Common Errors in Frequency-Severity Calculation Chains

Introduction

A frequency-severity calculation is a chain of linked choices: what counts as a claim, which payment is made for each claim, and how those payments are aggregated or shared with a reinsurer. An arithmetic error is not the only risk. A calculation can be internally consistent and still answer the wrong question because a deductible was applied at the wrong level or the frequency parameter no longer matches the severity distribution.

This tutorial develops a practical audit method for those errors. Let $N$ be the number of original claims, let $X_i$ be the loss on original claim $i$, and let $g$ be the contract’s per-claim payment function. The insurer’s payment on claim $i$ is $Z_i=g(X_i)$, and the aggregate payment is $S=\sum_{i=1}^{N}Z_i$. The central discipline is to keep the original loss, covered payment, and claim count distinct.

Definition: A frequency convention specifies what events are counted by $N$. Under the original-claim convention, $N$ counts all claims before coverage modifications, including claims that produce zero payment. Under the positive-payment convention, the count includes only claims for which $Z_i>0$. The severity distribution must match the chosen convention.

Start With the Contract and the Count

Before calculating an aggregate distribution, write down the sequence from original loss to final payment. A deductible and limit applied separately to each claim produce one payment $g(X_i)$ per original claim. An aggregate deductible instead acts on the sum of losses or payments; it is not interchangeable with a per-claim deductible.

For example, with an ordinary per-claim deductible $d$, no limit, and no coinsurance, the payment is $Z=(X-d)_+=\max(X-d,0)$. The deductible is applied to every claim before the payments are summed. An aggregate deductible $D$ on total loss instead produces $(\sum_i X_i-D)_+$. These expressions can give substantially different results.

A useful first audit is to make a small table of possible original losses, their covered payments, and whether each payment is positive. This exposes zero payments and helps determine whether the frequency parameter counts original claims or only payment-producing claims.

Worked Example: Per-Claim Deductible or Aggregate Deductible?

A hypothetical policy has exactly two claims, with losses of $80 and $80. There is no limit or coinsurance. Compare the total insurer payment under a $50 deductible applied per claim with the payment under a $50 aggregate deductible applied to total loss.

Method. Calculate each contract on its own terms. Do not sum losses first when the deductible is per claim.

With a $50 per-claim deductible, the first payment is $(80-50)_+=30$, and the second payment is also $(80-50)_+=30$. The aggregate payment is $30+30=60$.

With a $50 aggregate deductible, first sum the original losses: $80+80=160$. The payment is $(160-50)_+=110$.

The results are $60 and $110, respectively. As a check, each per-claim payment is less than its corresponding $80 loss, while the aggregate-deductible payment is less than the $160 total loss. The common error is to subtract the deductible only once from the total when the contract applies it separately to every claim.

Zero Payments and Frequency Consistency

A claim can be present in the original frequency count but produce no covered payment. This occurs, for example, when a loss is at or below a per-claim deductible. Whether that claim remains in the model depends on the convention used; it must not be decided implicitly by dropping zero values from a severity table.

If $N$ counts original claims and the original severity $X$ has a coverage-modified payment $Z=g(X)$, then zero payments may be included in the distribution of $Z$. If instead the model counts only positive payments, the frequency distribution and severity distribution must both be conditional on $Z>0$. For a Poisson original count, this adjustment has a particularly simple form.

Theorem: Suppose $N\sim\operatorname{Poisson}(\lambda)$, original losses are independent and identically distributed, and each claim independently produces a positive payment with probability $p=P(Z>0)$. Then the number $N_+$ of positive payments has distribution $\operatorname{Poisson}(\lambda p)$. Conditional on a positive payment, the severity distribution is that of $Z\mid Z>0$.

To justify the result, condition on the original claim count. Given $N=n$, the number of positive payments is binomial with parameters $n$ and $p$. For any nonnegative integer $k$, its probability is

\begin{aligned} P(N_+=k) &=\sum_{n=k}^{\infty}P(N=n)\binom{n}{k}p^k(1-p)^{n-k}\\ &=e^{-\lambda}\sum_{n=k}^{\infty} \frac{\lambda^n}{n!}\binom{n}{k}p^k(1-p)^{n-k}\\ &=e^{-\lambda p}\frac{(\lambda p)^k}{k!}. \end{aligned}

Thus the positive-payment count has mean $\lambda p$, not $\lambda$. This thinning result relies on a Poisson original count and independent claim-level payment eligibility with common probability $p$. Do not automatically use it for a binomial or negative binomial count; those frequencies require their own count adjustment.

Worked Example: Reconciling Zero Payments With Positive-Payment Frequency

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ original claims. Each original loss $X$ is $100 with probability 0.2, $300 with probability 0.5, and $500 with probability 0.3. The policy has a $200 per-claim deductible and no limit or coinsurance. Find the expected aggregate insurer payment, first using original claims and then using positive payments.

Method. Transform each possible original loss once, then match the count mean to the severity convention.

The per-claim payment is $Z=(X-200)_+$. The possible outcomes are:

If $X=100$, then $Z=0$, with probability 0.2.
If $X=300$, then $Z=100$, with probability 0.5.
If $X=500$, then $Z=300$, with probability 0.3.

Using the original-claim convention, the mean payment per original claim is $E[Z]=0(0.2)+100(0.5)+300(0.3)=0+50+90=140$. Therefore $E[S]=E[N]E[Z]=2(140)=\$280$.

For the positive-payment convention, $p=P(Z>0)=0.5+0.3=0.8$. Poisson thinning gives $N_+\sim\operatorname{Poisson}(2(0.8))=\operatorname{Poisson}(1.6)$. Conditional on a positive payment, the payment is $100 with probability $0.5/0.8=0.625$ and $300 with probability $0.3/0.8=0.375$. Its conditional mean is $100(0.625)+300(0.375)=62.5+112.5=\$175$. Thus $E[S]=1.6(175)=\$280$, agreeing with the original-claim calculation.

The check is the equality of the two aggregate means. A common mistake is to use the original mean frequency 2 with the conditional positive-payment severity, giving $2(175)=\$350$. Another is to use frequency 1.6 with the unconditional severity mean of $140, giving $224. Each calculation mixes conventions.

Transform Severity Once

Coverage-modified severity is already a transformed quantity. If the payment function is $g$, then the severity supplied to an aggregate calculation for the insurer’s payments is $Z=g(X)$. Applying the same deductible, limit, or coinsurance again to $Z$ describes a second contract modification, not the original one.

This distinction matters in direct expectation calculations and in aggregate methods such as Panjer recursion. The frequency and the severity inputs must describe the same modeled events and payments. For example, one may retain the original Poisson frequency with a severity distribution that includes zero payments, or thin to positive payments and use the conditional positive-payment severity. One must not thin the frequency and also retain a severity distribution that already assigns probability to zero payments.

Worked Example: Detecting a Second Deductible Application

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ original claims. Each loss $X$ is $100 or $300, each with probability 0.5. A $100 deductible applies once per claim, with no limit or coinsurance. Find the expected aggregate insurer payment and identify the result of mistakenly applying the deductible twice.

Method. Compute the contractual payment from the original loss before taking its expectation.

For $X=100$, the payment is $(100-100)_+=0$. For $X=300$, it is $(300-100)_+=200$. Thus $E[Z]=0(0.5)+200(0.5)=\$100$, and the expected aggregate payment is $2(100)=\$200$.

The positive-payment probability is 0.5, so an equivalent positive-payment model has $N_+\sim\operatorname{Poisson}(1)$. Every positive payment is $200, giving expected aggregate payment $1(200)=\$200$, which checks the first calculation.

If the deductible is mistakenly applied again to the already modified payment, the values become $(0-100)_+=0$ and $(200-100)_+=100$. This erroneous severity has mean $0(0.5)+100(0.5)=\$50$, producing an erroneous aggregate mean of $2(50)=\$100$. The error halves the answer because it subtracts the deductible a second time from the covered payment.

Reinsurance Layers and Order of Operations

A reinsurance calculation must follow the treaty wording: identify the loss base to which each layer applies, the attachment point, the limit, and the order in which covers inure. A per-claim excess layer applied after quota share is not generally the same as a layer applied to gross loss and then scaled by a quota-share percentage.

For a layer attaching at $a$ with limit $u$, the recovery on the stated loss base $W$ is $\min((W-a)_+,u)$. This expression does not by itself specify whether $W$ is the original gross loss or the loss remaining after another reinsurance cover. That is a contract-order question, not an algebraic convenience.

Worked Example: Applying Excess Cover to the Specified Layer Base

A hypothetical insurer has a $1,000 claim. A quota-share treaty first cedes 25% of the claim. A per-claim excess-of-loss treaty then covers the insurer’s remaining loss above $500, subject to a $300 limit. Calculate the quota-share recovery, excess-of-loss recovery, and final net retention.

Method. Follow the stated order: calculate the quota-share cession first, then apply the excess layer to the remaining loss.

The quota-share recovery is $0.25(1{,}000)=\$250$. The insurer’s loss after quota share is $1{,}000-250=\$750$. The excess layer applies to that $750 loss base, so its recovery is $\min((750-500)_+,300)=\min(250,300)=\$250$. The final net retention is $750-250=\$500$.

The conservation check is $250+250+500=\$1{,}000$, equal to the original claim. A common error is to apply the excess layer to the gross $1,000 loss and then multiply the recovery by 75%. That would give a recovery of $0.75\min((1{,}000-500)_+,300)=0.75(300)=\$225$, and a claimed net retention of $750-225=\$525$. It violates the treaty sequence because the excess layer was specified to apply after quota share, to a $750 base.

A Practical Audit Sequence

A reliable calculation chain can be checked in a fixed order. This does not replace the probability calculation; it helps ensure that the model represents the contract and the requested aggregate quantity.

Define the count.
State whether frequency counts original claims or positive payments, and record the distribution and parameter of that count.

Write the payment function.
Express the contract’s per-claim payment in terms of the original loss. Keep aggregate deductibles separate from per-claim modifications.

Build the matching severity.
Transform the original severity once. If using a positive-payment count, condition the severity on positive payment as well.

Follow the reinsurance order.
For each layer, identify its loss base, attachment, and limit before calculating recovery.

Check the result.
Compare expectations under equivalent frequency conventions when possible, check that payments do not exceed their stated loss bases, and reconcile retained loss plus recoveries.

These checks are useful because many errors can be found before a full aggregate distribution is calculated. For a Poisson model with independent claim payments, the mean aggregate payment must equal the original frequency mean times the unconditional mean payment per original claim. Under positive-payment thinning, the same mean must equal the thinned frequency mean times the conditional mean positive payment. A disagreement signals a mismatched count, severity, or transformation.

The same discipline applies when the objective is an aggregate distribution rather than its mean. The severity probabilities supplied to a calculation must represent the actual payment variable, and any probability at zero must be handled consistently with the frequency convention and the chosen calculation method. For reinsurance, aggregate calculations also require the contract’s actual order of application; a layer cannot be moved across another modification merely because the resulting expression looks simpler.

Key takeaway: Keep three objects separate throughout the calculation: the original claim count, the original loss, and the final payment on each claim. A valid frequency-severity model connects them exactly once and respects the stated order of every coverage and reinsurance modification. This diagnostic approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each question, identify the frequency convention and contract operation before calculating.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ original claims. Each claim has a positive insurer payment with probability 0.75, independently. What is the distribution and mean of the number of positive payments? (Answer: $N_+\sim\operatorname{Poisson}(3)$, with mean 3.)
Two claims have losses of $120 and $250. A $100 deductible applies separately to each claim, with no limit or coinsurance. Find the aggregate payment. (Answer: $(120-100)+(250-100)=\$170$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(5)$. Each original loss is $80 with probability 0.4 or $200 with probability 0.6. A $100 per-claim deductible applies, without a limit. Find the expected aggregate insurer payment. (Answer: $E[Z]=0(0.4)+100(0.6)=\$60$, so $E[S]=5(60)=\$300$.)
A $900 claim is first subject to a 30% quota-share cession. An excess layer then covers the insurer’s remaining loss above $500, with a $200 limit. Find the excess recovery and final net retention. (Answer: remaining loss is $0.70(900)=\$630$; excess recovery is $\min(630-500,200)=\$130$; net retention is $\$500$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(6)$, and each original claim produces a positive payment with probability 0.8. The conditional mean positive payment is $250. Find the expected aggregate payment using the positive-payment convention, and check it using an unconditional per-original-claim mean. (Answer: $E[N_+]=6(0.8)=4.8$, so the mean aggregate is $4.8(250)=\$1{,}200$; the unconditional mean payment is $0.8(250)=\$200$, giving $6(200)=\$1{,}200$.)

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