Choose the Observation Unit for Repeated Claims

Why the Observation Unit Matters

A repeated-claims data set can be arranged in several valid ways. One row might represent a policy, one claim, or one policy during one coverage year. The right choice depends on the decision the analysis is intended to support. Choosing rows for convenience, without aligning the outcome and predictors to that decision, can produce a precise answer to the wrong question.

For example, estimating the cost of a randomly selected claim is not the same task as estimating the annual cost for a randomly selected policy. A claim-level table gives more rows to policies with more claims; a policy-year table gives each policy-year its own outcome, including years with no claims. These are different populations of prediction targets, not just different layouts of the same table.

Definition: The observation unit is the entity represented by one row of an analysis table. A policy-level table has one row per policy for a specified observation window. A claim-level table has one row per claim. A policy-year table has one row per policy for each defined coverage year. The outcome and predictors in each row must be defined for that same unit and the intended prediction time.

Use $i$ to index policies, $j$ to index policy-years, and $k$ to index claims within a policy-year. Let $N_{ij}$ be the number of claims for policy $i$ in year $j$, and let $C_{ijk}$ be the amount for claim $k$. The total claim amount in that policy-year is $L_{ij}=\sum_{k=1}^{N_{ij}}C_{ijk}$. When there are no claims, $N_{ij}=0$ and $L_{ij}=0$; the sum over no claims is zero.

Match the Row to the Prediction

Start by stating what one future prediction will describe. If it is the cost of a new claim, each historical claim is a natural row and claim amount is a natural outcome. If it is the total cost for a policy-year, the row should represent a policy-year and the outcome should include all claims in that year, with zero for a year without claims. If the decision concerns a policy over a multi-year window, one policy row may be appropriate, but the outcome must be defined over that same window.

A policy-level total over several years is not automatically an annual prediction. Policies may have different observation periods or exposure. If the intended quantity is cost per unit of exposure, define the exposure $e_{ij}$ and use $L_{ij}/e_{ij}$ as the policy-year outcome when that ratio matches the business question. A portfolio-level cost per exposure unit is instead $\sum_{i,j}L_{ij}/\sum_{i,j}e_{ij}$. These quantities need not equal the unweighted average of policy-year ratios when exposures differ.

Worked Example: Predicting the Cost of a New Claim

Hypothetical scenario. A specialty insurer wants to estimate the average amount paid for a claim that is reported under its small-equipment coverage. In one completed period, Policy A had claims of $240, $360, and $600; Policy B had one claim of $800. Policy C had no claims. The target is the average amount per claim, not the cost per policy. Treat the listed amounts as the complete claim amounts for this exercise; no probability distribution is assumed.

Method: make claims the rows. There are $3+1=4$ claims, so the claim-level outcome values are $240, $360, $600, and $800. Their total is

$$ 240+360+600+800=2{,}000. $$

The average amount per claim is therefore $2{,}000/4=\$500.00$. Policy C contributes no claim row because the target is the amount conditional on a claim existing. Its zero-claim experience would matter for a policy-year cost or claim-occurrence target, but it is not a zero-dollar claim.

The weighting implied by the target is important. Policy A's average claim amount is $(240+360+600)/3=\$400$; Policy B's is $\$800$. The mean of those two policy averages is $(400+800)/2=\$600$. That answers an equal-policy-weighted question about average claim amount by policy, not the average amount for a randomly selected claim. The correct claim-level result is $500.00. The common mistake is to give each policy equal weight despite the target being a claim.

Construct Outcomes and Predictors on the Same Grain

For a policy-year target, construct one outcome from all claims assigned to that coverage year. Keep a row when the claim count is zero. Dropping zero-claim policy-years changes the target population to policy-years that had a claim; it no longer describes all policy-years. Conversely, assigning a zero outcome to each non-claim row in a claim-level severity table incorrectly treats “no claim” as a claim of zero severity.

Predictors also need to be defined at the row's grain and at the time the prediction is made. A policy attribute known at the start of a year can be repeated across that policy's year rows if it is valid for each year. A changing attribute should use the value available at the relevant prediction time, rather than a later value. If one policy row summarizes multiple years, changing information might require a specifically justified summary, such as a value at the start of the window or a total over a fully observed window. The choice must correspond to what would actually be known when the prediction is used.

Claim-level rows may repeat policy attributes across several claims. That can be appropriate when each claim is the target, but it does not make the claims statistically or operationally independent. Claim-specific predictors should refer to the claim and its relevant time. A policy-level summary attached to each claim should not be mistaken for new information on every repeated row.

Worked Example: Constructing Policy-Year Cost Outcomes

Hypothetical scenario. An insurer wants to predict total covered claim amount per one-year policy period. Each of the four policy-years below has a full year of coverage, so exposure is one policy-year for each row. Policy K had two claims of $120 and $280 in year 1, and no claims in year 2. Policy L had one claim of $900. Policy M had no claims. Assume the claim lists are complete, and use no probability distribution: this is a direct construction from observed amounts.

Method: aggregate claims within policy-year, retaining zero-claim years. For K in year 1, $N=2$ and $L=120+280=\$400$. For K in year 2, $N=0$ and $L=\$0$. For L, $N=1$ and $L=\$900$. For M, $N=0$ and $L=\$0$. The four policy-year outcomes are therefore $400, $0, $900, and $0.

The combined loss is $400+0+900+0=\$1{,}300$, across four policy-years. The unweighted average total cost per policy-year is $1{,}300/4=\$325.00$. By contrast, there are three claims, and their average amount is $1{,}300/3=\$433.33$, rounded to cents. That is a claim-severity summary, not the average annual cost across all policy-years.

The exam trap is to remove K's second year and M's year because they had no claims. The resulting calculation, $1{,}300/2=\$650.00$, is the average total cost among only the two policy-years with claims. It is not the stated all-policy-year target. A policy-year table also makes it possible to attach predictors known at the start of each year, rather than using information that became known only after claims occurred.

Repeated Rows Create Dependence

A policy-year table can contain multiple rows for the same policy when the observation window covers several years. Those rows share policy characteristics and may also share unobserved risk factors. As a result, outcomes from the same policy may be associated. Repeating a policy's attributes across rows does not create independent policies.

A simple calculation shows how this affects the variation of an average. Suppose each policy contributes $m$ rows. Assume every row outcome has variance $\sigma^2$, and every pair of outcomes from the same policy has correlation $\rho$, so their covariance is $\rho\sigma^2$. This equal-variance, common-correlation setup is a simplifying assumption for the calculation, not a claim that all real data follow it.

Justification: The variance of a sum is the sum of the individual variances plus twice the sum of the pairwise covariances. Under the assumptions above, the variance of the mean of the $m$ rows from one policy is

\operatorname{Var}\left(\frac{1}{m}\sum_{j=1}^{m}Y_j\right) =\frac{m\sigma^2+2\binom{m}{2}\rho\sigma^2}{m^2} =\frac{\sigma^2}{m}\left[1+(m-1)\rho\right].

The first equality follows by expanding the variance of the sum: there are $m$ variance terms, each equal to $\sigma^2$, and $\binom{m}{2}$ covariance pairs, each counted twice. Since $2\binom{m}{2}=m(m-1)$, dividing by $m^2$ gives the final expression. If $\rho=0$, the expression reduces to $\sigma^2/m$, the independent-row result. If $\rho>0$, the variance is larger. This does not prescribe a particular modeling method; it shows why the number of rows alone can overstate how much independent policy information is present.

Worked Example: Interpreting Three Years From One Policy

Hypothetical scenario. Consider a policy with three annual outcome rows. For illustration, assume each row has variance $\sigma^2=4$, and each pair of rows from that policy has correlation $\rho=0.5$. No distributional form is assumed; only these moments are specified. The target remains an annual policy-year outcome, so the rows are not collapsed into one multi-year claim outcome.

Method: calculate the variance of the within-policy row mean. The covariance for each pair is $\rho\sigma^2=0.5(4)=2$. There are three variance terms, contributing $3(4)=12$, and three distinct covariance pairs, contributing $2(3)(2)=12$ when counted twice. Thus the variance of the sum is $12+12=24$, and the variance of the mean is $24/3^2=24/9=8/3$.

If the three rows were incorrectly treated as independent, the calculated variance of their mean would be $4/3$. The dependence-aware result, $8/3$, is twice as large. The row unit is still policy-year, but the repeated policy identity matters when assessing how much information the rows provide. When the intended use is prediction for policies not represented in the development data, evaluation should also keep a policy's repeated rows together; otherwise, information about that policy can appear on both sides of the evaluation.

A Practical Choice Process

Name the future target.
Specify whether the prediction concerns a claim, a policy-year, or a policy over a defined window. State whether it is conditional on a claim occurring.

Define the outcome at that unit.
For claim cost, use claim amounts. For policy-year cost, aggregate the year's claims and retain zero-claim years. For a multi-year policy target, aggregate over the defined window and account for differing observation periods.

Align predictors to the prediction time.
Use values that would be available when the prediction is made. Choose explicit summaries for variables that change over the row's time span; do not use future information.

Check repeated identities and implied weighting.
Ask whether claim-rich policies receive more rows by design, whether policy-years from one policy are dependent, and whether the intended evaluation concerns new claims, new years, or new policies.

A useful final check is to describe one row aloud: “This row is one ___; its outcome is ___ over ___; and its predictors were known as of ___.” If that sentence is ambiguous, the table's grain or target needs clarification before analysis. The choice of observation unit is aligned with the SOA Exam PA (Predictive Analytics) syllabus because it connects data design to the meaning of a prediction and the evidence supporting it.

Common Mistakes

Calling every claim row an independent policy. Several claims from one policy remain repeated observations from that policy.
Using claim averages for an annual-cost question. Claim amounts alone omit policy-years with no claims and do not measure total annual cost.
Giving equal weight to each policy when the target is a random claim. A claim-level target naturally includes more rows for policies with more claims.
Dropping zero-claim policy-years. This changes an all-policy-year target to a target conditional on at least one claim.
Using a predictor measured after the prediction time. A value recorded after the year or claim begins may not have been available for the intended prediction.

Key takeaway: Choose rows from the prediction target outward. Define the outcome for that row, retain the observations required by the target, align predictors to the information available at prediction time, and account for policies that contribute repeated rows.

Check Your Understanding

For each problem, choose the row unit that matches the stated target and check the requested calculation.

A hypothetical cover has two claims of $200 and $400 for Policy A, one claim of $800 for Policy B, and no claim for Policy C. For the mean amount of a randomly selected claim, how many rows are needed and what is the mean? (Answer: 3 claim rows; $(200+400+800)/3=\$466.67$.)
Four equally exposed policy-years have claim amounts $100+500$, no claims, $300$, and no claims. What are the policy-year total outcomes and their mean? (Answer: $600, $0, $300, $0; mean $900/4=\$225.00$.)
Policy A has three claims of $100, $200, and $300; Policy B has one claim of $1,000. Find both the average amount per claim and the equal-policy-weighted average of each policy's claim average. Which matches a randomly selected claim? (Answer: claim average $1{,}600/4=\$400.00$; equal-policy average $(200+1{,}000)/2=\$600.00$; the claim average matches.)
Three annual rows from one policy have common variance $\sigma^2=9$ and pairwise correlation $\rho=1/3$. Under the equal-variance, common-correlation assumptions, calculate the variance of their mean and compare it with the independent-row calculation. (Answer: $(9/3)[1+2(1/3)]=5$; independence would give $9/3=3$.)
A policy has two periods with exposures 0.5 and 1.0 policy-years and losses of $300 and $900. If the target is cost per unit of exposure, calculate each period's loss per exposure and the combined exposure-weighted result. (Answer: $300/0.5=\$600$ and $900/1=\$900$ per policy-year; combined $1{,}200/1.5=\$800$ per policy-year.)

What You'll Learn