Layer Premiums from Two Stop-Loss Transforms

Why Two Stop-Loss Transforms Price a Layer

An aggregate-loss layer pays only for the part of a loss that lies between two thresholds. Its attachment point $a$ is where payment begins; its exhaustion point $b$ is where the layer has paid its full width, $b-a$. The expected layer payment can be found without calculating the layer payment separately for every possible loss: subtract the stop-loss transform at $b$ from the transform at $a$.

This connection is useful when a model or table already gives stop-loss transform values. It also gives a direct check on calculations from a probability distribution. The expected payment calculated this way is the pure expected layer loss; it is not necessarily the premium charged after expenses, risk margins, or other pricing adjustments.

Definition: Let $X$ be a nonnegative aggregate loss with finite mean. For an attachment point $d\geq 0$, its stop-loss transform is $\Pi(d)=E[(X-d)^+]$, where $z^+=\max(z,0)$. For $0\leq a<b$, the payment for the layer attaching at $a$ and exhausting at $b$ is $L_{a,b}=\min((X-a)^+,b-a)$.

The layer makes no payment when $X\leq a$. When $a<X<b$, it pays $X-a$. When $X\geq b$, it pays the full layer width $b-a$. Thus, the largest possible payment is $b-a$, regardless of how large the aggregate loss becomes.

The Layer-Premium Identity

The key identity follows by comparing the payment in each loss range. It does not require $X$ to have a density: it applies to discrete, continuous, and mixed distributions, including distributions with probability at $a$ or $b$.

Write the excess above each threshold.
At $a$, the excess is $(X-a)^+$; at $b$, the excess is $(X-b)^+$.

Subtract the excess above $b$.
If $X\leq a$, both excesses are zero. If $a<X\leq b$, only the first is positive, so the difference is $X-a$. If $X>b$, the difference is $(X-a)-(X-b)=b-a$.

Take expectations.
The pointwise difference is exactly the layer payment. Taking expectations gives the expected layer payment in terms of the two transforms.

L_{a,b}=(X-a)^+-(X-b)^+

E[L_{a,b}] =E[(X-a)^+]-E[(X-b)^+] =\Pi(a)-\Pi(b)

This is also a useful way to understand why the transforms are subtracted in this order. Raising the threshold from $a$ to $b$ removes the part of the excess loss that lies above $b$. The remaining excess is exactly what the layer pays. Since $b>a$, $\Pi(a)\geq\Pi(b)$, so the result cannot be negative.

Theorem: For a nonnegative aggregate loss $X$ with finite mean and layer thresholds $0\leq a<b$, the expected layer payment is $\Pi(a)-\Pi(b)$. The identity includes probability masses at either threshold: at $X=a$ the payment is zero, and at $X=b$ it is $b-a$.

Calculating a Layer from a Discrete Distribution

For a discrete loss, one method is to calculate both transforms by summing each possible excess times its probability, then subtract. A second method is to calculate the layer payment at each outcome and take its expectation. Agreement between the methods is a strong check on the treatment of losses above the exhaustion point.

Worked Example: A Layer on a Four-Outcome Aggregate Loss

A hypothetical portfolio has aggregate loss $X$, in thousands of dollars, with this distribution: $P(X=0)=0.20$, $P(X=4)=0.30$, $P(X=10)=0.30$, and $P(X=18)=0.20$. Find the expected payment, in thousands of dollars, for a layer attaching at $a=5$ and exhausting at $b=12$.

Method. Calculate $\Pi(5)$ and $\Pi(12)$, then subtract. The transform definition is $\Pi(d)=\sum_x P(X=x)(x-d)^+$.

At $d=5$, the excesses for $X=0,4,10,18$ are $0,0,5,13$, respectively. Therefore,

\Pi(5)=(0.20)(0)+(0.30)(0)+(0.30)(5)+(0.20)(13) =0+0+1.50+2.60=4.10

At $d=12$, the excesses are $0,0,0,6$. Thus,

\Pi(12)=(0.20)(0)+(0.30)(0)+(0.30)(0)+(0.20)(6) =1.20

The expected layer payment is the difference:

E[L_{5,12}]=\Pi(5)-\Pi(12)=4.10-1.20=2.90

Direct check. The layer pays $0$ at losses $0$ and $4$, $10-5=5$ at loss $10$, and the full width $12-5=7$ at loss $18$. Its expected payment is $(0.20)(0)+(0.30)(0)+(0.30)(5)+(0.20)(7)=0+0+1.50+1.40=2.90$. The direct calculation agrees with the transform calculation.

The result is $2.90$ thousand dollars, or $\$2,900$. The important check is that the loss of $18$ produces a layer payment of $7$, not an uncapped excess of $13$.

Using a Specified Aggregate Model

The identity works whether the distribution is supplied as a table or derived from a model. In a compound model, for example, the aggregate loss can be formed from a random number of claims and their severities. Once the aggregate distribution, or the required transform values, is available, the layer calculation itself is unchanged.

Worked Example: A Fixed Severity with Random Claim Count

In a hypothetical coverage period, the claim count $N$ has a binomial distribution with parameters $n=2$ and $q=0.5$: $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. Each claim has a fixed severity of $\$6$, so aggregate loss is $X=6N$, in dollars. Find the expected payment for a layer attaching at $\$4$ and exhausting at $\$9$.

Method. First list the aggregate-loss probabilities. The possible values are $X=0,6,12$, with probabilities $0.25,0.50,0.25$. Then calculate both transforms.

For $d=4$, the excesses at $X=0,6,12$ are $0,2,8$. Hence,

\Pi(4)=(0.25)(0)+(0.50)(2)+(0.25)(8) =0+1.00+2.00=\$3.00

For $d=9$, the excesses are $0,0,3$, so

\Pi(9)=(0.25)(0)+(0.50)(0)+(0.25)(3)=\$0.75

Subtract the transforms:

E[L_{4,9}]=\Pi(4)-\Pi(9)=\$3.00-\$0.75=\$2.25

Direct check. At $X=0$, the layer pays $\$0$. At $X=6$, it pays $6-4=\$2$. At $X=12$, it pays the full width $9-4=\$5$. Therefore, the expected payment is $(0.25)(\$0)+(0.50)(\$2)+(0.25)(\$5)=\$0+\$1.00+\$1.25=\$2.25$. The check agrees. In particular, the loss of $\$12$ does not produce a layer payment of $\$8$: payment is capped at the layer width of $\$5$.

Calculating Transforms in a Continuous Model

If $X$ has a continuous distribution with survival function $S_X(x)=P(X>x)$, its stop-loss transform can also be calculated by integrating the survival function above the attachment. For the exponential model below, this gives a simple closed form. The transform-difference identity itself does not depend on continuity.

Worked Example: An Exponential Aggregate Loss

Suppose a hypothetical aggregate loss $X$, measured in dollars, has an exponential distribution with rate $\lambda=0.2$ per dollar. Thus, $S_X(x)=e^{-0.2x}$ for $x\geq0$, and the mean aggregate loss is $1/\lambda=\$5$. Find the expected payment for a layer attaching at $\$5$ and exhausting at $\$10$.

Method. For a nonnegative loss, $(X-d)^+=\int_d^\infty \mathbf{1}_{\{X>x\}}\,dx$. Taking expectations gives $\Pi(d)=\int_d^\infty S_X(x)\,dx$. With the exponential survival function,

\Pi(d)=\int_d^\infty e^{-0.2x}\,dx =\left[-5e^{-0.2x}\right]_{d}^{\infty} =5e^{-0.2d}

At $d=5$, $\Pi(5)=5e^{-1}=1.839397\ldots$ dollars. At $d=10$, $\Pi(10)=5e^{-2}=0.676676\ldots$ dollars. Therefore,

E[L_{5,10}]=\Pi(5)-\Pi(10) =5e^{-1}-5e^{-2} =1.162721\ldots\text{ dollars}

Rounded to the nearest cent, the expected payment is $\$1.16$. As a check, the layer can pay only between $\$0$ and $\$5$, and the calculated expectation lies in that range. The integral over the layer itself gives the same value: $\int_5^{10}e^{-0.2x}\,dx=5(e^{-1}-e^{-2})=1.162721\ldots$.

Common Trap: Confusing a Conditional Excess with a Layer Payment

A conditional excess loss asks how much loss exceeds a threshold, given that the threshold has been crossed. A layer payment is an unconditional amount and is capped at its exhaustion point. These are different quantities.

For the discrete distribution in the first example, $P(X>5)=0.30+0.20=0.50$. Also, $\Pi(5)=4.10$, so the conditional expected excess above $5$, given $X>5$, is $4.10/0.50=8.20$ thousand dollars. Treating that conditional excess as the layer payment would give $8.20$, which is wrong: it conditions on crossing the attachment and does not account for the cap at $12$.

The correct unconditional layer expectation is $\Pi(5)-\Pi(12)=4.10-1.20=2.90$ thousand dollars. Equivalently, conditional on $X>5$, the layer payment has expectation $2.90/0.50=5.80$ thousand dollars; multiplying by $P(X>5)=0.50$ returns the unconditional expectation of $2.90$. The conditional figure and the unconditional layer payment answer different questions.

Practical Checks

Before accepting a transform-based layer calculation, check its setup and scale. For $0\leq a<b$, a correct result must satisfy

$\Pi(a)\geq\Pi(b)$, so the expected payment is nonnegative.
$E[L_{a,b}]\leq b-a$, because the payment cannot exceed the layer width.
The transform values and layer width use the same units as $X$.
The expected payment is an unconditional expected amount, not a conditional excess given that the attachment is crossed.

These checks do not replace the calculation, but they often reveal a reversed subtraction, a unit mismatch, or an uncapped payment at a loss above $b$. The identity $\Pi(a)-\Pi(b)$ is most useful when the attachment and exhaustion transforms are evaluated consistently for the same aggregate-loss model.

Check Your Understanding

For each problem, calculate the expected layer payment. Answers are stated in the units given.

A hypothetical aggregate loss $X$, in thousands of dollars, has probabilities $P(X=0)=0.40$, $P(X=8)=0.40$, and $P(X=20)=0.20$. Find the expected payment for a layer attaching at $5$ and exhausting at $14$. (Answer: $3.00$ thousand dollars.)
A claim count $N$ is binomial with parameters $n=3$ and $q=0.5$. Each claim has fixed severity $\$4$, so $X=4N$. Find the expected payment for a layer attaching at $\$5$ and exhausting at $\$9$. (Answer: $\$1.625$.)
A hypothetical aggregate loss $X$, in dollars, is exponential with rate $\lambda=0.25$ per dollar. Find the expected payment for a layer attaching at $\$4$ and exhausting at $\$12$, rounded to the nearest cent. (Answer: $\$1.27$.)
A hypothetical aggregate loss, in thousands of dollars, has $P(X=0)=0.50$, $P(X=10)=0.30$, and $P(X=25)=0.20$. Calculate the transforms at $8$ and $20$, then find the expected payment for the layer between those points. (Answer: $\Pi(8)=4.00$, $\Pi(20)=1.00$, and expected payment $=3.00$ thousand dollars.)
For a nonnegative aggregate loss with finite mean, suppose a model gives $\Pi(6)=7.40$ and $\Pi(16)=2.10$, both in thousands of dollars. Find the expected payment for the layer attaching at $6$ and exhausting at $16$, and verify that it does not exceed the layer width. (Answer: $5.30$ thousand dollars; the width is $10$ thousand dollars.)

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