Worked Case: Compound Loss with Aggregate Excess-of-Loss

Why the Aggregate Distribution Matters

Per-claim coverage changes the amount paid for each claim. An aggregate excess-of-loss treaty then operates on the sum of those modified payments over the treaty period. The order matters: first determine each claim’s covered payment, then construct the distribution of the total, and only then apply the aggregate attachment and limit.

This tutorial develops that calculation for discrete models and a Poisson-frequency example. All examples use one treaty period and dollar amounts, with no discounting. Interest and mortality assumptions are not relevant to these single-period loss models. The frequency and severity assumptions are stated in each example.

Definition: Let $N$ be the number of claims in a treaty period. Let $Y_j^L$ be the covered payment for claim $j$, after any per-claim coverage modification, and assume the claim payments are independent and identically distributed and independent of $N$. The aggregate covered loss is $$ T=\sum_{j=1}^{N}Y_j^L, $$ where the sum is zero when $N=0$. For aggregate attachment $A$ and aggregate limit $L$, the reinsurer’s payment is $R=\min((T-A)_+,L)$, where $x_+=\max(x,0)$. The insurer’s retained covered loss is $T-R$.

The attachment $A$ is the amount of aggregate covered loss the insurer bears before the treaty responds. The limit $L$ is the most the reinsurer pays. Thus a total below or equal to $A$ produces no recovery; a total between $A$ and $A+L$ produces a recovery equal to the excess over $A$; and a total at or above $A+L$ produces the full limit.

Constructing the Compound Aggregate Distribution

When $N$ is random, $T$ is a compound distribution: a random number of random claim payments are added. For discrete payments, one direct method is to calculate the total distribution conditional on each possible value of $N$, then weight each conditional distribution by $P(N=n)$. Equivalently, use convolution: the distribution of a sum of $n$ independent payments is the $n$-fold convolution of the distribution of $Y^L$.

A probability-generating function gives a compact version of the same calculation. For a nonnegative integer-valued random variable $W$, its probability-generating function is $G_W(z)=E[z^W]$. If the claim payments are recorded in integer multiples of a common unit, independence gives

G_T(z)=G_N(G_{Y^L}(z)).

For example, conditional on $N=n$, the generating function of the total is $(G_{Y^L}(z))^n$. Averaging over the possible claim counts gives $E[(G_{Y^L}(z))^N]=G_N(G_{Y^L}(z))$. This derivation requires independent, identically distributed claim payments independent of the frequency. If the payment distributions differ by claim, or depend on $N$, this substitution does not apply without modification.

Once the probabilities $P(T=t)$ are known, apply the treaty separately to each aggregate outcome. For a discrete aggregate distribution,

E[R]=\sum_t P(T=t)\min((t-A)_+,L), \qquad E[T-R]=E[T]-E[R].

The second equality follows because the insurer’s retained amount is $T-R$ for every outcome. It is also a useful check on the calculation: expected gross covered loss must equal expected reinsurer recovery plus expected retained loss.

Fix the payment basis.
Specify the per-claim payment $Y^L$ after any deductible or per-claim limit. Do not apply those modifications again when aggregating.

Combine frequency and severity.
Use conditional distributions, convolution, or a probability-generating function to find the distribution of $T$.

Apply the aggregate layer.
For each possible $T=t$, calculate $R=\min((t-A)_+,L)$, and then calculate $t-R$.

Average and check.
Weight the payments by their aggregate probabilities and verify that expected gross loss equals expected recovery plus expected retention.

Worked Aggregate Distribution and Treaty Recovery

Worked Example: Binomial Frequency with Deductible-Modified Payments

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$ claims in one period. Conditional on a claim, the ground-up loss $X$ is independently equal to $100, $200, or $300, each with probability $1/3$. The policy has a per-claim deductible $d=\$100$ and a maximum covered payment $u=\$200$, so $Y^L=\min((X-d)_+,u)$. An aggregate excess-of-loss treaty attaches at $A=\$200$ and has limit $L=\$150$. Find the aggregate distribution, expected recovery, and expected retained loss.

Method and assumptions. Use conditional convolution over the binomial claim count. Claim severities are independent of each other and of $N$. No other coverage modification applies. The claim payment outcomes are:

X=\$100,\ \$200,\ \$300 \quad\Longrightarrow\quad Y^L=\$0,\ \$100,\ \$200,

each with probability $1/3$. In particular, the deductible-modified payment can be zero even when a claim occurs.

Step 1: Condition on the claim count. The frequency probabilities are $P(N=0)=1/4$, $P(N=1)=1/2$, and $P(N=2)=1/4$. If $N=0$, then $T=0$. If $N=1$, the total is 0, 100, or 200 dollars, each with conditional probability $1/3$. If $N=2$, convolving the payment distribution with itself gives totals 0, 100, 200, 300, and 400 dollars with conditional probabilities $1/9,2/9,3/9,2/9,1/9$, respectively.

Step 2: Weight and combine. For example, the probability of a total of $200 is $P(N=1)(1/3)+P(N=2)(3/9)=1/6+1/12=1/4$. Combining all contributions gives:

\begin{aligned} P(T=0)&=\frac14+\frac12\left(\frac13\right)+\frac14\left(\frac19\right)=\frac49,\\ P(T=100)&=\frac12\left(\frac13\right)+\frac14\left(\frac29\right)=\frac29,\\ P(T=200)&=\frac12\left(\frac13\right)+\frac14\left(\frac39\right)=\frac14,\\ P(T=300)&=\frac14\left(\frac29\right)=\frac1{18},\\ P(T=400)&=\frac14\left(\frac19\right)=\frac1{36}. \end{aligned}

The probabilities sum to $16/36+8/36+9/36+2/36+1/36=1$.

Step 3: Apply the aggregate treaty. At totals of $0, $100, and $200, the recovery is zero. At $300, it is $\min(300-200,150)=\$100$; at $400, it is $\min(400-200,150)=\$150$. Therefore,

E[R]=100\left(\frac1{18}\right)+150\left(\frac1{36}\right) =\frac{350}{36} =\$9.72.

The expected gross covered loss is

E[T]=100\left(\frac29\right)+200\left(\frac14\right) +300\left(\frac1{18}\right)+400\left(\frac1{36}\right)=\$100.

Thus expected retained loss is $E[T]-E[R]=100-350/36=\$90.28$. As a separate check, the frequency-severity mean is $E[N]E[Y^L]=1(0+100+200)/3=\$100$, agreeing with the aggregate-distribution calculation. The method avoids treating the aggregate limit as a per-claim limit.

Further Worked Cases

Worked Example: Fixed Claim Count and a Capped Aggregate Layer

A hypothetical two-claim portfolio has exactly $N=2$ claims. Each claim’s payment after all per-claim coverage modifications is independently $100 or $200, each with probability $1/2$. An aggregate treaty attaches at $A=\$250$ and has limit $L=\$100$. Find the aggregate distribution, expected recovery, and expected retained amount. The amounts are one-period dollars, without discounting.

Method. Convolve the two modified-payment distributions. The possible totals and probabilities are $P(T=200)=1/4$, $P(T=300)=1/2$, and $P(T=400)=1/4$. The corresponding recoveries are $\min((200-250)_+,100)=0$, $\min(300-250,100)=\$50$, and $\min(400-250,100)=\$100$.

Therefore $E[R]=0(1/4)+50(1/2)+100(1/4)=\$50$. The expected gross total is $E[T]=2(100+200)/2=\$300$, so the expected retained amount is $300-50=\$250$. The check also follows outcome by outcome: the insurer retains $200, $250, or $300 at the three totals, and their probability-weighted mean is $200(1/4)+250(1/2)+300(1/4)=\$250$. This calculation avoids attaching the treaty separately to each of the two claims.

Worked Example: Poisson Frequency with Deterministic Modified Payment

A hypothetical portfolio has $N\sim\operatorname{Poisson}(\lambda=1)$ claims in a period. Each ground-up claim is exactly $150; a per-claim deductible of $50 and a maximum covered payment of $100 produce $Y^L=\$100$ for every claim. An aggregate treaty attaches at $A=\$200$ and has limit $L=\$200$. Find the expected recovery and expected retained amount. The model is undiscounted and has no mortality component.

Method. Since $T=100N$, the recovery is $0 for $N\leq2$, $100 for $N=3$, and $200 for $N\geq4$. The Poisson probabilities give $P(N=3)=e^{-1}/3!=1/(6e)$. Also,

P(N\geq4)=1-\sum_{n=0}^{3}\frac{e^{-1}}{n!} =1-\frac{8}{3e}.

Hence

E[R]=100\left(\frac{1}{6e}\right) +200\left(1-\frac{8}{3e}\right) =200-\frac{1550}{3e} =\$9.93.

Here $e$ is the base of the natural logarithm; rounding is to the nearest cent. Since $E[T]=E[N]E[Y^L]=1(\$100)=\$100$, expected retained loss is $100-9.93=\$90.07$. The recovery is small because the treaty responds only when at least three claims occur, and the full limit requires at least four. This method avoids replacing the Poisson frequency with its mean: using $N=1$ as though it were certain would incorrectly produce no recovery and would discard the frequency variation that drives the treaty payment.

Why the Order of Operations Is an Exam Trap

An aggregate treaty is applied to $T$, not to the original ground-up severities and not separately to each claim. Per-claim deductibles and limits determine $Y^L$; the frequency model determines how many such payments are added; the aggregate treaty then acts on their sum. Mixing these stages changes the distribution and therefore the expected recovery.

For example, in the first worked example, a common wrong calculation is to treat the $150 aggregate limit as a cap on the total, writing $R=\min(T,150)$. That gives a recovery of $150 at each of $T=\$200,\$300,\$400$, and an incorrect expected recovery of $150(1/4+1/18+1/36)=\$58.33$. The correct formula first subtracts the $200 attachment and then caps the excess: $R=\min((T-200)_+,150)$. It gives recoveries of $0, $0, $0, $100, and $150 at totals of $0, $100, $200, $300, and $400, respectively, and the correct mean is $9.72.

Other useful checks are to confirm that the aggregate probabilities sum to one, that $0\leq R\leq L$ for every outcome, and that $0\leq T-R\leq T$. If $E[Y^L]$ is finite and the frequency is independent of severity, then $E[T]=E[N]E[Y^L]$. This mean check does not determine an aggregate treaty recovery: the recovery also depends on the full distribution of $T$, especially the probability of exceeding the attachment.

Key takeaway: Build the distribution of the aggregate of contract-modified claim payments before evaluating an aggregate excess-of-loss treaty. Apply the attachment to the aggregate total, cap the excess at the aggregate limit, and average the resulting payments over that total distribution.

Check Your Understanding

For each problem, use the stated claim-payment basis and treaty terms. Amounts are undiscounted dollars for one period.

A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$, and each claim payment after per-claim modifications is exactly $100. An aggregate treaty attaches at $100 and has limit $100. Find the expected recovery. (Answer: $25.00)
There are exactly two independent claims. Each modified payment is $50 or $150 with equal probability. An aggregate treaty attaches at $150 and has limit $100. Find the expected recovery. (Answer: $50.00)
A portfolio has $N\sim\operatorname{Poisson}(2)$, and each modified claim payment is exactly $100. An aggregate treaty attaches at $100 and has limit $100. Find the expected recovery. (Answer: $100(1-3e^{-2})=\$59.40$, rounded to cents)
A portfolio has $N\sim\operatorname{Binomial}(3,0.5)$, and each modified payment is exactly $100. An aggregate treaty attaches at $100 and has limit $200. Find the expected recovery. (Answer: $62.50)
There are exactly two independent claims. Each payment after a per-claim deductible is $0 or $100 with equal probability. An aggregate treaty attaches at $50 and has limit $100. Find the expected recovery and expected retained amount. (Answer: expected recovery $50.00; expected retained amount $50.00)

What You'll Learn