When VaR Is Subadditive for Jointly Normal Losses

Why Dependence Matters for Normal-Loss VaR

Value at Risk (VaR) for a normal loss distribution has a simple form: a mean plus a multiple of a standard deviation. For a portfolio, however, the standard deviation depends on how the component losses move together. The covariance term is therefore central to deciding whether portfolio VaR is below the sum of stand-alone VaRs.

For jointly normal losses, the covariance-based standard deviation satisfies a triangle inequality. That fact gives a precise subadditivity result at confidence levels at or above 50%. The confidence-level condition matters: when the normal quantile is negative, multiplying the standard-deviation inequality by it reverses the inequality.

Definition: For a loss $X$ and $0<\alpha<1$, define $\operatorname{VaR}_{\alpha}(X)$ as the $\alpha$-quantile, $\inf\{x:F_X(x)\geq \alpha\}$. If $X\sim N(\mu_X,\sigma_X^2)$, then $$ \operatorname{VaR}_{\alpha}(X)=\mu_X+z_{\alpha}\sigma_X, $$ where $z_{\alpha}=\Phi^{-1}(\alpha)$ is the $\alpha$-quantile of the standard normal distribution. This expression also applies when $\sigma_X=0$, in which case the loss is constant at $\mu_X$.

Portfolio VaR for Jointly Normal Losses

Let $X$ and $Y$ be jointly normal losses, with means $\mu_X,\mu_Y$, standard deviations $\sigma_X,\sigma_Y$, and correlation $\rho$. Their sum $X+Y$ is normal, with mean $\mu_X+\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2+2\operatorname{Cov}(X,Y)$. Since $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$, portfolio VaR is

\operatorname{VaR}_{\alpha}(X+Y) = \mu_X+\mu_Y+ z_{\alpha}\sqrt{\sigma_X^2+\sigma_Y^2+2\rho\sigma_X\sigma_Y}.

The variance formula requires the covariance, not just the two stand-alone variances. Assuming independence is valid only when the losses are independent, which for jointly normal losses is equivalent to zero correlation. If the joint distribution is not normal, the sum need not be normal and this normal-quantile formula does not generally apply.

Worked Example: Independent Normal Losses and Diversification

A hypothetical insurer models two annual loss components, in thousands of dollars, as jointly normal and independent: $X\sim N(10,9)$ and $Y\sim N(6,16)$. Calculate each 95.00% stand-alone VaR, the 95.00% VaR of the combined loss, and the difference between the sum of stand-alone VaRs and portfolio VaR. Use $z_{0.9500}=1.6449$.

Method and assumptions. The means are 10 and 6 thousand dollars, and the standard deviations are $\sqrt{9}=3$ and $\sqrt{16}=4$ thousand dollars. Independence gives covariance zero. All VaR results below are in thousands of dollars.

For $X$, the stand-alone VaR is $10+1.6449(3)=10+4.9347=14.9347$. For $Y$, it is $6+1.6449(4)=6+6.5796=12.5796$. Their sum is $14.9347+12.5796=27.5143$.

The combined mean is $10+6=16$. Its variance is $9+16+2(0)=25$, so its standard deviation is $\sqrt{25}=5$. Therefore, the portfolio VaR is $16+1.6449(5)=16+8.2245=24.2245$.

The sum of stand-alone VaRs exceeds portfolio VaR by $27.5143-24.2245=3.2898$ thousand dollars. The calculation checks: the mean parts add to 16, while the stand-alone standard deviations sum to 7 and the portfolio standard deviation is 5.

Common mistake avoided: Adding the two means but also adding the standard deviations would give $16+1.6449(3+4)=27.5143$, the sum of stand-alone VaRs, not the portfolio VaR. For an independent portfolio, the standard deviation is $\sqrt{3^2+4^2}=5$, not $3+4$.

Why the Subadditivity Inequality Holds

The key step is the triangle inequality for standard deviations. To see it directly, center each loss by subtracting its mean. The variance of the sum is the second moment of the sum of those centered losses. The covariance cannot exceed the product of their standard deviations, so the portfolio standard deviation cannot exceed their sum.

Write the portfolio variance.
Using $\operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y$, $\operatorname{Var}(X+Y)=\sigma_X^2+\sigma_Y^2+2\operatorname{Cov}(X,Y)$.

Bound the covariance.
Cauchy–Schwarz gives $\operatorname{Cov}(X,Y)\leq \sigma_X\sigma_Y$. Thus $\operatorname{Var}(X+Y)\leq \sigma_X^2+\sigma_Y^2+2\sigma_X\sigma_Y=(\sigma_X+\sigma_Y)^2$.

Take nonnegative square roots.
Since standard deviations are nonnegative, $\sigma_{X+Y}\leq \sigma_X+\sigma_Y$. This is the covariance-based standard-deviation triangle inequality.

Joint normality now permits use of the normal VaR formula for $X$, $Y$, and $X+Y$. Subtracting the portfolio VaR from the sum of stand-alone VaRs cancels the means:

\operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y) -\operatorname{VaR}_{\alpha}(X+Y) = z_{\alpha}\bigl(\sigma_X+\sigma_Y-\sigma_{X+Y}\bigr).

The bracketed term is nonnegative by the triangle inequality. If $\alpha\geq 0.5000$, then $z_{\alpha}\geq 0$, so the whole right-hand side is nonnegative. This proves subadditivity in this setting:

Theorem: If $X$ and $Y$ are jointly normal losses and $\alpha\geq 0.5000$, then $\operatorname{VaR}_{\alpha}(X+Y)\leq \operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y)$. The result includes singular jointly normal distributions. At $\alpha=0.5000$, $z_{\alpha}=0$, so equality holds. For $\alpha>0.5000$, equality holds when $\sigma_{X+Y}=\sigma_X+\sigma_Y$; otherwise the inequality is strict.

Worked Example: Negative Dependence and Portfolio VaR

A hypothetical company models two annual loss components, in thousands of dollars, as jointly normal with means 5 and 7, standard deviations 3 and 3, and correlation $-1/9$. Calculate the 95.00% portfolio VaR and compare it with the sum of stand-alone VaRs. Use $z_{0.9500}=1.6449$.

Method and assumptions. The covariance is $(-1/9)(3)(3)=-1$ thousand-dollars squared. Apply the joint-normal variance formula, including this covariance; do not assume independence.

The stand-alone VaRs are $5+1.6449(3)=9.9347$ and $7+1.6449(3)=11.9347$. Their sum is $21.8694$ thousand dollars.

The portfolio mean is $5+7=12$. Its variance is $3^2+3^2+2(-1)=9+9-2=16$, giving standard deviation $\sqrt{16}=4$. The portfolio VaR is $12+1.6449(4)=12+6.5796=18.5796$ thousand dollars.

The difference is $21.8694-18.5796=3.2898$ thousand dollars, so portfolio VaR is lower. The variance is positive, as required, and the result satisfies the proved subadditivity inequality.

Common mistake avoided: Treating the components as independent would give variance $9+9=18$ and VaR $12+1.6449\sqrt{18}$, approximately $18.9787$ thousand dollars. That is not the model in the question: the negative covariance reduces the correct variance to 16.

Limits of the Result and an Exam Trap

The theorem is not a claim that VaR is subadditive for every loss distribution. It is a result for jointly normal losses, and its proof also uses $z_{\alpha}\geq 0$. At confidence levels below 50%, the quantile is negative. Multiplying the standard-deviation inequality by a negative number reverses its direction, so the preceding proof no longer establishes subadditivity.

Worked Example: Why the Confidence-Level Condition Matters

Consider two independent hypothetical normal losses, each measured in thousands of dollars, with $X\sim N(0,1)$ and $Y\sim N(0,1)$. At confidence level $\alpha=0.2500$, use $z_{0.2500}=-0.6745$. Compare the portfolio VaR with the sum of stand-alone VaRs.

Method and assumptions. Independence makes the losses jointly normal with covariance zero. Each standard deviation is 1, and the portfolio standard deviation is $\sqrt{1+1}=\sqrt{2}$.

Each stand-alone VaR is $0+(-0.6745)(1)=-0.6745$. The sum is $-0.6745+(-0.6745)=-1.3490$.

The portfolio VaR is $0+(-0.6745)\sqrt{2}$. Using $\sqrt{2}\approx1.4142$, this is approximately $-0.9539$. Hence $-0.9539\leq -1.3490$ is false: the portfolio VaR is greater than the sum of stand-alone VaRs at this confidence level.

Common mistake avoided: The standard-deviation inequality $\sqrt{2}\leq2$ is correct, but multiplying both sides by $-0.6745$ reverses the inequality. The counterexample checks the direction explicitly rather than assuming the usual high-confidence result applies.

A related calculation trap is omitting the covariance term when the correlation is given. For two losses, the correct portfolio variance is $\sigma_X^2+\sigma_Y^2+2\rho\sigma_X\sigma_Y$, not merely $\sigma_X^2+\sigma_Y^2$. Another trap is confusing the standard deviation with the variance: the normal VaR formula multiplies $z_{\alpha}$ by the standard deviation, the square root of the variance.

Practical Interpretation

For ordinary high-confidence risk measurement, where $\alpha$ exceeds 50%, jointly normal losses cannot have portfolio VaR greater than the sum of their stand-alone VaRs. The amount of reduction is determined by $z_{\alpha}(\sigma_X+\sigma_Y-\sigma_{X+Y})$; the means cancel. Dependence affects the portfolio standard deviation through covariance, while each stand-alone VaR remains determined by its own mean and standard deviation.

This result should be applied only after checking the model assumptions and the quantile sign. The joint-normal assumption makes the aggregate quantile calculable from its mean and standard deviation; the confidence-level condition makes the direction of the inequality valid. Neither condition should be silently dropped.

Key takeaway: For jointly normal losses, portfolio VaR equals the sum of portfolio means plus $z_{\alpha}$ times the covariance-based portfolio standard deviation. The triangle inequality bounds that standard deviation by the sum of stand-alone standard deviations. Therefore VaR is subadditive when $\alpha\geq0.5000$, but this argument does not extend automatically to nonnormal losses or to confidence levels below 50%.

Check Your Understanding

For each problem, use the stated normal quantile and keep monetary amounts in the stated units.

A hypothetical portfolio has independent normal losses $X\sim N(4,9)$ and $Y\sim N(6,16)$, in thousands of dollars. At $\alpha=0.9500$, use $z_{\alpha}=1.6449$. Find the portfolio VaR and the sum of stand-alone VaRs. (Answer: portfolio VaR $=18.2245$ thousand dollars; sum $=21.5143$ thousand dollars.)
Two jointly normal hypothetical losses have means 2 and 1, standard deviations 2 and 2, and correlation $0.2500$, in thousands of dollars. At $\alpha=0.9900$, use $z_{\alpha}=2.3263$. Find the portfolio variance and VaR. (Answer: variance $=10$ thousand-dollars squared; VaR $=10.3564$ thousand dollars.)
Two jointly normal losses have means 1 and 2, standard deviations 2 and 3, and correlation $1$. At $\alpha=0.9500$, use $z_{\alpha}=1.6449$. Find the portfolio VaR and compare it with the sum of stand-alone VaRs. (Answer: both are $11.2245$ in the same monetary units.)
Let $X$ and $Y$ be independent $N(0,1)$ losses. At $\alpha=0.2500$, use $z_{\alpha}=-0.6745$. Calculate the sum of stand-alone VaRs and the portfolio VaR, then state whether subadditivity holds. (Answer: sum of stand-alone VaRs $=-1.3490$; portfolio VaR $\approx-0.9539$; subadditivity does not hold.)

What You'll Learn