Constructing an Occurrence Loss Distribution

From Event Catalog to Annual Occurrence Loss

A catastrophe event catalog describes possible events, their modeled losses, and how often they are expected to occur. To assess occurrence risk over a year, those ingredients must be combined into a probability distribution. In this tutorial, the occurrence loss is the largest loss from any one modeled event during the year. It is not the sum of losses from multiple events.

That distinction matters whenever several events can occur in one year. For example, a year with two moderate events has an annual maximum equal to the larger event loss, even though its annual aggregate loss is their sum. The annual maximum is useful for questions about the size of the largest single occurrence. The construction below is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let event type $j$ have deterministic loss $l_j\geq 0$ and annual occurrence rate $\lambda_j$. Assume the number of occurrences of each type in a year is Poisson with mean $\lambda_j$, and that the counts for different event types are independent. The annual occurrence loss $M$ is the largest event loss in that year, with $M=0$ if no positive-loss event occurs.

The rate $\lambda_j$ is an expected number of occurrences per year; it is not generally the probability of at least one occurrence. Under a Poisson model, the probability of at least one occurrence of type $j$ in one year is $1-e^{-\lambda_j}$. A rate can exceed 1, even though a probability cannot.

If a catalog has multiple event entries with the same loss, their rates all contribute when calculating the chance of exceeding a threshold. The loss levels determine the possible values of $M$, while the rates determine the probability attached to those values.

Threshold Probabilities and the Mass at Zero

For a threshold $x\geq 0$, an annual occurrence loss exceeds $x$ if at least one event with loss greater than $x$ occurs. Define the total rate of events above that threshold as $\Lambda(x)$. Independent Poisson counts imply the following distribution formulas.

\Lambda(x)=\sum_{j:l_j>x}\lambda_j,\qquad P(M\leq x)=e^{-\Lambda(x)},\qquad P(M>x)=1-e^{-\Lambda(x)}.

The strict inequality in the event set is important: for $P(M>x)$, include only losses strictly greater than $x$. To calculate $P(M\geq x)$, include losses equal to $x$ as well. The probability mass at zero is the probability that there are no positive-loss events in the year.

P(M=0)=\exp\left(-\sum_{j:l_j>0}\lambda_j\right).

An event with zero modeled loss can occur without changing $M$, so its rate does not belong in the sum for the zero-loss probability. If every cataloged event has positive loss, the sum is simply the total annual event rate.

Group the event information.
Record each event loss and its annual rate. If several entries have the same loss, retain each rate or add the rates for that loss level.

Choose the probability being asked for.
For $P(M>x)$, identify event losses strictly greater than $x$. For $P(M\leq x)$, use the same set of rates to calculate the probability that none occurs.

Sum rates and apply the Poisson formula.
Calculate $\Lambda(x)$, then use $1-e^{-\Lambda(x)}$ for exceedance or $e^{-\Lambda(x)}$ for non-exceedance.

Check the distribution.
Confirm probabilities lie between 0 and 1, and that the masses across all possible values of $M$ sum to 1, allowing for rounding.

Why the Exponential Formula Holds

The formula follows directly from the probability of zero occurrences in a Poisson count. For each type $j$, let $N_j$ be the number of occurrences in the year. If $l_j>x$, the condition $M\leq x$ requires $N_j=0$. Event types with $l_j\leq x$ do not prevent $M\leq x$. Independence therefore lets us multiply the zero-count probabilities for just the event types above the threshold.

\begin{aligned} P(M\leq x) &=P(N_j=0\text{ for every }j\text{ with }l_j>x)\\ &=\prod_{j:l_j>x}P(N_j=0)\\ &=\prod_{j:l_j>x}e^{-\lambda_j}\\ &=e^{-\sum_{j:l_j>x}\lambda_j}. \end{aligned}

For distinct positive loss levels $0<l_1<l_2<\cdots<l_K$, the mass at a particular level can be found by subtracting adjacent cumulative probabilities. The maximum equals $l_k$ when it is at most $l_k$, but not less than $l_k$.

P(M=l_k)=P(M\leq l_k)-P(M<l_k).

This subtraction accounts for the possibility of more than one event in the year: the maximum is $l_k$ if at least one event at that level occurs and no event with a higher loss occurs. The formula does not say that exactly one event occurs at level $l_k$.

Worked Examples

Worked Example: Building the Distribution From Three Event Types

A hypothetical insurer models three positive-loss event types. Type A has a loss of $1.2 million and an annual rate of 0.30; Type B has a loss of $3.0 million and a rate of 0.12; Type C has a loss of $6.0 million and a rate of 0.08. Assume independent Poisson counts. Find the mass at zero, the probability that the annual maximum exceeds $2.0 million, and the probability that it equals $3.0 million.

Method. Use the annual maximum distribution. The total rate of positive-loss events is $0.30+0.12+0.08=0.50$, so $P(M=0)=e^{-0.50}=0.6065307$, or 0.6065 to four decimal places.

For $M>$2.0 million, the qualifying event types are B and C. Their rates sum to \(0.12+0.08=0.20$. Thus $P(M>$2.0\text{ million})=1-e^{-0.20}=0.1812692$, or 0.1813.

For $P(M=$3.0 million), first calculate \(P(M\leq $3.0\text{ million})$. Only Type C exceeds $3.0 million, so this probability is $e^{-0.08}=0.9231163$. Next, $M<$3.0 million$ requires no occurrences of either B or C, giving $e^{-(0.12+0.08)}=e^{-0.20}=0.8187308$. Subtract: $0.9231163-0.8187308=0.1043855$, or 0.1044.

As a check, the masses at 0, $1.2 million, $3.0 million, and $6.0 million are approximately 0.6065, 0.2122, 0.1044, and 0.0769. Their sum is 1.0000 after rounding. The common mistake avoided here is using the rate of Type B alone for $P(M>$2.0\text{ million})$; Type C also exceeds the threshold.

Worked Example: Combining Catalog Entries at the Same Loss

A hypothetical event catalog contains four event entries, with annual rates and insured losses shown below. Assume independent Poisson counts. Find the probability that the annual maximum is zero, the probability it exceeds $1.0 million, and the probability it equals $1.5 million. Losses are stated in millions of dollars.

Event entry	Loss	Annual rate
North	0.4	0.30
South	0.4	0.20
East	1.5	0.25
West	4.0	0.05

Method. Add rates across all positive-loss entries to find the total rate: $0.30+0.20+0.25+0.05=0.80$. Therefore, $P(M=0)=e^{-0.80}=0.4493290$, or 0.4493.

For an exceedance of $1.0 million, only East and West qualify. Their rates sum to $0.25+0.05=0.30$, giving $P(M>$1.0\text{ million})=1-e^{-0.30}=0.2591818$, or 0.2592.

For $M=$1.5 million, \(P(M\leq $1.5\text{ million})$ excludes only West, so it is $e^{-0.05}=0.9512294$. The probability $P(M<$1.5\text{ million})$ excludes both East and West; their rates sum to $0.30$, so it is $e^{-0.30}=0.7408182$. Hence the mass at $1.5 million is $0.9512294-0.7408182=0.2104112$, or 0.2104.

The two entries with loss $0.4 million contribute a combined rate of 0.50, even though they have different catalog labels. The answer is not obtained by counting catalog rows or by treating a catalog entry as equally likely.

Worked Example: Changing the Time Horizon

A hypothetical portfolio has event types with losses of $2.0 million and $7.0 million and annual rates of 0.20 and 0.05, respectively. Assume independent, stationary Poisson occurrence processes. Find the probability of at least one event exceeding $2.0 million in a five-year period, and the probability of no positive-loss event in that period.

Method. The event exceeding $2.0 million is the $7.0 million type, with annual rate 0.05. Over five years, its expected count is $5(0.05)=0.25$. The probability of one or more such events is $1-e^{-0.25}=0.2211992$, or 0.2212.

The total annual rate of positive-loss events is $0.20+0.05=0.25$. Over five years the expected count is $5(0.25)=1.25$, so the probability of no positive-loss event is $e^{-1.25}=0.2865048$, or 0.2865.

The horizon adjustment is multiplication of the annual rate by five, under the stated stationary Poisson assumption. The common mistake is to multiply the annual probability of at least one event by five. Probabilities do not scale linearly this way; the Poisson rate does.

Assumptions, Interpretation, and Exam Traps

The calculations require an event-rate interpretation and the stated Poisson assumptions. In particular, counts are independent across event types, and the rates describe occurrence frequencies over the relevant time horizon. If catalog entries instead represent mutually exclusive scenarios, or are probabilities for a single annual trial, treating them as independent Poisson rates may produce an incorrect distribution. Use the model that matches how the catalog weights were defined.

The event losses in these examples are deterministic conditional on the event type. If event loss varies for a given type, the calculation must account for that severity variation rather than assigning one fixed loss to every occurrence. Likewise, $M$ is a maximum, not a sum: the formulas here do not calculate annual aggregate loss.

Using event rates as probabilities. For rate $\lambda$, the Poisson probability of at least one occurrence is $1-e^{-\lambda}$, not $\lambda$. The rate and probability are close only when the rate is small.
Including an event at the threshold in a strict exceedance. For $P(M>x)$, sum rates for $l_j>x$, not $l_j\geq x$.
Ignoring event types above the threshold. All qualifying rates must be summed before applying the exponential formula.
Confusing occurrence maximum and annual aggregate. The maximum selects the largest single event loss; it does not add losses from different events.

A useful final check is monotonicity: as the threshold increases, $P(M>x)$ cannot increase. Also, $P(M\leq x)+P(M>x)=1$ for the same threshold and strictness convention.

Check Your Understanding

Assume independent Poisson event counts unless a question states otherwise. Report probabilities to four decimal places.

A positive-loss event type has annual rate 0.20 and loss $2 million; another has annual rate 0.10 and loss $5 million. Find $P(M=0)$ and $P(M>$2 million). (Answer: \(P(M=0)=0.7408$; $P(M>$2\text{ million})=0.0952$)
Event losses are $1 million and $3 million, with annual rates 0.40 and 0.20. Find the probability that the annual maximum equals $1 million. (Answer: 0.2699)
Three event types have losses of $0.5 million, $2 million, and $7 million, with annual rates 0.15, 0.10, and 0.05. Find $P(M>$1\text{ million})$ and $P(M\leq $2\text{ million})$. (Answer: 0.1393 and 0.9512)
The total annual rate of positive-loss event types is 0.35. Under a stationary Poisson model, find the probability of no positive-loss event over two years. (Answer: 0.4966)
A catalog has losses of $1 million, $4 million, and $9 million with annual rates 0.40, 0.12, and 0.03. Find the probability that the annual maximum is exactly $4 million. (Answer: 0.1097)

What You'll Learn