Diversification Benefit Under Dependence

Why Dependence Matters for Portfolio Risk

A portfolio’s risk depends not only on the loss distribution for each exposure, but also on how those losses occur together. Two policies may have unchanged individual loss distributions while a shared event makes their losses much more likely to happen at the same time. Portfolio risk measures therefore require a model for the joint distribution, not just the marginal distributions.

This tutorial compares portfolio value-at-risk and tail value-at-risk under independence, common-shock dependence, and a specified joint loss distribution. The losses below are hypothetical, one-period dollar amounts. There is no discounting or mortality assumption: neither interest nor life-contingency cash flows enter these calculations.

Definition: For losses $X_1,\ldots,X_n$, the aggregate loss is $S=\sum_{j=1}^n X_j$. At confidence level $\alpha$, the diversification benefit for a risk measure $\rho$ is $$ D_\rho=\sum_{j=1}^n \rho(X_j)-\rho(S). $$ A positive value means the portfolio risk measure is less than the sum of the stand-alone risk measures. A negative value means aggregation has increased the measure relative to that sum. The percentage diversification benefit, when the sum of stand-alone measures is positive, is $D_\rho/\sum_j\rho(X_j)$.

Risk Measures and Dependence

Use the quantile convention $VaR_\alpha(X)=\inf\{x:F_X(x)\geq\alpha\}$. For discrete losses, tail value-at-risk is defined here by the integral quantile formula:

TVaR_\alpha(X)=\frac{1}{1-\alpha}\int_\alpha^1 VaR_u(X)\,du.

This definition handles probability mass at the VaR point: the upper tail includes only the fraction of that mass needed to make a tail of size $1-\alpha$. Thus, for discrete losses, $TVaR_\alpha$ is not always the same as the conditional mean given $X>VaR_\alpha(X)$.

If $X_1$ and $X_2$ are independent discrete losses, their aggregate distribution is obtained by combining each pair of outcomes and multiplying their probabilities. In general, neither $VaR_\alpha(X_1+X_2)$ nor $TVaR_\alpha(X_1+X_2)$ can be determined from the marginal risk measures alone. Under dependence, the joint probabilities of outcome pairs replace those independence products.

Variance gives a useful contrast. For losses with finite variances,

\operatorname{Var}(X_1+X_2)=\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+2\operatorname{Cov}(X_1,X_2).

The covariance term makes the effect of dependence explicit for variance. But covariance or correlation alone does not generally determine VaR or tail value-at-risk: those tail measures depend on the full distribution of the aggregate.

Worked Example: Independent Losses With the Same Marginals

Two hypothetical coverages each have loss $X_j=100$ with probability $0.10$, and loss $0$ otherwise. The losses are independent. Calculate each stand-alone measure and the aggregate VaR and tail value-at-risk at $\alpha=0.95$, then find both diversification benefits.

Method and model. Each loss is a Bernoulli payment with amount $100$ and probability $0.10$. Independence means the probability of a pair of outcomes is the product of its marginal probabilities.

For either individual coverage, the loss is $0$ with probability $0.90$ and $100$ with probability $0.10$. Since $F(0)=0.90<0.95$ and $F(100)=1$, $VaR_{0.95}(X_j)=100$. The highest $5\%$ of outcomes all have loss $100$, so $TVaR_{0.95}(X_j)=100$.

For the aggregate $S=X_1+X_2$, the outcome probabilities are:

\begin{aligned} P(S=0)&=(0.90)(0.90)=0.81,\\ P(S=100)&=(0.10)(0.90)+(0.90)(0.10)=0.18,\\ P(S=200)&=(0.10)(0.10)=0.01. \end{aligned}

The cumulative probabilities are $F_S(0)=0.81$, $F_S(100)=0.99$, and $F_S(200)=1$. Therefore $VaR_{0.95}(S)=100$. For tail value-at-risk, the worst $5\%$ consists of all of the $1\%$ probability at loss $200$, plus $4\%$ of the probability at loss $100$:

TVaR_{0.95}(S)=\frac{(0.01)(200)+(0.04)(100)}{0.05} =\frac{2+4}{0.05}=120.

The sum of stand-alone VaRs is $100+100=200$, so $D_{VaR}=200-100=100$. The sum of stand-alone tail value-at-risk measures is $200$, so $D_{TVaR}=200-120=80$. The calculations check against the aggregate outcomes: its worst $5\%$ has mean loss $120$, below the stand-alone sum of $200$. The method avoids assuming that the sum of stand-alone quantiles is the quantile of the sum.

What the Dependence Assumption Changes

Under independence, simultaneous large losses may be less likely than under a common-shock arrangement. Under positive perfect dependence, losses move together in the strongest rank-aligned sense: in a common-event model, all covered losses occur on the same event. This can eliminate a diversification benefit. It does not change the marginal distribution of any one loss; it changes which outcomes occur together.

The next example uses exactly the same individual loss distributions as the independent example. Only the dependence changes. The confidence level is kept at $0.95$, so the comparison isolates the role of the joint model.

Worked Example: A Common Event Produces Perfect Dependence

Two hypothetical coverages each pay $100$ if a shared event occurs and $0$ otherwise. The event has probability $0.10$, and both coverages respond to it. Find the stand-alone and aggregate VaR and tail value-at-risk at $\alpha=0.95$, and calculate the diversification benefits.

Method and model. Each marginal loss is $100$ with probability $0.10$. The losses are perfectly aligned through the same event: with probability $0.10$, both are $100$, and with probability $0.90$, both are $0$.

For each coverage, $F(0)=0.90<0.95$, so $VaR_{0.95}(X_j)=100$. The worst $5\%$ lies entirely within the event probability of $10\%$; every loss in that tail is $100$. Hence $TVaR_{0.95}(X_j)=100$.

The aggregate is $S=200$ with probability $0.10$, and $S=0$ with probability $0.90$. Since $F_S(0)=0.90<0.95$, $VaR_{0.95}(S)=200$. The worst $5\%$ consists entirely of outcomes with aggregate loss $200$, giving $TVaR_{0.95}(S)=200$.

The stand-alone sum is $100+100=200$ for each measure. Thus $D_{VaR}=200-200=0$ and $D_{TVaR}=200-200=0$. These results check directly against the common-event aggregate: whenever a loss occurs, both payments occur. The method avoids treating the two event indicators as independent when the scenario explicitly makes them identical.

Calculating Risk From a Specified Joint Distribution

When the joint distribution is given explicitly, first add the losses in each joint outcome, then combine probabilities for equal aggregate amounts. Only after building the aggregate distribution should you apply the quantile and tail calculations. The next example also shows that positive dependence need not be perfect.

Worked Example: A Joint Table With Positive but Imperfect Dependence

Let $X$ be a hypothetical loss of $100$ when event A occurs, and $0$ otherwise. Let $Y$ be a loss of $150$ when event B occurs, and $0$ otherwise. The joint probabilities are $P(A,B)=0.10$, $P(A,\text{not }B)=0.10$, $P(\text{not }A,B)=0.10$, and $P(\text{not }A,\text{not }B)=0.70$. Calculate $VaR_{0.85}$ and $TVaR_{0.85}$ for each loss and their sum, then find the diversification benefits.

Method and model. This is a specified four-outcome joint distribution, not an independence model. The marginals are $P(A)=0.10+0.10=0.20$ and $P(B)=0.10+0.10=0.20$.

For $X$, the loss is $100$ with probability $0.20$ and $0$ with probability $0.80$. Since $F_X(0)=0.80<0.85$, $VaR_{0.85}(X)=100$. The upper $15\%$ is contained within the $20\%$ probability at loss $100$, so $TVaR_{0.85}(X)=100$.

For $Y$, the loss is $150$ with probability $0.20$ and $0$ with probability $0.80$. By the same quantile check, $VaR_{0.85}(Y)=150$; the upper $15\%$ has loss $150$ throughout, so $TVaR_{0.85}(Y)=150$.

The joint outcomes give the aggregate distribution:

\begin{array}{c|c|c} \text{Outcome}&S=X+Y&\text{Probability}\\ \text{Neither event}&0&0.70\\ A\text{ only}&100&0.10\\ B\text{ only}&150&0.10\\ A\text{ and }B&250&0.10 \end{array}

The aggregate cumulative probabilities are $0.70$ at $0$, $0.80$ at $100$, $0.90$ at $150$, and $1$ at $250$. Therefore $VaR_{0.85}(S)=150$. In the worst $15\%$, all of the $10\%$ probability at $250$ is included, together with $5\%$ probability at $150$:

TVaR_{0.85}(S)=\frac{(0.10)(250)+(0.05)(150)}{0.15} =\frac{25+7.5}{0.15}=216.6667.

The stand-alone VaR sum is $100+150=250$, so $D_{VaR}=250-150=100$. The stand-alone tail value-at-risk sum is also $250$, so $D_{TVaR}=250-216.6667=33.3333$, to four decimal places. The joint table gives $P(A,B)=0.10$, greater than the independence product $0.20(0.20)=0.04$; nevertheless, the risks are not perfectly aligned. The aggregate distribution and resulting measures confirm that the specified dependence, rather than the marginals alone, drives the answer.

Derivation: Why the Discrete Tail Calculation Uses a Fraction

For a discrete loss, the quantile function $VaR_u(X)$ is constant over probability intervals whose lengths equal the probabilities of the corresponding loss amounts. Integrating that step function from $\alpha$ to $1$ therefore adds the loss amounts weighted by the portions of their probability intervals that lie in the upper tail.

If the upper tail begins partway through the probability mass at $q=VaR_\alpha(X)$, only the remaining fraction of that mass belongs in the integral. Dividing the resulting weighted total by the tail probability $1-\alpha$ gives the average loss over the quantile tail. This is why the examples use a fractional amount of a boundary outcome when necessary, rather than including all of its probability or excluding it entirely.

Interpreting Diversification Benefits and Avoiding Traps

The examples illustrate two important distinctions. First, a portfolio’s VaR is not generally the sum of stand-alone VaRs. In the independent example, that sum was $200$, while aggregate VaR was $100$. Second, a dependence assumption can change the answer even when all marginal distributions stay fixed: the common-event example had aggregate VaR $200$ instead.

Tail value-at-risk is a coherent risk measure under the stated quantile-integral definition, so it is subadditive: $TVaR_\alpha(X+Y)\leq TVaR_\alpha(X)+TVaR_\alpha(Y)$ under the usual finite-tail-mean conditions. That property means its diversification benefit cannot be negative. VaR, by contrast, is not subadditive in general; a negative VaR diversification benefit is possible. A positive benefit in one example does not establish a universal guarantee for VaR.

A common wrong calculation is to take the two stand-alone VaRs in the independent example and report $VaR_{0.95}(S)=100+100=200$. That adds quantiles rather than finding the quantile of the aggregate. The correction is to construct the aggregate distribution: $P(S=0)=0.81$, $P(S=100)=0.18$, and $P(S=200)=0.01$. Since $F_S(0)=0.81<0.95\leq F_S(100)=0.99$, the correct aggregate VaR is $100$.

Do not infer independence from marginal distributions. Independence is an additional assumption about joint probabilities.
Do not apply a continuous-tail conditional-mean shortcut blindly. For discrete losses, the quantile-integral definition may require only part of the VaR-point probability.
Do not treat correlation as a complete tail model. A covariance or correlation value alone does not specify all joint outcome probabilities.
Check the direction of the comparison. Diversification benefit is the stand-alone sum minus the aggregate measure.

Key takeaway: To quantify diversification, specify the dependence, construct the distribution of the aggregate loss, and then calculate its risk measure using the stated quantile convention. The marginal risk measures alone are not enough.

Check Your Understanding

For each problem, calculate the requested aggregate measure from the stated joint model before finding the diversification benefit.

Two independent coverages each pay $80$ with probability $0.10$, and otherwise pay $0$. At $\alpha=0.95$, find aggregate VaR and its diversification benefit. (Answer: aggregate VaR $=80$; VaR diversification benefit $=80$.)
Two coverages pay $60$ and $140$, respectively, whenever the same event occurs; the event probability is $0.08$. At $\alpha=0.95$, find aggregate tail value-at-risk and its diversification benefit. (Answer: aggregate $TVaR_{0.95}=200$; diversification benefit $=0$.)
Loss $X$ is $50$ on event A and $0$ otherwise; loss $Y$ is $100$ on event B and $0$ otherwise. Let $P(A,B)=0.10$, $P(A,\text{not }B)=0.20$, $P(\text{not }A,B)=0.10$, and $P(\text{neither})=0.60$. At $\alpha=0.85$, find aggregate VaR and aggregate tail value-at-risk. (Answer: aggregate $VaR_{0.85}=100$; aggregate $TVaR_{0.85}=133.3333$.)
Two independent losses each equal $100$ with probability $0.10$, and otherwise equal $0$. At $\alpha=0.95$, calculate aggregate tail value-at-risk and its diversification benefit. (Answer: aggregate $TVaR_{0.95}=120$; diversification benefit $=80$.)

What You'll Learn