Covariance Corrections In Sums Of Dependent Risks

Why Dependence Changes Aggregate Risk

When several risks contribute to a total, adding their expected values is straightforward: expectation is additive whether or not the risks are independent. Variance is different. It measures spread around the mean, and the way two risks move together affects the spread of their sum. Ignoring that dependence can understate or overstate aggregate variability.

For example, two hypothetical insurance payments may both be more likely when the same severe weather affects separate regions. Their positive covariance increases the variance of the total. Conversely, two events that rarely occur together can have negative covariance, reducing the total variance relative to an independence calculation. The relevant adjustment is determined by the joint behavior of the risks, not merely by their separate distributions.

Definition: For random variables $X$ and $Y$ with finite second moments, their covariance is $$ \operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]. $$ Positive covariance indicates that deviations from the respective means tend to have the same sign; negative covariance indicates that they tend to have opposite signs. Covariance is measured in the product of the units of $X$ and $Y$.

The covariance is not a probability and is not restricted to the interval from $-1$ to $1$. Its magnitude depends on the units and scales of the variables. Correlation standardizes covariance, but for calculating a sum's variance the covariance itself is the direct quantity needed.

The Variance Formula for a Sum

Suppose $X_1,\ldots,X_n$ have finite variances. The variance of their sum includes each individual variance and a correction for every distinct pair. This does not require the variables to be identically distributed.

Theorem: For $S=\sum_{i=1}^n X_i$, $$ \operatorname{Var}(S)=\sum_{i=1}^n\operatorname{Var}(X_i)+2\sum_{1\leq i<j\leq n}\operatorname{Cov}(X_i,X_j). $$ The pairwise sum includes each unordered pair once. If all the variables are pairwise uncorrelated, every covariance term is zero and the variance is the sum of the individual variances. Mutual independence is sufficient for this, but is not required.

Here is why the formula holds. Center each variable by subtracting its mean, and expand the square of their sum. Each cross-product appears twice: once in each order. Since $\operatorname{Cov}(X_i,X_j)=\operatorname{Cov}(X_j,X_i)$, those two terms combine to give the factor of 2.

\begin{aligned} \operatorname{Var}\left(\sum_{i=1}^n X_i\right) &=E\left[\left(\sum_{i=1}^n(X_i-E[X_i])\right)^2\right]\\ &=\sum_{i=1}^n E[(X_i-E[X_i])^2] +2\sum_{1\leq i<j\leq n} E[(X_i-E[X_i])(X_j-E[X_j])]\\ &=\sum_{i=1}^n\operatorname{Var}(X_i) +2\sum_{1\leq i<j\leq n}\operatorname{Cov}(X_i,X_j). \end{aligned}

For two risks this reduces to $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)$. For three risks there are three distinct pairs: $(X_1,X_2)$, $(X_1,X_3)$, and $(X_2,X_3)$. A useful safeguard is to list those pairs before substituting numbers.

Worked Example: Overlapping Claims Reduce Total Variability

A hypothetical pool has payments $X=10{,}000I_A$ and $Y=15{,}000I_B$, where $I_A$ and $I_B$ are indicators of claim events $A$ and $B$. Suppose $P(A)=0.1000$, $P(B)=0.2000$, and $P(A\cap B)=0.0100$. Calculate $\operatorname{Var}(X+Y)$ in dollars squared. The events overlap, but less often than they would under independence.

Method. Find each indicator payment's variance, calculate the covariance from the joint event probability, and apply the two-variable variance formula.

For an event indicator $I_A$, $I_A^2=I_A$, so $\operatorname{Var}(I_A)=P(A)(1-P(A))$. Scaling a random variable by a constant $c$ multiplies its variance by $c^2$. Therefore,

\operatorname{Var}(X)=10{,}000^2(0.1000)(0.9000) =100{,}000{,}000(0.0900)=9{,}000{,}000.

Similarly,

\operatorname{Var}(Y)=15{,}000^2(0.2000)(0.8000) =225{,}000{,}000(0.1600)=36{,}000{,}000.

The covariance of the indicators is $P(A\cap B)-P(A)P(B)$. Scaling each indicator by its payment amount scales the covariance by the product of those amounts:

\begin{aligned} \operatorname{Cov}(X,Y) &=10{,}000(15{,}000)\bigl(0.0100-(0.1000)(0.2000)\bigr)\\ &=150{,}000{,}000(0.0100-0.0200)\\ &=-1{,}500{,}000. \end{aligned}

Thus the total variance is

\operatorname{Var}(X+Y) =9{,}000{,}000+36{,}000{,}000+2(-1{,}500{,}000) =42{,}000{,}000\text{ dollars}^2.

Check. The total payment is $0$, $10{,}000$, $15{,}000$, or $25{,}000$ with probabilities $0.7100$, $0.0900$, $0.1900$, and $0.0100$, respectively. Its mean is $4{,}000$, and its second moment is $58{,}000{,}000$. Hence its variance is $58{,}000{,}000-4{,}000^2=42{,}000{,}000$ dollars squared, as calculated. The common mistake is to call the events mutually exclusive or independent from their individual probabilities alone; the specified intersection probability must be used.

Interpreting the Covariance Correction

The sign of a covariance tells whether it raises or lowers the variance relative to the sum of individual variances. A positive covariance adds a positive correction; a negative covariance subtracts one. A zero covariance leaves the variance equal to the sum of individual variances, even if the variables are not independent.

For indicator payments $X=aI_A$ and $Y=bI_B$, where $a,b$ are fixed payment amounts, the covariance has a particularly useful form:

\operatorname{Cov}(X,Y)=ab\bigl(P(A\cap B)-P(A)P(B)\bigr).

This follows from $XY=abI_AI_B=abI_{A\cap B}$, so $E[XY]=abP(A\cap B)$, while $E[X]E[Y]=abP(A)P(B)$. If $P(A\cap B)>P(A)P(B)$, the covariance is positive. If the intersection probability is smaller, it is negative. Independence gives equality and thus zero covariance.

Worked Example: A Shared Hazard Raises Aggregate Variance

A hypothetical insurer models two regional payments as $X=5{,}000I_A$ and $Y=8{,}000I_B$. A shared hazard is reflected in $P(A)=0.1000$, $P(B)=0.2000$, and $P(A\cap B)=0.0500$. Find the variance of the total payment and compare it with the independence-based calculation.

Method. Calculate the two Bernoulli payment variances and the covariance from the supplied joint probability. The probability of both claims, $0.0500$, is greater than the product $0.1000(0.2000)=0.0200$, so the covariance should be positive.

\begin{aligned} \operatorname{Var}(X) &=5{,}000^2(0.1000)(0.9000) =25{,}000{,}000(0.0900)=2{,}250{,}000,\\ \operatorname{Var}(Y) &=8{,}000^2(0.2000)(0.8000) =64{,}000{,}000(0.1600)=10{,}240{,}000. \end{aligned}

The covariance is

\begin{aligned} \operatorname{Cov}(X,Y) &=5{,}000(8{,}000)(0.0500-0.0200)\\ &=40{,}000{,}000(0.0300)=1{,}200{,}000. \end{aligned}

Therefore,

\operatorname{Var}(X+Y) =2{,}250{,}000+10{,}240{,}000+2(1{,}200{,}000) =14{,}890{,}000\text{ dollars}^2.

Under an independence assumption, the covariance correction would be zero and the result would instead be $2{,}250{,}000+10{,}240{,}000=12{,}490{,}000$ dollars squared. That is not the variance under the stated joint model. As a check, the total payment has probabilities $0.7500$ for $0$, $0.0500$ for $5{,}000$, $0.1500$ for $8{,}000$, and $0.0500$ for $13{,}000$. Its mean is $2{,}100$, its second moment is $19{,}300{,}000$, and $19{,}300{,}000-2{,}100^2=14{,}890{,}000$ dollars squared. The common mistake is to use the marginal probabilities as though they establish independence; the joint probability contradicts that assumption.

Several Risks: Count Every Pair

With more than two risks, the method is unchanged, but missing or duplicating a pair becomes an easy source of error. The covariance between each pair can be positive, negative, or zero. A convenient calculation table keeps the individual variances separate from the covariance sum; remember that the final formula doubles the sum of the distinct covariances.

Worked Example: Combining Three Correlated Risk Components

Let $X_1,X_2,X_3$ be hypothetical risk components measured in thousands of dollars. Suppose their variances, in thousands of dollars squared, are $4$, $9$, and $16$, and their pairwise covariances are $\operatorname{Cov}(X_1,X_2)=2$, $\operatorname{Cov}(X_1,X_3)=-1$, and $\operatorname{Cov}(X_2,X_3)=3$. Find the variance of $S=X_1+X_2+X_3$.

Method. List all three unordered pairs, add their covariances, and double that sum. The signs matter: the negative covariance is subtracted within the pairwise total, not omitted.

\begin{aligned} \sum_{i=1}^3\operatorname{Var}(X_i)&=4+9+16=29,\\ \sum_{i<j}\operatorname{Cov}(X_i,X_j)&=2+(-1)+3=4,\\ \operatorname{Var}(S)&=29+2(4)=37. \end{aligned}

Thus $\operatorname{Var}(S)=37$ thousands of dollars squared, or $37{,}000{,}000$ dollars squared. If the components were incorrectly treated as independent, the result would be $29$ thousands of dollars squared. The correction is positive overall because the two positive covariances sum to more than the magnitude of the negative one. The supplied covariance values are individually within the allowable bound $|\operatorname{Cov}(X_i,X_j)|\leq\sqrt{\operatorname{Var}(X_i)\operatorname{Var}(X_j)}$; the total calculation uses all three as stated.

Common Exam Traps and a Reliable Procedure

The independence shortcut is valid only when the covariance correction is zero. Mutual independence guarantees zero covariance when the relevant variances are finite, but observing that the risks have separate descriptions or different labels does not establish independence. Also, zero covariance alone does not generally prove independence. Use the information supplied about joint behavior or covariance rather than inferring a relationship from the context.

A frequent arithmetic error is to add each covariance only once. For two variables, the cross-products in the square appear in both orders, so the covariance is multiplied by $2$. With three variables, add the three distinct covariances first, then double their total. Do not double each individual variance.

Exam procedure: Write the sum being analyzed; record each individual variance; list every distinct pair and its covariance; add the pairwise covariances with their signs; multiply that sum by $2$; and check that the result has squared units. If a covariance is not given, determine whether the problem supplies enough joint information to calculate it. Do not set it to zero without a stated or justified condition.

For indicator claims, the most common dependence error is replacing $P(A\cap B)$ with $P(A)P(B)$ without an independence assumption. That replacement is exactly the independence condition for the two events. A second trap is confusing a payment amount with a probability: when a fixed amount multiplies an indicator, its variance is multiplied by the square of the amount, while the covariance of two payments is multiplied by the product of their amounts.

Correct covariance adjustments are useful in aggregate-risk calculations whenever risks can respond to common conditions or otherwise co-vary. This is aligned with the SOA Exam P (Probability) syllabus: identify the relevant dependence, compute the covariance carefully, and use it to correct the variance of the total.

Check Your Understanding

For each problem, use the stated dependence information; do not assume independence unless it is given.

A hypothetical pair of payments is $X=4{,}000I_A$ and $Y=6{,}000I_B$, with $P(A)=0.2500$, $P(B)=0.4000$, and $P(A\cap B)=0.0800$. Find $\operatorname{Var}(X+Y)$ in dollars squared. (Answer: $10{,}680{,}000$ dollars squared)
Three risk components have variances $2$, $5$, and $7$, and pairwise covariances $0.5$, $-1$, and $1.5$, respectively, all in thousands of dollars squared. Find the variance of their sum. (Answer: $16$ thousands of dollars squared)
Two random variables have variances $4$ and $9$, and covariance $0$. What is the variance of their sum, and does the covariance information alone establish that they are independent? (Answer: variance $13$; no, zero covariance alone does not establish independence)
Hypothetical payments are $X=12{,}000I_A$ and $Y=20{,}000I_B$, where $P(A)=0.0500$, $P(B)=0.1000$, and $P(A\cap B)=0.0100$. Find $\operatorname{Var}(X+Y)$ in dollars squared. (Answer: $45{,}240{,}000$ dollars squared)
Suppose $\operatorname{Var}(X)=4$, $\operatorname{Var}(Y)=9$, and $\operatorname{Cov}(X,Y)=-2$, with all quantities measured in thousands of dollars squared. Find $\operatorname{Var}(3X-Y)$. (Answer: $57$ thousands of dollars squared)

What You'll Learn