Stop-Loss Transform of Aggregate Loss

Why the Stop-Loss Transform Matters

An aggregate stop-loss contract responds to the total loss over a specified period, not separately to each claim. If the aggregate loss is below an attachment point, the contract pays nothing; above that point, it pays the excess. The expected payment is a useful measure of the contract’s expected claim cost, but it is not automatically a quoted premium: expenses, risk margins, and other pricing provisions may also matter.

This tutorial develops the stop-loss transform as an expectation and shows how to calculate it from an aggregate-loss distribution. The calculations use hypothetical loss models. They assume the attachment applies once to the aggregate total; a per-claim deductible or a payment cap is a different contract feature.

Definition: Let $S$ be a nonnegative aggregate loss with finite expectation, and let $d\geq 0$ be an aggregate attachment. The stop-loss payment is $(S-d)^+=\max(S-d,0)$. The stop-loss transform at attachment $d$ is $$ \Pi(d)=E[(S-d)^+]. $$ It is the expected amount by which the aggregate loss exceeds $d$.

Calculating the Transform from a Distribution

For a discrete aggregate loss, list every possible value of $S$, determine the payment $(S-d)^+$ at each value, and weight that payment by its probability. Values at or below the attachment contribute zero. If losses are stated in thousands of dollars, the transform is first obtained in thousands and then converted to dollars if needed.

Worked Example: Direct Calculation and Expected Retention

A hypothetical annual aggregate loss $S$, measured in thousands of dollars, has the following distribution: $P(S=0)=0.30$, $P(S=2)=0.25$, $P(S=7)=0.20$, and $P(S=11)=0.25$. Find $\Pi(5)$, the expected retained loss under an aggregate attachment of 5, and verify their sum against $E[S]$.

Method. Evaluate the excess over 5 at each aggregate outcome. The payment is $(S-5)^+$, so the four possible payments, in thousands of dollars, are

(0-5)^+=0,\qquad (2-5)^+=0,\qquad (7-5)^+=2,\qquad (11-5)^+=6.

Weight each payment by its probability:

\Pi(5)=0.30(0)+0.25(0)+0.20(2)+0.25(6) =0+0+0.40+1.50=1.90.

Thus the expected stop-loss payment is $1.90$ thousand dollars, or $1,900.00.

The retained amount is $\min(S,5)$. Its possible values are 0, 2, 5, and 5, so

E[\min(S,5)] =0.30(0)+0.25(2)+0.20(5)+0.25(5) =0+0.50+1.00+1.25=2.75.

Expected retention is $2.75$ thousand dollars, or $2,750.00. As a check, the expected aggregate loss is

$$ E[S]=0.30(0)+0.25(2)+0.20(7)+0.25(11) =0+0.50+1.40+2.75=4.65 $$

thousand dollars, or $4,650.00. The payment and retention add to the mean: $1.90+2.75=4.65$. This check confirms that the two expected amounts reconcile with the expected total loss.

Why Payment and Retention Reconcile

For each realized aggregate loss $S$, the retained amount below the attachment and the amount above it partition the total. If $S\leq d$, the retained amount is $S$ and the stop-loss payment is zero. If $S>d$, the retained amount is $d$ and the payment is $S-d$. In either case, the pointwise identity is

S=\min(S,d)+(S-d)^+.

Taking expectations gives the corresponding mean identity. This derivation explains why the expected stop-loss payment can be checked by comparing expected total loss with expected retention; it does not require a continuous distribution or a special aggregate model.

E[S]=E[\min(S,d)]+\Pi(d), \qquad \Pi(d)=E[S]-E[\min(S,d)].

The identity is valid when $S$ is nonnegative and has finite expectation, as assumed here. It is a useful accounting check, not a substitute for calculating the correct payment at each aggregate outcome.

Building the Aggregate Outcomes First

When an aggregate loss is formed from several component losses, first determine the possible totals and their probabilities. Only then apply the aggregate attachment. In particular, an aggregate attachment is not separately deducted from each component loss. For independent components, joint outcome probabilities are products of the component probabilities.

Worked Example: Independent Losses and an Aggregate Attachment

A hypothetical portfolio has two independent losses, in thousands of dollars. The first loss $X_1$ is 0 or 8, each with probability 0.50. The second loss $X_2$ is 0 with probability 0.70 and 10 with probability 0.30. Let $S=X_1+X_2$, and calculate $\Pi(10)$.

Method. Enumerate the joint outcomes. Independence means each joint probability is the product of the two component probabilities:

$X_1=0,\ X_2=0$: $S=0$, probability $0.50(0.70)=0.35$.
$X_1=8,\ X_2=0$: $S=8$, probability $0.50(0.70)=0.35$.
$X_1=0,\ X_2=10$: $S=10$, probability $0.50(0.30)=0.15$.
$X_1=8,\ X_2=10$: $S=18$, probability $0.50(0.30)=0.15$.

These probabilities sum to $0.35+0.35+0.15+0.15=1.00$. At an attachment of 10, the respective stop-loss payments are 0, 0, 0, and 8, all in thousands of dollars. In particular, an outcome equal to the attachment has zero excess.

\Pi(10)=0.35(0)+0.35(0)+0.15(0)+0.15(8)=1.20.

The expected payment is $1.20$ thousand dollars, or $1,200.00. As a check, the mean aggregate loss is $E[S]=0.50(8)+0.30(10)=7.00$ thousand dollars. Expected retention is $0.35(0)+0.35(8)+0.15(10)+0.15(10)=5.80$ thousand dollars, and $5.80+1.20=7.00$.

Common mistake avoided. Applying a deductible of 10 to each component would not describe this contract. The contract compares the total $X_1+X_2$ with 10. Also, the outcome $S=10$ contributes no payment; only the joint outcome $S=18$ contributes here.

How the Attachment Changes the Transform

For a fixed loss distribution, raising the attachment cannot increase the payment at any realized loss: $(S-d)^+$ is nonincreasing in $d$. Consequently, $\Pi(d)$ is nonincreasing. The transform is also convex as a function of the attachment. For each fixed $S=s$, the function $(s-d)^+$ is the maximum of the two affine functions $s-d$ and 0, hence is convex in $d$. Taking an expectation preserves the convexity inequality, so $\Pi(d)$ is convex as well.

For a discrete aggregate distribution, the transform is found by summing outcome-by-outcome payments. As the attachment passes a possible aggregate-loss value, that outcome’s payment falls to zero. This makes it especially important to handle equality correctly: an outcome exactly equal to $d$ has payment $(d-d)^+=0$.

Worked Example: Comparing Two Attachments

A hypothetical aggregate loss $S$, in thousands of dollars, equals 0 with probability 0.20, 4 with probability 0.30, 9 with probability 0.30, and 15 with probability 0.20. Calculate the stop-loss transform at attachments 6 and 10, then check the lower-attachment result using expected retention.

At attachment 6. The payments at aggregate losses 0, 4, 9, and 15 are respectively 0, 0, 3, and 9. Therefore

\Pi(6)=0.20(0)+0.30(0)+0.30(3)+0.20(9) =0+0+0.90+1.80=2.70.

Thus $\Pi(6)=2.70$ thousand dollars, or $2,700.00.

At attachment 10. Only the outcome 15 exceeds 10, with payment 5:

\Pi(10)=0.20(0)+0.30(0)+0.30(0)+0.20(5)=1.00.

Thus $\Pi(10)=1.00$ thousand dollars, or $1,000.00. The result is smaller at the higher attachment, as expected.

For a check at attachment 6, the expected total loss is

$$ E[S]=0.20(0)+0.30(4)+0.30(9)+0.20(15) =0+1.20+2.70+3.00=6.90. $$

Expected retention at 6 is $0.20(0)+0.30(4)+0.30(6)+0.20(6)=0+1.20+1.80+1.20=4.20$ thousand dollars. Hence $4.20+2.70=6.90$, matching the expected total loss.

Common Exam Traps and Practical Checks

A reliable calculation keeps the contract definition, the aggregate distribution, and the units aligned. Before summing, confirm that the attachment applies to $S$, identify which aggregate outcomes strictly exceed it, and calculate the excess for each such outcome. Then weight by the probabilities. If the input losses are in thousands, retain that unit throughout the sum and convert to dollars only at the end.

Deducting before aggregation. For aggregate stop-loss, calculate the total loss first and apply the attachment to that total.
Paying at equality. If $S=d$, then $(S-d)^+=0$, not a positive amount.
Using a conditional average instead of an expected payment. $\Pi(d)$ averages payments over all outcomes, including zero-payment outcomes. It is not the average excess conditional on exceeding $d$.
Forgetting probabilities or units. An excess amount must be multiplied by the probability of its aggregate outcome; a result in thousands must be converted before reporting dollars.
Skipping a check. When the retained amount is $\min(S,d)$, the identity $E[S]=E[\min(S,d)]+\Pi(d)$ provides an independent arithmetic check.

The transform therefore summarizes expected payment for an aggregate excess cover at a specified attachment. Its value depends on the full distribution of the aggregate loss, not just on the attachment or the average loss. This is why a correctly constructed aggregate distribution is essential before calculating the expected payment.

Check Your Understanding

For each question, apply the attachment to the aggregate loss and keep track of the stated units.

A hypothetical aggregate loss, in thousands of dollars, is 0 with probability 0.40, 3 with probability 0.25, and 8 with probability 0.35. Find $\Pi(5)$. (Answer: $1.05$ thousand dollars, or $1,050.00.)
Independent losses in thousands are $X_1=0$ or 4, each with probability 0.50, and $X_2=0$ with probability 0.75 or 12 with probability 0.25. For $S=X_1+X_2$, find $\Pi(10)$. (Answer: $1.00$ thousand dollars, or $1,000.00.)
A hypothetical aggregate loss in thousands is 0 with probability 0.20, 5 with probability 0.30, and 12 with probability 0.50. Find the transform at attachment 7. (Answer: $2.50$ thousand dollars, or $2,500.00.)
A hypothetical aggregate loss in thousands is 0 with probability 0.20, 4 with probability 0.30, and 10 with probability 0.50. Find $\Pi(6)$ and expected retention. (Answer: $\Pi(6)=2.00$ thousand dollars and expected retention $=4.20$ thousand dollars.)
A hypothetical aggregate loss in thousands is 0 with probability 0.10, 2 with probability 0.20, 6 with probability 0.40, and 12 with probability 0.30. Find $\Pi(3)$, $\Pi(8)$, and the decrease between them. (Answer: $\Pi(3)=3.90$ thousand dollars, $\Pi(8)=1.20$ thousand dollars, and the decrease is $2.70$ thousand dollars.)

What You'll Learn