Retained and Ceded Aggregate Loss Decomposition

Introduction

An aggregate stop-loss contract divides a portfolio's total loss into two parts: the amount retained up to an attachment point and the amount ceded above it. The amounts must add to the original aggregate loss for every outcome, not merely on average. Once this pointwise identity is established, the expected retained and ceded amounts follow from the aggregate mean and the stop-loss transform.

This decomposition is useful for checking payment calculations and comparing the expected amounts borne by the insured party and the reinsurer. The methods here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $S$ be a nonnegative aggregate loss over a specified coverage period, and let $d\geq 0$ be the stop-loss attachment point. The retained loss is $R=\min(S,d)$, and the ceded loss is $C=(S-d)_+=\max(S-d,0)$. The stop-loss transform at $d$ is $\pi_S(d)=E[(S-d)_+]$, provided the expectation exists.

The contract described here has no upper limit on the ceded payment and no coinsurance percentage applied to that payment. Thus, once aggregate loss exceeds $d$, the reinsurer pays the entire amount above $d$. The attachment applies to the aggregate loss for the coverage period; it is not applied separately to each individual claim.

The Payment Identity

There are two cases. If $S\leq d$, then the retained amount is $S$ and the ceded amount is zero. If $S>d$, then the retained amount is $d$ and the ceded amount is $S-d$. In either case, the two payments add to $S$.

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

This is a pointwise identity: it holds for every value of $S$. In particular, if $S=d$, the retained amount is $d$ and the ceded amount is zero. A positive probability of loss exactly at the attachment does not change the identity.

Theorem: If $S$ is nonnegative and has a finite mean, then the expected retained and ceded losses satisfy $$ E[R]=E[S]-\pi_S(d), \qquad E[C]=\pi_S(d), \qquad E[R]+E[C]=E[S]. $$

Taking expectations of the pointwise payment identity gives the last equality. Since $C=(S-d)_+$, its expectation is the stop-loss transform $\pi_S(d)$; subtracting that expectation from $E[S]$ gives $E[R]$. This approach works whether the aggregate loss is discrete, continuous, or a mixture of the two.

Calculating the Stop-Loss Transform

The transform can be calculated directly from the distribution of $S$. For a discrete aggregate loss, multiply each possible ceded payment by its probability and add. For a continuous aggregate loss with density $f_S$, integrate the ceded payment against the density. In both cases, outcomes at or below $d$ contribute zero.

\pi_S(d)=E[(S-d)_+] =\sum_s (s-d)_+P(S=s) =\int_d^\infty (s-d)f_S(s)\,ds.

The sum applies when $S$ has a discrete distribution; the density integral applies when $S$ has a density. If a distribution has both discrete atoms and a continuous component, calculate the expectation by including the contributions from both. An atom at $S=d$ contributes zero because $(d-d)_+=0$.

For any nonnegative integrable $S$, there is also a tail-probability representation. For a fixed value $S=s$, the excess $(s-d)_+$ is the length of the interval of levels $x\geq d$ that fall below $s$. Averaging that interval length over all possible losses gives the integral of the probability that $S$ exceeds each level:

\pi_S(d)=\int_d^\infty P(S>x)\,dx.

This integral uses $P(S>x)$; at a discrete attachment, the distinction between strict and non-strict inequalities should be handled consistently. A single point in the integration variable has zero width, so an atom at a particular loss value does not add a separate area to the integral.

Identify the aggregate quantity.
Confirm that $S$ is the total loss over the period covered by the stop-loss contract, rather than an individual claim amount.

Apply the attachment to each outcome.
For each possible loss, calculate $C=(S-d)_+$ and $R=\min(S,d)$. Loss at or below $d$ produces no cession.

Find expected cession and retention.
Calculate $E[C]=\pi_S(d)$, then use $E[R]=E[S]-\pi_S(d)$. Check that their sum equals $E[S]$.

Worked Examples

Worked Example: A Discrete Aggregate Loss With an Atom at the Attachment

A hypothetical portfolio has aggregate loss $S$, measured in thousands of dollars, with probabilities $P(S=0)=0.20$, $P(S=6)=0.30$, and $P(S=14)=0.50$. The aggregate stop-loss attachment is $d=6$ thousand dollars. Find the expected ceded and retained losses. Use the discrete stop-loss transform and verify the decomposition.

Method. Calculate the ceded amount at each possible aggregate loss. The outcome $S=6$ is exactly at the attachment, so it produces no ceded payment.

At $S=0$, $C=(0-6)_+=0$. At $S=6$, $C=(6-6)_+=0$. At $S=14$, $C=(14-6)_+=8$. Therefore,

\pi_S(6)=E[C]=(0.20)(0)+(0.30)(0)+(0.50)(8)=4.

The expected aggregate loss is

$$ E[S]=(0.20)(0)+(0.30)(6)+(0.50)(14)=0+1.8+7=8.8. $$

Thus, the expected retained loss is

$$ E[R]=E[S]-E[C]=8.8-4=4.8. $$

Check. The expected payments add to $4.8+4=8.8$, equal to the expected aggregate loss. Directly, retained loss is $0$, $6$, or $6$, so $E[R]=(0.20)(0)+(0.30)(6)+(0.50)(6)=4.8$. All amounts are in thousands of dollars. The common mistake this method avoids is treating the atom at $S=d$ as a positive excess payment.

Worked Example: Building the Aggregate Distribution From Two Claims

A hypothetical one-period portfolio has two independent potential claims. Claim A pays \$5,000 with probability $0.40$, otherwise zero. Claim B pays \$9,000 with probability $0.20$, otherwise zero. There are no other losses. An aggregate stop-loss contract attaches at $d=\$7,000$. Calculate expected cession and retention.

Method. First combine the independent claim outcomes to obtain the possible values of $S$ and their probabilities. Then apply the aggregate attachment to those outcomes.

The probability of neither claim is $(1-0.40)(1-0.20)=0.60(0.80)=0.48$, giving $S=\$0$. The probability that only A occurs is $0.40(0.80)=0.32$, giving $S=\$5,000$. The probability that only B occurs is $0.60(0.20)=0.12$, giving $S=\$9,000$. The probability that both occur is $0.40(0.20)=0.08$, giving $S=\$14,000$. The probabilities sum to $0.48+0.32+0.12+0.08=1.00$.

The ceded amounts at $S=\$0,\$5,000,\$9,000,\$14,000$ are respectively $\$0,\$0,\$2,000,\$7,000$. Hence,

E[C]=(0.48)(0)+(0.32)(0)+(0.12)(2{,}000)+(0.08)(7{,}000) =240+560=\$800.

The expected aggregate loss is the sum of the expected claim amounts:

E[S]=(0.40)(5{,}000)+(0.20)(9{,}000)=2{,}000+1{,}800=\$3{,}800.

Therefore,

E[R]=E[S]-E[C]=\$3{,}800-\$800=\$3{,}000.

Check. The retained payments across the four outcomes are $\$0,\$5,000,\$7,000,\$7,000$. Their expectation is $(0.48)(0)+(0.32)(5{,}000)+(0.12)(7{,}000)+(0.08)(7{,}000)=1{,}600+840+560=\$3,000$. The expected retained and ceded amounts sum to $\$3,800$. This calculation avoids the mistake of applying the attachment separately to each claim: the contract attaches to $S$, the total loss.

Worked Example: An Exponential Aggregate Loss

For a hypothetical portfolio, suppose the aggregate loss $S$, in thousands of dollars, has an exponential distribution with mean $20$. Its density is $f_S(s)=\frac{1}{20}e^{-s/20}$ for $s\geq0$, and $E[S]=20$. A stop-loss contract attaches at $d=12$ thousand dollars. Find the expected ceded and retained amounts in dollars, rounded to the nearest cent.

Method. Integrate the ceded amount $s-12$ against the density above the attachment. Below the attachment, the ceded amount is zero.

\begin{aligned} E[C] &=\int_{12}^{\infty}(s-12)\frac{1}{20}e^{-s/20}\,ds\\ &=20e^{-12/20}\\ &=20e^{-0.6}\\ &\approx 10.9762327 \text{ thousand dollars}. \end{aligned}

The integral can be evaluated by setting $y=s-12$, so that $s=y+12$, and changing the lower limit to zero. It becomes $e^{-12/20}\int_0^\infty y\frac{1}{20}e^{-y/20}\,dy$. The remaining integral is the mean, $20$, of an exponential distribution with mean $20$; this gives $20e^{-12/20}$.

The expected retained loss is

E[R]=E[S]-E[C]=20-10.9762327=9.0237673 \text{ thousand dollars}.

Converting to dollars and rounding only at the final step gives expected cession $10.9762327(1{,}000)=\$10{,}976.23$ and expected retention $9.0237673(1{,}000)=\$9{,}023.77$, each rounded to the nearest cent.

Check. The rounded amounts add to $\$20,000.00$, the expected aggregate loss. The common mistake avoided here is integrating $s f_S(s)$ above the attachment without subtracting $12$; that would calculate the expected loss above the attachment, not the expected excess payment.

Common Mistakes and Checks

A stop-loss transform is an unconditional expected payment. It includes both the chance that the loss exceeds the attachment and the size of the excess when it does. For example, in the first worked example, conditional on $S>6$, the aggregate loss is $14$ and the excess is $8$ thousand dollars. The conditional excess is $8$, but the unconditional expected cession is $(0.50)(8)=4$ thousand dollars. Reporting $8$ as the expected cession would omit the probability of exceeding the attachment.

Do not cede the attachment itself. At $S=d$, the ceded payment is zero, not $d$.
Do not apply an aggregate attachment claim by claim. First form the total $S$, then calculate $(S-d)_+$.
Do not confuse expected retained loss with the attachment. Retention equals $d$ only when $S\geq d$; otherwise, it equals $S$.
Check units. If $S$ and $d$ are stated in thousands, the transform and expected retention are also in thousands until converted.
Use the identity as a check. If calculated expectations do not satisfy $E[R]+E[C]=E[S]$, revisit the outcome payments or the unit conversion.

Key takeaway: For an unlimited aggregate stop-loss contract, calculate ceded loss as $(S-d)_+$ and retained loss as $\min(S,d)$. Their sum equals $S$ for every outcome, so $E[C]=\pi_S(d)$ and $E[R]=E[S]-\pi_S(d)$.

Check Your Understanding

For each problem, find the requested expected amounts and check that expected retention plus expected cession equals the expected aggregate loss.

A hypothetical aggregate loss $S$, in thousands of dollars, equals $0$, $3$, or $8$ with probabilities $0.20$, $0.50$, or $0.30$, respectively. The attachment is $d=5$. Find expected cession and expected retention. (Answer: expected cession $0.9$ thousand dollars; expected retention $3.0$ thousand dollars.)
A hypothetical aggregate loss $S$, in thousands of dollars, equals $2$, $6$, or $12$ with probabilities $0.25$, $0.50$, or $0.25$, respectively. The attachment is $d=6$. Find expected cession and expected retention, accounting for the atom at the attachment. (Answer: expected cession $1.5$ thousand dollars; expected retention $5.0$ thousand dollars.)
Two independent potential claims pay \$5,000 with probability $0.40$ and \$9,000 with probability $0.20$, respectively; otherwise each pays zero. For an aggregate attachment of \$7,000, calculate expected cession and expected retention. (Answer: expected cession $\$800$; expected retention $\$3,000$.)
A hypothetical aggregate loss is uniform on the interval from $0$ to $80$ thousand dollars. With an attachment of $20$ thousand dollars, use the tail-probability integral to find expected cession and expected retention. (Answer: expected cession $22.5$ thousand dollars; expected retention $17.5$ thousand dollars.)
A hypothetical aggregate loss $S$, in thousands of dollars, is exponential with mean $10$, so $P(S>x)=e^{-x/10}$ for $x\geq0$. The attachment is $8$ thousand dollars. Use the tail-probability representation to find expected cession and expected retention, rounded to three decimal places in thousands. (Answer: expected cession $10e^{-0.8}\approx4.493$ thousand dollars; expected retention $10-10e^{-0.8}\approx5.507$ thousand dollars.)

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