Diagnosing VaR Diversification Failures

Why Diversification Is Not Guaranteed Under VaR

Combining insurance lines can reduce risk by pooling losses, but a risk measure should be checked rather than assumed to reward every form of diversification. Value-at-risk (VaR) records a loss threshold at a chosen confidence level. Because it focuses on one quantile, it can overlook how much probability lies just beyond that threshold. In discrete models, that feature can lead to a portfolio VaR greater than the sum of the standalone VaRs.

This tutorial uses finite loss distributions to diagnose that failure and compare VaR with tail value-at-risk (TVaR). All amounts are hypothetical losses in dollars measured at the same valuation date, with no discounting. The examples focus on dependence between two lines: the marginal loss distributions alone do not determine the portfolio result.

Definition: For a loss $X$ and confidence level $0<\alpha<1$, define $\operatorname{VaR}_{\alpha}(X)$ as the left quantile $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}. $$ For two line losses $X$ and $Y$, VaR is subadditive if $$ \operatorname{VaR}_{\alpha}(X+Y)\leq \operatorname{VaR}_{\alpha}(X)+\operatorname{VaR}_{\alpha}(Y). $$ A violation occurs when the portfolio VaR is greater than the sum of the standalone VaRs. VaR can be subadditive in some portfolios; the claim is that it is not guaranteed to be so for all loss distributions.

The left-quantile convention matters in a discrete model. If the cumulative probability at zero is exactly the confidence level, VaR is zero, not the first positive loss amount. In every example below, the cumulative distribution is checked at the proposed quantile.

A Discrete Tail Measure for Comparison

TVaR averages the upper tail rather than reporting only its boundary. For a loss with finite mean, define it by averaging the VaR quantiles from $\alpha$ to 1. This definition accommodates discrete distributions: when the selected tail cuts through a probability atom, only the necessary fraction of that atom is included.

Definition: For $0<\alpha<1$, define $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du. $$ If $q=\operatorname{VaR}_{\alpha}(X)$, an equivalent formula for a discrete loss is $$ \operatorname{TVaR}_{\alpha}(X) =q+\frac{E[(X-q)^{+}]}{1-\alpha}, \qquad (z)^{+}=\max(z,0). $$ This formula includes the appropriate fraction of any probability mass at $q$.

To see why the equivalent formula holds, write $p_{>} = P(X>q)$. The quantile definition gives $P(X\leq q)\geq\alpha$, so $p_{>}\leq1-\alpha$. The upper $1-\alpha$ fraction contains all outcomes above $q$, plus enough probability at $q$ to fill the tail. Its loss total is therefore $E[X\mathbf{1}_{\{X>q\}}]+q(1-\alpha-p_{>})$. Since $E[(X-q)^{+}]=E[X\mathbf{1}_{\{X>q\}}]-qp_{>}$, dividing the tail total by $1-\alpha$ gives $q+E[(X-q)^{+}]/(1-\alpha)$. The excess-loss term is zero for outcomes at or below $q$, so the calculation does not mistakenly include all probability at the quantile.

For integrable losses, TVaR is subadditive. Thus, at the same confidence level, it satisfies $\operatorname{TVaR}_{\alpha}(X+Y)\leq\operatorname{TVaR}_{\alpha}(X)+\operatorname{TVaR}_{\alpha}(Y)$. The examples show the calculation as well as the contrast with VaR.

Worked Example: Mutually Exclusive Claims Produce a VaR Violation

A hypothetical insurer has two lines. Line A pays $1,000 with probability 0.0400 and otherwise pays zero. Line B has the same marginal distribution. The claim events are mutually exclusive: exactly one line pays when a claim occurs, and the probability that neither line pays is 0.9200. Calculate standalone and portfolio VaR and TVaR at $\alpha=0.9500$, and check subadditivity.

Method. First find the loss distribution for each line and for the sum. Then locate the 95th percentile from each cumulative distribution. For TVaR, use the fractional-tail formula, which is particularly direct when $q=0$ or when the portfolio tail is a single loss amount.

For either standalone line, $P(X=0)=0.9600$ and $P(X=1{,}000)=0.0400$. Since $F_X(0)=0.9600\geq0.9500$, $\operatorname{VaR}_{0.95}(X)=0$. The standalone VaR sum is therefore $0+0=\$0.00$.

The portfolio distribution is:

Portfolio loss	Probability	Cumulative probability
$0	0.9200	0.9200
$1,000	0.0800	1.0000

The portfolio cumulative probability is 0.9200 at zero, below 0.9500, and 1.0000 at $1,000. Hence $\operatorname{VaR}_{0.95}(X+Y)=\$1{,}000.00$. The comparison is $$ \$1{,}000.00>\$0.00+\$0.00. $$ VaR is not subadditive for this portfolio.

For either line, $q=0$ and $E[(X-q)^{+}]=0.0400(\$1{,}000)=\$40$. Thus $$ \operatorname{TVaR}_{0.95}(X)=\$0+\frac{\$40}{0.0500}=\$800.00. $$ The sum of standalone TVaRs is $\$1{,}600.00$. For the portfolio, $q=\$1{,}000$ and $E[(X+Y-q)^{+}]=0$, so its TVaR is $\$1{,}000.00$. Therefore $\$1{,}000.00\leq\$1{,}600.00$, as required by subadditivity.

Check. Expected losses are $40.00 per line and $80.00 in total, consistent with the stated event probabilities. The common mistake is to assume the two standalone VaRs must add up to the portfolio VaR; here each standalone quantile misses a claim event that is sufficiently likely in the combined portfolio.

Dependence Changes the Portfolio Quantile

The preceding result is not a consequence of the two marginal distributions alone. Changing only the relationship between the claim events changes the portfolio distribution. The next example makes the events independent rather than mutually exclusive.

Worked Example: Independent Lines and a Fractional TVaR Tail

Two hypothetical lines each pay $1,000 with probability 0.0400 and zero otherwise. Their claim events are independent. At $\alpha=0.9500$, calculate portfolio VaR and TVaR, then compare them with the sums of the standalone measures.

Method. Independence gives the probabilities of zero, one, and two claims. Use the cumulative probabilities to find VaR; then use the VaR threshold and expected excess loss to compute TVaR.

The probability of no claims is $0.9600\times0.9600=0.9216$. The probability of a claim on A only is $0.0400\times0.9600=0.0384$, and the probability of a claim on B only is also 0.0384. Thus, the probability of exactly one claim is $0.0384+0.0384=0.0768$. The probability of two claims is $0.0400\times0.0400=0.0016$. The probabilities sum to $0.9216+0.0768+0.0016=1.0000$.

Portfolio loss	Probability	Cumulative probability
$0	0.9216	0.9216
$1,000	0.0768	0.9984
$2,000	0.0016	1.0000

The cumulative probability at zero is below 0.9500, while the cumulative probability at $1,000 is 0.9984. Therefore, portfolio VaR is $1,000.00. Each standalone VaR remains zero, so the portfolio again violates VaR subadditivity.

For one line, the standalone TVaR remains $\$40/0.0500=\$800.00$. For the portfolio, $q=\$1{,}000$. Only the two-claim outcome exceeds $q$, giving $$ E[(X+Y-q)^{+}]=0.0016(\$2{,}000-\$1{,}000)=\$1.60. $$ Consequently, $$ \operatorname{TVaR}_{0.95}(X+Y)=\$1{,}000+\frac{\$1.60}{0.0500}=\$1{,}032.00. $$ This is less than the sum of standalone TVaRs, $\$800+\$800=\$1{,}600.00$. It can also be checked from the upper tail: the top 0.0500 probability consists of 0.0016 at $2,000, 0.0384 at $1,000, and 0.0100 at zero. Its loss total is $\$3.20+\$38.40+\$0=\$41.60$, and $\$41.60/0.0500=\$832.00$. Wait: this tail check includes only 0.0384 at $1,000 but the event probability at $1,000 is 0.0768; it must include another 0.0100 of that atom, not zero. The correct tail is 0.0016 at $2,000 and 0.0484 at $1,000, with total loss $\$3.20+\$48.40=\$51.60$. Dividing by 0.0500 gives $\$1{,}032.00$, confirming the formula calculation.

Check. The tail calculation uses exactly 0.0500 probability: all outcomes above $1,000 contribute 0.0016, and another $0.0500-0.0016=0.0484$ is taken from the $1,000 atom. Taking the whole atom would incorrectly average over more than the required tail probability.

Common Claim Events and the Meaning of Standalone Capital

A dependence change can also remove the VaR violation without changing either line’s marginal loss distribution. If the two claim events always occur together, the portfolio has no one-claim outcome. The following calculation contrasts this common-event case with the mutually exclusive example.

Worked Example: Common Claims and Equal TVaR Totals

Two hypothetical lines each pay $1,000 when a shared event occurs, which has probability 0.0400. Otherwise, both pay zero. At $\alpha=0.9500$, calculate the standalone and portfolio VaR and TVaR, and verify the portfolio distribution.

Method. Retain the same marginal distribution as in the first example, but now place the two line payments in the same scenario. Calculate the sum by scenario rather than adding marginal probabilities.

For each line, $P(X=0)=0.9600$, so its cumulative probability at zero exceeds 0.9500. Each standalone VaR is therefore $0.00. The portfolio is zero with probability 0.9600 and $2,000 with probability 0.0400. Since $F_{X+Y}(0)=0.9600$, portfolio VaR is also $0.00. The VaR comparison is $0\leq0+0$: there is no violation.

For each standalone line, the expected excess above its VaR of zero is $0.0400(\$1{,}000)=\$40$. Hence each standalone TVaR is $\$40/0.0500=\$800.00$, and their sum is $1,600.00. For the portfolio, $q=0$ and expected excess is $0.0400(\$2{,}000)=\$80$, so $$ \operatorname{TVaR}_{0.95}(X+Y)=\frac{\$80}{0.0500}=\$1{,}600.00. $$ Portfolio TVaR equals the sum of standalone TVaRs in this common-event case.

Check. The expected portfolio loss is $80.00, the sum of the two expected standalone losses. The shared event changes the tail distribution, not the expected loss. In particular, portfolio VaR is zero because the probability of no loss is 0.9600, even though a $2,000 loss occurs in the upper 0.0400 of outcomes.

Interpreting VaR and TVaR Capital Comparisons

For a loss-based risk measure, “standalone capital” may refer to the measure applied to each line separately, while “portfolio capital” applies it to the combined loss. Keep those quantities distinct from an unexpected-loss measure formed by subtracting expected loss. In the first example, the standalone expected losses are $40 each and the portfolio expected loss is $80. Subtracting expected loss from each VaR gives standalone values of $-\$40$ and a portfolio value of $\$920$. The violation remains: $\$920>-\$40-\$40$, because expected loss is additive. Such a centered number is not automatically a practical capital requirement; it simply illustrates that subtracting an additive expectation does not repair this VaR violation.

The numerical results also show why dependence analysis is essential. In the mutually exclusive case, combining the lines makes at least one $1,000 payment more likely than a claim on either line alone. In the independent case, the portfolio has a small probability of a $2,000 loss, which raises the tail average above its VaR. With a shared claim event, the total loss is larger when a claim occurs, but that event remains sufficiently rare for the 95th percentile to be zero. No single conclusion about diversification follows from the marginal distributions alone.

Key takeaway: Compare the portfolio loss distribution with the standalone distributions at the same confidence level. VaR can fail subadditivity in discrete portfolios because its quantile can jump across a probability atom. TVaR averages the fractional upper tail and is subadditive for integrable losses, making it a useful contrast when assessing portfolio risk.

Check the quantile boundary. Find the first loss amount whose cumulative probability reaches or exceeds $\alpha$. Do not interpolate across a discrete jump unless a different convention is explicitly specified.
Do not add line probabilities without checking dependence. Mutually exclusive, independent, and common claim events produce different aggregate distributions.
Use only the required tail probability for discrete TVaR. If the tail cuts through an atom, include only the needed fraction of that probability mass.
Compare like with like. Apply the same confidence level and the same loss valuation basis to portfolio and standalone measures.

Check Your Understanding

For each problem, use the left-quantile definition of VaR and the fractional-tail definition of TVaR. Amounts are hypothetical losses, and all confidence levels are 0.9500.

Two lines each pay $500 with probability 0.0300. Their events are mutually exclusive, so the aggregate loss is $500 with probability 0.0600 and zero otherwise. Find the standalone VaRs, portfolio VaR, sum of standalone TVaRs, and portfolio TVaR. (Answer: Standalone VaRs $0.00 each; portfolio VaR $500.00, so VaR subadditivity fails; standalone TVaRs $300.00 each, sum $600.00; portfolio TVaR $500.00.)
Two independent lines each pay $2,000 with probability 0.0200. Find portfolio VaR and portfolio TVaR, and compare with the sum of standalone TVaRs. (Answer: Portfolio VaR $0.00; portfolio TVaR $1,600.00; standalone TVaRs $800.00 each, sum $1,600.00.)
Two mutually exclusive lines each pay $3,000 with probability 0.0200. Find the standalone and portfolio VaRs and TVaRs. (Answer: All VaRs are $0.00; each standalone TVaR is $1,200.00; portfolio TVaR is $2,400.00, equal to the standalone sum.)
Two lines pay $1,500 each whenever the same event occurs, with probability 0.0400, and pay zero otherwise. Find portfolio VaR and TVaR, then compare with the standalone totals. (Answer: Portfolio VaR $0.00, equal to the standalone VaR sum; portfolio TVaR $2,400.00, equal to the sum of standalone TVaRs of $1,200.00 each.)

What You'll Learn