The Quantile-Integral Formula for TVaR

Why Integrate Quantiles?

Value at Risk identifies a loss quantile at a chosen confidence level. Tail Value at Risk (TVaR) summarizes the upper tail by averaging quantiles across all confidence levels from that level to 1. The integral makes that averaging precise: it weights each increment in confidence level equally, rather than simply taking the mean of losses that exceed VaR.

The distinction matters for discrete losses. If a confidence level falls within a jump in the cumulative distribution, part of the quantile interval can correspond to the VaR loss itself. The quantile integral includes that part automatically. A conditional average over losses strictly greater than VaR does not.

Definition: For a loss $X$ with cumulative distribution function $F_X(x)=P(X\le x)$, define its quantile function by $$ Q_X(u)=\inf\{x:F_X(x)\ge u\}, \qquad 0<u<1. $$ For $0<\alpha<1$, the quantile-integral definition of TVaR is $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}Q_X(u)\,du, $$ provided the upper-tail integral is finite. Here $\alpha$ is the confidence level, $Q_X(u)$ is the loss quantile at confidence level $u$, and $1-\alpha$ is the length of the confidence-level interval being averaged.

The integration variable $u$ is a probability level, not a loss amount. The integral adds the areas under the quantile function from $u=\alpha$ to $u=1$; dividing by $1-\alpha$ turns that area into an average. If $Q_X$ is constant over a probability interval, its contribution is the constant loss times the interval's width. If it changes linearly, the contribution is the area of a trapezoid.

Why the Formula Is an Average

One way to justify the formula is to imagine choosing a confidence level uniformly from the upper interval. Let $U$ be uniformly distributed on $(0,1)$. The generalized inverse $Q_X$ has the property that $Q_X(U)$ has the same distribution as $X$: for any $x$, the probability that $Q_X(U)\le x$ is $F_X(x)$. If instead $U$ is chosen uniformly on $(\alpha,1)$, the average value of $Q_X(U)$ is exactly the integral average below.

E[Q_X(U)\mid \alpha<U<1] =\frac{1}{1-\alpha}\int_{\alpha}^{1}Q_X(u)\,du.

This gives the formula its interpretation: TVaR averages quantiles indexed by upper confidence levels. The average is over probability levels, not automatically over the event $X>\operatorname{VaR}_{\alpha}(X)$. In a discrete distribution, several confidence levels can have the same quantile, including the VaR value itself. Consequently, a conditional mean given strict exceedance need not equal this average.

Computing the Integral

For a stepwise quantile function, divide the integration interval wherever the quantile changes. For a linear segment, use the trapezoid area: interval width multiplied by the average of the quantile at its two endpoints. In every case, the final division is by $1-\alpha$, not by $\alpha$.

Locate the requested confidence level.
Identify $\alpha$ and calculate the length $1-\alpha$ of the upper confidence-level interval.

Write the quantile function on that interval.
For a distribution table, use the cumulative probability boundaries to identify each constant quantile segment.

Find the area from $\alpha$ to 1.
Add the areas of the constant or linear segments, including only the portions that fall inside the requested interval.

Divide by $1-\alpha$ and check.
Report the average loss in the stated units. For a nondecreasing quantile function, the answer must be at least $Q_X(\alpha)$.

Worked Example: A Confidence Level Inside a Quantile Step

A hypothetical portfolio has annual loss $X$, in dollars, with the following distribution: $P(X=0)=0.60$, $P(X=100)=0.25$, and $P(X=400)=0.15$. Calculate $\operatorname{TVaR}_{0.75}(X)$. Then compare it with the mean loss conditional on exceeding the 75th percentile. Use the quantile-integral method and state why strict exceedance gives a different answer.

Method: build the quantile steps, integrate only from 0.75 to 1, and divide by the interval length. The cumulative probabilities at losses 0, 100, and 400 are 0.60, 0.85, and 1.00. Therefore the quantile is 0 from probability levels 0 to 0.60, 100 from 0.60 to 0.85, and 400 from 0.85 to 1.

\operatorname{VaR}_{0.75}=Q_X(0.75)=100, \qquad 1-0.75=0.25.

From 0.75 to 0.85, the quantile is 100 over a width of $0.85-0.75=0.10$. From 0.85 to 1, it is 400 over a width of $1-0.85=0.15$. Thus the integrated area is

\int_{0.75}^{1}Q_X(u)\,du =(0.10)(100)+(0.15)(400) =10+60=70.

Dividing by the interval length gives

\operatorname{TVaR}_{0.75}(X)=\frac{70}{0.25}=280.

For comparison, losses strictly greater than 100 are 400, so $E[X\mid X>100]=400$. That is not the quantile-integral average: the integral also includes the portion from 0.75 to 0.85, where the quantile remains 100. The common mistake is to replace the quantile average with the strict-exceedance mean. The check is that the result, 280 dollars, is at least the VaR of 100 dollars and no greater than the maximum loss of 400 dollars.

Integrating a Piecewise-Linear Quantile Function

When quantiles change linearly with confidence level, the same method applies, but segment areas are trapezoids rather than rectangles. The confidence-level widths still determine the weights; the loss values are the heights.

Worked Example: Trapezoid Areas Across Two Segments

A hypothetical model reports losses in thousands of dollars through this quantile function: $Q_X(u)=50$ for $0<u\le0.50$; $Q_X(u)=100+200(u-0.50)$ for $0.50<u\le0.80$; and $Q_X(u)=160+500(u-0.80)$ for $0.80<u<1$. Calculate $\operatorname{TVaR}_{0.60}(X)$, in thousands of dollars.

Method: split the integral at 0.80 and use a trapezoid for each linear segment. The averaging interval has length $1-0.60=0.40$. On the first part, from 0.60 to 0.80, the quantile starts at $100+200(0.60-0.50)=120$ and ends at $100+200(0.80-0.50)=160$. Its area is

(0.80-0.60)\frac{120+160}{2} =(0.20)(140)=28.

On the second part, from 0.80 to 1, the quantile starts at 160 and approaches $160+500(1-0.80)=260$. Its area is

(1-0.80)\frac{160+260}{2} =(0.20)(210)=42.

The total area is $28+42=70$, so

\operatorname{TVaR}_{0.60}(X)=\frac{70}{0.40}=175 \quad\text{thousand dollars}.

The common error is to average the endpoint losses across the entire interval without accounting for where the slope changes. Splitting at 0.80 gives the correct area. The answer of 175 thousand dollars exceeds $Q_X(0.60)=120$ thousand dollars, as expected for an average of upper quantiles.

Comparing Confidence Levels

A higher confidence level shortens the integration interval to the most extreme quantiles. Since the quantile function is nondecreasing, this usually raises the average. A useful check is to compute two confidence levels from the same distribution and verify that the higher-level result reflects a more severe portion of the loss distribution.

Worked Example: Two TVaR Levels From One Distribution

For a hypothetical annual loss, let $P(X=20)=0.40$, $P(X=80)=0.30$, $P(X=200)=0.20$, and $P(X=500)=0.10$. Calculate $\operatorname{TVaR}_{0.50}(X)$ and $\operatorname{TVaR}_{0.80}(X)$, in dollars.

Method: use the cumulative-probability boundaries to identify the quantile steps. The quantile is 20 up to level 0.40, 80 from 0.40 to 0.70, 200 from 0.70 to 0.90, and 500 from 0.90 to 1.

For $\alpha=0.50$, the interval length is $1-0.50=0.50$. Its area is

$$ (0.70-0.50)(80)+(0.90-0.70)(200)+(1-0.90)(500) =16+40+50=106. $$

Therefore,

\operatorname{TVaR}_{0.50}(X)=\frac{106}{0.50}=212.

For $\alpha=0.80$, the interval length is $1-0.80=0.20$. The area is

(0.90-0.80)(200)+(1-0.90)(500) =20+50=70, \qquad \operatorname{TVaR}_{0.80}(X)=\frac{70}{0.20}=350.

The results are 212 dollars and 350 dollars. The common mistake is to use the 0.50 calculation's denominator for both levels; each result must be divided by its own interval length. The higher-confidence result is larger because it averages only the more severe quantiles.

Common Calculation Traps

Using the wrong interval length. The integral runs from $\alpha$ to 1, so the denominator is $1-\alpha$, not $\alpha$.
Integrating over losses instead of confidence levels. The horizontal variable in the quantile integral is $u$, and segment widths are probability differences.
Ignoring a segment boundary. If the quantile changes inside $(\alpha,1)$, split the integral there and include each segment's correct width.
Substituting a strict-exceedance mean without checking. A quantile average and a conditional mean above VaR are different calculations; a probability atom at VaR can make them unequal.
Forgetting units. If quantile values are in thousands of dollars, the integrated average is also in thousands of dollars.

Key takeaway: TVaR at level $\alpha$ is the area under $Q_X(u)$ from $\alpha$ to 1, divided by $1-\alpha$. Read the quantile function across confidence levels, include only the part of each segment inside that interval, and treat strict exceedance as a distinct calculation.

Check Your Understanding

For each problem, calculate the requested quantity using the quantile-integral formula. Loss amounts and quantiles are in dollars unless stated otherwise.

A hypothetical loss has probabilities $P(X=0)=0.50$, $P(X=60)=0.30$, and $P(X=240)=0.20$. Find $\operatorname{TVaR}_{0.70}(X)$. (Answer: 180 dollars)
A quantile function, in thousands of dollars, is $Q_X(u)=40$ for $0<u\le0.40$, $Q_X(u)=80+100(u-0.40)$ for $0.40<u\le0.80$, and $Q_X(u)=120+300(u-0.80)$ for $0.80<u<1$. Find $\operatorname{TVaR}_{0.60}(X)$. (Answer: 130 thousand dollars)
A hypothetical loss takes values 25, 75, and 175 with probabilities 0.20, 0.50, and 0.30, respectively. Find $\operatorname{TVaR}_{0.50}(X)$. (Answer: 135 dollars)
A quantile function is $Q_X(u)=10+100u$ for $0<u\le0.60$ and $Q_X(u)=70+200(u-0.60)$ for $0.60<u<1$. Find $\operatorname{TVaR}_{0.40}(X)$, rounding to the nearest cent. (Answer: 93.33 dollars)

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