Limited Stop-Loss Covers and Exhaustion

Introduction

A stop-loss contract begins paying when aggregate loss exceeds an attachment point. If the contract also has a payment limit, the insurer's payment cannot increase indefinitely: once aggregate loss reaches a particular threshold, the maximum payment has been reached. That threshold is the exhaustion point.

For an aggregate loss $S$, attachment point $d$, and payment limit $L$, the limited stop-loss payment is $\min((S-d)_+,L)$. Calculations require careful attention to the distinction between the attachment point, where payment starts, and the exhaustion point, where the payment reaches its limit. This tutorial develops the payment formula, its expected value, and the probabilities of zero, partial, and full-limit payments.

Definition: Let $S$ be a nonnegative aggregate loss, $d\geq 0$ an attachment point, and $L>0$ a maximum payment. The limited stop-loss payment is $$ Y=\min((S-d)_+,L), $$ where $x_+=\max(x,0)$. The exhaustion point is $d+L$. Equivalently, $$ Y= \begin{cases} 0, & S\leq d,\\ S-d, & d<S<d+L,\\ L, & S\geq d+L. \end{cases} $$

At exactly $S=d$, the payment is zero. At exactly $S=d+L$, the payment is $L$, so the cover is exhausted. These boundary conventions matter particularly for discrete aggregate losses, which can have positive probability at either threshold.

From the Attachment Point to Exhaustion

The payment increases dollar for dollar with aggregate loss while $S$ lies strictly between $d$ and $d+L$. Below or at $d$, the payment is zero; at or above $d+L$, it is capped at $L$. Thus increasing the loss beyond the exhaustion point does not increase the payment.

The limited payment can also be written as the difference between two uncapped stop-loss payments: $$ Y=(S-d)_+-(S-d-L)_+. $$ The second term removes the part of the excess loss above the exhaustion point. This representation makes it possible to calculate the expected limited payment using stop-loss transforms.

Worked Example: A Discrete Aggregate Loss and Its Exhaustion Point

A hypothetical benefit pool has aggregate loss $S$ with the following distribution. A limited stop-loss cover has attachment $d=\$50$ and payment limit $L=\$80$. Find the exhaustion point, the payment for each possible loss, the expected payment, and the probabilities of any payment and a full-limit payment.

Step 1: Find the exhaustion point. The exhaustion point is $$ d+L=\$50+\$80=\$130. $$

Step 2: Calculate each payment. Apply $Y=\min((S-50)_+,80)$ to each possible value of $S$.

Aggregate loss $S$	Probability	Payment $Y$	Calculation
\$0	0.15	\$0	At or below attachment
\$40	0.20	\$0	At or below attachment
\$90	0.25	\$40	\$90-\$50
\$130	0.25	\$80	At exhaustion
\$200	0.15	\$80	Above exhaustion; payment is capped

Step 3: Calculate the expected payment. The probabilities sum to $0.15+0.20+0.25+0.25+0.15=1$. Therefore, $$ E[Y]=(0.15)(0)+(0.20)(0)+(0.25)(40)+(0.25)(80)+(0.15)(80) =0+0+10+20+12=\$42.00. $$

Step 4: Calculate the payment probabilities. Any payment occurs when $S>d$, so $$ P(Y>0)=P(S>50)=0.25+0.25+0.15=0.65. $$ A full-limit payment occurs when $S\geq d+L=130$, so $$ P(Y=80)=P(S\geq130)=0.25+0.15=0.40. $$ The only partial payment is $\$40$, which occurs with probability $0.25$. As a check, the probabilities of no payment, a partial payment, and a full-limit payment sum to $0.35+0.25+0.40=1$.

Method and mistake avoided. Evaluate the capped payment at every possible loss before taking its expectation. In particular, losses of $\$130$ and $\$200$ both produce an $\$80$ payment; the payment does not continue increasing above exhaustion.

Expected Payment and Stop-Loss Transforms

Define the stop-loss transform of $S$ at $a$ by $\Pi_S(a)=E[(S-a)_+]$. This is the expected payment under an uncapped stop-loss contract with attachment $a$. The limited stop-loss identity gives a direct formula for the expected capped payment.

Theorem: For a nonnegative aggregate loss $S$, attachment $d\geq0$, and limit $L>0$, $$ E[\min((S-d)_+,L)]=\Pi_S(d)-\Pi_S(d+L). $$ If $S$ has a survival function $P(S>x)$, the same expectation can be written $$ E[\min((S-d)_+,L)]=\int_d^{d+L}P(S>x)\,dx. $$

Why the transform formula holds. For each possible value of $S$, consider the difference $(S-d)_+-(S-d-L)_+$. If $S\leq d$, both terms are zero. If $d<S<d+L$, the first term is $S-d$ and the second is zero. If $S\geq d+L$, subtracting $S-d-L$ from $S-d$ gives $L$. In all three cases, the difference equals $\min((S-d)_+,L)$. Taking expectations proves the transform formula.

For the tail-integral formula, write a nonnegative excess as the length of an interval: $$ (S-a)_+=\int_a^\infty \mathbf{1}_{\{S>x\}}\,dx. $$ Taking expectations gives $\Pi_S(a)=\int_a^\infty P(S>x)\,dx$. Subtracting the expressions at $a=d$ and $a=d+L$ leaves just the integral from $d$ to $d+L$. This argument applies to discrete as well as continuous distributions; point probabilities at individual boundaries do not change the integral.

Worked Example: Expected Payment for an Exponential Aggregate Loss

Suppose a hypothetical portfolio's aggregate loss $S$ has an exponential distribution with mean $\theta=\$100$, so $P(S>x)=e^{-x/100}$ for $x\geq0$. A limited stop-loss cover has $d=\$50$ and $L=\$100$. Find the exhaustion point, expected payment, and probabilities of any payment and a full-limit payment.

Step 1: Find the exhaustion point. $$ d+L=\$50+\$100=\$150. $$

Step 2: Calculate the expected payment. For an exponential loss with mean $\theta$, $\Pi_S(a)=\theta e^{-a/\theta}$. This follows by integrating its survival function: $$ \Pi_S(a)=\int_a^\infty e^{-x/\theta}\,dx=\theta e^{-a/\theta}. $$ Using the difference of stop-loss transforms, $$ E[Y]=100e^{-50/100}-100e^{-150/100} =100e^{-0.5}-100e^{-1.5}. $$ The two terms are approximately $60.653066$ and $22.313016$, respectively. Thus $$ E[Y]\approx 60.653066-22.313016=38.340050, $$ which rounds to $\$38.34$ to the nearest cent.

Step 3: Calculate the payment probabilities. Any payment requires $S>50$, so $$ P(Y>0)=P(S>50)=e^{-0.5}\approx0.6065. $$ A full-limit payment occurs at $S\geq150$. The exponential distribution is continuous, so $P(S\geq150)=P(S>150)$, giving $$ P(Y=100)=e^{-1.5}\approx0.2231. $$ The probability of a positive partial payment is $0.60653066-0.22313016=0.38340050$, or $0.3834$. As a check, the three probabilities are approximately $0.3935$ for no payment, $0.3834$ for a partial payment, and $0.2231$ for a full-limit payment; they sum to $1.0000$.

Method and mistake avoided. The two transform terms use different attachment points, $50$ and $150$. Subtracting the stop-loss expectation at the attachment alone would give the expected uncapped payment, not the expected payment under this limited cover.

Atoms at the Attachment and Exhaustion Points

For discrete losses, the distinctions between strict and non-strict inequalities determine whether boundary outcomes count. The relevant events are $$ \begin{aligned} Y=0 &\iff S\leq d,\\ 0<Y<L &\iff d<S<d+L,\\ Y=L &\iff S\geq d+L. \end{aligned} $$ Consequently, the probability of a full-limit payment includes an atom at the exhaustion point, while the probability of any payment excludes an atom at the attachment point. These definitions divide the possible outcomes into three disjoint categories.

Worked Example: Boundary Losses in a Discrete Distribution

A hypothetical group has aggregate loss $S$ equal to $\$25,\$65,\$105,\$145,$ or $\$205$, with respective probabilities $0.12,0.18,0.28,0.24,$ and $0.18$. A limited stop-loss cover has $d=\$65$ and $L=\$80$. Find the expected payment and the probabilities of any payment, a partial payment, and a full-limit payment.

Step 1: Locate both thresholds. The attachment is $\$65$, and the exhaustion point is $$ d+L=\$65+\$80=\$145. $$

Step 2: Apply the payment formula. At $S=\$25$ and $S=\$65$, payment is zero. At $S=\$105$, payment is $\$105-\$65=\$40$. At $S=\$145$, payment is exactly the limit, $\$80$; at $S=\$205$, it remains capped at $\$80$.

Step 3: Calculate the expectation. The probabilities sum to $0.12+0.18+0.28+0.24+0.18=1$. Hence $$ E[Y]=(0.12)(0)+(0.18)(0)+(0.28)(40)+(0.24)(80)+(0.18)(80) =0+0+11.20+19.20+14.40=\$44.80. $$

Step 4: Classify the outcomes. Any payment occurs for $S>65$, so $P(Y>0)=0.28+0.24+0.18=0.70$. A partial payment requires $65<S<145$, which here occurs only at $S=105$, so the probability is $0.28$. A full-limit payment requires $S\geq145$, including the outcome exactly at exhaustion; its probability is $0.24+0.18=0.42$. No payment has probability $0.12+0.18=0.30$, and $0.30+0.28+0.42=1$.

Method and mistake avoided. Compare the loss to the thresholds before classifying it. A loss exactly at the attachment produces no payment; a loss exactly at exhaustion produces a full-limit payment.

Common Exam Trap and Practical Checks

A frequent error is to use a strict inequality for a full-limit payment even when the aggregate distribution is discrete. For the distribution in the preceding example, an incorrect calculation would be $$ P(Y=80)=P(S>145)=0.18, $$ because it counts only the $\$205$ outcome. The correct event is $S\geq145$, so the atom at $\$145$ must be included: $$ P(Y=80)=P(S\geq145)=0.24+0.18=0.42. $$

Other useful checks are to verify that the exhaustion point is $d+L$, not $L$ alone; that every payment lies between $0$ and $L$; and that the probabilities of no payment, a partial payment, and a full-limit payment add to one. If using stop-loss transforms, check that the second transform is evaluated at the exhaustion point $d+L$.

The expected payment is an expected claim cost for the specified cover. It is not automatically a premium: a premium may also reflect expenses, risk margins, or other pricing assumptions. The limited stop-loss calculation itself depends only on the aggregate-loss distribution, attachment, and payment limit.

Key takeaway: A limited stop-loss cover attaches at $d$ and exhausts at $d+L$. Its payment is zero for $S\leq d$, increases as $S-d$ between the thresholds, and equals $L$ for $S\geq d+L$. Its expected payment is $\Pi_S(d)-\Pi_S(d+L)$.

Check Your Understanding

For each problem, identify the exhaustion point before calculating payments. Probabilities are exact unless a rounding instruction is given.

A hypothetical aggregate loss is $\$20,\$60,\$100,$ or $\$140$, with probabilities $0.30,0.40,0.20,$ and $0.10$. For $d=\$40$ and $L=\$60$, find the expected payment and probability of a full-limit payment. (Answer: exhaustion point $\$100$; expected payment $\$26.00$; full-limit probability $0.30$.)
A hypothetical aggregate loss is exponential with mean $\$120$. For $d=\$60$ and $L=\$120$, find the expected payment to the nearest cent and the probabilities of any payment and a full-limit payment to four decimals. (Answer: exhaustion point $\$180$; expected payment $\$46.01$; probabilities $0.6065$ and $0.2231$, respectively.)
A hypothetical aggregate loss is $\$0,\$50,\$90,\$120,$ or $\$180$, with probabilities $0.10,0.20,0.30,0.25,$ and $0.15$. For $d=\$50$ and $L=\$70$, find the expected payment, partial-payment probability, and full-limit probability. (Answer: exhaustion point $\$120$; expected payment $\$40.00$; partial-payment probability $0.30$; full-limit probability $0.40$.)
A hypothetical aggregate loss is $\$10,\$40,\$70,$ or $\$100$, with probabilities $0.20,0.30,0.30,$ and $0.20$. For $d=\$30$ and $L=\$50$, find the expected payment and probability of a positive partial payment. (Answer: exhaustion point $\$80$; expected payment $\$25.00$; partial-payment probability $0.60$.)
A hypothetical aggregate loss is exponential with mean $\$50$. For $d=\$25$ and $L=\$75$, find the expected payment to the nearest cent and the probabilities of any payment and a full-limit payment to four decimals. (Answer: exhaustion point $\$100$; expected payment $\$23.56$; probabilities $0.6065$ and $0.1353$, respectively.)

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