Why Estimate Stop-Loss Payments From Aggregate Data?
A stop-loss arrangement responds to the amount by which an aggregate loss exceeds a specified attachment point. When a data set records total losses for each observed period, the empirical estimate can be calculated directly from those totals. There is no need to reconstruct individual claims or fit a frequency and severity model to obtain this descriptive estimate.
The calculation is simple, but its interpretation depends on how the data were collected. Each observation should represent a comparable period with consistent coverage, exposure basis, and currency basis. The empirical average describes the observed periods; using it as an estimate of a future expected payment also requires that those periods be relevant to the future risk.
Here, \(d\) denotes the aggregate stop-loss attachment, not a claim-level deductible applied separately to each claim. The sum uses the loss in each period after subtracting the attachment only when that period's aggregate loss is greater than \(d\). A period with loss below or equal to \(d\) contributes zero payment.
Why the Empirical Formulas Work
For a fixed observed loss \(x_j\), the stop-loss payment is exactly the excess of that loss over the attachment, if any. Equivalently, it is zero unless the loss exceeds \(d\). Thus
Averaging these payments over all \(n\) periods gives the empirical stop-loss premium. To estimate the exceedance probability, count the periods with \(x_j>d\) and divide by the total number of periods. These are empirical averages: each observed period receives weight \(1/n\).
The probability formula is the proportion of observations strictly greater than \(d\). An observation equal to \(d\) contributes no stop-loss payment, and it is not counted as an exceedance under the event \(X>d\). This strict inequality matters whenever the data contain ties at the attachment.
Calculate the Payment Period by Period
A reliable approach is to make a payment column before summing. This makes it easier to detect whether the attachment was subtracted from every loss by mistake, or whether a payment was assigned to a loss that did not exceed the attachment.
Confirm that each value is an aggregate loss for one comparable period and that all values use the same currency and coverage basis.
For each observed loss \(x_j\), calculate \((x_j-d)^+=\max(x_j-d,0)\).
Add the \(n\) period payments and divide by \(n\) to obtain the empirical stop-loss premium per period.
Count observations strictly greater than \(d\), then divide by \(n\). Do not count an observation equal to \(d\) as exceeding it.
Worked Example: A Stop-Loss Estimate From Six Annual Totals
A hypothetical insurer reviews six years of aggregate covered losses. The years have equal exposure, the losses are complete and consistently measured, and all amounts below are in thousands of dollars: \(42, 65, 80, 80, 105, 150\). Estimate the empirical stop-loss premium per year and the probability of exceeding an attachment of \(80\) thousand dollars.
Method. Calculate the excess payment in each year, average those payments, and then count the years with losses strictly above \(80\). The equal-exposure condition makes an unweighted average across years appropriate for this calculation.
| Aggregate loss \(x_j\) ($000) | Payment \((x_j-80)^+\) ($000) | Exceeds 80? |
|---|---|---|
| 42 | 0 | No |
| 65 | 0 | No |
| 80 | 0 | No |
| 80 | 0 | No |
| 105 | 25 | Yes |
| 150 | 70 | Yes |
The total observed payment is \(0+0+0+0+25+70=95\) thousand dollars. Therefore,
There are two exceedances among six years, so
Rounded only at the final step, the estimate is \(\$15.83\) thousand per year, or \(\$15{,}833.33\) per year, and the empirical exceedance probability is 0.3333. As a check, only the two years above the attachment contribute positive payments; their total excess is \(25+70=95\) thousand dollars.
Common mistake avoided: Subtracting \(80\) from each of the six losses would produce negative amounts for some years and would not represent stop-loss payments. The positive-part operation sets each payment below or at the attachment to zero.
What the Two Estimates Tell You
The exceedance probability measures how often observed aggregate loss crossed the attachment. The stop-loss premium also reflects the sizes of the excesses. Consequently, two data sets can have the same exceedance probability but different average payments, or one can have more exceedances but a lower average payment.
For an empirical distribution that assigns probability \(1/n\) to each observed period, the stop-loss premium is the average of its \(n\) payments, while the exceedance probability is the fraction of its observations above \(d\). These are descriptive summaries of that empirical distribution. They do not by themselves quantify sampling uncertainty or guarantee that future losses will match the observed experience.
Worked Example: Separating Frequency of Exceedance From Payment Size
A hypothetical benefit plan has eight comparable annual periods of aggregate covered losses, in thousands of dollars: \(12, 18, 24, 31, 36, 46, 52, 68\). For an attachment of \(40\) thousand dollars, calculate the empirical stop-loss premium per year and the empirical exceedance probability.
Method. Subtract \(40\) from each annual total, retaining only positive results. Then average the eight payments and separately count totals greater than \(40\).
The period payments, in thousands, are
Their sum is \(6+12+28=46\) thousand dollars, so the empirical stop-loss premium is
Three of the eight annual losses exceed \(40\), giving
The results are \(\$5{,}750.00\) per year and an empirical exceedance probability of \(0.3750\). The payment calculation checks directly: the three positive excesses sum to \(46\) thousand dollars, and all five other periods contribute zero.
Common mistake avoided: Dividing the total excess by only the three exceeding periods would give \(46/3\), which is the average payment among exceeding periods, not the stop-loss premium averaged over all eight periods. The requested per-period estimate divides by the full number of observations.
Ties, Units, and Data Comparability
When an observed aggregate loss equals \(d\), its stop-loss payment is \((d-d)^+=0\). It also does not count toward the probability of \(X>d\). If a question instead asks for the empirical probability of \(X\ge d\), observations equal to \(d\) must be included in that count. The payment formula itself is unchanged.
Units must also stay consistent. If aggregate losses are recorded in thousands of dollars and \(d\) is stated in dollars, convert one of them before subtracting. The resulting average is still expressed in the units used for the losses. Reporting a premium estimate in dollars when the calculations were in thousands without converting would create a factor-of-one-thousand error.
The periods being averaged should be comparable. For example, totals from periods with different exposure, coverage limits, or inflation bases may not represent the same risk unit. Use the relevant exposure-standardized or otherwise adjusted data when those differences matter; do not assume that dividing by the number of periods automatically corrects them. If a period's total is censored, incomplete, or measured on a different coverage basis, the direct empirical average may not estimate the intended stop-loss payment without addressing that data issue.
Worked Example: Equal-to-Attachment Observations and Strict Exceedance
A hypothetical cooperative observes eight comparable annual aggregate losses, in thousands of dollars: \(0, 45, 70, 100, 100, 125, 160, 220\). Calculate the empirical stop-loss premium at \(d=100\) thousand dollars and compare the empirical probabilities of \(X>100\) and \(X\ge100\).
Method. Apply the positive-part formula to each observation, then use separate counts for the strict and inclusive events.
| Aggregate loss ($000) | Stop-loss payment at 100 ($000) | Counted for \(X>100\)? | Counted for \(X\ge100\)? |
|---|---|---|---|
| 0 | 0 | No | No |
| 45 | 0 | No | No |
| 70 | 0 | No | No |
| 100 | 0 | No | Yes |
| 100 | 0 | No | Yes |
| 125 | 25 | Yes | Yes |
| 160 | 60 | Yes | Yes |
| 220 | 120 | Yes | Yes |
The total stop-loss payment is \(25+60+120=205\) thousand dollars. Averaging over all eight years gives
There are three observations strictly above \(100\), so \(\widehat{P}(X>100)=3/8=0.3750\). There are five observations at least \(100\), so \(\widehat{P}(X\ge100)=5/8=0.6250\). The stop-loss premium is \(\$25{,}625.00\) per year. These calculations check because the two observations equal to the attachment contribute neither payment nor strict exceedance, but both count for the inclusive event.
Common mistake avoided: Treating \(X>100\) and \(X\ge100\) as the same event overcounts strict exceedances when there are ties. State the event as written before counting.
Practical Interpretation and Common Traps
An empirical stop-loss premium is not the same as the mean aggregate loss. It is the mean of payments after applying the attachment. Nor is it the average excess conditional on an exceedance: that conditional average divides the total excess by the number of exceedances, whereas the stop-loss premium per period divides by the number of all observed periods.
A useful check is to verify that the stop-loss premium is nonnegative and that it is zero if no observed loss exceeds the attachment. If \(k\) of the \(n\) observations exceed \(d\), then the empirical exceedance probability is \(k/n\), while the stop-loss premium averages the \(k\) positive excesses over \(n\), not over \(k\). The estimated premium should also be reported with the period and currency units clearly identified.
This empirical method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. Its main exam traps are applying the attachment at the wrong level, counting equality as a strict exceedance, changing units mid-calculation, and using the number of exceedances as the denominator for an average over all periods.
Check Your Understanding
For each problem, treat the listed periods as comparable and equally weighted. Amounts are in thousands of dollars unless otherwise stated.
- Observed annual aggregate losses are \(20, 40, 55, 80\). At an attachment of \(50\), find the empirical stop-loss premium per year and the empirical probability of exceeding the attachment. (Answer: \(\$8.75\) thousand per year; \(0.5000\).)
- Six annual totals are \(30, 45, 60, 60, 75, 105\). At \(d=60\), find the stop-loss premium and \( \widehat{P}(X>60)\). How many observations would count for \(X\ge60\)? (Answer: \(\$10.00\) thousand per year; \(0.3333\); 4 observations.)
- Five aggregate losses are \(0, 90, 110, 140, 160\). At an attachment of \(100\), calculate the empirical stop-loss premium and exceedance probability. (Answer: \(\$22.00\) thousand per period; \(0.6000\).)
- Eight annual losses are \(25, 50, 75, 100, 125, 150, 175, 200\). At \(d=125\), calculate the stop-loss premium per year and the probability of a strict exceedance. (Answer: \(\$18.75\) thousand per year; \(0.3750\).)
- Two hypothetical portfolios each have ten comparable annual observations. Portfolio A has losses \(20, 40, 50, 60, 80, 100, 120, 150, 180, 200\); portfolio B has \(0, 0, 0, 0, 0, 0, 0, 0, 250, 250\). At \(d=100\), calculate both stop-loss premiums and exceedance probabilities. Which portfolio has the higher empirical premium, and which has the higher exceedance probability? (Answer: A: \(\$25.00\) thousand per year and \(0.4000\); B: \(\$30.00\) thousand per year and \(0.2000\). B has the higher premium; A has the higher exceedance probability.)