Discrete Aggregate Losses and Stop-Loss Sums

Why Discrete Stop-Loss Sums Need Care

A stop-loss cover with attachment $d$ pays $(S-d)^+=\max(S-d,0)$ when aggregate loss is $S$. Its expected payment is therefore $E[(S-d)^+]$. If $S$ is discrete, this expectation can be calculated from the probability mass function (pmf), but it can also be written as a sum of tail probabilities. That second form is especially useful when the aggregate distribution is already available as a table of exceedance probabilities.

The main detail is the location of the attachment relative to the integer loss values. At an integer attachment, each strict tail probability contributes one unit of excess. At a noninteger attachment, the first tail contribution must be adjusted for the fraction between adjacent integers. Using the integer formula unchanged at a noninteger attachment gives the wrong expected payment.

Definition: Let $S$ be a nonnegative integer-valued aggregate loss, with $p_s=P(S=s)$ for integers $s\geq 0$. Define the strict tail probability at integer $r\geq 0$ by $q_r=P(S>r)=\sum_{s=r+1}^{\infty}p_s$. For attachment $d\geq 0$, the stop-loss payment is $(S-d)^+=\max(S-d,0)$, and the expected stop-loss payment is $E[(S-d)^+]$.

The formulas below apply to losses measured in whole units, such as whole dollars or thousands of dollars. If losses lie on a different regular grid, first express them in units of that grid. The pmf may have finite or infinite support. When $E[S]$ is finite, the expected stop-loss payment is finite for $d\geq 0$; otherwise, the sums are still valid as nonnegative sums but may be infinite.

Deriving the Discrete Tail-Sum Formula

First take an integer attachment $k\geq 0$. For any particular integer loss value $s$, the excess $ (s-k)^+ $ is the number of integers $r$ satisfying $k\leq r<s$. For example, if $s=k+3$, the three such integers are $k$, $k+1$, and $k+2$. This gives the pointwise identity

(s-k)^+=\sum_{r=k}^{\infty}\mathbf{1}\{s>r\}.

Taking expectations replaces each indicator by its probability. Since all terms are nonnegative, the order of summation and expectation can be interchanged. Thus,

E[(S-k)^+]=\sum_{r=k}^{\infty}P(S>r)=\sum_{r=k}^{\infty}q_r.

The strict inequality in $P(S>r)$ is essential: a loss exactly equal to the attachment makes no payment. The first term is $P(S>k)$, not $P(S\geq k)$.

For a noninteger attachment, write $d=k+\alpha$, where $k=\lfloor d\rfloor$ and $0<\alpha<1$. For integer $s$, the first unit interval above $k$ contributes only $1-\alpha$ to the excess. The remaining full units each contribute one. Pointwise,

(s-d)^+=(1-\alpha)\mathbf{1}\{s>k\} +\sum_{r=k+1}^{\infty}\mathbf{1}\{s>r\}.

Taking expectations gives the noninteger form. The equivalent expressions below make clear that the fractional adjustment applies to the first strict tail probability.

E[(S-d)^+] =(1-\alpha)P(S>k)+\sum_{r=k+1}^{\infty}P(S>r) =\sum_{r=k}^{\infty}P(S>r)-\alpha P(S>k).

Key result: For integer $S\geq 0$, an integer attachment $k$ gives $E[(S-k)^+]=\sum_{r=k}^{\infty}P(S>r)$. For $d=k+\alpha$, with $0<\alpha<1$, reduce that tail sum by $\alpha P(S>k)$. Equivalently, give the first tail probability the weight $1-\alpha$, and retain the later tails at full weight.

Calculating from a Probability Mass Function

A pmf calculation sums the payment at each possible loss value, weighted by the probability of that value:

E[(S-d)^+]=\sum_{s=0}^{\infty}(s-d)^+p_s.

For a finite table, terms at or below the attachment are zero and can be omitted. The tail-sum method reorganizes the same calculation: a loss several units above the attachment appears in several successive tail probabilities. Both methods must give the same expected payment.

Worked Example: Integer Attachment from a Finite Aggregate Table

A hypothetical portfolio has aggregate loss $S$, measured in thousands of dollars, with the following pmf. Calculate the expected payment under a stop-loss cover with attachment $k=2$ thousand dollars.

$s$	0	1	2	3	4	5
$p_s$	0.10	0.20	0.25	0.15	0.20	0.10

Method. Verify the probabilities, calculate the strict tails beginning at the attachment, and sum them. This avoids counting losses at the attachment as payments.

Check the pmf. The probabilities sum to $0.10+0.20+0.25+0.15+0.20+0.10=1.00$.

Calculate the needed tails. $P(S>2)=0.15+0.20+0.10=0.45$, $P(S>3)=0.20+0.10=0.30$, and $P(S>4)=0.10$. The tail beyond 5 is zero.

Apply the formula. The attachment is the integer $k=2$, so

$$ E[(S-2)^+]=0.45+0.30+0.10=0.85 $$

As a check, the direct pmf calculation is $1(0.15)+2(0.20)+3(0.10)=0.15+0.40+0.30=0.85$ thousand dollars. The two methods agree, so the expected payment is $850. The common mistake avoided is including $P(S=2)$, even though a loss of exactly 2 thousand dollars produces no excess payment.

Noninteger Attachments and the Fractional Adjustment

For $d=k+\alpha$, losses are still integer-valued even though the attachment is not. The first possible loss above the attachment is $k+1$; its payment is $1-\alpha$, not a full unit. One may calculate directly from the pmf or use the adjusted tail sum. The direct calculation is a useful independent check when a table has only a few possible loss values.

Worked Example: A Fractional Attachment

A hypothetical event portfolio has integer aggregate loss $S$, in thousands of dollars, with probabilities $P(S=0)=0.15$, $P(S=1)=0.10$, $P(S=2)=0.25$, $P(S=3)=0.20$, $P(S=5)=0.20$, and $P(S=6)=0.10$. Find the expected stop-loss payment for attachment $d=3.4$ thousand dollars.

Method. Write $d=3+0.4$, so $k=3$ and $\alpha=0.4$. Use the fractional tail formula, then verify by summing the positive pmf payments.

Find the tails. $P(S>3)=P(S=5)+P(S=6)=0.20+0.10=0.30$. Also $P(S>4)=0.30$ and $P(S>5)=0.10$; $P(S>6)=0$.

Apply the fractional adjustment.

$$ E[(S-3.4)^+] =(1-0.4)(0.30)+0.30+0.10 =0.18+0.30+0.10 =0.58 $$

Check directly from the pmf. Losses 0, 1, 2, and 3 make no payment. At loss 5, the payment is $5-3.4=1.6$, contributing $1.6(0.20)=0.32$. At loss 6, the payment is $6-3.4=2.6$, contributing $2.6(0.10)=0.26$. The total is $0.32+0.26=0.58$ thousand dollars, or $580.

The common mistake is to use the integer-attachment sum $0.30+0.30+0.10=0.70$ without adjusting its first term. That sum values each loss above 3 as though the attachment were 3, not 3.4. Subtracting $0.4P(S>3)=0.12$ gives the correct $0.70-0.12=0.58$.

Building the Aggregate Distribution Before Summing

Sometimes the aggregate pmf is not supplied directly. It may instead be obtained from a claim count and individual claim severities. For a simple model, let $N$ be the number of claims and let $X_1,\ldots,X_N$ be their severities; then $S=\sum_{j=1}^N X_j$, with $S=0$ when $N=0$. Once the probabilities of each possible aggregate loss have been established, the discrete stop-loss formulas apply as before. The calculation requires the claim-count distribution and severity assumptions to be stated, including any independence assumption used to form the aggregate probabilities.

Worked Example: Stop-Loss Payment from a Compound Binomial Model

A hypothetical coverage group has claim count $N\sim\operatorname{Binomial}(2,0.5)$. Conditional on a claim, its severity is 1 or 3 thousand dollars, each with probability 0.5. Claim severities are independent of one another and of $N$. Find the expected stop-loss payment at attachment $d=2.5$ thousand dollars.

Method. Construct the aggregate pmf by conditioning on the claim count. Then calculate the strict tail probabilities and apply the fractional-attachment formula.

For zero claims, $P(N=0)=0.25$, giving $S=0$ with probability 0.25.

For one claim, $P(N=1)=0.50$. The aggregate is 1 or 3, each with conditional probability 0.50, so $P(S=1)=0.25$ and $P(S=3)=0.25$.

For two claims, $P(N=2)=0.25$. The severity pairs give totals 2, 4, and 6 with conditional probabilities 0.25, 0.50, and 0.25. Therefore $P(S=2)=0.0625$, $P(S=4)=0.1250$, and $P(S=6)=0.0625$.

The aggregate pmf is $P(S=0)=0.25$, $P(S=1)=0.25$, $P(S=2)=0.0625$, $P(S=3)=0.25$, $P(S=4)=0.1250$, and $P(S=6)=0.0625$. These probabilities sum to $1.0000$.

Calculate the tails. With $d=2.5=2+0.5$, $P(S>2)=0.25+0.1250+0.0625=0.4375$. Also $P(S>3)=0.1250+0.0625=0.1875$, $P(S>4)=0.0625$, and $P(S>5)=0.0625$.

Apply the formula and check.

$$ E[(S-2.5)^+] =(1-0.5)(0.4375)+0.1875+0.0625+0.0625 =0.21875+0.3125 =0.53125 $$

Directly, the positive payments are $3-2.5=0.5$, $4-2.5=1.5$, and $6-2.5=3.5$, with probabilities 0.25, 0.125, and 0.0625. Their weighted sum is $0.5(0.25)+1.5(0.125)+3.5(0.0625)=0.125+0.1875+0.21875=0.53125$ thousand dollars. The expected payment is $531.25. The common mistake avoided is treating the first tail at $r=2$ as a full unit despite the half-unit gap from 2 to the attachment.

Exam Checks for Discrete Stop-Loss Sums

A reliable calculation follows the same sequence: identify the loss grid, write the attachment as an integer plus any fractional part, and use strict exceedance probabilities. For a finite pmf, a direct sum is an effective check, but it does not replace careful treatment of the attachment. In particular, $P(S>r)$ excludes the value $r$, while $P(S\geq r)$ includes it.

Confirm that the pmf probabilities sum to one before using them.
At integer attachment $k$, start with $P(S>k)$, not $P(S\geq k)$.
At noninteger attachment $k+\alpha$, weight $P(S>k)$ by $1-\alpha$.
For a loss several units above the attachment, include every successive tail term up to the last possible loss.
Keep the aggregate loss units consistent; a result in thousands of dollars must be converted if the requested answer is in dollars.

Check Your Understanding

Use either a direct pmf sum or the discrete tail formula. Each aggregate loss is measured in the units stated.

A hypothetical aggregate loss takes values $0,2,3,5$ with probabilities $0.20,0.30,0.30,0.20$, respectively. Find the expected stop-loss payment at attachment $2.5$. (Answer: $0.65$ loss units)
A discrete aggregate loss has probabilities $0.10,0.20,0.20,0.30,0.20$ at losses $0,1,2,4,6$, respectively. Calculate the expected payment at integer attachment $3$, showing which strict tails contribute. (Answer: $0.90$ loss units)
A hypothetical portfolio has probabilities $0.25,0.25,0.30,0.20$ at aggregate losses $0,1,2,3$. Find the expected payment at attachment $1.75$. (Answer: $0.325$ loss units)
A claim count is $N\sim\operatorname{Binomial}(2,0.4)$. Each independent claim has severity 2 or 5 thousand dollars with equal probability. Find the expected stop-loss payment at attachment 4 thousand dollars. (Answer: $0.72$ thousand dollars)
A hypothetical aggregate loss has probabilities $0.15,0.15,0.20,0.25,0.15,0.10$ at losses $0,1,2,3,5,7$, respectively. Calculate the expected payment at attachment $2.2$, and give the answer in loss units. (Answer: $1.10$ loss units)

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