Deriving Portfolio VaR from Covariance Inputs

From Factor Inputs to Portfolio VaR

A portfolio may respond to several risk factors at once: interest rates, exchange rates, equity prices, or other modeled quantities. Knowing each factor's volatility is not enough to determine portfolio risk. The result also depends on how the factors move together and on how sensitive the portfolio loss is to each one.

This tutorial derives the variance of a linear portfolio loss from a vector of factor sensitivities and a covariance matrix. It then uses that variance in the normal VaR formula. The result is exact when the factor vector is jointly normal and the loss is affine in those factors; it can also be used as a linear-normal approximation when a nonlinear portfolio is locally approximated by a linear one.

Definition: Let $Z=(Z_1,\ldots,Z_k)^\mathsf{T}$ be a vector of risk-factor changes, with mean vector $\mu_Z$ and covariance matrix $\Sigma$. A linear portfolio loss is $L=c+g^\mathsf{T}Z$, where $c$ is a constant and $g=(g_1,\ldots,g_k)^\mathsf{T}$ is the vector of loss sensitivities. The entries of $\Sigma$ are $\Sigma_{ij}=\operatorname{Cov}(Z_i,Z_j)$. The mean loss is $\mu_L=c+g^\mathsf{T}\mu_Z$.

The sensitivity $g_i$ measures the change in modeled loss for a one-unit change in factor $Z_i$, holding the other factors fixed. A positive sensitivity means the loss increases as that factor increases; a negative sensitivity means it decreases. The units of each sensitivity and factor must be compatible, and the resulting loss and variance units follow from those inputs.

A covariance matrix is symmetric and positive semidefinite: for every real vector $a$, $a^\mathsf{T}\Sigma a\geq 0$. It may have zero variance in a particular direction, so positive semidefiniteness guarantees nonnegative portfolio variance, not necessarily strictly positive variance.

Deriving the Portfolio Variance

The variance of the linear loss follows directly from the definition of covariance. Since subtracting a constant does not change variance,

\operatorname{Var}(L) =\operatorname{Var}(c+g^\mathsf{T}Z) =\operatorname{Var}(g^\mathsf{T}Z) =g^\mathsf{T}\Sigma g.

Expanding the quadratic form makes the role of the diagonal and off-diagonal entries explicit:

\operatorname{Var}(L) =\sum_{i=1}^{k}g_i^2\operatorname{Var}(Z_i) +2\sum_{i<j}g_i g_j\operatorname{Cov}(Z_i,Z_j).

To see why the cross terms appear, write the centered loss as $\sum_i g_i(Z_i-E[Z_i])$. Its variance is the expected square of that sum. Squaring produces a term $g_i^2(Z_i-E[Z_i])^2$ for each factor and a pair of terms for each distinct pair $i,j$. Those two terms have the same expectation, $g_i g_j\operatorname{Cov}(Z_i,Z_j)$, giving the factor of 2.

If factor standard deviations $\sigma_i$ and correlations $\rho_{ij}$ are supplied instead, convert them first using $\operatorname{Cov}(Z_i,Z_j)=\rho_{ij}\sigma_i\sigma_j$. The diagonal entries are $\sigma_i^2$. Correlation is dimensionless; it is not itself a covariance unless the factor standard deviations happen to equal one.

Theorem: If $Z$ is jointly multivariate normal and $L=c+g^\mathsf{T}Z$, then $L$ is normal with mean $\mu_L=c+g^\mathsf{T}\mu_Z$ and variance $\sigma_L^2=g^\mathsf{T}\Sigma g$. For $0<\alpha<1$, its VaR at confidence level $\alpha$ is $$ \operatorname{VaR}_{\alpha}(L)=\mu_L+z_{\alpha}\sigma_L, $$ where $z_\alpha$ is the $\alpha$-quantile of the standard normal distribution. This formula requires a normal loss model (or an explicitly stated linear-normal approximation); having only a mean and covariance does not in general make the loss normal.

This result follows because a linear combination of jointly normal variables is normal. Its mean is $c+g^\mathsf{T}\mu_Z$, and the variance calculation above gives $g^\mathsf{T}\Sigma g$. A normal variable with mean $\mu_L$ and standard deviation $\sigma_L$ has $\alpha$-quantile $\mu_L+z_\alpha\sigma_L$. For example, $z_{0.90}\approx1.2816$, $z_{0.95}\approx1.6449$, and $z_{0.99}\approx2.3263$.

Worked Example: Two Factors With a Negative Sensitivity

A hypothetical portfolio has mean one-period loss of $8,000. Its two factor changes are jointly normal, with covariance matrix, in factor-units squared, $\Sigma=\begin{pmatrix}4&1\\1&9\end{pmatrix}$. The loss sensitivities are $g=(2,-1)^\mathsf{T}$, in thousands of dollars per factor unit. Find the 95% VaR. Use $z_{0.95}=1.644854$, and report to the nearest dollar.

Method. Compute the variance with the quadratic form, take its square root, and substitute the resulting standard deviation and mean loss into the normal VaR formula. The sensitivities express loss in thousands of dollars, so the mean and final VaR are also in thousands of dollars until converted.

First calculate the matrix product:

\Sigma g = \begin{pmatrix}4&1\\1&9\end{pmatrix} \begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}8-1\\2-9\end{pmatrix} = \begin{pmatrix}7\\-7\end{pmatrix}.

Then the portfolio variance is

g^\mathsf{T}\Sigma g =(2,-1)\begin{pmatrix}7\\-7\end{pmatrix} =14+7=21 \quad\text{(thousands of dollars)}^2.

Thus $\sigma_L=\sqrt{21}\approx4.582576$ thousand dollars. The 95% VaR is

8+1.644854\sqrt{21} \approx 8+7.537668 =15.537668\text{ thousand dollars}.

Converting to dollars gives approximately $15,537.67, or $15,538 to the nearest dollar. As an arithmetic check, the expanded variance is $2^2(4)+(-1)^2(9)+2(2)(-1)(1)=16+9-4=21$. The negative sensitivity reduces the covariance contribution here; it does not remove either factor's diagonal variance. A common mistake is to omit the cross term, which would give 25 instead of 21 and overstate this portfolio's variance.

Worked Example: Three Factors and a Covariance Matrix

A hypothetical portfolio has mean loss of $20,000. Its jointly normal factor changes have covariance matrix $\Sigma=\begin{pmatrix}4&1&0\\1&1&0\\0&0&9\end{pmatrix}$, and its loss sensitivities are $g=(1,2,-1)^\mathsf{T}$, measured in thousands of dollars per factor unit. Find the 90% VaR using $z_{0.90}=1.281552$. Report to the nearest dollar.

Method. Multiply the covariance matrix by the sensitivity vector, then take the inner product with the sensitivity vector. This matrix-product check helps prevent sign and row-column errors.

\Sigma g = \begin{pmatrix}4&1&0\\1&1&0\\0&0&9\end{pmatrix} \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}4+2\\1+2\\-9\end{pmatrix} = \begin{pmatrix}6\\3\\-9\end{pmatrix}.

Therefore,

g^\mathsf{T}\Sigma g =(1,2,-1)\begin{pmatrix}6\\3\\-9\end{pmatrix} =6+6+9=21.

The variance is $21$ thousand-dollars squared, and the standard deviation is $\sqrt{21}\approx4.582576$ thousand dollars. Hence

\operatorname{VaR}_{0.90} =20+1.281552\sqrt{21} \approx20+5.872809 =25.872809\text{ thousand dollars}.

The 90% VaR is approximately $25,872.81, or $25,873 to the nearest dollar. The covariance matrix is symmetric and positive semidefinite: its upper-left block has positive diagonal entries and determinant $4(1)-1(1)=3>0$, while the remaining independent variance is 9. A common mistake is to treat the off-diagonal covariance as if it applied to every pair of factors; here the third factor has zero covariance with each of the first two, so its cross terms are zero.

Worked Example: Building Covariances From Correlations

For a hypothetical portfolio, three jointly normal factor changes have standard deviations $2,3,1$ factor units and correlations $\rho_{12}=0.5$, $\rho_{13}=-0.2$, and $\rho_{23}=0.4$. The mean portfolio loss is $5,000, and $g=(1,-2,3)^\mathsf{T}$ in thousands of dollars per factor unit. Find the 99% VaR using $z_{0.99}=2.326348$. Report to three decimals in thousands of dollars.

Method. Convert each correlation to covariance by multiplying by the two standard deviations, fill the diagonal with variances, and evaluate the quadratic form. The correlation inputs describe dependence, not the scale of factor movements.

The covariances are $\operatorname{Cov}(Z_1,Z_2)=0.5(2)(3)=3$, $\operatorname{Cov}(Z_1,Z_3)=-0.2(2)(1)=-0.4$, and $\operatorname{Cov}(Z_2,Z_3)=0.4(3)(1)=1.2$. The factor variances are $4,9,1$. Thus

\Sigma= \begin{pmatrix} 4&3&-0.4\\ 3&9&1.2\\ -0.4&1.2&1 \end{pmatrix}.

Using the expanded formula, the variance is

\begin{aligned} g^\mathsf{T}\Sigma g &=1^2(4)+(-2)^2(9)+3^2(1)\\ &\quad+2(1)(-2)(3)+2(1)(3)(-0.4)+2(-2)(3)(1.2)\\ &=4+36+9-12-2.4-14.4\\ &=20.2. \end{aligned}

The standard deviation is $\sqrt{20.2}\approx4.494441$ thousand dollars. Therefore,

\operatorname{VaR}_{0.99} =5+2.326348\sqrt{20.2} \approx5+10.455634 =15.455634\text{ thousand dollars}.

Rounded to three decimals in thousands of dollars, the 99% VaR is $15.456 thousand. The signs of the sensitivities matter: for instance, the first and second factors have positive covariance but opposite-signed sensitivities, so their portfolio cross term is negative. A common mistake is to insert the correlation $0.5$ directly into the covariance matrix; the correct covariance for factors 1 and 2 is $3$, not $0.5$.

Checks, Assumptions, and Interpretation

A reliable calculation has three checks. First, confirm the factor ordering in $g$ matches the row and column ordering in $\Sigma$. Second, check the units: if sensitivities are in thousands of dollars per factor unit, $g^\mathsf{T}\Sigma g$ is in thousands of dollars squared, and its square root is in thousands of dollars. Third, confirm that the covariance matrix is symmetric and gives a nonnegative quadratic form. A negative computed variance signals an invalid covariance input or an arithmetic or alignment error.

The covariance terms can either increase or reduce portfolio variance. Positively correlated factors with sensitivities of the same sign increase it; positively correlated factors with sensitivities of opposite signs reduce it. Negative covariance reverses those effects. Diversification therefore depends on both factor dependence and portfolio exposures, not on correlation alone.

The normal VaR result also has substantive assumptions. If the factor vector is jointly normal and the loss is affine, the result is exact. For nonlinear positions, a sensitivity-based loss is an approximation and may become inaccurate for large factor moves. If the factor distribution is not jointly normal, the mean and covariance still determine the variance of a linear loss, but they do not by themselves determine its VaR. In those cases, the normal formula is not an exact quantile calculation.

For a hypothetical portfolio, this method makes the dependence assumptions visible and turns them into a reproducible calculation. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus: a candidate should be able to derive the portfolio variance, track the units and signs, and apply the normal quantile only when its model assumptions are appropriate.

Check Your Understanding

For each problem, compute the requested variance or VaR from the supplied linear-normal inputs. Dollar amounts labeled “thousand dollars” are in thousands.

A hypothetical two-factor portfolio has mean loss of $3 thousand, sensitivities $g=(2,1)^\mathsf{T}$, and covariance matrix $\begin{pmatrix}1&0.5\\0.5&4\end{pmatrix}$. Find the 95% VaR using $z_{0.95}=1.644854$. (Answer: approximately $8.20 thousand)
Two factor changes have standard deviations 2 and 5 units and correlation $-0.3$. A portfolio has sensitivities $g=(1,2)^\mathsf{T}$. Find its loss variance. (Answer: 92 in squared loss units)
A hypothetical portfolio has mean loss of $12 thousand, sensitivities $g=(1,-1)^\mathsf{T}$, and covariance matrix $\begin{pmatrix}9&2\\2&4\end{pmatrix}$. Find its 90% VaR using $z_{0.90}=1.281552$, to the nearest dollar. (Answer: approximately $15,845)
A linear-normal portfolio has mean loss zero, sensitivities $g=(1,1)^\mathsf{T}$, and covariance matrix $\begin{pmatrix}1&0.8\\0.8&1\end{pmatrix}$. Find its 95% VaR using $z_{0.95}=1.644854$, to three decimals. (Answer: approximately 3.121 loss units)
A hypothetical portfolio has mean loss of $4 thousand. Factor standard deviations are 2 and 3 units, their correlation is 0.25, and sensitivities are $g=(2,-1)^\mathsf{T}$ thousand dollars per factor unit. Find the 99% VaR using $z_{0.99}=2.326348$, to the nearest dollar. (Answer: approximately $14,140)

What You'll Learn