Tail Value at Risk as an Average of Quantiles

TVaR as an Average Across the Upper Quantile Curve

Value at Risk identifies a loss threshold at a chosen confidence level. Tail Value at Risk (TVaR) adds information about the losses beyond that threshold by averaging the quantiles in the upper part of the probability scale. This definition is useful because it applies to continuous, discrete, and mixed loss distributions, including cases where the confidence level falls inside a probability mass.

The ideas here are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The convention in this tutorial is to define TVaR as a quantile average. This convention matters for discrete losses: if a probability atom straddles the confidence level, only the necessary fraction of that atom belongs to the upper quantile segment.

Definition: Let $X$ be a loss with finite expected value, and let $0\leq\alpha<1$. Define the quantile function by $q_u=\inf\{x:F_X(x)\geq u\}$, where $F_X$ is the cumulative distribution function of $X$. The Tail Value at Risk at confidence level $\alpha$ is the average of the quantiles from $\alpha$ to 1: $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}q_u\,du. $$

The interval from $\alpha$ to 1 has length $1-\alpha$. The integral adds up the quantile values over that interval, and dividing by its length gives their average. Equivalently, imagine moving along the horizontal probability axis from $\alpha$ to 1: TVaR is the average height of the quantile curve over that segment.

For a continuous loss distribution, the quantile curve may vary smoothly over the upper segment. For a discrete loss, it is a step function, so the average is a probability-weighted average of the loss values appearing there. If the segment begins partway through a step, only that part of the step is included. The same rule applies to a mixed distribution, which can have both jumps and continuously varying parts.

Why the Quantile Scale Gives an Average Loss

The probability axis is a way to index loss outcomes from lower to higher quantiles. A small interval of width $0.01$ on that axis represents one percentage point of probability. Thus an interval of width $r$ on which the quantile equals a constant loss $c$ contributes $rc$ to the integral. If the quantile varies linearly over an interval, its contribution is the interval width multiplied by the average of its endpoint values.

This explains why the definition handles probability atoms without a separate rule. An atom produces a flat segment in the quantile function whose width equals the atom's probability. When $\alpha$ cuts through that segment, the integral includes only the portion above $\alpha$. The final division by $1-\alpha$ makes the included probability weights sum to one.

More generally, if the upper segment consists of intervals of widths $r_1,\ldots,r_k$, and the quantile is constant at $x_j$ on interval $j$, then its contribution to the numerator is $\sum_j r_jx_j$. If a portion of the quantile curve is not constant, integrate that portion rather than treating it as a step. In each case, the total included width is $1-\alpha$.

Worked Example: A Continuous Uniform Loss

A hypothetical insurer models a claim loss $X$ as uniformly distributed from $20 to $120. Find $\operatorname{TVaR}_{0.80}(X)$. Use dollars throughout.

Method. Find the quantile function, then average it over probability levels $0.80$ to 1. For a uniform loss on $[20,120]$, the cumulative distribution increases linearly from 0 to 1, so $q_u=20+100u$ for $0\leq u\leq1$.

At the lower end of the required segment, $q_{0.80}=20+100(0.80)=100$. At the upper end, $q_1=20+100(1)=120$. The quantile is linear between these values, so its average over this segment is the average of the two endpoint values:

\operatorname{TVaR}_{0.80}(X) =\frac{100+120}{2} =110.

Answer: $\operatorname{TVaR}_{0.80}(X)=\$110.00$. The result is the average of the upper 20% of the quantile curve, not the upper endpoint loss of $120. The endpoint contributes no extra probability by itself.

Check. The quantile values on the relevant interval run from $100 to $120, so their average must lie between those values. The calculated $110 satisfies that check.

Discrete Losses and the Probability at VaR

For a discrete loss, the quantile function stays constant across probability intervals. The width of each interval is the probability assigned to that loss value. When $\alpha$ falls inside an interval, the part of the interval above $\alpha$—not the whole interval—is included in the TVaR average.

Worked Example: The Confidence Level Falls Inside a Discrete Atom

A hypothetical portfolio loss $X$ is $0 with probability 0.50, $20 with probability 0.30, and $100 with probability 0.20. Find $\operatorname{TVaR}_{0.75}(X)$, and explain why averaging all losses at least equal to VaR gives a different result.

Step 1: Locate the quantile intervals. The cumulative probability is 0.50 at $0 and 0.80 at $20. Therefore, $q_u=0$ for $0<u\leq0.50$, $q_u=20$ for $0.50<u\leq0.80$, and $q_u=100$ for $0.80<u<1$. Since the cumulative probability first reaches 0.75 at $20, $\operatorname{VaR}_{0.75}(X)=\$20$.

Step 2: Keep only the quantile segment above 0.75. The $20 step extends from probability 0.50 to 0.80, but only the portion from 0.75 to 0.80 is included. Its width is $0.80-0.75=0.05$. The $100 step has width $1-0.80=0.20$. The total included width is $0.05+0.20=0.25$, as required because $1-0.75=0.25$.

Step 3: Compute the average. The integral over the upper segment is $0.05(20)+0.20(100)=1+20=21$. Divide by the segment width:

\operatorname{TVaR}_{0.75}(X) =\frac{0.05(20)+0.20(100)}{0.25} =\frac{21}{0.25} =84.

Answer: $\operatorname{TVaR}_{0.75}(X)=\$84.00$. The calculation includes only 0.05 probability at $20, plus all 0.20 probability at $100.

Common mistake and correction. A tempting shortcut is to average the losses conditional on $X\geq\operatorname{VaR}_{0.75}(X)$. That includes the entire $20 atom, giving $\{0.30(20)+0.20(100)\}/0.50=(6+20)/0.50=\$52$. This is not the quantile average: it includes too much probability at $20. The upper quantile segment has width 0.25, not 0.50, so the correct calculation uses only 0.05 probability at $20 and gives $84.00.

Check. The upper quantile segment contains losses of $20 and $100, with more of its probability at $100. Its average of $84 lies between these loss values and is closer to $100, as expected.

Mixed Losses: Steps and Sloping Segments

A mixed distribution may place positive probability at particular loss values and also assign probability continuously over a range. Its quantile function then combines flat steps with sloping portions. The definition does not change: add the contributions over the probability levels from $\alpha$ to 1, then divide by $1-\alpha$.

Worked Example: A Mixed Loss Distribution

A hypothetical loss $X$ is $0 with probability 0.60, is uniformly distributed from $10 to $30 with probability 0.20, and is $30 with probability 0.20. Find $\operatorname{TVaR}_{0.70}(X)$.

Method. Write the quantile function on the relevant probability intervals and calculate the area above 0.70. The first 0.60 of probability is concentrated at $0. The next 0.20 covers the continuous uniform component, so the quantile rises from $10 to $30 over that interval. Its slope is $(30-10)/0.20=100$ dollars per unit of probability. Thus $q_u=10+100(u-0.60)$ for $0.60<u\leq0.80$. The remaining probability 0.20 is at $30.

The segment from 0.70 to 0.80 has width 0.10. Its endpoint quantiles are $q_{0.70}=10+100(0.10)=20$ and $q_{0.80}=30$. Since the quantile is linear there, its average height is $(20+30)/2=25$, and its area is $0.10(25)=2.50$. The flat segment from 0.80 to 1 has width 0.20 and height $30, so its area is $0.20(30)=6.00$.

The total area is $2.50+6.00=8.50$, and the total width is $1-0.70=0.30$. Therefore,

\operatorname{TVaR}_{0.70}(X) =\frac{8.50}{0.30} =28.3333\ldots.

Answer: Rounded to the nearest cent, $\operatorname{TVaR}_{0.70}(X)=\$28.33$. Rounding is done only after dividing the total quantile area by 0.30.

Check. The quantiles included range from $20 to $30. The answer, $28.33, is within that range and is nearer $30 because the final 0.20 of probability is entirely at $30.

Interpreting and Checking the Quantile Average

TVaR is measured in the same units as the loss. Its value is an average of the quantiles in the upper probability segment; it is not itself a probability, nor is it simply the largest possible loss. Several quick checks can help prevent errors:

Check the interval width. The quantile segment has width $1-\alpha$. For example, at confidence level 0.90, the relevant segment has width 0.10.
Check the probability weights. In a discrete calculation, the included widths must add to $1-\alpha$, including any partial step at the starting level.
Check the range. If the quantile is bounded between $a$ and $b$ over the whole segment, its average must also be between $a$ and $b$.
Keep the convention clear. At a probability atom, the quantile average uses only the portion of that atom above $\alpha$. Including the whole atom may produce a different number.

A key advantage of the quantile definition is that it treats continuous, discrete, and mixed losses within one framework. The shape of the quantile function changes from one model to another, but the interpretation remains the same: average the loss values along the upper segment of the probability scale. This makes the quantile curve—not only the VaR threshold—essential when describing the severity of upper-tail losses.

Key takeaway: Under the quantile-average convention, $\operatorname{TVaR}_{\alpha}(X)$ is the average of $q_u$ for $\alpha\leq u\leq1$. For discrete or mixed losses, include only the fraction of a probability step that lies above $\alpha$; do not automatically include the entire probability at VaR.

Check Your Understanding

For each problem, use the quantile average over probability levels from $\alpha$ to 1. Answers are in dollars.

A hypothetical continuous loss has quantile function $q_u=12+80u$ for $0\leq u\leq1$. Find $\operatorname{TVaR}_{0.75}(X)$. (Answer: $82.00)
A discrete loss is $0 with probability 0.30, $15 with probability 0.50, and $75 with probability 0.20. Find $\operatorname{TVaR}_{0.60}(X)$. (Answer: $45.00)
A mixed loss has quantile $q_u=0$ for $0<u\leq0.40$, $q_u=20+50(u-0.40)$ for $0.40<u\leq0.80$, and $q_u=40$ for $0.80<u<1$. Find $\operatorname{TVaR}_{0.70}(X)$, rounded to the nearest cent. (Answer: $39.17)
A loss has quantile $q_u=10$ for $0<u\leq0.60$, $q_u=10+250(u-0.60)$ for $0.60<u\leq0.80$, and $q_u=60$ for $0.80<u<1$. Find $\operatorname{TVaR}_{0.70}(X)$, rounded to the nearest cent. (Answer: $55.83)
A loss has quantile $q_u=0$ for $0<u\leq0.50$, $q_u=10$ for $0.50<u\leq0.70$, $q_u=10+200(u-0.70)$ for $0.70<u\leq0.90$, and $q_u=50$ for $0.90<u<1$. Find $\operatorname{TVaR}_{0.65}(X)$, rounded to the nearest cent. (Answer: $32.86)

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