Comparing Proportional and Excess-of-Loss Protection

Why Compare Reinsurance Structures?

Two treaties can protect an insurer against the same hypothetical portfolio in very different ways. A quota-share treaty cedes the same proportion of every claim, while excess-of-loss protection cedes loss only above a specified per-claim attachment. Comparing their expected cessions shows how much loss each transfers on average; comparing retained means and variances shows how their effects on the insurer’s aggregate loss differ.

The comparison below uses a compound Poisson model. Claim frequency is $N\sim\operatorname{Poisson}(\lambda)$, claim severities $X_1,X_2,\ldots$ are independent and identically distributed, and frequency is independent of severity. A treaty is applied separately to each claim, with no expenses, reinstatements, or other treaty interactions. These assumptions matter: the formulas below do not describe an annual aggregate deductible applied to total loss.

Definition: Under a quota-share treaty, let $\alpha$ be the insurer’s retained proportion, with $0\leq\alpha\leq1$. The retained payment per claim is $R_Q=\alpha X$, and the ceded payment is $C_Q=(1-\alpha)X$. Under a per-claim excess-of-loss treaty with deductible $d$ and limit $u$, the ceded payment is $C_X=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The retained payment is $R_X=X-C_X$. The limit $u$ is the maximum ceded amount per claim; $u=\infty$ denotes unlimited cover above $d$.

Aggregate Means and Variances

For either treaty, write $R$ for its retained payment on one claim. The retained aggregate loss is $T=\sum_{j=1}^{N}R_j$, and the aggregate cession is $K=\sum_{j=1}^{N}C_j$. Since the claim count is Poisson and independent of the claim amounts, the expected values and retained variance are

E[T]=\lambda E[R],\qquad E[K]=\lambda E[C],\qquad \operatorname{Var}(T)=\lambda E[R^2].

These formulas use moments of the per-claim payments. In particular, the retained variance uses the second moment $E[R^2]$, not just the variance of one retained claim. The result is specific to Poisson frequency; with another frequency model, the aggregate variance generally requires that model’s frequency moments.

Derivation: Conditional on $N$, the sum of retained payments has mean $N E[R]$ and variance $N\operatorname{Var}(R)$. By the law of total variance, $$ \operatorname{Var}(T) =E[N]\operatorname{Var}(R)+\operatorname{Var}(N)(E[R])^2. $$ For $N\sim\operatorname{Poisson}(\lambda)$, both $E[N]$ and $\operatorname{Var}(N)$ equal $\lambda$. Therefore $\operatorname{Var}(T)=\lambda\{\operatorname{Var}(R)+(E[R])^2\}=\lambda E[R^2]$. The extra term reflects variation in the number of claims.

For quota share, $R_Q=\alpha X$ and $C_Q=(1-\alpha)X$, so

E[K_Q]=\lambda(1-\alpha)E[X],\qquad E[T_Q]=\lambda\alpha E[X],\qquad \operatorname{Var}(T_Q)=\lambda\alpha^2 E[X^2].

For excess of loss, the retained mean is $E[R_X]=E[X]-E[C_X]$, and

E[R_X^2]=E[X^2]-2E[XC_X]+E[C_X^2],\qquad \operatorname{Var}(T_X)=\lambda E[R_X^2].

The cross-moment $E[XC_X]$ is essential: retained loss and ceded loss are linked because both are determined by the same claim. A calculation that subtracts only the ceded second moment from $E[X^2]$ is not valid.

Excess-of-Loss Moments for Exponential Severity

Suppose $X$ has an exponential distribution with mean $\theta$, so $S_X(x)=e^{-x/\theta}$ for $x\geq0$, $E[X]=\theta$, and $E[X^2]=2\theta^2$. For a finite per-claim layer with deductible $d$ and limit $u$, put $a=u/\theta$. Conditioning on a claim exceeding $d$, the excess above $d$ has the same exponential distribution, by the memoryless property. This gives

\begin{aligned} E[C_X]&=\theta\left(e^{-d/\theta}-e^{-(d+u)/\theta}\right),\\ E[C_X^2]&=2\theta^2 e^{-d/\theta}\left[1-(1+a)e^{-a}\right],\\ E[XC_X]&=e^{-d/\theta} \left[d\theta(1-e^{-a})+\theta^2\{2-(2+a)e^{-a}\}\right]. \end{aligned}

For example, the first identity follows because a claim must exceed $d$ to generate a cession, which occurs with probability $e^{-d/\theta}$. Given that event, the cession has the distribution of an exponential amount with mean $\theta$, capped at $u$. Its mean is $\int_0^u e^{-y/\theta}\,dy=\theta(1-e^{-u/\theta})$. Multiplying by the probability of exceeding $d$ gives the stated expression. The second and cross-moment identities follow by integrating the corresponding capped payment moments over that same conditional excess distribution. For unlimited cover, take $u$ to infinity in these expressions.

Worked Example: Comparing Quota Share With Unlimited Excess of Loss

A hypothetical insurer has $N\sim\operatorname{Poisson}(4)$ claims in a year. Each claim amount is exponential with mean $\theta=\$1{,}000$, independently of frequency. Compare a quota-share treaty retaining $\alpha=0.75$ with unlimited per-claim excess-of-loss cover above $d=\$1{,}000$. Calculate expected annual cessions, retained aggregate means, and retained aggregate variances.

Quota share. Here $E[X]=\$1{,}000$, $E[X^2]=2\theta^2=\$2{,}000{,}000$, and the ceded proportion is $1-\alpha=0.25$. Thus

E[K_Q]=4(0.25)(1{,}000)=\$1{,}000,\qquad E[T_Q]=4(0.75)(1{,}000)=\$3{,}000.

The retained second moment per claim is $\alpha^2E[X^2]=0.75^2(2{,}000{,}000)=\$1{,}125{,}000$. Therefore

\operatorname{Var}(T_Q)=4(1{,}125{,}000)=\$4{,}500{,}000.

Excess of loss. For unlimited cover, $C_X=(X-d)_+$, with $d=\theta$. The exponential tail gives $E[C_X]=\theta e^{-d/\theta}=1{,}000e^{-1}=\$367.8794$ per claim. Hence

E[K_X]=4(367.8794)=\$1{,}471.5178,\qquad E[T_X]=4(1{,}000-367.8794)=\$2{,}528.4822.

For this unlimited layer, $E[C_X^2]=2\theta^2e^{-d/\theta}$ and $E[XC_X]=e^{-d/\theta}(d\theta+2\theta^2)$. Substituting $d=\theta=1{,}000$ gives $E[C_X^2]=\$735{,}758.9$ and $E[XC_X]=\$1{,}103{,}638.3$. Therefore

\begin{aligned} E[R_X^2] &=2{,}000{,}000-2(1{,}103{,}638.3)+735{,}758.9\\ &=\$528{,}482.3,\\ \operatorname{Var}(T_X)&=4(528{,}482.3)=\$2{,}113{,}929.2. \end{aligned}

The check is that, under each treaty, expected cession plus expected retention equals the gross expected aggregate loss, $4(\$1{,}000)=\$4{,}000$. Here excess of loss has the larger expected cession and the lower retained mean and variance. That comparison is specific to these treaty parameters and this model; it is not a universal ranking of the two treaty forms.

Worked Example: A Finite Excess-of-Loss Layer

Suppose $N\sim\operatorname{Poisson}(2)$ and claim severity is exponential with mean $\theta=\$1{,}000$, independently of frequency. A hypothetical per-claim treaty has deductible $d=\$500$ and limit $u=\$1{,}000$. Find the expected annual cession, retained aggregate mean, and retained aggregate variance.

Method. This is a finite layer: the ceded amount is zero up to $\$500$, increases dollar for dollar through the next $\$1{,}000$, and is capped at $\$1{,}000$. Set $a=u/\theta=1$, $e^{-d/\theta}=e^{-0.5}=0.6065307$, and $e^{-a}=e^{-1}=0.3678794$. The per-claim expected cession is

E[C_X]=1{,}000(e^{-0.5}-e^{-1.5})=\$383.4005.

Thus $E[K_X]=2(383.4005)=\$766.8010$. The retained mean per claim is $1{,}000-383.4005=\$616.5995$, so $E[T_X]=2(616.5995)=\$1{,}233.1990$. The expected-value check is $766.8010+1{,}233.1990=\$2{,}000$, the gross expected aggregate loss.

For the variance, first calculate the per-claim moments. The finite-layer formulas give

\begin{aligned} E[C_X^2] &=2(1{,}000^2)e^{-0.5}\left(1-\frac{2}{e}\right) =\$320{,}540.7,\\ E[XC_X] &=e^{-0.5}\left[500(1{,}000)\left(1-\frac{1}{e}\right) +1{,}000^2\left(2-\frac{3}{e}\right)\right]\\ &=\$735{,}371.1. \end{aligned}

Using $E[X^2]=\$2{,}000{,}000$,

E[R_X^2]=2{,}000{,}000-2(735{,}371.1)+320{,}540.7 =\$849{,}798.5.

Therefore $\operatorname{Var}(T_X)=2(849{,}798.5)=\$1{,}699{,}597.0$. The finite limit matters: using an unlimited-layer formula would overstate the cession on claims above $\$1{,}500$ and would not give the correct retained moments.

Comparing at the Same Expected Cession

Treaty comparisons can also hold expected cession fixed. This separates differences in the payment patterns from differences in the average amount ceded. Matching expected cessions does not make the treaties equivalent: the per-claim retained payment distributions can still differ, so their retained second moments and aggregate variances can differ as well.

Worked Example: Matching Expected Cessions

Use a hypothetical portfolio with $N\sim\operatorname{Poisson}(4)$ and exponential claim severity with mean $\theta=\$1{,}000$. Compare unlimited excess-of-loss cover above $d=\$1{,}000$ with a quota-share treaty chosen to have the same expected cession. Calculate the retained means and variances.

For the excess-of-loss treaty, the expected cession per claim is $\theta e^{-1}=\$367.8794$. A quota share has expected cession per claim $(1-\alpha)\theta$, so matching the cession requires $1-\alpha=e^{-1}$, or $\alpha=1-e^{-1}=0.6321206$. Both treaties therefore have expected annual cession $4(367.8794)=\$1{,}471.5178$ and retained aggregate mean $4{,}000-1{,}471.5178=\$2{,}528.4822$.

For quota share, the retained second moment per claim is

\alpha^2E[X^2]=(1-e^{-1})^2(2{,}000{,}000)=\$799{,}152.8.

Thus $\operatorname{Var}(T_Q)=4(799{,}152.8)=\$3{,}196{,}611.2$. For the excess-of-loss treaty, the retained second moment per claim is $\$528{,}482.3$, as calculated from its payment structure, so $\operatorname{Var}(T_X)=4(528{,}482.3)=\$2{,}113{,}929.2$. The means and expected cessions match, but the retained aggregate variances do not. This check illustrates why expected cession alone does not determine the variance result.

Common Exam Trap: Omitting Frequency Variability

A frequent error is to calculate retained aggregate variance as $\lambda\operatorname{Var}(R)$. That expression omits variation in the number of claims and is not the compound Poisson variance. For the quota share in the first example, the severity variance is $\theta^2=\$1{,}000{,}000$, so the variance of one retained claim amount is $0.75^2(1{,}000{,}000)=\$562{,}500$. The incorrect calculation would be

4(562{,}500)=\$2{,}250{,}000.

The correct calculation uses $E[R^2]=\$1{,}125{,}000$, giving $4(1{,}125{,}000)=\$4{,}500{,}000$. The difference is the frequency contribution $\lambda(E[R])^2=4(750^2)=\$2{,}250{,}000$. Use $\lambda E[R^2]$ for a compound Poisson aggregate; do not substitute $\lambda\operatorname{Var}(R)$.

Key takeaway: To compare these treaties, define the retained and ceded payment per claim first. Then calculate expected aggregate cession and retention from the per-claim means, and calculate compound Poisson retained variance from the per-claim retained second moment. A matching expected cession does not imply matching retained variance.

Check Your Understanding

For each original scenario, use the stated frequency and severity assumptions. Give monetary answers to the nearest cent where applicable.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$ and exponential claim severity with mean $\$500$, independently. A quota-share treaty retains 80% of each claim. Find the retained aggregate mean and variance. (Answer: mean $\$1{,}200.00$; variance $\$960{,}000.00$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ and exponential severity with mean $\$2{,}000$. Unlimited per-claim excess-of-loss cover attaches at $d=\$2{,}000$. Find the expected aggregate cession and retained aggregate mean. (Answer: expected cession $\$1{,}471.52$; retained mean $\$2{,}528.48$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(5)$. Each claim is independently $\$100$ or $\$300$, each with probability $0.5$. A quota-share treaty retains 60%. Find the retained aggregate mean and variance. (Answer: mean $\$600.00$; variance $\$90{,}000.00$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ and exponential severity with mean $\$1{,}000$. A per-claim excess-of-loss treaty has $d=0$ and $u=\$1{,}000$. Find the expected aggregate cession. (Answer: $\$1{,}264.24$)

What You'll Learn