The Integral Quantile Formula for Tail Value-at-Risk

Why Average Quantiles?

Value-at-risk identifies a loss threshold at a chosen confidence level. Tail value-at-risk goes further: it summarizes the severity of losses in the upper tail. For continuous distributions, it can be calculated as the conditional mean above the threshold. For discrete distributions, however, the threshold may lie inside a probability mass. Averaging the quantiles above the confidence level gives a consistent definition in both cases.

Definition: For a loss $X$ with finite expected value, its quantile function is the generalized inverse $Q_X(u)=\inf\{x:F_X(x)\ge u\}$, for $0<u<1$. At confidence level $p$, where $0<p<1$, define $\operatorname{VaR}_p(X)=Q_X(p)$ and $\operatorname{TVaR}_p(X)=\dfrac{1}{1-p}\int_p^1 Q_X(u)\,du$.

The integral averages loss quantiles from level $p$ through level 1. The width of that interval is $1-p$, so dividing by $1-p$ produces an average rather than an unscaled area. The definition requires the relevant tail integral to be finite; finite $E[X]$ is a sufficient condition for an integrable loss.

The generalized inverse matters when $F_X$ has jumps. If a loss value has positive probability, the quantile function stays constant over an interval of probability levels. If $p$ falls inside that interval, the integral includes only the portion of that constant interval above $p$. In effect, it includes just enough of the probability mass at the cutoff to make the upper tail have probability $1-p$.

Why the Integral Formula Works

A useful justification comes from a uniform random variable $U$ on $(0,1)$. The generalized inverse transforms $U$ into a loss with the specified distribution: $Q_X(U)$ has the same distribution as $X$. This is the inverse-transform representation, including when the distribution has atoms.

Represent the loss by its quantile.
Because $Q_X(U)$ has the distribution of $X$, loss outcomes can be indexed by probability level $u$.

Select the upper probability levels.
The interval $p<u<1$ has length $1-p$. Since $Q_X$ is nondecreasing, it represents the highest $1-p$ probability of losses, with a partial interval at the cutoff if necessary.

Average those losses.
The mean quantile over that interval is its integral divided by its length, giving $\operatorname{TVaR}_p(X)=\dfrac{1}{1-p}\int_p^1 Q_X(u)\,du$.

The same reasoning gives a useful expression in terms of the cutoff $q_p=\operatorname{VaR}_p(X)$:

\operatorname{TVaR}_p(X) = \frac{E[X\,\mathbf{1}_{\{X>q_p\}}]+q_p\bigl(F_X(q_p)-p\bigr)}{1-p}.

To see why, all losses strictly greater than $q_p$ belong in the upper tail. Their probability is $1-F_X(q_p)$. The upper tail still needs probability $F_X(q_p)-p$ to reach its total size $1-p$; that remaining portion is taken at the cutoff value $q_p$. The numerator is therefore the total loss assigned to the upper tail, and division by $1-p$ gives its average.

For a continuous distribution with no probability mass at $q_p$, $F_X(q_p)=p$, so the cutoff term is zero and the expression reduces to $E[X\mid X>q_p]$. That conditional-mean shortcut does not generally apply to a discrete distribution: conditioning on losses above or at the cutoff can include the wrong amount of probability mass.

Continuous Distribution Example

Worked Example: Uniform Loss Above the 80th Percentile

A hypothetical equipment policy has loss $X\sim\operatorname{Uniform}(0,100)$, with amounts measured in thousands of dollars. Calculate $\operatorname{TVaR}_{0.80}(X)$. Use the integral quantile method and check the result using a conditional mean. The common mistake to avoid is averaging the entire loss distribution rather than only the upper 20%.

Method: find and integrate the quantile function. For a uniform loss on $[0,100]$, $F_X(x)=x/100$ on that interval. Solving $u=x/100$ gives $Q_X(u)=100u$, for $0<u<1$.

The value-at-risk cutoff is $Q_X(0.80)=100(0.80)=80$, in thousands of dollars. The upper-tail probability is $1-0.80=0.20$. Thus

\begin{aligned} \operatorname{TVaR}_{0.80}(X) &=\frac{1}{0.20}\int_{0.80}^{1}100u\,du\\ &=5\left[50u^2\right]_{0.80}^{1}\\ &=5\bigl(50-50(0.64)\bigr)\\ &=5(50-32)\\ &=5(18)=90. \end{aligned}

The result is $90$ thousand dollars, or $90,000. As a check, conditional on exceeding 80, a uniform loss is uniform on $(80,100)$. Its mean is $(80+100)/2=90$ thousand dollars, agreeing with the integral. The calculation uses only the upper 20%, not the overall mean of 50 thousand dollars.

Discrete Losses and a Partial Cutoff Mass

Suppose $X$ takes the values $x_1<x_2<\cdots<x_n$ with probabilities $r_1,\ldots,r_n$. The quantile function equals $x_j$ over the probability interval accumulated at that value. In the integral from $p$ to 1, an interval may be only partly included. That partial interval is essential: the upper tail must have probability exactly $1-p$.

Worked Example: A Confidence Level Inside a Discrete Mass

A hypothetical cargo policy has loss $X$, in thousands of dollars, with probabilities $P(X=0)=0.50$, $P(X=100)=0.30$, and $P(X=400)=0.20$. Calculate $\operatorname{TVaR}_{0.60}(X)$ by integrating quantiles. The common mistake is to include either all or none of the mass at the value-at-risk cutoff.

Method: integrate each quantile interval above $p=0.60$. The cumulative probabilities are $F_X(0)=0.50$, $F_X(100)=0.80$, and $F_X(400)=1.00$. Therefore $Q_X(u)=100$ for $0.50<u\le0.80$, and $Q_X(u)=400$ for $0.80<u<1$. The cutoff is $Q_X(0.60)=100$.

Only the interval from 0.60 to 0.80 at loss 100 is included, with width $0.20$. The interval from 0.80 to 1.00 at loss 400 has width $0.20$. The total tail probability is $0.20+0.20=0.40=1-0.60$. Hence

\begin{aligned} \operatorname{TVaR}_{0.60}(X) &=\frac{1}{0.40}\left(100(0.20)+400(0.20)\right)\\ &=\frac{20+80}{0.40}\\ &=250 \end{aligned}

The answer is $250,000. As a check, the selected upper tail contains probability $0.20$ at 100 and $0.20$ at 400, so its total probability is $0.40$ and its weighted mean is $100$. Wait: the weighted total is $100(0.20)+400(0.20)=100$, and $100/0.40=250$ thousand dollars, as calculated.

The cutoff-mass formula gives the same result: $E[X\mathbf{1}_{\{X>100\}}]=400(0.20)=80$, $F_X(100)-p=0.80-0.60=0.20$, and the cutoff contribution is $100(0.20)=20$. Thus $(80+20)/0.40=250$.

Mixed Distributions

A distribution can have both a probability mass and a continuous component. The quantile integral handles both in one calculation: constant sections of $Q_X$ represent atoms, while changing sections represent continuous outcomes. The same tail probability and partial-cutoff rules apply without needing a separate definition.

Worked Example: An Atom Followed by Continuous Losses

A hypothetical benefit has loss $X=0$ with probability $0.25$. With probability $0.75$, $X$ is uniform on $[0,80]$, in thousands of dollars. Calculate $\operatorname{TVaR}_{0.50}(X)$. Use the quantile integral and check the result from the selected conditional tail. The common mistake is to treat the positive-loss component as if it had total probability 1.

Method: construct the mixed-distribution quantile. For $x$ between 0 and 80, the cumulative distribution is $F_X(x)=0.25+0.75(x/80)$. For $u>0.25$, solving $u=0.25+0.75(x/80)$ gives $Q_X(u)=80(u-0.25)/0.75$. For $0<u\le0.25$, the quantile is 0.

Since $p=0.50$ is above the atom interval, the quantiles integrated in the tail follow the linear expression. The cutoff is $Q_X(0.50)=80(0.25)/0.75=26.6667$ thousand dollars. Keeping the exact fraction in the integral,

\begin{aligned} \operatorname{TVaR}_{0.50}(X) &=\frac{1}{0.50}\int_{0.50}^{1}\frac{80(u-0.25)}{0.75}\,du\\ &=2\cdot\frac{320}{3} \left[\frac{(u-0.25)^2}{2}\right]_{0.50}^{1}\\ &=2\cdot\frac{160}{3}\left(0.75^2-0.25^2\right)\\ &=2\cdot\frac{160}{3}(0.50)\\ &=\frac{160}{3}=53.3333\ldots . \end{aligned}

Rounded to the nearest cent, the result is $53,333.33. To check, the upper half of all outcomes lies entirely in the positive-loss component. It has probability 0.50 out of that component's total probability 0.75, so it comprises the upper $0.50/0.75=2/3$ of the uniform outcomes. Those outcomes run from $80(1-2/3)=26.6667$ to 80, whose average is $(26.6667+80)/2=53.3333$ thousand dollars, consistent with the integral.

Exam Traps and Practical Checks

For a discrete distribution, a common incorrect calculation is to condition on $X\ge\operatorname{VaR}_p(X)$ and average every outcome in that event. In the cargo example, $X\ge100$ has probability $0.50$, not the required tail probability $0.40$. Its conditional mean is

E[X\mid X\ge100] =\frac{100(0.30)+400(0.20)}{0.50} =\frac{110}{0.50} =220.

That value is not the required tail value-at-risk of 250 thousand dollars because it includes the entire 0.30 probability at 100. The quantile integral includes only 0.20 probability at 100 and all 0.20 probability at 400. Conversely, conditioning on $X>100$ gives 400, which excludes the cutoff contribution altogether. The integral avoids both errors.

Confirm that the integration interval is $[p,1]$, with width $1-p$.
For discrete losses, list the quantile intervals and include only the portion above $p$.
For continuous distributions without an atom at the cutoff, check the result against $E[X\mid X>\operatorname{VaR}_p(X)]$.
Keep monetary units consistent throughout, then convert to dollars if the losses were stated in thousands.

Key takeaway: Tail value-at-risk is the average of the highest $1-p$ fraction of quantiles. The quantile integral automatically handles continuous outcomes, atoms, and mixtures; a conditional expectation shortcut must be checked carefully when the cutoff has positive probability.

Check Your Understanding

For each problem, calculate tail value-at-risk at the stated confidence level using the quantile integral. Answers are given so you can check your calculation.

A hypothetical loss is uniform on $[0,60]$ dollars. Find $\operatorname{TVaR}_{0.75}(X)$. (Answer: $52.50)
A loss in dollars is 10 with probability 0.40, 30 with probability 0.35, and 90 with probability 0.25. Find $\operatorname{TVaR}_{0.60}(X)$, showing how much of the mass at the cutoff is included. (Answer: $67.50)
A loss is 0 with probability 0.20 and, with probability 0.80, is uniform on $[20,100]$ dollars. Find $\operatorname{TVaR}_{0.50}(X)$. (Answer: $75.00)
A loss is 0 with probability 0.70, 50 with probability 0.20, and 200 with probability 0.10. Find $\operatorname{TVaR}_{0.85}(X)$, accounting for a partial cutoff mass. (Answer: $150.00)
A loss has density $f_X(x)=2x$ for $0\le x\le1$, and zero elsewhere. Since $F_X(x)=x^2$ on $[0,1]$, its quantile is $Q_X(u)=\sqrt{u}$. Find $\operatorname{TVaR}_{0.36}(X)$, rounded to four decimals. (Answer: 0.8167)

What You'll Learn