Why the Unit of Coverage Matters
One severe event can generate many covered claims. A per-occurrence excess-of-loss contract responds to the combined covered loss assigned to that occurrence, rather than treating each claim as a separate attachment calculation. That difference can materially change how much the reinsurer pays.
The key steps are to identify which claims belong to an occurrence under the contract, add their covered amounts, and apply the occurrence retention and limit to that total. The occurrence grouping is a contractual input: the same underlying losses can produce different results if the agreement assigns them to different occurrences.
The layer attaches above \(R\) and extends \(L\) dollars above the retention. Its upper boundary is therefore \(R+L\). If the occurrence loss is at or below \(R\), the layer pays nothing. If it lies between \(R\) and \(R+L\), the layer pays the amount above \(R\). If it exceeds \(R+L\), the layer is exhausted and pays \(L\).
This calculation assumes that the stated \(X_j\) are the amounts that count toward the treaty’s covered loss, that the contract assigns the claims to the stated occurrence, and that the retention and limit are expressed on the same basis and in the same currency. It calculates the layer’s contractual loss payment, not the price of reinsurance or the timing of cash flows. If the agreement defines covered loss, occurrence grouping, or the limit differently, those terms govern; the formula must be adjusted to match them. No annual aggregate cap or other limit is implied by this one-occurrence formula.
From Occurrence Loss to the Two Shares
Once the recovery \(C\) is determined, the insurer’s net loss for the occurrence is \(N=S-C\). In this basic calculation, the covered gross loss is divided between the insurer and reinsurer: \(S=N+C\). The insurer retains the portion up to the attachment and any amount above the layer’s top.
The middle case follows because the reinsurer pays the part of the occurrence loss above the insurer’s retention. The first case follows because there is no loss above the retention. In the last case, the amount above the retention is at least \(L\), but the contract limits the payment to \(L\). These cases are exactly represented by \(\min\{\max(S-R,0),L\}\): the inner maximum prevents a negative recovery, and the outer minimum caps the payment.
Use the contract’s occurrence basis to decide which losses belong together. Do not infer the grouping solely from the order or number of claims.
Add the covered claim amounts assigned to that occurrence: \(S=\sum_j X_j\).
Subtract the retention, floor the result at zero, and cap it at the per-occurrence limit.
Calculate \(N=S-C\), then check that gross loss equals insurer net loss plus ceded recovery.
Worked Applications
In the examples below, dollar amounts are stated in thousands of dollars (so, for example, 100 means $100,000). The examples are hypothetical. Unless a probability distribution is explicitly given, each example is a deterministic calculation using the stated losses. There is no mortality assumption, interest assumption, or discounting in these loss-share calculations.
Worked Example: One Retention for the Occurrence
A hypothetical property treaty has a per-occurrence retention of \(R=100\) and a per-occurrence limit of \(L=300\). Three covered claims assigned to one occurrence are 120, 180, and 100. Calculate the occurrence loss, the reinsurer’s recovery, and the insurer’s net loss.
Method and assumption. Treat the three stated claim amounts as covered losses in thousands of dollars and group them as one occurrence. The same retention applies once to their combined occurrence loss.
Step 1: Aggregate the claims. The occurrence loss is \(S=120+180+100=400\).
Step 2: Apply the retention. Loss above the retention is \(S-R=400-100=300\).
Step 3: Apply the limit. The recovery is \(C=\min\{\max(400-100,0),300\}=\min(300,300)=300\). The layer is exactly exhausted.
Step 4: Calculate and check the insurer’s net loss. \(N=400-300=100\). The shares reconcile: \(N+C=100+300=400=S\).
Common mistake avoided. Applying the 100 retention separately to each claim would produce \(120-100=20\), \(180-100=80\), and \(\max(100-100,0)=0\), for a total of \(20+80+0=100\) in thousands. That is a per-claim calculation, not the stated per-occurrence calculation. The correct per-occurrence recovery is 300 in thousands, or $300,000.
Worked Example: Separate Occurrences and Layer Exhaustion
A hypothetical catastrophe contract provides a per-occurrence retention of \(R=200\) and a per-occurrence limit of \(L=300\), both in thousands of dollars. Under the contract’s stated grouping, Occurrence A produces covered claims of 90, 160, and 210; Occurrence B produces claims of 80 and 120. Find the recovery and insurer net loss for each occurrence, then the totals.
Method and assumption. Calculate each occurrence independently. Each has its own retention and its own limit; these two event totals are not combined for this per-occurrence calculation.
Occurrence A. Its loss is \(S_A=90+160+210=460\). The amount above the retention is \(460-200=260\), below the 300 limit. Thus \(C_A=\min(260,300)=260\), and \(N_A=460-260=200\). Check: \(200+260=460\).
Occurrence B. Its loss is \(S_B=80+120=200\). It does not exceed the retention, so \(C_B=\min\{\max(200-200,0),300\}=0\). The insurer retains \(N_B=200-0=200\). Check: \(200+0=200\).
Totals. Gross loss is \(460+200=660\), total recovery is \(260+0=260\), and total insurer net loss is \(200+200=400\). The reconciliation is \(260+400=660\).
Common mistake avoided. Combining both events first would give \(660\) above one retention of 200 and a capped recovery of 300. That incorrectly treats two contractually distinct occurrences as one. The correct total is 260 because the second occurrence does not penetrate its own retention.
Worked Example: Expected Recovery for a Random Occurrence Loss
For a hypothetical treaty, a single occurrence has aggregate covered loss \(S\), measured in thousands of dollars, with this discrete distribution: \(P(S=100)=0.20\), \(P(S=400)=0.50\), and \(P(S=800)=0.30\). The per-occurrence retention is \(R=250\) and the limit is \(L=400\). Calculate the expected recovery and expected insurer net loss.
Method and assumptions. The given probabilities are the complete distribution of the aggregate loss for one occurrence and sum to \(0.20+0.50+0.30=1.00\). Apply the treaty separately to each possible occurrence total, then probability-weight the outcomes. No interest or discounting is assumed.
At \(S=100\). Since \(100\le250\), recovery is \(C=0\), and net loss is \(N=100\).
At \(S=400\). The loss above retention is \(400-250=150\), below the 400 limit. Thus \(C=150\) and \(N=400-150=250\).
At \(S=800\). Loss above retention is \(800-250=550\), which exceeds the limit. Thus \(C=400\) and \(N=800-400=400\).
Expected recovery. \(E[C]=0.20(0)+0.50(150)+0.30(400)=0+75+120=195\) in thousands of dollars.
Expected net loss and check. \(E[N]=0.20(100)+0.50(250)+0.30(400)=20+125+120=265\). The expected gross loss is \(E[S]=0.20(100)+0.50(400)+0.30(800)=20+200+240=460\). Therefore \(E[N]+E[C]=265+195=460=E[S]\). Expected recovery is $195,000 and expected net loss is $265,000.
Common mistake avoided. Capping the loss itself at 400 would confuse the layer limit with the top of the layer. The limit caps the reinsurer’s payment, not the occurrence loss; at \(S=800\), the insurer still retains 400 above the layer.
Interpretation and Exam Traps
A per-occurrence layer responds to the size of the aggregate loss assigned to one occurrence. The claim count alone does not determine the recovery: several small claims can collectively penetrate the retention, while a different group of claims with the same total can produce a different result if the contract assigns them to separate occurrences. Always identify the aggregation unit before doing arithmetic.
| Occurrence loss | Layer response | Insurer’s net position |
|---|---|---|
| \(S\le R\) | No recovery | Retains all of \(S\) |
| \(R<S<R+L\) | Pays \(S-R\) | Retains \(R\) |
| \(S\ge R+L\) | Pays \(L\) | Retains \(S-L\), including loss above the layer |
A frequent limit error is to treat \(L\) as the maximum occurrence loss covered. It is instead the maximum recovery for the occurrence. The reinsurer’s layer spans from \(R\) to \(R+L\); the insured loss can exceed \(R+L\), in which case the insurer bears the amount above the layer.
Another error is to apply a retention per claim when the contract applies it per occurrence, or to apply one retention to losses from separate occurrences. A reliable check is to write the occurrence grouping and the units next to the loss amounts, calculate each event separately, and verify \(S=N+C\) for every event before summing results.
Check Your Understanding
For each hypothetical problem, amounts are in thousands of dollars. Assume the stated claim groupings and treaty terms apply as written.
- A per-occurrence layer has \(R=75\) and \(L=125\). One occurrence has covered claims of 30, 60, and 110. Find the recovery and insurer net loss. (Answer: \(S=200\), recovery \(=125\), net loss \(=75\).)
- A layer has \(R=150\) and \(L=250\). One occurrence has claims of 90, 100, and 210. Find the recovery and determine whether the layer is exhausted. (Answer: \(S=400\), recovery \(=250\), net loss \(=150\); the layer is exhausted.)
- A treaty has \(R=200\) and \(L=250\). Two separate occurrences have losses of 260 and 540. Calculate the total recovery across both occurrences. (Answer: \(60+250=310\).)
- For one occurrence, \(P(S=100)=0.25\), \(P(S=300)=0.50\), and \(P(S=700)=0.25\). A layer has \(R=200\) and \(L=300\). Find the expected recovery. (Answer: \(0.25(0)+0.50(100)+0.25(300)=125\).)
- Two covered losses of 180 each are assigned either to two separate occurrences or, under an alternative stated grouping, to one occurrence. The layer has \(R=200\) and \(L=300\). Find the total recovery under each grouping. (Answer: Separate occurrences: \(0+0=0\). One occurrence: \(S=360\), recovery \(=160\).)