How Reinsurance Changes Aggregate Variance

Why Reinsurance Changes Aggregate Variance

Reinsurance divides each claim, and therefore the portfolio’s aggregate loss, between the insurer and the reinsurer. The division changes more than expected payments: it also changes the severity moments that determine aggregate variability. The effect depends on both the treaty and the claim-count distribution. A proportional treaty scales each claim by a fixed share, while a per-claim excess-of-loss treaty applies different payment functions to claims of different sizes.

The calculations below use a frequency-severity model. Let $N$ be the number of claims, and let $X_1,X_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. The gross aggregate loss is $S=\sum_{i=1}^{N}X_i$, with the sum equal to zero when $N=0$. Assume the relevant moments are finite and there are no expenses, reinstatement premiums, or other payments.

Definition: For a claim-level payment function $g(X)$, define its aggregate payment by $T_g=\sum_{i=1}^{N}g(X_i)$. Write $\mu_g=E[g(X)]$ and $m_{2,g}=E[g(X)^2]$. For ceded and retained payments, use $C=\sum_{i=1}^{N}c(X_i)$ and $R=\sum_{i=1}^{N}r(X_i)$, respectively. When the treaty allocates each claim fully, $c(X)+r(X)=X$, so $C+R=S$.

The Random-Sum Variance Formula

The number of claims contributes to aggregate variability as well as the variability of individual claim payments. The formula for any claim-level payment $g(X)$ makes these two sources explicit.

\operatorname{Var}(T_g) =E[N]\operatorname{Var}(g(X))+\operatorname{Var}(N)\big(E[g(X)]\big)^2 =E[N]m_{2,g}+\big(\operatorname{Var}(N)-E[N]\big)\mu_g^2.

The first expression separates the average within-claim variance from the variation in claim count. The second follows by substituting $\operatorname{Var}(g(X))=m_{2,g}-\mu_g^2$. It is particularly convenient when treaty payments are expressed as functions of severity.

Condition on the claim count.
Given $N=n$, $T_g$ is a sum of $n$ independent payments, so its conditional mean is $n\mu_g$ and its conditional variance is $n\operatorname{Var}(g(X))$.

Apply the law of total variance.
$\operatorname{Var}(T_g)=E[\operatorname{Var}(T_g\mid N)]+\operatorname{Var}(E[T_g\mid N])$. The first term is $E[N]\operatorname{Var}(g(X))$; the second is $\operatorname{Var}(N)\mu_g^2$.

Use the second moment when helpful.
Replacing the severity variance by $m_{2,g}-\mu_g^2$ gives the second form of the formula. This form shows exactly which severity moments are needed.

For a Poisson claim count with mean $\lambda$, $\operatorname{Var}(N)=E[N]=\lambda$, and the formula simplifies to $\operatorname{Var}(T_g)=\lambda E[g(X)^2]$. For other frequency distributions, the mean alone is not enough: the variance of $N$ also matters.

The same conditioning argument gives a useful covariance identity for two claim-level payments $g(X)$ and $h(X)$:

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

Since $S=C+R$, a check on any ceded and retained variance calculation is $\operatorname{Var}(S)=\operatorname{Var}(C)+\operatorname{Var}(R)+2\operatorname{Cov}(C,R)$. Ceded and retained totals are generally not independent, so their covariance cannot usually be omitted.

Proportional Reinsurance

Suppose a fixed fraction $\alpha$, where $0\leq\alpha\leq1$, of every claim is ceded. The ceded and retained claim payments are $c(X)=\alpha X$ and $r(X)=(1-\alpha)X$. Thus $C=\alpha S$ and $R=(1-\alpha)S$, giving a direct variance result for any claim-count distribution.

Theorem: Under a proportional treaty with a constant ceded fraction $\alpha$, $\operatorname{Var}(C)=\alpha^2\operatorname{Var}(S)$ and $\operatorname{Var}(R)=(1-\alpha)^2\operatorname{Var}(S)$. Also, $\operatorname{Cov}(C,R)=\alpha(1-\alpha)\operatorname{Var}(S)$.

These identities follow because multiplying a random variable by a constant multiplies its variance by the square of that constant. They also satisfy the variance decomposition: the two scaled variances plus twice the covariance equal $\operatorname{Var}(S)$.

Worked Example: Proportional Shares with Poisson Frequency

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$. Claim severity is exponential with mean $\$10$, so $\operatorname{Var}(X)=100$ and $E[X^2]=100+10^2=200$, in dollars squared. A treaty cedes $25\%$ of each claim. Find the gross, ceded, and retained aggregate variances, and check the covariance decomposition.

Method. Use the second-moment formula for a compound Poisson aggregate. For an exponential severity with mean $\theta=10$, the second moment is $E[X^2]=2\theta^2=200$.

Gross. Since the frequency mean is $\lambda=4$, $\operatorname{Var}(S)=4(200)=800$ dollars squared.

Ceded. The per-claim ceded payment is $0.25X$, so its second moment is $0.25^2(200)=12.5$. Therefore, $\operatorname{Var}(C)=4(12.5)=50$ dollars squared.

Retained. The per-claim retained payment is $0.75X$, with second moment $0.75^2(200)=112.5$. Thus $\operatorname{Var}(R)=4(112.5)=450$ dollars squared.

Check. The covariance is $4E[(0.25X)(0.75X)]=4(0.1875)(200)=150$ dollars squared. Hence $50+450+2(150)=800$, which matches the gross variance. The calculation avoids the mistake of adding only ceded and retained variances: their covariance contributes the remaining $300$.

Per-Claim Excess-of-Loss Reinsurance

For a per-claim excess-of-loss layer, let $d$ be the attachment point and $u$ the layer width, or limit. The ceded payment on a claim of size $X$ is $c(X)=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The retained payment is $r(X)=X-c(X)$. The layer pays nothing at or below attachment, pays the amount above attachment within the layer, and is capped at width $u$. These are claim-level payments; an annual aggregate attachment is a different treaty basis.

The variance formula applies once the first and second moments of $c(X)$ and $r(X)$ have been found. In particular, for Poisson frequency, the ceded aggregate variance is $\lambda E[c(X)^2]$, not $\lambda\operatorname{Var}(c(X))$. For a non-Poisson count, use both $E[c(X)^2]$ and $E[c(X)]$ along with $E[N]$ and $\operatorname{Var}(N)$.

Worked Example: A Layer with Binomial Claim Frequency

A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.5)$. Each claim is $\$10$ or $\$30$, with probability $0.5$ each. A per-claim layer attaches at $\$10$ and has width $\$10$. Calculate the gross, ceded, and retained aggregate variances, then verify the covariance relationship.

Frequency moments. $E[N]=4(0.5)=2$, and $\operatorname{Var}(N)=4(0.5)(0.5)=1$.

Claim payments. For a $\$10$ claim, $c(X)=0$ and $r(X)=10$. For a $\$30$ claim, $c(X)=10$ and $r(X)=20$. Thus the ceded payment has mean $\mu_c=(0+10)/2=5$, second moment $m_{2,c}=(0^2+10^2)/2=50$, and variance $50-5^2=25$. The retained payment has mean $\mu_r=(10+20)/2=15$, second moment $m_{2,r}=(10^2+20^2)/2=250$, and variance $250-15^2=25$.

Ceded aggregate. Apply the random-sum formula: $\operatorname{Var}(C)=2(25)+1(5^2)=50+25=75$ dollars squared.

Retained aggregate. $\operatorname{Var}(R)=2(25)+1(15^2)=50+225=275$ dollars squared.

Gross aggregate. Gross severity has mean $(10+30)/2=20$, second moment $(10^2+30^2)/2=500$, and variance $500-20^2=100$. Therefore $\operatorname{Var}(S)=2(100)+1(20^2)=600$ dollars squared.

Check. The per-claim product $c(X)r(X)$ is $0$ for a $\$10$ claim and $200$ for a $\$30$ claim, so $E[cr]=100$ and $\operatorname{Cov}(c,r)=100-(5)(15)=25$. The aggregate covariance is $2(25)+1(5)(15)=125$. Then $75+275+2(125)=600$, as required.

Comparing Treaty Effects

The next example compares proportional and per-claim excess-of-loss protection on the same portfolio. Matching the expected ceded payment does not make the two treaties equivalent: their ceded payment distributions, and therefore their second moments, can differ.

Worked Example: Equal Expected Cessions, Different Variances

A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$. Each claim is $\$10$ or $\$30$, with probability $0.5$ each. Compare (a) a $25\%$ proportional cession and (b) a per-claim layer attaching at $\$10$ with width $\$10$. Find the ceded and retained aggregate variances and check each against the gross variance.

Gross benchmark. $E[X^2]=(10^2+30^2)/2=500$. Thus the gross aggregate variance is $3(500)=1{,}500$ dollars squared.

(a) Proportional treaty. The ceded payments are $\$2.50$ and $\$7.50$, so their second moment is $(2.5^2+7.5^2)/2=(6.25+56.25)/2=31.25$. The ceded variance is $3(31.25)=93.75$. The retained payments are $\$7.50$ and $\$22.50$, with second moment $(56.25+506.25)/2=281.25$; the retained variance is $3(281.25)=843.75$. The per-claim ceded-retained product has expectation $(18.75+168.75)/2=93.75$, so the aggregate covariance is $3(93.75)=281.25$. The check is $93.75+843.75+2(281.25)=1{,}500$.

(b) Per-claim layer. The ceded payments are $\$0$ and $\$10$, with second moment $(0+100)/2=50$; the ceded variance is $3(50)=150$. The retained payments are $\$10$ and $\$20$, with second moment $(100+400)/2=250$; the retained variance is $3(250)=750$. The expected ceded payment per claim is $\$5$ under both treaties. For the layer, the per-claim ceded-retained product has expectation $(0+200)/2=100$, giving aggregate covariance $3(100)=300$. The check is $150+750+2(300)=1{,}500$.

Both treaties have the same expected ceded payment per claim, but the layer has a larger ceded aggregate variance in this example: $150$ rather than $93.75$ dollars squared. Expected cession alone does not determine variability; the payment’s second moment matters.

What to Check in an Exam Calculation

For proportional reinsurance, scaling the gross variance is usually the shortest route. For a per-claim layer, first determine the ceded and retained payment for each possible severity, or derive their moments from the severity distribution. Then apply the frequency-severity formula with the actual $E[N]$ and $\operatorname{Var}(N)$.

For Poisson frequency, claim-level payment second moments alone determine aggregate variances. For a non-Poisson frequency, a change in the mean payment also affects the frequency-variation term. Consequently, although the retained payment is no larger than the gross payment claim by claim, it is not safe for every frequency model to assert an aggregate variance ordering without calculation.

Common exam trap: using the wrong variance for a fixed payment per claim. Suppose $N\sim\operatorname{Binomial}(4,0.5)$, every claim is $\$20$, and $30\%$ is ceded. The ceded payment is a constant $\$6$ per claim, so $C=6N$. Since $\operatorname{Var}(N)=4(0.5)(0.5)=1$, the correct result is $\operatorname{Var}(C)=6^2(1)=36$ dollars squared. Equivalently, $E[N]=2$, and the random-sum formula gives $2(0)+(1)(6^2)=36$. The incorrect value $84$ does not follow from either method. This case is a useful check that claim-count variability is included once, not replaced by a Poisson assumption.

Key takeaway: Calculate aggregate variance from the moments of the actual ceded or retained claim payment and the moments of claim frequency. Under proportional reinsurance, variance scales by the square of the share. Under per-claim excess-of-loss, the layer changes severity moments nonlinearly, and the frequency model determines how those moments translate into aggregate variability.

Check Your Understanding

For each problem, identify the payment per claim before applying the aggregate variance formula.

A hypothetical portfolio has Poisson claim frequency with mean $2$. Each claim is exactly $\$8$, and $40\%$ is ceded proportionally. Find the ceded aggregate variance. (Answer: $\$20.48$ squared.)
A hypothetical portfolio has Poisson claim frequency with mean $5$, and every claim is $\$12$. A per-claim layer attaches at $\$5$ and has width $\$4$. Find the ceded and retained aggregate variances. (Answer: ceded $\$80$ squared; retained $\$320$ squared.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(6,0.5)$. Every claim is $\$10$, with $20\%$ ceded proportionally. Find the ceded, retained, and gross aggregate variances. (Answer: ceded $\$6$ squared; retained $\$96$ squared; gross $\$150$ squared.)
A hypothetical portfolio has Poisson claim frequency with mean $2$. Severity is $\$5$ or $\$15$, each with probability $0.5$. A per-claim layer attaches at $\$5$ and has width $\$5$. Find the ceded and retained aggregate variances. (Answer: ceded $\$25$ squared; retained $\$125$ squared.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.25)$, and every claim is $\$10$. Half of each claim is ceded proportionally. Find the ceded and retained aggregate variances, and explain why the claim-count variance rather than its mean is the multiplier for each fixed per-claim payment. (Answer: ceded $\$18.75$ squared; retained $\$18.75$ squared; the fixed-payment aggregate is a constant times $N$, so its variance uses $\operatorname{Var}(N)=0.75$.)

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