Proving TVaR Coherence with Quantile Integrals

Why a Dual Representation Helps

TVaR summarizes the average loss in the upper tail of a loss distribution. Its quantile definition makes the tail interpretation direct, but proving that TVaR satisfies the coherence axioms is less immediate from the integral alone. A dual representation provides a concise route: it expresses TVaR as the largest expected loss under a class of bounded probability weights.

Throughout, $X$ denotes a loss with $E[|X|]<\infty$, and $\alpha$ is a fixed confidence level with $0<\alpha<1$. Losses are ordered in the usual way: $X\le Y$ almost surely means that $X$ is no larger than $Y$ in every outcome except possibly on an event of probability zero.

Definition: The quantile of $X$ at probability level $u$, for $0<u<1$, is $q_u(X)=\inf\{x:F_X(x)\ge u\}$. The Value at Risk at level $u$ is this quantile. Define TVaR at level $\alpha$ by the upper quantile integral $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}q_u(X)\,du. $$ Thus TVaR averages the quantiles from level $\alpha$ through the upper end of the distribution, rather than using only the single quantile at $\alpha$.

For the dual representation, consider random weights $Z$ on the same probability space as the losses. The weights are nonnegative, have expected value one, and are bounded above by $1/(1-\alpha)$. They can increase the importance of outcomes in the tail, but cannot assign an arbitrarily large weight to any one outcome.

Theorem: For integrable $X$ and $0<\alpha<1$, $$ \operatorname{TVaR}_{\alpha}(X) =\sup_{Z\in\mathcal{Z}_{\alpha}}E[XZ], \qquad \mathcal{Z}_{\alpha} =\left\{Z:0\le Z\le\frac{1}{1-\alpha}\ \text{almost surely},\ E[Z]=1\right\}. $$ The supremum is taken over admissible weights on the underlying probability space. The representation applies to losses, with larger losses receiving weakly greater importance in a maximizing choice of weights.

Why the Dual Representation Equals the Quantile Integral

A short derivation establishes the key result. Let $q=q_{\alpha}(X)$. The quantile definition gives $P(X>q)\le 1-\alpha\le P(X\ge q)$. For any admissible $Z$, write $X=q+(X-q)$. Since $E[Z]=1$, and $Z$ is nonnegative and bounded by $1/(1-\alpha)$,

E[XZ] =q+E[(X-q)Z] \le q+\frac{E[(X-q)^+]}{1-\alpha},

where $(X-q)^+=\max(X-q,0)$. The inequality follows because the contribution from $X<q$ is nonpositive, while the weight on $X>q$ cannot exceed $1/(1-\alpha)$.

The expression on the right is exactly the upper quantile integral. To see why, quantiles from $\alpha$ through $F_X(q)$ equal $q$; above $F_X(q)$, the quantiles account for the excess above $q$. Consequently,

\int_{\alpha}^{1}q_u(X)\,du =q(1-\alpha)+E[(X-q)^+].

The upper bound is attainable. Assign weight $1/(1-\alpha)$ to outcomes with $X>q$, zero weight to outcomes with $X<q$, and, if there is probability at $X=q$, assign an intermediate weight at $X=q$ so that $E[Z]=1$. The quantile inequalities ensure that such a weight is feasible. Thus the supremum equals the upper bound, which proves the dual representation. This fractional weighting at the threshold also makes the argument valid when the loss distribution has an atom at $q$.

Worked Applications

Worked Example: Computing a Discrete TVaR and Its Dual Weights

A hypothetical insurer models a one-year loss $X$, in thousands of dollars, as 0 with probability 0.50, 40 with probability 0.30, and 200 with probability 0.20. Calculate $\operatorname{TVaR}_{0.75}(X)$ from the quantile integral, then verify it using admissible dual weights.

Method. The upper tail has probability $1-\alpha=0.25$. The quantile is 40 from probability level 0.50 through 0.80, and 200 from 0.80 through 1. The part of the upper tail from 0.75 to 0.80 has width 0.05, while the part from 0.80 to 1 has width 0.20.

Quantile-integral calculation. The integrated loss over those two intervals is $0.05(40)=2$ and $0.20(200)=40$, respectively. Their total is $2+40=42$. Divide by the tail probability 0.25:

\operatorname{TVaR}_{0.75}(X)=\frac{42}{0.25}=168 \quad\text{thousand dollars}.

Dual check. The maximum allowed weight is $1/0.25=4$. Set $Z=4$ when $X=200$, $Z=2/3$ when $X=40$, and $Z=0$ when $X=0$. Then $E[Z]=0.20(4)+0.30(2/3)+0.50(0)=0.80+0.20=1$, so the weights are admissible. Their weighted expected loss is $0.20(200)(4)+0.30(40)(2/3)=160+8=168$ thousand dollars, confirming the result.

Common mistake. Averaging the two distinct positive loss amounts equally gives $(40+200)/2=120$, but that treats the two amounts as if they contributed equally to the upper tail. Only 0.05 probability at 40 is included in the 0.25 tail; the correct tail-weighted calculation is $[0.05(40)+0.20(200)]/0.25=168$.

Worked Example: Using Subadditivity for a Two-Risk Portfolio

A hypothetical portfolio has four equally likely scenarios. Losses $X$ from a property line are $0,0,100,100$ thousand dollars, and losses $Y$ from a separate equipment line are $0,100,0,100$ thousand dollars, in that scenario order. At $\alpha=0.50$, calculate the TVaR of each line and of their combined loss.

Method. At this confidence level the upper tail has probability 0.50, or two of the four equally likely scenarios. For $X$, the two largest losses are 100 and 100, so $\operatorname{TVaR}_{0.50}(X)=(100+100)/2=100$ thousand dollars. The two largest losses for $Y$ are also 100 and 100, giving $\operatorname{TVaR}_{0.50}(Y)=100$ thousand dollars.

Add the two line losses scenario by scenario. The combined losses are $0,100,100,200$ thousand dollars. The two largest are 100 and 200, so

\operatorname{TVaR}_{0.50}(X+Y) =\frac{100+200}{2} =150\ \text{thousand dollars} \le 100+100 =\operatorname{TVaR}_{0.50}(X)+\operatorname{TVaR}_{0.50}(Y).

The combined risk measure is strictly less than the sum of the stand-alone measures. This example also shows why one must add losses by scenario before calculating portfolio TVaR: the portfolio tail is determined by the combined outcomes, not by adding the two separate tail averages.

Worked Example: Checking Scaling and a Certain Loss Addition

A hypothetical loss $X$, in thousands of dollars, is 10 with probability 0.50, 30 with probability 0.30, and 90 with probability 0.20. At $\alpha=0.75$, calculate TVaR for $X$, then for a transformed loss $W=1.5X+12$, where 12 thousand dollars is a certain additional loss.

Method. The 0.25 upper tail contains 0.20 probability at 90 and 0.05 probability at 30. The tail-weighted total for $X$ is $0.20(90)+0.05(30)=18+1.5=19.5$ thousand dollars. Therefore, $\operatorname{TVaR}_{0.75}(X)=19.5/0.25=78$ thousand dollars.

Apply the positive scaling and certain addition to each outcome. The transformed losses are 27, 57, and 147 thousand dollars, with the same probabilities. Its upper-tail total is $0.20(147)+0.05(57)=29.4+2.85=32.25$ thousand dollars, giving

\operatorname{TVaR}_{0.75}(W)=\frac{32.25}{0.25}=129 \quad\text{thousand dollars}.

As a check, scaling and adding the certain loss to the original TVaR gives $1.5(78)+12=117+12=129$ thousand dollars. The check agrees with the direct tail calculation.

Deriving the Four Coherence Properties

A risk measure on losses is coherent when it satisfies monotonicity, translation invariance, positive homogeneity, and subadditivity. The dual representation proves all four for TVaR. For two losses $X$ and $Y$, use the same admissible set $\mathcal{Z}_{\alpha}$ on their common probability space.

Monotonicity.
If $X\le Y$ almost surely, then $XZ\le YZ$ for every admissible $Z$, because $Z\ge0$. Taking expectations and then suprema gives $\operatorname{TVaR}_{\alpha}(X)\le\operatorname{TVaR}_{\alpha}(Y)$.

Translation invariance.
For a deterministic amount $c$, $E[(X+c)Z]=E[XZ]+cE[Z]=E[XZ]+c$. The admissible set is unchanged, so $\operatorname{TVaR}_{\alpha}(X+c)=\operatorname{TVaR}_{\alpha}(X)+c$.

Positive homogeneity.
For $\lambda\ge0$, $\sup_{Z\in\mathcal{Z}_{\alpha}}E[\lambda XZ]=\lambda\sup_{Z\in\mathcal{Z}_{\alpha}}E[XZ]$. Hence $\operatorname{TVaR}_{\alpha}(\lambda X)=\lambda\operatorname{TVaR}_{\alpha}(X)$. For $\lambda=0$, both sides are zero.

Subadditivity.
For each admissible $Z$, $E[(X+Y)Z]=E[XZ]+E[YZ]\le\sup_{Q\in\mathcal{Z}_{\alpha}}E[XQ]+\sup_{R\in\mathcal{Z}_{\alpha}}E[YR]$. Taking the supremum over $Z$ proves $\operatorname{TVaR}_{\alpha}(X+Y)\le\operatorname{TVaR}_{\alpha}(X)+\operatorname{TVaR}_{\alpha}(Y)$.

In the subadditivity step, the maximizing weight for $X+Y$ need not maximize either $X$ or $Y$ separately. The inequality still holds because, for every single admissible weight, each term is bounded by its own supremum. No independence assumption is needed: the proof works for any joint dependence between the losses, provided they are defined on a common probability space and are integrable.

Exam Traps and Practical Interpretation

The dual representation describes TVaR as a worst weighted expectation, but the weights are constrained: they must average one and cannot exceed $1/(1-\alpha)$. Dropping either condition changes the optimization problem and no longer gives this TVaR representation. The bound also shows that a more extreme confidence level permits larger emphasis on tail outcomes.

Confusing quantile and tail average. $q_\alpha(X)$ is a single threshold; TVaR averages quantiles over the entire upper interval.
Using equal weights for distinct tail outcomes. The tail integral weights loss levels according to how much probability they contribute to the tail, including only the needed fraction at a cutoff.
Adding stand-alone TVaRs to obtain portfolio TVaR. Subadditivity gives an upper bound, not an equality in general. Aggregate the scenario losses first when the joint outcomes are known.
Assuming independence is required. The coherence proof uses neither independence nor a particular dependence model.

For actuarial risk management, these properties matter because they formalize how a tail risk measure responds to larger losses, cash additions, exposure scaling, and aggregation. The result is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus: TVaR's dual representation supplies a direct proof that it is a coherent risk measure on integrable losses.

Check Your Understanding

For each problem, use the upper quantile integral or the dual representation as indicated. Monetary amounts are in thousands of dollars unless stated otherwise.

A hypothetical loss is 0 with probability 0.60, 50 with probability 0.25, and 250 with probability 0.15. At $\alpha=0.80$, identify the portion of each loss level in the upper tail and calculate TVaR. (Answer: $200$ thousand dollars.)
A loss $X$ is $-20$ with probability 0.20, 40 with probability 0.50, and 160 with probability 0.30. At $\alpha=0.80$, calculate TVaR, accounting for the fact that the upper tail is smaller than the probability mass at the largest outcome. (Answer: $160$ thousand dollars.)
In three equally likely scenarios, losses $X$ are $20,60,100$, and losses $Y$ are $80,20,0$. At $\alpha=2/3$, calculate the two stand-alone TVaRs and the portfolio TVaR, then check subadditivity. (Answer: $\operatorname{TVaR}_{2/3}(X)=100$, $\operatorname{TVaR}_{2/3}(Y)=80$, and $\operatorname{TVaR}_{2/3}(X+Y)=100$ thousand dollars; $100\le180$.)
A hypothetical loss is 0 with probability 0.40, 50 with probability 0.40, and 150 with probability 0.20. At $\alpha=0.70$, compute TVaR and round to two decimal places. (Answer: $116.67$ thousand dollars.)
Four equally likely losses are $10,30,50,90$. At $\alpha=0.50$, the admissible dual weights must lie between 0 and 2 and average one. Find the maximum weighted expectation and give one maximizing set of weights in the listed outcome order. (Answer: $70$ thousand dollars, attained by weights $(0,0,2,2)$.)

What You'll Learn