Comparing Occurrence and Aggregate Reinsurance

Why the Cover Structure Matters

Two reinsurance contracts can respond very differently to the same year's catastrophe losses. A per-occurrence cover responds to each contract-defined event separately. An annual aggregate cover responds to the total eligible losses over the contract year. Consequently, an insurer's protection depends not only on how much loss occurs, but also on how those losses are divided among events.

The calculations in this tutorial use hypothetical contracts and losses. Amounts in the examples are in millions of dollars. Unless stated otherwise, there is no reinstatement premium, no other recovery, and no dispute about which losses are covered. Actual contract wording determines such matters as event definition, hours clauses, and the order in which multiple covers apply.

Definition: For a per-occurrence excess-of-loss cover with retention \(R\) and limit \(L\), the recovery on covered loss \(X_j\) from event \(j\) is \(C_j=\min\{(X_j-R)^+,L\}\), where \(z^+=\max(z,0)\). For a standalone annual aggregate excess-of-loss cover with annual attachment \(A\), annual limit \(G\), and annual eligible losses \(S=\sum_j X_j\), the recovery is \(C_A=\min\{(S-A)^+,G\}\). The occurrence retention applies separately to each event; the annual aggregate attachment applies once to the annual total.

How an Occurrence Layer Pays

An occurrence cover has a retention and a limit for each covered event. Loss below the retention remains with the insurer. Loss above the retention is recoverable, up to the per-occurrence limit. For a year with \(m\) events, absent reinstatements or other annual restrictions, total recovery is \(\sum_{j=1}^{m} C_j\).

The formula follows directly from the layer's payment boundaries. If \(X_j\le R\), there is no loss above the retention and recovery is zero. If \(R<X_j<R+L\), the layer pays the excess \(X_j-R\). If \(X_j\ge R+L\), the loss exhausts the layer, so recovery cannot exceed \(L\). Combining those cases gives \(\min\{(X_j-R)^+,L\}\).

C_j= \begin{cases} 0, & X_j\le R,\\ X_j-R, & R<X_j<R+L,\\ L, & X_j\ge R+L. \end{cases}

Worked Example: Applying an Occurrence Layer to Unequal Events

A hypothetical insurer buys a \(10\) million limit excess of \(5\) million per-occurrence cover. Three covered event losses are \(3\), \(8\), and \(18\) million. Calculate the recovery and the insurer's retained loss for each event and in total. Assume there is no reinstatement.

Method. Apply the retention and limit separately to each event using \(C_j=\min\{(X_j-R)^+,L\}\), with \(R=5\) and \(L=10\).

Event 1: \(X_1=3\). The amount above the retention is \((3-5)^+=0\), so \(C_1=\min(0,10)=0\). The insurer retains \(3-0=3\) million.

Event 2: \(X_2=8\). The amount above the retention is \(8-5=3\), so \(C_2=\min(3,10)=3\). The insurer retains \(8-3=5\) million.

Event 3: \(X_3=18\). The amount above the retention is \(18-5=13\), so \(C_3=\min(13,10)=10\). The insurer retains \(18-10=8\) million.

Total recovery is \(0+3+10=13\) million. Total gross loss is \(3+8+18=29\) million, so total retained loss is \(29-13=16\) million. The check also agrees with the event-level retained amounts: \(3+5+8=16\) million.

Common mistake avoided: The third event does not produce a \(13\) million recovery. That is the amount above the retention, but the \(10\) million limit caps the recovery.

How an Annual Aggregate Layer Pays

An annual aggregate cover applies its attachment to the sum of eligible losses during the contract year. Under the standalone structure defined here, it does not first subtract a separate occurrence retention from each event. It pays the part of annual loss above \(A\), subject to the annual limit \(G\).

The order of the operations matters: first total the eligible losses, then subtract the annual attachment, then cap the result at the annual limit. This cover can respond to many modest losses even when none of them would reach a per-occurrence retention. Conversely, a large annual total can exhaust the aggregate limit.

Worked Example: Comparing Covers on the Same Annual Losses

A hypothetical insurer has three covered event losses of \(3\), \(8\), and \(18\) million. Compare (a) a \(10\) million limit excess of \(5\) million per-occurrence cover with no reinstatement, and (b) a standalone annual aggregate cover attaching at \(20\) million with a \(10\) million annual limit.

Method for the occurrence cover. Calculate each event recovery separately. The recoveries are \(\min((3-5)^+,10)=0\), \(\min((8-5)^+,10)=3\), and \(\min((18-5)^+,10)=10\), all in millions. Total occurrence recovery is \(0+3+10=13\) million.

Method for the aggregate cover. First sum the annual losses: \(S=3+8+18=29\) million. The amount above the annual attachment is \(29-20=9\) million. The annual limit does not bind, because \(9<10\), so aggregate recovery is \(\min(9,10)=9\) million.

The occurrence cover therefore recovers \(13\) million, while the aggregate cover recovers \(9\) million on this loss pattern. Their respective retained totals are \(29-13=16\) million and \(29-9=20\) million. These results compare the stated contracts only; they do not imply that one type always pays more.

Common mistake avoided: Applying the annual attachment to each event separately would calculate a different contract. Here, the aggregate attachment is applied once to the annual sum.

Event Count, Loss Size, and Reinstatements

An occurrence cover is sensitive to how loss is divided among events because it applies the retention and limit repeatedly. An annual aggregate cover, as defined here, is sensitive to the annual total rather than the event-by-event division. This difference can be isolated by comparing loss patterns with the same annual total.

Worked Example: Repeated Losses Versus Fewer Large Events

Compare two hypothetical annual loss patterns, each totaling \(20\) million. Pattern A has five events of \(4\) million each. Pattern B has two events of \(14\) and \(6\) million. The insurer has a \(8\) million limit excess of \(5\) million per-occurrence cover, without reinstatement. It also considers a standalone annual aggregate cover attaching at \(12\) million with a \(10\) million limit. Calculate each cover's recovery under each pattern.

Occurrence cover, Pattern A: For each \(4\) million event, \((4-5)^+=0\). The five recoveries are \(0,0,0,0,0\), for a total of \(0\) million.

Occurrence cover, Pattern B: For the \(14\) million event, the amount above retention is \(14-5=9\), capped at \(8\), giving an \(8\) million recovery. For the \(6\) million event, the amount above retention is \(6-5=1\), giving a \(1\) million recovery. Total occurrence recovery is \(8+1=9\) million.

Aggregate cover, both patterns: Each pattern totals \(20\) million. The amount above the \(12\) million attachment is \(20-12=8\) million, below the \(10\) million limit. Thus aggregate recovery is \(8\) million for either pattern.

The aggregate cover produces the same recovery for the two patterns because their annual totals match. The occurrence cover produces different recoveries because its attachment is applied separately to every event. This example also shows why a sequence of losses that individually falls below an occurrence retention may still exceed an annual aggregate attachment.

Common mistake avoided: Do not apply the occurrence retention just once to the annual sum, or apply the aggregate attachment once per event. Those calculations switch the contract structures.

An occurrence limit can be exhausted by one event. A reinstatement restores some or all of that limit for later events, subject to the contract's terms. With one full reinstatement, the insurer generally has two per-occurrence limits available over the year in total; the restored amount is not an increase to the limit available for one event. Reinstatement provisions may also specify whether restoration is automatic, when it takes effect, and whether an additional premium is due.

Worked Example: Tracking a Full Reinstatement

A hypothetical insurer buys a \(5\) million limit excess of \(5\) million per-occurrence cover with one full reinstatement. For simplicity, assume the reinstatement is available immediately after the first limit is exhausted and has no additional premium. Three successive event losses are \(12\), \(12\), and \(8\) million. Calculate recoveries in order and compare with the same cover without reinstatement.

Method. For each event, first calculate its recovery under the per-occurrence layer. Then track the total available annual capacity. One initial limit plus one full reinstatement gives maximum annual recovery capacity of \(5+5=10\) million.

First event: The amount above retention is \(12-5=7\) million, but the per-occurrence limit is \(5\) million. Recovery is \(5\) million, exhausting the initial capacity. The full reinstatement restores \(5\) million for later events.

Second event: Its layer calculation is also \(\min(12-5,5)=5\) million. The reinstated capacity is exhausted, bringing cumulative recovery to \(5+5=10\) million.

Third event: Its layer calculation would be \(\min(8-5,5)=3\) million if capacity remained. No annual capacity remains, so actual recovery is \(0\) million. Total recovery is \(10\) million.

Without the reinstatement, only the initial \(5\) million capacity is available. The first event uses all of it, so the second and third recoveries are both zero; total recovery is \(5\) million. The insurer's total gross loss is \(12+12+8=32\) million, giving retained loss of \(32-10=22\) million with reinstatement and \(32-5=27\) million without it. Each result checks as gross loss minus recovery.

Common mistake avoided: The nominal event-by-event amounts \(5+5+3=13\) million exceed the \(10\) million total capacity. The reinstatement restores only one extra \(5\) million limit, so the third event cannot recover after both limits have been used.

Contract Structure and Calculation Checks

A reliable calculation begins by identifying exactly what the contract attaches to and what its limit caps. A standalone aggregate cover is not automatically equivalent to an occurrence cover with the same stated dollar attachment or limit. Some programs combine covers or define aggregate protection on losses remaining after other recoveries. In that case, the calculation must follow the contract's specified order; do not silently treat a combined program as the standalone aggregate cover used in the examples.

Identify the unit of application.
Determine whether the retention or attachment applies to each event or once to the annual total.

Calculate eligible loss.
Use the loss basis stated in the contract, and establish which events and amounts are covered before applying the layer.

Apply the attachment and limit.
For occurrence cover, calculate the layer event by event. For standalone aggregate cover, total eligible annual losses first.

Track annual capacity.
Account for reinstatements and any annual aggregate cap; do not allow total recovery to exceed capacity actually available.

Reconcile the result.
Check that recovery is nonnegative, respects all limits, and that gross loss equals recovery plus retained loss.

A frequent exam trap is to calculate the amount above an attachment and stop without applying the limit. Another is to treat a per-occurrence retention as if it applied once to the year. In every calculation, label the units—per event or per year—and state whether any reinstatement changes available capacity. Those checks make it easier to detect a plausible-looking but contractually incorrect answer.

Key takeaway: Occurrence covers apply a retention and limit event by event; annual aggregate covers apply an attachment and limit to an annual loss total. Reinstatements can restore occurrence capacity for later events, but do not remove the per-event limit or create unlimited annual recovery.

Check Your Understanding

For each hypothetical contract, state your calculation basis and track the applicable limits.

A \(9\) million limit excess of \(4\) million per-occurrence cover has no reinstatement. Covered event losses are \(3\), \(10\), and \(16\) million. Calculate each recovery and total recovery. (Answer: recoveries are \(0\), \(6\), and \(9\) million; total recovery is \(15\) million.)
A standalone annual aggregate cover attaches at \(12\) million and has a \(15\) million annual limit. Annual eligible losses are \(5\), \(4\), \(8\), and \(6\) million. Calculate recovery and retained loss. (Answer: recovery is \(11\) million and retained loss is \(12\) million.)
A \(6\) million limit excess of \(4\) million per-occurrence cover has no reinstatement. Event losses are \(9\), \(14\), and \(5\) million. Calculate total recovery, including the limit check on each event. (Answer: recoveries are \(5\), \(6\), and \(1\) million; total recovery is \(12\) million.)
Losses are \(12\), \(12\), and \(12\) million. Compare a \(8\) million limit excess of \(5\) million per occurrence with no reinstatement against a standalone annual aggregate cover attaching at \(18\) million with a \(12\) million limit. (Answer: occurrence recovery is \(7+7+7=21\) million; aggregate recovery is \(12\) million.)
A \(4\) million limit excess of \(3\) million per occurrence has one full reinstatement, available after the initial limit is exhausted. Three successive losses are each \(10\) million. Assuming no other annual cap, calculate the actual recovery for each event and in total. (Answer: recoveries are \(4\), \(4\), and \(0\) million; total recovery is \(8\) million.)

What You'll Learn