When Value-at-Risk Fails Subadditivity

Why VaR Can Fail to Reward Pooling

Value-at-Risk (VaR) summarizes a loss distribution at a chosen confidence level. It is tempting to expect that combining two risks cannot produce a larger VaR than the sum of their separate VaRs. That inequality is called subadditivity. However, VaR does not satisfy it for every pair of loss distributions. With discrete losses, the probability mass at and around the selected quantile can make pooling look worse by this measure—even when the risks do not produce large losses at the same time.

The key is to calculate the quantile of the aggregate loss from its own distribution. The component VaRs do not, in general, add up to an aggregate quantile. This tutorial uses discrete examples to show how to check the inequality and what features of the loss distribution drive a failure. The discussion is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For a loss $X$ and confidence level $0<\alpha<1$, define $\operatorname{VaR}_{\alpha}(X)$ as the lower $\alpha$-quantile: $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq \alpha\}. $$ For two losses $X$ and $Y$, VaR is subadditive at level $\alpha$ if $$ \operatorname{VaR}_{\alpha}(X+Y)\leq \operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y). $$

For a discrete loss, the quantile is the first loss amount at which the cumulative probability reaches or exceeds $\alpha$. If the cumulative probability just below a loss amount is less than $\alpha$, and the cumulative probability at that amount is at least $\alpha$, that amount is the VaR. The probability at the quantile—the atom—matters. A tail probability alone does not replace this cumulative-probability check.

A Mutually Exclusive Loss Counterexample

The simplest failure occurs when each component has a small chance of a loss, but the two losses cannot occur together. Each component's loss probability can be below the probability threshold relevant to VaR, while the probability of at least one loss in the portfolio is above it.

Worked Example: Two Mutually Exclusive Claims

A hypothetical portfolio contains two coverages. At the 95% confidence level, coverage $X$ pays $100 with probability 0.0400 and $0 otherwise. Coverage $Y$ also pays $100 with probability 0.0400 and $0 otherwise. The claims are mutually exclusive: $P(X=100,Y=0)=0.0400$, $P(X=0,Y=100)=0.0400$, and $P(X=0,Y=0)=0.9200$. Find both component VaRs and the VaR of the aggregate.

Method. Find each marginal CDF and the aggregate CDF, then apply the lower-quantile definition at $\alpha=0.95$.

For $X$, $P(X=0)=0.9600$. Thus $F_X(0)=0.9600\geq 0.95$, so $\operatorname{VaR}_{0.95}(X)=0$. The same calculation gives $F_Y(0)=0.9600$ and $\operatorname{VaR}_{0.95}(Y)=0$.

The aggregate $X+Y$ is $0 with probability 0.9200 and $100 with probability $0.0400+0.0400=0.0800$. Therefore $F_{X+Y}(0)=0.9200<0.95$, while $F_{X+Y}(100)=1.0000\geq0.95$. Hence $\operatorname{VaR}_{0.95}(X+Y)=100$.

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

Check. At $0, the aggregate CDF has not reached 0.95; at $100, it has. The aggregate VaR is therefore $100. Common mistake avoided: adding the two component VaRs and treating that sum as the portfolio VaR. Component quantiles do not determine the aggregate quantile.

This example also gives a short justification for the failure. Each component has a 0.0400 chance of a positive loss, so its CDF at zero exceeds 0.95. In the portfolio, however, the probability of zero loss is only 0.9200. The lower quantile must therefore move from zero to $100. The event “at least one coverage pays” has probability 0.0800, which is above the 0.0500 upper-tail probability corresponding to the 95% confidence level.

P(X+Y=0)=1-0.0400-0.0400=0.9200<0.95, \qquad F_{X+Y}(100)=1.0000\geq0.95.

The Same Failure Can Occur Under Independence

Mutual exclusivity makes the first example easy to see, but it is not necessary for a violation. Independent risks can also produce a portfolio loss probability large enough to cross the confidence threshold, even though each component's probability of a positive loss does not.

Worked Example: Independent Binary Losses

Two hypothetical, independent risks each have loss $100 with probability 0.0400 and loss $0 otherwise. Calculate the component VaRs and the aggregate VaR at the 95% confidence level.

Method. Use independence to multiply the probabilities for the aggregate outcomes, then locate the first cumulative probability at least 0.95.

For either component, the probability of a zero loss is 0.9600. Since $F_X(0)=F_Y(0)=0.9600\geq0.95$, both component VaRs are $0.

For the aggregate, a zero loss occurs only when both risks are zero: $P(X+Y=0)=0.9600(0.9600)=0.9216$. Exactly one risk has a loss with probability $0.0400(0.9600)+0.9600(0.0400)=0.0768$. Both risks have losses with probability $0.0400(0.0400)=0.0016$. These probabilities sum to $0.9216+0.0768+0.0016=1.0000$.

\begin{aligned} P(X+Y=0)&=0.9216,\\ P(X+Y=100)&=0.0768,\\ P(X+Y=200)&=0.0016. \end{aligned}

The aggregate CDF is $F_{X+Y}(0)=0.9216$, which is below 0.95. At $100 it is $0.9216+0.0768=0.9984$, which is at least 0.95. Thus $\operatorname{VaR}_{0.95}(X+Y)=100$, while the sum of the component VaRs is zero.

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

Check. The three aggregate probabilities total 1.0000, and the CDF crosses 0.95 at $100. Common mistake avoided: assuming independence prevents a subadditivity failure. Independence changes the joint probabilities, but does not guarantee that the portfolio quantile is bounded by the sum of the component quantiles.

Atoms, the Confidence Threshold, and the Size of a Violation

A discrete distribution has atoms: positive probability concentrated at particular loss amounts. At a confidence level $\alpha$, VaR is determined by where the cumulative probability first reaches $\alpha$. A modest change in probability can therefore leave VaR unchanged for a component but move aggregate VaR to a higher loss amount. The amount of the jump depends on the loss values; whether a jump occurs depends on the probabilities and the chosen confidence level.

The next example includes positive baseline losses, so the component VaRs are not zero. It also illustrates why the whole aggregate CDF, rather than a verbal description of dependence, should be checked.

Worked Example: Positive Component VaRs and a Larger Portfolio Quantile

A hypothetical two-coverage portfolio has the following joint loss distribution. With probability 0.9400, $(X,Y)=(\$5,\$7)$. With probability 0.0300, $(X,Y)=(\$105,\$7)$. With probability 0.0300, $(X,Y)=(\$5,\$87)$. Find the component and aggregate VaRs at the 95% confidence level.

Method. First combine states that give the same marginal loss, then form the aggregate loss in each joint state.

For $X$, the loss is $5 with probability $0.9400+0.0300=0.9700$, and $105 with probability 0.0300. Therefore $F_X(5)=0.9700\geq0.95$, giving $\operatorname{VaR}_{0.95}(X)=\$5$.

For $Y$, the loss is $7 with probability 0.9700 and $87 with probability 0.0300. Thus $F_Y(7)=0.9700\geq0.95$, giving $\operatorname{VaR}_{0.95}(Y)=\$7$. The sum of the component VaRs is $12.

The aggregate losses by state are:

$5+7=12$, with probability 0.9400;
$105+7=112$, with probability 0.0300;
$5+87=92$, with probability 0.0300.

In increasing order, the aggregate losses are $12, $92, and $112. The CDF at $12 is 0.9400, below 0.95. At $92 it is $0.9400+0.0300=0.9700$, at least 0.95. Thus $\operatorname{VaR}_{0.95}(X+Y)=\$92$.

\operatorname{VaR}_{0.95}(X+Y)=92>5+7 =\operatorname{VaR}_{0.95}(X)+\operatorname{VaR}_{0.95}(Y).

Check. The joint probabilities total 1.0000. In particular, the aggregate CDF is 0.9400 at $12 and 0.9700 at $92, so $92 is the first aggregate loss amount whose CDF reaches 0.95. Common mistake avoided: sorting aggregate outcomes incorrectly or treating the $112 outcome as the VaR simply because it is the largest. The quantile is $92, where the cumulative probability first crosses the threshold.

What the Counterexamples Do—and Do Not—Show

In each example, the component distributions put at least 95% of their probability at or below relatively low loss amounts. In the aggregate, the probability of staying at or below the sum of those amounts is less than 95%. The aggregate quantile must then include some higher loss amount. This is the distributional feature behind these violations: the location of probability mass relative to the chosen confidence threshold changes after aggregation.

A violation is not inevitable. Different loss amounts, probabilities, dependence, or confidence levels can produce a different result. The examples establish that VaR is not always subadditive; they do not show that every portfolio has a larger aggregate VaR than the sum of component VaRs. For any particular problem, calculate the component quantiles and the aggregate distribution under the specified joint model, then compare the two sides of the inequality.

Exam takeaway: For discrete losses, use the lower-quantile convention explicitly: find the smallest loss amount $x$ for which $F_X(x)\geq\alpha$. Then repeat the calculation for $X+Y$ using the joint distribution. A VaR subadditivity failure occurs if the aggregate quantile exceeds the sum of the separately calculated component quantiles.

Check Your Understanding

For each problem, use the lower-quantile definition of VaR and check the aggregate CDF at the relevant loss amounts.

At $\alpha=0.95$, two mutually exclusive hypothetical losses are each $40 when they occur. Each occurs with probability 0.0300, and neither occurs with probability 0.9400. Find each component VaR and the aggregate VaR. (Answer: component VaRs $0 and $0; aggregate VaR $40.)
At $\alpha=0.95$, two independent hypothetical losses are each $100 with probability 0.0400 and $0 otherwise. Find the component VaRs and aggregate VaR. (Answer: component VaRs $0 and $0; aggregate VaR $100.)
At $\alpha=0.95$, the joint outcomes $(X,Y)$ are $(\$10,\$15)$ with probability 0.9200, $(\$70,\$15)$ with probability 0.0400, and $(\$10,\$95)$ with probability 0.0400. Find the component VaRs and aggregate VaR. (Answer: component VaRs $10 and $15; aggregate VaR $85.)
At $\alpha=0.90$, two hypothetical losses are both $50 when a shared event occurs, which has probability 0.1200; otherwise both are $0. Find the component VaRs, aggregate VaR, and whether the subadditivity inequality is strict. (Answer: component VaRs $50 and $50; aggregate VaR $100; equality holds, so there is no strict violation.)
At $\alpha=0.975$, losses $X$ and $Y$ are mutually exclusive: $X=\$120$ with probability 0.0200, $Y=\$200$ with probability 0.0200, and both are $0 otherwise. Find the component VaRs and aggregate VaR. (Answer: component VaRs $0 and $0; aggregate VaR $120.)

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