uninsured.

Why the formula works: A layer covers dollars in the band from $a$ to $a+u$. If $x\leq a$, the loss does not reach the band and recovery is zero. If $a<x<a+u$, the loss enters the band by $x-a$, so the recovery is $x-a$. If $x\geq a+u$, the entire band is exhausted and recovery is $u$. These three cases are exactly $\min((x-a)_+,u)$.

For consecutive layers with a common lower attachment $a_1$, the total covered band runs from $a_1$ to the top of the last layer. The total recovery can then also be calculated as the amount of the loss within that combined band. This shortcut is useful as a check, but it does not replace calculating the individual layer recoveries when those amounts are required.

Worked Example: A Claim Enters Three Consecutive Layers

A hypothetical policy has a single claim of $385,000. Its per-claim reinsurance consists of three consecutive layers: $50,000 excess of $100,000; $150,000 excess of $150,000; and $100,000 excess of $300,000. Find the recovery from each layer and the insurer’s remaining amount. No interest, mortality, or severity-distribution assumption is involved: treat the claim amount as fixed at $385,000.

Method. Interpret “$u$ excess of $a$” as a layer attaching at $a$ with limit $u$. The first layer covers $100,000 to $150,000; the second covers $150,000 to $300,000; and the third covers $300,000 to $400,000. The attachment points confirm that the layers are consecutive.

First layer. The claim exceeds the $100,000 attachment by $285,000, but the layer limit is $50,000. Its recovery is $\min(\$385{,}000-\$100{,}000,\$50{,}000)=\$50{,}000$.

Second layer. The claim exceeds the $150,000 attachment by $235,000. Capping this amount at the $150,000 limit gives a recovery of $\min(\$235{,}000,\$150{,}000)=\$150{,}000$.

Third layer. The claim exceeds the $300,000 attachment by $85,000, which is below the $100,000 limit. The recovery is $\min(\$85{,}000,\$100{,}000)=\$85{,}000$.

Total and check. Total recovery is $\$50{,}000+\$150{,}000+\$85{,}000=\$285{,}000$. The insurer’s remaining amount is $\$385{,}000-\$285{,}000=\$100{,}000$. This is exactly the loss below the first attachment, so the calculation reconciles.

Common mistake avoided. The third layer’s recovery is not its full $100,000 limit: the claim penetrates that layer by only $85,000. Paying the full limit would overstate recovery by $15,000.

Layer Recoveries on Aggregate Amounts

The same layer formula can be used when the treaty applies to an aggregate amount rather than to an individual claim. In that case, $x$ denotes the specified aggregate loss, and each layer is applied to that amount. Do not first calculate per-claim recoveries unless the treaty is actually on a per-claim basis.

This distinction matters because adding claims together before applying the layers can change where the combined amount falls relative to each attachment point. The treaty’s stated basis determines the amount to which the formula applies; the arithmetic cannot decide that basis for you.

Worked Example: Aggregate Loss Through Three Layers

A hypothetical reinsurer covers an aggregate loss of $530,000 under three consecutive aggregate layers: $100,000 excess of $0; $150,000 excess of $100,000; and $200,000 excess of $250,000. Calculate each aggregate recovery and the remaining amount. The aggregate is treated as a fixed amount; there is no interest, mortality, or severity-distribution assumption.

Method. Because the treaty is explicitly aggregate, apply the layers to $x=\$530{,}000$, not separately to the underlying claims. The covered bands are $0 to $100,000, $100,000 to $250,000, and $250,000 to $450,000.

First layer. The amount above its $0 attachment is $530,000, which exceeds the $100,000 limit. Recovery is $\min(\$530{,}000,\$100{,}000)=\$100{,}000$.

Second layer. The amount above its $100,000 attachment is $430,000, which exceeds the $150,000 limit. Recovery is $\min(\$430{,}000,\$150{,}000)=\$150{,}000$.

Third layer. The amount above its $250,000 attachment is $280,000, which exceeds the $200,000 limit. Recovery is $\min(\$280{,}000,\$200{,}000)=\$200{,}000$.

Total and check. Total recovery is $\$100{,}000+\$150{,}000+\$200{,}000=\$450{,}000$. The remaining amount is $\$530{,}000-\$450{,}000=\$80{,}000$. The layers cover the full band from $0 to $450,000, so the $80,000 remaining is exactly the aggregate amount above the top of the last layer.

Common mistake avoided. The amount above an attachment is not always the recovery. For example, $280,000 lies above the third layer’s $250,000 attachment, but that layer pays only its $200,000 limit.

Per-Claim Versus Aggregate Application

When the treaty is per claim, apply every layer to each claim, then sum the recoveries for each layer and across all layers. When the treaty is aggregate, sum the losses into the specified aggregate amount first, then apply the layers once to that amount. A useful check is to reconcile original loss to total recovery plus amount remaining on the same basis.

Identify the treaty basis.
Determine whether the covered amount is an individual claim or a specified aggregate. This determines what value is used as $x$.

Write each covered band.
For a layer with attachment $a_j$ and limit $u_j$, its band runs from $a_j$ to $a_j+u_j$. Check whether successive bands are consecutive or whether a gap exists.

Calculate and cap each recovery.
Use $R_j(x)=\min((x-a_j)_+,u_j)$. A layer pays zero below attachment, the penetrated amount within its band, and its full limit once exhausted.

Reconcile the allocation.
Add the layer recoveries and subtract them from the covered loss amount. Confirm that retained amount plus total recovery equals the original amount.

Worked Example: Two Claims Under Per-Claim Layers

A hypothetical insurer has two claims, $90,000 and $230,000. Each claim is covered separately by consecutive per-claim layers of $100,000 excess of $0 and $150,000 excess of $100,000. Find the recovery by layer, total recovery, and insurer retention. Treat both claims as fixed amounts; no interest, mortality, or severity-distribution assumption is used.

Method. Apply the two layer formulas separately to each claim. The first layer covers $0 to $100,000, and the second covers $100,000 to $250,000.

$90,000 claim. The first layer pays $\min(\$90{,}000,\$100{,}000)=\$90{,}000$. The second layer pays $\min((\$90{,}000-\$100{,}000)_+,\$150{,}000)=\$0$. Retention on this claim is $\$90{,}000-\$90{,}000=\$0$.

$230,000 claim. The first layer pays its full $100,000 limit. The amount above the second layer’s $100,000 attachment is $130,000, so the second layer pays $\min(\$130{,}000,\$150{,}000)=\$130{,}000$. Retention on this claim is $\$230{,}000-\$100{,}000-\$130{,}000=\$0$.

Totals and check. The first layer pays $\$90{,}000+\$100{,}000=\$190{,}000$, and the second pays $\$0+\$130{,}000=\$130{,}000$. Total recovery is $320,000, equal to the total claims $\$90{,}000+\$230{,}000=\$320{,}000$. The insurer retains $0.

Common mistake avoided. Do not treat the $320,000 sum as a single loss under these per-claim layers. Doing so would give recoveries of $100,000 and $150,000, for total recovery $250,000 and retention $70,000. That is a different calculation on a different treaty basis.

Common Errors and Exam Checks

Layer calculations are straightforward when the bands and the basis are explicit, but small interpretation errors can affect several recoveries at once. In particular, distinguish a layer’s attachment point from its limit: the attachment determines where payment begins, while the limit determines the maximum payment. For a layer $u$ excess of $a$, the top of the layer is $a+u$, not $u$.

Paying the limit whenever a loss reaches a layer. A loss that only partly penetrates a layer receives only that penetrated amount, unless it exceeds the layer’s limit.
Adding uncapped amounts above attachments. Each recovery must be capped at its own limit; otherwise the same dollars can be over-allocated.
Forgetting retained amounts. Amounts below the lowest attachment and above the top layer are not recovered by these layers.
Changing the basis mid-calculation. Apply per-claim layers to each claim or aggregate layers to the specified aggregate. Do not use the other basis as a shortcut.
Failing to reconcile. On the basis specified by the treaty, total recovery plus remaining amount must equal the original covered loss amount.

Key takeaway: For each layer, calculate the amount above attachment and cap it at the layer limit. Sum the layer payments, then reconcile against the loss amount to find what remains with the insurer. Always apply the layers to the claim or aggregate specified by the treaty.

Check Your Understanding

For each question, identify the treaty basis before calculating. Amounts are hypothetical and fixed; no interest, mortality, or severity-distribution assumption is required.

A per-claim loss is $92,000. Consecutive layers cover $0 to $40,000, $40,000 to $70,000, and $70,000 to $120,000. Find the recovery from each layer and the insurer’s retention. (Answer: $40,000, $30,000, and $22,000 recovered; $0 retained.)
A per-claim loss is $275,000. The layers are $75,000 excess of $25,000, $100,000 excess of $100,000, and $50,000 excess of $200,000. Find total recovery and retention, including amounts below the first attachment and above the final layer. (Answer: $225,000 recovered; $50,000 retained.)
A per-claim loss is $165,000 and is covered by three consecutive layers: $60,000 excess of $20,000, $60,000 excess of $80,000, and $60,000 excess of $140,000. Calculate each layer’s recovery and the retained amount. (Answer: $60,000, $60,000, and $25,000 recovered; $20,000 retained.)
Two claims of $90,000 and $230,000 are covered by per-claim layers of $100,000 excess of $0 and $150,000 excess of $100,000. Find total recovery by layer and total retention. Then state the recovery and retention if, instead, the same two claims were combined and the layers applied once to the aggregate. (Answer: Per claim, $190,000 and $130,000 recovered, with $0 retained; on the $320,000 aggregate, $100,000 and $150,000 recovered, with $70,000 retained.)
A hypothetical aggregate loss of $680,000 is subject to consecutive aggregate layers of $120,000 excess of $0, $180,000 excess of $120,000, and $250,000 excess of $300,000. Find each recovery and the aggregate amount remaining. (Answer: $120,000, $180,000, and $250,000 recovered; $130,000 remaining.)

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