Resampling Policy-Level Frequency-Severity Records

Why the Resampling Unit Matters

A policy can have no claims, one claim, or several claims, and its claim count may be related to the sizes of its claims. For example, policies with more claims might tend to have smaller average claims, or they might tend to have larger total losses. A bootstrap intended to represent uncertainty across policies must retain each observed policy’s count and associated losses as a unit.

In this tutorial, assume the observed policies are independent and identically distributed, each represents the same exposure period, and all claim amounts and relevant policy attributes are completely observed. The bootstrap approximates repeated samples from the empirical distribution of those observed policy records. It does not assume that claim counts and losses are independent.

Definition: A policy-level frequency-severity record contains the claim count $N_i$ for policy $i$, together with its associated claim amounts $X_{i1},\ldots,X_{iN_i}$. It may also include other policy-level values, such as premium. A zero-claim policy has $N_i=0$ and an empty list of claim amounts; it remains a complete record and is eligible for resampling.

Write the total loss for policy $i$ as $T_i=\sum_{j=1}^{N_i}X_{ij}$, with $T_i=0$ when $N_i=0$. For $n$ equally exposed policies, the observed average claim frequency, pooled average claim severity, and average loss per policy are

\widehat{\lambda}=\frac{1}{n}\sum_{i=1}^n N_i,\qquad \widehat{\mu}=\frac{\sum_{i=1}^n T_i}{\sum_{i=1}^n N_i},\qquad \widehat{P}=\frac{1}{n}\sum_{i=1}^n T_i.

The pooled severity is defined when the observed total claim count is positive. The average loss per policy is the empirical pure premium for this equal-exposure setup. When at least one claim is present, these estimates satisfy $\widehat{P}=\widehat{\lambda}\widehat{\mu}$: average claims per policy multiplied by average dollars per claim equals average dollars of claims per policy.

How to Resample Complete Records

A policy-level bootstrap replicate draws $n$ records independently with replacement from the $n$ observed records. Equivalently, it draws $n$ indices $I_1^*,\ldots,I_n^*$, each uniformly from $1,\ldots,n$, and takes every attribute of a selected policy together. A selected policy may appear more than once; another observed policy may not appear in that replicate.

\widehat{\lambda}^{*}=\frac{1}{n}\sum_{k=1}^n N_{I_k^*},\qquad \widehat{\mu}^{*}= \frac{\sum_{k=1}^n T_{I_k^*}}{\sum_{k=1}^n N_{I_k^*}},\qquad \widehat{P}^{*}=\frac{1}{n}\sum_{k=1}^n T_{I_k^*}.

The formulas use the selected records, including repeated copies. In particular, the severity denominator in a replicate is its resampled claim count, not the number of policies. If that replicate has no claims, its pooled severity is undefined; the replicate’s average loss per policy is still defined and is zero.

Build the records.
Keep each policy’s claim count, individual claim amounts, and any other quantities needed for the target statistic together.

Draw policy indices.
Sample $n$ indices with replacement from the observed $n$ policies; do not draw a new sample size of claims.

Recalculate the target.
Use the selected records to recompute the statistic, applying its original numerator and denominator definitions.

Repeat and summarize.
Across many replicates, use the resulting estimates to assess bootstrap variability or form an interval using an appropriate bootstrap method.

Why Keeping the Record Intact Preserves Association

The observed policy records define an empirical joint distribution: each observed combination of claim count and claim amounts has probability $1/n$. Drawing whole records samples from that joint distribution. Drawing counts and claim amounts separately instead samples from separate empirical marginal distributions and combines values that may never have occurred together.

The distinction can be seen using policy claim count $N$ and total loss $T$. For the records $(N,T)=(1,120),(2,100),(0,0),(1,80)$, the empirical means are $\bar N=1$ and $\bar T=75$. The empirical mean of the products is

\overline{NT}=\frac{1(120)+2(100)+0(0)+1(80)}{4}=100, \qquad \widehat{\operatorname{Cov}}(N,T)=\overline{NT}-\bar N\bar T=100-75=25.

If $N$ and $T$ were resampled independently from their separate empirical distributions, the expected product would instead be $\bar N\bar T=75$, giving zero covariance in the resampled pairing. Whole-record resampling retains the observed pairing and its empirical association.

Justification: Let $Z_i$ denote a complete policy record. Its empirical distribution assigns probability $1/n$ to each observed $Z_i$. A bootstrap draw $Z^*$ therefore has the same joint empirical distribution as the observed records, and $n$ independent draws from it produce a bootstrap sample of policies. Since every component of $Z_i$ is selected using the same index, the count and the associated losses stay paired. Independent resampling of record components would replace this joint distribution with a product of marginal distributions, which generally changes their association.

Worked Examples

Worked Example: Recalculating Frequency, Severity, and Pure Premium

A hypothetical portfolio contains four equally exposed policies. Their records are: policy A, one claim of $120; policy B, two claims of $40 and $60; policy C, no claims; and policy D, one claim of $80. A bootstrap replicate selects A, A, C, and D. Calculate the replicate frequency, pooled severity, and average loss per policy.

Method and setup. Resample complete policy records. The selected claim counts are $1,1,0,1$, and the selected policy totals are $120,120,0,80$.

Frequency. The replicate has $1+1+0+1=3$ claims across four policies, so

\widehat{\lambda}^{*}=\frac{3}{4}=0.75\text{ claims per policy}.

Pooled severity. Its total claim amount is $120+120+0+80=\$320$, and it has three claims. Thus

\widehat{\mu}^{*}=\frac{\$320}{3}=\$106.666\ldots=\$106.67.

Average loss per policy. Divide the same total loss by the four selected policies:

\widehat{P}^{*}=\frac{\$320}{4}=\$80.00.

Check. The product $0.75(\$106.666\ldots)=\$80.00$ agrees with the direct average-loss calculation. The common mistake this method avoids is treating the four observed claims as the resampling units, which would omit policy C and would not reproduce the policy-level frequency variation.

Worked Example: Exact Bootstrap Standard Error for Average Policy Loss

Consider two hypothetical equally exposed policies. Record A has one claim of $100. Record B has two claims of $20 and $40, for a total of $60. Resample two complete records with replacement. Find the exact bootstrap standard error of the average loss per policy by enumerating all ordered bootstrap samples.

Method and setup. Each of the two draws can select A or B, so there are $2^2=4$ equally likely ordered samples. For each sample, calculate its two policy totals and average them.

The four possible samples and estimates are:

A, A: $(\$100+\$100)/2=\$100$.
A, B: $(\$100+\$60)/2=\$80$.
B, A: $(\$60+\$100)/2=\$80$.
B, B: $(\$60+\$60)/2=\$60$.

The mean of these four equally likely bootstrap estimates is $(100+80+80+60)/4=\$80$. Their squared deviations from $80 are $400,0,0,400$, respectively. Because this is the exact equally weighted bootstrap distribution, its variance is the mean squared deviation:

\operatorname{Var}^{*}(\widehat{P}^{*}) =\frac{400+0+0+400}{4}=200, \qquad \operatorname{SE}^{*}(\widehat{P}^{*})=\sqrt{200}=\$14.14.

Check. The estimate’s bootstrap mean, $80, equals the observed average policy loss $(\$100+\$60)/2$. The standard error is rounded to the nearest cent. The common mistake avoided here is using the sample standard deviation with denominator $4-1$ for an exact distribution whose four outcomes each have probability $1/4$; the exact distribution’s variance uses probability weights, or equivalently the denominator 4 in this enumeration.

Worked Example: Bootstrapping a Loss Ratio With Policy Premiums

A hypothetical three-policy portfolio has these records: A has premium $100 and total claims of $120; B has premium $200 and total claims of $100 from two claims; C has premium $100 and no claims. Calculate the loss ratio for two bootstrap replicates: A, A, C; and B, B, C.

Method and setup. The loss ratio is total claims divided by total premium. Resample each policy’s premium and claims together, so each selected policy contributes both its premium and its claim record.

For replicate A, A, C, total claims are $\$120+\$120+\$0=\$240$. Total premium is $\$100+\$100+\$100=\$300$. Therefore,

\widehat{\mathrm{LR}}^{*}_{AAC}=\frac{\$240}{\$300}=0.8000=80.00\%.

For replicate B, B, C, total claims are $\$100+\$100+\$0=\$200$. Total premium is $\$200+\$200+\$100=\$500$. Therefore,

\widehat{\mathrm{LR}}^{*}_{BBC}=\frac{\$200}{\$500}=0.4000=40.00\%.

Check. In each replicate, the numerator and denominator use the same three selected policies. The common mistake avoided is resampling claims but leaving the premium total fixed, or resampling premiums independently; either changes the policy-level pairing relevant to the ratio.

Common Exam Trap and Practical Checks

A frequent error is to resample individual claims and then treat the result as a bootstrap sample of policies. In the four-policy data from the first worked example, there are four claims but also four policies, which can make the two resampling schemes seem interchangeable. They are not: one claim amount of $40 and one of $60 belong to the same policy, while the zero-claim policy contributes no claim to a claim-level sample at all.

For example, drawing four claim amounts with replacement and calculating their mean estimates a claim-severity statistic. It does not create four policy records, does not vary the number of claims per policy, and cannot by itself produce the bootstrap distribution of average loss per policy. The correction is to sample four policies with replacement, retain each policy’s full claim bundle, and then divide the replicate’s total loss by four.

Before calculating a bootstrap statistic, check the observational unit and the denominator. A frequency estimate uses the number of policies or the relevant exposure; pooled severity uses the number of claims; average loss per policy uses the number of policies; and a loss ratio uses the premiums attached to the selected policies. Zero-claim records belong in the resample even though they contribute no claim amounts.

Key takeaway: When the target concerns policy-level experience, the complete policy record is the resampling unit. Resampling whole records preserves the observed frequency-severity relationship and keeps associated policy quantities, such as premium, aligned with the claims.

Check Your Understanding

For each question, use the stated policy-level resampling unit and retain all attributes of a selected policy together.

Three equally exposed records have claim counts and total losses $(0,\$0)$, $(1,\$90)$, and $(2,\$50)$. A bootstrap sample selects the first record once and the third record twice. Find its frequency, pooled severity, and average loss per policy. (Answer: $4/3$ claims per policy; $25.00 per claim; $33.33 per policy.)
Three policy records have premiums and total losses of A: $80 and $0; B: $120 and $150; C: $100 and $50. A replicate selects B, B, C. Calculate its loss ratio. (Answer: $\$350/\$340=1.0294$, or 102.94%.)
Three equally exposed policies have total losses of $40, $100, and $0. A replicate selects the first policy once and the second twice. What is its average loss per policy? (Answer: $(\$40+\$100+\$100)/3=\$80.00$.)
Three policy records have $(N,T)=(0,0),(1,100),(2,40)$. Using divisor $n=3$ for the empirical covariance, calculate the covariance between claim count and total loss. What covariance would independent resampling of the two marginals imply for their pairing? (Answer: $\overline{NT}=60$, $\bar N=1$, $\bar T=140/3$, so covariance is $40/3=\$13.33$ in count-dollar units; independent pairing implies covariance 0.)
A bootstrap replicate draws only policies with no claims. What are its average loss per policy and pooled claim severity? (Answer: Average loss per policy is $0.00; pooled severity is undefined because the replicate has zero claims.)

What You'll Learn