Selecting Deductibles for Different Loss Profiles

Why Deductible Selection Depends on the Loss Profile

A deductible does not have the same practical effect for every insured. A business with frequent, modest claims may retain a predictable stream of losses under a high deductible. A business with rare but severe claims may face much greater retained-loss volatility at that same deductible. Choosing between alternatives therefore requires more than comparing the deductible amounts: the decision should consider expected retained cost, retained volatility, and the insurer's share of expected losses.

The calculations below use hypothetical per-claim deductibles without an annual aggregate limit. They assume that each claim is allocated to the policy term and that the insurer reimburses the covered amount above the per-claim deductible. The earlier material on large deductibles establishes the basic gross-to-net allocation; here the emphasis is on comparing alternatives across different loss profiles.

Definition: For a ground-up claim loss $X$ and per-claim deductible $d$, the insured's retained payment per claim is $R_d=\min(X,d)$, and the insurer's reimbursement per claim is $I_d=(X-d)^+=\max(X-d,0)$. Thus $X=R_d+I_d$. For annual claims $X_1,\ldots,X_N$, aggregate retained loss is $R=\sum_{j=1}^{N}\min(X_j,d)$, and aggregate insurer reimbursement is $I=\sum_{j=1}^{N}(X_j-d)^+$.

These expressions describe a deductible applied separately to each claim. They do not describe an annual aggregate deductible, a policy limit, or a self-insured retention with different payment provisions. Those structures require their own allocation rules. The examples assume unlimited coverage above the deductible, no coinsurance, and no coverage disputes.

Measures for Comparing Alternatives

For each candidate deductible, calculate the expected annual retained loss $E[R]$, the standard deviation of annual retained loss $\sqrt{\operatorname{Var}(R)}$, and expected annual insurer reimbursement $E[I]$. Expected loss describes the average over many comparable years; standard deviation gives one measure of how much annual retained loss may vary around that average. Neither measure alone describes the full loss distribution.

A useful decomposition is $E[R]+E[I]=E[\sum_{j=1}^{N}X_j]$. It is a check on the allocation: increasing the deductible shifts expected loss from insurer to insured, but does not change expected ground-up loss under the same frequency and severity assumptions. A deductible may also change premium. To compare hypothetical program costs, a problem must specify how that premium is determined.

Model result: Suppose $N\sim\operatorname{Poisson}(\lambda)$, and claim severities $X_j$ are independent and identically distributed, independent of $N$. For a per-claim retained amount $Z=\min(X,d)$, annual retained loss $R=\sum_{j=1}^{N}Z_j$ has $$ E[R]=\lambda E[Z], \qquad \operatorname{Var}(R)=\lambda E[Z^2]. $$ The second expression uses the second raw moment $E[Z^2]$, not $\operatorname{Var}(Z)$.

The result assumes a Poisson claim count, independent severities with a common distribution, and independence between count and severities. If the count is not Poisson, or if claim amounts and frequency are dependent, the annual variance generally needs a different calculation. A per-claim deductible also makes $Z$ a capped version of each ground-up loss, so the retained amount for each claim must be calculated before applying the annual frequency model.

Why the Compound Poisson Variance Uses a Raw Moment

Condition on the number of claims. Given $N=n$, the annual retained loss is a sum of $n$ independent copies of $Z$. Its conditional mean is $nE[Z]$, and its conditional variance is $n\operatorname{Var}(Z)$. The law of total variance then gives

\begin{aligned} \operatorname{Var}(R) &=E[\operatorname{Var}(R\mid N)] +\operatorname{Var}(E[R\mid N])\\ &=E[N]\operatorname{Var}(Z) +\operatorname{Var}(N)(E[Z])^2\\ &=\lambda\operatorname{Var}(Z)+\lambda(E[Z])^2\\ &=\lambda E[Z^2], \end{aligned}

because a Poisson random variable has both mean and variance $\lambda$, and $E[Z^2]=\operatorname{Var}(Z)+(E[Z])^2$. The second variance term captures uncertainty in the number of claims. Omitting it understates annual volatility whenever the retained mean per claim is positive.

Worked Comparisons

The examples report money in thousands of dollars (denoted $ \$000 $). Standard deviations are also reported in $ \$000 $, rounded to three decimal places. All data are hypothetical, and any premium loading is explicitly stated for the example rather than treated as a universal pricing rule.

Worked Example: Frequent Small Claims Versus Rare Severe Claims

Two hypothetical risks each have expected annual ground-up losses of $ \$20{,}000 $. Risk A has $N_A\sim\operatorname{Poisson}(10)$ and a deterministic $ \$2{,}000 $ loss per claim. Risk B has $N_B\sim\operatorname{Poisson}(2)$ and a deterministic $ \$10{,}000 $ loss per claim. Compare deductibles of $ \$1{,}000 $ and $ \$5{,}000 $. For this comparison only, assume the insurer charges 20% above expected reimbursement, with no other premium components. Which alternative has lower expected retained loss plus premium, and what volatility is retained?

Method and assumptions. Use the compound Poisson formulas. Because each severity is deterministic, the retained amount $Z$ is also deterministic for a given deductible and risk. In each case, $E[R]=\lambda Z$, $\operatorname{Var}(R)=\lambda Z^2$, and expected reimbursement is $\lambda(X-Z)$.

Risk A, deductible $ \$1{,}000 $. In $ \$000 $, $Z=\min(2,1)=1$, so $E[R]=10(1)=10$, and $\operatorname{Var}(R)=10(1^2)=10$. The standard deviation is $\sqrt{10}=3.162$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $10(2-1)=10$ $ \$000 $; the hypothetical premium is $1.20(10)=12$ $ \$000 $. Expected retained loss plus premium is $10+12=22$ $ \$000 $.

Risk A, deductible $ \$5{,}000 $. Each $ \$2{,}000 $ claim is below the deductible, so $Z=2$. Thus $E[R]=10(2)=20$, $\operatorname{Var}(R)=10(2^2)=40$, and standard deviation is $\sqrt{40}=6.325$ $ \$000 $, rounded to three decimal places. Expected reimbursement and premium are both zero. Expected retained loss plus premium is $20$ $ \$000 $.

Risk B, deductible $ \$1{,}000 $. Each $ \$10{,}000 $ claim has $Z=1$, giving $E[R]=2$, $\operatorname{Var}(R)=2$, and standard deviation $\sqrt{2}=1.414$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $2(10-1)=18$ $ \$000 $, so premium is $1.20(18)=21.6$ $ \$000 $. Expected retained loss plus premium is $2+21.6=23.6$ $ \$000 $.

Risk B, deductible $ \$5{,}000 $. Now $Z=5$, so $E[R]=2(5)=10$, $\operatorname{Var}(R)=2(5^2)=50$, and standard deviation is $\sqrt{50}=7.071$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $2(10-5)=10$ $ \$000 $; premium is $12$ $ \$000 $. Expected retained loss plus premium is $10+12=22$ $ \$000 $.

Check and interpretation. For Risk A, expected retained loss plus reimbursement is $10+10=20$ under the lower deductible and $20+0=20$ under the higher one. For Risk B, the corresponding totals are $2+18=20$ and $10+10=20$. The higher deductible lowers the stated expected cost measure for both risks, but increases retained volatility. The frequency-severity profiles explain the different volatility levels: at a $ \$5{,}000 $ deductible, Risk A has standard deviation $6.325$ $ \$000 $, while Risk B has $7.071$ $ \$000 $. A decision-maker must weigh those trade-offs against the insured's risk tolerance. The common mistake this comparison avoids is assuming equal expected ground-up losses imply equal retained-loss volatility.

Worked Example: Deductible Layers for a Mixed Severity Profile

A hypothetical account has $N\sim\operatorname{Poisson}(4)$ claims per year. Each claim is $ \$2{,}000 $ with probability $0.75$, or $ \$8{,}000 $ with probability $0.25$, independently of the other claims and of $N$. Compare deductibles of $ \$3{,}000 $ and $ \$6{,}000 $. Assume a hypothetical premium equal to expected reimbursement plus a 25% loading. Calculate expected retained loss, standard deviation, expected reimbursement, and expected retained loss plus premium.

Method. Work in $ \$000 $. For each deductible, first find the possible retained claim amounts, then calculate their first and second raw moments.

Deductible $3$. Retained claim amounts are $2$ with probability $0.75$, and $3$ with probability $0.25$. Therefore,

E[Z]=0.75(2)+0.25(3)=2.25,\qquad E[Z^2]=0.75(2^2)+0.25(3^2)=3+2.25=5.25.

Annual expected retained loss is $4(2.25)=9$ $ \$000 $. Annual variance is $4(5.25)=21$ $(\$000)^2$, so standard deviation is $\sqrt{21}=4.583$ $ \$000 $, rounded to three decimal places. The insurer pays zero on the $ \$2{,}000 $ claims and $ \$5{,}000 $ on the $ \$8{,}000 $ claims. Expected annual reimbursement is $4[0.75(0)+0.25(5)]=5$ $ \$000 $. Premium is $1.25(5)=6.25$ $ \$000 $, and expected retained loss plus premium is $9+6.25=15.25$ $ \$000 $.

Deductible $6$. Retained claim amounts are $2$ and $6$, with the same respective probabilities. Hence $E[Z]=0.75(2)+0.25(6)=3$, and $E[Z^2]=0.75(4)+0.25(36)=3+9=12$. Annual expected retained loss is $4(3)=12$ $ \$000 $; variance is $4(12)=48$ $(\$000)^2$, giving standard deviation $\sqrt{48}=6.928$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $4[0.75(0)+0.25(2)]=2$ $ \$000 $; premium is $1.25(2)=2.5$ $ \$000 $. Expected retained loss plus premium is $12+2.5=14.5$ $ \$000 $.

Check and interpretation. Gross expected annual loss is $4[0.75(2)+0.25(8)]=4(3.5)=14$ $ \$000 $. The low-deductible allocation gives $9+5=14$, and the high-deductible allocation gives $12+2=14$. The higher deductible reduces the specified expected cost measure from $15.25$ to $14.5$ $ \$000 $, but increases the standard deviation of retained loss from $4.583$ to $6.928$ $ \$000 $. The trap is to treat the deductible as if it reduced the ground-up loss; it reallocates expected loss between the insured and insurer.

Worked Example: Selecting a Deductible Under a Volatility Limit

A hypothetical manufacturer has a Poisson claim count with mean $8$ per year and a deterministic claim severity of $ \$3{,}000 $. It is considering deductibles of $ \$1{,}000 $, $ \$2{,}000 $, and $ \$6{,}000 $. The insurer's hypothetical premium is expected reimbursement plus a 15% loading. The manufacturer will accept only alternatives with an annual retained-loss standard deviation no greater than $ \$6{,}000 $. Compare expected retained loss plus premium and select the lowest-cost alternative that satisfies the limit.

Method. Work in $ \$000 $, apply the compound Poisson formulas to the per-claim retention, then test the standard-deviation constraint. The full cost measure is expected retained loss plus the stated premium; the constraint applies to retained loss volatility, not to that cost measure.

Deductible $1$. Per-claim retention is $Z=1$. Thus $E[R]=8(1)=8$ $ \$000 $, variance is $8(1^2)=8$ $(\$000)^2$, and standard deviation is $\sqrt{8}=2.828$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $8(3-1)=16$ $ \$000 $, giving premium $1.15(16)=18.4$ $ \$000 $. Expected retained loss plus premium is $8+18.4=26.4$ $ \$000 $.

Deductible $2$. Per-claim retention is $Z=2$. Expected retained loss is $8(2)=16$ $ \$000 $, variance is $8(2^2)=32$ $(\$000)^2$, and standard deviation is $\sqrt{32}=5.657$ $ \$000 $, rounded to three decimal places. Expected reimbursement is $8(3-2)=8$ $ \$000 $; premium is $1.15(8)=9.2$ $ \$000 $. Expected retained loss plus premium is $16+9.2=25.2$ $ \$000 $.

Deductible $6$. Since the claim is only $ \$3{,}000 $, each claim is fully retained: $Z=3$. Expected retained loss is $8(3)=24$ $ \$000 $, variance is $8(3^2)=72$ $(\$000)^2$, and standard deviation is $\sqrt{72}=8.485$ $ \$000 $, rounded to three decimal places. Expected reimbursement and premium are zero, so expected retained loss plus premium is $24$ $ \$000 $.

Selection and check. The first two alternatives satisfy the $ \$6{,}000 $ standard-deviation limit; the $ \$6{,}000 $ deductible does not. Of the qualifying alternatives, the $ \$2{,}000 $ deductible has the lower expected retained loss plus premium, $ \$25{,}200 $ rather than $ \$26{,}400 $. As an allocation check, expected retained loss plus expected reimbursement equals $8(3)=24$ $ \$000 $ for each option. The common mistake is to select the lowest expected cost without first enforcing the stated risk constraint.

A Practical Selection Sequence

A disciplined comparison separates the loss allocation from the decision criterion. First calculate the expected loss transfer implied by each deductible. Then quantify retained volatility under the stated frequency-severity model. Finally, apply the decision rule supplied by the insured or the problem, such as a maximum retained standard deviation, a budget limit, or a cost objective subject to a risk constraint.

Specify the contract and model.
Confirm that the deductible is per claim, identify any applicable limits or aggregate features, and state the frequency-severity assumptions. Do not use a per-claim formula for an annual aggregate provision.

Calculate the per-claim allocation.
For each possible severity, calculate $Z=\min(X,d)$ and $I_d=(X-d)^+$. Keep the deductible applied to each claim rather than subtracting it once from annual losses.

Aggregate expected loss and volatility.
Under the independent Poisson model, use $E[R]=\lambda E[Z]$ and $\operatorname{Var}(R)=\lambda E[Z^2]$. Calculate insurer expected reimbursement on the same basis.

Apply the stated decision rule.
Include a premium only if its basis is given or explicitly assumed. Compare the resulting cost measure with retained-risk tolerances; a lower expected cost does not automatically make an alternative acceptable.

Check the allocation.
Verify that expected retained loss plus expected insurer reimbursement equals expected ground-up annual loss. Then check units, standard-deviation rounding, and whether every candidate satisfies the stated constraints.

Common Exam Traps and Practical Meaning

Deductible selection connects the structure of the insured's losses with its capacity to absorb risk. Two accounts with the same expected ground-up loss can have different annual volatility because one has frequent smaller claims and the other has fewer, larger claims. The deductible's expected transfer is therefore not enough to describe its financial effect. The same expected reimbursement can also accompany different distributional shapes, so a standard deviation is a comparison measure rather than a complete description of tail risk.

Using $\lambda\operatorname{Var}(Z)$ as annual variance. This omits claim-count uncertainty. Under a Poisson count, the correct expression is $\lambda E[Z^2]=\lambda\{\operatorname{Var}(Z)+(E[Z])^2\}$.
Applying the deductible annually instead of per claim. For a per-claim deductible, calculate $\min(X_j,d)$ separately for each claim and then sum. Deducting $d$ once from annual aggregate loss changes the contract.
Comparing expected retained loss alone. A higher deductible usually increases expected retained loss and reduces insurer reimbursement; a premium reduction may offset that change. Include only the premium basis specified in the problem.
Mixing dollars and thousands of dollars. If severities are entered in $ \$000 $, variances are in $(\$000)^2$, and standard deviations return to $ \$000 $. State the scale before reporting results.
Calling a lower mean a safer choice. Expected cost and volatility are distinct criteria. A deductible with lower expected cost can violate the insured's retained-risk limit.

The calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. In an applied decision, the selected deductible also depends on contract terms, premium provisions, and the insured's ability and willingness to retain volatility; those considerations should be stated rather than hidden inside a single expected-loss number.

Check Your Understanding

For each problem, use the stated assumptions and report dollar amounts in thousands unless otherwise specified. Round standard deviations to three decimal places.

A risk has $N\sim\operatorname{Poisson}(5)$ and deterministic claim severity $ \$4{,}000 $. Under a $ \$1{,}500 $ per-claim deductible, find expected annual retained loss, annual retained-loss standard deviation, and expected annual reimbursement. (Answer: expected retained loss $ \$7.5$ thousand; standard deviation $ \$3.354$ thousand; expected reimbursement $ \$12.5$ thousand.)
A risk has $N\sim\operatorname{Poisson}(3)$; severity is $ \$1{,}000 $ with probability $0.6$ and $ \$5{,}000 $ with probability $0.4$. Under a $ \$2{,}000 $ deductible, find annual expected retained loss, annual retained-loss variance, and expected annual reimbursement. (Answer: expected retained loss $ \$4.2$ thousand; variance $6.6$ $(\$000)^2$; expected reimbursement $ \$3.6$ thousand.)
A hypothetical risk has $N\sim\operatorname{Poisson}(4)$, with claim severity $ \$3{,}000 $ or $ \$9{,}000 $, each with probability $0.5$. Compare deductibles of $ \$3{,}000 $ and $ \$6{,}000 $. Assume premium is expected reimbursement plus a 10% loading. Which has lower expected retained loss plus premium, and what is the standard deviation under the $ \$6{,}000 $ deductible? (Answer: the $ \$6{,}000 $ deductible has lower expected retained loss plus premium, $ \$24.6$ thousand versus $ \$25.2$ thousand; its retained-loss standard deviation is $ \$9.487$ thousand.)
A hypothetical account has $N\sim\operatorname{Poisson}(2)$ and deterministic claim severity $ \$15{,}000 $. Under a $ \$5{,}000 $ deductible, what is expected annual insurer reimbursement, and what percentage of expected ground-up loss does it represent? (Answer: $ \$20$ thousand, or $66.67\%$ of expected ground-up loss.)
A risk has $N\sim\operatorname{Poisson}(6)$; severity is $ \$2{,}000 $ or $ \$6{,}000 $, each with probability $0.5$. Under a $ \$4{,}000 $ deductible, calculate expected annual retained loss, retained-loss standard deviation, and expected annual reimbursement. (Answer: expected retained loss $ \$18$ thousand; standard deviation $ \$7.746$ thousand; expected reimbursement $ \$6$ thousand.)

What You'll Learn