Common Errors in Applying Coherence Axioms

Apply the Axioms With a Consistent Sign Convention

Coherence calculations can go wrong before any probability is evaluated: the analyst may have defined the random variable as a loss in one line and as a profit in the next. Other errors arise from treating translation invariance as “risk is unchanged by cash,” or from using a conditional tail average at an atom without checking how much probability belongs in the tail. These are not cosmetic distinctions. Each can change a risk conclusion.

Throughout this tutorial, \(X\) denotes a random loss: larger values are worse. For \(0<\alpha<1\), define loss VaR using the lower quantile of the loss distribution. A risk measure \(\rho\) is coherent, under the usual loss convention, if it satisfies monotonicity, subadditivity, positive homogeneity, and translation invariance. These properties require attention to the sign and interpretation of the random variable being measured.

Definition: For a loss \(X\), \(\operatorname{VaR}_\alpha(X)=\inf\{x:F_X(x)\geq\alpha\}\). A risk measure \(\rho\) is monotone if \(X\leq Y\) implies \(\rho(X)\leq\rho(Y)\); subadditive if \(\rho(X+Y)\leq\rho(X)+\rho(Y)\); positively homogeneous if \(\rho(\lambda X)=\lambda\rho(X)\) for \(\lambda\geq0\); and translation invariant if \(\rho(X+c)=\rho(X)+c\) for a deterministic loss amount \(c\). Coherence is the conjunction of these four properties.

The convention is economically intuitive: adding a certain liability of \(c\) increases measured loss and therefore increases required risk by \(c\). If instead a variable \(\Pi\) records profit, the corresponding loss is \(-\Pi\). A convention that measures risk on profit directly will use signs that look different. Always identify the variable and convention before applying an axiom.

Worked Example: Converting Profit to Loss Before Measuring Risk

A hypothetical portfolio has profit \(\Pi=30\) with probability \(0.80\), and profit \(\Pi=-90\) with probability \(0.20\). Calculate its loss VaR at \(\alpha=0.90\). Then identify the result of incorrectly applying the same lower-quantile definition directly to profit.

Method. Define loss as \(X=-\Pi\). Thus \(X=-30\) with probability \(0.80\), and \(X=90\) with probability \(0.20\). The loss distribution has \(F_X(-30)=0.80\) and \(F_X(90)=1\). Since \(0.80<0.90\leq1\), \(\operatorname{VaR}_{0.90}(X)=90\).

For comparison, the profit distribution has \(F_\Pi(-90)=0.20\) and \(F_\Pi(30)=1\). Therefore, the lower \(0.90\)-quantile of profit is \(30\), not the loss risk. Reporting \(30\) as loss VaR would reverse the interpretation: the bad outcome is the loss of \(90\), which corresponds to profit \(-90\).

Check. At the stated loss quantile, \(P(X\leq90)=1\), while \(P(X<90)=0.80<0.90\). This confirms the generalized-quantile definition. The common mistake avoided is taking a quantile of a profit variable and labeling it as a quantile of loss.

Translation Invariance Is Not Invariance to Cash

Translation invariance describes how risk changes when a deterministic amount is added to the loss variable. If \(X\) is loss and \(c>0\), \(X+c\) is a larger loss by \(c\), so a coherent loss risk measure satisfies \(\rho(X+c)=\rho(X)+c\). If the company instead receives \(c\) in certain cash, its net loss becomes \(X-c\), and risk falls by \(c\): \(\rho(X-c)=\rho(X)-c\).

The quantile definition itself gives the VaR translation property. For any real \(x\), \(F_{X+c}(x)=P(X+c\leq x)=F_X(x-c)\). Thus the first point where \(F_{X+c}\) reaches \(\alpha\) is \(c\) above the first point where \(F_X\) reaches \(\alpha\). Consequently, \(\operatorname{VaR}_\alpha(X+c)=\operatorname{VaR}_\alpha(X)+c\), including when the distribution has atoms.

Worked Example: A Certain Cash Amount and a Certain Liability

A hypothetical insurer’s one-period loss \(X\) is \(0\) with probability \(0.70\) and \(50\) with probability \(0.30\). At \(\alpha=0.80\), calculate VaR before and after (a) receiving certain cash of \(12\), and (b) incurring a certain additional liability of \(12\). Use net loss as the risk variable.

Baseline. \(F_X(0)=0.70<0.80\) and \(F_X(50)=1\), so \(\operatorname{VaR}_{0.80}(X)=50\).

(a) Certain cash received. Net loss is \(X-12\), with outcomes \(-12\) and \(38\), still at probabilities \(0.70\) and \(0.30\). The cumulative probability at \(-12\) is \(0.70<0.80\), so \(\operatorname{VaR}_{0.80}(X-12)=38=50-12\).

(b) Additional liability. Net loss is \(X+12\), with outcomes \(12\) and \(62\). The cumulative probability at \(12\) is \(0.70<0.80\), so \(\operatorname{VaR}_{0.80}(X+12)=62=50+12\).

Check. Cash reduces the loss VaR by \(12\); a liability increases it by \(12\). The common mistake is to say that “cash invariance” means cash has no effect. Translation invariance is a signed change rule, not an assertion that risk is unchanged.

Atoms Require Care at the Quantile Boundary

A distribution has an atom at \(q\) when \(P(X=q)>0\). If the confidence level falls within that probability mass, VaR is still the boundary value \(q\), but a tail average must include only the fraction of the boundary atom needed to form a tail of probability \(1-\alpha\). Simply conditioning on \(X\geq q\) can include too much probability and produce the wrong average.

For integrable losses, define quantile-based TVaR by the quantile integral \(\operatorname{TVaR}_\alpha(X)=\frac{1}{1-\alpha}\int_\alpha^1\operatorname{VaR}_u(X)\,du\). If \(q=\operatorname{VaR}_\alpha(X)\) and \(F_X(q-)<\alpha\leq F_X(q)\), the equivalent boundary-atom calculation is

\operatorname{TVaR}_\alpha(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\bigl(F_X(q)-\alpha\bigr)}{1-\alpha}.

The term \(F_X(q)-\alpha\) is the portion of the atom at \(q\) included in the worst \(1-\alpha\) probability of outcomes. The rest of the numerator is the full expected contribution from losses strictly above \(q\).

Worked Example: TVaR When the Boundary Falls Inside an Atom

A hypothetical annual loss \(X\) is \(0\) with probability \(0.80\), \(10\) with probability \(0.15\), and \(100\) with probability \(0.05\). At \(\alpha=0.90\), calculate VaR and quantile-based TVaR. Also calculate the conditional mean \(E[X\mid X\geq\operatorname{VaR}_{0.90}(X)]\) to show why it is not the same quantity here.

Step 1: Find VaR. The cumulative probabilities are \(F_X(0)=0.80\), \(F_X(10)=0.95\), and \(F_X(100)=1\). Since \(0.80<0.90\leq0.95\), \(q=\operatorname{VaR}_{0.90}(X)=10\).

Step 2: Include the required part of the atom. Losses strictly greater than \(10\) contribute \(100(0.05)=5\). Since \(F_X(10)-\alpha=0.95-0.90=0.05\), the boundary contribution is \(10(0.05)=0.5\). The tail probability is \(1-\alpha=0.10\). Therefore,

\operatorname{TVaR}_{0.90}(X)=\frac{5+0.5}{0.10}=55.

Step 3: Compare the conditional mean. The event \(X\geq10\) has probability \(0.15+0.05=0.20\), and its expected loss contribution is \(10(0.15)+100(0.05)=1.5+5=6.5\). Hence \(E[X\mid X\geq10]=6.5/0.20=32.5\), not \(55\).

Check. The worst \(10\%\) of probability consists of all \(5\%\) at \(100\) and only \(5\%\) of the \(15\%\) mass at \(10\). Its total loss contribution is \(5.5\), so its average is \(5.5/0.10=55\). Conditioning on \(X\geq10\) incorrectly averages over \(20\%\) of probability. This is the atom-boundary trap.

Do Not Assume Every Quantile Is Subadditive

Subadditivity says that the risk of a combined loss is no greater than the sum of the separate risks. It formalizes one diversification property. It is not a consequence of taking a quantile, however: VaR can fail this axiom for discrete loss distributions. A calculation must use the joint distribution of the aggregate loss, not just the marginal VaRs.

Worked Example: A Discrete VaR Subadditivity Failure

Two hypothetical risks \(X\) and \(Y\) are independent. Each is a loss of \(100\) with probability \(0.04\), and \(0\) otherwise. At \(\alpha=0.95\), calculate each standalone VaR and the VaR of the aggregate. Determine whether subadditivity holds.

Standalone risks. For either risk, \(P(X=0)=0.96\), so \(F_X(0)=0.96\geq0.95\). Thus \(\operatorname{VaR}_{0.95}(X)=0\), and by the same calculation \(\operatorname{VaR}_{0.95}(Y)=0\). The sum of standalone VaRs is \(0\).

Aggregate distribution. Independence gives \(P(X+Y=0)=0.96^2=0.9216\). Exactly one loss occurs with probability \(2(0.04)(0.96)=0.0768\), giving aggregate loss \(100\). Both occur with probability \(0.04^2=0.0016\), giving aggregate loss \(200\). Therefore \(F_{X+Y}(0)=0.9216<0.95\), while \(F_{X+Y}(100)=0.9216+0.0768=0.9984\geq0.95\). Hence \(\operatorname{VaR}_{0.95}(X+Y)=100\).

The subadditivity test is \(100\leq0+0\), which is false. The common mistake is to infer aggregate VaR from standalone VaRs or to assume diversification guarantees the inequality for every risk measure. Here the aggregate has a positive 95th-percentile loss even though both standalone 95th-percentile losses are zero.

There is an important qualification for jointly normal losses. If \(X\) and \(Y\) are jointly normal with means \(\mu_X,\mu_Y\), standard deviations \(\sigma_X,\sigma_Y\), and finite variances, then

\operatorname{VaR}_\alpha(X)=\mu_X+z_\alpha\sigma_X, \qquad z_\alpha=\Phi^{-1}(\alpha).

The standard deviation of the sum satisfies \(\sigma_{X+Y}\leq\sigma_X+\sigma_Y\), since \(\operatorname{Cov}(X,Y)\leq\sigma_X\sigma_Y\). When \(\alpha\geq0.50\), \(z_\alpha\geq0\), so multiplying that standard-deviation inequality by \(z_\alpha\) preserves its direction. Consequently,

\operatorname{VaR}_\alpha(X+Y) =\mu_X+\mu_Y+z_\alpha\sigma_{X+Y} \leq \operatorname{VaR}_\alpha(X)+\operatorname{VaR}_\alpha(Y).

The confidence-level condition matters. For independent standard normal losses and \(\alpha=0.05\), \(z_{0.05}\approx-1.6449\). Each individual VaR is approximately \(-1.6449\), so their sum is approximately \(-3.2898\). The aggregate is normal with mean \(0\) and variance \(2\), giving \(\operatorname{VaR}_{0.05}(X+Y)\approx-1.6449\sqrt{2}=-2.3263\). Since \(-2.3263>-3.2898\), subadditivity fails at this confidence level. The correct statement is that VaR is subadditive for jointly normal losses when \(\alpha\geq0.50\), including the usual high confidence levels—not at every confidence level.

A Practical Axiom Check

Before accepting an axiom-based conclusion, use a short sequence of checks. It catches sign errors, boundary errors, and claims that rely on a condition that has not been established.

Fix the variable and direction.
State whether the random variable is a loss or profit. For a profit variable, convert to loss by changing sign if using the loss convention.

Write the relevant axiom in that convention.
For loss \(X\), adding deterministic loss \(c\) gives \(\rho(X+c)=\rho(X)+c\); certain cash received changes net loss to \(X-c\).

Inspect the quantile boundary.
Find \(F_X(q-)\) and \(F_X(q)\). If the confidence level lies within an atom, use the required fraction of that atom for quantile-based TVaR.

Test rather than presume subadditivity.
For VaR, calculate the aggregate quantile or verify a distributional condition that guarantees the inequality. Joint normality alone is not enough without the confidence-level qualification.

These checks are especially useful when a formula appears to produce an implausible sign, when a deterministic balance-sheet adjustment is involved, or when a diversification claim is based only on standalone risk estimates. They keep the axiom, the probability calculation, and the economic interpretation aligned. This treatment is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Key takeaway: Coherence axioms are applied to a specified loss variable, not to an ambiguous notion of “risk.” Translation invariance has a sign, atoms require boundary-mass treatment, and VaR subadditivity requires proof or appropriate conditions; it is not universal.

Check Your Understanding

For each problem, state the loss convention before calculating. Answers are given at the end of each item.

A portfolio profit is \(25\) with probability \(0.90\) and \(-75\) with probability \(0.10\). At \(\alpha=0.95\), find loss VaR. (Answer: \(75\))
A loss \(L\) is \(5\) with probability \(0.70\) and \(45\) with probability \(0.30\). At \(\alpha=0.80\), find VaR after certain cash of \(8\) is received, then after an additional liability of \(8\) is applied to the original loss. (Answer: cash-adjusted VaR \(37\); liability-adjusted VaR \(53\))
A loss is \(0\) with probability \(0.75\), \(20\) with probability \(0.20\), and \(80\) with probability \(0.05\). At \(\alpha=0.90\), find VaR and quantile-based TVaR. (Answer: VaR \(20\); TVaR \(50\))
Independent losses \(A\) and \(B\) are each \(100\) with probability \(0.03\), otherwise \(0\). At \(\alpha=0.95\), find the standalone VaRs and aggregate VaR, and state whether subadditivity holds. (Answer: standalone VaRs \(0\) and \(0\); aggregate VaR \(100\); subadditivity fails)
Let \(X,Y\) be independent standard normal losses. At \(\alpha=0.05\), use \(z_{0.05}\approx-1.6449\) to compare \(\operatorname{VaR}_\alpha(X+Y)\) with \(\operatorname{VaR}_\alpha(X)+\operatorname{VaR}_\alpha(Y)\). (Answer: aggregate VaR \(\approx-2.3263\), sum of standalone VaRs \(\approx-3.2898\); VaR is not subadditive here)

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