Reinsurance With Alternative Claim Frequencies

Why the Claim-Count Model Matters

A reinsurance arrangement determines how much of each loss is ceded, but the expected aggregate recovery and its variability also depend on how many claims occur. Two portfolios can have the same expected number of claims and the same severity distribution yet have different aggregate variances if their claim-count variances differ. This tutorial develops the calculation for binomial and negative binomial claim counts.

Assume claim severities are independent and identically distributed, and independent of the claim count. For each claim, let $X$ be the loss, $Y^L$ the payment per loss made by the reinsurer, and $X-Y^L$ the amount retained by the insurer. The formulas below apply to a specified treaty once the first two moments of each per-claim payment have been found. They do not require the treaty to be proportional.

Definition: Let $N$ be the number of claims. For a per-claim payment $W$, define the aggregate payment $T_W=\sum_{j=1}^{N}W_j$, where the $W_j$ are independent copies of $W$, independent of $N$. Write $\lambda=E[N]$, $\tau^2=\operatorname{Var}(N)$, $\mu_W=E[W]$, and $\sigma_W^2=\operatorname{Var}(W)$. Then $E[T_W]=\lambda\mu_W$ and $\operatorname{Var}(T_W)=\lambda\sigma_W^2+\tau^2\mu_W^2$.

Binomial and Negative Binomial Frequencies

A binomial count is appropriate when there are a fixed number $m$ of possible claim-producing exposures, each producing at most one claim during the period, independently with the same probability $q$. Thus $N\sim\operatorname{Bin}(m,q)$, with $E[N]=mq$ and $\operatorname{Var}(N)=mq(1-q)$. The fixed upper bound $m$ is part of the model; this is not a suitable count model if an exposure can produce multiple claims in the period.

For the negative binomial convention used here, $N$ is the number of failures observed before the $r$th success in independent Bernoulli trials, each with success probability $p$. Under this convention, $N$ is unbounded and

E[N]=\frac{r(1-p)}{p}, \qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}.

The convention matters: negative binomial notation is not universal, and some sources count successes rather than failures. Always translate the stated definition into the corresponding mean and variance before using an aggregate-loss formula.

Why the Aggregate Variance Formula Holds

Condition on the number of claims. Given $N=n$, $T_W$ is a sum of $n$ independent payments, so its conditional mean is $n\mu_W$ and its conditional variance is $n\sigma_W^2$. Applying the laws of total expectation and total variance gives the formulas.

E[T_W]=E\!\left[E[T_W\mid N]\right]=E[N\mu_W]=\lambda\mu_W,

\begin{aligned} \operatorname{Var}(T_W) &=E[\operatorname{Var}(T_W\mid N)] +\operatorname{Var}(E[T_W\mid N])\\ &=E[N\sigma_W^2]+\operatorname{Var}(N\mu_W)\\ &=\lambda\sigma_W^2+\tau^2\mu_W^2. \end{aligned}

The two variance terms have distinct sources. The first reflects variation in payment amounts across claims; the second reflects variation in the number of claims. Equivalently, since $E[W^2]=\sigma_W^2+\mu_W^2$,

\operatorname{Var}(T_W) =\lambda E[W^2]+(\tau^2-\lambda)\mu_W^2.

For a Poisson count, $\tau^2=\lambda$, so this reduces to $\lambda E[W^2]$. That special simplification should not be carried over to binomial or negative binomial counts unless their count variance happens to equal their mean.

Applying the Formula to Ceded and Retained Losses

For the ceded aggregate, use $W=Y^L$. For the retained aggregate, use $W=X-Y^L$. In particular, the expected ceded amount is the expected number of claims multiplied by the expected ceded payment per claim; it is not the expected payment for one claim alone.

\begin{aligned} E[C]&=\lambda E[Y^L], & \operatorname{Var}(C) &=\lambda\operatorname{Var}(Y^L)+\tau^2\big(E[Y^L]\big)^2,\\ E[R]&=\lambda E[X-Y^L], & \operatorname{Var}(R) &=\lambda\operatorname{Var}(X-Y^L) +\tau^2\big(E[X-Y^L]\big)^2. \end{aligned}

These expressions assume that the per-claim payments are independent across claims and independent of $N$. They require the moments of the treaty-adjusted payment, not necessarily the moments of the original loss alone. The earlier material on reinsurance treaty payments addresses how a specified contract determines those per-claim amounts.

Worked Example: Binomial Frequency With a Quota Share

A hypothetical book has 12 independent policies, each with claim probability $0.25$, and at most one claim per policy. Each claim loss $X$ is independently \$1,000 or \$3,000, each with probability $0.5$. A reinsurer pays 40% of every claim. Calculate the expected aggregate ceded payment and its variance, and then calculate the corresponding retained amounts.

Method and claim-count moments. The count is $N\sim\operatorname{Bin}(12,0.25)$, so $E[N]=12(0.25)=3$ and $\operatorname{Var}(N)=12(0.25)(0.75)=2.25$.

Per-claim ceded moments. The ceded payment is $Y^L=0.4X$, taking values \$400 and \$1,200 with equal probabilities. Therefore $E[Y^L]=(400+1{,}200)/2=\$800$. Its variance is $\big((400-800)^2+(1{,}200-800)^2\big)/2=\$160{,}000$.

Aggregate ceded payment. The mean is $3(800)=\$2{,}400$. The variance is $3(160{,}000)+2.25(800^2)=480{,}000+1{,}440{,}000=\$1{,}920{,}000$.

Retained payment. The insurer retains 60% of each claim, so the per-claim retained mean is $0.6(2{,}000)=\$1{,}200$. The original loss variance is $\big((1{,}000-2{,}000)^2+(3{,}000-2{,}000)^2\big)/2=\$1{,}000{,}000$, giving retained per-claim variance $0.6^2(1{,}000{,}000)=\$360{,}000$. Thus the retained aggregate mean is $3(1{,}200)=\$3{,}600$, and its variance is $3(360{,}000)+2.25(1{,}200^2)=1{,}080{,}000+3{,}240{,}000=\$4{,}320{,}000$.

Check. Ceded plus retained expected amounts equal $2{,}400+3{,}600=\$6{,}000$, which is $3E[X]=3(2{,}000)$. All monetary results are exact to the cent.

Worked Example: Negative Binomial Frequency With the Same Severity Model

For a hypothetical portfolio, let $N$ be the number of failures before the fourth success in independent Bernoulli trials with success probability $p=0.8$. Each claim loss $X$ is independently \$500 or \$1,500, each with probability $0.5$. A reinsurer pays 30% of each claim. Find the expected and variance of aggregate ceded and retained payments.

Method and frequency moments. Here $r=4$, and the stated convention gives $E[N]=4(0.2)/0.8=1$ and $\operatorname{Var}(N)=4(0.2)/(0.8)^2=1.25$.

Ceded payment per claim. The loss mean is $(500+1{,}500)/2=\$1{,}000$, and its variance is $\big((500-1{,}000)^2+(1{,}500-1{,}000)^2\big)/2=\$250{,}000$. The ceded payment $Y^L=0.3X$ has mean $\$300$ and variance $0.3^2(250{,}000)=\$22{,}500$. Consequently the aggregate ceded mean is $1(300)=\$300$, and its variance is $1(22{,}500)+1.25(300^2)=22{,}500+112{,}500=\$135{,}000$.

Retained payment per claim. The insurer retains 70%, giving per-claim mean $0.7(1{,}000)=\$700$ and variance $0.7^2(250{,}000)=\$122{,}500$. The aggregate retained mean is $1(700)=\$700$, and its variance is $1(122{,}500)+1.25(700^2)=122{,}500+612{,}500=\$735{,}000$.

Check. Expected ceded and retained payments sum to $\$1{,}000$, equal to $E[N]E[X]$. The frequency variance exceeds the frequency mean, so the aggregate variance includes more frequency-driven variation than a Poisson model with the same mean count would have.

Worked Example: Equal Mean Counts, Different Ceded Variances

Compare two hypothetical portfolios with the same claim severity and treaty. In both, each claim loss $X$ is \$1,000 or \$3,000 with equal probability, and the reinsurer pays 25% of each loss. Portfolio A has $N\sim\operatorname{Bin}(20,0.3)$. Portfolio B counts failures before the sixth success, with success probability $p=0.5$. Calculate the mean and variance of aggregate ceded payment for each portfolio.

Per-claim ceded moments. The loss mean is $\$2{,}000$, and its variance is $\$1{,}000{,}000$. Thus $Y^L=0.25X$ has mean $\$500$ and variance $0.25^2(1{,}000{,}000)=\$62{,}500$.

Portfolio A frequency. Its count mean is $20(0.3)=6$, and its variance is $20(0.3)(0.7)=4.2$. Therefore the aggregate ceded mean is $6(500)=\$3{,}000$, and the variance is $6(62{,}500)+4.2(500^2)=375{,}000+1{,}050{,}000=\$1{,}425{,}000$.

Portfolio B frequency. The negative binomial count has mean $6(0.5)/0.5=6$ and variance $6(0.5)/(0.5)^2=12$. Its aggregate ceded mean is also $\$3{,}000$, while its variance is $6(62{,}500)+12(500^2)=375{,}000+3{,}000{,}000=\$3{,}375{,}000$.

Check. The expected ceded payments agree because both portfolios have the same count mean and per-claim payment mean. Portfolio B has a larger aggregate variance because its count variance is larger. The same calculation for the retained payment, with per-claim mean $\$1{,}500$ and variance $\$562{,}500$, gives aggregate variances of $\$12{,}825{,}000$ for A and $\$30{,}375{,}000$ for B.

Common Exam Trap: Substituting a Poisson Formula

A frequent error is to use $\lambda E[W^2]$ for every frequency model. That expression is valid for a compound Poisson aggregate, not automatically for a compound binomial or negative binomial aggregate. The general formula uses both $\lambda$ and $\tau^2$.

For Portfolio A in the preceding example, $E[(Y^L)^2]=62{,}500+500^2=\$312{,}500$. The incorrect Poisson-style calculation gives $6(312{,}500)=\$1{,}875{,}000$. The correct binomial calculation is $6(62{,}500)+4.2(500^2)=\$1{,}425{,}000$. The error comes from implicitly replacing the actual count variance $4.2$ with the mean count $6$.

When comparing frequency models, first compute both count moments, then the per-claim payment moments, and only then calculate the aggregate moments. A higher frequency variance increases aggregate variance by $(\tau^2-\lambda)\mu_W^2$ relative to the Poisson-frequency result with the same mean count and per-claim distribution. This difference is especially consequential when the average payment per claim is large.

Specify the count convention.
Identify whether the model is binomial or negative binomial and calculate $\lambda=E[N]$ and $\tau^2=\operatorname{Var}(N)$.

Translate the treaty into a per-claim payment.
Use $Y^L$ for ceded payment and $X-Y^L$ for retained payment; find each payment's mean and variance.

Compound the moments.
For each aggregate, calculate its mean as $\lambda$ times the per-claim mean and its variance as $\lambda$ times the per-claim variance plus $\tau^2$ times the squared per-claim mean.

Key takeaway: Expected aggregate ceded loss depends on the mean claim count, but aggregate variance depends on both the mean and variance of the count. Binomial and negative binomial frequencies therefore cannot generally be treated as Poisson merely because their expected claim counts match. These methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

Use the stated negative binomial convention of counting failures before the specified number of successes. Monetary answers are in dollars.

A portfolio has $N\sim\operatorname{Bin}(10,0.2)$. Each loss is independently \$100 or \$300 with equal probability, and a quota-share reinsurer pays 50% of each loss. Find the expected aggregate ceded payment and its variance. (Answer: mean \$200; variance \$21,000.)
Let $N$ count failures before the third success, with success probability $0.5$. Each claim is exactly \$1,000, and the reinsurer pays 25% of each claim. Find the expected and variance of the aggregate ceded payment. (Answer: mean \$750; variance \$375,000.)
A count is $N\sim\operatorname{Bin}(8,0.5)$. The per-claim ceded payment has mean \$150 and variance $\$900$. Find the expected and variance of the aggregate ceded payment. (Answer: mean \$600; variance \$48,600.)
Two models have the same mean count. Model A is $N\sim\operatorname{Bin}(16,0.25)$; Model B counts failures before the second success with success probability $1/3$. For either model, per-claim ceded payment has mean \$200 and variance $\$1,600$. Calculate each aggregate variance and explain the difference. (Answer: Model A variance \$126,400; Model B variance \$486,400; Model B is larger because its count variance is 12 rather than 3.)

What You'll Learn