Integrated Workshop: Aggregate Loss With Coverage and Reinsurance

Introduction

An aggregate-loss calculation can involve several different amounts for each claim: the ground-up loss, the amount paid under the policy, the reinsurers’ recoveries, and the insurer’s final retained payment. Confusing these quantities—or applying a contract term at the wrong stage—can invalidate an otherwise correct frequency-severity calculation.

This workshop follows losses through coverage and reinsurance, then aggregates the resulting retained payments. It uses both moment calculations and Panjer recursion, and checks results by an independent expected-payment calculation. The methods and terminology are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $X$ be the ground-up loss on a claim. If a policy has deductible $d$ and a maximum payment $u$, its payment per loss is $Y^L=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. Let $Z$ be the insurer’s retained payment after reinsurance. For claim count $N$, the aggregate retained loss is $S=\sum_{i=1}^{N} Z_i$, with $S=0$ when $N=0$.

The model below assumes that claim amounts are independent and identically distributed, independent of $N$, and that each policy and reinsurance modification is applied separately to each claim. Unless specified otherwise, there is no aggregate policy limit or aggregate reinsurance limit. A per-claim limit and an aggregate limit are different contract terms.

A Reliable Calculation Chain

Write down the contract operations in the order they apply. A common arrangement is: apply the policy deductible and limit to $X$; apply quota share to the resulting policy payment; then apply per-claim excess reinsurance to the insurer’s amount after quota share. If a contract instead defines a different order, follow its wording.

Modify the loss under the policy.
Calculate the policy payment $Y^L$, applying the deductible and policy limit to the ground-up loss.

Apply proportional reinsurance.
If the reinsurer cedes a proportion $c$ of the policy payment, the insurer’s amount before any further treaty is $(1-c)Y^L$.

Apply per-claim excess reinsurance.
For attachment $A$ and limit $L$, the recovery on the amount $W$ before excess reinsurance is $\min((W-A)_+,L)$. Subtract this recovery to obtain $Z$.

Aggregate the retained payments.
Use the distribution of $N$ and the modified distribution of $Z$, not the original ground-up severity distribution.

Worked Example: From Ground-Up Loss to Retained Aggregate Moments

A hypothetical portfolio has $N\sim\operatorname{NegBin}(r=2,p=0.5)$, using the convention $P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n$. Thus $E[N]=r(1-p)/p=2$ and $\operatorname{Var}(N)=r(1-p)/p^2=4$. Claim amounts are independent of $N$ and have this distribution: $X=\$100,\$200,\$300$ with probabilities $0.2,0.5,0.3$. The policy has deductible $\$100$ and payment limit $\$150$. A quota-share reinsurer receives 50% of the policy payment. On the insurer’s remaining amount, a per-claim excess treaty pays losses above $\$40$, subject to a $\$30$ limit. Find the retained payment distribution and aggregate mean and variance.

Method. Transform each possible ground-up loss in contract order, then use compound-model moments. This avoids treating either the ground-up loss or policy payment as the insurer’s final severity.

For $X=\$100$, the policy payment is $\min((100-100)_+,150)=\$0$. For $X=\$200$, it is $\min(100,150)=\$100$. For $X=\$300$, it is $\min(200,150)=\$150$. After the 50% quota share, the insurer amounts are respectively $\$0,\$50,\$75$.

The excess recoveries are $\min((0-40)_+,30)=\$0$, $\min(50-40,30)=\$10$, and $\min(75-40,30)=\$30$. Therefore the final retained payments are $\$0,\$40,\$45$, with probabilities $0.2,0.5,0.3$.

The retained payment mean is

$$ E[Z]=0.2(0)+0.5(40)+0.3(45)=33.5. $$

Its second moment is $E[Z^2]=0.2(0^2)+0.5(40^2)+0.3(45^2)=800+607.5=1407.5$. Thus $\operatorname{Var}(Z)=1407.5-(33.5)^2=1407.5-1122.25=285.25$.

For independent severities and a frequency independent of severity, $\operatorname{Var}(S)=E[N]\operatorname{Var}(Z)+\operatorname{Var}(N)(E[Z])^2$. Hence $E[S]=2(33.5)=\$67$, and $\operatorname{Var}(S)=2(285.25)+4(1122.25)=570.5+4489=5059.5$ dollars squared.

Check. The expected policy payment per claim is $0.2(0)+0.5(100)+0.3(150)=\$95$. Expected quota-share recovery is $0.5(95)=\$47.50$. The expected excess recovery per claim is $0.5(10)+0.3(30)=\$14$. Subtracting both recoveries gives $95-47.50-14=\$33.50$, agreeing with $E[Z]$; multiplying by $E[N]=2$ again gives $\$67$. A common mistake is to subtract the deductible after quota share: for $X=\$200$, that would give $0.5(200)-100=0$, rather than the correct post-quota-share amount $0.5(200-100)=\$50$, before the excess recovery.

Choosing an Aggregate Calculation

The appropriate method depends on what is required and how the modified severity is represented. For an expected aggregate payment, moments usually suffice. For selected probabilities of a discrete aggregate, conditioning or recursion may be more useful. A continuous severity may need discretization before a discrete recursion is applied; in that case, the discretization choice affects the result.

Question	Useful method	Key check
Expected aggregate payment	$E[S]=E[N]E[Z]$	Independently calculate the expected retained amount per claim.
Aggregate variance	Frequency-severity moment formula	Use the second moment of $Z$ or the equivalent variance formula consistently.
Selected probabilities for discrete $Z$	Conditioning or Panjer recursion when applicable	Check that the zero probability and units match the model.
Continuous severity and an aggregate probability	Choose an approximation or discretize, as appropriate	State the approximation or discretization and assess its effect.

For example, if $N$ is Poisson with mean $\lambda$, then $E[S]=\lambda E[Z]$ and $\operatorname{Var}(S)=\lambda E[Z^2]$. If $N$ is not Poisson, do not use the Poisson variance shortcut: use $E[N]\operatorname{Var}(Z)+\operatorname{Var}(N)(E[Z])^2$.

Panjer Recursion for a Discrete Modified Severity

Panjer recursion is useful when the frequency distribution belongs to the $(a,b,0)$ class, whose probabilities $q_n=P(N=n)$ satisfy $q_n=(a+b/n)q_{n-1}$ for $n\geq1$. Suppose the retained payment is represented on a lattice with unit $h$: $P(Z=jh)=p_j$, including $p_0$. If $f_k=P(S=kh)$, then

f_0=G_N(p_0),\qquad f_k=\frac{1}{1-ap_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)p_j f_{k-j},\quad k\geq1,

where $G_N(s)=E[s^N]$ is the probability generating function of $N$. This formula requires $1-ap_0\ne0$. The severity probabilities must be on the chosen lattice, and $p_0$ is essential: claims with zero retained payment still occur and are already accounted for through the frequency model.

Derivation: Let $Q(s)=\sum_{n\geq0}q_ns^n$ and $P(z)=\sum_{j\geq0}p_jz^j$. The frequency recurrence implies $(1-as)Q'(s)=(a+b)Q(s)$. The aggregate probability generating function is $F(z)=Q(P(z))$. Substituting $s=P(z)$ and using the chain rule gives $(1-aP(z))F'(z)=(a+b)F(z)P'(z)$. Equating coefficients of $z^{k-1}$, isolating the term involving $p_0f_k$, and dividing by $k$ yields the displayed recurrence. This composition works because, conditional on $N=n$, the aggregate is the sum of $n$ independent payments with generating function $P(z)$.

Worked Example: Recursing to Selected Aggregate Probabilities

Use the portfolio in the first example. The frequency is negative binomial with $r=2,p=0.5$, for which $q_n/q_{n-1}=0.5+0.5/n$; hence $a=b=0.5$. The retained severity has probabilities $p_0=0.2$, $p_8=0.5$, and $p_9=0.3$ on a $\$5$ lattice. Find $P(S=0)$, $P(S=\$40)$, and $P(S=\$45)$.

Method. Apply the recursion directly on the $\$5$ lattice. Index 8 represents $\$40$, and index 9 represents $\$45$. First, $1-ap_0=1-(0.5)(0.2)=0.9$. The negative binomial probability generating function is $G_N(s)=[p/(1-(1-p)s)]^r$, so

f_0=G_N(0.2)=\left(\frac{0.5}{1-0.5(0.2)}\right)^2 =\left(\frac{0.5}{0.9}\right)^2 =\frac{25}{81}\approx0.3086.

For $k=8$, the only positive severity index not exceeding 8 is $j=8$. Its coefficient is $a+bj/k=0.5+0.5(8/8)=1$, giving

f_8=\frac{1}{0.9}(1)(0.5)f_0 =\frac{5}{9}\frac{25}{81} =\frac{125}{729}\approx0.1715.

For $k=9$, the only contributing positive severity index is $j=9$, with coefficient $0.5+0.5(9/9)=1$. Therefore

f_9=\frac{1}{0.9}(1)(0.3)f_0 =\frac{1}{3}\frac{25}{81} =\frac{25}{243}\approx0.1029.

Thus $P(S=0)=0.3086$, $P(S=\$40)=0.1715$, and $P(S=\$45)=0.1029$, with probabilities rounded to four decimal places. As a further recursion check, at $k=16$ only $j=8$ contributes: the coefficient is $0.5+0.5(8/16)=0.75$, so $f_{16}=(0.75)(0.5)f_8/0.9=(5/12)f_8=625/8748\approx0.0714$. The earlier probabilities use $f_0=25/81$; substituting $f_1=f_0$ would be wrong because the first positive aggregate lattice point is index 8, not index 1.

Independent Checks and Contract Traps

The same contract chain can be checked by separately calculating expected policy payments and expected reinsurance recoveries. This is valuable because an aggregate calculation can be internally consistent while using the wrong per-claim amount. In the first example, the check was $E[Y^L]-E[\text{quota-share recovery}]-E[\text{excess recovery}]=E[Z]$. For a frequency with finite mean, the corresponding aggregate expectation is $E[N]E[Z]$.

Worked Example: Verifying an Aggregate Expectation After Two Treaties

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2.4)$. Independent ground-up claims are $\$100,\$300,\$500$ with probabilities $0.25,0.50,0.25$. A policy has a $\$100$ deductible and a $\$300$ payment limit. A quota-share treaty cedes 20% of policy payments. A further per-claim excess treaty reimburses the insurer for amounts above $\$120$, up to $\$80$. Find the expected aggregate retained payment and its variance.

Method. Calculate the final retained severity for each ground-up amount, then use the compound Poisson mean and variance formulas. The excess treaty applies to the insurer’s 80% share after the policy modification.

The policy payments for losses $\$100,\$300,\$500$ are respectively $\min(0,300)=\$0$, $\min(200,300)=\$200$, and $\min(400,300)=\$300$. After quota share, the insurer amounts are $\$0,\$160,\$240$. The excess recoveries are $\$0,\min(40,80)=\$40,\min(120,80)=\$80$, leaving retained payments $Z=\$0,\$120,\$160$ with probabilities $0.25,0.50,0.25$.

The retained mean per claim is $E[Z]=0.25(0)+0.50(120)+0.25(160)=\$100$. The second moment is $E[Z^2]=0.50(120^2)+0.25(160^2)=7200+6400=13600$, so $\operatorname{Var}(Z)=13600-100^2=3600$. For Poisson frequency, $E[S]=2.4(100)=\$240$ and $\operatorname{Var}(S)=2.4(13600)=32640$ dollars squared.

Check. The expected policy payment per claim is $0.50(200)+0.25(300)=\$175$. The quota-share recovery is $0.20(175)=\$35$, leaving $\$140$ per claim for the insurer before excess reinsurance. Expected excess recovery is $0.50(40)+0.25(80)=\$40$, so the final expectation is $175-35-40=\$100$. At portfolio level, expected policy payments are $2.4(175)=\$420$; expected quota-share and excess recoveries are $2.4(35)=\$84$ and $2.4(40)=\$96$. Thus $420-84-96=\$240$, as calculated from retained severity.

A contract-order trap would be to apply the deductible after quota share. For a $\$300$ loss, this incorrectly gives $0.8(300)-100=\$140$ before excess reinsurance. The correct policy-first calculation is $0.8\min((300-100)_+,300)=\$160$. The deductible belongs to the policy calculation on the ground-up loss; it is not a deduction from the insurer’s post-quota-share amount.

What to Report

A complete solution should identify the frequency distribution and parameters, state the ground-up severity model, specify each contract operation, and give the final modified-severity distribution. For an aggregate calculation, also state whether the result is an expectation, a moment, an exact discrete probability, or an approximation. If using recursion, identify the lattice unit and include the zero-payment probability.

Check that probabilities in the modified severity distribution sum to 1.
Check that retained payments are nonnegative and do not exceed the applicable policy payment.
Do not confuse a per-claim treaty limit with an aggregate limit.
Use the frequency model’s actual mean and variance; do not substitute Poisson formulas for another frequency model.
When possible, verify an aggregate expectation by separately calculating expected policy payments and expected recoveries.

Key takeaway: Modify each claim in the contractual order before aggregating. Then select a method suited to the requested result, and use an independent expectation or moment calculation to check the aggregate output.

Check Your Understanding

For each problem, identify the relevant model and keep policy payments distinct from retained payments.

A portfolio has Poisson claim count with mean $1.2$. The final retained payment per claim is $\$0,\$50,\$100$ with probabilities $0.2,0.5,0.3$. Find the expected aggregate retained payment. (Answer: $\$66$)
There are $N\sim\operatorname{Bin}(3,0.4)$ potential claims. Each claim has retained payment $\$0$ with probability $0.25$ and $\$80$ with probability $0.75$, independently. Find the probability the aggregate is zero. (Answer: $0.3430$)
Claim count is Poisson with mean 2. Ground-up losses are $\$100,\$250,\$400$ with probabilities $0.2,0.5,0.3$. The policy has a $\$100$ deductible and $\$200$ payment limit; quota share cedes 30% of policy payments, with no further reinsurance. Find the expected aggregate retained payment. (Answer: $\$189$)
A compound Poisson model has frequency mean $2.5$, retained-payment mean $\$30$, and retained-payment second moment $1100$ dollars squared. Find the aggregate variance. (Answer: $2750$ dollars squared)
Two independent potential claims each occur with probability $0.5$. Conditional on occurrence, the retained payment is $\$0,\$40,\$60$ with probabilities $0.25,0.50,0.25$. Find the probability that the aggregate retained payment is exactly $\$40$. (Answer: $0.3125$)

Question	Useful method	Key check
Expected aggregate payment	\(E[S]=E[N]E[Z]\)	Independently calculate the expected retained amount per claim.
Aggregate variance	Frequency-severity moment formula	Use the second moment of \(Z\) or the equivalent variance formula consistently.
Selected probabilities for discrete \(Z\)	Conditioning or Panjer recursion when applicable	Check that the zero probability and units match the model.
Continuous severity and an aggregate probability	Choose an approximation or discretize, as appropriate	State the approximation or discretization and assess its effect.

What You'll Learn