Annual Layer Loss Under Frequency-Severity Models

From Per-Occurrence Payment to Annual Recovery

A per-occurrence layer responds separately to each claim. Its annual recovery is therefore the sum of the layer payments on all claims during the year—not the layer payment calculated on the sum of those claims. To estimate expected annual recovery, we need both a claim-frequency model and a severity model that determines the payment from one claim.

Let $N$ be the number of claims in a year, and let $X_j$ be the ground-up severity of claim $j$. For a layer attaching at $a$ and ending at $b$, where $0 \leq a < b$, the payment on claim $j$ is $Y_j=\min((X_j-a)_+,b-a)$, where $(z)_+=\max(z,0)$. The annual recovery is $R=\sum_{j=1}^{N}Y_j$; if $N=0$, the sum is zero.

Definition: In a frequency-severity model, $N$ describes the number of claims in the period and $X_1,X_2,\ldots$ describe their severities. When, conditional on $N$, claim severities are independent and identically distributed and are independent of $N$, write $m=E[Y]$ for the expected payment per claim and $q=E[Y^2]$ for its second moment. Then expected annual per-occurrence layer recovery is $E[R]=E[N]m$.

The layer payment is bounded between zero and $b-a$, even when ground-up severity has a long right tail. For a nonnegative severity distribution, the expected payment per claim can also be calculated from its survival function as $m=\int_a^b S_X(x)\,dx$. This is the survival-function form for a per-occurrence layer. In applications, the frequency and severity inputs must describe compatible claim populations, coverage terms, and time periods.

Why Expected Annual Recovery Has This Form

The product $E[N]m$ follows by conditioning on the number of claims. If $N=n$, there are $n$ claim payments, each with expectation $m$. Thus the conditional expected annual recovery is $E[R\mid N=n]=nm$. Taking the expectation over possible values of $N$ gives the result:

E[R]=E\!\left[E[R\mid N]\right] =E[Nm] =mE[N].

This argument does not require a Poisson count. It does require that the expected layer payment per claim is the same for each claim and does not vary with the count, after conditioning on the model being used. If claim severity differs by risk group, calculate the expected recovery for each group and add the results rather than applying one unexamined average.

Annual Variability and the Role of Frequency

Two frequency models can have the same mean claim count, and therefore the same expected annual recovery, but quite different count variability. Under the assumptions above, the variance of annual recovery is:

\operatorname{Var}(R) =E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)m^2 =E[N]q+\bigl(\operatorname{Var}(N)-E[N]\bigr)m^2.

To justify the first equality, conditional on $N=n$, the sum contains $n$ independent payments, so $\operatorname{Var}(R\mid N=n)=n\operatorname{Var}(Y)$. Also $E[R\mid N]=Nm$. The law of total variance then gives $\operatorname{Var}(R)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)m^2$. The second equality uses $q=\operatorname{Var}(Y)+m^2$.

For a Poisson count with mean $\lambda$, $\operatorname{Var}(N)=E[N]=\lambda$, so $\operatorname{Var}(R)=\lambda q$. An overdispersed count has variance greater than its mean. Holding the mean frequency and severity distribution fixed, it leaves expected annual recovery unchanged but increases its variance by $(\operatorname{Var}(N)-E[N])m^2$. This variance describes the aggregate recovery under the stated model; it is not, by itself, a risk load or a premium indication.

Worked Example: Poisson Frequency and a Capped Layer

A hypothetical insurer covers claims with a per-occurrence layer attaching at $50,000 and ending at $100,000. Claim severity $X$, measured in thousands of dollars, is 40 with probability 0.50, 70 with probability 0.30, and 130 with probability 0.20. Annual claim count is Poisson with mean 2.5, independent of the severities. Find expected annual recovery and its variance.

Method and layer payments. The layer has a $50,000 maximum payment per claim. At severity 40, the payment is $\min((40-50)_+,50)=0$. At severity 70, it is $\min(70-50,50)=20$. At severity 130, it is $\min(130-50,50)=50$. All amounts in this calculation are thousands of dollars.

Expected payment per claim. The mean payment is $m=0.50(0)+0.30(20)+0.20(50)=0+6+10=16$, or $16,000. The second moment is $q=0.50(0^2)+0.30(20^2)+0.20(50^2)=0+120+500=620$, in squared thousands of dollars.

Expected annual recovery. Since $E[N]=2.5$, $E[R]=2.5(16)=40$, or $40,000.

Variance and check. For a Poisson count, $\operatorname{Var}(R)=\lambda q=2.5(620)=1{,}550$ squared thousands of dollars. The standard deviation is $\sqrt{1{,}550}=39.370$ thousand dollars, or $39,370 to the nearest dollar. As a check on the mean, multiplying 2.5 expected claims by $16,000 expected payment per claim also gives $40,000. The common mistake avoided is applying the layer once to annual ground-up losses instead of summing individual claim payments.

Poisson and Overdispersed Frequency Compared

A Poisson model is a useful benchmark when the count is modeled as arising at a constant rate over the period, with independent increments and no additional variation in that rate. It imposes $\operatorname{Var}(N)=E[N]$. In contrast, an overdispersed model allows more count variation than the Poisson model. One example is a mixed-Poisson model: conditional on a random annual intensity $\Lambda$, $N$ is Poisson with mean $\Lambda$. Then $E[N]=E[\Lambda]$ and $\operatorname{Var}(N)=E[\Lambda]+\operatorname{Var}(\Lambda)$.

The mean recovery depends on $E[N]$, not on whether the count is Poisson or overdispersed. Frequency-model choice does matter for the variance and for the distribution of possible annual outcomes. It may also affect other pricing decisions, but those require additional assumptions and methods beyond the expected-loss calculation here.

Worked Example: Same Mean Frequency, Higher Count Variance

Use the layer and severity distribution from the previous example, so $m=16$ and $q=620$, both in thousands-based units. Compare a Poisson count with mean 4 to a hypothetical mixed-Poisson count with $E[\Lambda]=4$ and $\operatorname{Var}(\Lambda)=6$. Find expected recovery and recovery variance under each model.

Poisson model. Here $E[N]=\operatorname{Var}(N)=4$. Therefore $E[R]=4(16)=64$, or $64,000, and $\operatorname{Var}(R)=4(620)=2{,}480$ squared thousands of dollars.

Mixed-Poisson model. Its count mean is $E[N]=E[\Lambda]=4$, while its count variance is $\operatorname{Var}(N)=4+6=10$. Therefore $E[R]=4(16)=64$, again $64,000. Its recovery variance is $4(620)+(10-4)(16^2)=2{,}480+6(256)=2{,}480+1{,}536=4{,}016$ squared thousands of dollars.

Comparison and check. The mean is unchanged because both frequency models have mean 4. The mixed-Poisson recovery variance exceeds the Poisson variance by $4{,}016-2{,}480=1{,}536$, exactly $(10-4)(16^2)$. The common mistake avoided is treating a higher count variance as if it automatically increased the expected recovery.

Frequency Models With a Small Count Distribution

The same formulas apply when the count distribution is specified directly rather than through a named model. Computing its first two moments explicitly is useful both as a check and as a reminder that annual recovery is a random sum of per-claim payments.

Worked Example: A Discrete Overdispersed Claim Count

For a hypothetical portfolio, the number of claims $N$ is 0 with probability 0.25, 1 with probability 0.50, and 3 with probability 0.25. Each claim has the severity distribution in the first example and the same layer, so $m=16$ and $q=620$. Find expected annual recovery and its variance.

Count moments. $E[N]=0(0.25)+1(0.50)+3(0.25)=0+0.50+0.75=1.25$. Also, $E[N^2]=0^2(0.25)+1^2(0.50)+3^2(0.25)=0+0.50+2.25=2.75$. Hence $\operatorname{Var}(N)=2.75-(1.25)^2=2.75-1.5625=1.1875$.

Expected annual recovery. $E[R]=1.25(16)=20$, or $20,000.

Variance and check. $\operatorname{Var}(R)=1.25(620)+(1.1875-1.25)(16^2)=775+(-0.0625)(256)=775-16=759$ squared thousands of dollars. The result is positive, as a variance must be. The count is underdispersed relative to Poisson because $1.1875<1.25$; accordingly, its recovery variance is 16 squared thousands less than the Poisson variance with the same mean count. This calculation also checks the count moments directly rather than assuming a Poisson relationship between mean and variance.

Scope, Assumptions, and Common Traps

These calculations concern a per-occurrence layer with no aggregate cap on the insurer’s annual recovery. If the contract also has an annual aggregate limit, the recovery is no longer simply the unrestricted sum of per-claim layer payments: the aggregate cap must be applied to that sum, and $E[N]m$ will generally overstate expected capped recovery. Similarly, a contract with annual aggregate attachment requires modeling the annual total, not just the individual claim payments.

The identical-severity assumption is convenient, not universal. If claims come from distinct groups $g$, with count $N_g$ and expected payment $m_g$ for group $g$, then linearity of expectation gives $E[R]=\sum_g E[N_g]m_g$, provided the group-specific payments are defined on the appropriate layer. This mean calculation does not require the group counts to be independent. Calculating the variance of the combined recovery, however, also requires accounting for dependence between group recoveries.

Do not layer the annual total by accident. A per-occurrence layer is applied claim by claim; the annual recovery sums those individual payments.
Do not substitute the count variance for its mean. The expected recovery uses $E[N]$. The count variance enters the variance of recovery.
Keep the units consistent. If severities and layer payments are in thousands, the expected recovery is also in thousands, while the variance is in squared thousands.
Check the contract structure. An annual aggregate limit or attachment changes the recovery function and can invalidate the simple product formula.

Key takeaway: Under independent, identically distributed claim severities independent of frequency, expected annual per-occurrence layer recovery is expected claim count multiplied by expected layer payment per claim. With the mean count held fixed, overdispersion changes annual recovery variability, not this expected value.

This framework is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. In practice, the model inputs should reflect the covered exposure, time period, claim definition, and exact layer terms; the calculation is only as relevant as those inputs.

Check Your Understanding

For each problem, treat the stated per-claim moments or severity distribution as applying independently of the annual count, and keep all monetary amounts in the stated units.

A hypothetical layer attaches at $40,000 and ends at $100,000. Claim severity, in thousands of dollars, is 30 with probability 0.40, 70 with probability 0.35, and 120 with probability 0.25. If $N$ is Poisson with mean 1.8, calculate expected annual recovery. (Answer: $45,900)
A layer payment has $m=12$ thousand dollars and $q=250$ squared thousands of dollars. Annual claim count has mean 3 and variance 8. Calculate expected annual recovery and its variance. (Answer: $36,000; 1,470 squared thousands of dollars)
Annual claim count is 0, 1, or 4 with probabilities 0.20, 0.50, and 0.30, respectively. Per-claim layer payments have $m=9$ thousand dollars and $q=145$ squared thousands of dollars. Calculate the count mean and variance, then the mean and variance of annual recovery. (Answer: $E[N]=1.7$, $\operatorname{Var}(N)=2.41$; $E[R]=$15,300$, $\operatorname{Var}(R)=304.01$ squared thousands of dollars)
Two independent claim groups have Poisson count with mean 2 and mixed-Poisson count with mean 1 and variance 3, respectively. Their layer payments have $(m_A,q_A)=(8,100)$ and $(m_B,q_B)=(20,600)$, in thousands-based units. Find total expected recovery and variance. (Answer: $36,000; 1,600 squared thousands of dollars)
A Poisson frequency model with mean 2.4 and an overdispersed model with mean 2.4 and variance 5.4 use the same severity distribution, for which $m=10$ thousand dollars and $q=180$ squared thousands of dollars. Compare their expected annual recoveries and variances. (Answer: both means are $24,000; Poisson variance is 432 and overdispersed variance is 732 squared thousands of dollars)

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