Compound Poisson Moments After Reinsurance

Why Reinsurance Changes Aggregate Moments

A reinsurance treaty transforms each claim before the claim payments are added across a portfolio. To calculate the aggregate ceded loss or aggregate retained loss, first find the first and second moments of the corresponding per-claim payment. The frequency model then converts those moments into aggregate moments.

The key point is that a compound Poisson aggregate variance uses the second raw moment of the transformed claim payment, not just its variance. This distinction is easy to miss because the claim variance is needed in one step of the derivation, but the frequency variation contributes an additional term.

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$ be the number of claims in a period. Let $X_i$ be the severity of claim $i$, where the severities are independent and identically distributed and are independent of $N$. For a per-claim treaty, let $Y_i^L$ be the ceded payment for claim $i$, and let $Z_i=X_i-Y_i^L$ be the retained payment. The aggregate ceded and retained losses are $$ C=\sum_{i=1}^{N}Y_i^L,\qquad R=\sum_{i=1}^{N}Z_i. $$ If $\lambda$ is finite and the relevant per-claim second moments are finite, then $$ E[C]=\lambda E[Y^L],\qquad \operatorname{Var}(C)=\lambda E[(Y^L)^2], $$ and $$ E[R]=\lambda E[Z],\qquad \operatorname{Var}(R)=\lambda E[Z^2]. $$

These formulas apply when each payment is determined by transforming an individual claim and the claims follow the stated compound Poisson model. They do not automatically apply to a treaty whose payment depends on the annual aggregate, such as an annual aggregate deductible: that payment is not generally a sum of independent per-claim transformations.

Why the Variance Uses a Raw Second Moment

Let $W_i$ be the payment for claim $i$, after applying any per-claim transformation. Write $m_1=E[W]$ and $m_2=E[W^2]$, and define $S=\sum_{i=1}^{N}W_i$. Conditional on $N=n$, the sum contains $n$ independent payments.

Condition on the claim count.
$E[S\mid N]=Nm_1$, and $\operatorname{Var}(S\mid N)=N(m_2-m_1^2)$, because $\operatorname{Var}(W)=m_2-m_1^2$.

Take the unconditional expectation.
Since $E[N]=\lambda$, the conditional mean gives $E[S]=E[Nm_1]=\lambda m_1$.

Apply the law of total variance.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=\lambda(m_2-m_1^2)+\lambda m_1^2=\lambda m_2$, using $\operatorname{Var}(N)=\lambda$ for a Poisson count.

The last term, $\lambda m_1^2$, reflects variation in the number of claims. It combines with the conditional variance to produce $\lambda E[W^2]$. The argument applies to the ceded payment $Y^L$ and to the retained payment $Z=X-Y^L$, each considered as its own transformed claim amount.

Calculating the Transformed Claim Moments

For a per-claim treaty, define the ceded payment as a function of the original severity: $Y^L=g(X)$. The retained payment is then $Z=X-g(X)$. The required inputs are

E[Y^L],\quad E[(Y^L)^2],\quad E[Z],\quad E[Z^2].

For a discrete severity distribution, multiply each possible transformed payment by its probability to obtain its mean, and multiply each squared payment by its probability to obtain its second raw moment. For a continuous severity, the same moments are expectations under the severity distribution. Atoms at zero—for example, claims below an excess-of-loss attachment—must remain in the calculation with their full probability; a zero payment contributes zero to the payment moments, but its probability cannot be discarded.

Worked Example: A Limited Per-Claim Layer

A hypothetical insurer has $N\sim\operatorname{Poisson}(4)$ claims in a year. Claim severities are independent of $N$ and have the distribution $P(X=500)=0.75$, $P(X=1{,}500)=0.25$, with amounts in dollars. A per-claim layer pays the amount above an attachment of $800, subject to a maximum payment of $600 per claim. Find the mean and variance of aggregate ceded and retained losses.

Method. Transform each severity into a ceded payment and a retained payment, find each payment’s first and second raw moments, then apply the compound Poisson formulas. The per-claim ceded amount is $Y^L=\min((X-800)_+,600)$, where $(a)_+=\max(a,0)$.

For $X=500$, the claim is below attachment, so $Y^L=0$ and $Z=500-0=500$. For $X=1{,}500$, the amount above attachment is $1{,}500-800=700$; the $600 payment cap gives $Y^L=600$, and $Z=1{,}500-600=900$.

Thus,

E[Y^L]=0.75(0)+0.25(600)=150,\qquad E[(Y^L)^2]=0.75(0^2)+0.25(600^2)=90{,}000. $$ $$ E[Z]=0.75(500)+0.25(900)=375+225=600, $$ $$ E[Z^2]=0.75(500^2)+0.25(900^2) =0.75(250{,}000)+0.25(810{,}000) =187{,}500+202{,}500=390{,}000.

With $\lambda=4$, the aggregate ceded moments are

E[C]=4(150)=\$600,\qquad \operatorname{Var}(C)=4(90{,}000)=\$360{,}000.

The aggregate retained moments are

E[R]=4(600)=\$2{,}400,\qquad \operatorname{Var}(R)=4(390{,}000)=\$1{,}560{,}000.

Check. The original mean severity is $0.75(500)+0.25(1{,}500)=750$. The mean ceded plus mean retained payment per claim is $150+600=750$, so the aggregate means also reconcile: $600+2{,}400=4(750)=3{,}000$. This checks the means, not the variances.

Common mistake avoided: The ceded claim variance is $90{,}000-150^2=67{,}500$, not the quantity to multiply by $\lambda$ for the aggregate variance. The correct aggregate variance is $4E[(Y^L)^2]=360{,}000$, not $4(67{,}500)=270{,}000$.

Proportional Transformation: A Second Application

The same moment method applies when a treaty pays a fixed proportion of each claim. If the ceded fraction is $q$, then $Y^L=qX$ and $Z=(1-q)X$. Consequently, $E[Y^L]=qE[X]$, $E[(Y^L)^2]=q^2E[X^2]$, $E[Z]=(1-q)E[X]$, and $E[Z^2]=(1-q)^2E[X^2]$. These are per-claim moment relations; multiplying by $\lambda$ gives the aggregate moments under the compound Poisson assumptions.

Worked Example: Ceded and Retained Shares

A hypothetical portfolio has $N\sim\operatorname{Poisson}(10)$ claims. Each severity is $100 or $300 with probability $0.5$ each, independently across claims and independently of $N$. A quota share cedes 40% of every claim. Find the mean and variance of both aggregate payments.

Method. Find the severity’s first and second raw moments, scale them by the appropriate per-claim payment proportions, and apply the compound Poisson formulas.

The severity moments are

E[X]=0.5(100)+0.5(300)=200, $$ $$ E[X^2]=0.5(100^2)+0.5(300^2)=0.5(10{,}000)+0.5(90{,}000)=50{,}000.

The ceded payment is $Y^L=0.4X$. Therefore,

E[Y^L]=0.4(200)=80,\qquad E[(Y^L)^2]=0.4^2(50{,}000)=8{,}000.

The retained payment is $Z=0.6X$. Therefore,

E[Z]=0.6(200)=120,\qquad E[Z^2]=0.6^2(50{,}000)=18{,}000.

Multiplying each per-claim moment by $\lambda=10$ gives

E[C]=\$800,\quad \operatorname{Var}(C)=\$80{,}000, $$ $$ E[R]=\$1{,}200,\quad \operatorname{Var}(R)=\$180{,}000.

Check. The per-claim means add to $80+120=200=E[X]$, and the aggregate means add to $800+1{,}200=2{,}000=10E[X]$. As in the first example, this check does not imply that aggregate variances add in the same way.

Common mistake avoided: The ceded second moment is $0.4^2E[X^2]$, not $0.4E[X^2]$. Squaring the payment proportion is necessary because the moment is based on the squared payment.

A Discrete Treaty Transformation With a Zero Payment

A layer can produce several payment amounts, including zero for claims below attachment. A useful calculation layout lists the original severity, its probability, the ceded payment, and the retained payment. This makes it easier to ensure that every severity outcome is included and that each retained amount is $X-Y^L$.

Worked Example: Three Severity Outcomes

A hypothetical insurer’s annual claim count is $N\sim\operatorname{Poisson}(3.5)$. Independent severities are $200 with probability $0.2$, $700 with probability $0.5$, and $1,600 with probability $0.3$. A per-claim layer attaches at $500 and has a maximum payment of $700. Calculate the mean and variance of the aggregate ceded and retained losses.

Method. Apply $Y^L=\min((X-500)_+,700)$ at each severity value, then calculate the first and second raw moments for both payment types.

$X$	Probability	$Y^L$	$Z=X-Y^L$
$200	0.2	$0	$200
$700	0.5	$200	$500
$1,600	0.3	$700	$900

For the ceded payment,

E[Y^L]=0.2(0)+0.5(200)+0.3(700)=0+100+210=310, $$ $$ E[(Y^L)^2]=0.2(0^2)+0.5(200^2)+0.3(700^2) =0+20{,}000+147{,}000=167{,}000.

For the retained payment,

E[Z]=0.2(200)+0.5(500)+0.3(900)=40+250+270=560, $$ $$ E[Z^2]=0.2(200^2)+0.5(500^2)+0.3(900^2) =8{,}000+125{,}000+243{,}000=376{,}000.

The aggregate results are

E[C]=3.5(310)=\$1{,}085,\qquad \operatorname{Var}(C)=3.5(167{,}000)=\$584{,}500, $$ $$ E[R]=3.5(560)=\$1{,}960,\qquad \operatorname{Var}(R)=3.5(376{,}000)=\$1{,}316{,}000.

Check. The original mean severity is $0.2(200)+0.5(700)+0.3(1{,}600)=40+350+480=870$. The transformed means satisfy $310+560=870$, and the aggregate means satisfy $1{,}085+1{,}960=3.5(870)=3{,}045$.

Common mistake avoided: The $1,600 claim generates a $700 ceded payment because the layer reaches its maximum, leaving $900 retained. Using $1,100 as the cession would ignore the treaty limit.

Interpretation and Exam Checks

The procedure is deliberately modular: determine the treaty payment for each claim, calculate its first and second raw moments, and then use the frequency model. It also helps keep three different quantities separate: severity $X$, per-claim payment $Y^L$ or $Z$, and aggregate payment $C$ or $R$. The compound Poisson formulas are applied to the transformed claim payment, not directly to the original severity unless there is no transformation.

Check the treaty at each severity. Respect attachment points, payment limits, and the definition of the payment function.
Check retained payment claim by claim. For a treaty that partitions each claim, verify $Z=X-Y^L$.
Use raw second moments. The aggregate variance is $\lambda E[W^2]$, not $\lambda\operatorname{Var}(W)$.
Check the model boundary. The formulas here assume Poisson claim frequency and per-claim payment transformations; they are not a general formula for every aggregate-level treaty.

Key takeaway: Under compound Poisson frequency, transform each claim first. For either ceded or retained payments $W$, the aggregate mean is $\lambda E[W]$ and the aggregate variance is $\lambda E[W^2]$, provided the stated assumptions and finite second moment hold.

Check Your Understanding

For each problem, identify the transformed per-claim payment and apply the compound Poisson moment formulas.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(5)$. A per-claim layer pays $300 with probability $0.2$ and zero otherwise. Find the aggregate ceded mean and variance. (Answer: mean $300; variance $90,000.)
Claim severity is $100 or $500 with equal probability, and $N\sim\operatorname{Poisson}(2)$. A quota share cedes 25% of each claim. Find the aggregate retained mean and variance. (Answer: mean $450; variance $146,250.)
Let $N\sim\operatorname{Poisson}(1.5)$, and let severity be $400 with probability $0.4$ or $1,000 with probability $0.6$. A per-claim layer attaches at $600 with no payment cap. Find the aggregate ceded mean and variance. (Answer: mean $360; variance $144,000.)
Let $N\sim\operatorname{Poisson}(8)$. Each claim’s ceded payment is $50 with probability $0.1$ and zero with probability $0.9$, independently across claims and of $N$. Find the aggregate ceded mean and variance. (Answer: mean $40; variance $2,000.)
Let $N\sim\operatorname{Poisson}(4)$, with severity $100 or $900, each with probability $0.5$. A per-claim layer attaches at $400 and has a maximum payment of $300. Find the aggregate retained mean and variance. (Answer: mean $1,400; variance $740,000.)

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