Derive Per-Claim Excess-of-Loss Layer Distributions

Introduction

A per-claim excess-of-loss treaty divides each covered claim between the insurer and reinsurer according to the claim’s position relative to an attachment and a layer limit. The key is to apply the treaty to each claim amount, not to an aggregate of claims. The resulting payment distributions can contain atoms even when the original claim amount is continuous: some claims produce zero reinsurance payment, and a whole interval of claim amounts can produce the same insurer retention.

Let $X$ be the nonnegative covered amount of one claim before this treaty is applied. Assume an attachment $a\geq 0$ and a layer limit $L>0$. The layer covers the part of a claim above $a$, up to a maximum payment of $L$. No interest or mortality assumption is needed for this per-claim allocation.

Definition: The reinsurer’s payment per loss and the insurer’s retained payment per loss are $$ Y_R^L=\min\bigl((X-a)_+,L\bigr),\qquad Y_I^L=X-Y_R^L, $$ where $(z)_+=\max(z,0)$. Thus $Y_R^L$ is the amount in the layer, while $Y_I^L$ is the amount left with the insurer after this treaty. The attachment is $a$; the layer is exhausted at claim amount $a+L$, not at $L$.

Equivalently, the allocation can be written by ranges of $X$:

(Y_R^L,Y_I^L)= \begin{cases} (0,X), & X\leq a,\\ (X-a,a), & a<X\leq a+L,\\ (L,X-L), & X>a+L. \end{cases}

At the attachment, the reinsurer pays zero. At the exhaustion point $X=a+L$, the reinsurer pays exactly $L$ and the insurer retains exactly $a$. These endpoint conventions matter when $X$ has point masses.

Deriving the Payment Distributions

Write $F_X(x)=P(X\leq x)$. For a general severity distribution, $F_X$ may have discrete masses as well as a continuous component. The formulas below therefore use left limits where necessary; $F_X(c-)=P(X<c)$.

Theorem: For the reinsurer’s payment, the distribution has a mass at zero of $F_X(a)$ and a mass at the layer limit of $$ P(Y_R^L=L)=P(X\geq a+L)=1-F_X((a+L)-). $$ For $0\leq r<L$, $$ F_{Y_R^L}(r)=P(Y_R^L\leq r)=F_X(a+r). $$ For the insurer’s retention, assuming $X\geq0$, $$ F_{Y_I^L}(y)= \begin{cases} F_X(y), & y<a,\\ F_X(y+L), & y\geq a. \end{cases} $$ In particular, the insurer has a mass at $a$ equal to $$ P(Y_I^L=a)=P(a\leq X\leq a+L)=F_X(a+L)-F_X(a-). $$

To see why the reinsurer CDF has this form, take $0\leq r<L$. A claim produces a payment no greater than $r$ exactly when $X\leq a+r$: below the attachment the payment is zero, and above it the payment increases with $X$ until the layer is exhausted. Hence $P(Y_R^L\leq r)=F_X(a+r)$. At $r=L$, every claim is covered by at most the limit, and the probability mass at exactly $L$ comes from all claims at or above $a+L$, including a claim exactly at exhaustion.

For the insurer CDF below $a$, a claim amount $y<a$ is retained unchanged, so $P(Y_I^L\leq y)=F_X(y)$. At or above $a$, every claim up to $a+L$ is retained at most $a$; a larger claim $X$ is retained as $X-L$. Thus $Y_I^L\leq y$ is equivalent to $X\leq y+L$, giving the second branch. The retention atom at $a$ arises because every claim in the closed interval from $a$ to $a+L$ leaves exactly $a$ with the insurer.

If $X$ has a density $f_X$, with no point masses, the reinsurer has a density $f_X(a+r)$ for $0<r<L$, as well as the endpoint masses $F_X(a)$ and $S_X(a+L)$. The insurer has density $f_X(y)$ below $a$, density $f_X(y+L)$ above $a$, and an atom at $a$ of size $F_X(a+L)-F_X(a)$. Here $S_X(x)=P(X>x)$; in the continuous case the limit atom is $S_X(a+L)$. For a distribution with point masses, use the general CDF and left-limit formulas rather than relying on density alone.

Worked Allocations and Boundary Cases

Worked Example: A Discrete Claim With a Layer Interior Payment

A hypothetical covered claim $X$ takes values $1,5,9$ dollars with probabilities $0.25,0.50,0.25$, respectively. A per-claim treaty has attachment $a=4$ dollars and limit $L=3$ dollars. Find the joint payment pairs and their probabilities. Check each allocation against the claim amount.

Method. For each possible $X$, calculate $Y_R^L=\min((X-4)_+,3)$, then calculate $Y_I^L=X-Y_R^L$. This is a discrete distribution, so each possible claim value transfers its probability to the resulting payment pair.

For $X=1$, the claim is below the attachment: $$ Y_R^L=\min((1-4)_+,3)=\min(0,3)=0,\qquad Y_I^L=1-0=1. $$ The pair is $(0,1)$, with probability $0.25$, and $0+1=1$.

For $X=5$, the amount above the attachment is $5-4=1$, which is below the layer limit: $$ Y_R^L=\min((5-4)_+,3)=1,\qquad Y_I^L=5-1=4. $$ The pair is $(1,4)$, with probability $0.50$, and $1+4=5$.

For $X=9$, the amount above the attachment is $9-4=5$, so the reinsurer pays the full limit: $$ Y_R^L=\min((9-4)_+,3)=3,\qquad Y_I^L=9-3=6. $$ The pair is $(3,6)$, with probability $0.25$, and $3+6=9$.

The joint distribution is therefore $(0,1)$ with probability $0.25$, $(1,4)$ with probability $0.50$, and $(3,6)$ with probability $0.25$. The probabilities sum to $1$, and every payment pair sums to its corresponding claim. A common mistake is to calculate the insurer payment at $X=1$ as $\min(1,4)=1$ and then incorrectly treat that expression as the reinsurance calculation. The treaty instead gives $Y_R^L=0$ below attachment; the insurer retains $1$.

Worked Example: Continuous Severity and Atoms at Both Reinsurer Endpoints

Suppose a hypothetical claim amount has an exponential distribution with rate $\lambda=0.01$ per dollar, so $F_X(x)=1-e^{-0.01x}$ for $x\geq0$ and $S_X(x)=e^{-0.01x}$. A per-claim layer has attachment $a=100$ dollars and limit $L=200$ dollars. Derive the reinsurer and insurer payment distributions and report probabilities to four decimal places.

The layer is exhausted at $a+L=300$ dollars. For the reinsurer, the zero-payment probability is $$ P(Y_R^L=0)=F_X(100)=1-e^{-1}=0.6321206\ldots\approx 0.6321. $$ The payment-limit probability is $$ P(Y_R^L=200)=S_X(300)=e^{-3}=0.0497871\ldots\approx 0.0498. $$ For $0<r<200$, the reinsurer payment density is $$ f_{Y_R^L}(r)=f_X(100+r)=0.01e^{-0.01(100+r)}. $$ Its probability in the interior is $$ \int_0^{200}0.01e^{-0.01(100+r)}\,dr =e^{-1}-e^{-3} =0.3180924\ldots\approx0.3181. $$ The interior probability and the two endpoint masses add to $0.6321+0.3181+0.0498=1.0000$, subject to rounding.

For the insurer, the atom at $100$ dollars is $$ P(100\leq X\leq300)=e^{-1}-e^{-3} =0.3180924\ldots\approx0.3181. $$ Its CDF is $F_X(y)$ for $y<100$ and $F_X(y+200)$ for $y\geq100$. Since $X$ is continuous, the insurer density is $0.01e^{-0.01y}$ for $0<y<100$ and $0.01e^{-0.01(y+200)}$ for $y>100$. The lower continuous portion has probability $1-e^{-1}\approx0.6321$, the upper portion has probability $e^{-3}\approx0.0498$, and the atom has probability $e^{-1}-e^{-3}\approx0.3181$; these total $1.0000$. The atom belongs to the insurer’s retention at the attachment, even though the original claim distribution is continuous.

Worked Example: Point Masses at the Attachment and Exhaustion Point

A hypothetical covered claim has the following distribution: $X=2$ with probability $0.20$, $X=5$ with probability $0.25$, $X=7$ with probability $0.30$, $X=8$ with probability $0.10$, and $X=12$ with probability $0.15$. A per-claim layer has $a=5$ dollars and $L=3$ dollars. Find both marginal payment distributions, treating the boundary values exactly.

The exhaustion point is $a+L=8$. At $X=2$, the claim is below attachment, so $(Y_R^L,Y_I^L)=(0,2)$, with probability $0.20$. At $X=5$, the claim is exactly at attachment, so the pair is $(0,5)$, with probability $0.25$. At $X=7$, the reinsurer pays $7-5=2$, leaving $5$; the pair is $(2,5)$, with probability $0.30$. At $X=8$, the layer is exactly exhausted, so the pair is $(3,5)$, with probability $0.10$. Finally, at $X=12$, the reinsurer pays $3$ and the insurer retains $12-3=9$, giving $(3,9)$ with probability $0.15$.

Combining equal payments, the reinsurer pays $0$ with probability $0.20+0.25=0.45$, $2$ with probability $0.30$, and $3$ with probability $0.10+0.15=0.25$. The insurer retains $2$ with probability $0.20$, $5$ with probability $0.25+0.30+0.10=0.65$, and $9$ with probability $0.15$. Each marginal totals probability $1$. The boundary masses are handled as intended: $X=5$ produces zero reinsurance payment, while $X=8$ produces the full limit and leaves the insurer with $a=5$.

Checks, Assumptions, and Exam Traps

The allocation rule assumes the layer is applied directly to the specified covered claim amount $X$. If $X$ is already a payment amount after another coverage modification, do not silently apply that modification again. The treaty’s attachment and limit are measured on the same monetary basis as $X$. A per-claim layer is evaluated separately for each claim; it is not a cap on the sum of all claims.

Identify the amount being layered.
Confirm that $X$ is the covered amount to which the treaty terms apply, rather than an aggregate or an amount already adjusted a second time.

Find the exhaustion point.
Compute $a+L$. The layer limit is a payment amount; it is not the claim amount at which the layer is exhausted.

Apply endpoint conventions.
Claims $X\leq a$ produce zero reinsurance payment. Claims $X\geq a+L$ produce the full limit. The insurer retains $a$ for claims from $a$ through $a+L$, inclusive.

Check the allocation.
For every possible claim amount, verify $Y_R^L+Y_I^L=X$, and check that the probabilities in each marginal distribution sum to one.

A frequent exam error is to treat $L$ as the exhaustion claim amount and place the reinsurer’s limit atom at $X=L$. That is only correct when $a=0$. With a positive attachment, full payment begins at claim amount $a+L$. Another error is to omit the insurer’s atom at $a$: claims throughout the layer, not only claims exactly at $a$, leave the insurer retaining $a$.

Key takeaway: The reinsurer payment is zero through the attachment, increases across the layer, and is capped at $L$ from claim amount $a+L$ onward. The insurer’s retention is unchanged below the attachment, equals $a$ throughout the layer, and equals $X-L$ above exhaustion. Atoms follow from these flat portions of the allocation, together with any atoms already present in $X$.

These distributional details support later calculations involving layer costs and comparisons of treaty structures. The essential preliminary work is always to define the claim-level payment variables, derive their distributions with correct boundary treatment, and verify that the two payments reconcile to the covered claim. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, apply the layer to the stated claim amount and pay attention to the attachment and exhaustion point.

A hypothetical claim takes values $3,6,10$ dollars with probabilities $0.20,0.50,0.30$. A layer has attachment $4$ and limit $4$. Give the joint payment pairs with probabilities. (Answer: $(0,3)$ with $0.20$, $(2,4)$ with $0.50$, and $(4,6)$ with $0.30$.)
Let $X$ be uniform on $[0,12]$, with attachment $a=3$ and limit $L=5$. Give the reinsurer’s zero-payment probability, limit-payment probability, and interior density. (Answer: zero mass $0.2500$, limit mass $0.3333$, and density $1/12$ on $0<r<5$.)
A claim takes values $4,6,7,8,11$ with probabilities $0.10,0.20,0.30,0.15,0.25$. A layer has $a=6$ and $L=2$. Give the insurer’s retention distribution. (Answer: $4$ with probability $0.10$, $6$ with probability $0.65$, and $9$ with probability $0.25$.)
Suppose $X$ is exponential with rate $0.02$ per dollar, attachment $50$, and limit $50$. Find the reinsurer’s zero and limit atoms and the insurer’s atom at the attachment. (Answer: reinsurer zero atom $1-e^{-1}=0.6321$, reinsurer limit atom $e^{-2}=0.1353$, insurer atom at $50$ equal to $e^{-1}-e^{-2}=0.2325$.)
A hypothetical claim has values $9,12,15,16$ dollars with probabilities $0.10,0.20,0.30,0.40$. Under a layer with attachment $10$ and limit $5$, find the reinsurer payment distribution and insurer retention distribution. (Answer: reinsurer payments $0,2,5$ with probabilities $0.10,0.20,0.70$; insurer retentions $9,10,11$ with probabilities $0.10,0.50,0.40$.)

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