Why Component Covariances Determine Aggregate VaR
An insurer or risk manager often models losses by risk component, such as separate hypothetical lines of business or risk categories. The aggregate loss is the sum of those component losses. If the components are dependent, their covariance affects the variability of the sum—and therefore its VaR. Adding component VaRs, or calculating aggregate variability as if the components were independent, can give the wrong result.
This tutorial derives the aggregate VaR formula under a joint normal model and applies it at several confidence levels. The key inputs are the component means and the full covariance matrix. All loss amounts in the examples are hypothetical and refer to a common time horizon, currency, and loss basis.
The Joint Normal Model
Assume that the vector of component losses is jointly multivariate normal: \( \mathbf{X}=(X_1,\ldots,X_n)^\mathsf{T}\sim N_n(\boldsymbol{\mu},\boldsymbol{\Sigma}) \). Here, \(\boldsymbol{\mu}\) contains the component means and \(\boldsymbol{\Sigma}\) is their covariance matrix. Its diagonal entries are the component variances, and its off-diagonal entries are the pairwise covariances. Joint normality is an assumption about the whole vector, not just about each component viewed separately.
Let \(\mathbf{1}\) be the vector of \(n\) ones, so that \(L=\mathbf{1}^\mathsf{T}\mathbf{X}\). A linear combination of jointly normal variables is normal. Its mean and variance are therefore:
Since \( \operatorname{Cov}(X_i,X_i)=\operatorname{Var}(X_i) \), the variance can also be written as:
The factor of 2 appears because each covariance between distinct components occurs twice in the double sum: once as \(\operatorname{Cov}(X_i,X_j)\) and once as \(\operatorname{Cov}(X_j,X_i)\). A covariance matrix is symmetric, so these entries are equal.
If \(\sigma_L>0\), standardizing \(L\) gives \( (L-\mu_L)/\sigma_L\sim N(0,1) \). Let \(z_p=\Phi^{-1}(p)\), where \(\Phi\) is the standard normal cumulative distribution function. Then:
For example, commonly used standard normal quantiles are \(z_{0.90}=1.2816\), \(z_{0.95}=1.6449\), and \(z_{0.99}=2.3263\). If \(\sigma_L=0\), the aggregate loss is constant under the model and its VaR is \(\mu_L\) at every confidence level. The VaR formula above is not a general formula for non-normal aggregate losses.
Why the Formula Holds
The distribution and VaR formulas follow from the behavior of linear combinations of a multivariate normal vector. The covariance expression can also be verified by expanding the variance of a sum:
\(L=\sum_i X_i\). The mean is additive, so \(E[L]=\sum_i E[X_i]=\sum_i\mu_i\).
By bilinearity of covariance, \(\operatorname{Var}(L)=\operatorname{Cov}(\sum_i X_i,\sum_j X_j)=\sum_i\sum_j\operatorname{Cov}(X_i,X_j)\).
A linear combination of a jointly normal vector is normal, so \(L\sim N(\mu_L,\sigma_L^2)\).
For \(\sigma_L>0\), \(P(L\leq \mu_L+z_p\sigma_L)=\Phi(z_p)=p\). Thus the \(p\)-quantile is \(\mu_L+z_p\sigma_L\).
This justification depends on the joint normal model for the quantile step. The mean and variance formulas themselves hold whenever the component means and covariances exist, but those two moments alone do not generally determine VaR.
Worked Calculations
Worked Example: VaR From a Two-Component Covariance Matrix
Hypothetical scenario. Two component losses, \(X_1\) and \(X_2\), are measured in millions of dollars over the same annual horizon. Assume they are jointly normal with mean vector \((8,12)^\mathsf{T}\) and covariance matrix below. Calculate aggregate VaR at 95% and 99%.
Method. First add the means. Then use the diagonal variances and both off-diagonal covariance entries to calculate aggregate variance.
The aggregate mean is \(8+12=20\) million dollars. The component variances are 9 and 16, and their covariance is 6. Therefore:
in millions of dollars. At 95% confidence, using \(z_{0.95}=1.6449\):
At 99% confidence, using \(z_{0.99}=2.3263\):
Rounded to two decimal places, the VaRs are $30.01 million at 95% and $34.15 million at 99%. As a check, both exceed the aggregate mean of $20 million, and the higher confidence level gives the higher VaR. The common mistake avoided here is dropping the covariance: it contributes \(2(6)=12\) to the aggregate variance.
Worked Example: Converting Correlations to Covariances
Hypothetical scenario. Three jointly normal annual loss components are measured in millions of dollars. Their means are 2.4, 3.1, and 1.5, and their standard deviations are 2, 3, and 4. The pairwise correlations are \(\rho_{12}=0.25\), \(\rho_{13}=0.10\), and \(\rho_{23}=0.20\). Calculate aggregate VaR at 90% and 99%.
Method. Convert each correlation to a covariance using \(\operatorname{Cov}(X_i,X_j)=\rho_{ij}\sigma_i\sigma_j\). The covariances are:
The aggregate mean is \(2.4+3.1+1.5=7.0\) million dollars. The component variances are \(2^2=4\), \(3^2=9\), and \(4^2=16\). Include each pairwise covariance twice:
At 90%, using \(z_{0.90}=1.2816\):
At 99%, using \(z_{0.99}=2.3263\):
Rounded to two decimal places, the results are $14.94 million at 90% and $21.42 million at 99%. The covariance inputs have units of squared millions of dollars, as do the variances; after taking the square root, the aggregate standard deviation is back in millions of dollars. This unit check helps catch a frequent error: using a correlation directly as though it were a covariance.
Worked Example: A Negative Covariance and Its Effect on VaR
Hypothetical scenario. Two jointly normal annual loss components have means of $5 million and $2 million, standard deviations of $5 million and $3 million, and correlation \(-0.40\). Calculate aggregate VaR at 95% and 99%, and assess the effect of the negative dependence.
Method. First calculate the covariance, then use it in the variance of the sum. The covariance is:
in squared millions of dollars. The aggregate mean and variance are:
At 95%:
At 99%:
Rounded to two decimal places, the VaRs are $14.72 million at 95% and $17.91 million at 99%. For comparison, omitting covariance would give variance \(25+9=34\), an overstatement of aggregate variance in this example. A negative covariance reduces the variance of the sum, but it does not make the aggregate variance negative: here it remains 22. Check that the specified covariance matrix is valid before relying on the joint normal model.
Interpretation, Scope, and Common Traps
The formula reports the loss quantile itself. It is not the same as the amount above expected loss, sometimes called unexpected loss at confidence level \(p\). Under the joint normal model, that amount is \(z_p\sigma_L\), while VaR is \(\mu_L+z_p\sigma_L\). State which quantity is requested and whether the answer should include the mean.
The covariance matrix must describe the same component losses on the same time horizon, currency, and measurement basis. Variances have squared loss units; covariances do too. If correlations are given instead, convert each one using the corresponding standard deviations before forming aggregate variance. A valid covariance matrix is symmetric and positive semidefinite; a matrix that fails this condition cannot be the covariance matrix of a random vector.
- Omitting dependence: The independent-components formula applies only when the relevant covariances are zero. Otherwise, include all covariance terms.
- Counting a covariance once: In the variance-of-a-sum expression using \(i<j\), each distinct covariance is multiplied by 2.
- Adding component VaRs: In general, the sum of component quantiles is not the quantile of the aggregate. Calculate the aggregate distribution under the model instead.
- Assuming marginal normality is enough: Normal marginal distributions and a covariance matrix do not, by themselves, establish joint normality. The exact Gaussian VaR formula requires a jointly normal vector.
- Over-interpreting Gaussian tails: If the joint loss distribution is not adequately represented by a normal model, this formula need not give an accurate tail quantile. Its result is conditional on the model assumption.
Check Your Understanding
For each hypothetical problem, assume joint normality where stated. Round monetary VaRs to two decimal places in the units given.
- Two annual loss components, in millions of dollars, have means 6 and 4, variances 4 and 9, and covariance 2. Calculate aggregate VaR at 95%. (Answer: $16.78 million)
- Three jointly normal annual losses, in millions, have means 1.2, 2.8, and 1.0; standard deviations 1, 2, and 3; and correlations \(\rho_{12}=0.30\), \(\rho_{13}=0\), and \(\rho_{23}=0.20\). Calculate aggregate VaR at 90%. (Answer: $10.38 million)
- Two jointly normal components have means $3 million and $6 million, variances 16 and 25 million dollars squared, and covariance \(-4\) million dollars squared. Calculate aggregate VaR at 99%. (Answer: $22.36 million)
- Two jointly normal losses, in thousands of dollars, have means 10 and 20, standard deviations 2 and 4, and correlation 0.50. Calculate aggregate VaR at 95%. (Answer: $38.70 thousand)
- Two loss components each have a normal marginal distribution, but their dependence structure is unspecified. Their means, variances, and covariance are known. Is that information sufficient to claim the exact aggregate VaR is the Gaussian formula? State the additional assumption needed. (Answer: No; joint normality of the component vector is also required.)