Multi-Layer Stop-Loss Pricing Workshop

Why Price Aggregate Layers Separately?

An aggregate stop-loss contract pays when the total loss over a specified period exceeds an attachment point. A reinsurer may divide that protection into adjacent layers, each with its own attachment and limit. Pricing the layers separately makes their costs visible, while a check against the combined protection can reveal a reversed subtraction, a misplaced attachment, or a missing layer.

Throughout, let $X$ be a nonnegative aggregate loss for one exposure period, and let $0\leq a\leq b<\infty$. A layer from $a$ to $b$ attaches at $a$ and has width $b-a$. Its pure premium is the expected indemnity, with no expense, risk, or profit loading. The loss amounts and premiums below are in dollars for the same exposure period; no interest adjustment is being made.

Definition: The stop-loss transform of $X$ at attachment $d\geq0$ is $\Pi_X(d)=E[(X-d)^+]$, when this expectation is finite, where $z^+=\max(z,0)$. The payment from the layer attaching at $a$ and exhausting at $b$ is $\min((X-a)^+,b-a)$. Its pure premium is the expected value of that payment.

The Layer-Premium Formula

For any realized loss $x$, the layer payment can also be written as the difference between two excess losses. If $x\leq a$, neither excess is positive. If $a<x\leq b$, only $x-a$ is positive. If $x>b$, the difference is $(x-a)-(x-b)=b-a$, the full layer width. Thus the pointwise identity is

\min((x-a)^+,b-a)=(x-a)^+-(x-b)^+.

This identity holds for every $x$, including $x=a$ and $x=b$, and does not depend on whether the loss distribution is discrete, continuous, or mixed. The layer payment is bounded between zero and $b-a$, so its expectation is finite. However, using the identity to subtract the two expected excess losses requires both transforms to be finite: subtracting two infinite expectations is not defined.

Theorem: If $\Pi_X(a)$ and $\Pi_X(b)$ are finite, then the pure premium for the layer $[a,b]$ is \[ E[\min((X-a)^+,b-a)]=\Pi_X(a)-\Pi_X(b). \] For nonnegative $X$, the condition $E[X]<\infty$ is sufficient to make both transforms finite at every finite attachment. The layer’s expected payment itself remains finite even when the transforms are infinite.

The formula follows by taking expectations of the pointwise identity. Under the stated finiteness condition, linearity of expectation gives

E[(X-a)^+-(X-b)^+] =E[(X-a)^+]-E[(X-b)^+] =\Pi_X(a)-\Pi_X(b).

The order matters: a lower attachment has at least as large a stop-loss transform as a higher one, so a layer premium is the transform at the lower attachment minus the transform at the upper attachment.

Pricing Adjacent Layers

Suppose attachments satisfy $0\leq a_0<a_1<\cdots<a_m$, with a layer $[a_{j-1},a_j]$ at each step. If the transforms at these attachments are finite, the pure premium for layer $j$ is $\Pi_X(a_{j-1})-\Pi_X(a_j)$. Adding the layer prices cancels all intermediate transform values:

\sum_{j=1}^{m}\bigl(\Pi_X(a_{j-1})-\Pi_X(a_j)\bigr) =\Pi_X(a_0)-\Pi_X(a_m).

This telescoping check has a direct insurance interpretation. The adjacent layers together pay $\min((X-a_0)^+,a_m-a_0)$, the capped excess loss from $a_0$ up to $a_m$. Thus their expected payments must sum to the expected payment for that combined layer. This check applies to pure premiums; it does not by itself combine different expense or loading arrangements.

List the attachment points.
Include every lower and upper endpoint, in increasing order, and identify the layer associated with each adjacent pair.

Find each needed transform.
Use the supplied transform values or calculate $E[(X-d)^+]$ from the loss distribution. Confirm the relevant expectations are finite before subtracting.

Difference in the correct direction.
For a layer $[a,b]$, calculate $\Pi_X(a)-\Pi_X(b)$. The result must be nonnegative and no greater than $b-a$.

Check the set of layers.
Add the prices and compare with $\Pi_X(a_0)-\Pi_X(a_m)$, or calculate the combined capped excess payment directly.

Worked Example: Three Discrete Layers

Method. Calculate the stop-loss transform at each attachment by summing the positive excess loss times its probability, then difference successive values.

A hypothetical portfolio has aggregate loss $X$ with the following distribution: $P(X=0)=0.20$, $P(X=10)=0.30$, $P(X=20)=0.30$, and $P(X=30)=0.20$. Price layers $[5,15]$, $[15,25]$, and $[25,35]$.

The probability total is $0.20+0.30+0.30+0.20=1.00$. At attachment 5, the positive excesses are 5, 15, and 25 for losses 10, 20, and 30, respectively:

\Pi_X(5)=5(0.30)+15(0.30)+25(0.20) =1.50+4.50+5.00=11.00.

At attachment 15, only losses 20 and 30 contribute. Their excesses are 5 and 15:

\Pi_X(15)=5(0.30)+15(0.20)=1.50+3.00=4.50.

At attachment 25, only loss 30 contributes, with excess 5. At attachment 35, no loss exceeds the attachment. Therefore $\Pi_X(25)=5(0.20)=1.00$ and $\Pi_X(35)=0$.

The layer pure premiums are $\Pi_X(5)-\Pi_X(15)=11.00-4.50=\$6.50$, $\Pi_X(15)-\Pi_X(25)=4.50-1.00=\$3.50$, and $\Pi_X(25)-\Pi_X(35)=1.00-0=\$1.00$.

As a direct check, the layers pay, respectively: $5$ on loss 10 and $10$ each on losses 20 and 30, giving $5(0.30)+10(0.30)+10(0.20)=\$6.50$; $5$ on loss 20 and $10$ on loss 30, giving $5(0.30)+10(0.20)=\$3.50$; and $5$ on loss 30, giving $5(0.20)=\$1.00$. Their sum is $\$11.00=\Pi_X(5)-\Pi_X(35)$. Each premium is nonnegative and no larger than its layer width of 10. The method avoids treating the transform at an attachment as the price of the layer beginning there.

Using a Continuous Distribution

For a continuous distribution with density $f_X$, a transform can be calculated by integrating the excess over losses above the attachment:

\Pi_X(d)=\int_d^\infty (x-d)f_X(x)\,dx,

when the expectation is finite. The same layer-difference rule then applies. A tail-probability form can also be useful for a bounded layer: its expected payment is $\int_a^b P(X>t)\,dt$. This expression depends only on the probabilities within the layer’s attachment-to-exhaustion interval, and the integral is finite because the interval has finite width.

Worked Example: Layers Under a Uniform Aggregate Model

Method. Integrate the excess loss to find the transform at each endpoint, then difference the transforms and check the total against the combined capped excess.

Assume a hypothetical aggregate loss $X$ is uniformly distributed on $[0,40]$ dollars, with density $1/40$ on that interval and zero elsewhere. Price layers $[8,20]$, $[20,32]$, and $[32,40]$. The distribution has finite mean, so all transforms used here are finite.

For $0\leq d\leq40$, integrate only over $x>d$:

\Pi_X(d)=\frac{1}{40}\int_d^{40}(x-d)\,dx =\frac{1}{40}\left[\frac{(x-d)^2}{2}\right]_{d}^{40} =\frac{(40-d)^2}{80}.

At attachment 8, $\Pi_X(8)=(40-8)^2/80=32^2/80=1024/80=\$12.80$. At 20, $\Pi_X(20)=20^2/80=400/80=\$5.00$. At 32, $\Pi_X(32)=8^2/80=64/80=\$0.80$. At 40, $\Pi_X(40)=0$.

The three pure premiums are therefore $12.80-5.00=\$7.80$, $5.00-0.80=\$4.20$, and $0.80-0=\$0.80$. Their total is $\$12.80$, equal to $\Pi_X(8)-\Pi_X(40)$. As an additional check, each price is no greater than its layer width of 12, and the top layer price is no greater than its width of 8. The method avoids assuming that a continuous model is needed for the layer formula; the formula is also valid for discrete losses when the transforms are finite.

Endpoint Masses and a Complete Layer Check

A probability mass exactly at an attachment does not create ambiguity. At $X=a$, the layer starting at $a$ pays zero. At $X=b$, the layer $[a,b]$ pays its full width $b-a$, while a layer starting at $b$ pays zero. These outcomes follow directly from the positive-part definition; no special adjustment for inclusive or exclusive endpoints is needed.

Worked Example: Adjacent Layers with Masses at Attachments

Method. Compute the transforms at the common endpoints, then verify the prices from the individual possible layer payments, including losses equal to those endpoints.

A hypothetical aggregate loss has $P(X=0)=0.25$, $P(X=12)=0.25$, $P(X=25)=0.30$, and $P(X=40)=0.20$. These probabilities sum to 1.00. Price layers $[0,12]$, $[12,25]$, and $[25,40]$.

The mean loss, which is $\Pi_X(0)$, is $0(0.25)+12(0.25)+25(0.30)+40(0.20)=0+3.00+7.50+8.00=\$18.50$. At attachment 12, losses 25 and 40 have excesses 13 and 28, so $\Pi_X(12)=13(0.30)+28(0.20)=3.90+5.60=\$9.50$. At attachment 25, only loss 40 contributes: $\Pi_X(25)=15(0.20)=\$3.00$. Finally, $\Pi_X(40)=0$.

The prices are $[0,12]$: $18.50-9.50=\$9.00$; $[12,25]$: $9.50-3.00=\$6.50$; and $[25,40]$: $3.00-0=\$3.00$. Directly, the first layer pays 12 at $X=12,25,40$, with expected payment $12(0.25+0.30+0.20)=\$9.00$. The second pays zero at $X=12$, 13 at $X=25$, and 13 at $X=40$, giving $13(0.30+0.20)=\$6.50$. The last pays zero at $X=25$ and 15 at $X=40$, giving $15(0.20)=\$3.00$. The total is $\$18.50=\Pi_X(0)-\Pi_X(40)$. In particular, the mass at 25 exhausts the middle layer but does not begin a payment in the final layer.

Common Mistake: Reversing the Transform Difference

For the first layer in the discrete example, subtracting in the wrong direction gives $\Pi_X(15)-\Pi_X(5)=4.50-11.00=-\$6.50$. A negative pure premium is impossible. The correction is to subtract the upper-attachment transform from the lower-attachment transform: $\Pi_X(5)-\Pi_X(15)=\$6.50$. More generally, check that every layer price is between zero and its width, then verify that adjacent prices telescope to the expected combined payment.

A separate technical trap is applying the transform-difference formula when both transforms are infinite. For example, a heavy-tailed loss can have an infinite mean even though any finite-width layer payment is bounded. In that case, the layer expectation can be well-defined while “infinite minus infinite” is not. Use the layer payment directly, or its finite-interval tail-probability integral, rather than subtracting infinite transforms.

Key takeaway: For finite stop-loss transforms, the pure premium for layer $[a,b]$ is $\Pi_X(a)-\Pi_X(b)$. Price adjacent layers using successive differences, preserve the lower-minus-upper order, and check their sum against the combined capped excess payment.

Check Your Understanding

For each problem, calculate pure expected layer payments unless the question asks for a finiteness check.

A hypothetical aggregate loss has probabilities $P(X=0)=0.40$, $P(X=10)=0.40$, and $P(X=20)=0.20$. Find the pure premiums for layers $[5,15]$ and $[15,25]$, using transform differences. (Answer: $\$4.00$ and $\$1.00$, respectively.)
Suppose $X$ is uniformly distributed on $[0,30]$ dollars. Find the pure premiums for layers $[6,18]$ and $[18,30]$, and check their sum against $\Pi_X(6)-\Pi_X(30)$. (Answer: $\$7.20$ and $\$2.40$; total $\$9.60$.)
A hypothetical aggregate loss has $P(X=0)=0.20$, $P(X=8)=0.30$, $P(X=16)=0.30$, and $P(X=28)=0.20$. Price the three adjacent layers $[4,12]$, $[12,20]$, and $[20,28]$. (Answer: $\$5.20$, $\$2.80$, and $\$1.60$, respectively; total $\$9.60$.)
A supplied finite stop-loss transform has values $\Pi_X(0)=22$, $\Pi_X(10)=14$, $\Pi_X(25)=5$, $\Pi_X(40)=1$, and $\Pi_X(50)=0$, all in dollars. Find the prices of layers $[0,10]$, $[10,25]$, $[25,40]$, and $[40,50]$, then check their total. (Answer: $\$8$, $\$9$, $\$4$, and $\$1$; total $\$22$.)
Let $X$ have Pareto survival function $P(X>x)=x^{-1/2}$ for $x\geq1$, with $P(X>x)=1$ for $0\leq x<1$. Explain why $\Pi_X(4)-\Pi_X(9)$ is not a valid subtraction of finite transforms, and find the expected payment for layer $[4,9]$. (Answer: Both transforms are infinite, so their difference is undefined; the layer expectation is $\int_4^9 x^{-1/2}\,dx=2$.)

What You'll Learn