TVaR Allocation Through Conditional Tail Losses

Why Allocate TVaR Through Tail Losses?

A portfolio TVaR describes the average loss in the worst part of its distribution. An actuarial capital model usually needs more than the portfolio total: it also needs a defensible way to assign that tail capital to business units or risk components. A direct approach is to examine each component’s average loss in the scenarios that make up the aggregate tail.

This approach is transparent, but the tail event must be defined consistently with the TVaR measure. If aggregate loss has no probability mass at the VaR threshold, conditioning on aggregate loss exceeding VaR gives the upper-tail average directly. If there is an atom at the threshold, taking every loss strictly above VaR can include too little probability mass. In that case, the boundary losses must be included in a consistent fractional amount.

Definition: Let $X_1,\ldots,X_n$ be integrable component losses and let $S=\sum_{i=1}^n X_i$ be aggregate loss. At confidence level $\alpha$, define $\operatorname{VaR}_{\alpha}(S)$ as the $\alpha$-quantile of $S$. Define $\operatorname{TVaR}_{\alpha}(S)$ as the average of the upper $1-\alpha$ fraction of the aggregate-loss distribution, equivalently $\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{p}(S)\,dp$. When $S$ has no atom at $q=\operatorname{VaR}_{\alpha}(S)$, this equals $E[S\mid S>q]$.

For a continuous aggregate distribution, the conditional-tail allocation to component $i$ is $C_i=E[X_i\mid S>\operatorname{VaR}_{\alpha}(S)]$. This is an allocation of TVaR dollars, not a percentage allocation: its units are the same as the component losses. The method requires the component losses to add to the aggregate loss being measured, and the conditioning event must use that same aggregate loss and confidence level.

Conditional Component Contributions and Full Allocation

Suppose $S$ has no atom at its VaR threshold and has a well-defined upper tail of probability $1-\alpha$. Then the component contributions sum to aggregate TVaR simply because the component losses sum to $S$. This is the full-allocation property for the conditional-tail method.

\sum_{i=1}^n C_i =\sum_{i=1}^n E[X_i\mid S>q] =E\left[\sum_{i=1}^n X_i\mid S>q\right] =E[S\mid S>q] =\operatorname{TVaR}_{\alpha}(S), \qquad q=\operatorname{VaR}_{\alpha}(S).

The second equality follows from linearity of conditional expectation: once the tail event is fixed, the conditional expectation of a sum is the sum of the conditional expectations. The final equality uses the no-atom condition at $q$. Thus full allocation is not a separate adjustment or balancing step; it follows from applying the same tail event to every component and defining TVaR on that same tail.

Worked Example: Comonotonic Components with a Uniform Driver

A hypothetical insurer models two component losses, in thousands of dollars, as $X_1=40+60U$ and $X_2=20+40U$, where $U$ is uniformly distributed on $[0,1]$. Calculate the component TVaR contributions and aggregate TVaR at $\alpha=0.90$.

Method. Write the aggregate as a function of $U$, find its 90th percentile, and then condition each component on the corresponding upper-tail values of $U$. Because $U$ is continuous and the aggregate is strictly increasing in $U$, the aggregate has no atom at VaR.

First add the losses: $S=X_1+X_2=(40+60U)+(20+40U)=60+100U$. Since $U$ is uniform, its 90th percentile is 0.90, so $\operatorname{VaR}_{0.90}(S)=60+100(0.90)=150$ thousand dollars. The event $S>150$ is equivalent to $U>0.90$.

Given $U>0.90$, $U$ is uniform on the interval from 0.90 to 1.00. Its conditional mean is $(0.90+1.00)/2=0.95$. Therefore, $C_1=E[40+60U\mid U>0.90]=40+60(0.95)=97$ thousand dollars, and $C_2=E[20+40U\mid U>0.90]=20+40(0.95)=58$ thousand dollars.

The aggregate tail mean is $E[60+100U\mid U>0.90]=60+100(0.95)=155$ thousand dollars. The check is $97+58=155$, equal to aggregate TVaR. Common mistake: using $U=0.90$, the threshold value, for both conditional component means would give 94 and 56. Those values sum to the VaR of 150, not the average over the upper 10% of losses.

Worked Example: Independent Components and a Triangular Aggregate

Worked Example: Tail Allocation with Independent Uniform Losses

A hypothetical portfolio has losses $X_1=100U_1$ and $X_2=100U_2$ dollars, where $U_1$ and $U_2$ are independent uniform random variables on $[0,1]$. Find the VaR, each component contribution, and TVaR at $\alpha=0.50$.

Method. Use the symmetry of the sum to identify its median, then integrate a component loss over the upper-tail region. The aggregate $S=100(U_1+U_2)$ has a continuous distribution, so conditional tail means apply without a boundary adjustment.

The transformation $(U_1,U_2)\mapsto(1-U_1,1-U_2)$ maps $U_1+U_2$ to $2-(U_1+U_2)$. The line $U_1+U_2=1$ divides the unit square into two equal-area triangles. Thus $P(S>100)=0.50$, and $\operatorname{VaR}_{0.50}(S)=100$ dollars.

For a fixed $U_1=u$, the condition $U_1+U_2>1$ requires $U_2>1-u$. Because $U_2$ is uniform, the length of the allowable interval for $U_2$ is $u$. The unnormalized conditional moment is therefore $E[U_1\,1_{\{U_1+U_2>1\}}]=\int_0^1 u\cdot u\,du=1/3$. The probability of the event is $1/2$, so $E[U_1\mid U_1+U_2>1]=(1/3)/(1/2)=2/3$. By symmetry, the conditional mean of $U_2$ is also $2/3$.

Each component contribution is $100(2/3)=66.666\ldots$ dollars. Aggregate TVaR is their sum, $133.333\ldots$ dollars. To the nearest cent, each contribution is $66.67 and aggregate TVaR is $133.33. The rounded contributions sum to $133.34, a one-cent rounding difference; before rounding, the contributions sum exactly to TVaR. Common mistake: assuming the conditional mean of each uniform variable remains $0.50$ in the tail. The event $U_1+U_2>1$ selects larger values of both variables, so their conditional means are $2/3$.

When the VaR Threshold Has an Atom

If aggregate loss has a positive probability at $q=\operatorname{VaR}_{\alpha}(S)$, the event $S>q$ may have probability less than $1-\alpha$. Conditional expectation on that event alone then averages over a smaller, more severe portion of the distribution than the upper $1-\alpha$ tail used by TVaR.

A consistent definition gives full weight to losses above $q$, no weight to losses below $q$, and just enough weight to losses equal to $q$ to make the total tail probability $1-\alpha$. Write $p_>=P(S>q)$ and $p_= = P(S=q)$. When $p_> \le 1-\alpha \le p_>+p_=$ and $p_=>0$, the required fraction of threshold probability is $\gamma=((1-\alpha)-p_>)/p_=$. Define the tail weight $W$ to be 1 when $S>q$, $\gamma$ when $S=q$, and 0 when $S<q$.

\operatorname{TVaR}_{\alpha}(S)=\frac{E[SW]}{1-\alpha}, \qquad C_i=\frac{E[X_iW]}{1-\alpha}, \qquad \sum_{i=1}^n C_i =\frac{E[\left(\sum_{i=1}^n X_i\right)W]}{1-\alpha} =\operatorname{TVaR}_{\alpha}(S).

This weighting is the discrete or mixed-distribution counterpart of conditioning on a continuous upper tail. It preserves full allocation because the same $W$ is applied to every component and because $\sum_i X_i=S$. The simple conditional-mean formula is recovered when there is no atom at VaR: then $P(S>q)=1-\alpha$, and $W$ is just the indicator of $S>q$.

Worked Example: Allocating the Boundary Mass

A hypothetical portfolio has the following aggregate losses and component breakdowns. At a 20% probability, the losses are $X_1=30$ and $X_2=10$, so $S=40$. At a 20% probability, they are $X_1=5$ and $X_2=15$, so $S=20$. At a 60% probability, they are $X_1=2$ and $X_2=8$, so $S=10$. Calculate the 80% VaR and the TVaR component contributions.

Method. Order the aggregate outcomes, identify the probability at and above the quantile, and include only the fraction of threshold mass needed to form the upper 20% tail.

The cumulative probability at loss 10 is 0.60, and at loss 20 it is 0.80. Thus $\operatorname{VaR}_{0.80}(S)=20$. The probability strictly above 20 is 0.20, which already equals the required tail probability $1-\alpha=0.20$. Hence $\gamma=(0.20-0.20)/0.20=0$: none of the probability mass at 20 is needed.

The upper-tail TVaR is therefore $40$. The component contributions are $C_1=30$ and $C_2=10$, since the entire upper 20% consists of the outcome with aggregate loss 40. They sum to $30+10=40$, as required. Common mistake: including threshold losses automatically just because they equal VaR. The tail contains exactly 20% probability already, so adding any of the 20-loss outcome would overstate its probability and would not match the quantile-integral definition of TVaR.

Practical Checks and Exam Traps

For conditional-tail allocation, first identify the aggregate loss and its VaR threshold, then verify the probability of the event used to calculate component means. The equality $C_i=E[X_i\mid S>q]$ is appropriate only when that event represents exactly the upper $1-\alpha$ fraction. When an atom makes the strict exceedance probability too small or too large for that tail, use boundary weighting instead.

Do not condition on each component’s own VaR. The allocation event is determined by aggregate loss $S$, not by separate component thresholds.
Do not confuse VaR with TVaR. VaR is the threshold; TVaR averages losses across the upper tail.
Use consistent component definitions. If components do not sum to the aggregate used for VaR and TVaR, their contributions will not reconcile by this method.
Do not discard an atom without checking its probability. Strict exceedance gives the correct conditional-mean formula only when its event has probability $1-\alpha$.
Keep rounding until the end. Individually rounded contributions may differ from the rounded portfolio total by a small amount, even when the unrounded allocation fully reconciles.

Conditional-tail allocation is useful because it links each assigned amount to the component’s actual average loss in the portfolio’s adverse scenarios. Its interpretation depends on the chosen confidence level and on the modeled dependence among components; it is not a standalone capital measure for each component. This treatment of TVaR allocation is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, identify the tail event or boundary weighting before calculating component contributions.

A hypothetical portfolio has $X_1=30+20U$ and $X_2=10+40U$, where $U$ is uniform on $[0,1]$. At $\alpha=0.90$, find aggregate VaR and TVaR and the two component contributions. (Answer: VaR = $94; contributions = $49 and $48; TVaR = $97.)
A portfolio has $X_1=45$ and $X_2=15$ with probability 0.20, $X_1=8$ and $X_2=16$ with probability 0.30, and $X_1=2$ and $X_2=10$ with probability 0.50. At $\alpha=0.75$, find VaR and the boundary fraction, then calculate TVaR contributions. (Answer: VaR = $24; boundary fraction $\gamma=1/6$; contributions = $37.60 and $15.20; TVaR = $52.80.)
Let $X_1=60U_1$ and $X_2=60U_2$, where $U_1,U_2$ are independent uniform variables on $[0,1]$. At $\alpha=0.50$, calculate the two conditional-tail contributions and portfolio TVaR. (Answer: each contribution = $40; TVaR = $80.)
A portfolio has $P(S=50)=0.10$, $P(S=20)=0.20$, and $P(S=5)=0.70$. At $\alpha=0.80$, what is VaR, what fraction of the VaR atom belongs to the upper tail, and what is TVaR? (Answer: VaR = $20; $\gamma=1/2$; TVaR = $35.)
In a continuous aggregate model, the component conditional means given $S>\operatorname{VaR}_{0.95}(S)$ are $12 thousand, $9 thousand, and $4 thousand. What is portfolio TVaR, and what probability should the conditioning event have? (Answer: TVaR = $25 thousand; the event probability is 0.05.)

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