Coherent Risk Measure Axioms

Why Coherence Matters

A risk measure assigns a number to a loss distribution, often to describe the capital needed for risk. But a number alone does not tell us whether the measure responds sensibly to a change in the loss scale, a guaranteed additional payment, or the combination of two losses. Coherent risk measure axioms make these expectations precise.

Throughout, $X$ and $Y$ are losses, so larger values mean worse outcomes. A risk measure $\rho$ maps a loss to a real-valued risk amount. Unless otherwise specified, assume losses are integrable and that constants are measured in the same units as losses. The four axioms are properties a measure must satisfy for every eligible pair of losses, not just for one example.

Definition: A risk measure $\rho$ is coherent if it satisfies all four properties below:

Translation invariance: $\rho(X+c)=\rho(X)+c$ for every constant $c$. Adding a certain loss of $c$ increases measured risk by exactly $c$.
Positive homogeneity: $\rho(aX)=a\rho(X)$ for every $a\geq 0$. Scaling all losses by a nonnegative factor scales risk by that factor.
Monotonicity: if $X\leq Y$ with probability 1, then $\rho(X)\leq\rho(Y)$. A loss that is never larger should not have greater risk.
Subadditivity: $\rho(X+Y)\leq\rho(X)+\rho(Y)$. Measuring the combined loss should not exceed the sum of the separate risk amounts.

These properties are distinct. For instance, a measure can respond properly to adding a certain loss and still fail subadditivity. To establish coherence, all four properties must hold on the stated domain.

Variance Is Not a Coherent Risk Measure

Variance describes spread around the mean, rather than the dollar amount of loss. Its formula is $\operatorname{Var}(X)=E[(X-E[X])^2]$. Variance is unchanged by adding a constant, while translation invariance requires the risk measure to increase by that constant. It also scales quadratically: $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$, not $a\operatorname{Var}(X)$ in general.

Worked Example: Testing Variance Under a Shift and a Scale Change

A hypothetical claim loss $X$ is $0$ or $\$2$, each with probability $0.5$. Compare its variance with the variance after adding a certain $\$5$ loss and after multiplying every loss by 3. Use dollars throughout.

Method. Calculate the mean and variance from the two outcomes, then compare the transformed variances with the corresponding axioms.

First, $E[X]=0(0.5)+2(0.5)=1$. Therefore, $\operatorname{Var}(X)=0.5(0-1)^2+0.5(2-1)^2=0.5+0.5=1$ dollar squared.

For $X+5$, the outcomes are 5 and 7, each with probability 0.5. Its mean is $0.5(5)+0.5(7)=6$, so $\operatorname{Var}(X+5)=0.5(5-6)^2+0.5(7-6)^2=1$. Translation invariance would require $\operatorname{Var}(X+5)=\operatorname{Var}(X)+5=6$, so it fails.

For $3X$, the outcomes are 0 and 6, and the mean is 3. Thus, $\operatorname{Var}(3X)=0.5(0-3)^2+0.5(6-3)^2=9$. Positive homogeneity would require $3\operatorname{Var}(X)=3$, not 9.

Check. The calculations agree with $\operatorname{Var}(X+c)=\operatorname{Var}(X)$ and $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. Variance fails translation invariance and positive homogeneity, so it is not coherent.

Common mistake avoided: Do not treat the invariance of variance under a shift as translation invariance. Translation invariance requires the risk amount itself to increase by $c$.

Variance also fails monotonicity in general. On the same two equally likely outcomes, let $X$ be 0 or 1, and let $Y=1$ in both outcomes. Then $X\leq Y$ always, but $\operatorname{Var}(X)=0.25$ and $\operatorname{Var}(Y)=0$. It is not a general subadditive measure either: if $X=Y$ is 0 or 2 with equal probability, then $\operatorname{Var}(X)=1$ and $\operatorname{Var}(X+Y)=\operatorname{Var}(2X)=4>2$. A property that happens to hold for some pairs of losses does not make variance coherent.

Value-at-Risk: Three Axioms, but Not Always Four

Use the left-quantile convention for value-at-risk at level $\alpha$, where $0<\alpha<1$:

Definition: For loss $X$ with cumulative distribution function $F_X$, $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}$. This convention applies to both continuous and discrete losses; at a probability mass, the quantile is the smallest loss value whose cumulative probability reaches $\alpha$.

Under this convention, VaR satisfies translation invariance, positive homogeneity, and monotonicity. For translation, shifting every outcome by $c$ shifts every quantile by $c$. For $a>0$, multiplying every outcome by $a$ multiplies the quantile by $a$; at $a=0$, the scaled loss is identically zero and its VaR is zero. If $X\leq Y$ always, then $F_X(t)\geq F_Y(t)$ for every $t$, so the $\alpha$-quantile of $X$ cannot exceed that of $Y$.

The fourth axiom, subadditivity, does not hold for VaR in general. A discrete example makes the issue explicit.

Worked Example: A Discrete Failure of VaR Subadditivity

Two hypothetical policies have losses $X$ and $Y$. With probability 0.06, policy 1 has a loss of $\$100$ and policy 2 has no loss; with probability 0.06, policy 2 has a loss of $\$100$ and policy 1 has no loss. With probability 0.88, neither has a loss. Find each policy’s VaR and the VaR of their combined loss at $\alpha=0.90$.

Method. Apply the left-quantile definition to each discrete distribution, checking the cumulative probability at each loss value.

For $X$, the probability of zero loss is $0.94$. Since $F_X(0)=0.94\geq0.90$, $\operatorname{VaR}_{0.90}(X)=0$. The same calculation gives $\operatorname{VaR}_{0.90}(Y)=0$.

The combined loss $X+Y$ is 0 with probability 0.88 and 100 with probability $0.12$. Thus $F_{X+Y}(0)=0.88<0.90$, while $F_{X+Y}(100)=1\geq0.90$. Therefore, $\operatorname{VaR}_{0.90}(X+Y)=100$.

The subadditivity comparison is $\operatorname{VaR}_{0.90}(X+Y)=100>0+0=\operatorname{VaR}_{0.90}(X)+\operatorname{VaR}_{0.90}(Y)$. VaR fails subadditivity for this pair, so VaR is not coherent in general.

Check. The combined loss has exactly two possible values, 0 and 100, with probabilities summing to $0.88+0.12=1$.

Common mistake avoided: Do not add the two separate VaRs and assume the result is the VaR of the sum. The quantile of a sum depends on the joint outcomes, not only on the two marginal quantiles.

Tail Value-at-Risk and Coherence

For integrable losses, define tail value-at-risk using the integral of the left quantile:

Definition: For $0<\alpha<1$, $\operatorname{TVaR}_{\alpha}(X)=\dfrac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du$. This definition includes a fractional contribution from probability mass at the quantile when the integration interval cuts through that mass.

Unlike VaR, this definition averages quantiles across the upper tail. That averaging is important for coherence. For a discrete loss, one must not simply average all outcomes that are strictly greater than VaR: the upper-tail probability may include only part of the probability mass at VaR.

Worked Example: Discrete TVaR with Fractional Quantile Mass

A hypothetical equipment loss $X$ is $\$0$ with probability 0.80, $\$10$ with probability 0.15, and $\$30$ with probability 0.05. Calculate $\operatorname{TVaR}_{0.90}(X)$, then calculate it after adding a certain $\$7$ loss and after scaling the original loss by 2.

Method. Use the quantile integral, interpreting its interval as the upper 10% of probability.

The upper 10% consists of all of the 5% mass at 30 and 5% of the 15% mass at 10. The quantile integral over $[0.90,1]$ is therefore $0.05(30)+0.05(10)=1.5+0.5=2$. Dividing by $1-\alpha=0.10$ gives $\operatorname{TVaR}_{0.90}(X)=2/0.10=\$20$.

After adding 7, the outcomes become 7, 17, and 37 with the same probabilities. The upper 10% integral is $0.05(37)+0.05(17)=1.85+0.85=2.70$; hence $\operatorname{TVaR}_{0.90}(X+7)=2.70/0.10=\$27$.

After scaling the original loss by 2, the upper outcomes are 60 and 20. The upper 10% integral is $0.05(60)+0.05(20)=3+1=4$; dividing by 0.10 gives $\operatorname{TVaR}_{0.90}(2X)=\$40$.

Check. The results are $20+7=27$ and $2(20)=40$, as translation invariance and positive homogeneity require. The fractional mass is 5% at 10, not the full 15%.

Common mistake avoided: Averaging only the outcomes strictly greater than VaR would use just the 5% mass at 30 and incorrectly report 30. The upper 10% also uses 5% of the mass at 10.

Why TVaR Is Subadditive

A useful justification comes from a representation of TVaR as a supremum of weighted expected losses. For $0<\alpha<1$, let $Z$ range over nonnegative random variables satisfying $E[Z]=1$ and $Z\leq 1/(1-\alpha)$. For integrable $X$,

\operatorname{TVaR}_{\alpha}(X) = \sup_{\substack{0\leq Z\leq 1/(1-\alpha)\\E[Z]=1}} E[ZX].

To see why, the weight $Z$ can assign at most $1/(1-\alpha)$ times the usual weight to any outcome, and its mean must remain 1. The largest weighted expectation is obtained by assigning the greatest permitted weight to the largest losses, zero weight to sufficiently small losses, and an intermediate weight at the quantile if needed to meet $E[Z]=1$. This selects precisely the upper fraction $1-\alpha$ of the loss distribution, counting only the required fraction of any probability mass at the boundary. The resulting weighted expectation is the integral-quantile definition of TVaR.

For any admissible $Z$, $E[Z(X+Y)]=E[ZX]+E[ZY]$. Taking a supremum over the same set of admissible weights gives

\operatorname{TVaR}_{\alpha}(X+Y) \leq \sup_Z E[ZX]+\sup_Z E[ZY] = \operatorname{TVaR}_{\alpha}(X)+\operatorname{TVaR}_{\alpha}(Y).

The inequality follows because the supremum of a sum over one common set cannot exceed the sum of the two separate suprema. The same representation also explains the other axioms: adding $c$ adds $cE[Z]=c$, scaling a loss by $a\geq0$ scales every weighted expectation by $a$, and $X\leq Y$ implies $E[ZX]\leq E[ZY]$ for every admissible nonnegative $Z$. Thus TVaR is coherent on the stated domain.

Exam Checks and Common Traps

When asked whether a measure is coherent, test the universal statements in the definition rather than relying on intuition about what a measure is intended to capture. A single valid counterexample disproves an axiom. Conversely, calculations on one convenient distribution do not prove that an axiom holds for every loss pair.

Keep the loss convention fixed. Here larger $X$ means a worse loss, so monotonicity is $\rho(X)\leq\rho(Y)$ when $X\leq Y$.
Separate scaling laws. Standard deviation scales linearly for nonnegative factors, but variance scales quadratically; neither fact makes variance translation invariant.
Check atoms carefully. With discrete VaR or TVaR, evaluate cumulative probabilities and account for the required fraction of probability mass at a quantile.
Do not infer subadditivity from the other axioms. VaR satisfies three coherence axioms under the stated convention but can fail the fourth.
State the domain and level. The TVaR result here is for integrable losses and a level $0<\alpha<1$; the quantile convention is the left quantile.

Key takeaway: Variance is not coherent: it fails translation invariance and positive homogeneity, among other possible failures. VaR satisfies translation invariance, positive homogeneity, and monotonicity, but not subadditivity in general. Integral-quantile TVaR is coherent for integrable losses under the stated convention.

Check Your Understanding

For each question, use the loss-positive convention and the definitions stated above. Treat all unspecified outcomes as having probability zero.

A loss $X$ is 2 or 6 with equal probability. What are $\operatorname{Var}(X)$ and $\operatorname{Var}(X+4)$, and which axiom does the comparison test? (Answer: $\operatorname{Var}(X)=4$, $\operatorname{Var}(X+4)=4$; translation invariance fails because it would require 8.)
A discrete loss is 0 with probability 0.92 and 50 with probability 0.08. Find its $\operatorname{VaR}_{0.90}$. (Answer: 0.)
Two losses are mutually exclusive: each is 80 with probability 0.07, while both are zero with probability 0.86. Find their individual and combined $\operatorname{VaR}_{0.90}$. (Answer: Each individual VaR is 0; the combined VaR is 80.)
A loss is 0 with probability 0.70, 12 with probability 0.20, and 32 with probability 0.10. Find $\operatorname{TVaR}_{0.80}$. (Answer: 22; the upper 20% contains all of the 10% mass at 32 and 10% mass at 12.)
For the loss in question 4, calculate $\operatorname{TVaR}_{0.80}(3X+5)$. (Answer: 71; translation and positive homogeneity give $3(22)+5=71$.)

What You'll Learn