TVaR for Continuous Losses by Tail Conditioning

Introduction

Value at Risk identifies a threshold at a chosen confidence level. Tail Value at Risk (TVaR) goes further: for a continuous loss, it is the average loss among outcomes that exceed that threshold. This conditional-mean interpretation is especially useful when a model gives a simple survival function or density.

The key calculation has two parts. First, find the VaR threshold $q$. Then calculate the mean excess loss above $q$ and add it to $q$. This tutorial develops that calculation and applies it to uniform, exponential, and Pareto-type loss models.

Definition: Let $X$ be a nonnegative loss with finite mean, and let $0<\alpha<1$. Define $\operatorname{VaR}_{\alpha}(X)$ as the $\alpha$-quantile of $X$. If $X$ has a continuous distribution and $q=\operatorname{VaR}_{\alpha}(X)$ satisfies $P(X>q)=1-\alpha$, then $$ \operatorname{TVaR}_{\alpha}(X)=E[X\mid X>q]. $$ The condition on the tail probability holds, in particular, when the distribution function is continuous and strictly increasing at the quantile, so that $F_X(q)=\alpha$. This conditional-mean formula is not a formula for losses with an atom at $q$: conditioning on $X>q$ may then omit some of the probability mass needed for the upper tail.

From Conditional Mean to Tail Integral

For a continuous loss, the event $X>q$ has probability $1-\alpha$. Conditional expectation therefore gives

\operatorname{TVaR}_{\alpha}(X) =\frac{E[X\mathbf{1}_{\{X>q\}}]}{1-\alpha}.

A second form is often easier to calculate from a survival function. It expresses TVaR as the threshold plus the mean excess above the threshold, conditional on exceeding it.

Theorem: Under the definition’s assumptions, $$ \operatorname{TVaR}_{\alpha}(X) =q+\frac{1}{1-\alpha}\int_q^\infty S_X(x)\,dx, \qquad q=\operatorname{VaR}_{\alpha}(X), $$ where $S_X(x)=P(X>x)$. If the loss has a finite upper endpoint, the integrand is zero beyond that endpoint, so the integral can be stopped there.

The integral in this expression is an unconditional expected excess loss, not yet the average excess among losses above $q$. The division by $1-\alpha$ converts it to that conditional average. The justification follows directly from the positive-part identity.

Write the excess above the threshold.
For each outcome, $(X-q)_+$ is zero if $X\leq q$, and is $X-q$ if $X>q$. Equivalently, $(X-q)_+=\int_q^\infty \mathbf{1}_{\{X>x\}}\,dx$.

Take expectations.
The integrand is nonnegative, so expectation and integration can be interchanged. Thus $E[(X-q)_+]=\int_q^\infty P(X>x)\,dx=\int_q^\infty S_X(x)\,dx$.

Condition on exceeding the threshold.
On $X>q$, $X=q+(X-q)$. Therefore $E[X\mid X>q]=q+E[(X-q)\mathbf{1}_{\{X>q\}}]/P(X>q)$. The numerator is $E[(X-q)_+]$, and the denominator is $1-\alpha$, giving the theorem.

If a density is more convenient, the same conditional mean can be written as $\int_q^\infty x f_X(x)\,dx/(1-\alpha)$. The survival-integral form instead avoids integrating $x f_X(x)$ directly. The two calculations agree because they express the same conditional expectation.

Uniform Losses: A Bounded Tail

For $X\sim\operatorname{Uniform}(0,b)$, the distribution function on $0\leq x\leq b$ is $F_X(x)=x/b$, and the survival function is $S_X(x)=1-x/b$. Thus the $\alpha$-quantile is $q=\alpha b$, and the remaining probability above it is $1-\alpha$. The tail integral is the area under a straight-line survival curve from $q$ to $b$.

Worked Example: TVaR for a Uniform Loss

A hypothetical portfolio has annual loss $X\sim\operatorname{Uniform}(0,120)$, with amounts measured in thousands of dollars. Find $\operatorname{TVaR}_{0.75}(X)$. Give the VaR and TVaR in thousands to two decimal places, then state the TVaR in dollars.

Method. Find the quantile, then use the conditional tail expectation or the survival integral. Since this is a continuous uniform model, $P(X>q)=0.25$.

The distribution function is $F_X(x)=x/120$ for $0\leq x\leq120$. Solve $F_X(q)=0.75$:

$$ q=0.75(120)=90. $$

The survival function is $S_X(x)=1-x/120$ on the support. The excess-loss integral is

\begin{aligned} \int_{90}^{\infty} S_X(x)\,dx &=\int_{90}^{120}\left(1-\frac{x}{120}\right)\,dx\\ &=\left[x-\frac{x^2}{240}\right]_{90}^{120}\\ &=(120-60)-(90-33.75)\\ &=60-56.25=3.75. \end{aligned}

The integrand is zero beyond 120 because the loss cannot exceed 120. Divide the excess integral by the tail probability, 0.25, and add the threshold:

\operatorname{TVaR}_{0.75}(X) =90+\frac{3.75}{0.25} =90+15 =105.

As a check, conditioning a uniform loss on $X>90$ leaves a uniform distribution over $(90,120)$, whose mean is $(90+120)/2=105$. The answer is $105.00$ thousand dollars, or $\$105{,}000.00$. The common mistake here is to treat the excess integral, 3.75, as the conditional excess mean; it must first be divided by 0.25.

Exponential Losses: A Tail with No Upper Endpoint

For an exponential loss with rate $\lambda>0$, the survival function is $S_X(x)=e^{-\lambda x}$ for $x\geq0$. Solving $S_X(q)=1-\alpha$ gives the VaR. Integrating the survival function above $q$ gives an especially simple excess expectation, so this model illustrates why the tail-integral form can be efficient.

Worked Example: TVaR for an Exponential Loss

A hypothetical insurer models a claim amount $X$ in dollars by an exponential distribution with rate $\lambda=0.005$ per dollar. Find the 90% VaR and TVaR. Round dollar amounts to the nearest cent.

Method. Use $S_X(q)=1-\alpha$ to find the VaR, then integrate the exponential survival function. The rate is $0.005$ per dollar, so the mean is $1/\lambda=200$ dollars.

At $\alpha=0.90$, the upper-tail probability is 0.10. Set the survival probability equal to 0.10 and solve:

e^{-0.005q}=0.10, \qquad -0.005q=\ln(0.10), \qquad q=\frac{-\ln(0.10)}{0.005}=460.517\ldots.

Thus the VaR is $\$460.52$. Next calculate the tail integral:

\begin{aligned} \int_q^\infty S_X(x)\,dx &=\int_q^\infty e^{-0.005x}\,dx\\ &=\frac{e^{-0.005q}}{0.005}\\ &=\frac{0.10}{0.005}=20. \end{aligned}

This is the unconditional expected excess loss contributed by the upper tail. The mean excess conditional on being in that tail is $20/0.10=200$ dollars. Therefore

\operatorname{TVaR}_{0.90}(X) =460.517\ldots+200 =660.517\ldots \approx \$660.52.

The result can also be checked using the exponential memoryless property: conditional on exceeding $q$, the expected additional amount beyond $q$ is $1/\lambda=200$. Hence the conditional mean is $q+200$, agreeing with the tail-integral calculation. The method avoids confusing the rate with the mean: using 0.005 as the mean would give an incorrect scale by a large factor.

A Pareto-Type Example Using the Survival Function

A continuous loss need not have an exponential tail for the formula to work. Consider the Pareto-type (Lomax) survival function $S_X(x)=(1+x/\theta)^{-k}$ for $x\geq0$, where $\theta>0$ is a scale parameter and $k>1$. The condition $k>1$ ensures a finite mean, and hence a finite conditional tail mean at any finite threshold.

Worked Example: Tail Conditioning for a Pareto-Type Loss

A hypothetical loss has survival function $S_X(x)=(1+x/100)^{-2}$ for $x\geq0$, with loss measured in thousands of dollars. Find its 75% VaR and TVaR, reporting both to two decimal places.

Method. Solve $S_X(q)=1-\alpha=0.25$, then integrate the survival function from $q$ to infinity.

\left(1+\frac{q}{100}\right)^{-2}=0.25 \quad\Longrightarrow\quad 1+\frac{q}{100}=2 \quad\Longrightarrow\quad q=100.

The VaR is 100 thousand dollars. For the excess integral, use $u=1+x/100$, so $dx=100\,du$. When $x=100$, $u=2$:

\begin{aligned} \int_{100}^\infty S_X(x)\,dx &=100\int_2^\infty u^{-2}\,du\\ &=100\left[-u^{-1}\right]_2^\infty\\ &=100\left(0-\left(-\frac12\right)\right)=50. \end{aligned}

The upper-tail probability is 0.25, so the conditional mean excess is $50/0.25=200$ thousand dollars. It follows that

\operatorname{TVaR}_{0.75}(X) =100+200=300

in thousands of dollars. Thus the VaR is $\$100{,}000.00$ and the TVaR is $\$300{,}000.00$. As a check, the excess integral is positive and finite, and the TVaR exceeds the VaR, as an average of losses strictly above the threshold should.

Common Error and a Reliable Calculation Sequence

The most frequent error is to stop after calculating $\int_q^\infty S_X(x)\,dx$. That integral is $E[(X-q)_+]$, the expected excess over all outcomes, including the many outcomes for which the excess is zero. It is not the expected excess conditional on being in the tail.

Common mistake: In the uniform example, taking $90+3.75=93.75$ treats the unconditional excess integral as if it were a conditional mean. The correction is to divide by the tail probability: $90+3.75/0.25=105$. Always identify whether a calculated quantity is unconditional or conditional before adding it to VaR.

A disciplined calculation makes the distinction explicit:

State the continuous loss model and its parameter values.
Find $q=\operatorname{VaR}_{\alpha}(X)$ and verify that $P(X>q)=1-\alpha$.
Calculate the excess integral $\int_q^\infty S_X(x)\,dx$, using the support to set any finite upper endpoint.
Divide that integral by $1-\alpha$, then add $q$.
Check that the TVaR is at least the VaR and that units and rounding are consistent.

The conditioning interpretation is useful because it describes exactly which outcomes are averaged: the fraction $1-\alpha$ of losses above the quantile. This tail-conditioning method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each continuous model, find the requested VaR and TVaR. Answers are given so you can check the tail threshold, tail probability, and conditional excess.

A hypothetical loss is uniform on $[10,70]$ dollars. Find its 80% VaR and TVaR. (Answer: VaR $=\$58.00$; TVaR $=\$64.00$.)
A hypothetical claim amount is exponential with rate $0.02$ per dollar. Find its 95% VaR and TVaR, rounded to cents. (Answer: VaR $=\$149.79$; TVaR $=\$199.79$.)
A loss has survival function $S_X(x)=(1+x/50)^{-3}$ for $x\geq0$, with amounts in dollars. Find its 87.5% VaR and TVaR. (Answer: VaR $=\$50.00$; TVaR $=\$100.00$.)
A loss has survival function $S_X(x)=(1+x/100)^{-2}$ for $x\geq0$. Find its 80% VaR and TVaR, rounded to four decimals. (Answer: VaR $=123.6068$; TVaR $=347.2136$, in the same units as $X$.)

What You'll Learn