Why VaR Can Fail to Reward Diversification
Diversification is often expected to reduce the capital required for a group of risks. But that expectation does not hold for every risk measure and every loss distribution. In particular, aggregate VaR can exceed the sum of the standalone VaRs. This is not a calculation inconsistency: it is a consequence of how a quantile responds to probability mass near the selected confidence level.
The key exam skill is to construct or read the aggregate loss distribution, then find its quantile directly. Adding component VaRs, averaging possible losses, or looking only at the most likely outcome can all give the wrong conclusion. The examples use hypothetical insurance loss models and specify their dependence assumptions.
For two component losses \(X\) and \(Y\), the aggregate loss is \(X+Y\). At a common confidence level, the comparison of interest is between \(\operatorname{VaR}_q(X+Y)\) and \(\operatorname{VaR}_q(X)+\operatorname{VaR}_q(Y)\). A strict inequality in the first direction means that VaR has failed to reward diversification in that example. The aggregate quantile must be calculated from the joint loss outcomes, so dependence matters.
A Discrete Quantile Must Be Read From the Cumulative Distribution
In a continuous distribution, a quantile is often found by solving an equation involving a smooth cumulative distribution function. A discrete distribution instead has jumps. If the cumulative probability just below a loss value is less than \(q\), but at that value it reaches or exceeds \(q\), that value is the VaR. The size of the jump matters; the VaR is not an average of adjacent loss amounts.
Worked Example: Mutually Exclusive Rare Losses
A hypothetical insurer considers two lines, \(A\) and \(B\), each of which can produce a loss of $100 or no loss. Use 100 equally likely scenarios. Line \(A\) has a $100 loss in scenarios 1 through 4; line \(B\) has a $100 loss in scenarios 5 through 8. Neither line loses in the other scenarios. Find each standalone 95% VaR and the 95% VaR of the aggregate. Assume the stated scenario set is the full joint model.
Method and standalone distributions. Each line has a $100 loss in 4 of 100 scenarios and a zero loss in 96 of 100. Thus, for either component, \(P(X=0)=0.96\) and \(P(X=100)=0.04\). Since \(F_X(0)=0.96\ge0.95\), the smallest loss reaching the 95% threshold is zero. Therefore, \(\operatorname{VaR}_{0.95}(A)=\operatorname{VaR}_{0.95}(B)=\$0\).
Aggregate distribution. The losses are mutually exclusive: there are 8 scenarios with an aggregate loss of $100, and the other 92 have an aggregate loss of zero. Therefore, \(P(A+B=0)=0.92\) and \(P(A+B=100)=0.08\). The cumulative probabilities are \(F_{A+B}(0)=0.92\) and \(F_{A+B}(100)=1.00\). Since 0.92 is below 0.95, but 1.00 is at least 0.95, \(\operatorname{VaR}_{0.95}(A+B)=\$100\).
Comparison and check. The sum of standalone VaRs is \(\$0+\$0=\$0\), while aggregate VaR is $100. The aggregate distribution assigns 8% probability to a positive loss, which is enough to push the 95% quantile above zero. The probabilities sum to \(0.92+0.08=1.00\), as required. The calculation avoids the mistake of treating two separate 4% loss probabilities as if the aggregate also had only a 4% chance of loss.
Why Component Quantiles Do Not Generally Bound Aggregate VaR
Let \(a=\operatorname{VaR}_q(X)\) and \(b=\operatorname{VaR}_q(Y)\). By the quantile definition, \(P(X\le a)\ge q\) and \(P(Y\le b)\ge q\). If both events occur, then \(X+Y\le a+b\). However, the probability that both occur is not necessarily at least \(q\). The general lower bound from the union bound is
For example, at \(q=0.95\), this guarantees only a lower bound of 0.90 for the probability that both component losses are at or below their VaRs. A 0.90 lower bound does not establish that \(P(X+Y\le a+b)\ge0.95\), which is what would be needed to conclude that aggregate VaR is no greater than \(a+b\). The bound does not prove that aggregate VaR must exceed the sum; it explains why the component quantiles alone cannot rule out that outcome. The aggregate distribution must still be evaluated.
Use the same \(q\) for each component and the aggregate, and identify VaR as the smallest loss whose cumulative probability reaches \(q\).
State how component losses occur together. Marginal distributions alone do not determine the aggregate distribution.
Add component losses within each joint outcome, combine probabilities for equal totals, and order the resulting aggregate losses.
Find the first aggregate loss where cumulative probability is at least \(q\), then compare that VaR with the sum of the component VaRs.
Worked Example: Same Marginals, Independent Losses
Now suppose two hypothetical lines each have a $100 loss with probability 0.04 and no loss with probability 0.96. Unlike the previous example, assume the two losses are independent. Find the component and aggregate 95% VaRs, and explain how the joint model changes the aggregate distribution.
Method and component VaRs. For each line, \(F(0)=0.96\), which is at least 0.95. Thus each component has 95% VaR of $0, and the sum of standalone VaRs is $0.
Aggregate probabilities. Independence gives probability \(0.96\times0.96=0.9216\) of no losses. Exactly one line loses with probability \(2(0.04)(0.96)=0.0768\). Both lines lose with probability \(0.04\times0.04=0.0016\). The aggregate distribution is therefore $0 with probability 0.9216, $100 with probability 0.0768, and $200 with probability 0.0016. These probabilities sum to \(0.9216+0.0768+0.0016=1.0000\).
Aggregate VaR and check. The cumulative probability at $0 is 0.9216, below 0.95. At $100 it is \(0.9216+0.0768=0.9984\), at least 0.95. Therefore, \(\operatorname{VaR}_{0.95}(X+Y)=\$100\), which exceeds the $0 sum of standalone VaRs. The 0.16% probability of a $200 aggregate loss does not set the 95% quantile: the relevant threshold is first crossed at $100. Unlike the mutually exclusive model, independence permits both losses in the same scenario, but the 95% VaR remains $100.
When Standalone VaRs Are Positive
The effect is not limited to examples where component VaRs are zero. When each standalone distribution has a substantial middle outcome and a less frequent severe outcome, an aggregate can cross the confidence threshold only after one component takes its severe value. The sorted aggregate outcomes and their cumulative probabilities reveal exactly where this happens.
Worked Example: Positive Component VaRs but a Larger Aggregate VaR
Two hypothetical risks are independent and have the same loss distribution: $0 with probability 0.20, $40 with probability 0.70, and $200 with probability 0.10. Calculate the 90% VaR for each risk and for their aggregate. Then compare the aggregate VaR with the sum of standalone VaRs.
Method and component quantile. For either risk, the cumulative probability at $0 is 0.20, and at $40 it is \(0.20+0.70=0.90\). The smallest loss at which the cumulative probability reaches 0.90 is $40. Thus each standalone 90% VaR is $40, and their sum is $80.
Aggregate distribution. By independence, the joint outcomes have the following aggregate losses and probabilities:
| Component outcomes | Aggregate loss | Probability | Cumulative probability |
|---|---|---|---|
| $0 and $0 | $0 | 0.20 × 0.20 = 0.04 | 0.04 |
| One $0 and one $40 | $40 | 2 × 0.20 × 0.70 = 0.28 | 0.32 |
| $40 and $40 | $80 | 0.70 × 0.70 = 0.49 | 0.81 |
| One $0 and one $200 | $200 | 2 × 0.20 × 0.10 = 0.04 | 0.85 |
| One $40 and one $200 | $240 | 2 × 0.70 × 0.10 = 0.14 | 0.99 |
| $200 and $200 | $400 | 0.10 × 0.10 = 0.01 | 1.00 |
The cumulative probability is 0.85 at $200, which is below 0.90, and 0.99 at $240, which is at least 0.90. Therefore, the aggregate 90% VaR is $240. It exceeds the $80 sum of standalone VaRs by $160. The probabilities in the table sum to \(0.04+0.28+0.49+0.04+0.14+0.01=1.00\), and the cumulative probabilities reach 1.00 at the largest aggregate loss.
The reason is visible in the distribution: losses up to $200 account for only 85% of outcomes. The aggregate must include the $240 outcome before its cumulative probability reaches 90%. The standalone quantile of $40 does not imply that both risks will be at or below $40 in at least 90% of joint outcomes.
Common Exam Trap: Averaging or Skipping Probability Mass
Consider the independent $100-loss example above. A tempting but incorrect calculation is to call the aggregate 95% VaR $96, using the probability-weighted average \(0(0.9216)+100(0.0768)+200(0.0016)=7.52\) for the expected loss—or to interpolate between $0 and $100 because the 95% threshold falls between cumulative probabilities 0.9216 and 0.9984. Neither operation calculates a discrete VaR. The expected loss is $8.00, not $7.52: the correct arithmetic is \(100(0.0768)+200(0.0016)=7.68+0.32=8.00\). In either case, an expectation is not a quantile.
The correct quantile calculation is \(F_{X+Y}(0)=0.9216<0.95\) and \(F_{X+Y}(100)=0.9984\ge0.95\), so the 95% VaR is exactly $100. Do not interpolate across a jump unless a different quantile convention has explicitly been specified. Also do not combine marginal probabilities without considering whether losses are independent, mutually exclusive, or otherwise dependent.
Interpreting the Result
VaR is a threshold measure: it identifies the loss level at which the cumulative distribution reaches a chosen confidence level. A small change in probability around that threshold can move VaR from one discrete loss amount to another, sometimes a large amount. Consequently, an aggregate VaR greater than the sum of standalone VaRs is possible even when the aggregate loss model is internally consistent.
This does not mean diversification is always ineffective or that aggregate VaR must exceed standalone totals. The direction and size of the comparison depend on the marginal distributions, the joint dependence structure, the confidence level, and the location of probability mass. For a discrete model, a reliable exam solution should show the relevant aggregate outcomes and cumulative probabilities rather than rely on intuition about diversification.
Check Your Understanding
For each hypothetical model, use the stated left-quantile definition and show the cumulative probability that determines the requested VaR.
- Two independent risks each pay a loss of $60 with probability 0.03 and zero otherwise. At the 95% confidence level, find each standalone VaR and the aggregate VaR. (Answer: Each standalone VaR is $0; aggregate VaR is $60.)
- Two independent risks each have loss $50 with probability 0.08 and zero otherwise. At the 90% confidence level, find the sum of standalone VaRs and the aggregate VaR. (Answer: Sum of standalone VaRs is $0; aggregate VaR is $50.)
- Two independent risks each have loss $0 with probability 0.25, $30 with probability 0.65, and $150 with probability 0.10. At the 90% confidence level, find the standalone VaR for each risk and the aggregate VaR. (Answer: Each standalone VaR is $30; aggregate VaR is $180.)
- Two risks each have a $100 loss in 4% of outcomes and zero otherwise. Their losses are mutually exclusive, so they never occur together. At the 95% confidence level, find the component VaRs and aggregate VaR. (Answer: Each component VaR is $0; aggregate VaR is $100.)
- Two independent risks each have loss $20 with probability 0.20 and zero otherwise. At the 95% confidence level, calculate the aggregate loss distribution and VaR. (Answer: Aggregate losses are $0 with probability 0.64, $20 with probability 0.32, and $40 with probability 0.04; aggregate VaR is $20.)