Why Dependence Matters in a Loss Portfolio
A portfolio’s aggregate loss depends on both the size of its component losses and the way those losses move together. A portfolio split across two hypothetical insurance coverages may have the same component means and variances under two models, yet a different aggregate variance if the dependence between the coverages changes. Portfolio weights matter as well: a larger weight on a volatile component can increase risk, while a larger weight on a component that offsets others can reduce it.
This tutorial studies portfolio risk through the mean and variance of a weighted sum. The calculations require finite second moments and a specified vector of component means and covariances. They do not require the component losses to be independent or normally distributed. Unless a problem states otherwise, treat each loss as a hypothetical one-period amount in the stated currency units: there is no discounting, so the effective annual interest rate is \(i=0\), and the losses are not life-contingent, so no mortality assumption is used. A distribution is not assumed when only means and covariances are supplied.
The Covariance-Matrix Calculation
Write \(\boldsymbol{X}=(X_1,\ldots,X_n)^\mathsf{T}\) and \(\boldsymbol{w}=(w_1,\ldots,w_n)^\mathsf{T}\). The covariance matrix \(\boldsymbol{\Sigma}\) has entry \(\operatorname{Cov}(X_j,X_k)\) in row \(j\), column \(k\). In particular, its diagonal entries are the component variances. Then \(L=\boldsymbol{w}^\mathsf{T}\boldsymbol{X}\), and
Here \(\boldsymbol{\mu}\) is the vector of component means. The weights may be portfolio shares that sum to one, but the formula also applies to general fixed multipliers. A negative weight has a mathematical interpretation as an offsetting position; for insurance-loss allocations, weights are often nonnegative. The variance formula applies in either case.
Why the Variance Formula Has Cross Terms
The covariance terms appear because the component losses can move together. To see the formula directly, use \(L-E[L]=\sum_j w_j(X_j-E[X_j])\), square this expression, and take expectations. The terms involving a single component give the weighted variances. For each distinct pair \(j,k\), the square contains both the \(j,k\) product and the \(k,j\) product; each has expectation \(w_jw_k\operatorname{Cov}(X_j,X_k)\). Those equal contributions combine to give the factor of 2.
This justification uses only finite second moments and the definition of covariance. Independence is not required. If the components are independent, their pairwise covariances are zero, so the cross terms vanish; zero covariance alone does not generally establish independence.
Worked Example: A Two-Coverage Portfolio Under Three Dependence Assumptions
A hypothetical insurer considers a one-period portfolio with losses \(X_1\) and \(X_2\), measured in thousands of dollars. The stated means are \(E[X_1]=40\) and \(E[X_2]=60\); standard deviations are 10 and 20. The portfolio uses weights \(w_1=0.6\) and \(w_2=0.4\). First calculate risk when the correlation is \(0.3\), then compare it with independence and perfect positive correlation. Interest is \(i=0\), there is no mortality assumption, and no joint distribution is specified or needed for these mean-and-variance calculations.
Method. Convert the stated correlation into a covariance, then apply the weighted-sum formulas. The covariance is \(\operatorname{Cov}(X_1,X_2)=0.3(10)(20)=60\) thousand-dollars squared. The covariance matrix is therefore \(\begin{pmatrix}100&60\\60&400\end{pmatrix}\).
The expected portfolio loss is \(0.6(40)+0.4(60)=24+24=48\) thousand dollars.
Its variance is \(0.6^2(100)+0.4^2(400)+2(0.6)(0.4)(60)\). The three contributions are \(36\), \(64\), and \(28.8\), respectively, so the variance is \(36+64+28.8=128.8\) thousand-dollars squared. The standard deviation is \(\sqrt{128.8}=11.349\ldots\), or \(11.35\) thousand dollars to two decimal places.
Under independence, the covariance is zero, so the variance would be \(36+64=100\) and the standard deviation would be \(10.00\) thousand dollars. Under perfect positive correlation, the covariance is \(10(20)=200\), giving variance \(36+64+2(0.6)(0.4)(200)=196\) and standard deviation \(14.00\) thousand dollars. The positive dependence in the stated case raises variance above the independence result, but not to the perfect-positive-correlation result.
Check. The mean is a weighted average of the component means, \(48\) thousand dollars, and the covariance contribution is positive. The computed variance is consequently greater than the independence variance. The common mistake this method avoids is treating correlation \(0.3\) as the covariance; the covariance is \(60\), not \(0.3\).
Weights and Several Component Losses
For a portfolio with more than two components, the matrix form organizes the same calculation without omitting pairs. The diagonal terms describe each component’s contribution after weighting. Each off-diagonal covariance contributes twice in the expanded variance. A covariance matrix must be positive semidefinite, which ensures that every weighted portfolio variance \(\boldsymbol{w}^\mathsf{T}\boldsymbol{\Sigma}\boldsymbol{w}\) is nonnegative.
The mean and variance should be kept distinct. Changing weights can change both, so a lower variance alone does not establish that one portfolio is preferable if its expected loss also differs. A risk comparison should state the weights and, where relevant, compare portfolios with an appropriate common objective such as the same expected loss.
Worked Example: Comparing Weights Using a Three-Loss Covariance Matrix
For a hypothetical one-period portfolio of three coverages, losses \(X_1,X_2,X_3\) are measured in thousands of dollars. Their means are \(20,30,50\), and their covariance matrix, in squared-thousands units, is \(\begin{pmatrix}9&6&3\\6&16&8\\3&8&25\end{pmatrix}\). Compare weights \((0.5,0.3,0.2)\) with equal weights \((1/3,1/3,1/3)\). Interest is \(i=0\); the losses are not life-contingent; and no distributional form is assumed.
Method. Calculate each portfolio’s mean by weighting the component means and its variance by multiplying the covariance matrix on both sides by the weight vector. The matrix’s diagonal entries imply standard deviations \(3,4,5\) thousand dollars. Its covariances are \(6,3,8\). For example, the first correlation is \(6/(3\cdot4)=0.5\); the covariance matrix, rather than correlations alone, is used in the variance calculation.
For weights \((0.5,0.3,0.2)\), the mean is \(0.5(20)+0.3(30)+0.2(50)=10+9+10=29\) thousand dollars.
The diagonal variance contributions are \(0.5^2(9)=2.25\), \(0.3^2(16)=1.44\), and \(0.2^2(25)=1.00\), totaling \(4.69\). The three covariance contributions, including the factor of 2, are \(2(0.5)(0.3)(6)=1.80\), \(2(0.5)(0.2)(3)=0.60\), and \(2(0.3)(0.2)(8)=0.96\). Thus the variance is \(4.69+1.80+0.60+0.96=8.05\) squared-thousands, and the standard deviation is \(\sqrt{8.05}=2.837\ldots\), or \(2.84\) thousand dollars to two decimal places.
For equal weights, the mean is \((20+30+50)/3=33.333\ldots\) thousand dollars. The variance contributions are the diagonal total \((9+16+25)/9=50/9\) and the covariance total \(2(6+3+8)/9=34/9\). The variance is therefore \(84/9=9.333\ldots\) squared-thousands, and the standard deviation is \(\sqrt{84/9}=3.055\ldots\), or \(3.06\) thousand dollars to two decimal places.
Check. The first weighting has lower variance, \(8.05<9.333\ldots\), but it also has a lower mean, \(29<33.333\ldots\). These results compare the stated weighted loss measures; they do not by themselves show that either portfolio is preferable. The common mistake this calculation avoids is comparing variances while overlooking the change in expected loss.
Holding Marginal Risk Fixed and Changing Dependence
A useful comparison isolates dependence by holding component means, variances, and weights fixed. In that comparison, changing the covariance changes aggregate variance through the cross terms. Positive covariance increases variance when both weights are positive; negative covariance decreases it. These directional statements depend on the signs of the weights as well as the covariances.
Worked Example: Equal Allocation Under Positive and Negative Dependence
A hypothetical portfolio equally weights three one-period losses \(X_1,X_2,X_3\), each measured in thousands of dollars. Each has mean \(25\) and standard deviation \(10\). Compare three dependence models: all pairwise correlations are \(0\), all are \(0.25\), or all are \(-0.25\). Interest is \(i=0\), there is no mortality assumption, and the joint distribution is otherwise unspecified.
Method. With equal weights, the aggregate loss is \(L=(X_1+X_2+X_3)/3\). Its mean in every model is \((25+25+25)/3=25\) thousand dollars. Each variance is \(100\), and each pairwise covariance is \(100\rho\), where \(\rho\) is the common correlation.
There are three component variances and three distinct covariance pairs, so \(\operatorname{Var}(L)=\{3(100)+2[3(100\rho)]\}/9 =100(3+6\rho)/9\). For \(\rho=0\), the variance is \(300/9=33.333\ldots\), giving standard deviation \(5.77\) thousand dollars to two decimal places.
For \(\rho=0.25\), the variance is \(100(3+1.5)/9=50\), giving standard deviation \(\sqrt{50}=7.07\) thousand dollars to two decimal places. For \(\rho=-0.25\), the variance is \(100(3-1.5)/9=16.666\ldots\), giving standard deviation \(\sqrt{16.666\ldots}=4.08\) thousand dollars to two decimal places.
Check. All three models have the same mean and component standard deviations, but aggregate standard deviation ranges from \(4.08\) to \(7.07\) thousand dollars. Each equicorrelation model is valid: for three variables, the equicorrelation matrix is positive semidefinite when \(-1/2\leq\rho\leq1\), and all three stated values satisfy this condition. The common mistake this comparison avoids is attributing the risk difference to changed component volatility; here only dependence changes.
Interpretation and Exam Traps
The variance calculation measures dispersion around the expected aggregate loss. It does not, on its own, describe the entire loss distribution, a tail probability, or a worst-case loss. Two portfolios can have the same mean and variance but different distributions. No normal approximation or probability of exceeding a threshold follows merely from a covariance matrix.
When using the formula, keep units consistent: if losses are in thousands of dollars, variances and covariances are in squared-thousands of dollars. Also verify that a supplied covariance matrix is symmetric and positive semidefinite before treating it as a possible joint covariance structure. A negative portfolio variance signals an arithmetic error or an invalid covariance matrix, not a possible risk outcome.
- Using correlation as covariance. Convert with \(\operatorname{Cov}(X_j,X_k)=\rho_{jk}\sigma_j\sigma_k\).
- Omitting a factor of 2. In a sum over distinct pairs \(j<k\), each covariance term is multiplied by 2.
- Assuming independence without being told. Independence makes the cross-covariances zero, but a nonzero correlation requires them to be included.
- Comparing only variances. If weights change, expected loss may change too; state both when comparing portfolios.
- Inferring a full distribution from moments. Means and covariances determine the mean and variance of a weighted sum, not generally its complete distribution.
Check Your Understanding
For each hypothetical one-period loss problem, use the stated means, weights, and covariance information. Unless stated otherwise, values are in thousands of dollars, interest is \(i=0\), losses are not life-contingent, and no distributional form is assumed.
- Two losses have standard deviations \(8\) and \(12\), correlation \(0.5\), and weights \(0.25\) and \(0.75\). Find the aggregate variance and standard deviation. (Answer: variance \(103\) squared-thousands; standard deviation \(10.15\) thousand dollars.)
- A three-loss portfolio has weights \((0.2,0.5,0.3)\) and covariance matrix \(\begin{pmatrix}4&2&-1\\2&9&3\\-1&3&16\end{pmatrix}\). Calculate its variance, including every cross term. (Answer: \(5.51\) squared-thousands.)
- Two losses have standard deviations \(5\) and \(7\), correlation \(-0.2\), and weights \(0.4\) and \(0.6\). Find their covariance and the aggregate variance. (Answer: covariance \(-7\) squared-thousands; aggregate variance \(18.28\) squared-thousands.)
- Two losses have means \(30\) and \(50\), and covariance matrix \(\begin{pmatrix}25&15\\15&100\end{pmatrix}\). For weights \((0.8,0.2)\), find the correlation, portfolio mean, and portfolio variance. (Answer: correlation \(0.3\); mean \(34\) thousand dollars; variance \(24.8\) squared-thousands.)
- Three equally weighted losses each have variance \(64\); every pair has covariance \(-16\). Find the aggregate mean if each component mean is \(18\), and find the aggregate variance and standard deviation. (Answer: mean \(18\) thousand dollars; variance \(32/3=10.6667\) squared-thousands; standard deviation \(3.27\) thousand dollars.)