Calculate Insurer and Reinsurer Aggregate Loss Moments

Why Aggregate Payment Moments Require Care

A reinsurance treaty divides claim payments between an insurer and a reinsurer. Because both parties’ payments arise from the same claims, their aggregate losses are generally dependent. Correctly calculating each party’s expected aggregate payment and variance therefore requires both a claim-count model and a clear description of how each covered loss is allocated.

This tutorial considers per-claim payment rules. It does not treat aggregate stop-loss payments, which depend on the total loss over a period rather than on each claim separately. Assume that claim amounts are nonnegative, claim severities are independent and identically distributed, claim count is independent of severities, and the required second moments are finite. There are no expenses, premiums, or timing adjustments in the calculations.

Definition: Let $N$ be the number of claims in a period, and let $X_j$ be the covered loss on claim $j$. For a specified treaty, write $I(X_j)$ for the insurer’s payment and $R(X_j)$ for the reinsurer’s payment on that claim. The aggregate payments are $$ S_I=\sum_{j=1}^{N} I(X_j), \qquad S_R=\sum_{j=1}^{N} R(X_j). $$ An empty sum is zero when $N=0$. If the treaty allocates the full covered loss with no gap or overlap, then $I(X_j)+R(X_j)=X_j$ for every claim.

General Aggregate Moment Formulas

For a single claim, let $\mu_I=E[I(X)]$, $\mu_R=E[R(X)]$, and let $\sigma_I^2=\operatorname{Var}(I(X))$, $\sigma_R^2=\operatorname{Var}(R(X))$. Denote the single-claim covariance by $\sigma_{IR}=\operatorname{Cov}(I(X),R(X))$. Under the stated independence assumptions, for a random claim count with finite mean and variance,

\begin{aligned} E[S_I]&=E[N]\mu_I, & \operatorname{Var}(S_I)&=E[N]\sigma_I^2+\operatorname{Var}(N)\mu_I^2,\\ E[S_R]&=E[N]\mu_R, & \operatorname{Var}(S_R)&=E[N]\sigma_R^2+\operatorname{Var}(N)\mu_R^2,\\ \operatorname{Cov}(S_I,S_R)&=E[N]\sigma_{IR} +\operatorname{Var}(N)\mu_I\mu_R. \end{aligned}

These formulas separate two sources of aggregate variability. The terms involving $E[N]$ reflect variation in the payment amounts on claims. The terms involving $\operatorname{Var}(N)$ reflect variation in the number of claims. The covariance formula also has both parts: insurer and reinsurer payments can vary together on an individual claim, and a high claim count tends to raise both aggregate totals.

Derivation: Conditional on $N=n$, the insurer’s aggregate is a sum of $n$ independent claim payments. Thus $E[S_I\mid N]=N\mu_I$ and $\operatorname{Var}(S_I\mid N)=N\sigma_I^2$. The law of total variance gives $$ \operatorname{Var}(S_I) =E[\operatorname{Var}(S_I\mid N)] +\operatorname{Var}(E[S_I\mid N]) =E[N]\sigma_I^2+\operatorname{Var}(N)\mu_I^2. $$ For the covariance, conditional on $N=n$, the $n$ paired insurer and reinsurer payments have covariance $n\sigma_{IR}$. Applying the law of total covariance gives $$ \operatorname{Cov}(S_I,S_R) =E[N]\sigma_{IR}+\operatorname{Var}(N)\mu_I\mu_R. $$ The second term follows because the conditional means are $N\mu_I$ and $N\mu_R$, both driven by the same $N$.

When $N\sim\operatorname{Poisson}(\lambda)$, its mean and variance are both $\lambda$. The formulas then simplify to

\begin{aligned} E[S_I]&=\lambda E[I(X)],& \operatorname{Var}(S_I)&=\lambda E[I(X)^2],\\ E[S_R]&=\lambda E[R(X)],& \operatorname{Var}(S_R)&=\lambda E[R(X)^2],\\ \operatorname{Cov}(S_I,S_R)&=\lambda E[I(X)R(X)]. \end{aligned}

For example, $\operatorname{Var}(S_I)=\lambda E[I(X)^2]$, not $\lambda\operatorname{Var}(I(X))$. The latter omits the variability from the random claim count. These Poisson formulas require independent Poisson frequency and independent, identically distributed claim severities; they do not automatically apply if frequency and severity are dependent or if the treaty payment on a claim depends on other claims in the period.

Worked Examples

Worked Example: Quota Share with Poisson Frequency

A hypothetical insurer has $N\sim\operatorname{Poisson}(4)$ covered claims in a year. The claim amounts are independent exponential losses with mean $\$1{,}000$, independent of $N$. A reinsurer pays 25% of each covered claim, and the insurer retains 75%. Calculate both expected aggregate payments, both aggregate variances, and the covariance between the aggregate payments.

Method. Apply the Poisson compound-moment formulas to the insurer and reinsurer’s per-claim shares. For an exponential loss with mean $\theta$, $E[X]=\theta$ and $E[X^2]=2\theta^2$. Here $\theta=\$1{,}000$, so

E[X]=\$1{,}000,\qquad E[X^2]=2(\$1{,}000)^2=\$2{,}000{,}000.

On each claim, $I(X)=0.75X$ and $R(X)=0.25X$. Therefore,

\begin{aligned} E[I(X)]&=0.75(\$1{,}000)=\$750,\\ E[R(X)]&=0.25(\$1{,}000)=\$250,\\ E[I(X)^2]&=(0.75)^2(\$2{,}000{,}000)=\$1{,}125{,}000,\\ E[R(X)^2]&=(0.25)^2(\$2{,}000{,}000)=\$125{,}000,\\ E[I(X)R(X)]&=(0.75)(0.25)(\$2{,}000{,}000)=\$375{,}000. \end{aligned}

Multiplying each relevant per-claim moment by $\lambda=4$ gives

\begin{aligned} E[S_I]&=4(\$750)=\$3{,}000,& \operatorname{Var}(S_I)&=4(\$1{,}125{,}000)=\$4{,}500{,}000,\\ E[S_R]&=4(\$250)=\$1{,}000,& \operatorname{Var}(S_R)&=4(\$125{,}000)=\$500{,}000,\\ \operatorname{Cov}(S_I,S_R)&=4(\$375{,}000)=\$1{,}500{,}000. \end{aligned}

Check. The expected aggregate payments add to $\$4{,}000$, which is $4E[X]$. The aggregate total equals the gross aggregate loss, whose Poisson variance is $4E[X^2]=\$8{,}000{,}000$. The component check agrees: $\$4{,}500{,}000+\$500{,}000+2(\$1{,}500{,}000)=\$8{,}000{,}000$. The covariance is positive because both parties’ payments increase with the same claims. A common mistake is to assume the quota-share percentages make the aggregate payments independent; they do not.

Worked Example: Per-Claim Excess-of-Loss Layer

A hypothetical insurer has $N\sim\operatorname{Poisson}(3)$ claims. Each covered loss is $\$80$ or $\$200$, with probability $0.5$ for each amount. The insurer pays the first $\$100$ of each loss, and the reinsurer pays the amount above $\$100$, with no upper limit. Calculate the aggregate means, variances, and covariance.

Method. First apply the treaty to each possible claim amount. The insurer’s payment is $I(X)=\min(X,\$100)$; the reinsurer’s is $R(X)=\max(X-\$100,0)$.

\begin{array}{c|cc} X&I(X)&R(X)\\ \hline \$80&\$80&\$0\\ \$200&\$100&\$100 \end{array}

The per-claim means and second moments are

\begin{aligned} E[I(X)]&=0.5(\$80)+0.5(\$100)=\$90,\\ E[R(X)]&=0.5(\$0)+0.5(\$100)=\$50,\\ E[I(X)^2]&=0.5(\$80^2)+0.5(\$100^2)=\$8{,}200,\\ E[R(X)^2]&=0.5(\$0^2)+0.5(\$100^2)=\$5{,}000,\\ E[I(X)R(X)]&=0.5(\$80)(\$0)+0.5(\$100)(\$100)=\$5{,}000. \end{aligned}

Use $\lambda=3$:

\begin{aligned} E[S_I]&=3(\$90)=\$270,& \operatorname{Var}(S_I)&=3(\$8{,}200)=\$24{,}600,\\ E[S_R]&=3(\$50)=\$150,& \operatorname{Var}(S_R)&=3(\$5{,}000)=\$15{,}000,\\ \operatorname{Cov}(S_I,S_R)&=3(\$5{,}000)=\$15{,}000. \end{aligned}

Check. The parties’ means add to $\$420$. The gross claim has mean $\$140$ and second moment $0.5(\$80^2)+0.5(\$200^2)=\$23{,}200$, so the gross aggregate variance is $3(\$23{,}200)=\$69{,}600$. The component identity gives $\$24{,}600+\$15{,}000+2(\$15{,}000)=\$69{,}600$. The positive covariance arises because both parties are paid on a $\$200$ claim; it would be wrong to assume that an excess-of-loss layer makes the parties’ totals independent.

Worked Example: Binomial Frequency and Quota Share

A hypothetical portfolio has $N\sim\operatorname{Binomial}(5,0.4)$ claims in a period. Each claim is independently $\$100$ or $\$300$, equally likely, and claim amounts are independent of $N$. The reinsurer pays 20% of each claim. Find both expected aggregate payments, both aggregate variances, and their covariance.

Method. Because the count is binomial rather than Poisson, use the general random-count formulas. The count moments are $E[N]=5(0.4)=2$ and $\operatorname{Var}(N)=5(0.4)(0.6)=1.2$.

For gross severity, $E[X]=0.5(\$100)+0.5(\$300)=\$200$ and $E[X^2]=0.5(\$100^2)+0.5(\$300^2)=\$50{,}000$. Thus $\operatorname{Var}(X)=\$50{,}000-(\$200)^2=\$10{,}000$. The insurer pays $I(X)=0.8X$, and the reinsurer pays $R(X)=0.2X$. Consequently,

\begin{aligned} \mu_I&=\$160,& \sigma_I^2&=(0.8)^2(\$10{,}000)=\$6{,}400,\\ \mu_R&=\$40,& \sigma_R^2&=(0.2)^2(\$10{,}000)=\$400,\\ \sigma_{IR}&=(0.8)(0.2)(\$10{,}000)=\$1{,}600. \end{aligned}

The expected aggregate payments are

E[S_I]=2(\$160)=\$320,\qquad E[S_R]=2(\$40)=\$80.

The variances and covariance follow by including both count-variation terms:

\begin{aligned} \operatorname{Var}(S_I) &=2(\$6{,}400)+1.2(\$160)^2 =\$12{,}800+\$30{,}720=\$43{,}520,\\ \operatorname{Var}(S_R) &=2(\$400)+1.2(\$40)^2 =\$800+\$1{,}920=\$2{,}720,\\ \operatorname{Cov}(S_I,S_R) &=2(\$1{,}600)+1.2(\$160)(\$40) =\$3{,}200+\$7{,}680=\$10{,}880. \end{aligned}

Check. The means add to $\$400=E[N]E[X]$. The gross aggregate variance is $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2=2(\$10{,}000)+1.2(\$200)^2=\$68{,}000$. The parties’ moments give $\$43{,}520+\$2{,}720+2(\$10{,}880)=\$68{,}000$. This check confirms that the binomial frequency variation has been included.

Interpreting the Covariance and Checking Results

The insurer’s and reinsurer’s aggregate payments are not independent merely because a treaty assigns separate payment amounts to them. For per-claim allocations, both payments are functions of the same loss, and both aggregate totals are also affected by the same claim count. The covariance can be positive, zero, or—in more general payment arrangements—negative. Its sign and size should be calculated from the payment rules, not assumed.

If the treaty fully allocates each loss, then $S_I+S_R=\sum_{j=1}^N X_j$. This gives two useful checks:

The expected aggregate payments must add to the expected gross aggregate loss.
The gross aggregate variance must equal $\operatorname{Var}(S_I)+\operatorname{Var}(S_R)+2\operatorname{Cov}(S_I,S_R)$.

A frequent exam trap is to calculate the two variances and then add them to get the variance of the total. That calculation is only valid when the covariance is zero. With shared claims, use the covariance identity. Another trap is to use the Poisson shortcut for a non-Poisson count: when the count distribution is different, retain the separate $E[N]$ and $\operatorname{Var}(N)$ terms. These moment formulas also do not apply directly to an aggregate treaty whose payment depends on the annual total rather than on separate per-claim payments.

Key takeaway: Define both parties’ payment functions on the same claim loss, calculate their per-claim moments and cross-product, and then combine those quantities with the correct claim-count moments. For a fully allocated treaty, finish by checking the mean and variance of the combined aggregate loss.

Check Your Understanding

For each problem, calculate the requested aggregate moments using the stated frequency and treaty assumptions.

A portfolio has $N\sim\operatorname{Poisson}(2)$ claims, each with a deterministic covered loss of $\$50$. The reinsurer pays 40% of each claim. Find both aggregate means, both aggregate variances, and their covariance. (Answer: $E[S_I]=\$60$, $E[S_R]=\$40$, $\operatorname{Var}(S_I)=\$1{,}800$, $\operatorname{Var}(S_R)=\$800$, and $\operatorname{Cov}(S_I,S_R)=\$1{,}200$.)
A portfolio has $N\sim\operatorname{Poisson}(1.5)$ claims, each loss being $\$100$ or $\$300$ with equal probability. The insurer pays up to $\$150$ per claim and the reinsurer pays the excess. Find both aggregate means, variances, and covariance. (Answer: $E[S_I]=\$187.50$, $E[S_R]=\$112.50$, $\operatorname{Var}(S_I)=\$24{,}375$, $\operatorname{Var}(S_R)=\$16{,}875$, and $\operatorname{Cov}(S_I,S_R)=\$16{,}875$.)
A portfolio has $N\sim\operatorname{Binomial}(4,0.5)$ claims, each with a deterministic loss of $\$80$. The reinsurer pays 25% of each loss. Calculate both aggregate means, variances, and covariance. (Answer: $E[S_I]=\$120$, $E[S_R]=\$40$, $\operatorname{Var}(S_I)=\$3{,}600$, $\operatorname{Var}(S_R)=\$400$, and $\operatorname{Cov}(S_I,S_R)=\$1{,}200$.)
A portfolio has $N\sim\operatorname{Poisson}(2)$ claims. Each loss is $\$50$ or $\$250$, equally likely. The insurer pays the first $\$100$ of each claim and the reinsurer pays any excess. Calculate both aggregate means, variances, and covariance, then verify the gross aggregate variance. (Answer: $E[S_I]=\$150$, $E[S_R]=\$150$, $\operatorname{Var}(S_I)=\$12{,}500$, $\operatorname{Var}(S_R)=\$22{,}500$, $\operatorname{Cov}(S_I,S_R)=\$15{,}000$, and gross aggregate variance $=\$65{,}000$.)

What You'll Learn