Why Loss Ordering Matters
Suppose two hypothetical insurance portfolios face the same underlying scenarios, and one portfolio pays no more than the other in any scenario. A risk measure that assigns greater risk to the less costly portfolio would contradict that outcome-by-outcome comparison. Monotonicity formalizes this basic consistency requirement. It applies to Value at Risk (VaR) and Tail Value at Risk (TVaR), but the comparison must be made using the same loss convention and probability level.
The key condition is stronger than saying one loss has a smaller mean: the losses must be ordered in every outcome, except possibly on an event of probability zero. A related distributional condition lets us compare losses even when no particular pairing of outcomes has been specified. The distinction is useful when portfolio scenarios are given directly in a table in one problem and only marginal distributions are supplied in another.
Quantiles and the Monotonicity Result
Use the lower-quantile convention. For a loss \(X\) with cumulative distribution function \(F_X(x)=P(X\le x)\), its quantile at probability \(u\in(0,1)\) is \(Q_X(u)=\inf\{x:F_X(x)\ge u\}\). At confidence level \(p\), \(\operatorname{VaR}_p(X)=Q_X(p)\). For losses with a finite tail mean, TVaR is the average of the quantiles above level \(p\):
This quantile-integral definition also handles atoms. If the distribution has a large probability at VaR, the integral includes just the fraction of that probability needed to fill the upper tail of size \(1-p\). It is not always correct to calculate TVaR as \(E[X\mid X\ge\operatorname{VaR}_p(X)]\).
The same conclusion holds if the losses are ordered by their distributions: \(F_X(x)\ge F_Y(x)\) for every real \(x\). This condition says that the lower-loss distribution \(X\) has at least as much probability at or below each threshold. In particular, an outcome-wise ordering implies this distributional ordering: if \(X\le Y\), then the event \(Y\le x\) is contained, up to probability-zero outcomes, in the event \(X\le x\). Therefore \(P(X\le x)\ge P(Y\le x)\).
Here is why the quantile comparison follows. If \(F_X(x)\ge F_Y(x)\) for every \(x\), then whenever \(F_Y(x)\ge u\), we also have \(F_X(x)\ge u\). Thus the set of thresholds that reach probability \(u\) for \(Y\) is contained in the corresponding set for \(X\), giving \(Q_X(u)\le Q_Y(u)\). Setting \(u=p\) proves the VaR inequality. Integrating the quantile inequality over \(p\le u\le1\), then dividing by the positive quantity \(1-p\), proves the TVaR inequality. This argument relies on the quantile ordering, not on a comparison of the means.
Worked Examples
Worked Example: Ordered Outcomes With a VaR Atom
A hypothetical insurer evaluates four equally likely scenarios for two coverage designs. The losses, in dollars, are \(X=(0,100,100,300)\) for Design A and \(Y=(20,120,180,400)\) for Design B, with entries paired by scenario. Calculate VaR and TVaR at \(p=0.75\), and check monotonicity.
Method. First compare the paired outcomes: \(0\le20\), \(100\le120\), \(100\le180\), and \(300\le400\). Thus \(X\le Y\) in every scenario. Each scenario has probability \(0.25\).
VaR. For \(X\), the cumulative probabilities at losses \(0,100,300\) are \(0.25,0.75,1\), so \(\operatorname{VaR}_{0.75}(X)=100\). For \(Y\), the cumulative probabilities at \(20,120,180,400\) are \(0.25,0.50,0.75,1\), so \(\operatorname{VaR}_{0.75}(Y)=180\). Therefore \(100\le180\).
TVaR. The upper tail above the 75th percentile has probability \(1-p=0.25\), exactly the probability of the largest scenario. Its loss is \(300\) for \(X\) and \(400\) for \(Y\). Hence \(\operatorname{TVaR}_{0.75}(X)=300\) and \(\operatorname{TVaR}_{0.75}(Y)=400\).
Check and trap. Both risk measures preserve the scenario ordering. A tempting but wrong calculation for \(X\) is \(E[X\mid X\ge100]=(100+100+300)/3=166.67\). That conditions on all three scenarios at or above VaR, even though the required upper tail has probability only \(0.25\). The quantile-integral definition takes only the top scenario here, giving TVaR \(300\).
Worked Example: Dominance From Two Discrete Distributions
Two hypothetical claim portfolios have the following loss distributions. Portfolio A has losses \(0,40,100\) with probabilities \(0.5,0.3,0.2\). Portfolio B has losses \(10,50,120\) with probabilities \(0.2,0.4,0.4\). Compare their VaR and TVaR at \(p=0.70\), and establish which distribution is larger in the quantile ordering.
Method. The quantile of A is \(0\) for \(0<u\le0.5\), \(40\) for \(0.5<u\le0.8\), and \(100\) for \(0.8<u<1\). The quantile of B is \(10\) for \(0<u\le0.2\), \(50\) for \(0.2<u\le0.6\), and \(120\) for \(0.6<u<1\). Comparing these intervals shows \(Q_A(u)\le Q_B(u)\) throughout: on the successive intervals cut by \(0.2,0.5,0.6,0.8\), the values are respectively \(0\le10\), \(0\le50\), \(40\le50\), \(40\le120\), and \(100\le120\).
VaR. Portfolio A has cumulative probability \(0.5\) at 0 and \(0.8\) at 40, so \(\operatorname{VaR}_{0.70}(A)=40\). Portfolio B has cumulative probability \(0.6\) at 50 and \(1\) at 120, so \(\operatorname{VaR}_{0.70}(B)=120\).
TVaR. The upper tail has probability \(0.30\). For A, it contains all \(0.20\) probability at loss 100 and \(0.10\) probability at loss 40. The tail loss total is \(0.20(100)+0.10(40)=20+4=24\), giving \(\operatorname{TVaR}_{0.70}(A)=24/0.30=80\). For B, the upper \(0.30\) probability lies within the \(0.40\) probability at loss 120, so \(\operatorname{TVaR}_{0.70}(B)=120\).
Check and trap. The results satisfy \(40\le120\) for VaR and \(80\le120\) for TVaR. A common error is to include all \(0.20\) probability at 100 and all \(0.30\) probability at 40 in A’s tail. That totals \(0.50\), not the required \(0.30\); only \(0.10\) probability at 40 belongs in the tail.
Worked Example: Comparing Losses Built From a Common Uniform Scenario
Let \(U\) be uniformly distributed on \((0,1)\). A hypothetical contract has loss \(X=100U\). A revised contract has loss \(Y=100U\) when \(U<0.5\), and \(Y=100U+30\) when \(U\ge0.5\). Find both risk measures at \(p=0.75\) and verify the ordering.
Method and VaR. For every value of \(U\), \(Y\ge X\): they are equal below \(0.5\), and \(Y\) exceeds \(X\) by 30 at and above \(0.5\). Since \(X\) is uniform from 0 to 100, \(Q_X(u)=100u\). For \(u>0.5\), the upper half of \(Y\)'s distribution corresponds to \(U>0.5\), where \(Y=100U+30\); hence \(Q_Y(u)=100u+30\). In particular, \(\operatorname{VaR}_{0.75}(X)=100(0.75)=75\), while \(\operatorname{VaR}_{0.75}(Y)=100(0.75)+30=105\).
TVaR. The average value of \(u\) from \(0.75\) to 1 is \((0.75+1)/2=0.875\). Therefore \(\operatorname{TVaR}_{0.75}(X)=100(0.875)=87.5\). Over that same quantile interval, \(Q_Y(u)=100u+30\), so \(\operatorname{TVaR}_{0.75}(Y)=100(0.875)+30=117.5\).
Check and trap. The revised loss is never lower, and both results are larger for \(Y\). The upper-tail interval lies entirely above \(u=0.5\), so the relevant quantile formula for \(Y\) is \(100u+30\), not the lower-half formula \(100u\).
What Monotonicity Does—and Does Not—Say
Monotonicity is an ordering rule for losses. It does not say that a comparison of expected losses alone determines which portfolio has greater VaR or TVaR. Nor does it compare two differently defined quantities, such as losses in dollars for one portfolio and losses in thousands of dollars for another. Before comparing risk measures, identify the loss variables, align the probability level \(p\), and check that the units agree.
When outcomes are paired, the direct test is to check \(X\le Y\) in each scenario. When only distributions are given, compare their cumulative probabilities or quantiles. If \(F_X(x)\ge F_Y(x)\) at every threshold, the quantile argument establishes the VaR and TVaR ordering. A few selected thresholds may suggest dominance, but do not establish it over the entire distribution.
Atoms require particular care. VaR uses the stated quantile convention, while TVaR averages exactly the upper fraction \(1-p\) of probability mass. If that fraction cuts through an atom, only the needed portion of the atom is included. This prevents an apparent exception to monotonicity caused by mistakenly conditioning on every outcome tied at VaR.
Check Your Understanding
For each problem, compare the losses or quantiles first, then calculate the requested risk measures.
- Three equally likely scenarios give losses \(X=(2,8,15)\) and \(Y=(5,8,20)\), paired in order. At \(p=2/3\), find both VaRs and TVaRs. (Answer: VaR values are 8 and 8; TVaR values are 15 and 20, respectively.)
- Portfolio A has losses \(0,30,90\) with probabilities \(0.4,0.4,0.2\). Portfolio B has losses \(10,40,100\) with probabilities \(0.2,0.5,0.3\). Compare their quantiles and calculate VaR and TVaR at \(p=0.6\). (Answer: \(Q_A(u)\le Q_B(u)\); VaR values are 30 and 40; TVaR values are 60 and 85, respectively.)
- Let \(U\) be uniform on \((0,1)\), \(X=50U\), and \(Y=50U\) for \(U\le0.8\), while \(Y=50U+25\) for \(U>0.8\). At \(p=0.9\), calculate VaR and TVaR for each loss. (Answer: VaR values are 45 and 70; TVaR values are 47.5 and 72.5, respectively.)
- Loss A takes values \(0,20,60\) with probabilities \(0.25,0.50,0.25\). Loss B takes values \(5,30,80\) with probabilities \(0.25,0.25,0.50\). At \(p=0.6\), check the quantile ordering and calculate both risk measures. (Answer: \(Q_A(u)\le Q_B(u)\); VaR values are 20 and 80; TVaR values are 45 and 80, respectively.)
- Suppose two losses satisfy \(X\le Y\) almost surely and \(F_X(x)\ge F_Y(x)\) for all \(x\). A proposed conclusion is that their TVaRs must be ordered at \(p=0.95\), even if they have atoms. State whether it follows and identify the definition that justifies the conclusion. (Answer: Yes, \(\operatorname{TVaR}_{0.95}(X)\le\operatorname{TVaR}_{0.95}(Y)\); the quantile-integral definition applies with fractional tail mass at atoms.)