Why VaR Can Fail Subadditivity

Why a Portfolio Quantile Can Exceed the Sum of Its Parts

VaR summarizes a loss distribution at one selected confidence level. That makes it useful for describing a portfolio, but it also means the result depends on how much probability lies below a particular loss threshold. When separate risks have losses in different states, their combined loss can exceed its own VaR threshold more often than either individual loss exceeds its threshold. The portfolio VaR can then be greater than the sum of the individual VaRs.

The issue is not that individual VaRs were calculated incorrectly. It is that a quantile of a sum is not generally the sum of the quantiles. To test subadditivity, calculate or bound the aggregate distribution at the sum of the individual VaRs, taking account of how the losses occur together.

Definition: For a loss random variable $X$ and confidence level $0<\alpha<1$, define $\operatorname{VaR}_{\alpha}(X)$ by the lower-quantile convention $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq \alpha\}, $$ where $F_X(x)=P(X\leq x)$. VaR is subadditive for a pair of losses $X,Y$ at level $\alpha$ if $$ \operatorname{VaR}_{\alpha}(X+Y)\leq \operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y). $$

Here $X$ and $Y$ are losses over the same period, measured in the same units, and defined on the same probability space. Their dependence matters: it describes which loss values can occur together. In a discrete distribution, the lower-quantile convention means that VaR is the smallest loss level whose cumulative probability reaches or exceeds $\alpha$. In particular, probability exactly equal to $\alpha$ at a threshold is enough for that threshold to be the VaR.

The Probability Condition Behind the Test

Write $q_X=\operatorname{VaR}_{\alpha}(X)$ and $q_Y=\operatorname{VaR}_{\alpha}(Y)$. If both individual losses are at or below their VaRs, their sum is at or below $q_X+q_Y$. This inclusion of events gives a useful sufficient condition for subadditivity.

\{X\leq q_X,\;Y\leq q_Y\}\subseteq\{X+Y\leq q_X+q_Y\}.

Consequently, if the joint probability of the two individual threshold events is at least $\alpha$, then the aggregate loss has at least $\alpha$ probability at or below the sum of the individual VaRs. Its $\alpha$-quantile cannot be larger than that sum.

P(X\leq q_X,\;Y\leq q_Y)\geq\alpha \quad\Longrightarrow\quad F_{X+Y}(q_X+q_Y)\geq\alpha \quad\Longrightarrow\quad \operatorname{VaR}_{\alpha}(X+Y)\leq q_X+q_Y.

For the converse test, if $F_{X+Y}(q_X+q_Y)<\alpha$, the aggregate distribution has not yet reached the selected confidence level at the sum of the individual VaRs. Under the lower-quantile definition, this means $\operatorname{VaR}_{\alpha}(X+Y)>q_X+q_Y$, so subadditivity fails for that pair.

The two individual events $\{X\leq q_X\}$ and $\{Y\leq q_Y\}$ each have probability at least $\alpha$. But that alone does not say their intersection has probability at least $\alpha$. The general probability bound is

P(X\leq q_X,\;Y\leq q_Y) \geq \max\!\left(0,\;P(X\leq q_X)+P(Y\leq q_Y)-1\right) \geq \max(0,2\alpha-1).

For any $\alpha<1$, $2\alpha-1<\alpha$. Thus this general bound does not guarantee subadditivity. It is only a lower bound: an intersection smaller than $\alpha$ does not by itself prove failure, because other combinations of losses might still keep $X+Y$ below $q_X+q_Y$. The aggregate probability at that threshold is the decisive check.

Worked Example: Separate Loss Events at a 75% Confidence Level

A hypothetical insurer considers two one-period losses, $X$ and $Y$. The following four states are equally likely, and the two losses occur in mutually exclusive states. Calculate each 75% VaR and the VaR of the combined loss. Determine whether subadditivity holds.

State	Probability	Loss $X$	Loss $Y$	Combined loss $X+Y$
1	1/4	$0	$0	$0
2	1/4	$0	$0	$0
3	1/4	$40	$0	$40
4	1/4	$0	$40	$40

Method. Apply the lower-quantile definition to the individual and aggregate distributions. The common mistake this avoids is assuming that two individual VaRs of zero must imply a portfolio VaR of zero.

For $X$, three of the four states have loss zero, so $F_X(0)=3/4=0.75$. Therefore $\operatorname{VaR}_{0.75}(X)=\$0$. The same calculation applies to $Y$: $F_Y(0)=3/4$, giving $\operatorname{VaR}_{0.75}(Y)=\$0$.

The combined loss is zero in two states and $40 in two states. Hence $F_{X+Y}(0)=2/4=0.50$, which is below 0.75, while $F_{X+Y}(40)=4/4=1$. Thus $\operatorname{VaR}_{0.75}(X+Y)=\$40$.

The subadditivity comparison is $\$40>\$0+\$0$, so VaR is not subadditive for this pair at the 75% confidence level. Each separate loss is zero in at least 75% of states, but their zero-loss events overlap in only two states, or 50% of the probability.

Check. The aggregate CDF at the sum of the individual VaRs is $F_{X+Y}(0)=0.50<0.75$. This confirms that the aggregate VaR must exceed that threshold.

Failure Can Occur Even When Individual VaRs Are Positive

A zero individual VaR is not required for a failure. The same probability mechanism can operate above positive thresholds: each loss can exceed its own VaR in a different state, making the sum exceed the sum of those VaRs more often than the selected tail probability allows.

Worked Example: Positive Individual VaRs and a Larger Portfolio VaR

A hypothetical risk manager models two losses over four equally likely scenarios. Calculate their 75% VaRs and the aggregate VaR, then test subadditivity.

State	Probability	Loss $X$	Loss $Y$	Combined loss $X+Y$
1	1/4	$10	$10	$20
2	1/4	$10	$10	$20
3	1/4	$30	$0	$30
4	1/4	$0	$30	$30

Method. Sort each set of outcomes, or equivalently inspect its CDF at the candidate threshold. The common mistake this avoids is checking only the marginal loss distributions while ignoring which scenario produces each loss.

For $X$, three states have losses at or below $10, so $F_X(10)=3/4=0.75$. At any amount below $10, the CDF is only $1/4$. Therefore $\operatorname{VaR}_{0.75}(X)=\$10$. By the same calculation, $F_Y(10)=3/4$, so $\operatorname{VaR}_{0.75}(Y)=\$10$.

For $X+Y$, two states have loss $20 and two have loss $30. Thus $F_{X+Y}(20)=2/4=0.50<0.75$, whereas $F_{X+Y}(30)=1$. Therefore $\operatorname{VaR}_{0.75}(X+Y)=\$30$.

The sum of the individual VaRs is $\$10+\$10=\$20$. Since $\$30>\$20$, subadditivity fails. At the individual threshold sum of $20, only half the probability lies at or below the aggregate threshold; in the other half, one loss is above its own VaR.

Check. The aggregate CDF crosses 0.75 at $30, so the calculated aggregate VaR is consistent with the lower-quantile definition.

Dependence Can Change the Outcome

Marginal distributions determine the individual VaRs, but they do not specify how losses pair across scenarios. To isolate the role of dependence, consider the same marginal loss distribution for each risk in two different joint arrangements. Each individual loss is $40 in one of four equally likely states and zero in the other three. At the 75% confidence level, each individual VaR is $0.

Worked Example: Same Marginals, Different Joint Outcomes

Compare two hypothetical ways to pair these marginal losses. In the first arrangement, the $40 losses occur in the same state. In the second, they occur in different states. Find the aggregate VaR for each arrangement.

Arrangement A: losses occur together. The joint outcomes for $(X,Y)$ are $(40,40),(0,0),(0,0),(0,0)$. The aggregate loss is $80 in one state and zero in three. Its CDF at zero is $3/4=0.75$, so $\operatorname{VaR}_{0.75}(X+Y)=\$0$. The sum of the individual VaRs is also zero, and the subadditivity inequality holds with equality.

Arrangement B: losses occur in different states. The joint outcomes are $(40,0),(0,40),(0,0),(0,0)$. The aggregate loss is $40 in two states and zero in two. Its CDF at zero is $2/4=0.50$, and at $40 it is 1. Therefore $\operatorname{VaR}_{0.75}(X+Y)=\$40$, which exceeds the sum of the individual VaRs, $0.

The individual distributions, and therefore the individual VaRs, are identical in both arrangements. Only the dependence differs. This comparison does not establish a universal rule for either arrangement; it shows why marginal VaRs alone cannot settle the portfolio subadditivity question.

Check. Each arrangement gives $F_X(0)=F_Y(0)=0.75$. The aggregate CDF at zero is 0.75 in Arrangement A but 0.50 in Arrangement B, accounting for the different aggregate VaRs.

A Practical Test and Common Trap

For a particular pair of losses, the direct test is to evaluate the aggregate CDF at the sum of the individual VaRs. If that probability is below $\alpha$, the aggregate VaR is strictly larger than the sum. If it is at least $\alpha$, the aggregate VaR is no larger than the sum. A joint threshold-event calculation can provide a convenient sufficient condition for the latter result, but it need not give the exact aggregate CDF.

A common incorrect calculation is: “$\operatorname{VaR}_{0.75}(X)=0$ and $\operatorname{VaR}_{0.75}(Y)=0$, so $\operatorname{VaR}_{0.75}(X+Y)=0+0=0$.” This substitutes the sum of individual quantiles for the quantile of the sum. In the first worked example, the correction is to calculate $F_{X+Y}(0)=0.50$. Because $0.50<0.75$, the portfolio VaR is $40, not zero.

Another trap is to treat “each individual threshold event has probability at least $\alpha$” as if their intersection also had probability at least $\alpha$. At $\alpha=0.95$, the general intersection bound is only $2(0.95)-1=0.90$, which is below 0.95. The actual dependence structure or aggregate distribution must be examined; the bound does not prove that the intersection is exactly 0.90, nor does it by itself prove a VaR failure.

Key takeaway: VaR can fail subadditivity when the aggregate loss has less than $\alpha$ probability at or below the sum of the individual $\alpha$-VaRs. Separate losses in different scenarios can create this pattern. The joint probability of the individual threshold events is informative, but only the aggregate CDF at the candidate threshold decides the comparison.

Check Your Understanding

Use the lower-quantile convention. For equally likely states, a probability of $k/n$ corresponds to $k$ of the $n$ states.

At confidence level $\alpha=0.80$, five equally likely states have losses $X=(0,0,0,0,25)$ and $Y=(0,0,25,0,0)$. Find both individual VaRs, the aggregate VaR, and whether subadditivity holds. (Answer: $\operatorname{VaR}_{0.80}(X)=\operatorname{VaR}_{0.80}(Y)=\$0$; aggregate VaR $=\$25$; subadditivity fails.)
At $\alpha=0.75$, four equally likely states have $X=(2,2,9,0)$ and $Y=(3,3,0,9)$. Find the two individual VaRs and the aggregate VaR. (Answer: Individual VaRs are $2 and $3; aggregate VaR is $9, so subadditivity fails because $9>2+3$.)
At $\alpha=0.60$, five equally likely states have $X=(0,0,4,4,9)$ and $Y=(0,7,0,7,12)$. Find the individual VaRs, the probability that both losses are at or below their VaRs, and the aggregate VaR. (Answer: Individual VaRs are $4 and $7; the joint threshold probability is $4/5=0.80$; aggregate VaR is $7, which is no greater than the sum $11.)
For $\alpha=0.90$, what lower bound does the general intersection inequality give for $P(X\leq\operatorname{VaR}_{0.90}(X),\,Y\leq\operatorname{VaR}_{0.90}(Y))$? Does that bound alone guarantee subadditivity? (Answer: The bound is $2(0.90)-1=0.80$; no, because 0.80 is below 0.90.)

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