TVaR Coherence from Its Dual Representation

Why the Dual Representation Matters

TVaR describes the average severity of losses in the upper tail. Its dual representation gives another way to calculate and reason about that tail: instead of selecting the worst fraction of outcomes directly, it chooses a probability weight that can emphasize bad outcomes, subject to a bound. This formulation makes the coherence properties of TVaR especially transparent.

Throughout, $X$ denotes an integrable loss, and $\alpha$ is a confidence level with $0<\alpha<1$. Losses are measured on a common probability space when comparing or combining them. Write $r=1-\alpha$ for the upper-tail probability. Use the quantile convention $\operatorname{VaR}_\alpha(X)=\inf\{x:F_X(x)\geq\alpha\}$. At an atom, TVaR averages exactly the worst fraction $r$ of the probability mass, taking only the required fraction of the mass at the quantile if needed.

Definition: For an integrable loss $X$, the dual representation of TVaR is $$ \operatorname{TVaR}_\alpha(X) = \sup\left\{E[ZX]: 0\leq Z\leq \frac{1}{1-\alpha}\ \text{almost surely},\ E[Z]=1\right\}. $$ The random variable $Z$ is an admissible weight. It reweights outcomes without changing total probability, since $E[Z]=1$, and cannot assign more than $1/(1-\alpha)$ times their original weight to any outcome.

Equivalently, $Z$ can be viewed as a density defining a reweighted probability distribution: the probability assigned to an event $A$ becomes $E[Z\mathbf{1}_A]$. The upper bound prevents the reweighting from concentrating all probability on an arbitrarily small set. For TVaR, a maximizing weight puts as much emphasis as allowed on the largest losses, with a fractional weight at the threshold when the threshold outcome has positive probability.

Why the Representation Holds

Let $q=\operatorname{VaR}_\alpha(X)$. The quantile convention implies $P(X>q)\leq r\leq P(X\geq q)$. For any admissible $Z$, use $E[Z]=1$ to write $E[ZX]=q+E[Z(X-q)]$. On outcomes where $X-q$ is positive, $Z\leq 1/r$; where $X-q$ is negative, $Z\geq0$. Therefore,

E[ZX] =q+E[Z(X-q)] \leq q+\frac{1}{r}E[(X-q)_+],

where $(x)_+=\max(x,0)$. The bound can be attained. Set $Z=1/r$ when $X>q$, $Z=0$ when $X<q$, and choose a value between $0$ and $1/r$ when $X=q$ so that $E[Z]=1$. The quantile inequalities ensure such a value exists. This is precisely the weighting that takes all losses strictly above $q$ and just enough probability at $q$ to fill the upper tail. Thus,

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

This argument also explains why the atom convention matters: if $P(X=q)>0$, using the whole atom may include more than the required tail probability. The maximizing weight uses only the needed fraction.

Worked Example: Finding the Tail Weights at an Atom

A hypothetical benefit plan has a one-year loss $X$ with the following distribution: $P(X=0)=0.4$, $P(X=100)=0.4$, and $P(X=300)=0.2$, with amounts in dollars. Find $\operatorname{TVaR}_{0.70}(X)$ from the dual representation and identify a maximizing weight. The model is this stated discrete distribution; no independence assumption is needed.

Method. Here $r=1-0.70=0.30$. The cumulative probability at 0 is 0.4, and at 100 it is 0.8, so $q=\operatorname{VaR}_{0.70}(X)=100$. The loss 300 has probability 0.2, leaving $0.30-0.20=0.10$ probability to take from the atom at 100.

The quantile formula gives $E[(X-q)_+]=0.2(300-100)+0.4(0)+0.4(0)=40$. Consequently, $\operatorname{TVaR}_{0.70}(X)=100+40/0.30=233.333\ldots$ dollars.

To form a maximizing weight, assign $Z=1/0.30=10/3$ at loss 300. At loss 100, only probability 0.10 of its 0.40 mass is needed, so assign a constant weight $0.10/(0.40\times0.30)=5/6$ there. Assign $Z=0$ at loss 0. Check its mean: $0.2(10/3)+0.4(5/6)+0.4(0)=2/3+1/3=1$. It respects $0\leq Z\leq10/3$, and $E[ZX]=0.2(10/3)(300)+0.4(5/6)(100)=200+33.333\ldots=233.333\ldots$ dollars.

The direct tail check is $(0.2(300)+0.1(100))/0.30=70/0.30=233.333\ldots$ dollars. The two calculations agree. The common mistake is to include the entire 0.40 probability at loss 100; the upper tail has probability 0.30, not 0.60.

Deriving the Coherence Properties

A risk measure $\rho$ on losses is coherent when it satisfies monotonicity, translation invariance, positive homogeneity, and subadditivity. For TVaR, each property follows from the same admissible set of weights. The following arguments apply to integrable losses; for comparisons, inequalities between losses are understood almost surely.

Monotonicity.
If $X\leq Y$, then for every admissible $Z\geq0$, $E[ZX]\leq E[ZY]$. Taking the supremum over the same admissible set gives $\operatorname{TVaR}_\alpha(X)\leq\operatorname{TVaR}_\alpha(Y)$.

Translation invariance.
For any constant $c$, $E[Z(X+c)]=E[ZX]+cE[Z]=E[ZX]+c$. Taking the supremum gives $\operatorname{TVaR}_\alpha(X+c)=\operatorname{TVaR}_\alpha(X)+c$.

Positive homogeneity.
For any constant $k\geq0$, $E[Z(kX)]=kE[ZX]$. Taking the supremum gives $\operatorname{TVaR}_\alpha(kX)=k\operatorname{TVaR}_\alpha(X)$; at $k=0$, both sides are zero.

Subadditivity.
For any two integrable losses $X,Y$, use the same admissible set for their sum: $\sup_Z E[Z(X+Y)]=\sup_Z(E[ZX]+E[ZY])\leq\sup_ZE[ZX]+\sup_ZE[ZY]$. Thus $\operatorname{TVaR}_\alpha(X+Y)\leq\operatorname{TVaR}_\alpha(X)+\operatorname{TVaR}_\alpha(Y)$.

The subadditivity inequality does not require $X$ and $Y$ to be independent. It follows because each single weight $Z$ produces an objective $E[ZX]+E[ZY]$ no greater than the sum of the two separate suprema. The weight that maximizes one loss need not maximize the other, or their sum. That distinction is important in portfolio calculations.

Worked Example: Translation and Scaling Through the Weights

Suppose a hypothetical portfolio has the same loss distribution as in the preceding worked example: losses of 0, 100, and 300 dollars with probabilities 0.4, 0.4, and 0.2. At confidence level $0.70$, find the TVaR after the portfolio is replaced by a loss $W=1.5X+40$, where 40 dollars is a fixed additional loss. State the rate-free probability model and check the result using a maximizing weight.

The transformation is increasing, so the outcomes for $W$ are $1.5(0)+40=40$, $1.5(100)+40=190$, and $1.5(300)+40=490$, with the same respective probabilities 0.4, 0.4, and 0.2. The quantile at level 0.70 is 190 dollars, and the upper-tail probability is $r=0.30$.

The amount above the quantile is 300 dollars in the highest-loss state and zero in the other two states. Thus $E[(W-190)_+]=0.2(490-190)=60$ dollars, and $\operatorname{TVaR}_{0.70}(W)=190+60/0.30=390$ dollars.

The maximizing weights from the first example remain admissible: assign $10/3$ at the original loss-300 state, $5/6$ at the original loss-100 state, and 0 at the original loss-0 state. Their mean is still 1. Their weighted expectation of $W$ is $0.2(10/3)(490)+0.4(5/6)(190)=326.666\ldots+63.333\ldots=390$ dollars.

As a second check, the original TVaR was $233.333\ldots$ dollars. Translation and positive homogeneity give $1.5(233.333\ldots)+40=350+40=390$ dollars. The common mistake is to scale the fixed 40-dollar addition as well; the specified transformation is $1.5X+40$, not $1.5(X+40)$.

Worked Example: Subadditivity for Two Dependent Losses

A hypothetical insurer considers two loss components $X$ and $Y$ across four equally likely scenarios. Their dollar losses are respectively $X=(100,0,0,0)$ and $Y=(0,100,0,0)$. Find their individual and combined TVaR at confidence level $0.75$, and verify subadditivity using the dual representation. These are joint scenario outcomes; no independence is assumed.

Here $r=0.25$. For $X$, the worst quarter of outcomes is the first scenario, with loss 100, so $\operatorname{TVaR}_{0.75}(X)=100$. A maximizing weight is $Z=(4,0,0,0)$: its mean is $(4+0+0+0)/4=1$, its largest value is $1/r=4$, and $E[ZX]=100$. For $Y$, the second scenario is the worst quarter. The weight $Z=(0,4,0,0)$ similarly gives $E[ZY]=100$, so $\operatorname{TVaR}_{0.75}(Y)=100$.

The combined outcomes are $X+Y=(100,100,0,0)$. Their largest quarter has loss 100, so $\operatorname{TVaR}_{0.75}(X+Y)=100$. In the dual form, any admissible weight has four entries between 0 and 4 with mean 1, so their sum is 4. The objective is $E[Z(X+Y)]=25(Z_1+Z_2)$, which is at most $25(4)=100$, and this bound is attained by $Z=(4,0,0,0)$.

Therefore $100\leq100+100=200$, as subadditivity requires. The common mistake is to treat the supremum of the sum as the sum of the separate suprema and report 200 as the combined TVaR. The two individual maximizing weights emphasize different scenarios; the combined loss itself has TVaR 100.

What to Check in an Exam Calculation

The dual representation is most useful when a proof or a discrete distribution makes the tail weighting clearer than a direct quantile calculation. Before applying it, verify the following:

Use the upper-tail probability. At confidence level $\alpha$, the weight cap is $1/(1-\alpha)$, not $1/\alpha$.
Keep the weights normalized. An admissible $Z$ must satisfy $E[Z]=1$, as well as its lower and upper bounds.
Handle quantile atoms fractionally. Include only enough probability at the quantile to make the selected tail mass equal $1-\alpha$.
Do not impose independence for subadditivity. The proof uses the common admissible set and linearity of expectation, not an assumption about dependence.
Do not exchange separate suprema as an equality. In general, $\sup_Z(E[ZX]+E[ZY])$ is at most, and may be strictly less than, $\sup_ZE[ZX]+\sup_ZE[ZY]$.

These checks connect the tail interpretation of TVaR with its risk-measure properties. The dual argument establishes coherence for integrable losses while also showing why subadditivity is an inequality: one probability weighting must evaluate the portfolio total, even when its separate components have different worst-tail scenarios. This result is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated discrete model and confidence level. Amounts are hypothetical dollar losses.

A loss is 0 with probability 0.8 and 50 with probability 0.2. At confidence level 0.90, identify a maximizing weight for the loss-50 outcome and calculate TVaR. (Answer: Weight 5 at loss 50; TVaR is $50.)
A loss is 10 with probability 0.5, 40 with probability 0.3, and 100 with probability 0.2. Find TVaR at confidence level 0.80, noting the upper-tail probability. (Answer: $100.)
A loss $X$ is 0 with probability 0.5, 20 with probability 0.3, and 80 with probability 0.2. At confidence level 0.60, calculate $\operatorname{TVaR}_{0.60}(X)$, then calculate TVaR for $1.5X+15$. (Answer: $50 for $X$; $90 for $1.5X+15$.)
Four equally likely scenarios have $X=(5,20,0,30)$ and $Y=(10,25,4,31)$. At confidence level 0.75, find both TVaRs and check monotonicity scenario by scenario. (Answer: $\operatorname{TVaR}_{0.75}(X)=\$30$, $\operatorname{TVaR}_{0.75}(Y)=\$31$, and $X\leq Y$ in every scenario.)
Three equally likely scenarios have component losses $X=(0,40,0)$ and $Y=(0,0,40)$. At confidence level $2/3$, find each component TVaR, the TVaR of the sum, and verify subadditivity. (Answer: Component TVaRs are $40 each; the sum has TVaR $40, so $40 $\leq$ $80.)

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