A Discrete VaR Counterexample to Subadditivity

Why a Discrete Counterexample Matters

VaR is a quantile of a loss distribution. For many familiar loss models it is useful and interpretable, but that does not mean it always satisfies the desirable properties of a coherent risk measure. In particular, subadditivity requires that the risk measure of an aggregate position be no greater than the sum of the risks assigned to its components. A discrete distribution can violate this requirement, even when the component losses are independent.

The key is to distinguish a quantile from a tail average. VaR records the loss threshold at a selected confidence level; it does not directly account for how large losses may be beyond that threshold. In a discrete model, a probability jump can move the aggregate quantile above the individual quantiles. The calculations below show exactly where that jump occurs.

Definition: For a loss random variable $X$ with cumulative distribution function $F_X(x)=P(X\le x)$, the lower-quantile Value-at-Risk at confidence level $\alpha$, where $0<\alpha<1$, is $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}. $$ For two loss positions $X$ and $Y$, subadditivity is the inequality $$ \operatorname{VaR}_{\alpha}(X+Y)\le \operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y). $$

The infimum convention matters at jumps. The VaR is the first loss value at which the cumulative probability reaches or exceeds $\alpha$. It is not necessarily the value where the cumulative probability equals $\alpha$. For a finite discrete distribution, a reliable method is to list every possible loss in increasing order, accumulate its probability, and select the first loss whose cumulative probability is at least $\alpha$.

How the Violation Arises

Suppose two positions each have a small probability $p$ of a fixed positive loss $L$, and otherwise have loss zero. Assume the positions are independent, so their joint probabilities are products of their marginal probabilities. Let $q=1-p$. If $q\ge\alpha$, each individual position has VaR zero: the probability of a zero loss already reaches the confidence level. However, the aggregate is zero only if neither loss occurs, with probability $q^2$. If $q^2<\alpha$, aggregate VaR must be positive.

For equal loss amounts $L$, the aggregate takes values $0,L,2L$. Its probability of being at most $L$ is $1-p^2$, because only the event that both positions incur losses produces an aggregate greater than $L$. Therefore, whenever $q^2<\alpha\le 1-p^2$, the aggregate VaR is exactly $L$, while both individual VaRs are zero. These conditions show how a probability mass that is adequate for each individual position can be inadequate for the joint no-loss outcome.

Find each marginal quantile.
Check whether the probability of a loss at or below the no-loss amount is at least $\alpha$. If so, the individual VaR is zero.

Calculate the aggregate probabilities.
Use the specified dependence assumption to obtain the probabilities of each aggregate loss.

Apply the quantile convention to the aggregate.
Accumulate probabilities in order and select the first loss value whose cumulative probability reaches $\alpha$.

Compare the two sides.
If aggregate VaR is greater than the sum of the individual VaRs, subadditivity fails for this distribution and confidence level.

Worked Counterexamples

Worked Example: Two Independent Equal-Severity Positions

A hypothetical insurer considers two independent one-period loss positions. Each position has loss $100$ dollars with probability $0.04$, and loss $0$ dollars with probability $0.96$. Find each position's 95% VaR and the 95% VaR of their aggregate. All amounts are in dollars.

Method. First calculate the marginal quantiles, then form the joint distribution using independence. The common mistake to avoid is assuming that because each position has a 96% chance of no loss, the aggregate also has at least a 95% chance of no loss.

For either position, $F(0)=0.96$. Since $0.96\ge0.95$, the first loss value reaching the 95% confidence level is zero:

\operatorname{VaR}_{0.95}(X)=\operatorname{VaR}_{0.95}(Y)=0.

By independence, the aggregate probabilities are

$$ P(X+Y=0)=0.96(0.96)=0.9216, $$

$$ P(X+Y=100)=0.04(0.96)+0.96(0.04)=0.0384+0.0384=0.0768, $$

$$ P(X+Y=200)=0.04(0.04)=0.0016. $$

The probabilities sum to $0.9216+0.0768+0.0016=1.0000$. The cumulative probability at aggregate loss $0$ is $0.9216$, which is below $0.95$. At aggregate loss $100$, it is $0.9216+0.0768=0.9984$, which reaches the confidence level. Thus

\operatorname{VaR}_{0.95}(X+Y)=100>0+0 =\operatorname{VaR}_{0.95}(X)+\operatorname{VaR}_{0.95}(Y).

The subadditivity inequality fails. The check is that the aggregate cumulative probability crosses $0.95$ at $100$, not at $0$; the probability of no aggregate loss is only $0.9216$.

Worked Example: Baseline Losses and Unequal Increments

Consider two independent hypothetical positions at the 95% confidence level. Position $X$ always incurs a baseline loss of $50$ dollars and incurs an additional $50$ dollars with probability $0.04$. Position $Y$ always incurs a baseline loss of $80$ dollars and incurs an additional $60$ dollars with probability $0.04$. Find both stand-alone VaRs and the aggregate VaR.

Method. Treat each baseline amount as the minimum possible loss, and then calculate the cumulative probabilities at each possible aggregate amount. A zero stand-alone VaR would be incorrect here because neither position can lose zero dollars.

For $X$, the loss is $50$ with probability $0.96$ and $100$ with probability $0.04$. Since $F_X(50)=0.96\ge0.95$, $\operatorname{VaR}_{0.95}(X)=50$. For $Y$, the loss is $80$ with probability $0.96$ and $140$ with probability $0.04$. Since $F_Y(80)=0.96\ge0.95$, $\operatorname{VaR}_{0.95}(Y)=80$. Their sum is $130$ dollars.

Independence gives four aggregate outcomes. If neither additional loss occurs, the aggregate is $50+80=130$, with probability $0.96(0.96)=0.9216$. If only $X$'s additional loss occurs, the aggregate is $100+80=180$, with probability $0.04(0.96)=0.0384$. If only $Y$'s occurs, the aggregate is $50+140=190$, also with probability $0.96(0.04)=0.0384$. If both occur, the aggregate is $100+140=240$, with probability $0.04(0.04)=0.0016$. The probabilities sum to $1.0000$.

The cumulative probability at $130$ is $0.9216$, below $0.95$. At $180$ it is $0.9216+0.0384=0.9600$, so the aggregate VaR is $180$ dollars. Therefore

\operatorname{VaR}_{0.95}(X+Y)=180>130 =\operatorname{VaR}_{0.95}(X)+\operatorname{VaR}_{0.95}(Y).

This verifies the violation with nonzero baselines. The first aggregate amount above the baseline is sufficient to reach the confidence level, even though neither marginal VaR includes its additional-loss amount.

Worked Example: A Counterexample From a Joint Probability Table

Two hypothetical loss positions $X$ and $Y$ have the following joint distribution. No independence assumption is made. Find their individual VaRs and the VaR of their sum at confidence level $0.90$.

$X$	$Y$	Joint probability
0	0	0.88
30	0	0.08
0	50	0.03
30	50	0.01

Method. Obtain each marginal distribution by summing the relevant joint probabilities, then aggregate each row's two losses. The trap is to infer an independence structure that the table does not specify.

For $X$, the probability of zero is $0.88+0.03=0.91$, so $F_X(0)=0.91\ge0.90$ and $\operatorname{VaR}_{0.90}(X)=0$. For $Y$, the probability of zero is $0.88+0.08=0.96$, so $F_Y(0)=0.96\ge0.90$ and $\operatorname{VaR}_{0.90}(Y)=0$.

The aggregate losses from the four rows are $0,30,50,80$, with probabilities $0.88,0.08,0.03,0.01$, respectively. The cumulative probabilities are $0.88$ at $0$, $0.88+0.08=0.96$ at $30$, $0.99$ at $50$, and $1.00$ at $80$. Therefore $\operatorname{VaR}_{0.90}(X+Y)=30$, and

30>0+0 =\operatorname{VaR}_{0.90}(X)+\operatorname{VaR}_{0.90}(Y).

The joint probabilities sum to $1.00$, and the marginal and aggregate quantiles follow directly from the stated joint distribution. Independence is not needed for a discrete counterexample; the dependence structure must simply be handled correctly.

Why the Formula Works—and What It Does Not Say

For two independent equal-severity positions that each lose $L$ with probability $p$ and zero otherwise, the aggregate can be $0,L,$ or $2L$. The probability of an aggregate loss no greater than $L$ is the probability that the two positions do not both incur the loss. Independence makes the probability of both losses $p^2$, giving

P(X+Y\le L)=1-p^2.

Meanwhile, the probability of an individual loss no greater than zero is $1-p$. Thus, if

1-p\ge\alpha,\qquad (1-p)^2<\alpha,\qquad 1-p^2\ge\alpha,

each stand-alone VaR is zero, the aggregate's probability of being zero is too small to reach $\alpha$, and the aggregate's probability of being at most $L$ is large enough to reach it. The quantile definition then gives individual VaRs of zero and aggregate VaR $L$. In the first worked example, these three probabilities are $0.96$, $0.9216$, and $0.9984$, respectively, at $\alpha=0.95$.

This argument establishes a counterexample, not a claim that VaR always fails subadditivity. The result depends on the distribution, confidence level, and dependence between the positions. In particular, a quantile calculation for an aggregate cannot generally be inferred by adding the marginal quantiles. It must be computed from the aggregate distribution, using the dependence information available.

Common Exam Trap

A tempting but incorrect calculation in the first example is to say: “Each position has a 96% chance of zero loss, so the aggregate has a 96% chance of zero loss; therefore its 95% VaR is zero.” This incorrectly transfers a marginal probability to the joint event. Under independence, the joint no-loss probability is

$$ P(X=0,Y=0)=P(X=0)P(Y=0)=0.96(0.96)=0.9216, $$

not $0.96$. Since $0.9216<0.95$, aggregate VaR cannot be zero. Accumulating the aggregate probabilities shows that it is $100$ dollars. Always distinguish $P(X=0)$ from $P(X=0,Y=0)$, and apply the quantile definition to the correct distribution.

A second error is using a strict crossing rule and selecting a loss only when the cumulative probability is greater than $\alpha$. The definition here uses $F_X(x)\ge\alpha$. At an exact boundary, a cumulative probability equal to $\alpha$ already qualifies. This matters in discrete distributions and is one reason to state the quantile convention before calculating.

Key takeaway: To test subadditivity for discrete VaR, calculate the marginal quantiles and the aggregate distribution separately. A discrete jump can make the aggregate quantile exceed the sum of the component quantiles, even for independent losses.

Check Your Understanding

For each problem, use the lower-quantile convention $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}$. All losses are hypothetical. Check the cumulative probabilities, including at the confidence boundary.

Two independent positions each lose $25$ dollars with probability $0.03$, and otherwise lose zero. At confidence level $0.95$, calculate the two individual VaRs and the VaR of their aggregate. (Answer: Individual VaRs $0$ dollars each; aggregate VaR $25$ dollars.)
Two independent positions each lose $200$ dollars with probability $0.08$, and otherwise lose zero. Find the individual and aggregate VaRs at confidence level $0.90$. (Answer: Individual VaRs $0$ dollars each; aggregate VaR $200$ dollars.)
At confidence level $0.95$, $X$ always loses $20$ dollars and loses an additional $30$ dollars with probability $0.05$. Independently, $Y$ always loses $40$ dollars and loses an additional $80$ dollars with probability $0.05$. Find the two individual VaRs and the aggregate VaR. (Answer: Individual VaRs $20$ and $40$ dollars; aggregate VaR $90$ dollars.)
At confidence level $0.90$, the joint probabilities for $(X,Y)$ are $P(0,0)=0.87$, $P(40,0)=0.07$, $P(0,90)=0.05$, and $P(40,90)=0.01$. Find each marginal VaR and the aggregate VaR. (Answer: Individual VaRs $0$ dollars each; aggregate VaR $40$ dollars.)
For two independent positions that each lose $L>0$ with probability $p$ and otherwise lose zero, state conditions on $p$ and $\alpha$ that ensure each individual VaR is zero while aggregate VaR is $L$. (Answer: $1-p\ge\alpha$, $(1-p)^2<\alpha$, and $1-p^2\ge\alpha$.)

What You'll Learn