Why Retention Selection Is a Trade-Off
A higher retention generally leaves more loss with the insurer and reduces the amount ceded to a reinsurer. That can lower the reinsurance premium, but it can also increase the insurer’s exposure to volatility and severe losses. A lower retention generally transfers more loss, at a price. Retention selection therefore requires more than comparing premiums: the actuary should also examine expected retained loss, earnings variability, and the amount of loss that can remain with the insurer.
This tutorial compares per-occurrence retentions under explicit loss and pricing assumptions. The calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. They do not determine a universally best retention: the appropriate choice depends on the insurer’s objectives, constraints, and view of the cost of risk.
For candidate retentions, compare at least three different quantities. The expected retained loss \(E[Z_r]\) is the average loss left with the insurer. The variance \(\operatorname{Var}(Z_r)\) measures dispersion around that average, and its square root is the standard deviation. Severe-loss measures, such as the maximum retained loss or the probability that retained loss exceeds a chosen threshold, describe aspects of the distribution that the mean alone cannot show.
Expected reinsurance cost is a separate component. In the examples, the quoted reinsurance premium is a stated multiple of expected ceded loss. Actual prices may also reflect expenses, uncertainty, market conditions, and contract features. A premium is not itself a measure of how much volatility the contract removes.
Calculating Retained Loss and Its Variability
For a discrete loss distribution, list the possible gross losses, apply \(Z_r=\min(X,r)\) to each one, and retain the original probabilities. Then calculate the first and second moments. The variance is the second moment minus the square of the mean:
where \(z_j=\min(x_j,r)\), and \(p_j\) is the probability of gross loss \(x_j\). The second moment is \(E[Z_r^2]\), not the square of \(E[Z_r]\). Keeping these quantities separate is essential when calculating variance.
Worked Example: Comparing Two Retentions for One Occurrence
A hypothetical insurer models one occurrence loss as \(X=0\) with probability 0.50, \(X=60\) with probability 0.30, and \(X=180\) with probability 0.20. Compare retentions \(r=60\) and \(r=120\). The reinsurer charges a premium equal to 1.25 times expected ceded loss. Ignore all other expenses. Calculate retained-loss means and variances, expected total loss-plus-reinsurance-premium cost, and the probability that retained loss exceeds 100.
Method and assumption. Apply the per-occurrence unlimited-cover model separately at each retention. Treat the reinsurance premium as deterministic, so it changes expected cost but not retained-loss variance.
At \(r=60\), the retained losses corresponding to gross losses \(0,60,180\) are \(0,60,60\). Thus \(E[Z_{60}]=0(0.50)+60(0.30)+60(0.20)=18+12=30\). The second moment is \(E[Z_{60}^2]=0^2(0.50)+60^2(0.30)+60^2(0.20)=1{,}080+720=1{,}800\). Therefore \(\operatorname{Var}(Z_{60})=1{,}800-30^2=900\).
At \(r=120\), retained losses are \(0,60,120\). The mean is \(E[Z_{120}]=0(0.50)+60(0.30)+120(0.20)=18+24=42\). The second moment is \(E[Z_{120}^2]=60^2(0.30)+120^2(0.20)=1{,}080+2{,}880=3{,}960\). Hence \(\operatorname{Var}(Z_{120})=3{,}960-42^2=3{,}960-1{,}764=2{,}196\).
For \(r=60\), ceded loss is 120 only when \(X=180\), so \(E[C_{60}]=120(0.20)=24\). The reinsurance premium is \(1.25(24)=30\), and expected retained loss plus premium is \(30+30=60\). For \(r=120\), expected ceded loss is \(60(0.20)=12\); the premium is \(1.25(12)=15\), giving expected cost \(42+15=57\).
At \(r=60\), retained loss cannot exceed 60, so \(P(Z_{60}>100)=0\). At \(r=120\), retained loss exceeds 100 only when \(X=180\), so \(P(Z_{120}>100)=0.20\). The higher retention has lower expected cost in this pricing model, but greater variance and a positive chance of retaining more than 100. The calculations check against \(X=Z_r+C_r\) in every outcome, and the expected total loss-plus-premium costs are 60 and 57.
Common mistake avoided: Do not subtract the reinsurance premium from retained loss or include a deterministic premium in the retained-loss variance. The premium is part of expected cost; it is not a random claim payment in this model.
Why the Mean Formula Uses the Survival Function
For a nonnegative loss, the expected value can be obtained by adding the probabilities that the variable exceeds each possible level. For a continuous loss distribution, the corresponding tail-integral identity gives a useful way to understand how a retention affects expected retained loss. Define \(S_X(x)=P(X>x)\). Since \(Z_r=\min(X,r)\), it exceeds a level \(x\) exactly when \(X>x\), for \(0\leq x<r\); it cannot exceed \(r\).
The first equality follows by writing a nonnegative value \(z\) as the area under the indicator \(1_{\{z>x\}}\) over \(x\geq0\), then taking expectations. The second follows from \(P(Z_r>x)=P(X>x)\) below the retention and zero above it. For distributions with point masses, the analogous tail-sum identity or direct probability calculation should be used with consistent endpoint conventions.
This identity explains why expected retained loss increases as the retention increases: expanding the upper integration limit adds nonnegative area. It does not imply that variance must increase at every step for every loss distribution. Variance depends on both the first and second moments, so compare it directly for the candidate retentions.
Worked Example: Correctly Calculating Variance Across Three Retentions
A hypothetical loss has values 0, 30, and 120 with probabilities 0.60, 0.25, and 0.15, respectively. Calculate the retained-loss mean and variance at retentions 20, 40, and 100. Use the results to describe the volatility trade-off.
Method and assumption. Apply the retention to each possible loss, then calculate \(E[Z_r]\) and \(E[Z_r^2]\) separately. The loss distribution is unchanged across the three comparisons.
At \(r=20\), retained losses are \(0,20,20\). The mean is \(20(0.25)+20(0.15)=5+3=8\). The second moment is \(20^2(0.25)+20^2(0.15)=100+60=160\). Thus the variance is \(160-8^2=96\).
At \(r=40\), retained losses are \(0,30,40\). The mean is \(30(0.25)+40(0.15)=7.5+6=13.5\). The second moment is \(30^2(0.25)+40^2(0.15)=225+240=465\). The variance is \(465-13.5^2=465-182.25=282.75\), or 282.8 to one decimal place.
At \(r=100\), retained losses are \(0,30,100\). The mean is \(30(0.25)+100(0.15)=7.5+15=22.5\). The second moment is \(30^2(0.25)+100^2(0.15)=225+1{,}500=1{,}725\). The variance is \(1{,}725-22.5^2=1{,}725-506.25=1{,}218.75\).
The means rise from 8 to 13.5 to 22.5, as expected when more loss is retained. For this distribution, variance also rises, from 96 to 282.75 to 1,218.75. Each variance is nonnegative, and each second moment minus squared mean gives the stated result.
Common mistake avoided: At retention 40, the second moment is \(30^2(0.25)+40^2(0.15)=465\), not 296.25. Subtracting the squared mean gives \(465-182.25=282.75\). Confusing the probability-weighted second moment with another quantity produces a materially wrong variance.
From Retained Loss to Earnings Protection
A retention affects earnings through the random retained claims and, in a pricing comparison, through the reinsurance premium. Consider a simplified annual model with a fixed number \(n\) of independent, identically distributed occurrences. Let \(R\) be deterministic premium income before losses, and let \(Q_r\) be the deterministic reinsurance premium for the selected retention. Annual earnings in this simplified model are \(E_r=R-Q_r-\sum_{k=1}^n Z_{r,k}\). Here \(Z_{r,k}\) is retained loss on occurrence \(k\). This model ignores expenses, investment income, taxes, and changes in premium income.
The mean follows by taking expectations term by term. For the variance, \(R\) and \(Q_r\) are constants and therefore contribute no variance; independence makes the variance of the sum equal to the sum of the individual variances. If occurrence losses are dependent, or if annual claim count is random, this variance calculation does not apply as written. Those cases require a model that reflects the dependence or frequency uncertainty.
Worked Example: Annual Earnings and a Severe-Loss Threshold
An insurer expects exactly two independent occurrences during a year. Each occurrence loss is 0 with probability 0.50, 80 with probability 0.30, and 200 with probability 0.20. Compare retentions 80 and 160. The reinsurance premium is 1.20 times expected ceded loss per occurrence, paid for each of the two occurrences. Annual premium income is a fixed 300. Calculate expected annual earnings, earnings standard deviation, and the probability that annual retained claims exceed 200.
Method and assumption. First calculate per-occurrence retained moments and ceded expectations. Then use independence and the fixed two-claim count. The premium income and reinsurance premium are deterministic.
At \(r=80\), retained losses are \(0,80,80\). Per occurrence, the mean is \(80(0.30)+80(0.20)=40\). The second moment is \(80^2(0.30)+80^2(0.20)=1{,}920+1{,}280=3{,}200\), so variance is \(3{,}200-40^2=1{,}600\). Ceded loss is 120 when gross loss is 200, so expected ceded loss is \(120(0.20)=24\), and the per-occurrence premium is \(1.20(24)=28.8\). For two occurrences, expected retained claims are \(2(40)=80\), total reinsurance premium is \(2(28.8)=57.6\), and expected earnings are \(300-80-57.6=162.4\). Earnings variance is \(2(1{,}600)=3{,}200\), so standard deviation is \(\sqrt{3{,}200}=56.57\).
At \(r=160\), retained losses are \(0,80,160\). The per-occurrence mean is \(80(0.30)+160(0.20)=24+32=56\). The second moment is \(80^2(0.30)+160^2(0.20)=1{,}920+5{,}120=7{,}040\), so variance is \(7{,}040-56^2=7{,}040-3{,}136=3{,}904\). Expected ceded loss is \(40(0.20)=8\), and the per-occurrence reinsurance premium is \(1.20(8)=9.6\). Annual expected retained claims are \(112\), total reinsurance premium is \(19.2\), and expected earnings are \(300-112-19.2=168.8\). Earnings variance is \(2(3{,}904)=7{,}808\), giving standard deviation \(\sqrt{7{,}808}=88.36\).
At \(r=80\), aggregate retained claims cannot exceed \(2(80)=160\), so the probability of exceeding 200 is zero. At \(r=160\), an aggregate above 200 occurs if one occurrence retains 160 and the other retains 80, or if both retain 160. By independence, the probability is \(2(0.20)(0.30)+(0.20)^2=0.12+0.04=0.16\).
The higher retention produces greater expected earnings in this simplified model because its lower reinsurance premium more than offsets its higher expected retained claims. It also raises earnings standard deviation from 56.57 to 88.36 and changes the probability of annual retained claims exceeding 200 from zero to 0.16. These results check against the two-occurrence maximum retained totals of 160 and 320. The comparison illustrates why expected earnings alone does not describe earnings protection.
Common mistake avoided: Do not multiply a one-occurrence variance by \(2^2\). For two independent occurrences, variances add, so the annual variance is \(2\operatorname{Var}(Z_r)\), not \(4\operatorname{Var}(Z_r)\).
Practical Comparison and Common Traps
A disciplined retention comparison separates the components before combining them. For a stated reinsurance pricing rule, calculate expected ceded loss and the resulting premium, then add that premium to expected retained loss if the objective is expected loss-plus-premium cost. Separately report a volatility measure and at least one severe-loss measure relevant to the decision.
| Measure | What it describes | Interpretation caution |
|---|---|---|
| \(E[Z_r]\) | Average loss retained per occurrence | Does not describe the spread or tail by itself |
| \(\operatorname{Var}(Z_r)\) or its square root | Dispersion of retained loss | Variance is in squared currency units; standard deviation is in currency units |
| \(E[C_r]\) and quoted premium | Expected ceded loss and its stated price | Expected ceded loss need not equal the reinsurance premium |
| \(P(Z_r>a)\) or the maximum \(Z_r\) | Exposure above a selected threshold or the per-occurrence cap | State whether the threshold is per occurrence or annual aggregate |
- Applying the retention in the wrong direction. The insurer retains \(\min(X,r)\); the reinsurer pays \((X-r)^+\). The retention is not a deductible subtracted from every loss regardless of size.
- Using the mean in place of the second moment. Variance requires \(E[Z_r^2]-(E[Z_r])^2\). Squaring each loss before weighting is different from squaring the mean after weighting.
- Confusing per-occurrence and annual measures. An occurrence cap does not by itself give the maximum annual retained total when several occurrences can happen.
- Treating a lower expected cost as complete protection. A lower average cost can accompany higher variance or a greater probability of severe retained outcomes.
Check Your Understanding
For each problem, apply the stated per-occurrence retention model. Treat occurrences as independent where specified.
- A loss is 10 with probability 0.40, 50 with probability 0.40, and 100 with probability 0.20. At retention 30, calculate the retained-loss mean and variance. (Answer: \(E[Z_{30}]=22\); \(\operatorname{Var}(Z_{30})=96\).)
- A loss is 0 with probability 0.70, 40 with probability 0.20, and 90 with probability 0.10. At retention 50, calculate the retained-loss mean and variance. (Answer: \(E[Z_{50}]=13\); \(\operatorname{Var}(Z_{50})=401\).)
- There are exactly three independent occurrences, each with retained-loss mean 12 and variance 25 at the selected retention. Calculate the annual retained-loss mean, variance, and standard deviation. (Answer: mean \(=36\), variance \(=75\), standard deviation \(=8.6603\).)
- Two independent losses each have values 0, 70, and 150 with probabilities 0.50, 0.30, and 0.20. At retention 100, find the probability that aggregate retained loss exceeds 150. (Answer: \(0.1600\).)
- A loss is 0 with probability 0.60, 50 with probability 0.25, and 200 with probability 0.15. Compare expected retained loss plus reinsurance premium at retentions 50 and 100 if the premium is 1.10 times expected ceded loss. (Answer: at retention 50, expected retained loss \(=20\), premium \(=24.75\), total \(=44.75\); at retention 100, expected retained loss \(=27.5\), premium \(=16.5\), total \(=44.00\).)