Common-Factor Aggregation of Multiple Risks

Why Use a Common-Factor Model?

Several risks can respond to the same underlying condition. For example, a hypothetical insurer's claims from different product lines might all be affected by the severity of a regional disruption. Treating the losses as independent can miss the way a bad factor state raises several loss probabilities at once. A common-factor model represents this shared influence explicitly, while still allowing residual randomness specific to each risk.

The central modeling distinction is between dependence given the factor and dependence without conditioning on it. Components may be independent after the factor is known, yet dependent overall because they share uncertainty about the factor. This structure supports aggregation by calculating each conditional aggregate distribution and then averaging across factor states.

Definition: Let \(Z\) be a common factor and let \(X_1,\ldots,X_n\) be component losses on the same time period and loss basis. A common-factor model specifies the conditional distribution of each \(X_i\) given \(Z\), and assumes \(X_1,\ldots,X_n\) are conditionally independent given \(Z\). The aggregate loss is \(S=\sum_{i=1}^n X_i\). Conditional independence means that, for any component outcomes \(x_1,\ldots,x_n\), their joint conditional probability factors into the product of their conditional probabilities.

For a finite-valued factor with states \(z_1,\ldots,z_m\), write \(w_j=P(Z=z_j)\). Conditional independence gives

P(X_1=x_1,\ldots,X_n=x_n) =\sum_{j=1}^{m} w_j\prod_{i=1}^{n}P(X_i=x_i\mid Z=z_j).

The summation over factor states is essential: it combines the component distributions within each state, then mixes those results using the state probabilities. For continuous factors, the corresponding operation is integration over the distribution of \(Z\).

Constructing the Aggregate Distribution

For discrete component losses, a direct method is to calculate the conditional distribution of \(S\) for each factor state. Within a state, the component probability distributions are convolved. The unconditional aggregate distribution is the probability-weighted mixture of these conditional distributions:

P(S\le s)=\sum_{j=1}^{m}w_jP(S\le s\mid Z=z_j).

This formula follows by partitioning the event \(\{S\le s\}\) according to the mutually exclusive factor states. In each state, conditional independence permits the component distributions to be combined. The conditional distribution must be mixed only after this conditional aggregation; mixing each component's probabilities first and then treating the components as independent generally gives the wrong joint distribution.

Worked Example: A Strict Threshold With Two Conditional Losses

A hypothetical portfolio has two losses, \(X_1=40I_1\) and \(X_2=90I_2\), where each \(I_i\) is an indicator. A common factor \(Z\) has states A and B with probabilities \(P(A)=0.25\) and \(P(B)=0.75\). Given A, the occurrence probabilities are \(P(I_1=1\mid A)=0.2\) and \(P(I_2=1\mid A)=0.1\); given B, they are 0.6 and 0.3. The indicators are conditionally independent given \(Z\). Find the distribution of \(S=X_1+X_2\) and \(P(S>90)\).

Method. First enumerate the four indicator outcomes within each factor state. Then mix the resulting aggregate probabilities. In state A, the probabilities for \((I_1,I_2)=(0,0),(1,0),(0,1),(1,1)\) are \(0.8(0.9)=0.72\), \(0.2(0.9)=0.18\), \(0.8(0.1)=0.08\), and \(0.2(0.1)=0.02\). These outcomes produce aggregate losses 0, 40, 90, and 130, respectively.

In state B, the four outcome probabilities in the same order are \(0.4(0.7)=0.28\), \(0.6(0.7)=0.42\), \(0.4(0.3)=0.12\), and \(0.6(0.3)=0.18\). Now weight each state distribution by its probability:

\(P(S=0)=0.25(0.72)+0.75(0.28)=0.18+0.21=0.39.\)
\(P(S=40)=0.25(0.18)+0.75(0.42)=0.045+0.315=0.36.\)
\(P(S=90)=0.25(0.08)+0.75(0.12)=0.02+0.09=0.11.\)
\(P(S=130)=0.25(0.02)+0.75(0.18)=0.005+0.135=0.14.\)

The probabilities sum to \(0.39+0.36+0.11+0.14=1.00\). Since a loss of exactly 90 does not exceed 90, \(S>90\) occurs only when both indicators equal 1. Therefore \(P(S>90)=P(S=130)=0.14\). As a further check, the unconditional mean is \(40(0.36)+90(0.11)+130(0.14)=14.4+9.9+18.2=42.5\). The conditional means are \(40(0.2)+90(0.1)=17\) in A and \(40(0.6)+90(0.3)=51\) in B, giving \(0.25(17)+0.75(51)=42.5\).

Common mistake avoided: The marginal occurrence probabilities are \(P(I_1=1)=0.25(0.2)+0.75(0.6)=0.5\) and \(P(I_2=1)=0.25(0.1)+0.75(0.3)=0.25\). Multiplying these gives 0.125, but that incorrectly assumes unconditional independence. The correct joint probability is \(0.25(0.2)(0.1)+0.75(0.6)(0.3)=0.14\).

Why Conditional Independence Does Not Mean Independence

A shared factor can create positive or negative unconditional association even when the residual risks are conditionally independent. For two indicators \(I_1,I_2\), conditional independence implies

\operatorname{Cov}(I_1,I_2) =\operatorname{Cov}\big(P(I_1=1\mid Z),P(I_2=1\mid Z)\big).

To see why, use the conditional expectation identity \(E[I_i\mid Z]=P(I_i=1\mid Z)\). Conditional independence gives \(E[I_1I_2\mid Z]=E[I_1\mid Z]E[I_2\mid Z]\). Taking expectations yields \(E[I_1I_2]=E[P(I_1=1\mid Z)P(I_2=1\mid Z)]\). Subtracting \(E[I_1]E[I_2]\) gives the covariance formula above. If both conditional probabilities tend to be high in the same factor states, their covariance is positive.

The same reasoning extends to the variance of a sum. For any component losses with finite second moments, the law of total variance gives

\operatorname{Var}(S) =E[\operatorname{Var}(S\mid Z)] +\operatorname{Var}(E[S\mid Z]).

This identity follows by writing \(S-E[S]=(S-E[S\mid Z])+(E[S\mid Z]-E[S])\). On squaring and taking expectations, the cross-product has expectation zero because \(E[S-E[S\mid Z]\mid Z]=0\). The first term in the variance identity captures residual variation within factor states; the second captures variation in conditional aggregate means across states. Under conditional independence, \(\operatorname{Var}(S\mid Z)=\sum_i\operatorname{Var}(X_i\mid Z)\). Do not omit the second term merely because the components are conditionally independent.

Worked Example: Three Risks and a Two-State Factor

A hypothetical portfolio contains three losses, each equal to \(100I_i\). A common factor is in state L with probability 0.4 and state H with probability 0.6. Given L, each indicator has probability 0.1; given H, each has probability 0.5. The indicators are conditionally independent given the factor. Find \(P(S\ge 200)\), \(E[S]\), and \(\operatorname{Var}(S)\).

For \(S\ge200\), at least two indicators must occur. In L, the probability of exactly two occurrences is \(\binom{3}{2}(0.1)^2(0.9)=0.027\), and the probability of three is \((0.1)^3=0.001\). Thus \(P(S\ge200\mid L)=0.028\). In H, the corresponding probabilities are \(\binom{3}{2}(0.5)^2(0.5)=0.375\) and \((0.5)^3=0.125\), so \(P(S\ge200\mid H)=0.500\). Mixing gives \(P(S\ge200)=0.4(0.028)+0.6(0.500)=0.0112+0.3000=0.3112\).

The conditional aggregate means are \(E[S\mid L]=3(100)(0.1)=30\) and \(E[S\mid H]=3(100)(0.5)=150\). Hence \(E[S]=0.4(30)+0.6(150)=12+90=102\).

Within L, the conditional variance is \(3(100^2)(0.1)(0.9)=2{,}700\); within H, it is \(3(100^2)(0.5)(0.5)=7{,}500\). Therefore \(E[\operatorname{Var}(S\mid Z)]=0.4(2{,}700)+0.6(7{,}500)=1{,}080+4{,}500=5{,}580\). The variance of conditional means is \(0.4(30-102)^2+0.6(150-102)^2=0.4(5{,}184)+0.6(2{,}304)=2{,}073.6+1{,}382.4=3{,}456\). The total variance is \(5{,}580+3{,}456=9{,}036\) dollars squared.

Check. Each indicator has unconditional probability \(0.4(0.1)+0.6(0.5)=0.34\), so the expected number of occurrences is \(3(0.34)=1.02\), consistent with \(E[S]=102\). The common factor contributes variance beyond the within-state variance; omitting it would give only 5,580.

Conditional Simulation

When component distributions are too complex for exact convolution, simulation can approximate the aggregate distribution. The simulation must preserve the model's dependence: draw one common factor for a trial, then draw every component loss from its distribution conditional on that same factor. Drawing a separate factor for each component would remove the modeled common influence.

Draw the common factor.
Sample \(Z\) from its specified distribution for one simulated period.

Draw conditional component losses.
Given the sampled \(Z\), generate each \(X_i\) using its conditional model. Use independent residual draws if the model assumes conditional independence.

Aggregate.
Record \(S=\sum_i X_i\) for that period. Repeat the first two steps for every simulation trial.

Estimate the quantity of interest.
For a threshold \(s\), estimate \(P(S>s)\) by the fraction of simulated aggregate losses exceeding \(s\), or estimate a quantile from the ordered aggregate losses.

A useful refinement is conditional Monte Carlo. For a tail probability, calculate \(q(Z)=P(S>s\mid Z)\) when that conditional probability is tractable, and average these values over simulated factor draws. This replaces a simulated zero-or-one tail indicator with its conditional expectation. It is unbiased because \(E[q(Z)]=P(S>s)\), and it typically has lower simulation variance.

Worked Example: Estimating a Tail Probability by Conditioning

A hypothetical portfolio has \(X_1=80I_1\) and \(X_2=120I_2\). The common factor has states A, B, and C with probabilities 0.2, 0.5, and 0.3. Conditional occurrence probabilities \((P(I_1=1\mid Z),P(I_2=1\mid Z))\) are \((0.1,0.2)\) in A, \((0.4,0.5)\) in B, and \((0.8,0.7)\) in C. The indicators are conditionally independent. Estimate \(P(S\ge200)\) using 10,000 conditional simulations, with 2,000 factor draws in A, 5,000 in B, and 3,000 in C. Suppose the observed numbers of simulated joint occurrences are 38, 1,012, and 1,686, respectively.

The threshold \(S\ge200\) requires both indicators to occur, since \(80+120=200\). Conditional independence therefore gives conditional tail probabilities \(0.1(0.2)=0.02\) in A, \(0.4(0.5)=0.20\) in B, and \(0.8(0.7)=0.56\) in C. The exact probability implied by the model is \(0.2(0.02)+0.5(0.20)+0.3(0.56)=0.004+0.100+0.168=0.272\).

The crude simulation proportions by state are \(38/2{,}000=0.019\), \(1{,}012/5{,}000=0.2024\), and \(1{,}686/3{,}000=0.562\). Weighting them by the factor probabilities gives \(0.2(0.019)+0.5(0.2024)+0.3(0.562)=0.0038+0.1012+0.1686=0.2736\). Thus this run's estimated tail probability is 0.2736, compared with the exact model probability 0.2720. The difference is simulation variation, not a contradiction.

If the conditional probabilities are evaluated instead of simulating the indicators, the conditional Monte Carlo estimate for these fixed factor-state counts is \(0.2(0.02)+0.5(0.20)+0.3(0.56)=0.2720\). This illustration has fixed, probability-proportional state counts; in general, factor draws are random or are allocated using a suitable stratified design, and the estimator must use the corresponding weights.

Common mistake avoided: Multiplying the marginal occurrence probabilities gives \(P(I_1=1)=0.2(0.1)+0.5(0.4)+0.3(0.8)=0.46\) and \(P(I_2=1)=0.2(0.2)+0.5(0.5)+0.3(0.7)=0.51\), whose product is 0.2346. That is not the joint probability. The shared factor makes the correct joint probability 0.272.

Practical Checks and Common Traps

A common-factor representation is valuable because it makes the source of dependence visible and gives a practical route to an aggregate distribution. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus, where a sound aggregation result depends on the model assumptions and the calculation, not just on a reported capital number.

Use one factor realization per simulated period. Re-drawing the factor separately for each component does not represent a shared systematic condition.
Condition before multiplying. Conditional independence permits multiplication of conditional probabilities. It does not imply that marginal probabilities can be multiplied.
Respect strict and non-strict thresholds. If a question asks for \(S>s\), losses exactly equal to \(s\) are excluded. If it asks for \(S\ge s\), they are included.
Check probability totals and moments. A discrete aggregate distribution must sum to one, and its mean should agree with the weighted conditional means.
State the model's limits. Conditional independence is an assumption, not a consequence of using a common factor. Additional residual dependence requires a model that represents it.

Key takeaway: Aggregate within each common-factor state, then average across states. Conditional independence simplifies the first step; the shared factor still creates unconditional dependence and contributes to aggregate uncertainty.

Check Your Understanding

For each problem, use the stated conditional independence assumption and mix results across factor states only after conditioning.

Two losses are \(25I_1\) and \(60I_2\). A factor has states A and B with probabilities 0.4 and 0.6. The conditional occurrence-probability pairs are \((0.2,0.5)\) and \((0.5,0.2)\). Given the factor, the indicators are independent. Find \(P(S>60)\). (Answer: 0.1000)
Two losses are each \(100I_i\). A factor has states with probabilities 0.3 and 0.7, and each indicator has conditional probabilities 0.1 and 0.4 in those states, respectively. Given the factor, the indicators are independent. Find \(P(S=0)\). (Answer: 0.4950)
Three losses are each \(50I_i\). A factor has states with probabilities 0.25 and 0.75; each indicator has conditional occurrence probability 0.2 and 0.6, respectively. Given the factor, the indicators are independent. Find \(P(S\ge100)\). (Answer: 0.5120)
Two conditionally independent indicators have conditional probability pairs \((0.1,0.3)\) and \((0.5,0.7)\) in two equally likely factor states. Their losses are \(200I_1\) and \(50I_2\). Find \(\operatorname{Cov}(200I_1,50I_2)\). (Answer: 400 dollars squared)
Three losses are each \(100I_i\). A factor has states with probabilities 0.4 and 0.6, and each indicator has conditional occurrence probabilities 0.1 and 0.5, respectively. Given the factor, the indicators are independent. Find \(\operatorname{Var}(S)\). (Answer: 9,036 dollars squared)

What You'll Learn