Why a Deductible Changes the Data
A claim database may contain the ground-up loss, the amount paid after a deductible, or only claims that generated a positive payment. Those are different observations of risk. If the deductible is ignored, a calculated mean or empirical distribution can describe the wrong quantity, even when every arithmetic step is correct.
This tutorial focuses on ordinary per-loss deductibles. It distinguishes a payment recorded as zero from a loss that is absent from the database, and shows how to summarize positive payments when the data include only losses above the deductible.
The distinction between “zero” and “omitted” is about how the sample was assembled, not about the payment formula. If a database contains every reported claim, including claims whose payment is zero, its payment data include zeros. If it contains only claims with positive payments, it is a sample conditional on exceeding the deductible. The positive payments alone do not show how many below-deductible claims were omitted.
Three Data Views
For a fixed deductible, the same underlying experience can be represented in three ways. A ground-up loss record gives \(X\). A payment record gives \(Y^L\), which may be zero. A positive-payment record gives \(Y^L\) only when \(X>d\). The third representation is left-truncated: its observed losses come from the part of the ground-up distribution above \(d\).
| Database contents | What one row represents | What can be summarized directly |
|---|---|---|
| All ground-up claims | A loss \(X\), including losses at or below \(d\) | Ground-up loss summaries and, after applying \(d\), payment summaries |
| All reported claims with payments | A payment \(Y^L\), including zeros | Payments per reported claim, with zero payments retained |
| Positive payments only | A payment conditional on \(X>d\) | Payment summaries conditional on a positive payment |
When all ground-up losses are available, apply the deductible to each loss before calculating payment summaries. When the database contains positive payments only, adding \(d\) to each payment reconstructs the corresponding ground-up loss in the observed sample. It does not reconstruct the omitted losses or the complete ground-up distribution.
What a Positive-Payment Sample Estimates
Suppose losses are observed only when \(X>d\). In that sample, the observed payment is \(Y^L=X-d\). For \(y\geq 0\), its conditional distribution is
Here \(F_X(x)=P(X\leq x)\), and the expression assumes \(P(X>d)>0\). The subtraction in the numerator removes losses that do not exceed \(d\); the denominator renormalizes the remaining probability to one. For a discrete loss model, the event \(X>d\) must match the actual inclusion rule. If claims exactly at \(d\) are included or excluded differently, the numerator and denominator must be adjusted to that rule.
For a finite positive-payment sample \(y_1,\ldots,y_{n_+}\), the empirical conditional distribution assigns mass \(1/n_+\) to each observed payment, counting repeated values with their multiplicities. Its empirical mean payment is \( \sum y_i/n_+ \). The corresponding empirical mean ground-up loss among the observed claims is \(d+\sum y_i/n_+\), because each observed loss equals its payment plus \(d\).
Determine whether the data include all claims, all payments including zeros, or positive payments only.
For complete ground-up data, calculate \(Y^L_i=\max(X_i-d,0)\). For a positive-payment record, recover the observed ground-up loss as \(X_i=d+Y^L_i\).
Include zeros when the target is payment per claim across all reported claims; exclude them only when the target is explicitly conditional on a positive payment.
Positive payments alone do not reveal the number or severity distribution of omitted losses at or below \(d\).
Worked Examples
Worked Example: Retaining Zero Payments in a Complete Claim List
A hypothetical insurer has the ground-up losses \(180, 400, 520, 760, 1{,}040,\) and \(1{,}300\) dollars for six reported claims. The ordinary deductible is \(d=\$400\), and the database includes all six claims. Find the payment for each claim, the mean payment per reported claim, the mean positive payment, and the empirical proportion of claims with a positive payment.
Method. Apply \(Y^L=\max(X-400,0)\) to each loss. Since all six claims are present, calculate the first requested mean using all six payments, including zeros.
The payments in the same order as the losses are
Their total is \(0+0+120+360+640+900=\$2{,}020\). Thus the mean payment per reported claim is \(\$2{,}020/6=\$336.67\), rounded to the nearest cent. There are four positive payments, so the mean positive payment is \(\$2{,}020/4=\$505.00\). The empirical proportion with a positive payment is \(4/6=0.6667\), to four decimals.
Check. The four ground-up losses above the deductible sum to \(520+760+1{,}040+1{,}300=\$3{,}620\), giving mean \(\$3{,}620/4=\$905.00\). This equals \(d+\) mean positive payment, or \(\$400+\$505=\$905\).
Common mistake avoided. Dividing \(\$2{,}020\) by four gives the mean conditional on a positive payment, not the mean payment per reported claim. The latter uses six as the denominator because the two zero payments are observed claims.
Worked Example: Summarizing a Left-Truncated Payment Sample
A hypothetical claim extract contains only positive payments under an ordinary deductible of \(d=\$500\). Its five payments are \(\$180,\$320,\$320,\$780,\) and \(\$1{,}200\). Find the empirical mean payment, the corresponding mean ground-up loss among these observed claims, and the empirical conditional proportion of observed losses no greater than \(\$820\).
Method. Because the extract includes positive payments only, all summaries describe claims conditional on exceeding the deductible. First add the deductible to each payment to recover the observed ground-up losses:
The payments sum to \(180+320+320+780+1{,}200=\$2{,}800\). The empirical mean payment is \(\$2{,}800/5=\$560.00\). Therefore the mean ground-up loss among these five observed claims is \(\$500+\$560=\$1{,}060.00\). Directly, the reconstructed losses sum to \(\$5{,}300\), and \(\$5{,}300/5=\$1{,}060.00\), which checks the calculation.
Two of the five reconstructed losses are no greater than \(\$820\) only if there are two such values—but the value \(\$680\) and both occurrences of \(\$820\) qualify. Thus three of five qualify, and the empirical conditional proportion is \(3/5=0.6000\).
Common mistake avoided. The answer \(3/5\) is conditional on appearing in this positive-payment extract. It is not the proportion of all ground-up claims that are at or below \(\$820\), because losses at or below \(\$500\) are absent from the extract.
Worked Example: A Model for the Positive Payment
For a hypothetical portfolio, assume each ground-up claim severity \(X\) has an exponential distribution with mean \(\$2{,}000\), so its rate is \(1/2{,}000\) per dollar. The deductible is \(d=\$600\). Find the probability a claim produces a positive payment, the conditional mean payment given a positive payment, and the expected payment per ground-up claim.
Method. For an exponential loss with mean \(\theta=\$2{,}000\), the survival probability is \(P(X>x)=e^{-x/\theta}\) for \(x\geq0\). A positive payment occurs exactly when \(X>d\), so
Conditional on \(X>600\), the payment \(Y^L=X-600\) has an exponential distribution with the same mean \(\$2{,}000\). To justify this, for \(y\geq0\),
This is the survival function of an exponential distribution with mean \(\$2{,}000\). Hence \(E[Y^L\mid Y^L>0]=\$2{,}000.00\). For a payment recorded per ground-up claim, the payment is zero when \(X\leq600\), so
Check. The result is smaller than the conditional mean positive payment, as it must be: the per-claim expectation also gives weight to claims producing zero payment. The probability is rounded to four decimals and dollar amounts to the nearest cent.
Common mistake avoided. The mean \(\$2{,}000\) describes the payment conditional on a positive payment under this exponential model. It is not the expected payment per ground-up claim; that expectation must also account for the probability of exceeding the deductible.
What Positive Payments Cannot Tell You
A positive-payment extract can describe the empirical severity distribution conditional on exceeding the deductible. Adding \(d\) to its payments recovers the observed ground-up losses, and empirical summaries can be calculated for those observed losses. But the extract contains no direct information about the number of claims at or below \(d\), their ground-up amounts, or the fraction of all claims that exceed \(d\).
In particular, a positive-payment count is not automatically a total claim count. If the data are used to estimate a claim rate, the analyst must know whether below-deductible claims were recorded and whether the exposure denominator corresponds to the same observation process. Otherwise, a rate based only on positive payments describes observed paid claims, not necessarily all ground-up claims.
If all reported claims are present, zeros should be retained whenever the target summary is payment per reported claim. If only paid claims are present, the resulting mean is a conditional mean. State the conditioning and the denominator explicitly; these choices are part of the statistical definition of the quantity being estimated, not merely formatting details.
Check Your Understanding
For each problem, identify whether the requested quantity is unconditional over recorded claims or conditional on a positive payment.
- A complete list of four ground-up losses is \(\$90,\$250,\$510,\$900\), with deductible \(\$300\). Find the four payments and the mean payment per reported claim. (Answer: payments \(\$0,\$0,\$210,\$600\); mean \(\$202.50\))
- Only positive payments are retained for a deductible of \(\$400\). The payments are \(\$100,\$300,\$700\). Find the mean positive payment and mean ground-up loss among the observed claims. (Answer: mean payment \(\$366.67\); mean ground-up loss \(\$766.67\))
- A complete database contains ten reported claims, of which six have positive payments. What is the empirical proportion with a positive payment, and what additional information is needed to calculate the mean positive payment? (Answer: \(0.6000\); the positive payment amounts, or their sum)
- Only positive payments are observed under a deductible of \(\$250\). Three observed payments are \(\$150,\$450,\$650\). What is the empirical conditional proportion of reconstructed ground-up losses at most \(\$700\)? (Answer: \(2/3=0.6667\))
- Assume \(X\) is exponential with mean \(\$1{,}000\), and \(d=\$500\). Find the probability of a positive payment and the mean payment conditional on a positive payment. (Answer: \(P(Y^L>0)=e^{-0.5}=0.6065\); conditional mean payment \(\$1{,}000.00\))