How Deductibles Change Retained Loss Variance

Why Variance Depends on Both the Deductible and Claim Count

A per-claim deductible changes how each loss is divided between the insured and the insurer. It changes not only the expected amount paid by each party, but also the variability of those payments. To calculate annual variance correctly, distinguish two questions: how a deductible transforms an individual claim, and how the number of claims is generated.

This tutorial considers a fixed per-claim deductible $d$, no policy limit, and claim severity $X\geq 0$. It treats each claim’s insured-retained amount and insurer payment as separate random variables. Results for a fixed claim count and a Poisson claim count differ, so the count assumption must be stated before calculating annual variance.

Definition: For a claim with loss $X$ and a per-claim deductible $d$, the insured-retained amount is $R=\min(X,d)$, and the insurer payment is $I=(X-d)_+=\max(X-d,0)$. Thus $R+I=X$. These definitions assume no limit above the deductible and no other coverage features. They describe the allocation of loss, not any separate reimbursement, fee, or expense arrangement.

For example, if $X=7$ thousand dollars and $d=4$ thousand dollars, then $R=4$, $I=3$, and $R+I=7$, with amounts in thousands. If $X=2$ thousand dollars, then $R=2$ and $I=0$. Because both amounts are determined by the same claim severity, their variances should not be treated as independent components.

From Claim-Level Variance to Annual Variance

Let $Z_j$ be the amount of one loss component on claim $j$, where $Z_j$ could be $R_j$ or $I_j$. Assume claim severities are independent and identically distributed, with finite second moments. For a fixed number $n$ of claims, the annual component total is $S_n=\sum_{j=1}^{n}Z_j$, and independence gives

\operatorname{Var}(S_n)=n\operatorname{Var}(Z).

When the claim count $N$ is random and independent of the severities, the total is $S_N=\sum_{j=1}^{N}Z_j$. Conditioning on $N$ gives a formula that covers many count models:

\operatorname{Var}(S_N) =E[N]\operatorname{Var}(Z)+\operatorname{Var}(N)\big(E[Z]\big)^2.

For a Poisson count with mean $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$. Therefore,

\operatorname{Var}(S_N) =\lambda\left(\operatorname{Var}(Z)+(E[Z])^2\right) =\lambda E[Z^2].

Derivation: Conditional on $N$, the total has $E[S_N\mid N]=NE[Z]$ and $\operatorname{Var}(S_N\mid N)=N\operatorname{Var}(Z)$. The law of total variance gives $\operatorname{Var}(S_N)=E[\operatorname{Var}(S_N\mid N)]+\operatorname{Var}(E[S_N\mid N])$. Substituting the conditional expressions yields $E[N]\operatorname{Var}(Z)+\operatorname{Var}(N)(E[Z])^2$. Under a Poisson count, the two count moments are both $\lambda$; using $E[Z^2]=\operatorname{Var}(Z)+(E[Z])^2$ gives $\lambda E[Z^2]$.

The distinction between fixed and Poisson counts matters. For a fixed count, the severity variance is multiplied by $n$. For a Poisson count, the raw second moment is multiplied by $\lambda$, because fluctuations in the count also contribute to aggregate variance. Apply the appropriate formula separately to $R$ and $I$.

Worked Applications

Worked Example: Exponential Severity With a Deductible Equal to the Mean

A hypothetical policy has exactly two independent claims. Each claim severity is exponential with mean $\theta=\$5{,}000$, so its variance is $\theta^2$. The per-claim deductible is $d=\$5{,}000$, and there is no limit. Find the annual variance of the insured-retained total and the insurer-payment total.

Method and assumptions. Use $R=\min(X,d)$ and $I=(X-d)_+$ for each claim. The severity distribution is exponential with survival function $S_X(x)=e^{-x/\theta}$ for $x\geq 0$. Since the count is fixed at two and claims are independent, each annual variance is twice the corresponding single-claim variance.

For the retained amount, tail integration gives $E[R]=\theta(1-e^{-d/\theta})$ and $E[R^2]=2\theta^2[1-(1+d/\theta)e^{-d/\theta}]$. Here $d/\theta=1$, so

E[R]=5{,}000(1-e^{-1})=\$3{,}160.6028, \qquad E[R^2]=50{,}000{,}000(1-2e^{-1}) =\$13{,}212{,}055.88.

Thus the single-claim retained variance is

\begin{aligned} \operatorname{Var}(R) &=E[R^2]-(E[R])^2\\ &=25{,}000{,}000\left[2(1-2e^{-1})-(1-e^{-1})^2\right]\\ &=\$3{,}222{,}645.86. \end{aligned}

For the insurer payment, the exponential memoryless property, or direct integration above $d$, gives $E[I]=\theta e^{-d/\theta}$ and $E[I^2]=2\theta^2e^{-d/\theta}$. Therefore,

\begin{aligned} \operatorname{Var}(I) &=2\theta^2e^{-1}-\theta^2e^{-2}\\ &=25{,}000{,}000(2e^{-1}-e^{-2})\\ &=\$15{,}010{,}589.98. \end{aligned}

Multiplying each single-claim variance by two gives annual variances of $\$6{,}445{,}291.72$ for the insured-retained total and $\$30{,}021{,}179.96$ for the insurer-payment total, in dollars squared. These are twice the single-claim variances, as required for exactly two independent claims.

Worked Example: Discrete Severity With a Poisson Claim Count

In a hypothetical book, claim severity is 2, 5, or 9 thousand dollars with respective probabilities 0.25, 0.50, and 0.25. The claim count is Poisson with mean $\lambda=3$, independent of severity. Each claim has a 4-thousand-dollar deductible and no limit. Calculate the annual retained and insurer-payment variances in thousands of dollars squared.

Method and assumptions. First allocate each possible severity to $R$ and $I$. Then use $\operatorname{Var}(S_N)=\lambda E[Z^2]$ for each component, since the count is Poisson.

The per-claim allocations, in thousands of dollars, are $R=2,4,4$ and $I=0,1,5$. Their raw second moments are

$$ E[R^2]=0.25(2^2)+0.50(4^2)+0.25(4^2) =1+8+4=13, $$

$$ E[I^2]=0.25(0^2)+0.50(1^2)+0.25(5^2) =0+0.5+6.25=6.75. $$

Therefore, the annual variances are

\operatorname{Var}(S_R)=3(13)=\mathbf{39}, \qquad \operatorname{Var}(S_I)=3(6.75)=\mathbf{20.25}.

As a check on the allocation, the expected retained and insurer amounts per claim are $0.25(2)+0.50(4)+0.25(4)=3.5$ and $0.25(0)+0.50(1)+0.25(5)=1.75$. Their sum is 5.25, which equals the expected severity $0.25(2)+0.50(5)+0.25(9)=5.25$. The annual variances are 39 and 20.25 thousand-dollars-squared, respectively.

Worked Example: Poisson Frequency and Exponential Severity

A hypothetical portfolio has a Poisson claim count with mean $\lambda=1.5$. Claim severities are exponential with mean $\theta=\$8{,}000$, and the per-claim deductible is $\$8{,}000$, with no limit. Find the annual variance of each loss component.

Method and assumptions. Since the count is Poisson and independent of identically distributed severities, each annual component variance is $\lambda$ times that component’s single-claim raw second moment. At $d/\theta=1$,

E[R^2]=2\theta^2(1-2e^{-1}) =128{,}000{,}000(1-2e^{-1}) =\$33{,}822{,}863.06.

Consequently, the annual retained variance is

\operatorname{Var}(S_R) =1.5E[R^2] =\$50{,}734{,}294.59.

For the insurer payment, $E[I^2]=2\theta^2e^{-1}=128{,}000{,}000e^{-1}=\$47{,}088{,}568.47$. Hence

\operatorname{Var}(S_I) =1.5E[I^2] =\$70{,}632{,}852.71.

Both results are annual variances in dollars squared. This example uses raw second moments in the Poisson formula; replacing them with single-claim variances would omit the contribution from claim-count variability.

What the Variances Do—and Do Not—Add Up To

For each claim, $R+I=X$, but this identity does not imply that the retained and insurer variances sum to the severity variance. The variance identity is

\operatorname{Var}(R+I) =\operatorname{Var}(R)+\operatorname{Var}(I)+2\operatorname{Cov}(R,I).

The retained and insurer amounts arise from the same claim, so their covariance generally is not zero. At the annual level, the same warning applies: a claim that reaches above the deductible contributes to both components. Calculate the variance of each component using its own distribution and the correct count model; do not infer one component’s variance by subtracting the other component’s variance from total variance.

These formulas describe an ordinary per-claim deductible with independent claim severities and the stated claim-count assumptions. An annual aggregate deductible, a cap on reimbursement, dependence among claims, or additional limits changes the allocation or the dependence structure and may require a different calculation.

Common Exam Trap

A frequent error is to use $\lambda\operatorname{Var}(Z)$ as the compound Poisson annual variance. That expression omits variability from the random number of claims. For the discrete example above, the retained mean is 3.5 and its raw second moment is 13, so its single-claim variance is $13-3.5^2=0.75$. The incorrect calculation would give $3(0.75)=2.25$, rather than the correct $3(13)=39$ thousand-dollars-squared. For insurer payments, the single-claim variance is $6.75-1.75^2=3.6875$; the incorrect result is $3(3.6875)=11.0625$, rather than the correct $3(6.75)=20.25$.

The correction is to identify the count model before applying a variance formula. A fixed count uses $n\operatorname{Var}(Z)$; a Poisson count uses $\lambda E[Z^2]$. Also check that the deductible is applied to each claim before aggregating, and keep the units of variance as squared currency units.

Key takeaway: First transform each claim into retained loss and insurer payment. Then calculate each component’s single-claim moments and apply the formula for the actual claim-count model. For a Poisson count, annual variance is the frequency mean times the per-claim raw second moment.

Check Your Understanding

For each problem, treat the stated severity and count assumptions as independent where applicable, and report variances in the requested units.

A policy has exactly four independent claims. Severity is 0.5, 2, or 6 thousand dollars with probabilities 0.20, 0.50, and 0.30, respectively. The per-claim deductible is 3 thousand dollars, with no limit. Find the annual retained and insurer-payment variances in thousands of dollars squared. (Answer: retained 3; insurer payment 7.56.)
Claim count is Poisson with mean 2.5. Severity is 1, 4, or 7 thousand dollars with probabilities 0.20, 0.50, and 0.30, respectively. The deductible is 4 thousand dollars per claim. Find both annual component variances in thousands of dollars squared. (Answer: retained 32.5; insurer payment 6.75.)
Claim count is Poisson with mean 2. Claim severity is exponential with mean $\$6{,}000$; the per-claim deductible is $\$3{,}000$, with no limit. Find both annual component variances in dollars squared, rounded to the nearest cent. (Answer: retained $\$12{,}989{,}377.50$; insurer payment $\$87{,}340{,}415.00$.)
A hypothetical book has claim count $N$, independent of severities, with $E[N]=4$ and $\operatorname{Var}(N)=6$. Severity is 1, 3, or 6 thousand dollars with probabilities 0.50, 0.25, and 0.25, and the per-claim deductible is 3 thousand dollars. Use the general random-count formula to find the annual retained and insurer-payment variances in thousands of dollars squared. (Answer: retained 28; insurer payment 10.125.)

What You'll Learn