Aggregate Excess-of-Loss Treaty Payments

What an Annual Aggregate Treaty Covers

An insurer may want protection against the total claims it pays over a year, rather than protection against any one unusually large claim. An annual aggregate excess-of-loss treaty responds when the portfolio’s aggregate loss for the treaty year exceeds a specified attachment. Its payment is determined from the total for the year, so several moderate claims can trigger a recovery even when none of them is large on its own.

Let $N$ be the number of claims in the year, and let $X_j$ be the amount of the $j$th claim before this treaty. The annual aggregate loss is $Z=\sum_{j=1}^{N}X_j$, with $Z=0$ when $N=0$. Let $A$ be the annual aggregate attachment and $L$ the maximum payment the reinsurer will make under the treaty. The attachment and limit apply to $Z$, not separately to each $X_j$.

Definition: Under an annual aggregate excess-of-loss treaty with attachment $A$ and payment limit $L$, the ceded payment for annual aggregate loss $Z$ is $C(Z)=\min\{(Z-A)_+,L\}$, where $(x)_+=\max(x,0)$. The retained loss is $R(Z)=Z-C(Z)$. The treaty begins to pay above $A$ and reaches its maximum payment $L$ when $Z$ reaches the exhaustion point $A+L$.

In words, the insurer retains the first $A$ dollars of annual aggregate loss. Above that amount, the reinsurer pays dollar for dollar until its total payment reaches $L$. Any further aggregate loss is retained by the insurer. The formulas presume that the treaty is applied to the stated annual losses without other adjustments, such as reinstatements, exclusions, or a separate coinsurance percentage.

A contract may instead describe its limit as an exhaustion point, the aggregate loss at which the cover is fully used. If that point is $E$, then the payment capacity is $L=E-A$, provided $E\geq A$. Always establish whether a stated limit caps the reinsurer’s payment or gives the aggregate loss at which the layer is exhausted.

Why the Payment Formula Has This Form

The payment is zero while the annual loss is at or below the attachment. Once the loss passes the attachment, the amount in the layer is the part above $A$, but the reinsurer cannot pay more than $L$. This gives three regions:

C(Z)= \begin{cases} 0, & Z\leq A,\\ Z-A, & A<Z<A+L,\\ L, & Z\geq A+L. \end{cases}

The positive-part expression $(Z-A)_+$ combines the first two regions: it is zero below the attachment and equals $Z-A$ above it. Taking the minimum with $L$ then caps the payment. Since the original loss is divided between insurer and reinsurer, retained loss is the original total less the recovery, $R(Z)=Z-C(Z)$. Equivalently, its piecewise form is:

R(Z)= \begin{cases} Z, & Z\leq A,\\ A, & A<Z<A+L,\\ Z-L, & Z\geq A+L. \end{cases}

Within the layer, the insurer’s retained loss is fixed at $A$; after the layer is exhausted, each additional dollar of loss is retained again. This also gives a useful check for any calculation: $C(Z)+R(Z)=Z$, with $0\leq C(Z)\leq L$.

Calculating Ceded and Retained Losses

Worked Example: Several Claims Trigger an Annual Layer

A hypothetical insurer has a fixed three-claim portfolio for a treaty year. Claim severities are independent and have a discrete distribution: $P(X=80{,}000)=0.25$, $P(X=140{,}000)=0.50$, and $P(X=220{,}000)=0.25$. The three realized claim amounts are $80{,}000$, $140{,}000$, and $220{,}000$. An annual aggregate treaty has attachment $A=\$250{,}000$ and payment limit $L=\$180{,}000$. Find the ceded and retained losses. Assume no discounting; mortality is not relevant.

Method. First aggregate all claims for the year, then apply the treaty formula to that annual total.

Step 1: Find annual aggregate loss.

Z=80{,}000+140{,}000+220{,}000=\$440{,}000.

Step 2: Find the amount above attachment. The excess over the annual attachment is $440{,}000-250{,}000=\$190{,}000$.

Step 3: Apply the payment limit.

C(Z)=\min(190{,}000,180{,}000)=\$180{,}000.

The exhaustion point is $A+L=250{,}000+180{,}000=\$430{,}000$. Because the aggregate loss of $\$440{,}000$ is above that point, the treaty is exhausted.

Step 4: Calculate retained loss and check.

R(Z)=440{,}000-180{,}000=\$260{,}000, \qquad C(Z)+R(Z)=180{,}000+260{,}000=\$440{,}000.

The result is a $\$180{,}000$ ceded payment and a $\$260{,}000$ retained loss. The check reproduces the original annual aggregate.

Worked Example: The Annual Total, Not Each Claim, Determines Recovery

For a hypothetical portfolio, the annual claim count $N$ has a Poisson distribution with parameter $\lambda=2.5$. Claim severities are independent of $N$ and independent of one another, with $P(X=40{,}000)=0.20$, $P(X=90{,}000)=0.50$, and $P(X=160{,}000)=0.30$. In one realized year, there are four claims of $\$40{,}000$, $\$90{,}000$, $\$160{,}000$, and $\$160{,}000$. The aggregate treaty has $A=\$300{,}000$ and $L=\$100{,}000$. Find the recovery and retained loss. Assume no discounting; mortality is not relevant.

Step 1: Aggregate the realized claims. The frequency distribution describes the claim count model, but the treaty payment for this realized year uses the actual count and claim amounts:

N=4,\qquad Z=40{,}000+90{,}000+160{,}000+160{,}000=\$450{,}000.

Step 2: Apply the annual attachment. The annual aggregate loss exceeds the attachment by $450{,}000-300{,}000=\$150{,}000$.

Step 3: Cap the recovery at the limit.

C(Z)=\min(150{,}000,100{,}000)=\$100{,}000.

The exhaustion point is $300{,}000+100{,}000=\$400{,}000$, so the treaty is exhausted. The insurer retains $450{,}000-100{,}000=\$350{,}000$. The excess above the exhaustion point is $\$50{,}000$, which is retained in addition to the attachment: $300{,}000+50{,}000=\$350{,}000$. Thus the allocation checks.

Worked Example: Reading an Exhaustion Point Correctly

In a hypothetical book of business, the annual aggregate loss $Z$ has the following discrete distribution: $P(Z=0)=0.10$, $P(Z=200{,}000)=0.20$, $P(Z=450{,}000)=0.30$, $P(Z=700{,}000)=0.25$, and $P(Z=900{,}000)=0.15$. A treaty has an annual attachment of $\$200{,}000$ and is described as exhausting at an aggregate loss of $\$500{,}000$. For each possible aggregate loss, determine the treaty payment and the insurer’s retained loss. Assume no discounting; mortality is not relevant.

Method. The stated $\$500{,}000$ is the exhaustion point, not the maximum payment. The payment limit is the width of the layer:

L=500{,}000-200{,}000=\$300{,}000.

Step 1: Calculate each payment. At $Z=0$ and $Z=200{,}000$, the loss does not exceed the attachment, so recovery is zero. At $Z=450{,}000$, the loss is $\$250{,}000$ above attachment, which is below the $\$300{,}000$ payment limit. At $Z=700{,}000$ and $Z=900{,}000$, the layer is exhausted, so recovery is $\$300{,}000$.

Step 2: Subtract each recovery from the corresponding aggregate loss.

Annual aggregate loss $Z$	Ceded payment $C(Z)$	Retained loss $R(Z)$	Check $C(Z)+R(Z)$
$\$0$	$\$0$	$\$0$	$\$0$
$\$200{,}000$	$\$0$	$\$200{,}000$	$\$200{,}000$
$\$450{,}000$	$\$250{,}000$	$\$200{,}000$	$\$450{,}000$
$\$700{,}000$	$\$300{,}000$	$\$400{,}000$	$\$700{,}000$
$\$900{,}000$	$\$300{,}000$	$\$600{,}000$	$\$900{,}000$

At $\$450{,}000$, the insurer retains exactly the $\$200{,}000$ attachment because the loss is inside the layer. At $\$700{,}000$, the insurer retains the attachment plus the $\$200{,}000$ above the $\$500{,}000$ exhaustion point. Each row satisfies the loss-allocation identity.

Common Exam Trap: Applying the Attachment Claim by Claim

An annual aggregate attachment is not a per-claim deductible. In the second worked example, the incorrect calculation would apply the $\$300{,}000$ attachment separately to each claim:

(40{,}000-300{,}000)_+ +(90{,}000-300{,}000)_+ +(160{,}000-300{,}000)_+ +(160{,}000-300{,}000)_+=0.

That calculation incorrectly gives a zero recovery because it treats the annual aggregate treaty as a per-claim treaty. The correct calculation first sums the claims, obtaining $Z=\$450{,}000$, and then applies the annual attachment and limit:

C(Z)=\min\{(450{,}000-300{,}000)_+,100{,}000\}=\$100{,}000.

Another frequent error is to subtract the attachment but forget the payment cap. If $Z>A+L$, the recovery is $L$, not $Z-A$. Conversely, if a contract gives an exhaustion point, do not treat that point itself as the payment limit; subtract the attachment to obtain the layer’s capacity.

Practical Calculation Sequence

Establish the treaty basis.
Confirm that the attachment and limit apply to the annual aggregate, and identify the claims included in that aggregate.

Find the annual loss.
Sum all covered claim amounts for the treaty year to obtain $Z$.

Interpret the stated limit.
If it is a payment cap, use it as $L$. If it is an exhaustion point $E$, calculate $L=E-A$.

Calculate the recovery and retention.
Use $C(Z)=\min\{(Z-A)_+,L\}$ and $R(Z)=Z-C(Z)$.

Check the result.
Verify $0\leq C(Z)\leq L$, $R(Z)\geq0$, and $C(Z)+R(Z)=Z$.

These calculations concern treaty payments for specified aggregate losses. They do not, by themselves, calculate the probability of reaching the attachment or the expected recovery over all possible years; those questions require working with the distribution of annual aggregate loss.

Key takeaway: For annual aggregate excess-of-loss, aggregate the year’s covered claims first. The reinsurer pays the part of the annual total above the attachment, capped at the payment limit; the insurer retains the balance.

Check Your Understanding

For each hypothetical treaty, apply the annual aggregate attachment to the total annual loss and check the allocation.

A fixed three-claim portfolio has realized losses of $\$55{,}000$, $\$85{,}000$, and $\$120{,}000$. Its severity model is discrete with probabilities $0.25, 0.50, 0.25$ at $\$55{,}000$, $\$85{,}000$, and $\$120{,}000$, respectively. Find ceded and retained losses with attachment $\$200{,}000$ and payment limit $\$75{,}000$. (Answer: Ceded $\$60{,}000$; retained $\$200{,}000$.)
A hypothetical year has $N=3$ claims, each with severity distribution $P(X=\$70{,}000)=0.4$, $P(X=\$130{,}000)=0.6$, and realized amounts $\$70{,}000,\$130{,}000,\$130{,}000$. With annual attachment $\$250{,}000$ and payment limit $\$120{,}000$, find the exhaustion point, recovery, and retained loss. (Answer: Exhaustion point $\$370{,}000$; ceded $\$80{,}000$; retained $\$250{,}000$.)
A discrete annual aggregate loss has possible values $\$100{,}000$, $\$350{,}000$, and $\$800{,}000$, with probabilities $0.2$, $0.5$, and $0.3$. A treaty has attachment $\$300{,}000$ and payment cap $\$250{,}000$. Find ceded and retained loss for each possible aggregate. (Answer: For the three aggregate losses in order, ceded losses are $\$0,\ \$50{,}000,\ \$250{,}000$, and retained losses are $\$100{,}000,\ \$300{,}000,\ \$550{,}000$.)
A hypothetical treaty attaches at $\$450{,}000$ and is described as exhausting at $\$700{,}000$. For an annual aggregate loss of $\$820{,}000$, find the payment limit, recovery, and retained loss. (Answer: Payment limit $\$250{,}000$; ceded $\$250{,}000$; retained $\$570{,}000$.)
A portfolio has a Poisson claim count with parameter $\lambda=3$ and independent severities with $P(X=\$60{,}000)=0.5$ and $P(X=\$180{,}000)=0.5$. In a realized year, there are five claims: $\$60{,}000,\ \$60{,}000,\ \$180{,}000,\ \$180{,}000,\ \$180{,}000$. Calculate the recovery and retained loss for attachment $\$500{,}000$ and payment limit $\$200{,}000$. (Answer: Annual aggregate $\$660{,}000$; ceded $\$160{,}000$; retained $\$500{,}000$.)

Annual aggregate loss \(Z\)	Ceded payment \(C(Z)\)	Retained loss \(R(Z)\)	Check \(C(Z)+R(Z)\)
\(\$0\)	\(\$0\)	\(\$0\)	\(\$0\)
\(\$200{,}000\)	\(\$0\)	\(\$200{,}000\)	\(\$200{,}000\)
\(\$450{,}000\)	\(\$250{,}000\)	\(\$200{,}000\)	\(\$450{,}000\)
\(\$700{,}000\)	\(\$300{,}000\)	\(\$400{,}000\)	\(\$700{,}000\)
\(\$900{,}000\)	\(\$300{,}000\)	\(\$600{,}000\)	\(\$900{,}000\)

What You'll Learn