Allocating Recoveries Across Excess-of-Loss Layers

Why Layer-by-Layer Allocation Matters

A large claim may pass through several excess-of-loss layers, each with its own attachment point and limit. The allocation is not necessarily continuous: layers may have gaps, and a loss may exceed the highest layer’s limit. To find the insurer’s final retained amount, calculate each layer’s recovery and then account for every part of the loss that no layer covers.

This tutorial considers a single ground-up loss applied to several per-occurrence excess-of-loss layers. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. Assume the stated loss is the amount subject to the layers, and ignore reinstatements, aggregate limits, coinsurance, and other reinsurance unless a problem explicitly includes them.

Definition: A layer described as $L$ excess of $A$, or $L$ xs $A$, has attachment point $A$ and limit $L$. It covers loss amounts above $A$ up to $A+L$. Thus, its covered interval on the loss-dollar scale is from $A$ to $A+L$. For a ground-up loss $X$, the layer’s recovery is the portion of that loss falling within this interval. A layer does not pay loss below its attachment or above its exhaustion point $A+L$.

For calculation, compare the loss with each layer’s attachment and exhaustion point. A layer pays nothing if the loss does not exceed its attachment; it pays its full limit if the loss reaches or exceeds its exhaustion point; and it pays only the part of the loss above attachment if the loss ends within the layer. In symbols, for a layer $L$ xs $A$, the recovery is:

R(A,L;X)=\max\bigl(0,\min(X,A+L)-A\bigr).

Use this calculation separately for each layer. The insurer’s retained amount is the ground-up loss less the sum of the recoveries. That retained amount can include loss below the lowest attachment, gaps between layers, and loss above the highest layer’s exhaustion point.

A Reliable Allocation Procedure

A useful way to prevent omissions is to mark every attachment and exhaustion point on the loss scale, then identify which intervals are covered and which remain with the insurer. In particular, do not assume that layers are contiguous merely because they appear in a list.

Write each layer’s endpoints.
For a layer $L$ xs $A$, calculate its exhaustion point $A+L$. Its covered interval runs from $A$ to $A+L$.

Calculate each recovery.
Compare the ground-up loss with that layer’s attachment and exhaustion point. Do not carry unused limit from one layer to another.

Reconcile the allocation.
Add the layer recoveries, then subtract their total from the ground-up loss. Alternatively, add the uninsured portions below, between, or above the layers.

Worked Allocations

Worked Example: A Gap Between Two Layers

A hypothetical insurer has a ground-up per-occurrence loss of $760,000. Its reinsurance consists of a $250,000 xs $100,000 layer and a $300,000 xs $400,000 layer. Calculate both recoveries and the insurer’s total retained amount. Assume the layers respond to the same loss and there are no other adjustments.

Method. Find the endpoints of both layers, calculate each recovery separately, and reconcile with the gross loss. This avoids treating the gap between the layers as covered.

First layer. The attachment is $100,000 and the limit is $250,000, so the exhaustion point is $100{,}000+250{,}000=350{,}000$. Since the loss of $760,000 exceeds $350,000, the layer pays its full $250,000 limit.

Second layer. The attachment is $400,000 and the limit is $300,000, so the exhaustion point is $400{,}000+300{,}000=700{,}000$. The loss exceeds $700,000, so this layer pays its full $300,000 limit.

Total recovery. Add the layer recoveries: $250{,}000+300{,}000=550{,}000$.

Retained amount. Subtract total recovery from the ground-up loss: $760{,}000-550{,}000=210{,}000$.

Check by intervals. The insurer retains $100,000 below the first attachment, $50,000 in the gap from $350,000 to $400,000, and $60,000 above the second layer’s exhaustion point. These amounts total $100{,}000+50{,}000+60{,}000=210{,}000$, agreeing with the subtraction check. The total allocation is $550{,}000+210{,}000=760{,}000$.

Worked Example: A Loss That Partly Enters the Highest Layer

A hypothetical equipment loss is $925,000. Three per-occurrence layers apply: $200,000 xs $150,000, $350,000 xs $350,000, and $250,000 xs $700,000. Calculate each recovery and the insurer-retained amount. Assume the listed attachments and limits are in dollars and there are no other terms affecting payment.

Method. Calculate each layer’s endpoints and compare them with the $925,000 loss. This makes clear that the highest layer is only partly used.

First layer. Its endpoints are $150,000 and $150{,}000+200{,}000=350{,}000$. The loss exceeds $350,000, so recovery is $200,000.

Second layer. Its endpoints are $350,000 and $350{,}000+350{,}000=700{,}000$. The loss exceeds $700,000, so recovery is $350,000.

Third layer. Its endpoints are $700,000 and $700{,}000+250{,}000=950{,}000$. The loss ends at $925,000, within this layer. Its recovery is $925{,}000-700{,}000=225{,}000$, which is less than the $250,000 limit.

Total recovery and retention. Total recovery is $200{,}000+350{,}000+225{,}000=775{,}000$. The insurer retains $925{,}000-775{,}000=150{,}000$, the loss below the first attachment.

Check. The covered intervals meet at $350,000 and $700,000, so there is no gap between them. Recovery plus retention equals $775{,}000+150{,}000=925{,}000$, the original loss.

Worked Example: Gaps and a Loss Below the Highest Exhaustion Point

A hypothetical per-occurrence loss is $1,480,000. The insurer has a $400,000 xs $200,000 layer, a $500,000 xs $700,000 layer, and a $300,000 xs $1,250,000 layer. Find each recovery and the total amount retained. Assume all amounts are ground-up dollars and the layers have no special sharing provisions.

Method. Map the layer intervals first, then allocate the loss through those intervals. This prevents the gaps from being mistaken for covered loss.

First layer. Its endpoints are $200,000 and $200{,}000+400{,}000=600{,}000$. The loss exceeds $600,000, so recovery is $400,000.

Second layer. Its endpoints are $700,000 and $700{,}000+500{,}000=1{,}200{,}000$. The loss exceeds $1,200,000, so recovery is $500,000.

Third layer. Its endpoints are $1,250,000 and $1{,}250{,}000+300{,}000=1{,}550{,}000$. The loss ends at $1,480,000, so recovery is $1{,}480{,}000-1{,}250{,}000=230{,}000$.

Total recovery and retention. Total recovery is $400{,}000+500{,}000+230{,}000=1{,}130{,}000$. The retained amount is $1{,}480{,}000-1{,}130{,}000=350{,}000$.

Check by intervals. The insurer retains $200,000 below the first attachment, $100,000 in the gap from $600,000 to $700,000, and $50,000 in the gap from $1,200,000 to $1,250,000. The loss does not exceed the highest exhaustion point of $1,550,000, so there is no retained amount above that point. The uncovered amounts total $200{,}000+100{,}000+50{,}000=350{,}000$, and $1{,}130{,}000+350{,}000=1{,}480{,}000$.

Why the Reconciliation Works

The recovery calculation assigns a loss dollar to a layer only when that dollar lies within the layer’s covered interval. Every other loss dollar remains with the insurer under the assumptions used here. Thus, the covered portions and the retained portions together partition the original ground-up loss; there is no loss dollar to count twice or omit.

X=\sum_j R_j+\text{retained amount}, \qquad \text{retained amount}=X-\sum_j R_j.

This identity is both a calculation method and a useful check. It applies when the recoveries are calculated on the same ground-up loss basis and the stated layers are the only sources of recovery. If additional terms change the amount eligible for a layer, those terms must be applied before using the reconciliation.

Common Mistakes and Practical Checks

A frequent error is to subtract an attachment from the insurer’s retained amount as though the attachment were a ceded payment. An attachment determines where the layer starts; it is not itself a recovery. For the $760,000 loss above, the incorrect calculation $760{,}000-250{,}000-300{,}000-100{,}000=110{,}000$ wrongly subtracts the first attachment as if it were paid by the reinsurers. The correct retained amount is $760{,}000-250{,}000-300{,}000=210{,}000$. The $100,000 below the first attachment is part of what the insurer retains, not an amount to remove from that retention.

Check the layer notation. In $L$ xs $A$, $A$ is the attachment and $L$ is the maximum payment. The exhaustion point is $A+L$.
Check for gaps. A layer attaching at $400,000 does not cover the interval from $350,000 to $400,000 simply because another layer ends at $350,000.
Check for partial use. If the loss ends within a layer, its recovery is less than the limit.
Check the highest layer. Loss above its exhaustion point remains with the insurer unless additional protection is specified.
Check the total. Under the stated assumptions, total recovery plus retained loss must equal the ground-up loss.

Key takeaway: Calculate every layer on its own attachment-to-exhaustion interval. Then reconcile the total recovery against the ground-up loss, remembering that retained loss can occur below, between, and above the layers.

Check Your Understanding

For each hypothetical loss, calculate the layer recoveries and the insurer-retained amount. Assume the stated layers apply to the same ground-up loss with no other adjustments.

A $430,000 loss is subject to a $150,000 xs $100,000 layer and a $200,000 xs $300,000 layer. Find both recoveries and retention. (Answer: $150,000 and $130,000 recoveries; $150,000 retained.)
A $1,050,000 loss is subject to a $250,000 xs $200,000 layer and a $400,000 xs $500,000 layer. Find both recoveries and retention, including the amounts in any gap or above a layer. (Answer: $250,000 and $400,000 recoveries; $400,000 retained.)
A $1,280,000 loss is subject to layers of $300,000 xs $100,000, $250,000 xs $450,000, and $500,000 xs $700,000. Find all recoveries and retention. (Answer: $300,000, $250,000, and $500,000 recoveries; $230,000 retained.)
A $2,400,000 loss is subject to layers of $600,000 xs $300,000, $700,000 xs $1,000,000, and $500,000 xs $1,800,000. Identify all uncovered intervals and calculate total recovery and retention. (Answer: $1,800,000 total recovery; $600,000 retained, comprising $300,000 below the first layer, two $100,000 gaps, and $100,000 above the highest layer.)
A $875,000 loss is subject to layers of $275,000 xs $125,000, $350,000 xs $425,000, and $200,000 xs $800,000. Calculate each recovery and retention; verify the total allocation. (Answer: $275,000, $350,000, and $75,000 recoveries; $175,000 retained.)

What You'll Learn