Deriving The Quantile Integral Formula For TVaR

From an Upper-Tail Average to an Integral

TVaR summarizes the size of losses in the upper tail, not just the loss at a selected confidence level. The quantile-integral formula makes that summary calculable by averaging loss quantiles from the confidence level $p$ through the upper end of the probability scale. The length of this interval is $1-p$, the probability assigned to the tail.

The formula is especially direct for a continuous loss distribution: each probability level corresponds to a unique loss quantile, and the losses above the $p$-quantile correspond to probability levels above $p$. This tutorial derives that connection and shows how to compute the integral, including when the quantile function changes slope.

Definition: For a loss $X$ with cumulative distribution function $F_X(x)=P(X\leq x)$, define its quantile function by $$ Q_X(u)=\inf\{x:F_X(x)\geq u\}, \qquad 0<u<1. $$ For a continuous loss distribution with finite upper-tail mean, the TVaR at confidence level $p$, where $0<p<1$, is $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 Q_X(u)\,du. $$

The denominator is the probability mass of the upper tail. The integral accumulates the quantile values over that tail, and division by $1-p$ turns the accumulated amount into an average loss. For example, at $p=0.95$, the formula averages quantiles over the final $0.05$ of the probability scale; it does not average over the first $0.95$.

For the derivation below, assume $X$ has a continuous distribution, has no probability mass at its $p$-quantile, and has a finite expected loss above that quantile. Write $q_p=Q_X(p)$. In this setting, $P(X>q_p)=1-p$, and $F_X(q_p)=p$. If the distribution has a density $f_X$, the upper-tail loss total can be written as an integral against that density.

Why the Quantile Integral Holds

The key step is a change of variable from a loss amount $x$ to its cumulative probability $u=F_X(x)$. For a continuous, strictly increasing distribution, $Q_X$ is the inverse of $F_X$, so $x=Q_X(u)$. Where the density exists, $du=f_X(x)\,dx$. The upper-tail integral therefore changes from an integral over losses to one over quantiles:

Start with the upper-tail loss total.
For $q_p=Q_X(p)$, the total expected loss contributed by losses above $q_p$ is $\int_{q_p}^{\infty}x f_X(x)\,dx$.

Change from loss to cumulative probability.
Set $u=F_X(x)$. Then $x=Q_X(u)$, $du=f_X(x)\,dx$, and the endpoints $x=q_p$ and $x\to\infty$ become $u=p$ and $u\to1$.

Divide by the probability of the tail.
The tail has probability $1-p$, so its average loss is the upper-tail loss total divided by $1-p$.

\int_{q_p}^{\infty}x f_X(x)\,dx = \int_p^1 Q_X(u)\,du \quad\Longrightarrow\quad \operatorname{TVaR}_p(X) = \frac{1}{1-p}\int_p^1 Q_X(u)\,du.

This derivation explains both the integration limits and the normalization. The lower endpoint is $p$ because $q_p$ is the loss at cumulative probability $p$. The upper endpoint is $1$ because it includes the rest of the distribution. Dividing by $1-p$ averages the quantiles over precisely that probability interval.

The change of variable can also be understood through probability levels: if $U$ is uniform on $(0,1)$, then $Q_X(U)$ has distribution $F_X$. Restricting $U$ to the interval from $p$ to $1$ selects the upper $1-p$ fraction of quantile levels. For distributions with atoms, the exact treatment at the quantile requires care; the formula here is stated for continuous losses, where there is no point mass to allocate at the boundary.

Worked Applications

Worked Example: Uniform Loss and an Upper-Tail Average

A hypothetical insurer models a one-year claim loss $X$ as uniformly distributed from $0 to $100. Calculate $\operatorname{TVaR}_{0.90}(X)$, first using the quantile integral and then by checking the upper-tail interval. State the answer in dollars to the cent.

Model and method. Assume $X\sim\operatorname{Uniform}(0,100)$, so $F_X(x)=x/100$ for $0\leq x\leq100$. Inverting gives $Q_X(u)=100u$. The quantile-integral method averages this function over $0.90\leq u\leq1$.

Step 1: Integrate the quantile.

\int_{0.90}^1 100u\,du = \left[50u^2\right]_{0.90}^{1} = 50(1^2-0.90^2) = 50(1-0.81) = 9.5.

Step 2: Divide by the tail probability. Here $1-p=1-0.90=0.10$, so $$ \operatorname{TVaR}_{0.90}(X)=\frac{9.5}{0.10}=95. $$ The answer is $95.00.

Check. The 90th percentile is $Q_X(0.90)=90$. For a uniform loss on $[0,100]$, the upper 10% of losses runs uniformly from $90 to $100. Its average is $(90+100)/2=95$, agreeing with the integral.

Common mistake avoided. Dividing the integral by $0.90$ would average over the wrong probability mass. The upper-tail interval has length $0.10$, not $0.90$.

Worked Example: Exponential Loss Quantiles

A hypothetical portfolio has annual loss $X$ following an exponential distribution with rate $\lambda=0.05$ per dollar, so its mean scale parameter is $1/\lambda=$20$. Find $\operatorname{TVaR}_{0.80}(X)$ from the quantile integral. Give the answer to the nearest cent.

Model and method. Assume an exponential loss with $F_X(x)=1-e^{-0.05x}$ for $x\geq0$. Solving $u=1-e^{-0.05x}$ for $x$ gives $$ Q_X(u)=-20\ln(1-u). $$ We integrate this quantile over the upper $20\%$ probability interval and divide by $0.20$.

Step 1: Set up the integral.

\operatorname{TVaR}_{0.80}(X) = \frac{1}{0.20}\int_{0.80}^1 -20\ln(1-u)\,du.

Step 2: Substitute $t=1-u$. The endpoints change from $u=0.80$ and $u=1$ to $t=0.20$ and $t=0$. Reversing the order of the endpoints yields $$ \int_{0.80}^1 -20\ln(1-u)\,du = 20\int_0^{0.20}-\ln(t)\,dt. $$ An antiderivative of $-\ln(t)$ is $-t\ln(t)+t$, since its derivative is $-\ln(t)$. At $t=0$, the limit of $-t\ln(t)+t$ is $0$. Therefore, $$ 20\left[-t\ln(t)+t\right]_0^{0.20} = 20\bigl[-0.20\ln(0.20)+0.20\bigr] = 4\bigl(1-\ln(0.20)\bigr). $$

Step 3: Normalize and evaluate. Dividing by $0.20$ gives $$ \operatorname{TVaR}_{0.80}(X) = 20\bigl(1-\ln(0.20)\bigr) = 20(1+\ln 5) \approx 52.1888. $$ Thus, to the nearest cent, $\operatorname{TVaR}_{0.80}(X)=\mathbf{$52.19}$.

Check. The 80th-percentile loss is $-20\ln(0.20)\approx$32.19$, and the calculated upper-tail average is larger, as expected. The scale parameter is $20, and the resulting tail average is $20(1+\ln 5)\), with the units of dollars.

Common mistake avoided. The exponential quantile contains $-\ln(1-u)$, not $-\ln(u)$. Replacing $1-u$ by $u$ reverses which end of the loss distribution is being represented.

Worked Example: Integrating a Piecewise Quantile Function

A hypothetical loss model is specified by the continuous quantile function $$ Q_X(u)= \begin{cases} 20u, & 0<u\leq0.80,\\ 16+270(u-0.80), & 0.80<u<1. \end{cases} $$ Calculate $\operatorname{TVaR}_{0.80}(X)$. Then calculate $\operatorname{TVaR}_{0.75}(X)$, showing how the integral is split when the quantile function changes slope. Report losses in dollars to the cent.

Model and method. The stated function is increasing and continuous at $u=0.80$: both pieces give $Q_X(0.80)=$16$. Use the quantile-integral formula, splitting the second calculation at the point where the formula changes.

Step 1: Calculate the 80% TVaR. On the interval $0.80\leq u\leq1$, the quantile increases linearly from $16 to $$ Q_X(1)=16+270(1-0.80)=16+54=70. $$ The integral of a linear function over an interval is the interval width times the average of its endpoint values. Thus $$ \int_{0.80}^1 Q_X(u)\,du = (1-0.80)\frac{16+70}{2} = 0.20(43) = 8.6. $$ Dividing by $1-0.80=0.20$ gives $\operatorname{TVaR}_{0.80}(X)=8.6/0.20=\mathbf{$43.00}$. The tail average is $43, between the endpoint quantiles $16 and $70.

Step 2: Integrate for the 75% TVaR. The interval from $0.75$ to $1$ includes part of each segment. At $u=0.75$, $Q_X(0.75)=20(0.75)=$15$. At $u=0.80$, it is $16; at $u=1$, it is $70.

From $0.75$ to $0.80$, the interval width is $0.05$, and the endpoint average is $(15+16)/2=15.5$. That area is $0.05(15.5)=0.775$. From $0.80$ to $1$, the area is $8.6$, as calculated above. The total integral is $0.775+8.6=9.375$. Therefore, $$ \operatorname{TVaR}_{0.75}(X) = \frac{9.375}{1-0.75} = \frac{9.375}{0.25} = \mathbf{$37.50}. $$

Check. The second result lies above $Q_X(0.75)=$15$ and below the largest quantile, $70. It is also below $\operatorname{TVaR}_{0.80}(X)=$43$, because lowering the confidence level adds a segment with lower quantiles to the average.

Common mistake avoided. Applying only the second piece of $Q_X$ to the whole interval from $0.75$ to $1$ would ignore the change in slope at $0.80$. Splitting at that point preserves the correct quantile values and areas.

Calculation Checks and Interpretation

A quantile integral is an average over probability levels, not an unnormalized area and not an average over loss values chosen without regard to their probability. Before evaluating it, identify the confidence level, write the corresponding tail probability, and use the same $p$ in both the lower integration limit and the normalization.

Check	Correct interpretation	Common error
Lower integration limit	$p$, the confidence level	Integrating from $0$ to $p$
Upper integration limit	$1$, the end of the probability scale	Stopping at $p$
Normalization	Divide by $1-p$, the upper-tail probability	Divide by $p$
Piecewise quantile function	Split the integral wherever the formula changes	Using one segment over the entire interval

For a continuous loss distribution, the formula provides a direct bridge between the quantile description of risk and the average loss represented by the upper tail. It also makes the role of confidence level visible: increasing $p$ narrows the probability interval being averaged and focuses the calculation on more extreme quantiles.

A final reasonableness check is to compare the calculated TVaR with the quantiles being averaged. For an increasing quantile function, the average over $[p,1]$ cannot be less than $Q_X(p)$. If it is, check for a reversed integration limit, an incorrect tail probability, or an arithmetic error.

Key takeaway: For a continuous loss with a finite upper-tail mean, TVaR is the average of $Q_X(u)$ over $p\leq u\leq1$. The integral accumulates the upper-tail quantiles, and division by $1-p$ converts that area into the tail average.

Check Your Understanding

For each problem, identify the quantile interval and use its length as the denominator. All loss amounts are in dollars.

A hypothetical loss is uniform on $[20,120]$. Find its 75th-percentile loss and $\operatorname{TVaR}_{0.75}$. (Answer: $Q_X(0.75)=$95$; $\operatorname{TVaR}_{0.75}=$107.50$)
A hypothetical exponential loss has rate $0.10$ per dollar. Use $Q_X(u)=-10\ln(1-u)$ to find $\operatorname{TVaR}_{0.90}$, to the nearest cent. (Answer: $\operatorname{TVaR}_{0.90}=$33.03$)
A continuous loss model has quantile function $Q_X(u)=20u$ for $0<u\leq0.80$, and $Q_X(u)=16+270(u-0.80)$ for $0.80<u<1$. Find $\operatorname{TVaR}_{0.80}$. (Answer: $\operatorname{TVaR}_{0.80}=$43.00$)
A quantile function is linear from $40 at $u=0.75$ to $60 at $u=0.85$, then linear from $60 at $u=0.85$ to $90 at $u=1$. Calculate $\operatorname{TVaR}_{0.75}$, splitting the integral at $0.85$. (Answer: $\operatorname{TVaR}_{0.75}=$65.00$)
A hypothetical continuous loss has $Q_X(u)=30+40u$ for $0<u<1$. Find $Q_X(0.60)$ and $\operatorname{TVaR}_{0.60}$, and state whether the TVaR is above or below the 60th-percentile loss. (Answer: $Q_X(0.60)=$54$; $\operatorname{TVaR}_{0.60}=$62$, above the 60th-percentile loss)

Check	Correct interpretation	Common error
Lower integration limit	\(p\), the confidence level	Integrating from \(0\) to \(p\)
Upper integration limit	\(1\), the end of the probability scale	Stopping at \(p\)
Normalization	Divide by \(1-p\), the upper-tail probability	Divide by \(p\)
Piecewise quantile function	Split the integral wherever the formula changes	Using one segment over the entire interval

What You'll Learn