Tail Value At Risk As An Average Of Upper Quantiles

Why Average the Upper Quantiles?

Value at Risk identifies a loss threshold at a chosen confidence level. For example, a 90% VaR is the loss level at the 90th percentile: it describes a boundary, not the size of losses beyond that boundary. Tail Value at Risk (TVaR) adds information about those more severe outcomes by averaging the loss quantiles from the selected confidence level through the most extreme quantile.

The confidence level $p$ determines where this upper-tail range begins. If $p=0.95$, the range of quantile levels being averaged is from 0.95 to 1, a span of 0.05. TVaR is therefore an average over that span, not an average over all losses and not a loss at a single percentile.

Definition: For a loss $X$ with finite upper-tail quantile integral and confidence level $0<p<1$, define the quantile at level $q$ by $\operatorname{VaR}_q(X)=\inf\{x:F_X(x)\ge q\}$. Tail Value at Risk at level $p$ is the average of these quantiles from $p$ through 1:

\operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq.

Here $F_X$ is the cumulative distribution function of the loss. For a continuous distribution with a strictly increasing cumulative distribution function, $\operatorname{VaR}_q(X)=F_X^{-1}(q)$.

Reading the Definition

The integration variable $q$ is a probability level, not a loss amount. At each level $q$, $\operatorname{VaR}_q(X)$ is a loss measured in dollars. The integral accumulates those loss quantiles across the interval from $p$ to 1; dividing by the interval length $1-p$ turns that accumulation into an average loss.

This gives an intuitive interpretation: imagine moving through percentile ranks in the upper tail, giving equal weight to each rank. TVaR is the average loss level encountered. A higher TVaR indicates greater average severity in that selected tail, though it does not by itself describe the full shape of the loss distribution.

Why divide by $1-p$? The interval of quantile levels being averaged has width $1-p$. An average over an interval equals the area under the function divided by the interval's width. Thus the factor $1/(1-p)$ normalizes the upper-quantile integral; without it, the result would be an accumulated area, not an average loss.

The definition is particularly direct when the quantile function is known. For a uniform loss on $[a,b]$, the cumulative distribution function is $F_X(x)=(x-a)/(b-a)$ for $a\le x\le b$. Solving $q=(x-a)/(b-a)$ gives the quantile $a+(b-a)q$. That linear quantile curve makes the upper-tail average straightforward to compute.

Worked Example: TVaR for a Uniform Loss

A hypothetical portfolio has an annual loss $X$, measured in thousands of dollars, that is uniformly distributed on $[40,170]$. Assume the distribution is continuous and that this model applies to the annual loss being measured. Find $\operatorname{VaR}_{0.80}(X)$ and $\operatorname{TVaR}_{0.80}(X)$, then report both in dollars.

Method. Use the uniform quantile function $F_X^{-1}(q)=40+(170-40)q$, then average that function over the interval from 0.80 to 1.

First, the quantile at 0.80 is

\operatorname{VaR}_{0.80}(X)=40+130(0.80)=40+104=144

The VaR is $144$ thousand dollars, or $144,000.

For a linear function, its average over an interval equals the average of its endpoint values. The quantile at level 1 is the upper bound, 170. Therefore,

\operatorname{TVaR}_{0.80}(X)=\frac{144+170}{2}=157

The TVaR is $157$ thousand dollars, or $157,000. It exceeds the 80% VaR because it averages quantiles above that threshold, up to the maximum loss of $170,000. The calculation is checked by noting that 157 lies between the two endpoint quantiles, 144 and 170, as an average must.

Common mistake avoided: dividing the integral by $p=0.80$ instead of by the upper-tail width $1-p=0.20$. The integral is $0.20(157)=31.4$ in thousands of dollars. Dividing that area by 0.80 gives 39.25, which is not the average of quantiles in the interval from 0.80 to 1. The correct denominator is 0.20, giving 157.

Calculating an Average from a Quantile Curve

A quantile curve need not be linear over the whole upper tail. If its values are supplied at several levels and the model assumes linear interpolation between them, calculate the area over each segment and add those areas. Divide the total by the full interval width $1-p$. This is a weighted average in which each segment's contribution depends on its width as well as its loss levels.

Worked Example: Combining Unequal Quantile Segments

A hypothetical risk model specifies losses in thousands of dollars and gives the following quantiles: $\operatorname{VaR}_{0.75}(X)=80$, $\operatorname{VaR}_{0.90}(X)=140$, and $\operatorname{VaR}_{1}(X)=200$. Assume the quantile curve is linear from 0.75 to 0.90 and linear from 0.90 to 1. Find $\operatorname{TVaR}_{0.75}(X)$.

Method. Calculate the area under the quantile curve on each segment as a trapezoid, add the areas, and divide by the full tail width, $1-0.75=0.25$.

From 0.75 to 0.90, the segment width is $0.90-0.75=0.15$, and its endpoint quantiles are 80 and 140. Its area is

0.15\left(\frac{80+140}{2}\right)=0.15(110)=16.5.

From 0.90 to 1, the width is $1-0.90=0.10$, and the endpoint quantiles are 140 and 200. Its area is

0.10\left(\frac{140+200}{2}\right)=0.10(170)=17.

The total area is $16.5+17=33.5$. The tail interval has width 0.25, so

\operatorname{TVaR}_{0.75}(X)=\frac{33.5}{0.25}=134.

Thus $\operatorname{TVaR}_{0.75}(X)$ is $134,000. As a check, the average is above the threshold quantile of $80,000 and below the maximum quantile of $200,000. The wider first segment and narrower second segment have both been included according to their actual widths.

Common mistake avoided: taking the unweighted average of 80, 140, and 200. Those three values are not equally spaced in quantile level, so averaging them directly would not represent the average over percentile ranks from 0.75 to 1.

A Nonlinear Quantile Curve

When the quantile function is curved, averaging just its endpoint values is generally not sufficient. Integrate the function over the required probability interval, or use a suitable numerical approximation if the curve is known only at selected points. The next example uses a specified distribution so that the quantile curve and its integral can both be stated explicitly.

Worked Example: Averaging a Curved Quantile Function

For a hypothetical annual loss in thousands of dollars, let $U$ be uniformly distributed on $[0,1]$, and define $X=20+80U+40U^2$. The loss is measured in thousands of dollars, with no interest adjustment. Find the 50% VaR and TVaR at 50%.

Method. The function $20+80u+40u^2$ is strictly increasing on $[0,1]$, since its derivative is $80+80u>0$. Therefore, the quantile at level $q$ is $20+80q+40q^2$.

At the 50% level,

\operatorname{VaR}_{0.50}(X)=20+80(0.50)+40(0.50)^2 =20+40+10=70.

Thus the 50% VaR is $70,000. For TVaR, integrate the quantile function from 0.50 to 1 and divide by $1-0.50=0.50$:

\begin{aligned} \operatorname{TVaR}_{0.50}(X) &=2\int_{0.50}^{1}(20+80q+40q^2)\,dq\\ &=2\left[20q+40q^2+\frac{40}{3}q^3\right]_{0.50}^{1}. \end{aligned}

At $q=1$, the antiderivative is $20+40+40/3=220/3$. At $q=0.50$, it is $10+10+5/3=65/3$. Their difference is $155/3$, and multiplying by 2 gives $310/3=103.333\ldots$ thousand dollars. Rounded to the nearest cent, the TVaR is $103,333.33.

The result is consistent with the quantile curve: it increases from $70,000 at the start of the averaged range to $140,000 at level 1, and the average, approximately $103,333.33, lies between those endpoint values. The curved quantile function is why the integral, rather than an endpoint midpoint, is needed here.

Common mistake avoided: treating the curved quantile function as if it were linear and taking $(70+140)/2=105$ thousand dollars. That shortcut would be valid for a linear quantile curve over the interval, but the quadratic term makes this curve nonlinear.

Interpretation and Exam Traps

TVaR summarizes average loss severity across an upper range of quantiles. Raising $p$ narrows that range to more extreme quantiles; the resulting TVaR depends on the distribution's tail shape. It is not enough to know the VaR at one confidence level to determine TVaR: the quantile curve above that level matters.

When solving a problem, keep the confidence level and tail width distinct. A level of 95% means the integral begins at 0.95, while the range being averaged has width 0.05. Also keep track of units: if quantiles are stated in thousands of dollars, the integral-based average is in thousands of dollars and must be converted only when reporting in dollars.

Using the wrong interval. TVaR at $p$ averages quantiles from $p$ to 1, not from 0 to $p$.
Using the wrong divisor. Divide the upper-quantile integral by $1-p$, the width of the interval being averaged.
Ignoring the quantile curve's shape. An endpoint midpoint is valid for a linear quantile function, not automatically for a curved one.
Mixing probability levels with loss units. The variable of integration is a probability level; the quantile values themselves carry the loss units.

For continuous loss models, this average-quantile definition gives a direct interpretation of the upper tail. For discrete models, how probability mass at a VaR threshold is handled requires additional care; do not assume that every alternative tail-based definition is automatically identical. The central calculation here remains the average of the quantiles over the specified probability-level interval.

Key takeaway: TVaR at confidence level $p$ is the area under the loss-quantile curve from $p$ to 1, divided by $1-p$. It measures the average loss severity across those upper quantiles, rather than reporting only the boundary loss at VaR.

Check Your Understanding

For each problem, use TVaR as the average quantile from the stated confidence level to 1. Unless otherwise specified, losses are in dollars.

A hypothetical loss is uniformly distributed on $[25,125]$. Find $\operatorname{VaR}_{0.90}(X)$ and $\operatorname{TVaR}_{0.90}(X)$. (Answer: VaR = $115; TVaR = $120.)
A hypothetical loss is uniformly distributed on $[60,210]$. Find $\operatorname{VaR}_{0.60}(X)$ and $\operatorname{TVaR}_{0.60}(X)$. (Answer: VaR = $150; TVaR = $180.)
A hypothetical model gives quantiles, in thousands of dollars, of 80 at level 0.75, 140 at level 0.90, and 200 at level 1. Assume linear interpolation on each segment. Find $\operatorname{TVaR}_{0.75}(X)$. (Answer: $134,000.)
Let $U$ be uniform on $[0,1]$ and let a hypothetical loss in thousands of dollars be $X=30+50U+20U^2$. Find $\operatorname{VaR}_{0.50}(X)$ and $\operatorname{TVaR}_{0.50}(X)$, rounding the TVaR to the nearest cent. (Answer: VaR = $60,000; TVaR = $79,166.67.)
A hypothetical loss is uniformly distributed on $[0,300]$. Calculate TVaR at 0.80 and 0.90, then state how much TVaR increases when the confidence level rises from 0.80 to 0.90. (Answer: TVaR at 0.80 = $270; TVaR at 0.90 = $285; increase = $15.)

What You'll Learn