VaR Additivity for Comonotonic Losses

Why Comonotonicity Matters for VaR

A portfolio loss is a sum of individual losses, but its VaR cannot usually be found by simply adding the component VaRs. There is an important special case: when the losses are comonotonic, they move together in an ordered way, and VaR is additive. The condition matters; the formula is not a general rule for every portfolio.

To state and prove the result precisely, first fix the quantile convention. This is particularly important for discrete losses, where the cumulative distribution function can jump over a confidence level.

Definition: For a real-valued loss $X$ and confidence level $p$, where $0<p<1$, define $$ \operatorname{VaR}_p(X)=Q_X(p)=\inf\{x:F_X(x)\ge p\}. $$ This is the left-quantile convention: VaR is the smallest loss level at which the cumulative probability is at least $p$. The quantile function $Q_X$ is nondecreasing.

A useful way to describe comonotonic losses is by their common quantile construction. Let $U\sim\operatorname{Uniform}(0,1)$. Losses $X_1,\ldots,X_n$ are comonotonic when their joint distribution can be represented as $X_j=Q_{X_j}(U)$ for every $j$. In this representation, each loss is an increasing function of the same variable. When $U$ is high, none of these losses is selected from a lower quantile while another is selected from a higher one.

More generally, one may describe the same structure by saying the losses are nondecreasing functions of a common random variable. For exact calculations at a confidence level, however, the common-quantile representation is especially useful: it fixes the quantile convention, including at jumps.

Theorem: If $X_1,\ldots,X_n$ are comonotonic and VaR uses the left-quantile convention above, then, for every $0<p<1$, $$ \operatorname{VaR}_p\!\left(\sum_{j=1}^{n}X_j\right) = \sum_{j=1}^{n}\operatorname{VaR}_p(X_j). $$

Why the Additivity Formula Holds

The key idea is to form the portfolio loss using the same value of $U$ for every component. Each component is evaluated at the same quantile level, so their sum is also an increasing function of $U$. Its $p$-quantile is the value of that sum at $U=p$.

Use a common quantile variable.
Write $X_j=Q_{X_j}(U)$, where $U\sim\operatorname{Uniform}(0,1)$, for every component $j$.

Form the portfolio quantile function.
Set $h(u)=\sum_{j=1}^{n}Q_{X_j}(u)$. Each quantile function is nondecreasing, so $h$ is nondecreasing. Under the left-quantile convention, these quantile functions, and therefore their finite sum $h$, are left-continuous on $0<u<1$.

Identify the portfolio’s $p$-quantile.
For $u\le p$, monotonicity gives $h(u)\le h(p)$, so $P(h(U)\le h(p))\ge p$. If $x<h(p)$, left-continuity gives some $u_0<p$ with $h(u_0)>x$. For $u\ge u_0$, monotonicity then gives $h(u)>x$, so $P(h(U)\le x)\le u_0<p$. Thus the smallest $x$ with $P(h(U)\le x)\ge p$ is $h(p)$.

The portfolio has the same distribution as $h(U)$. The argument therefore shows $$ \operatorname{VaR}_p\!\left(\sum_{j=1}^{n}X_j\right) =h(p) =\sum_{j=1}^{n}Q_{X_j}(p) =\sum_{j=1}^{n}\operatorname{VaR}_p(X_j). $$ The shared value of $U$ is essential: it ensures that all component losses are being evaluated at the same rank in their distributions.

Worked Applications

Worked Example: Adding Continuous Linear Losses

A hypothetical portfolio has two losses, measured in thousands of dollars. Let $U\sim\operatorname{Uniform}(0,1)$, and suppose $X=40+60U$ and $Y=10+30U$. Find each component’s 90% VaR and the 90% VaR of $X+Y$. State why the additivity result applies.

Method and condition. Both losses are increasing functions of the same uniform variable, so they are comonotonic. For a uniform variable on $(0,1)$, the $p$-quantile is $p$. At $p=0.9$, use $U=0.9$.

First component. $$ \operatorname{VaR}_{0.9}(X)=40+60(0.9)=40+54=94 $$ Thus $X$'s 90% VaR is \$94,000.

Second component. $$ \operatorname{VaR}_{0.9}(Y)=10+30(0.9)=10+27=37 $$ Thus $Y$'s 90% VaR is \$37,000.

Portfolio check. The combined loss is $$ X+Y=(40+10)+(60+30)U=50+90U. $$ Therefore, $$ \operatorname{VaR}_{0.9}(X+Y)=50+90(0.9)=50+81=131. $$ The component VaRs add to $94+37=131$, also \$131,000. The two calculations agree, as the theorem requires. The common mistake avoided here is applying the sum formula without first checking that the losses share an increasing common variable.

Worked Example: Discrete Losses at a Quantile Atom

Let $U\sim\operatorname{Uniform}(0,1)$. In a hypothetical portfolio, set $X=0$ when $U\le0.6$ and $X=120$ when $U>0.6$, in thousands of dollars. Set $Y=0$ when $U\le0.3$ and $Y=80$ when $U>0.3$, also in thousands. Find the component VaRs and portfolio VaR at confidence levels $p=0.6$ and $p=0.7$.

Method. Both loss functions are nondecreasing in $U$, so the losses are comonotonic. Use the left-quantile definition: at a probability jump, select the smallest loss for which the cumulative probability is at least $p$.

At $p=0.6$, component $X$. The probability of $X=0$ is $0.6$, so $F_X(0)=0.6$. Hence $\operatorname{VaR}_{0.6}(X)=0$.

At $p=0.6$, component $Y$. The probability of $Y=0$ is $0.3$, which is below $0.6$, while $F_Y(80)=1$. Hence $\operatorname{VaR}_{0.6}(Y)=80$.

At $p=0.6$, portfolio. The sum is 0 for $U\le0.3$, 80 for $0.3<U\le0.6$, and 200 for $U>0.6$. Thus $P(X+Y\le0)=0.3$ and $P(X+Y\le80)=0.6$, giving $\operatorname{VaR}_{0.6}(X+Y)=80=0+80$.

At $p=0.7$, components and portfolio. Since $F_X(0)=0.6<0.7$, $\operatorname{VaR}_{0.7}(X)=120$. Since $F_Y(80)=1$, $\operatorname{VaR}_{0.7}(Y)=80$. For the sum, $P(X+Y\le80)=0.6<0.7$, while $P(X+Y\le200)=1$, so $\operatorname{VaR}_{0.7}(X+Y)=200=120+80$.

Trap and correction. At $p=0.6$, it would be wrong to assign $X$'s VaR as 120 merely because $X$ can jump to 120 immediately above the 0.6 threshold. The definition asks for the smallest $x$ with $F_X(x)\ge0.6$; $x=0$ already meets that condition. Using 120 would incorrectly give a portfolio VaR of 200 rather than the correct 80 at $p=0.6$.

Worked Example: Nonlinear Increasing Loss Functions

Let $Z\sim\operatorname{Uniform}(0,5)$, with amounts measured in thousands of dollars. A hypothetical portfolio has losses $X=2Z^2$ and $Y=10Z$. Find their 80% VaRs and the 80% VaR of the total loss.

Method and condition. On $0\le Z\le5$, both functions are increasing: if $z_2>z_1\ge0$, then $z_2^2>z_1^2$ and $10z_2>10z_1$. Thus both losses increase with the same $Z$, so they are comonotonic. The 80th percentile of a uniform variable on $(0,5)$ is $0.8(5)=4$.

Component calculations. $$ \operatorname{VaR}_{0.8}(X)=2(4)^2=2(16)=32, \qquad \operatorname{VaR}_{0.8}(Y)=10(4)=40. $$ The component VaRs are \$32,000 and \$40,000.

Portfolio calculation. The total loss as a function of $Z$ is $2Z^2+10Z$, which is increasing on $0\le Z\le5$ because both terms increase there. Consequently, its 80% VaR is its value at $Z=4$: $$ \operatorname{VaR}_{0.8}(X+Y)=2(4)^2+10(4)=32+40=72. $$ The check $32+40=72$ confirms additivity; the total 80% VaR is \$72,000. This example shows that the functions need not be linear; their common increasing direction is what matters.

Conditions, Interpretation, and Exam Traps

The result is a statement about a particular joint arrangement of losses, not just about their separate distributions. Comonotonicity means that a single rank variable orders all the losses in the same direction. The proof uses this structure to put each component at its $p$-quantile at the same time. If that shared ordering is not part of the assumptions, the theorem does not justify adding component VaRs.

A second trap is switching quantile conventions mid-calculation. For losses with atoms, a value exactly at the confidence threshold can satisfy $F_X(x)=p$. Under the left-quantile definition used here, that value is VaR even if the distribution has a larger possible loss immediately above it. State the convention first, then apply it consistently to each component and to their sum.

The practical value of the theorem is that it gives an exact portfolio VaR calculation when losses are comonotonic: calculate each component quantile at the same confidence level and add. It is also a useful structural benchmark. It identifies a dependence arrangement under which all losses line up by rank, but it does not make VaR additive for arbitrary losses.

Key takeaway: With the left-quantile convention, comonotonic losses have additive VaR at every confidence level $0<p<1$. Verify the common increasing or common-quantile structure before using the formula, and handle probability atoms using $\inf\{x:F_X(x)\ge p\}$.

Check Your Understanding

For each problem, use the left-quantile convention and check whether the losses have the common increasing structure needed for additivity.

Let $U\sim\operatorname{Uniform}(0,1)$, with losses $X=5+10U$ and $Y=2+3U$, in thousands of dollars. Find both 80% component VaRs and the 80% VaR of their sum. (Answer: $13$, $4.4$, and $17.4$ thousand dollars.)
Let $U\sim\operatorname{Uniform}(0,1)$. Set $X=0$ for $U\le0.4$ and $X=70$ otherwise; set $Y=0$ for $U\le0.75$ and $Y=30$ otherwise. Find the 60% VaRs of $X$, $Y$, and $X+Y$. (Answer: $70$, $0$, and $70$.)
Let $Z\sim\operatorname{Uniform}(0,1)$, and let $X=100Z$ and $Y=50Z^2$, in dollars. Find both 90% component VaRs and the 90% VaR of the total. (Answer: $\$90$, $\$40.50$, and $\$130.50$.)
Let $U\sim\operatorname{Uniform}(0,1)$. Set $X=0$ for $U\le0.25$ and $X=100$ otherwise; set $Y=0$ for $U\le0.5$ and $Y=40$ otherwise. Find the 50% VaR of each loss and their total. (Answer: $100$, $0$, and $100$.)
Let $Z\sim\operatorname{Uniform}(0,4)$, with losses $X=Z$ and $Y=Z^2$, in thousands of dollars. Find the 75% component VaRs and the 75% VaR of their sum. (Answer: $3$, $9$, and $12$ thousand dollars.)

What You'll Learn