Evaluating Reinsurance Effects on Capital and Ruin

Why Evaluate Reinsurance Across Several Measures?

Reinsurance changes more than the amount an insurer pays on a large claim. It changes the distribution of retained aggregate losses, requires a premium outflow, and therefore affects the surplus available to absorb losses. A sound comparison keeps those effects separate: lower retained claims do not, by themselves, establish that the insurer’s overall financial position or ruin probability improves.

This tutorial evaluates a specified per-claim excess-of-loss treaty. The treaty is treated as fixed rather than optimized. The comparisons are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For gross claim amounts $X_1,\ldots,X_N$ and a per-claim retention $M>0$, the insurer retains $Y_j=\min(X_j,M)$ on claim $j$, and the reinsurer pays $Z_j=(X_j-M)_+=\max(X_j-M,0)$. Thus $X_j=Y_j+Z_j$. The gross, retained, and ceded aggregate claims are $S_G=\sum_{j=1}^N X_j$, $S_R=\sum_{j=1}^N Y_j$, and $S_C=\sum_{j=1}^N Z_j$, so $S_G=S_R+S_C$.

For the compound Poisson calculations below, assume $N\sim\operatorname{Poisson}(\lambda)$, that claim amounts are independent and identically distributed, and that they are independent of $N$. The treaty applies separately to each claim. All monetary amounts in an example are stated in dollars or in thousands of dollars, as specified.

Expected Claims and Variance After Reinsurance

Under the compound Poisson assumptions, if each payment per loss $Y$ has a finite second moment, then the aggregate payment $S=\sum_{j=1}^N Y_j$ satisfies $E[S]=\lambda E[Y]$ and $\operatorname{Var}(S)=\lambda E[Y^2]$. Apply these formulas separately to gross, retained, and ceded payments. In particular, retained payments are calculated claim by claim before aggregating.

Important distinction: The compound Poisson aggregate variance is $\lambda E[Y^2]$, not $\lambda\operatorname{Var}(Y)$. The randomness in the claim count contributes to aggregate variability as well as the randomness in claim amounts.

Worked Example: Expected Cost and Variance Under a Retention

A hypothetical insurer models annual claim count as $N\sim\operatorname{Poisson}(2)$. Each claim amount $X$, in thousands of dollars, is $1$ with probability $0.5$, $3$ with probability $0.3$, and $5$ with probability $0.2$. A per-claim excess-of-loss treaty has retention $M=4$ thousand dollars. The reinsurer charges an annual premium equal to 125% of expected ceded claims. Calculate gross and retained aggregate expected claims and variances, and the reinsurance premium.

Method and claim-level payments. The retained amount is $Y=\min(X,4)$, so it is $1,3,4$ with probabilities $0.5,0.3,0.2$. The ceded amount is $Z=(X-4)_+$, so it is $0,0,1$ with those same probabilities.

Gross severity moments. The mean claim is $E[X]=0.5(1)+0.3(3)+0.2(5)=0.5+0.9+1.0=2.4$ thousand dollars. Its second moment is $E[X^2]=0.5(1^2)+0.3(3^2)+0.2(5^2)=0.5+2.7+5.0=8.2$ thousand dollars squared. Therefore $E[S_G]=2(2.4)=4.8$ thousand dollars and $\operatorname{Var}(S_G)=2(8.2)=16.4$ thousand dollars squared.

Retained severity moments. $E[Y]=0.5(1)+0.3(3)+0.2(4)=0.5+0.9+0.8=2.2$ thousand dollars. Also, $E[Y^2]=0.5(1)+0.3(9)+0.2(16)=0.5+2.7+3.2=6.4$ thousand dollars squared. Thus $E[S_R]=2(2.2)=4.4$ thousand dollars and $\operatorname{Var}(S_R)=2(6.4)=12.8$ thousand dollars squared.

Ceded cost and premium. $E[Z]=0.2(1)=0.2$ thousand dollars per claim, so expected annual ceded claims are $E[S_C]=2(0.2)=0.4$ thousand dollars. The stated premium is $1.25(0.4)=0.5$ thousand dollars, or $500.

Check. Expected aggregate claims reconcile: $E[S_R]+E[S_C]=4.4+0.4=4.8=E[S_G]$. The gross variance is $16.4$, not $2(2.44)=4.88$: $2.44=E[X^2]-(E[X])^2$ is the individual claim variance, whereas the compound Poisson aggregate variance uses $E[X^2]=8.2$.

Why the Compound Poisson Variance Formula Holds

The variance formula follows by conditioning on the number of claims. Given $N=n$, the sum of $n$ independent payments has conditional mean $nE[Y]$ and conditional variance $n\operatorname{Var}(Y)$. The law of total variance then adds the variation in the conditional mean as the claim count varies.

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])\\ &=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2\\ &=\lambda\operatorname{Var}(Y)+\lambda(E[Y])^2\\ &=\lambda E[Y^2]. \end{aligned}

The last equality uses $\operatorname{Var}(Y)=E[Y^2]-(E[Y])^2$, and the preceding equality uses $E[N]=\operatorname{Var}(N)=\lambda$ for a Poisson count. This result requires the stated compound Poisson assumptions and a finite second moment. It should not be transferred unchanged to a different frequency model.

Comparing Tail Capital

A tail measure such as Value-at-Risk (VaR) summarizes a selected quantile of aggregate claims. For confidence level $p$, define $\operatorname{VaR}_p(S)=\inf\{s:P(S\le s)\ge p\}$. In a capital comparison, one common convention is to call $\operatorname{VaR}_p(S)-E[S]$ capital above expected claims. State the convention: VaR itself and VaR less the expected loss are different quantities.

For a fixed treaty, each retained claim is no greater than its corresponding gross claim. Under the same claim-count and claim-severity outcomes, $S_R\le S_G$ for every outcome. This implies that the retained aggregate distribution is no larger in the usual quantile ordering, so its VaR cannot exceed the gross VaR at the same confidence level. However, the difference between VaR and expected claims need not fall in exactly the same proportion as either quantity.

Worked Example: Gross and Retained 99% VaR

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims per year. Each claim is exactly $4,000. A treaty has a per-claim retention of $2,500, so the insurer retains $2,500 and the reinsurer pays $1,500 on every claim. Find gross and retained 99% VaR and capital measured as VaR less expected aggregate claims.

Method. Since claim counts are Poisson, $P(N=k)=e^{-1}/k!$. The cumulative probabilities needed to locate the 99th percentile are $P(N\le3)=e^{-1}(1+1+1/2+1/6)=0.9810$ and $P(N\le4)=e^{-1}(1+1+1/2+1/6+1/24)=0.9963$. Therefore the 99% quantile of $N$ is 4.

Gross results. Gross aggregate claims equal $4{,}000N$, so $\operatorname{VaR}_{0.99}(S_G)=4(4{,}000)=\$16{,}000$. The expected gross aggregate is $E[S_G]=1(4{,}000)=\$4{,}000$. Capital above the mean is $16{,}000-4{,}000=\$12{,}000$.

Retained results. Retained aggregate claims equal $2{,}500N$, so $\operatorname{VaR}_{0.99}(S_R)=4(2{,}500)=\$10{,}000$. The expected retained aggregate is $E[S_R]=1(2{,}500)=\$2{,}500$. Capital above the mean is $10{,}000-2{,}500=\$7{,}500$.

Check and trap avoided. Both VaRs use the same claim-count quantile, but each is multiplied by its own per-claim amount. Using the gross $4,000 severity for retained claims would incorrectly give a retained VaR of $16,000. The values are discrete quantiles: $P(N\le3)<0.99$ and $P(N\le4)\ge0.99$, which verifies the selected count.

Premium Income and Ruin Must Be Assessed Together

Let $c_G$ denote gross premium income over a specified period and $\pi_R$ the reinsurance premium paid for that period. The insurer’s premium income net of reinsurance is $c_N=c_G-\pi_R$. If initial surplus is $u$, retained claims are paid from surplus while net premium is received. For a one-period model with all premiums and claims settled at the horizon, the terminal surplus is $U(1)=u+c_N-S_R$.

A ruin calculation must specify what counts as ruin, the time horizon, and the timing of premiums and claims. The example below defines ruin as negative terminal surplus at the end of one year. Because the model settles all claims at that point, this is an exact one-period terminal-ruin probability for the stated assumptions; it is not a general continuous-time path-ruin calculation.

Worked Example: Net Premium and One-Year Terminal Ruin

A hypothetical insurer starts with surplus of $5,000. Its annual gross premium income is $3,000. It buys a treaty with a $2,000 per-claim retention and pays a reinsurance premium of $600 for the year, leaving net premium income of $2,400. Annual claim count is $N\sim\operatorname{Poisson}(0.5)$, and every gross claim is exactly $4,000. Claims and premiums are settled at year-end. Compare gross and retained terminal-ruin probabilities, where ruin means negative year-end surplus.

Retained calculation. Under the treaty, the insurer pays $2,000 per claim. Its terminal surplus is $U_R(1)=5{,}000+2{,}400-2{,}000N=7{,}400-2{,}000N$. Ruin occurs when $7{,}400-2{,}000N<0$, or $N>3.7$, hence when $N\ge4$.

For a Poisson random variable with mean $0.5$, $P(N\ge4)=1-P(N\le3)$. Therefore $P(N\ge4)=1-e^{-0.5}(1+0.5+0.5^2/2+0.5^3/6)=1-e^{-0.5}(1+0.5+0.125+0.0208333)=0.0018$, to four decimal places.

Gross comparison. Without the treaty, the insurer receives the full $3,000 gross premium and pays $4,000 per claim. Its terminal surplus is $U_G(1)=5{,}000+3{,}000-4{,}000N=8{,}000-4{,}000N$. Ruin occurs when $N\ge3$. Thus $P(U_G(1)<0)=1-e^{-0.5}(1+0.5+0.125)=0.0144$, to four decimal places.

Check and interpretation. Expected retained claims are $0.5(2{,}000)=\$1{,}000$; expected gross claims are $0.5(4{,}000)=\$2{,}000$. Expected terminal surplus is therefore $6,400 with the treaty and $6,000 without it. In this particular model, both the lower retained claim per event and the resulting ruin threshold reduce terminal-ruin probability. The calculation uses the net premium of $2,400, not the gross premium of $3,000, in the retained surplus equation.

Common Traps in Reinsurance Comparisons

Using the wrong variance formula. For a compound Poisson aggregate, use $\lambda E[Y^2]$, not $\lambda\operatorname{Var}(Y)$. The latter omits the claim-count contribution.
Applying the treaty to the aggregate instead of each claim. A per-claim retention gives $Y_j=\min(X_j,M)$ for every claim. It is not generally the same as applying one retention to the annual total.
Ignoring the reinsurance premium. Retained claims fall, but premium income available to the insurer also falls by $\pi_R$. Use net premium in the retained surplus process.
Equating lower VaR with a proportional capital reduction. State whether capital means VaR or VaR less the expected loss, and calculate each quantity from the relevant distribution.
Calling a terminal calculation path ruin. A one-year negative terminal surplus probability is not automatically the probability surplus became negative at any time during the year. Timing assumptions determine which event is being measured.

Key takeaway: Evaluate a treaty by transforming each claim into retained and ceded payments, then compare expected cost, tail capital, net premium, and a clearly defined ruin event. Reinsurance reduces claim payments pathwise under a per-claim limit, but its effect on financial outcomes depends on both the reinsurance premium and the surplus model.

Check Your Understanding

For each problem, use the assumptions stated and keep the gross, retained, and ceded quantities distinct.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$. Each claim is $2,000 with probability 0.75 and $6,000 with probability 0.25. Find the expected aggregate claims and aggregate variance. (Answer: $E[S]=\$9{,}000$; $\operatorname{Var}(S)=\$36{,}000{,}000$.)
A portfolio has $N\sim\operatorname{Poisson}(0.2)$, and each gross claim is $10,000. A per-claim retention is $6,000. Find expected annual ceded claims and the reinsurance premium if the reinsurer charges 20% above expected ceded claims. (Answer: expected ceded claims $=\$800$; reinsurance premium $=\$960$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims per year, each exactly $5,000. A treaty retains $3,000 per claim. Find gross and retained 95% VaR, and capital above expected claims under each basis. (Answer: gross VaR $=\$15{,}000$, gross capital $=\$10{,}000$; retained VaR $=\$9{,}000$, retained capital $=\$6{,}000$.)
An insurer has initial surplus $10,000, annual gross premium income $4,000, and a reinsurance premium of $1,000. Claim count is $\operatorname{Poisson}(0.5)$, each gross claim is $5,000, and the per-claim retention is $2,000. If all cash flows settle at year-end and ruin means negative terminal surplus, find the retained terminal-ruin probability. (Answer: ruin occurs for $N\ge7$, so $P(\text{ruin})=1-e^{-0.5}\sum_{k=0}^{6}0.5^k/k!\approx0.0000$ to four decimals.)

What You'll Learn