Expected Aggregate Stop-Loss Recoveries

Why Aggregate Stop-Loss Expectations Matter

An annual aggregate excess-of-loss treaty responds to the portfolio’s total covered loss for a year, not separately to each claim. Its expected recovery therefore depends on how the aggregate loss is distributed around the treaty’s attachment point and limit. A convenient way to calculate that expectation is to use the portfolio’s stop-loss expectation at two thresholds.

The key idea is that a treaty with a finite limit pays the excess above its attachment point, but only up to the limit. The expected capped payment can be calculated directly from the aggregate loss distribution, or as the difference between two uncapped stop-loss expectations. Both methods should agree when the inputs and treaty terms are interpreted consistently.

Definition: Let $S$ be the insurer’s annual aggregate loss on the treaty’s covered basis. For an attachment point $a$, the stop-loss expectation is $\pi_S(a)=E[(S-a)^+]$, where $(z)^+=\max(z,0)$. It is the expected amount by which aggregate loss exceeds $a$, without an upper limit on the payment.

Aggregate Treaty Payment and Stop-Loss Difference

Suppose the treaty attaches at $A$ and has a limit $L$, where $A\geq 0$ and $L\geq 0$. Here $L$ is the maximum recovery, or equivalently the width of the layer above $A$. The annual recovery is

R=\min\big((S-A)^+,L\big).

If $S\leq A$, there is no recovery. If $A<S<A+L$, the recovery is $S-A$. If $S\geq A+L$, the limit is exhausted and the recovery is $L$. The expectation of this payment is

E[R]=\pi_S(A)-\pi_S(A+L).

This form is useful when stop-loss expectations are already available, or when they are easier to compute than a separate distribution of treaty payments. It also makes clear that using only $\pi_S(A)$ gives the expected payment for an unlimited aggregate layer, not for a capped treaty.

Conditions: The identity assumes a nonnegative annual aggregate loss $S$ with finite expectation, a common covered-loss basis for both stop-loss expectations, and a treaty recovery of $\min((S-A)^+,L)$. It does not include expenses, reinstatement premiums, recovery delays, or discounting. Those features require additional terms or a different valuation.

Why the Difference Formula Holds

The identity follows by comparing the two stop-loss payments for each possible value of $S$, before taking expectations. For $S\leq A$, both payments are zero. For $A<S<A+L$, only the first payment is positive. For $S\geq A+L$, both payments are positive, and their difference is exactly the treaty limit.

(S-A)^+-(S-A-L)^+ = \begin{cases} 0, & S\leq A,\\ S-A, & A<S<A+L,\\ L, & S\geq A+L. \end{cases}

The right-hand side is precisely $\min((S-A)^+,L)$ in each case. Thus the equality holds for every value of $S$, not just on average. Taking expectations of both sides gives

E[R] =E[(S-A)^+]-E[(S-(A+L))^+] =\pi_S(A)-\pi_S(A+L).

When the aggregate loss is discrete, a stop-loss expectation can be found by summing the excess over the threshold, weighted by the probability of each outcome:

\pi_S(a)=\sum_s P(S=s)(s-a)^+.

Alternatively, direct calculation of $R$ outcome by outcome is often the clearest approach. The stop-loss difference then provides an efficient calculation or an independent check.

Worked Example: A Capped Treaty From a Discrete Aggregate Distribution

A hypothetical insurer has the following annual aggregate loss distribution. Loss amounts are in thousands of dollars. The annual aggregate excess-of-loss treaty attaches at $A=500$ thousand dollars and has a limit $L=700$ thousand dollars. Find the expected recovery both directly and using stop-loss expectations.

Assumptions and data. The probabilities describe the entire annual aggregate loss $S$: $P(S=100)=0.20$, $P(S=600)=0.50$, and $P(S=1{,}500)=0.30$, with losses in thousands of dollars. They sum to $1.00$. There is no mortality assumption because this is a non-life aggregate loss model. Interest is $i=0\%$ for this calculation: the recovery is valued without discounting.

Method: calculate the treaty payment for each outcome. For $S=100$, the aggregate loss is below $500$, so the recovery is $0$. For $S=600$, the loss exceeds the attachment by $600-500=100$, which is below the limit, so the recovery is $100$. For $S=1{,}500$, the excess above attachment is $1{,}500-500=1{,}000$, but the treaty pays at most $700$, so the recovery is $700$. All figures in this calculation are in thousands of dollars.

$$ E[R]=0.20(0)+0.50(100)+0.30(700)=0+50+210=260 $$

The expected recovery is $260$ thousand dollars, or $260,000.00.

Stop-loss check. At the attachment $500$, the three excess amounts are $0$, $600-500=100$, and $1{,}500-500=1{,}000$. Therefore,

\pi_S(500)=0.20(0)+0.50(100)+0.30(1{,}000)=350

The limit is exhausted at $A+L=500+700=1{,}200$. At that threshold, only the $1{,}500$ outcome has a positive excess, equal to $1{,}500-1{,}200=300$. Hence

\pi_S(1{,}200)=0.20(0)+0.50(0)+0.30(300)=90

The difference is $350-90=260$ thousand dollars, agreeing with the direct calculation. This check avoids the common mistake of treating the layer as unlimited.

Worked Example: Constructing Aggregate Outcomes From Claim Count

A hypothetical portfolio has a deterministic claim severity of $400$ thousand dollars per claim. The annual claim count $N$ has probabilities $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. An annual aggregate treaty attaches at $300$ thousand dollars and has a limit of $250$ thousand dollars. Find the expected recovery.

Assumptions and parameters. Each claim is $400$ thousand dollars, so $S=400N$ thousand dollars. The specified claim-count distribution is the only randomness; severity is fixed. There is no mortality assumption, and interest is $i=0\%$ because the calculation is for an undiscounted annual recovery.

Method: map claim counts to aggregate losses, then apply the treaty. When $N=0$, $S=0$, and the recovery is $0$. When $N=1$, $S=400$, so the excess over $300$ is $100$, below the $250$ limit; the recovery is $100$. When $N=2$, $S=800$, so the excess is $800-300=500$, capped at $250$.

$$ E[R]=0.25(0)+0.50(100)+0.25(250)=0+50+62.5=112.5 $$

Thus the expected recovery is $112.5$ thousand dollars, or $112,500.00.

Stop-loss check. At $300$, the excesses for $S=0,400,800$ are $0,100,500$. At $A+L=300+250=550$, the excesses are $0,0,250$. Therefore,

\pi_S(300)=0.25(0)+0.50(100)+0.25(500)=175

\pi_S(550)=0.25(0)+0.50(0)+0.25(250)=62.5

The stop-loss difference is $175-62.5=112.5$ thousand dollars. It matches the direct expectation. The calculation also shows why the claim count alone is not the treaty payment: first convert $N$ into the portfolio aggregate $S$, then apply the annual attachment and limit.

Worked Example: Atoms Below, Within, and Above the Layer

A hypothetical portfolio’s annual aggregate loss $S$, measured in thousands of dollars, has this distribution: $P(S=0)=0.10$, $P(S=250)=0.20$, $P(S=500)=0.30$, $P(S=900)=0.25$, and $P(S=1{,}400)=0.15$. A treaty attaches at $A=400$ and has limit $L=600$. Calculate the expected recovery.

Assumptions and parameters. The probabilities specify the aggregate loss distribution and sum to $1.00$. There is no mortality assumption; interest is $i=0\%$, so all amounts are undiscounted. Both attachment and limit are in thousands of dollars.

Method: calculate recoveries at every possible aggregate loss. For $S=0$ and $S=250$, the loss does not exceed $400$, so recovery is zero. For $S=500$, the excess is $100$. For $S=900$, the excess is $500$, still below the $600$ limit. For $S=1{,}400$, the excess is $1{,}000$, so the recovery is capped at $600$.

$$ E[R]=0.10(0)+0.20(0)+0.30(100)+0.25(500)+0.15(600) $$

$$ E[R]=0+0+30+125+90=245 $$

The expected recovery is $245$ thousand dollars, or $245,000.00.

Stop-loss check. At $400$, the positive excesses are $100$ for $S=500$, $500$ for $S=900$, and $1{,}000$ for $S=1{,}400$. Thus $\pi_S(400)=0.30(100)+0.25(500)+0.15(1{,}000)=30+125+150=305$ thousand dollars. The upper threshold is $A+L=1{,}000$; only $S=1{,}400$ exceeds it, by $400$. Hence $\pi_S(1{,}000)=0.15(400)=60$ thousand dollars. Subtracting gives $305-60=245$, which checks the direct answer.

This example includes probability mass exactly below, inside, and above the treaty layer. The payment formula handles each point without special adjustments: an atom at the attachment produces zero recovery, while an atom above the exhaustion point produces the full limit.

Interpreting the Result and Avoiding Common Errors

The expected recovery is an average over possible annual aggregate losses. It is not generally the treaty payment evaluated at the expected aggregate loss. In general, the positive-part and cap operations are nonlinear, so $E[\min((S-A)^+,L)]$ cannot be replaced by $\min((E[S]-A)^+,L)$. The distribution of $S$, particularly its probability of exceeding the attachment and exhaustion points, matters.

A frequent error is to calculate only $\pi_S(A)$ and report it as the expected recovery. In the first worked example, that incorrect approach would give $350$ thousand dollars. It is the expected payment for an unlimited layer attaching at $500$, not for the treaty limited to $700$. The correction is to subtract the stop-loss expectation at the exhaustion point:

350-\pi_S(500+700)=350-90=260

Other checks help prevent mistakes:

Confirm what the limit measures. In this tutorial, $L$ is the maximum recovery, so the upper stop-loss threshold is $A+L$.
Keep the treaty basis consistent. Apply the annual attachment to the annual aggregate $S$, not separately to each claim.
Check the bounds. For a finite limit, $0\leq E[R]\leq L$. Also, $E[R]\leq \pi_S(A)$, since the cap cannot increase a payment.
Track units. If aggregate losses and stop-loss expectations are in thousands of dollars, the final recovery is in thousands until converted.
Use only covered losses. The distribution of $S$ must reflect the loss definition and exclusions in the hypothetical treaty terms.

The approach is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus: identify the annual aggregate basis, translate the attachment and limit correctly, and calculate the expected recovery with a transparent calculation that can be checked by the stop-loss difference.

Check Your Understanding

For each problem, calculate an expected annual aggregate recovery. Unless otherwise stated, assume no discounting and no mortality model.

A hypothetical portfolio has aggregate loss $S=0,200,500$ thousand dollars with probabilities $0.40,0.30,0.30$, respectively. A treaty has $A=150$ and $L=200$, in thousands. Find the expected recovery. (Answer: $75,000.00)
A hypothetical aggregate loss, in thousands of dollars, is $100,400,800$ with probabilities $0.20,0.30,0.50$. For $A=350$ and $L=300$, find the expected recovery by direct calculation. (Answer: $165,000.00)
For a hypothetical portfolio, $\pi_S(250)=120$ thousand dollars and $\pi_S(700)=35$ thousand dollars. A treaty attaches at $250$ thousand dollars and has a $450$-thousand-dollar limit. Find the expected recovery. (Answer: $85,000.00)
A hypothetical portfolio has deterministic severity $250$ thousand dollars per claim. Its annual claim count satisfies $P(N=0)=0.20$, $P(N=1)=0.50$, and $P(N=2)=0.30$. A treaty attaches at $300$ thousand dollars and has a $400$-thousand-dollar limit. Find the expected recovery. (Answer: $60,000.00)
A hypothetical aggregate loss distribution, in thousands of dollars, is $S=0,300,600,1{,}000,1{,}600$ with probabilities $0.10,0.20,0.30,0.25,0.15$, respectively. A treaty attaches at $500$ and has limit $700$. Calculate the expected recovery directly and check it using two stop-loss expectations. (Answer: $260,000.00)

What You'll Learn