Covariance in Aggregate Insurance Losses

Why Dependence Matters for Aggregate Claims

An insurer’s total claims over a period are the sum of the claims from its individual risks. Knowing each claim’s average and variance is not enough to determine the variance of that total: the claims may rise or fall together. For example, a shared event could increase claims on several policies at once, while a particular portfolio structure could make some claims move in opposite directions.

This tutorial develops the aggregate-variance calculation from the individual claim variances and their pairwise covariances. The claim amounts below are hypothetical. Unless stated otherwise, amounts are undiscounted and measured on the same basis and over the same period; no interest rate or mortality assumption enters the calculations. Each random claim amount is assumed to have a finite variance.

Definition: Let $X_1,\ldots,X_n$ be the claim amounts for $n$ risks, and define the aggregate claim amount by $S=\sum_{i=1}^n X_i$. The covariance of claims $X_i$ and $X_j$ is $\operatorname{Cov}(X_i,X_j)=E[(X_i-E[X_i])(X_j-E[X_j])]$. The covariance matrix has the individual variances on its diagonal and pairwise covariances off the diagonal.

The Aggregate Variance Formula

For the unweighted aggregate $S$, its expected value is the sum of the individual expected claim amounts. Its variance includes both each individual variance and the dependence between every distinct pair of claims.

Theorem: If $X_1,\ldots,X_n$ have finite variances, then $$ E[S]=\sum_{i=1}^n E[X_i], \qquad \operatorname{Var}(S)=\sum_{i=1}^n \operatorname{Var}(X_i) +2\sum_{1\le i<j\le n}\operatorname{Cov}(X_i,X_j). $$ The second sum contains each unordered pair once, and the factor 2 accounts for the two corresponding terms in the variance expansion.

Here is a direct justification. Write each claim as its mean plus its deviation from that mean, then expand the square of the sum of deviations. Taking expectations turns each squared deviation into an individual variance and each product of two different deviations into a covariance.

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

If every pair of claims is independent, each pairwise covariance is zero and the aggregate variance is the sum of the individual variances. Independence is sufficient for zero covariance, but zero covariance alone does not establish independence. Therefore, for dependent claims, use the covariance information supplied or derived from the model rather than assuming that the claims are independent.

For two claims, correlation can be used to obtain covariance when the standard deviations are known: $\operatorname{Cov}(X_1,X_2)=\rho_{12}\sigma_1\sigma_2$. Correlation is dimensionless; covariance has squared-money units when the claims are measured in money. Keep the units consistent throughout the calculation.

Worked Example: Two Claims with Negative Correlation

A hypothetical insurer models two undiscounted claim amounts, $X_1$ and $X_2$, over the same period. Their means are $\$1{,}000$ and $\$2{,}000$; their standard deviations are $\$2{,}000$ and $\$3{,}000$; and their correlation is $-0.40$. Find the aggregate mean, covariance, and variance. The joint distribution is not specified, so use only the supplied moments, assume finite second moments, and make no independence assumption. No interest rate or mortality assumption applies.

Method. Add means for the aggregate mean. Convert correlation to covariance, then apply the two-claim variance formula.

First, the aggregate mean is

E[X_1+X_2]=1{,}000+2{,}000=\$3{,}000.

The individual variances are the squares of the standard deviations:

\operatorname{Var}(X_1)=2{,}000^2=4{,}000{,}000, \qquad \operatorname{Var}(X_2)=3{,}000^2=9{,}000{,}000.

Next, use the correlation to calculate the covariance:

\operatorname{Cov}(X_1,X_2)=(-0.40)(2{,}000)(3{,}000) =-2{,}400{,}000\ \text{dollars}^2.

Finally, include twice this covariance in the aggregate variance:

\begin{aligned} \operatorname{Var}(X_1+X_2) &=4{,}000{,}000+9{,}000{,}000 +2(-2{,}400{,}000)\\ &=13{,}000{,}000-4{,}800{,}000\\ &=8{,}200{,}000\ \text{dollars}^2. \end{aligned}

Check. The negative covariance reduces the variance below the independent-claims value of $\$13{,}000{,}000$ dollars squared, as expected. The aggregate mean is $\$3{,}000$, the covariance is $-\$2{,}400{,}000$ dollars squared, and the aggregate variance is $\$8{,}200{,}000$ dollars squared. A common mistake is to include the covariance only once, which would give $\$10{,}600{,}000$; the expansion contains two cross-products, so the correct result includes twice the covariance.

Using a Covariance Matrix

With several risks, a covariance matrix organizes the calculation. Add the diagonal entries to get the sum of individual variances, add the distinct off-diagonal entries once, and double that off-diagonal sum. This procedure avoids omitting a pair or counting it more than once.

List the individual moments.
Record each claim’s mean and variance in consistent units.

List each distinct pair once.
Use the supplied covariances, or calculate them from correlations and standard deviations.

Apply the aggregate formula.
Add the individual variances and twice the sum of the distinct pairwise covariances.

Check scale and direction.
Confirm that all variance terms have squared-money units and that positive or negative dependence changes the result in the expected direction.

Worked Example: Three Claims from Pairwise Covariances

A hypothetical portfolio has three undiscounted claim amounts measured in thousands of dollars. Their means are $6$, $8$, and $10$; their variances are $4$, $9$, and $16$ in squared-thousand-dollar units. The pairwise covariances are $\operatorname{Cov}(X_1,X_2)=2$, $\operatorname{Cov}(X_1,X_3)=-1$, and $\operatorname{Cov}(X_2,X_3)=3$, in the same squared-thousand-dollar units. Find the aggregate mean and variance. The joint distribution is unspecified; the supplied moments and finite variances are the model information required. No interest rate or mortality assumption applies.

The covariance matrix, with rows and columns ordered $X_1,X_2,X_3$, is

\begin{pmatrix} 4&2&-1\\ 2&9&3\\ -1&3&16 \end{pmatrix}.

The means add:

E[S]=6+8+10=24\ \text{thousand dollars}.

The individual variances sum to $4+9+16=29$. The three distinct covariances sum to $2+(-1)+3=4$. Therefore,

\begin{aligned} \operatorname{Var}(S) &=29+2(4)\\ &=37\ (\text{thousand dollars})^2\\ &=37{,}000{,}000\ \text{dollars}^2. \end{aligned}

Check. There are three distinct pairs, and each covariance has been included twice in the variance expansion. The one negative covariance offsets some of the positive dependence but does not outweigh it. The aggregate mean is $24$ thousand dollars and the variance is $37$ squared thousand dollars.

How Dependence Changes Aggregate Risk

A useful comparison holds the individual claim distributions fixed and changes only the dependence between claims. Consider two hypothetical policies, each paying $\$1{,}000$ if a claim event occurs and zero otherwise. Let $I_1$ and $I_2$ indicate whether the respective claim events occur, so $X_1=1{,}000I_1$ and $X_2=1{,}000I_2$. Each indicator has probability $0.5$ of being one, so each claim has mean $\$500$ and variance $\$250{,}000$. The joint claim-event probabilities below fully specify each distribution. There is no discounting or mortality model.

Dependence case	$P(I_1=1,I_2=1)$	$\operatorname{Cov}(X_1,X_2)$	$\operatorname{Var}(X_1+X_2)$
Positive dependence	0.40	$1{,}000{,}000(0.40-0.25)=150{,}000$	$250{,}000+250{,}000+2(150{,}000)=800{,}000$
Independence	0.25	$1{,}000{,}000(0.25-0.25)=0$	$250{,}000+250{,}000=500{,}000$
Negative dependence	0.10	$1{,}000{,}000(0.10-0.25)=-150{,}000$	$250{,}000+250{,}000-2(150{,}000)=200{,}000$

To see the covariance calculation, $E[I_1I_2]=P(I_1=1,I_2=1)$, while $E[I_1]E[I_2]=0.5(0.5)=0.25$. Thus $\operatorname{Cov}(X_1,X_2)=1{,}000^2\{P(I_1=1,I_2=1)-0.25\}$. The aggregate mean remains $\$1{,}000$ in all three cases. The joint probability changes the covariance, and hence changes aggregate variance, even though each policy’s marginal claim distribution stays the same.

Worked Example: Comparing Three Dependence Structures

For the two hypothetical policies just described, calculate the aggregate variance when the probability of both claims occurring is $0.40$, $0.25$, or $0.10$. Show that each case has the same marginals and state which case produces the greatest variance. The joint distribution is determined by the stated joint probability and the two marginal probabilities of $0.5$; amounts are undiscounted, with no interest rate or mortality assumption.

Method. For each joint probability $q$, the four probabilities for $(I_1,I_2)=(1,1),(1,0),(0,1),(0,0)$ are $q,0.5-q,0.5-q,q$. For $q=0.40$, these are $0.40,0.10,0.10,0.40$; for $q=0.25$, all four are $0.25$; and for $q=0.10$, they are $0.10,0.40,0.40,0.10$. Each set is valid and preserves both marginal probabilities at $0.5$.

Each individual claim has variance $(1{,}000)^2(0.5)(0.5)=\$250{,}000$ dollars squared. For $q=0.40$, covariance is $1{,}000{,}000(0.40-0.25)=\$150{,}000$, giving aggregate variance $250{,}000+250{,}000+2(150{,}000)=\$800{,}000$ dollars squared.

For $q=0.25$, covariance is $1{,}000{,}000(0.25-0.25)=0$, giving aggregate variance $\$500{,}000$ dollars squared. For $q=0.10$, covariance is $1{,}000{,}000(0.10-0.25)=-\$150{,}000$, giving aggregate variance $250{,}000+250{,}000-300{,}000=\$200{,}000$ dollars squared.

Check. The individual means and variances are unchanged across the cases, while the aggregate variances are $\$800{,}000$, $\$500{,}000$, and $\$200{,}000$ dollars squared. Positive dependence gives the greatest aggregate variance; negative dependence gives the least.

Common Exam Traps

Forgetting the factor of 2. In the sum over distinct pairs, each covariance appears once, but the aggregate variance uses twice that sum.
Using correlation as if it were covariance. Correlation must be multiplied by both standard deviations to obtain covariance. It has no money units; covariance has squared-money units.
Assuming independence without evidence. Independence makes covariances zero, but separate policies are not automatically independent. A shared peril or other common influence may create dependence.
Misreading a negative covariance. A negative covariance reduces aggregate variance; it does not mean the aggregate variance itself is negative. A valid model’s variance cannot be negative.
Mixing units. If claim amounts are in thousands of dollars, their variances and covariances are in squared thousands of dollars. Convert the final variance by squaring the scale factor.

Key takeaway: Aggregate claim variance equals the sum of individual variances plus twice the sum of all distinct pairwise covariances. Positive covariance increases aggregate variance, negative covariance decreases it, and independence is only one special case.

Check Your Understanding

For each problem, use consistent units and include every distinct pairwise covariance in the aggregate-variance formula.

Two hypothetical claims have means of $2.4$ and $1.6$ thousand dollars, variances of $1.44$ and $0.81$ squared thousand dollars, and covariance $0.30$ squared thousand dollars. Find the aggregate mean and variance. (Answer: Mean $4.0$ thousand dollars; variance $2.85$ squared thousand dollars.)
Three claim variances are $5$, $7$, and $8$ squared thousand dollars. Their distinct pairwise covariances are $1$, $-2$, and $0.5$ squared thousand dollars. Find the aggregate variance. (Answer: $19$ squared thousand dollars.)
A portfolio amount is $T=2X_1+X_2$, where $\operatorname{Var}(X_1)=9$, $\operatorname{Var}(X_2)=16$, and $\operatorname{Cov}(X_1,X_2)=3$, all in squared thousand dollars. Find $\operatorname{Var}(T)$. (Answer: $64$ squared thousand dollars.)
Two hypothetical claims pay $\$5{,}000$ and $\$8{,}000$ when their respective events occur. The event probabilities are $0.30$ and $0.40$, and the probability both occur is $0.20$. Find the aggregate variance. (Answer: $\$27{,}010{,}000$ dollars squared.)
Three claims have standard deviations $1$, $2$, and $3$ thousand dollars. Their pairwise correlations are $\rho_{12}=0.5$, $\rho_{13}=-0.2$, and $\rho_{23}=0.1$. Find the aggregate variance. (Answer: $16$ squared thousand dollars.)

Dependence case	\(P(I_1=1,I_2=1)\)	\(\operatorname{Cov}(X_1,X_2)\)	\(\operatorname{Var}(X_1+X_2)\)
Positive dependence	0.40	\(1{,}000{,}000(0.40-0.25)=150{,}000\)	\(250{,}000+250{,}000+2(150{,}000)=800{,}000\)
Independence	0.25	\(1{,}000{,}000(0.25-0.25)=0\)	\(250{,}000+250{,}000=500{,}000\)
Negative dependence	0.10	\(1{,}000{,}000(0.10-0.25)=-150{,}000\)	\(250{,}000+250{,}000-2(150{,}000)=200{,}000\)

What You'll Learn