Discrete Stop-Loss Expectations From Aggregate Probabilities

From an Aggregate Distribution to a Stop-Loss Expectation

An aggregate stop-loss contract pays when the total loss in a period exceeds a specified deductible. The payment is the amount of the excess, not the entire aggregate loss. When the aggregate loss is discrete, the expected payment can be calculated directly from its probability mass function. The key detail is to compare each attainable aggregate loss with the deductible, even when the deductible itself is not an attainable value.

Let $Z$ denote the nonnegative aggregate loss for a period, and let $d\geq 0$ be the aggregate deductible. This tutorial assumes a discrete distribution with attainable values $z_j$ and probabilities $p_j=P(Z=z_j)$, where the probabilities sum to one. There is no discounting in the examples; interest and mortality assumptions are not needed for this aggregate-loss calculation.

Definition: The aggregate stop-loss payment is $(Z-d)_+=\max(Z-d,0)$. Its expected value is $$ E[(Z-d)_+]=\sum_j \max(z_j-d,0)p_j =\sum_{j:z_j>d}(z_j-d)p_j. $$ An outcome equal to the deductible contributes zero. An outcome below the deductible also contributes zero.

The strict inequality $z_j>d$ is useful when selecting outcomes for the sum. If $z_j=d$, then $z_j-d=0$, so including or excluding that outcome makes no numerical difference; using $z_j>d$ makes clear that only positive payments are being summed. If the deductible lies between two attainable losses, the first contributing outcome is the first support value above the deductible, not the closest value on either side.

Why the Discrete Sum Gives the Expected Payment

For a discrete random variable, the expectation of a function of that variable is the probability-weighted sum of the function's values. Here the function is $g(z)=\max(z-d,0)$. At an outcome $z_j\leq d$, $g(z_j)=0$; at an outcome $z_j>d$, $g(z_j)=z_j-d$. Substituting those two cases into the expectation gives the stop-loss formula.

\begin{aligned} E[(Z-d)_+] &=\sum_j g(z_j)P(Z=z_j)\\ &=\sum_{j:z_j\leq d}0\cdot p_j +\sum_{j:z_j>d}(z_j-d)p_j\\ &=\sum_{j:z_j>d}(z_j-d)p_j. \end{aligned}

This also explains why summing $z_jp_j$ over losses above $d$ is not sufficient: that sum weights each event by its gross aggregate loss, while the contract pays only the excess $z_j-d$. The deductible must be subtracted separately for each outcome that exceeds it.

Direct Calculation From Aggregate Probabilities

The most reliable hand calculation is to list the attainable aggregate outcomes, their probabilities, and their excess payments. Outcomes at or below the deductible have a zero payment. Keeping those columns separate helps prevent the deductible from being applied incorrectly.

Worked Example: A Deductible Between Aggregate Outcomes

A hypothetical annual aggregate loss $Z$, in dollars, has the following distribution: $P(Z=0)=0.10$, $P(Z=80)=0.20$, $P(Z=160)=0.30$, $P(Z=240)=0.25$, and $P(Z=320)=0.15$. Calculate the expected aggregate stop-loss payment for a deductible of $d=\$190$. The stated probabilities define the complete aggregate distribution; no discounting applies.

Method. Apply the discrete stop-loss sum to each attainable outcome. Since $\$190$ lies between $\$160$ and $\$240$, only $\$240$ and $\$320$ produce positive payments.

\begin{aligned} E[(Z-190)_+] &=(240-190)(0.25)+(320-190)(0.15)\\ &=50(0.25)+130(0.15)\\ &=12.50+19.50\\ &=\$32.00. \end{aligned}

The outcome probabilities sum to $0.10+0.20+0.30+0.25+0.15=1.00$, so the distribution is complete. The result is checked by its component payments: the $\$240$ outcome contributes an expected $\$12.50$, and the $\$320$ outcome contributes $\$19.50$, for $\$32.00$ in total.

Common mistake avoided. Summing gross losses above the deductible gives $240(0.25)+320(0.15)=108$, which is not the stop-loss expectation. That calculation ignores the deductible on each payment. Subtracting $190$ from each exceeding outcome before probability-weighting gives the correct $\$32.00$.

When Aggregate Probabilities Come From a Compound Model

The same calculation applies when the aggregate probabilities have been obtained from a frequency-severity model. For example, if $N$ is the number of claims and $X_1,X_2,\ldots$ are claim severities, the aggregate loss is $Z=\sum_{i=1}^{N}X_i$, with $Z=0$ when $N=0$. First determine the probability of each aggregate outcome; then apply the stop-loss sum. The next example works through both stages for a small discrete model.

Worked Example: Building a Small Aggregate Distribution

A hypothetical policy has claim count $N\sim\operatorname{Binomial}(2,0.5)$. Given a claim, its severity is independently $\$100$ or $\$200$, each with probability $0.5$. Claim severities are independent of $N$ and of one another. Calculate the expected aggregate stop-loss payment for $d=\$250$. There is no discounting.

Step 1: Find the count probabilities. For $N\sim\operatorname{Binomial}(2,0.5)$, $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$.

Step 2: Find the aggregate outcomes and probabilities. With no claims, $Z=0$ with probability $0.25$. With one claim, the aggregate is $\$100$ or $\$200$; each has probability $0.50(0.50)=0.25$. With two claims, the four equally likely severity pairs are $(100,100)$, $(100,200)$, $(200,100)$, and $(200,200)$, each with joint probability $0.25(0.25)=0.0625$. The resulting aggregate probabilities are:

\begin{aligned} P(Z=0)&=0.25,\\ P(Z=100)&=0.25,\\ P(Z=200)&=0.25+0.0625=0.3125,\\ P(Z=300)&=0.0625+0.0625=0.125,\\ P(Z=400)&=0.0625. \end{aligned}

The probability at $\$200$ combines the one-claim outcome of $\$200$ and the two-claim outcome $(100,100)$. The probabilities sum to $0.25+0.25+0.3125+0.125+0.0625=1.00$.

Step 3: Apply the deductible. The deductible is $\$250$, which is between the attainable outcomes $\$200$ and $\$300$. Thus only $\$300$ and $\$400$ contribute:

\begin{aligned} E[(Z-250)_+] &=(300-250)(0.125)+(400-250)(0.0625)\\ &=50(0.125)+150(0.0625)\\ &=6.25+9.375\\ &=\$15.625. \end{aligned}

Rounded to the nearest cent, the expected payment is $\$15.63$. Before rounding, the two contributing terms sum to $\$15.625$, which confirms the result. The method avoids treating a two-claim aggregate of $\$200$ as exceeding the deductible merely because it is the largest outcome below $\$250$.

A Tail-Probability Check on an Equally Spaced Lattice

When aggregate outcomes are multiples of a fixed amount $h$, a tail-probability calculation provides an independent check. This is especially useful when probabilities $P(Z\geq jh)$ are available from an aggregate distribution calculation. Write $d=(m+r)h$, where $m$ is a nonnegative integer and $0\leq r<1$. For $Z$ on the lattice $0,h,2h,\ldots$, the stop-loss expectation is:

Result: For $Z$ supported on nonnegative multiples of $h$, and $d=(m+r)h$ with $0\leq r<1$, $$ E[(Z-d)_+] =h\sum_{j=m+1}^{\infty}P(Z\geq jh) -rh\,P(Z\geq(m+1)h). $$ Only finitely many terms are nonzero when the aggregate distribution has finite support.

To see why this holds, consider one possible outcome $Z=kh$. If $k\leq m$, the payment is zero, and none of the tail indicators $1_{\{Z\geq jh\}}$ for $j\geq m+1$ equals one. If $k\geq m+1$, exactly $k-m$ of those indicators equal one, for $j=m+1,\ldots,k$. The expression inside the expectation therefore equals $h(k-m)-rh=(k-m-r)h=kh-d$, the correct excess. Taking expectations of these indicators gives the stated formula.

Worked Example: Checking a Between-Lattice Deductible With Tail Probabilities

A hypothetical aggregate loss $Z$ is supported on $0,50,100,150,200,250$ dollars, with probabilities $0.05,0.15,0.25,0.25,0.20,0.10$, respectively. Calculate the expected stop-loss payment at $d=\$125$, first by direct summation and then using lattice tail probabilities. The given probabilities sum to $1.00$; there is no discounting.

Direct sum. The deductible lies between $\$100$ and $\$150$, so the positive payments are $\$25$, $\$75$, and $\$125$:

\begin{aligned} E[(Z-125)_+] &=(150-125)(0.25)+(200-125)(0.20)+(250-125)(0.10)\\ &=25(0.25)+75(0.20)+125(0.10)\\ &=6.25+15.00+12.50\\ &=\$33.75. \end{aligned}

Tail check. Here $h=50$ and $125=(2+0.5)50$, so $m=2$ and $r=0.5$. The required tail probabilities are $P(Z\geq150)=0.25+0.20+0.10=0.55$, $P(Z\geq200)=0.20+0.10=0.30$, and $P(Z\geq250)=0.10$. Therefore:

\begin{aligned} E[(Z-125)_+] &=50(0.55+0.30+0.10)-0.5(50)(0.55)\\ &=50(0.95)-25(0.55)\\ &=47.50-13.75\\ &=\$33.75. \end{aligned}

The two methods agree. The correction term is essential: the tail sum in increments of $\$50$ measures excess above $\$100$, while the actual deductible is $\$125$, an additional $\$25$ for outcomes at or above $\$150$. Omitting the correction would overstate the expected payment by $\$13.75$.

Exam Checks and Common Traps

A discrete stop-loss calculation is short, but errors often arise before the arithmetic. Confirm the aggregate support, use probabilities of aggregate outcomes rather than claim-level probabilities, and compute the payment for each outcome before taking its expectation. A deductible between attainable outcomes does not require interpolation: the payment is determined by the actual support values.

Using gross losses: $zP(Z=z)$ is not the stop-loss contribution; use $(z-d)P(Z=z)$ when $z>d$.
Using claim-level rather than aggregate outcomes: a per-claim severity distribution is not itself the distribution of the total loss.
Misreading the threshold: an aggregate loss below the deductible produces no payment, even if it is close to the deductible.
Rounding too early: retain exact probabilities and intermediate products until the final reported amount.
Forgetting the lattice adjustment: the tail formula's correction is needed when the deductible is not itself a lattice point.

These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In a longer problem, the aggregate probabilities may come from convolution or a compound-distribution calculation; once they are known, the stop-loss expectation is a probability-weighted sum of positive excesses.

Check Your Understanding

For each problem, identify the aggregate outcomes that exceed the deductible before calculating the expectation.

A hypothetical aggregate loss has probabilities $0.20,0.30,0.30,0.20$ at $0,100,200,300$ dollars, respectively. Find the expected stop-loss payment for a deductible of $\$150$. (Answer: $\$45.00$)
A discrete aggregate loss has probabilities $0.10,0.20,0.40,0.30$ at $0,40,80,120$ dollars, respectively. Find the expected stop-loss payment for a deductible of $\$80$. (Answer: $\$12.00$)
An aggregate loss has probabilities $0.10,0.20,0.30,0.25,0.15$ at $0,50,100,150,200$ dollars, respectively. Calculate the expected payment for a deductible of $\$125$. (Answer: $\$17.50$)
A hypothetical aggregate distribution has $P(Z=0)=0.25$, $P(Z=60)=0.35$, $P(Z=120)=0.25$, and $P(Z=180)=0.15$. Find the expected stop-loss payment at $d=\$90$. (Answer: $\$21.00$)
An aggregate loss is a multiple of $\$25$. The tail probabilities are $P(Z\geq75)=0.60$, $P(Z\geq100)=0.40$, and $P(Z\geq125)=0.10$, with no possible outcomes above $\$125$. Find the expected stop-loss payment at $d=\$62.50$. (Answer: $\$20.00$)

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