Deriving the Variance of a Sum

Why Covariance Is Part of Aggregate Variance

An aggregate loss is the sum of losses from its component risks, measured on a common basis. Its variance depends not only on how variable each component is, but also on whether components tend to be high or low together. Positive covariance adds to aggregate variance; negative covariance subtracts from it. Thus, component variances alone are not enough unless the covariances are also known or can be justified as zero.

For this tutorial, let $X_1,\ldots,X_n$ be loss random variables measured over the same period, in the same currency, and using consistent definitions of gross or net loss. Assume each has a finite second moment, so its variance and the pairwise covariances exist. The aggregate loss is $S=\sum_{i=1}^{n}X_i$. The results below do not require independence.

Definition: For random variables $X_i$ and $X_j$ with finite second moments, their covariance is $\operatorname{Cov}(X_i,X_j)=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j]$. The variance of $X_i$ is $\operatorname{Var}(X_i)=\operatorname{Cov}(X_i,X_i)$. Covariance is positive when the variables tend to move together relative to their means, and negative when they tend to move in opposite directions.

If standard deviations and a correlation are supplied instead, use $\operatorname{Cov}(X_i,X_j)=\rho_{ij}\sigma_i\sigma_j$, where $\sigma_i=\sqrt{\operatorname{Var}(X_i)}$ and $\rho_{ij}$ is the correlation between the two losses. Correlation is dimensionless; covariance has squared loss units. A zero covariance means no linear co-movement, but does not in general establish independence.

The key identity follows directly from the definition of variance. Center each loss by subtracting its expectation, then expand the square of the centered aggregate. Each distinct pair appears twice in that expansion.

Center the aggregate.
Since $E[S]=\sum_{i=1}^{n}E[X_i]$, the deviation of the total from its mean is $S-E[S]=\sum_{i=1}^{n}(X_i-E[X_i])$.

Square and take expectations.
By definition, $\operatorname{Var}(S)=E[(S-E[S])^2]$. Expanding the square produces the individual squared deviations and all cross-products.

Recognize variances and covariances.
The expected squared deviation for component $i$ is $\operatorname{Var}(X_i)$; the expected cross-product for $i,j$ is $\operatorname{Cov}(X_i,X_j)$. Each unordered pair occurs in both orders.

\begin{aligned} \operatorname{Var}(S) &=E\left[\left(\sum_{i=1}^{n}(X_i-E[X_i])\right)^2\right]\\ &=\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 $\operatorname{Var}(X_1+X_2)=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+2\operatorname{Cov}(X_1,X_2)$. The factor of two is essential: expanding the square includes both the $X_1$-by-$X_2$ cross-product and the $X_2$-by-$X_1$ cross-product. For $n$ risks, the summation over $i<j$ includes each distinct pair once, so the outside factor of two accounts for both orders.

If the components are independent, their pairwise covariances are zero, and aggregate variance is the sum of component variances. Independence is sufficient for this conclusion, but it is not necessary: uncorrelated components also have zero covariance. If dependence is present, the covariance terms must be included. A deterministic amount added to the aggregate changes its mean but not its variance.

Applying the Formula to Two Risks

Worked Example: Two Risks With a Given Correlation

A hypothetical insurer models annual net losses $X_1$ and $X_2$, both in thousands of dollars. Their standard deviations are 20 and 30, and their modeled correlation is 0.25. Calculate the variance and standard deviation of the combined loss. Treat the supplied standard deviations and correlation as model inputs for this same annual basis.

Method. Convert the correlation to covariance, then apply the two-risk variance identity. The common mistake is to add the two variances while ignoring positive dependence.

First, the component variances are $20^2=400$ and $30^2=900$, in thousands-of-dollars squared. The covariance is $\operatorname{Cov}(X_1,X_2)=0.25(20)(30)=150$, also in thousands-of-dollars squared.

\begin{aligned} \operatorname{Var}(X_1+X_2) &=400+900+2(150)\\ &=1{,}300+300\\ &=1{,}600\quad (\text{thousands of dollars})^2. \end{aligned}

The standard deviation is $\sqrt{1{,}600}=40$ thousand dollars. Therefore the aggregate variance is $1{,}600$ thousand-dollars squared and its standard deviation is $40,000. As a check, omitting covariance would give variance $1{,}300$, which is too small because the modeled covariance is positive.

Finding Covariance From a Joint Loss Model

Sometimes the dependence input is not supplied as a correlation. A joint probability model can be used to calculate $E[X_1X_2]$, and then covariance follows from $E[X_1X_2]-E[X_1]E[X_2]$. The product expectation must use the joint outcomes; multiplying the marginal means instead would assume the desired result rather than calculate it.

Worked Example: Covariance From Joint Event Probabilities

Consider two hypothetical annual loss components, in dollars. Loss $X_1$ is $100 when event A occurs and zero otherwise. Loss $X_2$ is $80 when event B occurs and zero otherwise. The joint probabilities are: both events occur, 0.10; A occurs but B does not, 0.10; B occurs but A does not, 0.15; neither occurs, 0.65. Find the variance of $X_1+X_2$.

Method. Calculate each mean and variance from its marginal outcomes, then use the joint outcome to find covariance. A common mistake is to treat the events as independent even though the joint probability is specified as 0.10.

The probability of A is $0.10+0.10=0.20$, so $E[X_1]=100(0.20)=20$. Also, $E[X_1^2]=100^2(0.20)=2{,}000$, and therefore $\operatorname{Var}(X_1)=2{,}000-20^2=1{,}600$.

The probability of B is $0.10+0.15=0.25$, so $E[X_2]=80(0.25)=20$. Also, $E[X_2^2]=80^2(0.25)=1{,}600$, and therefore $\operatorname{Var}(X_2)=1{,}600-20^2=1{,}200$.

Both losses are positive together only in the joint event. Thus $E[X_1X_2]=(100)(80)(0.10)=800$, and $\operatorname{Cov}(X_1,X_2)=800-(20)(20)=400$.

\operatorname{Var}(X_1+X_2)=1{,}600+1{,}200+2(400)=3{,}600\text{ dollars}^2.

Check using the aggregate outcomes: the total is $180 with probability 0.10, $100 with probability 0.10, $80 with probability 0.15, and zero with probability 0.65. Its mean is $180(0.10)+100(0.10)+80(0.15)=40$. Its second moment is $180^2(0.10)+100^2(0.10)+80^2(0.15)=5{,}200$. Hence its variance is $5{,}200-40^2=3{,}600$ dollars squared, agreeing with the covariance calculation.

Aggregating Several Risks

For a portfolio with more than two components, make a list of every distinct pair and include its covariance once inside the pairwise sum. The calculation is manageable if the variance contributions and covariance contributions are kept separate. Covariances can be negative, so preserve their signs when summing.

Worked Example: Four Components With Mixed Covariances

A hypothetical insurer tracks four annual net loss components, $X_1,\ldots,X_4$, in thousands of dollars. Their variances, in thousands-of-dollars squared, are 64, 100, 144, and 196, respectively. The modeled pairwise covariances are $\operatorname{Cov}(X_1,X_2)=12$, $\operatorname{Cov}(X_1,X_3)=-8$, $\operatorname{Cov}(X_1,X_4)=20$, $\operatorname{Cov}(X_2,X_3)=5$, $\operatorname{Cov}(X_2,X_4)=-10$, and $\operatorname{Cov}(X_3,X_4)=14$, in the same squared units. Calculate the variance and standard deviation of the total.

Method. Sum the four variances, then sum the six distinct covariances, retaining the negative signs. The common mistake is to omit a pair, count one covariance twice in the pair sum, or lose the sign of a negative covariance.

The sum of component variances is $64+100+144+196=504$. The sum of distinct pairwise covariances is $12-8+20+5-10+14=33$. Therefore,

\begin{aligned} \operatorname{Var}(X_1+X_2+X_3+X_4) &=504+2(33)\\ &=570\quad (\text{thousands of dollars})^2. \end{aligned}

The aggregate standard deviation is $\sqrt{570}=23.874672\ldots$ thousand dollars, or $23,874.67 when rounded to the nearest cent. The variance has units of squared thousands of dollars; its square root returns to thousands of dollars. The final result includes all six pairs and both contributions from each covariance through the factor of two.

Interpretation and Common Traps

The covariance terms describe how component deviations combine. A positive covariance increases the variance of the total relative to the sum of standalone variances; a negative covariance decreases it. This is an algebraic property of the specified joint model, not a conclusion about whether a portfolio is adequately capitalized or how a tail risk measure behaves.

Dropping the factor of two. For two risks, $ \operatorname{Var}(X_1+X_2)$ contains $2\operatorname{Cov}(X_1,X_2)$, not just one covariance.
Adding variances as if risks were independent. This is justified only when the relevant covariances are zero; independence is one sufficient condition.
Confusing correlation and covariance. Correlation must be multiplied by both standard deviations to obtain covariance, with the correct loss units.
Mixing loss bases. Components must represent comparable periods, currency, and gross or net definitions before their sum and covariance terms have the intended meaning.
Discarding negative signs. Negative covariances reduce the pairwise contribution; replacing them with absolute values changes the model calculation.

Key takeaway: For finite-variance component losses on a consistent basis, aggregate variance equals the sum of component variances plus twice the sum of each distinct pairwise covariance. Use independence only when it is an appropriate justification for zero covariance; do not infer it merely from having separate risk labels.

Check Your Understanding

For each problem, use the variance-of-a-sum formula and retain covariance signs and units.

Two annual loss components have variances 49 and 81 thousand-dollars squared and covariance 21 thousand-dollars squared. Find their aggregate variance. (Answer: 172 thousand-dollars squared)
Three loss components have variances 25, 64, and 100 dollars squared. Their pairwise covariances, in the order (1,2), (1,3), and (2,3), are 8, -5, and 12 dollars squared. Find the variance of their sum. (Answer: 219 dollars squared)
Two losses have standard deviations of $6,000 and $10,000, and correlation -0.20. Find the covariance and the variance of their sum. (Answer: covariance = -$12,000,000; aggregate variance = $112,000,000)
Three components have variances 16, 25, and 36. Their covariances are $\operatorname{Cov}(X_1,X_2)=4$ and $\operatorname{Cov}(X_1,X_3)=-3$. If the aggregate variance is 91, find $\operatorname{Cov}(X_2,X_3)$. All quantities are in squared loss units. (Answer: 6 squared loss units)
Four risks have variances 9, 16, 25, and 36. Their nonzero pairwise covariances are $\operatorname{Cov}(X_1,X_2)=3$, $\operatorname{Cov}(X_1,X_3)=-2$, $\operatorname{Cov}(X_2,X_4)=4$, and $\operatorname{Cov}(X_3,X_4)=1$; all other pairwise covariances are zero. Find the variance of the total. (Answer: 98 squared loss units)

Deriving the Variance of a Sum

What You'll Learn

Why Covariance Is Part of Aggregate Variance