Deriving Aggregate Stop-Loss Identities

Why Derive Aggregate Stop-Loss Identities?

An aggregate excess-of-loss treaty pays only when total covered loss exceeds an attachment point. If the payment is also capped by a treaty limit, its expected recovery depends on the aggregate loss distribution across a range of thresholds, not just on the probability of one particular loss value. Two identities make that calculation systematic: a tail-integral formula and a difference of limited means.

The formulas apply to an aggregate loss that is nonnegative, whether its distribution is discrete, continuous, or mixed. In particular, they do not require a continuous distribution. This matters when the aggregate loss has atoms, including a positive probability of no claims or a positive probability of landing exactly at the attachment point.

Definition: Let $S\geq 0$ be aggregate loss, $d\geq 0$ the treaty attachment point, and $u>0$ the treaty limit measured as the maximum payment. The aggregate excess-of-loss recovery is $R_{d,u}=\min\{(S-d)_+,u\}$, where $(z)_+=\max(z,0)$. The layer exhausts at $d+u$. With no cap, write $R_{d,\infty}=(S-d)_+$.

The recovery is zero when $S\leq d$, rises dollar for dollar when $d<S<d+u$, and equals $u$ when $S\geq d+u$. For a finite limit, the expected payment is always finite because $0\leq R_{d,u}\leq u$. The unlimited recovery has finite expectation only if $E[(S-d)_+]$ is finite; a finite mean $E[S]$ is sufficient, though not necessary for every choice of $d$.

The Tail-Integral Identity

For any nonnegative aggregate loss $S$, the expected capped recovery is the area under the aggregate survival probability between attachment and exhaustion:

E[R_{d,u}]=\int_d^{d+u}P(S>x)\,dx.

For an unlimited treaty, the corresponding identity is

E[(S-d)_+]=\int_d^\infty P(S>x)\,dx.

The integral can be interpreted as an ordinary integral of the survival function. Atoms in the distribution can create jumps in $P(S>x)$, but do not invalidate the identity. The value of the integrand at a single endpoint does not change the integral.

Condition: The finite-layer identity holds for any nonnegative $S$, $d\geq0$, and finite $u>0$. The unlimited identity holds as an equality of nonnegative extended expectations; it gives a finite expected recovery when the tail integral is finite. In particular, it is finite if $E[S]<\infty$.

Why the Tail Integral Works

The derivation represents a payment as the total length of thresholds that the aggregate loss exceeds. For a realized value $S=s$, the layer pays for thresholds $x$ above $d$ and below both $s$ and $d+u$. Thus, up to endpoint values that do not affect an integral,

\min\{(s-d)_+,u\} =\int_d^{d+u}\mathbf{1}_{\{s>x\}}\,dx.

Taking expectations and interchanging expectation and integration is valid because the integrand is nonnegative (Tonelli's theorem). The expected value of the indicator $\mathbf{1}_{\{S>x\}}$ is $P(S>x)$. Therefore,

\begin{aligned} E[R_{d,u}] &=E\left[\int_d^{d+u}\mathbf{1}_{\{S>x\}}\,dx\right]\\ &=\int_d^{d+u}E[\mathbf{1}_{\{S>x\}}]\,dx\\ &=\int_d^{d+u}P(S>x)\,dx. \end{aligned}

For an unlimited treaty, replace the upper endpoint $d+u$ by infinity. The same argument applies to nonnegative integrands, although the resulting expectation may be infinite.

The Limited-Mean Identity

A second route uses the limited expected value of aggregate loss. For $t\geq0$, define

L_S(t)=E[\min(S,t)].

This is the expected amount of aggregate loss counted up to threshold $t$. It is finite for every finite $t$, even if $E[S]$ is infinite. The limited-mean identity for a finite layer is

E[R_{d,u}]=L_S(d+u)-L_S(d).

To see why, consider a fixed aggregate loss $S=s$. The increase in the amount counted when the threshold rises from $d$ to $d+u$ is $\min(s,d+u)-\min(s,d)$. If $s\leq d$, this difference is zero. If $d<s<d+u$, it is $s-d$. If $s\geq d+u$, it is $u$. These are exactly the three possible recovery amounts. Hence the equality holds for every $s$, and taking expectations proves the identity.

Theorem: For $S\geq0$, $d\geq0$, and finite $u>0$, the following expressions are equal: $E[\min\{(S-d)_+,u\}]$, $\int_d^{d+u}P(S>x)\,dx$, and $L_S(d+u)-L_S(d)$. The identities remain valid when $S$ has atoms.

The tail-integral form is useful when survival probabilities are easy to evaluate over the treaty's layer. The limited-mean form is useful when limited expected values at the attachment and exhaustion points are available. For an unlimited layer, $L_S(t)$ alone does not give a finite upper endpoint: one must use $E[S]-L_S(d)$ if $E[S]<\infty$, or evaluate the unlimited tail integral.

Worked Examples

Worked Example: A Discrete Aggregate Loss and a Limited Layer

A hypothetical portfolio has aggregate loss $S$ with the following distribution: $P(S=0)=0.20$, $P(S=50)=0.30$, $P(S=100)=0.30$, and $P(S=180)=0.20$. An annual aggregate treaty attaches at $d=\$60$ and has a payment limit $u=\$70$. Find its expected recovery using both identities.

Method 1: Calculate recoveries by aggregate-loss outcome. At $S=0$, the recovery is $\min\{(0-60)_+,70\}=0$. At $S=50$, it is also zero. At $S=100$, it is $\min\{40,70\}=40$. At $S=180$, it is $\min\{120,70\}=70$. Therefore,

E[R_{60,70}] =(0.20)(0)+(0.30)(0)+(0.30)(40)+(0.20)(70) =12+14 =\$26.

Method 2: Use the tail integral. For $60\leq x<100$, $P(S>x)=0.30+0.20=0.50$. For $100\leq x<180$, $P(S>x)=0.20$. The layer's integration interval is $[60,130]$, so

\int_{60}^{130}P(S>x)\,dx =(100-60)(0.50)+(130-100)(0.20) =40(0.50)+30(0.20) =20+6 =\$26.

Method 3: Use limited means. First, $L_S(130)=(0.20)(0)+(0.30)(50)+(0.30)(100)+(0.20)(130)=15+30+26=\$71$. Next, $L_S(60)=(0.20)(0)+(0.30)(50)+(0.30)(60)+(0.20)(60)=15+18+12=\$45$. Thus $L_S(130)-L_S(60)=71-45=\$26$, agreeing with both calculations.

Check and common mistake. The largest possible recovery is $\$70$, and only $30\%$ of outcomes pay $\$40$ while $20\%$ pay $\$70$, so an expected recovery of $\$26$ is consistent. A common error is to treat the limit as the exhaustion point: the treaty exhausts at $d+u=\$130$, not at $\$70$.

Worked Example: A Corrected Discrete Portfolio With a Layer Limit

For a hypothetical portfolio, claim count $N$ is 0, 1, or 2, with probabilities $0.5000$, $0.3000$, and $0.2000$, respectively. Each claim costs $\$65$, so aggregate loss is $S=65N$. An annual aggregate treaty attaches at $d=\$70$ and has a payment limit $u=\$60$. Find the expected recovery.

Method: Apply the tail-integral identity. The possible aggregate losses are $S=0$ with probability $0.5000$, $S=65$ with probability $0.3000$, and $S=130$ with probability $0.2000$. Only the last outcome exceeds the attachment. Its recovery is $\min\{130-70,60\}=\$60$. Directly, the expected recovery is

(0.5000)(0)+(0.3000)(0)+(0.2000)(60)=\$12.

For $70\leq x<130$, $P(S>x)=0.2000$, and above $130$ the probability is zero. Consequently,

\int_{70}^{130}P(S>x)\,dx =(130-70)(0.2000) =60(0.2000) =\$12.

As a limited-mean check, $L_S(130)=(0.3000)(65)+(0.2000)(130)=19.50+26.00=\$45.50$, while $L_S(70)=(0.3000)(65)+(0.2000)(70)=19.50+14.00=\$33.50$. Their difference is $\$45.50-\$33.50=\$12.00$.

Check and common mistake. The expected recovery cannot exceed $0.2000(\$60)=\$12$, since only the two-claim outcome triggers the layer and its payment is capped at $\$60$. Do not multiply the triggering probability by the claim amount: the treaty payment is determined by aggregate loss above attachment, subject to its limit.

Worked Example: A Continuous Aggregate Loss

Suppose a hypothetical portfolio's aggregate loss $S$ has an exponential distribution with mean $\$100$, so its survival function is $P(S>x)=e^{-x/100}$ for $x\geq0$. A treaty attaches at $d=\$50$ and has a limit $u=\$100$. Find the expected recovery, rounded to the nearest cent.

Method: Integrate the survival function across the layer. Exhaustion is at $d+u=\$150$. Therefore,

\begin{aligned} E[R_{50,100}] &=\int_{50}^{150}e^{-x/100}\,dx\\ &=\left[-100e^{-x/100}\right]_{50}^{150}\\ &=100(e^{-0.5}-e^{-1.5})\\ &=100(0.6065307-0.2231302)\\ &=38.34005\ldots \end{aligned}

Thus the expected recovery is $\$38.34$, rounded to two decimal places. As a limited-mean check, an exponential variable with mean $\$100$ has $L_S(t)=100(1-e^{-t/100})$. Hence $L_S(150)=100(1-e^{-1.5})=\$77.68698$, and $L_S(50)=100(1-e^{-0.5})=\$39.34693$. Their difference is $\$77.68698-\$39.34693=\$38.34005$, or $\$38.34$ to the nearest cent.

Check and common mistake. The recovery is positive and cannot exceed the $\$100$ limit. Integrating from zero would include losses below the treaty attachment; integrating to infinity would instead calculate the unlimited recovery.

Atoms, Boundaries, and Exam Traps

When $S$ has an atom at the attachment $d$, that probability produces no recovery: $(S-d)_+=0$ when $S=d$. The tail-integral identity uses $P(S>x)$, but using $P(S\geq x)$ instead gives the same integral for ordinary integration because the two functions differ only at points with positive probability atoms. In discrete calculations, however, it is safer to calculate each recovery directly from $\min\{(S-d)_+,u\}$ or to follow the stated strict inequality consistently.

An atom at exhaustion $d+u$ pays exactly $u$. Losses above exhaustion also pay $u$. The probability at or above exhaustion therefore matters when calculating expected recovery from outcomes, even though a single endpoint does not affect the tail integral.

Identify the treaty payment.
Write $R_{d,u}=\min\{(S-d)_+,u\}$, and distinguish the layer limit $u$ from the exhaustion point $d+u$.

Select the identity.
Use $\int_d^{d+u}P(S>x)\,dx$ when the survival function is convenient; use $L_S(d+u)-L_S(d)$ when limited means are available.

Check conditions and boundaries.
Confirm $S\geq0$, $d\geq0$, and that the layer limit is interpreted as a maximum payment. For an unlimited layer, check whether the tail integral is finite.

Check the result.
A finite-layer expected recovery must lie between zero and $u$. For a discrete model, compare with the probability-weighted outcome recoveries.

These identities are valuable because they turn an expectation of a nonlinear treaty payment into either an integral of exceedance probabilities or a difference of two standard limited means. The same logic works for mixed aggregate distributions; no continuous-density assumption is needed. The central exam trap is to confuse the treaty's payment limit with its exhaustion point, or to apply the formula to claim-level losses when the treaty is written on aggregate loss.

Check Your Understanding

For each problem, identify the aggregate layer correctly before applying an identity. Answers are provided for checking.

A hypothetical aggregate loss is $0$ with probability $0.40$, $\$40$ with probability $0.35$, and $\$100$ with probability $0.25$. An unlimited aggregate treaty attaches at $\$50$. Find its expected recovery using the stop-loss identity. (Answer: $\$12.50$)
A hypothetical aggregate loss has an exponential distribution with mean $\$80$. A treaty attaches at $\$40$ and has a limit of $\$60$. Write the tail-integral expression and calculate the expected recovery to the nearest cent. (Answer: $80(e^{-0.5}-e^{-1.25})=\$25.60$)
Aggregate loss is $\$0$, $\$80$, or $\$140$ with probabilities $0.20$, $0.30$, and $0.50$, respectively. A treaty attaches at $\$60$ and has a $\$50$ limit. Calculate the expected recovery by outcomes and verify it using the limited-mean identity. (Answer: $\$31.00$)
A hypothetical portfolio's aggregate loss is $\$20$, $\$60$, or $\$100$ with probabilities $0.20$, $0.50$, and $0.30$. A layer attaches at $\$20$ and has a limit of $\$50$. Calculate $L_S(70)$, $L_S(20)$, and the expected recovery. (Answer: $L_S(70)=\$55.00$, $L_S(20)=\$20.00$, expected recovery $=\$35.00$)
Aggregate loss is $\$0$, $\$50$, or $\$150$ with probabilities $0.20$, $0.50$, and $0.30$. A treaty attaches at $\$50$ and has a $\$75$ limit. Account for the atom at attachment and calculate the expected recovery. (Answer: $\$22.50$)

What You'll Learn