Dependence Effects on Portfolio VaR

Why Dependence Matters for Portfolio VaR

Knowing the loss distribution for each risk in a portfolio does not, by itself, determine the distribution of the total loss. The risks may tend to produce large losses together, or a large loss in one component may tend to occur when another component’s loss is small. Those patterns affect the aggregate loss and can change its value-at-risk, even when every component’s marginal distribution stays fixed.

This tutorial compares portfolio VaR under independence, perfect positive dependence, and other joint distributions. The examples use discrete losses so that the joint probabilities and the quantile crossing can be inspected directly. All losses are hypothetical, measured in dollars, and are not discounted.

Definition: For a loss $X$ with cumulative distribution function $F_X(x)=P(X\leq x)$, the value-at-risk at confidence level $\alpha$, where $0<\alpha<1$, is the lower quantile $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq \alpha\}. $$ For a discrete loss, this is the smallest loss amount at which the cumulative probability is at least $\alpha$. In particular, if the cumulative probability equals $\alpha$ at a loss amount, that amount is the VaR.

A portfolio with component losses $X_1$ and $X_2$ has aggregate loss $S=X_1+X_2$. The marginal distributions specify the probabilities of values of $X_1$ and $X_2$ separately. To find the distribution of $S$, we also need their joint distribution: the probabilities of each pair of outcomes occurring together.

How the Joint Distribution Determines the Aggregate

Suppose $X_1$ takes values $x_i$, $X_2$ takes values $y_j$, and $p_{ij}=P(X_1=x_i,X_2=y_j)$. Every pair contributes its joint probability to the aggregate value $x_i+y_j$. Thus, for any threshold $s$, the aggregate cumulative distribution is

P(S\leq s)=\sum_{i,j:\,x_i+y_j\leq s}p_{ij}.

This is the key calculation: the set of pairs whose sum is at most $s$ is determined by the loss values, but their probabilities $p_{ij}$ depend on the dependence assumption. Under independence, $p_{ij}=P(X_1=x_i)P(X_2=y_j)$. Under another dependence structure, those same marginal probabilities can be paired differently.

Key distinction: Independence is a particular joint distribution, not a consequence of knowing the marginal distributions. Perfect positive dependence also preserves the marginals, but aligns high outcomes with high outcomes. Other joint distributions can preserve the same marginals while producing different aggregate probabilities and therefore different VaR.

For continuous or discrete marginals, one standard construction of perfect positive dependence is to take a common uniform random variable $U$ on $(0,1)$ and set each component to its marginal quantile at $U$. The components then move together through their respective distributions. For discrete losses, this means aligning the low outcomes with one another and the high outcomes with one another, subject to the marginal probabilities.

Worked Comparisons

Worked Example: Two Rare Losses Under Three Dependence Structures

Two hypothetical coverage layers each have loss $0$ with probability $0.90$ and loss $\$100$ with probability $0.10$. Find the portfolio VaR at confidence level $0.95$ under independence, perfect positive dependence, and a joint model in which the two $\$100$ losses cannot occur together.

Method. For each joint model, list the possible pairs of component losses, add the two losses in each pair, and accumulate the corresponding probabilities until the cumulative probability reaches $0.95$. The marginal loss distribution for each component remains the stated $0/\$100$ distribution.

Independence. The four pairs have probabilities $P(0,0)=0.90(0.90)=0.81$, $P(100,0)=0.10(0.90)=0.09$, $P(0,100)=0.90(0.10)=0.09$, and $P(100,100)=0.10(0.10)=0.01$. The aggregate distribution is therefore

P(S=0)=0.81,\qquad P(S=100)=0.09+0.09=0.18,\qquad P(S=200)=0.01.

The cumulative probabilities are $F_S(0)=0.81$ and $F_S(100)=0.81+0.18=0.99$. Since $0.81<0.95\leq0.99$, the independent portfolio has $\operatorname{VaR}_{0.95}(S)=\$100$.

Perfect positive dependence. The two losses are zero together with probability $0.90$, and are both $\$100$ with probability $0.10$. Hence $P(S=0)=0.90$ and $P(S=200)=0.10$. Since $F_S(0)=0.90<0.95$ and $F_S(200)=1$, the portfolio VaR is $\$200$.

No simultaneous high losses. One valid joint model assigns probability $0.80$ to $(0,0)$, probability $0.10$ to $(100,0)$, and probability $0.10$ to $(0,100)$. Each component still has probability $0.10$ of a $\$100$ loss, and neither pair has two high losses. Thus $P(S=0)=0.80$ and $P(S=100)=0.20$. Here $F_S(0)=0.80<0.95$ and $F_S(100)=1$, so VaR is $\$100$.

Check. In each joint model, the probability that the first component is high is $0.10$, and the probability that the second component is high is also $0.10$. The VaR values differ because the probabilities of the pairs differ. The common mistake is to assume that component marginals alone imply a unique portfolio VaR.

Worked Example: Different Loss Sizes and Opposite Rank Pairing

Two components have marginal losses $X_1=0$ or $\$10$, each with probability $0.50$, and $X_2=0$ or $\$20$, each with probability $0.50$. Find the portfolio VaR at confidence level $0.75$ under independence, perfect positive dependence, and an opposite-rank pairing.

Independence. Each of the four pairs has probability $0.50(0.50)=0.25$. The pairs $(0,0),(10,0),(0,20),(10,20)$ give aggregate losses $0,10,20,30$, respectively. Therefore each aggregate value has probability $0.25$. The cumulative probability at $20$ is $0.25+0.25+0.25=0.75$, so $\operatorname{VaR}_{0.75}(S)=\$20$.

Perfect positive dependence. Pair the lower outcome of each component together and the higher outcome of each component together. The pair $(0,0)$ occurs with probability $0.50$, while $(10,20)$ occurs with probability $0.50$. The aggregate is $0$ or $30$, each with probability $0.50$. Since $F_S(0)=0.50<0.75$ and $F_S(30)=1$, VaR is $\$30$.

Opposite-rank pairing. Pair $(0,20)$ with probability $0.50$ and $(10,0)$ with probability $0.50$. The first component is zero half the time and $\$10$ half the time; the second is zero half the time and $\$20$ half the time, so both marginals are preserved. The aggregate losses are $\$20$ and $\$10$, each with probability $0.50$. Thus $F_S(10)=0.50$, $F_S(20)=1$, and the $0.75$ VaR is $\$20$.

Check. The same marginal loss distributions produce VaR values of $\$20$, $\$30$, and $\$20$ under the three pairings. In the independent case, the quantile is exactly at a cumulative probability of $0.75$; it is not the next larger aggregate value.

Worked Example: A Joint Probability Changes VaR While Marginals Stay Fixed

Two components each have loss $0$ with probability $0.70$ and loss $\$100$ with probability $0.30$. Compare the portfolio VaR at confidence level $0.80$ under independence with a specified alternative joint distribution. The alternative assigns probabilities $0.65,0.05,0.05,0.25$ to the pairs $(0,0),(100,0),(0,100),(100,100)$, respectively.

Verify the alternative marginals first. For the first component, the probability of a $\$100$ loss is $0.05+0.25=0.30$. For the second component it is also $0.05+0.25=0.30$. Thus the alternative joint distribution has the required marginals.

Under independence. The pair probabilities are $P(0,0)=0.70(0.70)=0.49$, $P(100,0)=0.30(0.70)=0.21$, $P(0,100)=0.70(0.30)=0.21$, and $P(100,100)=0.30(0.30)=0.09$. The aggregate probabilities are $P(S=0)=0.49$, $P(S=100)=0.21+0.21=0.42$, and $P(S=200)=0.09$. Therefore $F_S(0)=0.49$, $F_S(100)=0.49+0.42=0.91$, and $\operatorname{VaR}_{0.80}(S)=\$100$.

Under the specified alternative. The aggregate is zero for $(0,0)$, $\$100$ for either single-high pair, and $\$200$ for $(100,100)$. Hence $P(S=0)=0.65$, $P(S=100)=0.05+0.05=0.10$, and $P(S=200)=0.25$. The cumulative probability through $\$100$ is $0.65+0.10=0.75$, which is less than $0.80$. Adding the probability at $\$200$ brings the cumulative probability to $1$. Thus $\operatorname{VaR}_{0.80}(S)=\$200$.

Check. Both models have the same $0/\$100$ marginal losses with high-loss probability $0.30$, but the alternative makes high losses occur together more often than independence. Its VaR is therefore higher at this confidence level. The calculation uses the specified joint probabilities, not an assumption that every positive association has the same effect at every quantile.

A Short Justification and a Reliable Calculation Method

The aggregate CDF formula follows by partitioning outcomes into mutually exclusive pairs. For a fixed $s$, the event $S\leq s$ consists exactly of those pairs $(X_1=x_i,X_2=y_j)$ for which $x_i+y_j\leq s$. Since these pair events are disjoint, their probabilities add:

P(S\leq s)=\sum_{i,j:\,x_i+y_j\leq s}P(X_1=x_i,X_2=y_j).

Independence replaces each joint probability in this sum with the product of the corresponding marginal probabilities. A different dependence assumption changes the joint probabilities in the sum, while leaving the marginal probabilities unchanged. Since VaR is determined by where this aggregate CDF first reaches $\alpha$, changing dependence can move the quantile.

Record the marginals.
List each component’s possible losses and probabilities. Check that the probabilities for each component sum to one.

Specify the joint model.
For independence, multiply marginal probabilities. For a stated alternative, use its pair probabilities and verify that summing across rows and columns reproduces the required marginals.

Aggregate the pairs.
Add the component losses for every pair, then combine probabilities for pairs that produce the same total.

Locate the quantile crossing.
Arrange aggregate losses from smallest to largest and find the first one whose cumulative probability is at least $\alpha$. Include probability mass at that loss amount.

Common Traps and Why the Comparison Matters

Dependence comparisons are useful because a portfolio is exposed to joint outcomes, not just to each risk in isolation. Marginal VaRs do not specify the distribution of the aggregate, and adding marginal VaRs is not a general method for calculating portfolio VaR. The aggregate distribution must be built using a joint model.

Assuming independence without being told. Multiplying marginal probabilities is justified only when the components are independent.
Changing a marginal by mistake. In an alternative joint table, check each row and column total against the stated component probabilities. A table can represent a valid dependence comparison only if it preserves the marginals.
Reading the wrong quantile point. VaR is the first loss whose cumulative probability is at least the confidence level. If the cumulative probability equals the level exactly, that loss is the VaR.
Assuming the VaR changes monotonically with a vague description of dependence. Calculate the aggregate CDF at the relevant confidence level. The particular marginal shapes, joint probabilities, and quantile location all matter.

Perfect positive dependence is an important comparison because it aligns component outcomes in rank order, but it should not be confused with independence or treated as an unstated assumption. A sound portfolio VaR calculation names the joint model, checks that its marginals are correct, and then applies the quantile definition to the resulting aggregate distribution. This method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, keep the component marginals fixed when comparing joint models, and use the lower-quantile convention defined above.

Two components each have loss $0$ with probability $0.80$ and loss $\$50$ with probability $0.20$. Find the portfolio VaR at level $0.90$ under independence and perfect positive dependence. (Answer: Independence, $\$50$; perfect positive dependence, $\$100$.)
Component $A$ has loss $0$ or $\$8$, each with probability $0.50$. Component $B$ has loss $0$ or $\$12$, each with probability $0.50$. Find the VaR at level $0.60$ under independence and under perfect positive dependence. (Answer: Independence, $\$12$; perfect positive dependence, $\$20$.)
Two components each have loss $0$ with probability $0.60$ and loss $\$40$ with probability $0.40$. Their joint model gives probability $0.30$ to $(0,0)$, $0.30$ to each of $(40,0)$ and $(0,40)$, and $0.10$ to $(40,40)$. Find portfolio VaR at level $0.95$. (Answer: $\$80$.)
Component $A$ has loss $\$60$ with probability $0.25$; component $B$ has loss $\$100$ with probability $0.40$. Each is otherwise zero. Under an alternative joint model, $P(60,100)=0.20$, $P(60,0)=0.05$, $P(0,100)=0.20$, and $P(0,0)=0.55$. Verify the marginals and find VaR at level $0.85$. (Answer: The high-loss probabilities are $0.25$ and $0.40$; VaR is $\$160$.)
Two components each have loss $0$ with probability $0.70$ and loss $\$100$ with probability $0.30$. A joint model has probabilities $0.55,0.15,0.15,0.15$ for $(0,0),(100,0),(0,100),(100,100)$, respectively. Find VaR at level $0.80$, and compare it with the independent-model VaR. (Answer: The joint model gives $\$100$; independence also gives $\$100$.)

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