Common-Shock Models Of Dependent Losses

Why Model a Shared Shock?

Losses from separate coverages or business units need not arise independently. A single event—such as a storm affecting several locations—can contribute to each loss, while each location may also experience its own individual loss. A common-shock model represents these two sources separately. It makes explicit how the shared component can cause losses to move together, even when their individual components are independent.

The model is deliberately simple: it is useful for understanding the mechanism of dependence, not a claim that every real portfolio has only one shared driver. Its results depend on clear assumptions about which random quantities are independent.

Definition: A two-loss additive common-shock model is $$ X_1=a_1W+U_1,\qquad X_2=a_2W+U_2, $$ where $W$ is a shared random shock, $U_1$ and $U_2$ are individual shocks, and $a_1,a_2$ are fixed exposure coefficients. Assume $W,U_1,U_2$ are mutually independent, have finite means and variances, and $a_1,a_2\geq 0$. The individual shocks are therefore independent of each other and of the shared shock. Losses and shock amounts are measured in consistent units.

A frequent special case is an event that either occurs or does not. Write $W=JH$, where $J$ is an indicator of the common event and $H$ is its severity. If $J$ is Bernoulli with occurrence probability $p$, and is independent of $H$, then

E[W]=pE[H],\qquad \operatorname{Var}(W)=pE[H^2]-p^2(E[H])^2.

For a fixed event severity $c$, this reduces to $E[W]=pc$ and $\operatorname{Var}(W)=p(1-p)c^2$. The common-shock model does not require a Bernoulli event; $W$ may have any distribution with finite variance.

How the Shared Component Creates Covariance

Under the stated independence assumptions, the two losses have means and variances

E[X_i]=a_iE[W]+E[U_i], \qquad \operatorname{Var}(X_i)=a_i^2\operatorname{Var}(W)+\operatorname{Var}(U_i), \quad i=1,2.

Their covariance is

\operatorname{Cov}(X_1,X_2)=a_1a_2\operatorname{Var}(W).

To see why, expand covariance using its linearity in each argument:

\begin{aligned} \operatorname{Cov}(X_1,X_2) &=\operatorname{Cov}(a_1W+U_1,a_2W+U_2)\\ &=a_1a_2\operatorname{Var}(W) +a_1\operatorname{Cov}(W,U_2) +a_2\operatorname{Cov}(U_1,W) +\operatorname{Cov}(U_1,U_2). \end{aligned}

Mutual independence makes each of the last three covariances zero, leaving the shared component. This also explains the assumptions’ importance: if individual shocks are dependent, or are related to $W$, the extra covariance terms need not vanish and the simplified result does not apply.

When both exposure coefficients are positive and $W$ is nonconstant, the covariance is positive. If $W$ is constant, it contributes no variation and creates no covariance, even if its constant value is large. Covariance measures co-movement in squared loss units; correlation puts that co-movement on a standardized scale:

\operatorname{Corr}(X_1,X_2)= \frac{a_1a_2\operatorname{Var}(W)} {\sqrt{\left(a_1^2\operatorname{Var}(W)+\operatorname{Var}(U_1)\right) \left(a_2^2\operatorname{Var}(W)+\operatorname{Var}(U_2)\right)}}.

This formula requires both loss variances to be positive. If either loss has zero variance, its correlation with the other loss is undefined.

Worked Example: A Fixed-Severity Shared Event

A hypothetical insurer models losses for two coverages, in thousands of dollars. A shared event produces a loss of 100 in each coverage with probability 0.10 and produces no common loss otherwise. Each coverage also has its own individual claim: $U_i=20$ with probability 0.20 and $U_i=0$ otherwise. Assume the event and both individual claims are mutually independent. Find the mean and variance of each loss, their covariance, and their correlation.

Method. Define $W$ as the shared-event loss, then apply the mean, variance, and covariance formulas. The individual claims have Bernoulli loss distributions.

The common component has $E[W]=0.10(100)=10$. Its variance is $0.10(0.90)(100^2)=900$. Each individual claim has mean $0.20(20)=4$ and variance $0.20(0.80)(20^2)=64$.

Thus $X_i=W+U_i$, so each loss has mean $10+4=14$ and variance $900+64=964$, in thousands of dollars and squared thousands of dollars, respectively. The covariance is $\operatorname{Var}(W)=900$. Since the two variances are equal, the correlation is $900/\sqrt{964(964)}=900/964=0.9336$, to four decimal places.

Check. The covariance is positive and less than either variance, and the correlation is between zero and one. A common mistake is to use $E[W^2]=1{,}000$ as the covariance. The correction is to use $\operatorname{Var}(W)=E[W^2]-(E[W])^2=1{,}000-100=900$.

Aggregate Losses and Exposure Differences

The same reasoning extends directly to the sum of two losses. Let $S=X_1+X_2$. The variance identity for a sum gives

\operatorname{Var}(S) =\operatorname{Var}(X_1)+\operatorname{Var}(X_2) +2\operatorname{Cov}(X_1,X_2).

Substituting the common-shock results gives an equivalent expression:

\operatorname{Var}(S) =(a_1+a_2)^2\operatorname{Var}(W) +\operatorname{Var}(U_1)+\operatorname{Var}(U_2).

The shared component is multiplied by the sum of the exposures before its variance is calculated. This is why the covariance term matters: adding separate loss variances alone would fail to account for the shock affecting both losses at once.

Worked Example: Unequal Exposures to a Common Shock

In a hypothetical portfolio, $W=10$ thousand dollars with probability 0.20 and is zero otherwise. The losses are $X_1=2W+U_1$ and $X_2=3W+U_2$. Independently, $U_1=5$ with probability 0.40 and otherwise is zero; $U_2=4$ with probability 0.50 and otherwise is zero. All amounts are in thousands of dollars, and $W,U_1,U_2$ are mutually independent. Calculate the means, variances, covariance, correlation, and variance of $S=X_1+X_2$.

Method. First calculate the component moments, then scale the common component by each exposure coefficient. This avoids treating a shared event as two independent shocks.

For $W$, $E[W]=0.20(10)=2$ and $\operatorname{Var}(W)=0.20(0.80)(10^2)=16$. For $U_1$, $E[U_1]=0.40(5)=2$ and $\operatorname{Var}(U_1)=0.40(0.60)(5^2)=6$. For $U_2$, $E[U_2]=0.50(4)=2$ and $\operatorname{Var}(U_2)=0.50(0.50)(4^2)=4$.

The means are $E[X_1]=2(2)+2=6$ and $E[X_2]=3(2)+2=8$. Their variances are $\operatorname{Var}(X_1)=2^2(16)+6=70$ and $\operatorname{Var}(X_2)=3^2(16)+4=148$. Their covariance is $2(3)(16)=96$, so their correlation is $96/\sqrt{70(148)}=96/\sqrt{10{,}360}=0.9432$, to four decimal places.

The aggregate variance is $70+148+2(96)=410$. As a check, $S=5W+U_1+U_2$, so its variance is $5^2(16)+6+4=410$, the same result.

Common mistake avoided. Adding only the two individual variances gives $70+148=218$, not the aggregate variance. The missing amount is $2\operatorname{Cov}(X_1,X_2)=192$, because the shared shock appears in both losses.

Joint Exceedance Probabilities by Conditioning

A shared event can also affect the probability that both losses exceed specified thresholds. For a discrete common component, condition on its possible values. Within each case, evaluate the joint event using the individual-shock distributions and their independence, then weight by the probability of that common-component value. The common event’s probability must not be counted twice as though each loss had a separate event.

Worked Example: Probability Both Losses Exceed a Threshold

Consider hypothetical losses, in thousands of dollars, $X_1=W+U_1$ and $X_2=W+U_2$. The common component $W$ is 40 with probability 0.15, 100 with probability 0.05, and zero with probability 0.80. Independently, $U_1$ is 10 with probability 0.30 and otherwise zero; $U_2$ is 10 with probability 0.50 and otherwise zero. Find $P(X_1\geq 10,\ X_2\geq 10)$, and compare it with the marginal exceedance probabilities.

Method. Split the calculation according to whether the common component is positive or zero. When $W=40$ or $100$, both losses exceed 10 regardless of the individual claims. This contributes $0.15+0.05=0.20$.

When $W=0$, both losses reach 10 only if both individual claims occur. Their independence gives conditional probability $0.30(0.50)=0.15$. This case contributes $0.80(0.15)=0.12$. Therefore $P(X_1\geq10,\ X_2\geq10)=0.20+0.12=0.32$.

For the first marginal, a positive common component guarantees exceedance, while when $W=0$, $U_1$ must occur. Thus $P(X_1\geq10)=0.20+0.80(0.30)=0.44$. Similarly, $P(X_2\geq10)=0.20+0.80(0.50)=0.60$.

Check. The joint probability, 0.32, is no greater than either marginal probability. The calculation also shows why the common component matters: both losses exceed the threshold together whenever that shared component is positive.

Assumptions and a Common Exam Trap

The model’s compact covariance formula depends on the assumptions, not just on the fact that the losses are exposed to the same event. In particular, it assumes the individual components are independent of each other and of the common component. If those assumptions are not justified, expand the covariance and retain any nonzero terms rather than automatically setting them to zero.

For a common component with positive exposure coefficients, a useful check is that its contribution to covariance is $a_1a_2\operatorname{Var}(W)$, not $a_1a_2E[W^2]$. Also, aggregate variance includes twice the covariance, not just once. For two losses, both checks follow from the formulas derived above.

Key takeaway: Under mutual independence, a shared random shock contributes $a_1a_2\operatorname{Var}(W)$ to the covariance of two losses. The individual shocks add to each loss’s variance but do not add to the covariance. Aggregate variance must include twice that covariance.

Check Your Understanding

Use the common-shock formulas and state the result in the units given.

In thousands of dollars, $W=30$ with probability 0.10 and otherwise is zero. Two losses are $X_i=W+U_i$, with mutually independent individual shocks and common component. What is $\operatorname{Cov}(X_1,X_2)$? (Answer: 81 squared thousands of dollars.)
$W=20$ with probability 0.30 and otherwise is zero. Let $X_1=W+U_1$, $X_2=W+U_2$, where $U_1$ is 5 with probability 0.20 and $U_2$ is 10 with probability 0.40, independently. Find $E[X_1]$, $E[X_2]$, and $\operatorname{Cov}(X_1,X_2)$. All amounts are in thousands of dollars. (Answer: 7 thousand dollars, 10 thousand dollars, and 84 squared thousands of dollars.)
$W=25$ with probability 0.25 and otherwise is zero. Independently, $U_1=10$ with probability 0.20 and $U_2=10$ with probability 0.40; otherwise each is zero. For $X_i=W+U_i$, find $P(X_1\geq10,\ X_2\geq10)$. (Answer: 0.3100.)
$W=12$ with probability 0.25 and otherwise is zero. Losses in thousands of dollars are $X_1=1.5W+U_1$ and $X_2=2W+U_2$. Independently, $U_1=4$ with probability 0.50 and $U_2=6$ with probability 0.25; otherwise each is zero. Find both loss variances, their covariance, and the variance of $X_1+X_2$. (Answer: 64.75, 114.75, 81, and 341.5, respectively, in squared thousands of dollars.)
$W$ is 0, 10, or 30 with probabilities 0.80, 0.10, and 0.10. Let $X_1=W+U_1$ and $X_2=W+U_2$, where $U_1=6$ with probability 0.50 and $U_2=8$ with probability 0.50, independently of each other and $W$. Find the correlation of the losses. (Answer: 0.8710.)

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