An Independent Bernoulli Counterexample to VaR Subadditivity

Why Independence Does Not Guarantee VaR Subadditivity

A loss can be zero most of the time and take a fixed positive value when an event occurs. Such a two-outcome loss is simple to model, but it can expose a limitation of value at risk (VaR): even when two losses are independent, the VaR of their sum can exceed the sum of their separate VaRs. The key is to calculate the probability of the aggregate loss being zero, rather than combining the individual VaR values.

This tutorial uses the lower-quantile convention for VaR. For a loss $X$ and confidence level $\alpha$, VaR is the smallest loss amount whose cumulative probability is at least $\alpha$. This convention is important for discrete losses, because the cumulative distribution function can jump over the confidence level at a probability atom.

Definition: For a loss $X$ with cumulative distribution function $F_X(x)=P(X\le x)$, the VaR at confidence level $\alpha$, where $0<\alpha<1$, is $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}. $$ A Bernoulli loss of amount $b>0$ with event probability $p$ is $X=bI$, where $I=1$ with probability $p$ and $I=0$ with probability $1-p$. Thus $P(X=0)=1-p$ and $P(X=b)=p$.

Individual and Aggregate VaR for Two Losses

Let $X_1$ and $X_2$ be independent Bernoulli losses with positive amounts $b_1,b_2$ and event probabilities $p_1,p_2$. Independence means that the probability of both events occurring is $p_1p_2$, and the probability of neither occurring is $(1-p_1)(1-p_2)$. The sum $S=X_1+X_2$ can take the values $0,b_1,b_2,b_1+b_2$, combining equal values if the amounts happen to be equal.

For an individual loss $X_j$, the VaR is particularly direct. If $1-p_j\ge\alpha$, the probability of zero loss already reaches the confidence level, so $\operatorname{VaR}_{\alpha}(X_j)=0$. If $1-p_j<\alpha$, the cumulative probability at zero is too small, and the next possible loss is $b_j$, so $\operatorname{VaR}_{\alpha}(X_j)=b_j$.

For the aggregate, the probability of zero loss is the product of the two probabilities of no loss. It can fall below $\alpha$ even when each individual probability of no loss is at least $\alpha$. In that case, each individual VaR can be zero while the portfolio VaR is positive.

Key result: If $X_1$ and $X_2$ are independent Bernoulli losses, and $P(X_1=0)\ge\alpha$ and $P(X_2=0)\ge\alpha$, then both individual VaRs are zero. If, in addition, $P(X_1=0,X_2=0)<\alpha$, the aggregate VaR is positive. Therefore, $$ \operatorname{VaR}_{\alpha}(X_1+X_2)> \operatorname{VaR}_{\alpha}(X_1)+\operatorname{VaR}_{\alpha}(X_2)=0. $$

Why the Bernoulli Counterexample Works

The equal-amount case gives a concise condition for a strict violation. Suppose each independent loss is $b$ with probability $p$ and zero otherwise. For either individual loss, VaR is zero if $1-p\ge\alpha$. For the aggregate, the probability of zero is $(1-p)^2$. If that is below $\alpha$, the aggregate VaR cannot be zero. The probability of an aggregate loss no greater than $b$ is $1-p^2$: this includes no event or exactly one event. If $1-p^2\ge\alpha$, the aggregate VaR is exactly $b$.

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

These inequalities explain the mechanism. Individually, a zero-loss outcome is sufficiently probable to meet the confidence level. Jointly, the probability of no loss is smaller, and the portfolio crosses the confidence threshold at a positive loss amount. For equal Bernoulli losses, the conditions can be written as $1-\sqrt{\alpha}<p\le1-\alpha$; this interval gives individual VaRs of zero and an aggregate VaR of $b$.

Calculate each individual zero-loss probability.
Compare $1-p_j$ with $\alpha$. Do not assume that an event probability must exceed $\alpha$ for the loss VaR to be positive; the relevant comparison is the cumulative probability at zero.

Use independence to form joint probabilities.
Multiply the appropriate event and no-event probabilities to find the probabilities for the aggregate outcomes.

Accumulate aggregate probabilities in loss order.
Find the first possible loss value where the aggregate cumulative probability is at least $\alpha$. That value is the aggregate VaR.

Worked Calculations

Worked Example: Equal Independent Losses at 95% Confidence

A hypothetical portfolio has two independent coverages. Each produces a loss of $100,000 with probability $0.04$, and otherwise produces no loss. Calculate the 95% VaR of each coverage and of their combined loss, then compare the results.

Method. Model each coverage as $X_j=100{,}000I_j$, with $P(I_j=1)=0.04$, and use independence for the aggregate distribution. The confidence level is $\alpha=0.95$.

For either coverage, the probability of zero loss is $1-0.04=0.96$. Since $0.96\ge0.95$, the 95% VaR of each individual loss is $0.

For the combined loss $S=X_1+X_2$, the possible amounts and probabilities are:

No losses: $P(S=0)=0.96(0.96)=0.9216$.
Exactly one loss: $P(S=100{,}000)=2(0.04)(0.96)=0.0768$.
Both losses: $P(S=200{,}000)=(0.04)(0.04)=0.0016$.

The cumulative probability at zero is $0.9216<0.95$, so the aggregate VaR is not zero. At $100,000, the cumulative probability is $0.9216+0.0768=0.9984$, which is at least $0.95$. Therefore, the aggregate 95% VaR is $100,000.

Check. The probabilities sum to $0.9216+0.0768+0.0016=1.0000$. The portfolio VaR of $100,000 exceeds the sum of the individual VaRs, which is $0. The mistake to avoid is assuming that independent losses must have aggregate VaR equal to the sum of their individual VaRs.

Worked Example: Unequal Loss Amounts and Ordered Outcomes

Two independent hypothetical risks have the following loss distributions. Risk 1 produces $80 with probability $0.03$, and risk 2 produces $150 with probability $0.04$; either risk otherwise produces zero. Find the individual and combined 95% VaRs.

Method. For each individual loss, first compare its zero-loss probability with $0.95$. Then list and accumulate the aggregate probabilities in order of loss amount. All amounts are in dollars.

For risk 1, $P(X_1=0)=1-0.03=0.97\ge0.95$, so $\operatorname{VaR}_{0.95}(X_1)=0$. For risk 2, $P(X_2=0)=1-0.04=0.96\ge0.95$, so $\operatorname{VaR}_{0.95}(X_2)=0$.

For $S=X_1+X_2$, independence gives:

$P(S=0)=0.97(0.96)=0.9312$.
$P(S=80)=0.03(0.96)=0.0288$, because only risk 1 incurs a loss.
$P(S=150)=0.97(0.04)=0.0388$, because only risk 2 incurs a loss.
$P(S=230)=0.03(0.04)=0.0012$, because both risks incur losses.

The cumulative probability at zero is $0.9312<0.95$. At $80 it is $0.9312+0.0288=0.9600\ge0.95$. Since $80 is the first aggregate loss amount at which the cumulative probability reaches the confidence level, $\operatorname{VaR}_{0.95}(S)=\$80$.

Check. The four probabilities sum to $0.9312+0.0288+0.0388+0.0012=1.0000$. The combined VaR is $80, greater than the sum of the individual VaRs, $0. The ordering matters: the $80 outcome reaches the threshold before the $150 outcome.

Worked Example: A Higher Confidence Level with Smaller Event Probabilities

Consider two independent hypothetical risks. Each has a loss of $250,000 with probability $0.006$, and otherwise has no loss. Calculate the 99% VaR of each risk and of their sum.

Method. Apply the same quantile test at $\alpha=0.99$, keeping enough precision in the probability calculations to determine which side of the threshold each cumulative probability lies.

For an individual risk, the zero-loss probability is $1-0.006=0.994$, which is at least $0.99$. Thus each individual 99% VaR is $0.

For the aggregate, the probability of no losses is $(0.994)^2=0.988036$, or approximately $0.9880$. This is below $0.99$. The probability of at least one loss is $1-0.988036=0.011964$, approximately $0.0120$. Since the two losses have the same amount, the aggregate loss is either $0, $250,000, or $500,000. The cumulative probability through $250,000 is the probability of no more than one event: $1-(0.006)^2=0.999964$, approximately $1.0000$. This exceeds $0.99$.

Therefore, $\operatorname{VaR}_{0.99}(X_1)=\operatorname{VaR}_{0.99}(X_2)=\$0$, while $\operatorname{VaR}_{0.99}(X_1+X_2)=\$250{,}000$.

Check. The aggregate probability at zero is below $0.99$, but its cumulative probability at $250,000 is above $0.99$. The small event probability does not make the aggregate VaR zero: the relevant comparison is the probability of no losses across both risks.

Common Quantile Errors and Interpretation

A frequent error is to calculate each individual VaR correctly and then treat their sum as the portfolio VaR. VaR is not generally additive, and independence does not make it additive. Independence determines the joint probabilities; the aggregate quantile must still be calculated from the resulting aggregate distribution.

Another error is to use $F_X(x)>\alpha$ instead of $F_X(x)\ge\alpha$. Under the lower-quantile convention, equality is enough: the VaR is the first value at which the cumulative probability reaches or exceeds $\alpha$. This distinction can matter at a discrete probability atom. A related mistake is to inspect only the probability of at least one event. For an unequal-loss portfolio, that probability alone does not identify the quantile; the outcomes must be ordered by their aggregate loss amounts and accumulated.

Key takeaway: For two independent Bernoulli losses, the individual VaRs can both be zero because each has a sufficiently large probability of no loss. The portfolio can nevertheless have positive VaR when the joint probability of no loss falls below the confidence level. This explicit counterexample shows that VaR is not subadditive in general, even for independent losses, and is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the lower-quantile convention and calculate the aggregate distribution where needed.

Two independent risks each produce a loss of $500 with probability $0.02$, and otherwise produce zero. Find both individual 97% VaRs and the 97% VaR of their sum. (Answer: individual VaRs are $0 and $0; aggregate VaR is $500.)
Independent losses are $40 with probability $0.08$ and $90 with probability $0.05$, each otherwise zero. At 90% confidence, find the sum of individual VaRs and the portfolio VaR. (Answer: sum of individual VaRs is $0; portfolio VaR is $40.)
Independent losses are $30 with probability $0.12$ and $70 with probability $0.15$, each otherwise zero. At 90% confidence, find the individual VaRs and portfolio VaR. (Answer: individual VaRs are $30 and $70; portfolio VaR is $70.)
Independent losses are $100 with probability $0.02$ and $300 with probability $0.06$, each otherwise zero. At 95% confidence, calculate the individual VaRs and the portfolio VaR. (Answer: individual VaRs are $0 and $300; portfolio VaR is $300.)

What You'll Learn