From Records to a Stop-Loss Price
A stop-loss price is only as credible as the data and assumptions behind it. A complete calculation starts by defining the exposure unit, estimating how often claims occur, and describing the size of a claim. Those fitted inputs determine a distribution for aggregate loss; the stop-loss premium is then the expected amount by which that aggregate exceeds the attachment point.
This tutorial follows that chain with a hypothetical portfolio. The example uses a simple frequency model and an empirical severity model so that every stage of the calculation can be checked directly. The resulting net premium is a model-based expected payment, not a promise about actual future claims.
Define the Exposure Unit Before Fitting Frequency
Frequency estimates depend on what counts as an exposure. In the main example, each hypothetical contract covers two separate units for one year. A claim count of two for a contract means one claim on each unit; the model will treat the two units as having the same claim probability and as independent. Thus the contract-year count has a binomial distribution with two trials, not a Poisson distribution with an unspecified exposure base.
Worked Example: Estimating Frequency From Contract Records
A hypothetical insurer observes 100 contract-years. Each contract covers two units for the full year. The records show 25 contracts with no claims, 50 with one claim, and 25 with two claims. Estimate the claim probability per unit-year and check the fitted count model.
Step 1: Count exposure units. There are \(100\times2=200\) unit-years of exposure.
Step 2: Count claims. The contracts with one claim contribute \(50\times1=50\) claims. Those with two claims contribute \(25\times2=50\) claims. The total is \(50+50=100\).
Step 3: Fit the claim probability. Under the assumption that each unit-year has the same claim probability \(q\), the maximum-likelihood estimate is total claims divided by total unit-years:
Step 4: Check the implied contract count distribution. With two independent units and \(\hat q=0.5\), the model is \(N\sim\operatorname{Binomial}(2,0.5)\). It gives \(P(N=0)=0.5^2=0.25\), \(P(N=1)=2(0.5)(0.5)=0.50\), and \(P(N=2)=0.5^2=0.25\). These match the observed proportions \(25/100\), \(50/100\), and \(25/100\).
Check. The fitted expected number of claims per contract is \(E[N]=2(0.5)=1\), which agrees with the observed average \(100/100=1\). The common mistake is to divide the 100 claims by 100 contracts and call that quantity the per-unit claim probability. It is the average number of claims per contract; the probability per unit-year is \(100/200=0.5\).
The matching proportions are a useful arithmetic check, but they do not prove that the binomial assumptions are true. The model presumes a fixed two-unit exposure for every contract, a common claim probability, and independence between units. If unit probabilities differ or claims are dependent, this fitted model may not describe the data well.
Fit Severity From Claim Amounts
Frequency describes the number of claims; severity describes the amount of one claim. Let \(X\) denote the gross amount of a claim. When an empirical severity model is used, each observed claim amount is assigned probability equal to its observed proportion among the claim records. This approach makes few distributional shape assumptions, but it can be sensitive to a small or unrepresentative sample.
Worked Example: Constructing an Empirical Severity Model
In the hypothetical portfolio above, a complete claim-record review finds 100 claims: 50 have gross amounts of $500 and 50 have gross amounts of $1,500. Construct the empirical severity distribution and calculate its mean.
Step 1: Convert record counts to probabilities. The empirical model assigns probability \(50/100=0.5\) to each observed amount:
Step 2: Calculate the mean severity. Weight each amount by its empirical probability:
Check. The total of the claim records is \(50(500)+50(1{,}500)=25{,}000+75{,}000=\$100{,}000\). Dividing by 100 claims gives \(\$1{,}000\), the same mean. This is an empirical model, not a claim that all future claims can only be $500 or $1,500. Its use here is an explicit modeling choice that makes the aggregate calculation transparent.
A key modeling decision is whether the claim records represent the same claim amount that will enter the aggregate calculation. The main example uses gross claim amounts without deductibles, limits, or other coverage adjustments. If the records were already adjusted payments, the severity model would describe payments rather than gross losses, and applying the same adjustments again would double-count them.
Build and Check the Aggregate Distribution
For each of the two covered units, define the unit loss \(Z\) to be zero when there is no claim and to equal the claim severity when there is a claim. With the fitted frequency and empirical severity assumptions, the unit-loss distribution is
The two unit losses are assumed independent and identically distributed. The contract aggregate is \(S=Z_1+Z_2\). Its distribution follows by multiplying the probabilities for each pair of unit outcomes and adding probabilities when different pairs give the same aggregate.
The aggregate probabilities can be checked explicitly. For example, \(S=500\) arises from \((0,500)\) or \((500,0)\), giving \(2(0.5)(0.25)=0.25\). An aggregate of $2,000 arises from \((500,1{,}500)\) or \((1{,}500,500)\), giving \(2(0.25)(0.25)=0.125\). Applying the same method to every possible pair gives:
| Aggregate loss \(s\) | How it arises | \(P(S=s)\) |
|---|---|---|
| $0 | \((0,0)\) | 0.25 |
| $500 | \((0,500)\), \((500,0)\) | 0.25 |
| $1,000 | \((500,500)\) | 0.0625 |
| $1,500 | \((0,1,500)\), \((1,500,0)\) | 0.25 |
| $2,000 | \((500,1,500)\), \((1,500,500)\) | 0.125 |
| $3,000 | \((1,500,1,500)\) | 0.0625 |
The probabilities sum to \(0.25+0.25+0.0625+0.25+0.125+0.0625=1\). The mean aggregate is also a check:
This equals two units multiplied by the expected loss per unit, \(2(0.5)(\$1{,}000)=\$1{,}000\). Agreement between the two calculations is a strong check against an omitted outcome or a probability error.
Calculate the Stop-Loss Premium
For a discrete aggregate distribution, calculate the stop-loss premium by multiplying each possible excess above the attachment point by its probability. Outcomes at or below the attachment contribute zero. In particular, the attachment is applied to the contract aggregate here, not separately to each claim.
Worked Example: Pricing a Per-Contract Aggregate Layer
Using the fitted models and aggregate distribution above, calculate the net stop-loss premium per contract-year for an attachment point of $1,000. Then calculate the expected total payment for 100 independent contracts, each with its own $1,000 attachment.
Step 1: Apply the attachment to each aggregate outcome. The payment is zero at aggregate losses of $0, $500, and $1,000. At $1,500 it is $500; at $2,000 it is $1,000; and at $3,000 it is $2,000.
Step 2: Take the probability-weighted average.
The net stop-loss premium is therefore $375 per contract-year. For 100 contracts with separate attachments, linearity of expectation gives \(100(\$375)=\$37{,}500\) as the expected total payment.
Step 3: Check using the aggregate mean. For every nonnegative \(S\), \(S=\min(S,d)+(S-d)_+\). At \(d=\$1{,}000\), the expected retained amount is
Thus \(E[\min(S,1{,}000)]+E[(S-1{,}000)_+]=\$625+\$375=\$1{,}000=E[S]\), as required. A common mistake is to use the mean aggregate loss of $1,000 as the stop-loss premium. That is the expected loss before the attachment, not the expected amount above it.
Interpretation, Assumptions, and Common Traps
The calculation is internally consistent, but its interpretation depends on the model. It assumes the frequency probability and severity distribution estimated from the hypothetical records apply to the contracts being priced. It also assumes two fixed, independent unit exposures per contract, identical unit claim probabilities, independent claim severities, and no coverage changes between the records and the priced contract. If the portfolio has unequal exposures, a common probability may not be appropriate; if claims are dependent, convolving independent unit losses can understate or overstate aggregate variation.
The $37,500 total is the expected payment across 100 separately attached contract layers. It is not the expected payment for a single portfolio-level layer with a $1,000 attachment on the combined losses of all 100 contracts. Those are different insurance arrangements with different payment functions. Similarly, the $375 figure is a net expected payment, not a loaded commercial premium. Expenses, uncertainty margins, reinsurance terms, and any desired risk load would need separate treatment.
- Use the correct exposure denominator. Divide by unit-years when estimating a per-unit claim probability; do not confuse that probability with average claims per contract.
- Keep claim amounts and aggregate losses distinct. The severity model describes one claim, while the stop-loss attachment in this example applies to the sum of claims on a contract.
- Check the distribution before pricing. Aggregate probabilities should sum to one, and the aggregate mean should agree with the frequency-severity calculation.
- State the layer basis. A per-contract attachment and a portfolio aggregate attachment generally do not have the same expected payment.
Check Your Understanding
For each problem, identify the model assumptions that matter and check the arithmetic before using the final answer.
- A hypothetical group has 80 contracts, each covering two independent unit-years. There are 72 claims in total. Estimate the common claim probability per unit-year and the expected number of claims per contract. (Answer: \(72/160=0.45\) per unit-year; \(2(0.45)=0.90\) claims per contract.)
- A complete hypothetical claim sample contains four amounts: $2,000, $2,000, $4,000, and $8,000. Under the empirical severity model, calculate the mean claim amount and the expected payment per claim after a $3,000 per-claim deductible. (Answer: mean severity \((2{,}000+2{,}000+4{,}000+8{,}000)/4=\$4{,}000\); expected payment \((0+0+1{,}000+5{,}000)/4=\$1{,}500\).)
- Each of two independent units has probability 0.25 of one claim, and every claim is exactly $800. Find the aggregate-loss probabilities and the stop-loss premium at an aggregate attachment of $1,000. (Answer: \(P(S=0)=0.75^2=0.5625\), \(P(S=800)=2(0.25)(0.75)=0.3750\), \(P(S=1{,}600)=0.25^2=0.0625\); premium \(=600(0.0625)=\$37.50\).)
- For the main example's fitted model, calculate the expected payment for 40 independent contracts if each contract has its own $1,000 attachment. Explain why multiplying the expected payment per contract is valid. (Answer: \(40(\$375)=\$15{,}000\); linearity of expectation applies, whether or not contract payments are independent.)
- For the main example, suppose instead that the attachment remains $1,000 separately for each of 100 contracts. Give the expected total payment, and state one reason this is not the same contract as a single portfolio layer attached at $1,000. (Answer: $37,500; a portfolio layer applies its attachment to the combined portfolio loss rather than separately to each contract.)