Deriving Aggregate Stop-Loss Expectations

Why Aggregate Stop-Loss Expectations Use the Tail

An aggregate stop-loss contract pays only when the total covered loss for a period exceeds an attachment point. Its expected payment depends not just on how large losses can be, but on the probabilities that the aggregate loss exceeds successive thresholds. The aggregate survival function captures exactly that information.

Let $S$ denote a nonnegative aggregate loss and let $d\geq 0$ be an aggregate attachment point. The payment under unlimited aggregate stop-loss cover is $(S-d)_+$, where $z_+=\max(z,0)$. A contract with a further limit $u\geq 0$ pays no more than $u$: its recovery is $\min((S-d)_+,u)$. In this tutorial, the expected recovery is the pure expected claim cost, without expenses or a risk loading.

Definition: The aggregate distribution function is $F_S(x)=P(S\leq x)$, and its survival function is $P(S>x)=1-F_S(x)$. The unlimited stop-loss expectation at attachment $d$ is $E[(S-d)_+]$. The limited aggregate recovery with attachment $d$ and limit $u$ is $\min((S-d)_+,u)$.

For an aggregate loss with a finite mean, the expected unlimited recovery is finite. A finite limit gives a bounded recovery, so its expectation is finite even when $S$ does not have a finite mean. The formulas below assume a nonnegative aggregate loss and a nonnegative attachment; the limited-layer formula also assumes a nonnegative limit.

Deriving the Tail-Integral Formula

The key idea is to express a payment amount as the total length of the thresholds it exceeds. For any fixed realized aggregate loss $S=s$, $(s-d)_+$ is the length of the interval of $x$-values from $d$ up to $s$, if $s>d$, and is zero otherwise. Equivalently,

(S-d)_+=\int_d^\infty \mathbf{1}_{\{S>x\}}\,dx.

Taking expectations and interchanging the nonnegative integral and expectation gives the result. The interchange is valid for a nonnegative integrand; if the mean is finite, the resulting expectation is finite as well.

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

This derivation also explains why the entire tail above the attachment matters: each threshold $x\geq d$ contributes its probability of being exceeded. For a limited layer, the same argument stops at $d+u$, because no payment can exceed $u$.

Theorem: For $S\geq 0$, $d\geq 0$, and $u\geq 0$,

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

The second equality follows pointwise from $\min((S-d)_+,u)=\int_d^{d+u}\mathbf{1}_{\{S>x\}}\,dx$. It also follows by subtracting two unlimited stop-loss expectations:

\min((S-d)_+,u)=(S-d)_+-(S-d-u)_+.

Thus the expected limited recovery is the expected excess above $d$, less the expected excess above $d+u$. This difference is useful when stop-loss expectations at both attachment points are already available.

Working with Discrete Aggregate Losses

For a discrete aggregate distribution, there are two convenient approaches. One is to calculate the payment at every possible value of $S$, multiply by its probability, and sum. The other is to construct the aggregate survival function and integrate its stepwise-constant values. The direct sum is usually simplest for a short probability table; the tail calculation makes the theorem visible and can be efficient for a long distribution.

E[(S-d)_+]=\sum_s (s-d)_+P(S=s), \qquad E[\min((S-d)_+,u)] =\sum_s \min((s-d)_+,u)P(S=s).

If $S$ takes integer values and $d$ is an integer, the unlimited expectation can also be written as a sum of tail probabilities: $E[(S-d)_+]=\sum_{k=d}^{\infty}P(S>k)$. Each unit interval contributes the probability that the loss reaches beyond that interval. For a loss on a grid with spacing $h$, the corresponding sum uses intervals of width $h$, so the tail probabilities must be multiplied by $h$. Do not use the unit-step formula unchanged for losses on a different grid.

Worked Example: A Binomial Claim Count and a Limited Aggregate Layer

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$ claims in a year. Each claim costs exactly $4, independently of the claim count, so the aggregate loss is $S=4N$. Calculate the expected unlimited stop-loss recovery at attachment $d=\$3$, and the expected recovery with the same attachment and limit $u=\$3$. Treat both expected recoveries as pure premiums.

Method. First obtain the distribution of the aggregate loss from the specified frequency model, then calculate the recovery for each possible aggregate outcome. The common mistake to avoid is applying the attachment to each claim separately: the contract here is defined on the aggregate $S$.

For $N\sim\operatorname{Binomial}(2,0.5)$, $P(N=0)=0.5^2=0.25$, $P(N=1)=2(0.5)(0.5)=0.50$, and $P(N=2)=0.5^2=0.25$. Since $S=4N$, the aggregate outcomes are $0, $4, and $8 with those same probabilities.

Aggregate loss $S$	$P(S)$	Unlimited payment $(S-3)_+$	Limited payment $\min((S-3)_+,3)$
$0	0.25	$0	$0
$4	0.50	$1	$1
$8	0.25	$5	$3

The unlimited expected recovery is $0.25(0)+0.50(1)+0.25(5)=0+0.50+1.25=\$1.75$. The limited expected recovery is $0.25(0)+0.50(1)+0.25(3)=0+0.50+0.75=\$1.25$.

As a check using the tail formula, $P(S>x)=0.75$ for $3\leq x<4$, and $P(S>x)=0.25$ for $4\leq x<8$. The unlimited expectation is $0.75(4-3)+0.25(8-4)=0.75+1.00=\$1.75$. For the limited layer, integrate only from 3 to 6: $0.75(4-3)+0.25(6-4)=0.75+0.50=\$1.25$, agreeing with the direct sums.

Continuous Aggregate Distributions

When the aggregate loss has a continuous distribution with density $f_S$, the survival function can be integrated directly. It is also possible to integrate the payment against the density. The tail form emphasizes the probabilities of exceeding thresholds; the density form emphasizes the payment at each realized loss.

E[(S-d)_+]=\int_d^\infty (s-d)f_S(s)\,ds, \qquad E[\min((S-d)_+,u)] =\int_d^{d+u}(s-d)f_S(s)\,ds+uP(S>d+u).

In the limited formula, losses between $d$ and $d+u$ produce partial payments, while losses above $d+u$ produce the full limit $u$. The formula assumes a density for the displayed integral representation. The tail-integral formula remains valid for distributions with atoms as well as continuous distributions.

Worked Example: Stop-Loss and Limited Recovery for a Uniform Aggregate Loss

A hypothetical annual aggregate loss $S$ is modeled as uniform on $[0,10]$ thousand dollars. Thus its density is $f_S(s)=0.1$ for $0\leq s\leq10$, and its survival function on that interval is $P(S>x)=(10-x)/10$. Calculate the pure expected recovery for attachment $d=4$ thousand dollars, both with unlimited cover and with limit $u=3$ thousand dollars.

Method. Integrate the aggregate survival function from the attachment to infinity for unlimited cover, and only to $d+u=7$ for limited cover. The common mistake to avoid is integrating the limited layer beyond its exhaustion point.

For unlimited cover, the survival probability is zero above 10. Therefore,

\begin{aligned} E[(S-4)_+] &=\int_4^{10}\frac{10-x}{10}\,dx\\ &=\frac{1}{10}\left[10x-\frac{x^2}{2}\right]_4^{10}\\ &=\frac{1}{10}\left((100-50)-(40-8)\right)\\ &=\frac{18}{10}=1.8. \end{aligned}

The unlimited pure expected recovery is $1.80 thousand, or $1,800.00. For the limited layer, the integral ends at $4+3=7$:

\begin{aligned} E[\min((S-4)_+,3)] &=\int_4^7\frac{10-x}{10}\,dx\\ &=\frac{1}{10}\left((70-24.5)-(40-8)\right)\\ &=\frac{13.5}{10}=1.35. \end{aligned}

The limited pure expected recovery is $1.35 thousand, or $1,350.00. It is less than the unlimited expectation, as it must be. As a second check, the limited payment is $s-4$ for $4<s<7$ and 3 for $s\geq7$; integrating those payments against the density gives $0.1\int_4^7(s-4)\,ds+0.1(3)(10-7)=0.45+0.90=1.35$ thousand.

Reading Atoms and Choosing the Right Calculation

At a discrete mass point, the value of $P(S>x)$ changes abruptly. This does not invalidate the tail-integral formula: changing the integrand at a single point does not change an integral. The strict inequality in $P(S>x)$ is consistent with $F_S(x)=P(S\leq x)$. For direct payment calculations, however, use the actual possible aggregate values, including any value exactly equal to the attachment. At $S=d$, the recovery is zero; at $S=d+u$, the limited recovery is exactly $u$.

A useful exam check is to calculate a discrete expectation in two ways when the support is short: directly from payments and probabilities, and from the tail areas. The two calculations should agree. For a larger distribution obtained by convolution or a frequency recursion, first identify the aggregate probabilities or tail probabilities needed; the stop-loss formula does not require a new frequency model.

Worked Example: A Tabulated Aggregate Distribution with an Attachment Between Outcomes

A hypothetical aggregate loss $S$, measured in thousands of dollars, has the following distribution: $P(S=0)=0.20$, $P(S=2)=0.30$, $P(S=5)=0.30$, and $P(S=9)=0.20$. Calculate the expected unlimited recovery at attachment $d=3$, and the expected limited recovery when $u=3$.

Method. Apply the payment function to each aggregate outcome, then verify the results by summing tail areas. The common mistake to avoid is treating the attachment as though it had to coincide with a possible loss value.

The probabilities sum to $0.20+0.30+0.30+0.20=1$. The unlimited payments at $S=0,2,5,9$ are respectively $0,0,2,6$ thousand. Thus the expected unlimited recovery is $0.20(0)+0.30(0)+0.30(2)+0.20(6)=0+0+0.60+1.20=1.80$ thousand.

With limit $u=3$, the payments at the same outcomes are $0,0,2,3$ thousand. The expected limited recovery is $0.20(0)+0.30(0)+0.30(2)+0.20(3)=0+0+0.60+0.60=1.20$ thousand.

For the tail check, $P(S>x)=0.50$ for $3\leq x<5$, and $P(S>x)=0.20$ for $5\leq x<9$. The unlimited expectation is $0.50(5-3)+0.20(9-5)=1.00+0.80=1.80$ thousand. The limited layer ends at $d+u=6$, so its expected recovery is $0.50(5-3)+0.20(6-5)=1.00+0.20=1.20$ thousand. The answers agree with the direct calculation.

Common Errors and Practical Checks

The tail-integral formula is compact, but several mistakes can produce plausible-looking answers. In the first example, the unconditional mean aggregate loss is $E[S]=4$ dollars and $P(S>3)=0.75$. A tempting but incorrect calculation is $P(S>3)(E[S]-3)=0.75(1)=\$0.75$. It multiplies the exceedance probability by the excess of the unconditional mean, rather than using the actual conditional distribution above the attachment. The correct expected recovery is $\$1.75$, as shown by the payment calculation or tail integral.

Other checks help catch errors before interpreting a premium:

The limited expected recovery cannot exceed the unlimited expected recovery, and cannot exceed $u$.
Increasing the attachment cannot increase the expected recovery; increasing the limit cannot decrease it.
If $d=0$, the unlimited stop-loss expectation is $E[S]$, when that mean exists.
Keep monetary units consistent. If losses are measured in thousands, the integral and expected recovery are also in thousands until converted.
Use an aggregate distribution for aggregate cover. Applying the attachment to individual claims instead changes the contract being valued.

These expectations are pure expected recovery amounts. A commercial reinsurance premium may include a loading, expenses, or other pricing adjustments, none of which is determined by the tail-expectation formula alone. The formula supplies the expected covered loss component.

Check Your Understanding

For each problem, calculate the requested expected recovery in the stated units.

A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.5)$ claims, each of fixed amount $2, independently of $N$. Find the unlimited aggregate stop-loss expectation at attachment $3. (Answer: $0.75)
A continuous aggregate loss is uniform on $[0,12]$ thousand dollars. Find the unlimited expected recovery at attachment 5 thousand dollars. (Answer: 2.0417 thousand dollars)
A discrete aggregate loss in thousands has values $0,1,4,7$ with probabilities $0.10,0.20,0.40,0.30$, respectively. Find the limited expected recovery with attachment 2 and limit 3. (Answer: 1.70 thousand dollars)
A hypothetical aggregate loss has an exponential distribution with mean 4 thousand dollars. Find its unlimited expected recovery at attachment 2 thousand dollars. (Answer: 2.4261 thousand dollars)
A discrete aggregate loss in dollars takes values $0,3,8$ with probabilities $0.40,0.35,0.25$, respectively. Find the limited expected recovery with attachment $3 and limit $2. (Answer: $0.50)

What You'll Learn