Why the Severity Target Must Match the Recorded Amount
A severity model is only interpretable when its response matches the amount being modeled. A claim’s ground-up loss is not generally the same as the insurer’s payment: under a deductible, the insurer pays nothing on losses at or below the deductible and pays only the excess on larger losses. Thus a model of ground-up loss and a model of positive insurer payments have different targets and describe different quantities.
The distinction is important when severity data contain only claims that generated payments. Those observations are not ground-up losses, and their average is not the expected payment per reported claim. The first step is to define the loss variable, the deductible, and the population to which the response refers.
The notation \(Y^L\) emphasizes the payment measured per underlying loss, including zero payments. The notation \(Y^P\) emphasizes the positive amount measured among paid claims. In contrast, \(X\) remains the ground-up loss. If the data record ground-up amounts, the response is \(X\); if they record positive insurer payments, the response is \(Y^P\). If the question concerns average insurer cost per underlying claim, the relevant quantity is \(E[Y^L]\), which includes claims with zero payment.
| Target | What is averaged? | Typical interpretation |
|---|---|---|
| \(X\) | Ground-up losses | Mean loss before the deductible |
| \(Y^P\) | Positive insurer payments only | Mean payment conditional on a payment |
| \(Y^L\) | Payments per underlying loss, including zeros | Expected insurer payment per loss |
These are different targets, even though they are connected. In particular, the average positive payment cannot be used as the expected payment per loss without accounting for the probability that a loss exceeds the deductible.
Connecting the Two Payment Measures
Write \(q=P(X>d)\), the probability of a positive payment. Provided \(q>0\), the expected payment per underlying loss is the probability of payment multiplied by the average payment conditional on payment:
This follows by splitting the expectation according to whether the loss exceeds the deductible. If \(X\leq d\), then \(Y^L=0\); if \(X>d\), then \(Y^L=X-d\). Therefore,
For a nonnegative continuous ground-up loss with survival function \(S_X(x)=P(X>x)\), the same distinction can be expressed using the tail of the loss distribution. For \(y\geq0\), conditional on a payment, \(Y^P>y\) exactly when \(X>d+y\). Hence
For any nonnegative random variable, its expectation is the integral of its survival function. Applying that identity to the conditional payment gives
The change of variable \(x=d+y\) yields the final equality. Without the denominator, the integral is the expected payment per underlying loss, \(E[Y^L]=\int_d^\infty S_X(x)\,dx\). Dividing by \(S_X(d)\) conditions on a payment having occurred. These survival-integral formulas assume a nonnegative loss and apply directly to continuous severity distributions; for distributions with atoms, the conditional expectation identity \(E[Y^L]=P(X>d)E[Y^P]\) remains the basic relationship.
Worked Examples
Worked Example: Discrete Losses With a Deductible
A hypothetical insurer considers ground-up losses with distribution \(P(X=\$1{,}000)=0.25\), \(P(X=\$5{,}000)=0.35\), \(P(X=\$9{,}000)=0.25\), and \(P(X=\$13{,}000)=0.15\). The per-loss deductible is \(d=\$4{,}000\), with no limit. Find the probability of a payment, the average positive payment, and the expected payment per underlying loss.
Assumptions. The stated discrete distribution is the full ground-up severity distribution; the deductible applies to each loss. There is no discounting or mortality assumption because this is a single-loss severity calculation.
Step 1: Calculate the payment on each possible loss. The four insurer payments are \(\max(X-\$4{,}000,0)\): respectively, \(\$0\), \(\$1{,}000\), \(\$5{,}000\), and \(\$9{,}000\). A payment occurs for the last three loss values, so
Step 2: Average only the positive payments. The probability-weighted total payment per underlying loss is
Divide by the probability of payment to obtain the conditional average:
Step 3: State both payment measures. The expected payment per underlying loss, including zero payments, is \(\$2{,}950.00\). As a check, \(0.75(\$3{,}933.33)\) is approximately \(\$2{,}950.00\); the small difference before rounding is less than one cent. The method is to calculate payment amounts first, then distinguish the conditional average from the zero-inclusive average. A common mistake is to report the conditional average as the expected cost per underlying loss, omitting the 25% of losses with no payment.
Worked Example: Exponential Ground-Up Severity
For a hypothetical book, suppose ground-up loss \(X\) has an exponential distribution with mean \(\$8{,}000\), so its rate is \(\lambda=1/8{,}000\) per dollar. The deductible is \(d=\$4{,}000\), with no limit. Calculate the payment probability, the average payment conditional on payment, and the expected payment per underlying loss.
Assumptions. \(X\) is continuous, nonnegative, and exponentially distributed with the stated mean; the same deductible applies to every loss. There is no discounting or mortality assumption.
Step 1: Find the probability of a positive payment. For this exponential distribution, \(S_X(x)=e^{-x/8{,}000}\), so
Step 2: Find the conditional payment mean. The exponential distribution is memoryless: conditional on \(X>d\), the excess \(X-d\) has the same exponential distribution as \(X\). Therefore,
Step 3: Find expected payment per underlying loss. Multiply by the probability of a payment:
The result is consistent with the survival-integral formula: for an exponential loss with mean \(1/\lambda\), \(E[Y^L]=e^{-\lambda d}/\lambda\). Here that is \(e^{-0.5}(\$8{,}000)=\$4{,}852.25\). The method avoids treating the mean ground-up loss as the mean payment: the deductible changes the payment target, while the exponential memoryless property gives the mean excess conditional on payment.
Worked Example: Uniform Losses and a Double-Subtraction Trap
Suppose a hypothetical ground-up loss is uniformly distributed from \(\$0\) to \(\$18{,}000\), and the per-loss deductible is \(\$6{,}000\), with no limit. Find the probability of payment, the mean positive payment, and expected payment per underlying loss.
Assumptions. \(X\sim\operatorname{Uniform}(0,\$18{,}000)\); the deductible is applied to each loss, and there is no policy limit. There is no discounting or mortality assumption.
Step 1: Calculate the payment probability. The part of the ground-up range above the deductible has length \(\$12{,}000\), out of a total length of \(\$18{,}000\). Thus
Step 2: Find the conditional payment distribution. Given \(X>\$6{,}000\), the ground-up loss is uniform from \(\$6{,}000\) to \(\$18{,}000\). Subtracting the deductible shifts that interval to a payment uniformly distributed from \(\$0\) to \(\$12{,}000\). Its mean is therefore
Step 3: Find expected payment per underlying loss.
The same result follows directly because \(Y^L\) is zero below \(\$6{,}000\) and rises linearly from zero to \(\$12{,}000\) above it. The common mistake is to calculate the mean positive payment as \(\$6{,}000\) and then subtract the deductible again. That would count the deductible twice: the payment interval has already been shifted to \(\$0\) through \(\$12{,}000\), so its average payment is \(\$6{,}000.00\). Multiplying by the payment probability gives the distinct zero-inclusive amount, \(\$4{,}000.00\).
Choosing a Response for a Severity Model
For a claim-level severity model, first ask what dollar amount each observation represents and which claims are included. If each response is a ground-up loss, the target is \(X\), even if a deductible is part of the insurance contract. If each response is the insurer’s positive payment and only paid claims are included, the target is \(Y^P\). Those two responses are not interchangeable.
If the requested measure is average insurer payment per underlying loss, the target must also reflect zero payments. One can obtain the expected payment from a ground-up model using \(E[(X-d)_+]\), or combine the probability of payment with the conditional mean payment. Simply averaging positive payments answers a different question. Likewise, including zero payments in a response changes the target and the population represented; it should not be confused with modeling the positive payment amounts.
A deductible changes not only the dollar amount but also which claims can appear in a paid-claim sample. The appropriate probability model and likelihood depend on how the data were selected. Here the essential target distinction is that a paid-claim response is conditional on \(X>d\); the detailed likelihood treatment for deductible-selected observations is a separate issue.
Check Your Understanding
For each problem, identify the target carefully and distinguish positive-payment averages from zero-inclusive expected payments.
- A hypothetical discrete ground-up loss is \(\$2{,}000\) with probability \(0.4\), \(\$6{,}000\) with probability \(0.4\), and \(\$10{,}000\) with probability \(0.2\). With a \(\$5{,}000\) deductible and no limit, find the probability of payment, mean positive payment, and expected payment per underlying loss. (Answer: \(0.6000\), \(\$2{,}333.33\), and \(\$1{,}400.00\), respectively.)
- Ground-up loss is uniform from \(\$0\) to \(\$20{,}000\). With a \(\$5{,}000\) deductible and no limit, find the payment probability, mean positive payment, and expected payment per underlying loss. (Answer: \(0.7500\), \(\$7{,}500.00\), and \(\$5{,}625.00\), respectively.)
- Ground-up loss is exponential with mean \(\$12{,}000\), and the deductible is \(\$6{,}000\), with no limit. Find the probability of payment and both payment means. (Answer: \(0.6065\), \(\$12{,}000.00\), and \(\$7{,}278.37\), respectively.)
- Ground-up loss is uniform from \(\$4{,}000\) to \(\$16{,}000\), and the deductible is \(\$7{,}000\), with no limit. Find the probability of payment and both payment means. (Answer: \(0.7500\), \(\$4{,}500.00\), and \(\$3{,}375.00\), respectively.)
- Ground-up loss is uniform from \(\$0\) to \(\$24{,}000\), and the deductible is \(\$8{,}000\), with no limit. Find the probability of payment, the mean payment conditional on payment, and the expected payment per underlying loss. (Answer: \(0.6667\), \(\$8{,}000.00\), and \(\$5{,}333.33\), respectively.)