Measuring Risk Transfer Under Large Deductibles

Risk Transfer Is About More Than Expected Loss

A large deductible changes who pays each claim, but comparing only expected payments gives an incomplete picture. Two deductible options can be compared by measuring both the insured’s and insurer’s expected annual losses and the variability of those annual totals. The variability comparison matters because loss amounts are uncertain, and the annual claim count is uncertain as well.

The calculations here use an uncapped per-claim deductible: the insured retains the first $d$ dollars of each loss, and the insurer pays the amount above $d$. There is no policy limit, aggregate deductible, reimbursement cap, expense adjustment, or discounting. Unless stated otherwise, money is measured in thousands of dollars for one policy year.

Definition: For a nonnegative claim loss $X$ and a per-claim deductible $d\ge 0$, the insured’s retained payment is $R=\min(X,d)$, and the insurer’s payment is $I=(X-d)^+=\max(X-d,0)$. Thus $X=R+I$. When $X>d>0$, both $R$ and $I$ are positive; when $X\le d$, $I=0$.

For a portfolio with $N$ claims, define the annual retained and insurer-paid totals as $T_R=\sum_{j=1}^{N}R_j$ and $T_I=\sum_{j=1}^{N}I_j$. The annual gross loss is $T_X=T_R+T_I$. The same claim count and claim severities determine both totals, so the two parties’ annual losses are generally dependent.

Compound Poisson Results for Annual Losses

Assume $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda$ is the expected number of claims in the policy year. Given $N$, the claim severities $X_1,\ldots,X_N$ are independent and identically distributed, independent of $N$, and have the same distribution as $X$. For any payment function $g(X)$, let $T_g=\sum_{j=1}^{N}g(X_j)$. The expectation formula requires $E[|g(X)|]<\infty$; the variance formula requires $E[g(X)^2]<\infty$. For two payment functions $g$ and $h$, the covariance formula below applies when their means and cross-moment are finite; finite second moments for both functions are a sufficient condition.

Theorem: Under the stated compound Poisson assumptions, $$ E[T_g]=\lambda E[g(X)],\qquad \operatorname{Var}(T_g)=\lambda E[g(X)^2],\qquad \operatorname{Cov}(T_g,T_h)=\lambda E[g(X)h(X)]. $$ For retained and insurer-paid losses, use $g(X)=R$ and $h(X)=I$.

The variance formula uses a second moment, not just the variance of a single claim payment. This distinction accounts for random claim frequency. The covariance formula also reflects that an individual claim can produce payments to both parties and that the annual claim count affects both totals.

Why the Annual Variance Formula Holds

Condition on the claim count $N=n$. The sum $T_g$ then has conditional mean $nE[g(X)]$ and conditional variance $n\operatorname{Var}(g(X))$, by independence of claim severities. Applying the laws of total expectation and total variance, and using $E[N]=\operatorname{Var}(N)=\lambda$, gives

E[T_g]=E[N]E[g(X)]=\lambda E[g(X)],

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

For two payment functions, conditional covariance given $N$ is $N\operatorname{Cov}(g(X),h(X))$, while the covariance of their conditional means is $\lambda E[g(X)]E[h(X)]$. Adding these terms gives

\operatorname{Cov}(T_g,T_h) =\lambda\operatorname{Cov}(g(X),h(X)) +\lambda E[g(X)]E[h(X)] =\lambda E[g(X)h(X)].

Since $R+I=X$, the total-loss variance provides an important check:

\operatorname{Var}(T_X) =\operatorname{Var}(T_R)+\operatorname{Var}(T_I) +2\operatorname{Cov}(T_R,T_I) =\lambda E[X^2].

Worked Comparisons Under a Large Deductible

Worked Example: Comparing Two Deductibles for a Discrete Severity Model

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ claims per year. Claim severity $X$, in thousands of dollars, equals 2 with probability 0.5, 8 with probability 0.3, and 20 with probability 0.2. Compare deductibles $d=5$ and $d=12$. Calculate each party’s expected annual loss, annual variance, and covariance, then check the gross-loss variance.

Method and severity moments. First calculate the severity moments that are independent of the deductible:

$$ E[X]=0.5(2)+0.3(8)+0.2(20)=1+2.4+4=7.4, $$

$$ E[X^2]=0.5(2^2)+0.3(8^2)+0.2(20^2) =2+19.2+80=101.2. $$

For $d=5$, the pairs $(R,I)$ at losses $2,8,20$ are $(2,0),(5,3),(5,15)$. Therefore

E[R]=0.5(2)+0.3(5)+0.2(5)=3.5,\quad E[I]=0.5(0)+0.3(3)+0.2(15)=3.9,

E[R^2]=0.5(4)+0.3(25)+0.2(25)=14.5,\quad E[I^2]=0.5(0)+0.3(9)+0.2(225)=47.7,

$$ E[RI]=0.5(0)+0.3(15)+0.2(75)=19.5. $$

The expected annual retained and insurer-paid losses are $4(3.5)=14.0$ and $4(3.9)=15.6$ thousand dollars. Their variances are $4(14.5)=58.0$ and $4(47.7)=190.8$ squared thousands of dollars, and their covariance is $4(19.5)=78.0$ squared thousands of dollars. The expected amounts sum to $29.6=4(7.4)$.

For $d=12$, the pairs are $(2,0),(8,0),(12,8)$. Thus

E[R]=0.5(2)+0.3(8)+0.2(12)=5.8,\quad E[I]=0.2(8)=1.6,

E[R^2]=0.5(4)+0.3(64)+0.2(144)=50,\quad E[I^2]=0.2(64)=12.8,

$$ E[RI]=0.2(12)(8)=19.2. $$

The expected annual totals are $4(5.8)=23.2$ and $4(1.6)=6.4$ thousand dollars. The annual variances are $4(50)=200$ and $4(12.8)=51.2$ squared thousands of dollars; the covariance is $4(19.2)=76.8$ squared thousands of dollars.

Check and interpretation. For $d=5$, $58.0+190.8+2(78.0)=404.8$. For $d=12$, $200+51.2+2(76.8)=404.8$. Both equal $4E[X^2]=4(101.2)=404.8$. Raising the deductible shifts expected loss to the insured and increases the insured’s annual variance, while reducing the insurer’s variance. The gross-loss variance does not change.

Common mistake avoided: using the per-claim variance instead of the second moment for annual variance. At $d=5$, $\operatorname{Var}(R)=14.5-(3.5)^2=2.25$, so multiplying by 4 gives 9.0, not the correct annual variance of 58.0. The omitted amount arises from the variability in claim count.

Worked Example: Measuring Variability When the Insurer Pays Less Often

For a hypothetical policy year, let $N\sim\operatorname{Poisson}(10)$. Each claim is 1, 4, or 10 thousand dollars with probabilities 0.4, 0.4, and 0.2, respectively. The per-claim deductible is $d=3$ thousand dollars. Find expected annual payments and annual variances for each party, along with their covariance.

Method. The retained and insurer-paid amounts for the three severities are $(1,0),(3,1),(3,7)$. The claim-level expectations are

E[R]=0.4(1)+0.4(3)+0.2(3)=2.2,\qquad E[I]=0.4(0)+0.4(1)+0.2(7)=1.8.

The second moments and cross-moment are

$$ E[R^2]=0.4(1)+0.4(9)+0.2(9)=5.8, $$

E[I^2]=0.4(0)+0.4(1)+0.2(49)=10.2,\qquad E[RI]=0.4(0)+0.4(3)+0.2(21)=5.4.

Multiplying each claim-level expectation by $\lambda=10$, the expected annual retained loss is $22.0$ thousand dollars and the expected annual insurer payment is $18.0$ thousand dollars. The variances are $10(5.8)=58.0$ and $10(10.2)=102.0$ squared thousands of dollars. Their covariance is $10(5.4)=54.0$ squared thousands of dollars.

Check and interpretation. The expected amounts sum to $40.0$ thousand dollars. Directly, $E[X]=0.4(1)+0.4(4)+0.2(10)=4$, so the gross expectation is $10(4)=40.0$. Also, $E[X^2]=0.4(1)+0.4(16)+0.2(100)=26.8$, giving gross annual variance $10(26.8)=268.0$. The components check: $58+102+2(54)=268$. The insurer pays only above the deductible, but its annual payments remain positively associated with the insured’s because both are affected by the same claim count and large claims.

Common mistake avoided: treating the insured’s and insurer’s totals as independent and adding only their variances. That would give $58+102=160$, omitting $2(54)=108$; the correct variance of gross loss is 268 squared thousands of dollars.

Worked Example: Comparing Variance Reallocation Across Deductible Levels

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$. Severity $X$, in thousands of dollars, is 3, 6, or 15 with probabilities 0.5, 0.25, and 0.25. Compare $d=4$ with $d=10$, and identify how the parties’ expected losses and variances change.

Method. First calculate the gross severity moments:

E[X]=0.5(3)+0.25(6)+0.25(15)=6.75,\qquad E[X^2]=0.5(9)+0.25(36)+0.25(225)=69.75.

At $d=4$, the payment pairs are $(3,0),(4,2),(4,11)$. Therefore

E[R]=0.5(3)+0.25(4)+0.25(4)=3.5,\quad E[I]=0.25(2)+0.25(11)=3.25,

$$ E[R^2]=0.5(9)+0.25(16)+0.25(16)=12.5, $$

E[I^2]=0.25(4)+0.25(121)=31.25,\quad E[RI]=0.25(8)+0.25(44)=13.

Annual expected losses are $2(3.5)=7.0$ and $2(3.25)=6.5$ thousand dollars. Annual variances are $2(12.5)=25.0$ and $2(31.25)=62.5$ squared thousands; covariance is $2(13)=26.0$ squared thousands.

At $d=10$, the payment pairs are $(3,0),(6,0),(10,5)$. Hence

E[R]=0.5(3)+0.25(6)+0.25(10)=5.5,\quad E[I]=0.25(5)=1.25,

E[R^2]=0.5(9)+0.25(36)+0.25(100)=38.5,\quad E[I^2]=0.25(25)=6.25,\quad E[RI]=0.25(50)=12.5.

Annual expected losses are $11.0$ and $2.5$ thousand dollars; annual variances are $77.0$ and $12.5$ squared thousands; and covariance is $25.0$ squared thousands.

Check and interpretation. Expected annual losses sum to $13.5=2(6.75)$ thousand dollars under either deductible. For $d=4$, $25+62.5+2(26)=139.5$; for $d=10$, $77+12.5+2(25)=139.5$. Both match $2E[X^2]=139.5$. The higher deductible shifts both expected loss and annual variance toward the insured, while the insurer’s variance decreases.

Common mistake avoided: calling each party’s variance a share of gross variance and expecting those percentages to sum to 100%. Variances are not additive when the totals have positive covariance; the covariance term must be included.

Interpreting Risk Transfer Correctly

Expected-loss allocation is additive: $E[T_R]+E[T_I]=E[T_X]$. The insurer’s share of expected loss can therefore be reported as $E[T_I]/E[T_X]$, provided $E[T_X]>0$. Variance allocation is different. Since

\operatorname{Var}(T_X)=\operatorname{Var}(T_R)+\operatorname{Var}(T_I) +2\operatorname{Cov}(T_R,T_I),

the two standalone variances do not, in general, sum to total variance. A ratio such as $\operatorname{Var}(T_I)/\operatorname{Var}(T_X)$ can be reported as a comparison, but it is not an additive “percentage of total variance.” State the covariance or make clear what ratio is being used.

For the uncapped per-claim deductible, $R$ and $I$ are nondecreasing functions of $X$, so their claim-level covariance is nonnegative. The compound Poisson annual covariance $\lambda E[RI]$ is also nonnegative. If $d=0$, then $R=0$; if all possible losses are at or below $d$, then $I=0$. Between these extremes, a claim above the deductible can generate positive payments to both parties.

Key takeaway: To describe deductible risk transfer, calculate expected annual retained and insurer-paid losses, their annual variances, and their covariance. Under a compound Poisson model, use $\lambda E[R]$, $\lambda E[I]$, $\lambda E[R^2]$, $\lambda E[I^2]$, and $\lambda E[RI]$, respectively. The results are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, assume claim severities are independent and identically distributed and independent of the stated Poisson claim count. Use thousands of dollars where specified.

A portfolio has $N\sim\operatorname{Poisson}(5)$. Claim severity is 1 or 5 thousand dollars, with probabilities 0.6 and 0.4. With a deductible of 2 thousand dollars, calculate expected annual retained and insurer-paid losses. (Answer: 7.0 and 6.0 thousand dollars, respectively.)
Let $N\sim\operatorname{Poisson}(3)$, and let severity be 2 or 10 thousand dollars with probabilities 0.75 and 0.25. At $d=4$, calculate both annual variances and the covariance. (Answer: retained variance $21.0$, insurer variance $27.0$, and covariance $18.0$, all in squared thousands of dollars.)
A portfolio has $N\sim\operatorname{Poisson}(2)$ and severities 1, 7, or 13 thousand dollars with probabilities 0.5, 0.3, and 0.2. Compare $d=3$ with $d=8$: give the annual expected losses for the insured and insurer under each deductible, and the insured’s annual variance. (Answer: at $d=3$, expected losses are 4.0 and 6.4 thousand dollars and retained variance is 10.0 squared thousands; at $d=8$, they are 8.4 and 2.0 thousand dollars and retained variance is 56.0 squared thousands.)
For a compound Poisson portfolio with $\lambda=4$, suppose $E[R^2]=9$, $E[I^2]=16$, and $E[RI]=5$, with all amounts in thousands of dollars. Calculate the retained and insurer annual variances, their covariance, and the gross annual variance. (Answer: variances $36$ and $64$, covariance $20$, and gross variance $140$, all in squared thousands of dollars.)

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