Why Shared Events Need Joint Scenarios
A single event can generate losses in several risk categories at once. A severe storm, for example, might damage insured buildings, interrupt business operations, and create liability claims. Adding losses across categories is straightforward; preserving the fact that they arose together is not. If the components are modeled separately and then treated as independent, the resulting aggregate distribution can assign the wrong probabilities to joint outcomes, even when the component loss distributions look reasonable on their own.
The key is to describe each event scenario by a vector of component losses. The vector records the losses that occur together under that scenario. The total for the scenario is the sum of its components. This approach is useful for both an occurrence loss (the total from one event) and an annual loss (the total from all events in a year).
The scenario is the joint outcome, not a separate draw for each component. If one scenario says that both property and business-interruption losses occur, those outcomes remain together when the aggregate distribution is constructed. The scenario probabilities may come from an event model, a portfolio study, or an explicitly assumed hypothetical model; the aggregation arithmetic does not establish that the scenario assumptions are realistic.
From Scenario Vectors to an Occurrence Distribution
To construct an occurrence-loss distribution, calculate the component sum within every scenario and attach the scenario probability to that sum. If distinct scenarios produce the same aggregate amount, their probabilities combine. For example, if two scenarios each produce an aggregate loss of 50, then the probability assigned to an aggregate loss of 50 is the sum of those scenario probabilities.
The mean and variance follow from the resulting distribution. In particular, the mean aggregate occurrence loss is the probability-weighted sum of scenario totals. The variance can be calculated from the probability-weighted squared totals minus the square of the mean. These are properties of the aggregate \(Z\); they are not found by assuming the component losses are independent.
The reason is direct: \(Z\) takes the value \(z_j\) with probability \(p_j\), so its expectation is the weighted average of those values. Applying the definition of variance gives \(\operatorname{Var}(Z)=E[Z^2]-(E[Z])^2\), and \(E[Z^2]=\sum_jp_jz_j^2\). This justification requires a complete set of mutually exclusive scenarios covering the event outcomes being modeled.
Worked Example: An Occurrence Loss From a Shared Storm
A hypothetical insurer models the losses from one storm to two coverage components: property, \(X_1\), and business interruption, \(X_2\). The storm has three mutually exclusive scenarios: no covered impact with probability 0.80 and losses \((0,0)\); a localized impact with probability 0.15 and losses \((40,20)\), in thousands of dollars; and a severe impact with probability 0.05 and losses \((100,80)\), in thousands of dollars. Find the aggregate occurrence-loss distribution, its mean and variance, and its 95% VaR. Use the convention \(\operatorname{VaR}_{0.95}(Z)=\inf\{z:F_Z(z)\geq0.95\}\).
Method. Sum the two component losses within each storm scenario, then preserve that scenario’s probability. The scenario probabilities check: \(0.80+0.15+0.05=1.00\).
Aggregate amounts. No impact gives \(0+0=0\). Localized impact gives \(40+20=60\). Severe impact gives \(100+80=180\). Thus \(Z\) takes values 0, 60, and 180 with probabilities 0.80, 0.15, and 0.05, respectively, all in thousands of dollars.
Mean. In thousands of dollars, \(E[Z]=0.80(0)+0.15(60)+0.05(180)=0+9+9=18\).
Variance. First calculate \(E[Z^2]=0.80(0^2)+0.15(60^2)+0.05(180^2)=0+540+1{,}620=2{,}160\), in squared thousands of dollars. Therefore \(\operatorname{Var}(Z)=2{,}160-18^2=2{,}160-324=1{,}836\), in squared thousands of dollars.
95% VaR. The cumulative probabilities are \(F_Z(0)=0.80\), \(F_Z(60)=0.95\), and \(F_Z(180)=1.00\). Hence \(\operatorname{VaR}_{0.95}(Z)=60\) thousand dollars under the stated quantile convention.
Check. The distribution probabilities sum to 1, and its mean lies between its smallest and largest possible losses. A common mistake is to put probability 0.15 on a property-only loss of 40 and probability 0.05 on a business-interruption-only loss of 80. That discards the stated joint vectors. The correct aggregate outcomes are 60 and 180, because both components are affected in each of those scenarios.
Keep Shared Occurrence Separate From Component Marginals
A marginal probability describes one component by itself. It does not identify how often two components are affected together. The joint scenario table does. For an aggregate threshold, calculate the total in each scenario first and then add the probabilities of the scenarios meeting the threshold. For the probability that two components are both positive, identify scenarios in which both entries of the vector are positive.
This distinction matters even when marginal loss probabilities are known exactly. In general, multiplying the marginals to get a joint probability assumes independence. A common event may instead make joint occurrence much more likely. The event scenarios supply that dependence directly, so a separate independence assumption is neither needed nor appropriate unless it is actually part of the model.
Worked Example: Annual Loss With One Shared Trigger
In a hypothetical one-year model, at most one covered trigger occurs. The mutually exclusive annual outcomes are: no trigger, probability 0.70, with losses \((0,0)\); trigger A, probability 0.20, with losses \((50,30)\); and trigger B, probability 0.10, with losses \((20,90)\). Losses are in thousands of dollars. Calculate the annual aggregate expected loss, the probability both components have positive losses, and the probability that aggregate loss is at least 100 thousand dollars.
Method. Treat the listed outcomes as annual joint scenarios. Since the model permits at most one trigger, the three probabilities are mutually exclusive and exhaustive: \(0.70+0.20+0.10=1.00\).
Aggregate outcomes. No trigger gives \(0\). Trigger A gives \(50+30=80\). Trigger B gives \(20+90=110\). Thus the annual aggregate loss distribution assigns probabilities 0.70, 0.20, and 0.10 to 0, 80, and 110, respectively.
Expected annual loss. The calculation is \(0.70(0)+0.20(80)+0.10(110)=0+16+11=27\) thousand dollars.
Joint component occurrence. Both components are positive in trigger A and trigger B, so the probability is \(0.20+0.10=0.30\).
Threshold probability. Only trigger B produces an aggregate loss of at least 100 thousand dollars. Therefore \(P(Z\geq100)=0.10\).
Check and trap. The marginal probabilities of positive loss are 0.30 for the first component and 0.30 for the second. Multiplying them gives \(0.30(0.30)=0.09\), but this is an independence calculation, not the model’s joint probability. The scenario table gives 0.30, because both components are positive under either trigger. The expected loss check also works component by component: the first component’s mean is \(0.20(50)+0.10(20)=12\), the second’s is \(0.20(30)+0.10(90)=15\), and their sum is 27 thousand dollars.
From Event Losses to Annual Losses With Multiple Events
When several events can occur in a year, aggregate each event’s component vector first. If event \(n\) produces vector \((X_{1n},\ldots,X_{kn})\), its total event loss is \(Z_n=\sum_r X_{rn}\). The annual aggregate is the sum of those event totals, \(S=\sum_{n=1}^{N}Z_n\). This is a two-level calculation: component losses are combined within an event, and event totals are then combined over the year.
One useful model assumes that the number of events \(N\) in a year has a Poisson distribution with parameter \(\lambda\), and that event totals \(Z_n\) are independent and identically distributed, independent of \(N\). Under these assumptions, \(S\) is a compound Poisson loss. Its mean and variance depend on the event-total severity distribution, which is formed from the shared scenarios.
To see why the variance formula holds, condition on the number of events. Given \(N=n\), the sum has mean \(nE[Z]\) and variance \(n\operatorname{Var}(Z)\). The law of total variance then gives \(\operatorname{Var}(S)=E[N]\operatorname{Var}(Z)+\operatorname{Var}(N)(E[Z])^2\). For a Poisson variable, \(E[N]=\operatorname{Var}(N)=\lambda\). Substitution yields \(\lambda\operatorname{Var}(Z)+\lambda(E[Z])^2=\lambda E[Z^2]\). This result relies on the stated Poisson frequency and independent, identically distributed event totals; it does not automatically apply to clustered or otherwise dependent event arrivals.
Worked Example: Compound Poisson Annual Loss From Shared Events
A hypothetical portfolio has a Poisson number \(N\) of covered events per year with parameter \(\lambda=0.5\). Each event independently has either an ordinary scenario, probability 0.80, with component losses \((30,10)\), or a severe scenario, probability 0.20, with component losses \((100,40)\), in thousands of dollars. Every event has one of these two outcomes, and event scenarios are independent across events. Calculate the mean and variance of annual aggregate loss and the probability of zero annual loss.
Method and event totals. First aggregate within each event. An ordinary event has total \(30+10=40\); a severe event has total \(100+40=140\). The scenario probabilities sum to \(0.80+0.20=1.00\), so these event totals define the full severity distribution.
Event-severity moments. In thousands of dollars, \(E[Z]=0.80(40)+0.20(140)=32+28=60\). Also \(E[Z^2]=0.80(40^2)+0.20(140^2)=0.80(1{,}600)+0.20(19{,}600)=1{,}280+3{,}920=5{,}200\). (The severe-event squared total is \(140^2=19{,}600\).)
Annual mean and variance. The compound Poisson formulas give \(E[S]=0.5(60)=30\) thousand dollars and \(\operatorname{Var}(S)=0.5(5{,}200)=2{,}600\) squared thousands of dollars.
Zero-loss probability. Each event has a positive total, so annual loss is zero exactly when \(N=0\). Thus \(P(S=0)=P(N=0)=e^{-0.5}=0.6065\), to four decimal places.
Check and trap. The event-severity variance is \(5{,}200-60^2=1{,}600\). Using only \(\lambda\operatorname{Var}(Z)\) would give 800, which omits variation from the random event count. The correct annual variance is \(\lambda E[Z^2]=2{,}600\). The annual mean of 30 thousand dollars also equals the expected event count, 0.5, times the mean event total, 60 thousand dollars.
Practical Checks and Common Errors
Before relying on an aggregate result, verify that the event scenarios describe the same event basis and that their probabilities form a complete distribution. Then check that component totals are added in consistent units and that losses assigned to the same scenario are genuinely joint outcomes. When extending to annual loss, confirm whether the frequency model permits multiple events and whether its assumptions about event counts and event severities are appropriate.
- Separating a shared event into independent component events: This changes the joint outcomes and can distort exceedance and co-occurrence probabilities.
- Adding scenario probabilities incorrectly: Probabilities can be added for mutually exclusive scenarios, such as distinct event types; they cannot simply be added when outcomes overlap.
- Confusing occurrence and annual totals: One event’s component sum is not necessarily the year’s total if multiple events can occur.
- Using the wrong severity moment: Under compound Poisson frequency, annual variance is \(\lambda E[Z^2]\), not \(\lambda\operatorname{Var}(Z)\).
- Ignoring units: If losses are stated in thousands of dollars, means are in thousands and variances are in squared thousands.
Shared-event aggregation makes the joint structure explicit before calculating totals. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus because it connects dependence modeling to practical risk aggregation: the aggregate result reflects not only how much each component can lose, but which losses occur together under the modeled events.
Check Your Understanding
For each problem, aggregate the component losses within each stated scenario before calculating the requested quantity.
- A hypothetical one-event model has scenarios: no loss, probability 0.60, vector \((0,0)\); local event, probability 0.25, vector \((12,8)\); major event, probability 0.15, vector \((60,40)\), in thousands of dollars. Find expected aggregate loss and the probability aggregate loss is at least 100. (Answer: 20 thousand dollars; 0.1500.)
- A one-year model has at most one event: no event with probability 0.80 and vector \((0,0)\); event X with probability 0.12 and vector \((20,50)\); event Y with probability 0.08 and vector \((40,0)\), in thousands. Find expected aggregate loss and the probability both components are positive. (Answer: 11.6 thousand dollars; 0.1200.)
- Annual event count is Poisson with parameter 0.2. Each event has aggregate total 30 thousand dollars with probability 0.90 or 130 thousand dollars with probability 0.10. Find the annual mean and variance. (Answer: 8 thousand dollars; 500 squared thousands of dollars.)
- In a one-year shared-scenario model, the outcomes are no loss with probability 0.50 and vector \((0,0)\), scenario A with probability 0.30 and vector \((10,20)\), and scenario B with probability 0.20 and vector \((30,70)\), in thousands. Find the expected aggregate loss, variance, and probability aggregate loss is at least 30. (Answer: 29 thousand dollars; 1,429 squared thousands of dollars; 0.5000.)
- Annual event count is Poisson with parameter 0.3. Event component vectors are \((25,15)\) with probability 0.80 and \((80,40)\) with probability 0.20, in thousands. Find annual mean, annual variance, and probability of zero annual loss. (Answer: 16.8 thousand dollars; 1,248 squared thousands of dollars; 0.7408.)