Pricing Directly from an Aggregate-Loss Table
When an aggregate loss is given as a probability table, a stop-loss premium can be calculated by evaluating the insurer’s payment at each listed loss and weighting it by that outcome’s probability. The essential discipline is to apply the contract terms to each outcome before multiplying by its probability. This makes the method transparent even when the table has a large zero-loss probability, an outcome exactly at the attachment, or a payment cap.
The examples use hypothetical aggregate-loss distributions. Unless stated otherwise, the probabilities are unconditional, the table lists every possible outcome, and the premium is the expected payment with no expenses, risk loading, or discounting. It is therefore an expected annual payment, not a gross premium.
The attachment \(d\) is the amount of aggregate loss the insured retains before the stop-loss cover pays. The cap \(u\) is the maximum amount the cover pays; it is a layer width, so the cover is exhausted when \(X\geq d+u\). A loss equal to \(d\) produces no payment, while a loss equal to \(d+u\) produces the full capped payment \(u\).
A probability table should first be checked as a distribution. Its probabilities must be nonnegative and sum to one. If the probabilities instead describe only a selected subset of outcomes, the missing probability mass must be addressed before treating the table as a complete distribution. Also check that all loss amounts, the attachment, and the cap use the same units.
Why the Table Calculation Works
For a discrete random variable, an expected value is the sum of each possible value multiplied by the probability of that value. Here the quantity whose expectation is wanted is not \(X\) itself, but the payment dictated by the contract. Define \(g(x)=\max(x-d,0)\) for an unlimited cover, or \(g(x)=\min(\max(x-d,0),u)\) for a capped cover. Since \(g(X)=g(x_j)\) whenever \(X=x_j\), the definition of expectation gives:
This also explains the correct order of operations: calculate the contractual payment \(g(x_j)\) at each outcome, then weight it by \(p_j\). Averaging the aggregate loss first and applying the attachment afterward generally gives a different answer, because the payment function is nonlinear.
Confirm the entries are aggregate-loss amounts, the probabilities sum to one, and the units are consistent.
For each \(x_j\), calculate \(\max(x_j-d,0)\); if there is a cap, limit that result to \(u\).
Multiply each payment by its table probability and sum the products. Keep the stated loss units through the sum, then convert if needed.
Worked Calculations
Worked Example: Unlimited Stop-Loss from a Dollar Table
A hypothetical annual aggregate claim \(X\), measured in dollars, has the following complete probability table. A reinsurer covers the amount above an attachment of \(d=\$2{,}500\), with no payment cap. Calculate the expected annual payment.
| Aggregate loss \(x_j\) | Probability \(p_j\) | Payment \(\max(x_j-2{,}500,0)\) | Probability-weighted payment |
|---|---|---|---|
| $0 | 0.1000 | $0 | $0.00 |
| $1,000 | 0.2000 | $0 | $0.00 |
| $2,500 | 0.3000 | $0 | $0.00 |
| $4,000 | 0.2500 | $1,500 | $375.00 |
| $6,000 | 0.1500 | $3,500 | $525.00 |
Method. Apply the unlimited stop-loss payment to each listed loss, then sum the weighted payments. The table probabilities total \(0.1000+0.2000+0.3000+0.2500+0.1500=1.0000\), so they form a complete distribution.
At losses of $0, $1,000, and $2,500, the excess over the attachment is zero. At $4,000, the payment is \($4{,}000-\$2{,}500=\$1{,}500\), contributing \(0.2500\times \$1{,}500=\$375.00\). At $6,000, the payment is \($6{,}000-\$2{,}500=\$3{,}500\), contributing \(0.1500\times \$3{,}500=\$525.00\).
Thus the expected payment is \(\$0.00+\$0.00+\$0.00+\$375.00+\$525.00=\$900.00\). No rounding is needed. Check: only losses above $2,500 contribute, and the two positive contributions sum to $900.00. A common mistake avoided here is counting the probability at exactly $2,500 as a positive payment; the excess at the attachment is zero.
Worked Example: A Capped Layer with Losses in Thousands
A hypothetical aggregate-loss table is stated in thousands of dollars. A layer attaches at \(d=\$3{,}000\) and has a payment cap of \(u=\$3{,}000\). Find its expected payment in dollars.
| Aggregate loss (thousands of dollars) | Probability | Layer payment (thousands of dollars) | Weighted payment (thousands of dollars) |
|---|---|---|---|
| 0 | 0.1000 | 0 | 0.0000 |
| 2 | 0.2000 | 0 | 0.0000 |
| 4 | 0.2500 | 1 | 0.2500 |
| 6 | 0.2500 | 3 | 0.7500 |
| 8 | 0.2000 | 3 | 0.6000 |
Method. Use \(\min(\max(x-d,0),u)\), with \(d=3\) and \(u=3\) in thousands of dollars. The probabilities sum to \(1.0000\). The layer exhausts at \(d+u=6\) thousand dollars.
For loss 0 or 2, the loss does not exceed the attachment, so payment is zero. For loss 4, the payment is \(\min(4-3,3)=1\) thousand. For loss 6, it is \(\min(6-3,3)=3\) thousand. For loss 8, the uncapped excess is 5 thousand, but the payment is limited to 3 thousand. The weighted sum is \(0.2500(1)+0.2500(3)+0.2000(3)=0.2500+0.7500+0.6000=1.6000\) thousand dollars.
The expected payment is therefore \(\$1{,}600.00\). Check: every payment is between zero and the $3,000 cap. The key mistake to avoid is using the entire excess at the $8,000 outcome; the contract pays $3,000 there, not $5,000.
Worked Example: An Atom at Attachment and a Fractional-Thousand Cap
A hypothetical annual aggregate loss \(X\) is recorded in thousands of dollars. Its complete probability mass function is shown below. A layer attaches at 3 thousand dollars and is capped at \(u=2.5\) thousand dollars. Calculate the expected layer payment in dollars, including the effect of the outcome exactly at attachment.
| \(x_j\) (thousands of dollars) | \(p_j\) | \(\min(\max(x_j-3,0),2.5)\) (thousands) | Weighted payment (thousands) |
|---|---|---|---|
| 0 | 0.1200 | 0 | 0.0000 |
| 1 | 0.1800 | 0 | 0.0000 |
| 3 | 0.2500 | 0 | 0.0000 |
| 4 | 0.2000 | 1 | 0.2000 |
| 9 | 0.2500 | 2.5 | 0.6250 |
Method. Evaluate the capped payment at each loss, using the layer’s exhaustion point \(d+u=3+2.5=5.5\) thousand dollars. The probabilities total \(0.1200+0.1800+0.2500+0.2000+0.2500=1.0000\).
The payment at \(x=3\) is zero, even though that outcome has probability 0.2500, because it is exactly the attachment. At \(x=4\), the excess is 1 thousand and is below the cap, so the weighted contribution is \(0.2000\times 1=0.2000\) thousand. At \(x=9\), the excess is 6 thousand but the cap restricts payment to 2.5 thousand, contributing \(0.2500\times 2.5=0.6250\) thousand.
Adding all contributions gives \(0.2000+0.6250=0.8250\) thousand dollars, or \(\$825.00\). Check: the capped outcome’s contribution is \(0.6250\) thousand, which is less than its maximum possible weighted contribution of \(0.2500\times 2.5=0.6250\) thousand (in fact, it reaches that maximum). The common mistake is to multiply the uncapped 6-thousand excess at \(x=9\) by its probability.
Exam Checks and Common Traps
A direct probability-table calculation is simple, but small interpretation errors can affect several rows or the whole result. Before finalizing a premium, check the following:
- Attachment versus payment: the cover pays only the amount above \(d\), not the full aggregate loss whenever \(X\geq d\). At \(X=d\), payment is zero.
- Cap versus attachment: a cap \(u\) limits the payment, so the exhaustion level is \(d+u\). It is not generally the loss level \(u\).
- Apply terms before averaging: \(E[\max(X-d,0)]\) is not generally \(\max(E[X]-d,0)\). The payment must be calculated at every table outcome.
- Use the stated units: if losses and the cap are in thousands, the weighted sum is also in thousands until converted to dollars.
- Do not renormalize a complete table: if its probabilities already sum to one, dividing them by their total changes the stated distribution. If they do not sum to one, investigate whether outcomes are omitted or the entries are conditional probabilities.
These checks also help distinguish a stop-loss payment from a conditional excess loss. The table calculation here averages the payment over all outcomes, including those with zero payment; it does not condition on \(X>d\). This is why the probabilities of losses at or below the attachment remain in the table but contribute zero to the premium.
Check Your Understanding
For each hypothetical problem, calculate the expected payment using the complete aggregate-loss table provided. Amounts are in dollars unless stated otherwise.
- An unlimited stop-loss cover attaches at $2. Aggregate losses are 0, 1, 3, and 5 with probabilities 0.2000, 0.3000, 0.3000, and 0.2000, respectively. What is the expected payment? (Answer: $0.90)
- An unlimited cover attaches at $3. Aggregate losses are 0, 2, 4, and 7 with probabilities 0.1000, 0.2500, 0.4000, and 0.2500. Calculate the expected payment. (Answer: $1.40)
- A layer attaches at $5 and has a payment cap of $4. Aggregate losses are 0, 4, 6, 9, and 12 with probabilities 0.1000, 0.2000, 0.2500, 0.3000, and 0.1500. Find the expected payment. (Answer: $2.05)
- A layer attaches at $4 and has a cap of $3. Aggregate losses are 0, 3, 5, 8, and 11 with probabilities 0.1500, 0.2000, 0.2500, 0.2500, and 0.1500. Find the expected payment. (Answer: $1.45)
- Loss amounts are in thousands of dollars. A layer attaches at 4 thousand and has a cap of 5 thousand. Aggregate losses are 0, 2, 5, 8, and 10 thousand, with probabilities 0.1000, 0.2000, 0.3000, 0.2500, and 0.1500. Calculate the expected payment in dollars. (Answer: $2,050.00)