Conditional Tail Expectation And Expected Shortfall

Why Tail-Risk Names Need Definitions

A tail-risk report may use the terms TVaR, conditional tail expectation (CTE), and expected shortfall (ES) for a measure of average loss in the worst part of a distribution. For a continuous loss distribution, several common definitions produce the same number. For a distribution with a point mass at the VaR threshold, they may not. The name alone is therefore not enough to identify the calculation.

This distinction matters when comparing models, evaluating capital, or interpreting a reported tail measure. The first step is to identify the definition in use; the second is to check whether the loss distribution has probability concentrated exactly at its VaR. Throughout, $X$ denotes a nonnegative or signed loss, larger values are worse, and $p$ is a confidence level with $0<p<1$. We assume $E[|X|]<\infty$ when an expected tail loss is calculated.

Definition: The lower $p$-quantile, or $\operatorname{VaR}_p(X)$, is $v_p=\inf\{x:F_X(x)\ge p\}$. The quantile-based tail measure is $$ \operatorname{ES}_p(X)=\operatorname{TVaR}_p(X) =\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq. $$ This definition averages quantiles from $p$ up to 1. It is also called quantile-based TVaR or expected shortfall.

The integral gives an average across quantile levels, not necessarily the ordinary conditional mean given that $X$ exceeds $v_p$. That distinction is invisible for many continuous models, but it is essential when $X$ can equal $v_p$ with positive probability.

Three Definitions That May Be Called CTE, TVaR, or ES

The quantile-integral definition is one precise convention. A second convention defines CTE as the strict conditional mean $E[X\mid X>v_p]$. A third uses the inclusive conditional mean $E[X\mid X\ge v_p]$. These conditional quantities require their conditioning events to have positive probability. Terminology varies across actuarial and risk-management sources, so a calculation should state the formula, not merely the label.

A convenient expression for the quantile-based measure is

\operatorname{ES}_p(X) =v_p+\frac{E[(X-v_p)_+]}{1-p} =\frac{E[X\,\mathbf{1}_{\{X>v_p\}}]+v_p(F_X(v_p)-p)}{1-p},

where $(z)_+=\max(z,0)$ and $\mathbf{1}_{\{X>v_p\}}$ is 1 when $X>v_p$ and 0 otherwise. The added term involving $F_X(v_p)-p$ accounts for the part of the probability at the threshold that belongs in the upper $1-p$ tail. It is what makes this formula agree with the quantile integral even when there is an atom at VaR.

When the definitions agree: If $P(X=v_p)=0$ and $F_X(v_p)=p$, then $P(X>v_p)=1-p$. In that case the quantile-based measure, the strict conditional mean, and the inclusive conditional mean agree: $$ \operatorname{ES}_p(X) =E[X\mid X>v_p] =E[X\mid X\ge v_p]. $$ A continuous loss distribution satisfies these conditions at its $p$-quantile.

If there is positive probability at $v_p$, the strict and inclusive conditional means use different conditioning events and neither should automatically be substituted for the quantile-based measure. In particular, conditioning on $X\ge v_p$ includes all the mass at the threshold, even if only part of that mass lies in the worst $1-p$ portion of the distribution.

Why the Tail-Excess Formula Holds

The identity using expected excess loss gives a direct way to calculate the quantile-based measure when the threshold and expected excess are known. Here is a justification. Write $Q(q)=\operatorname{VaR}_q(X)$. The quantile-integral definition gives

\int_p^1 Q(q)\,dq =(1-p)v_p+\int_p^1 (Q(q)-v_p)\,dq.

For $q$ below the quantile threshold, $Q(q)-v_p$ contributes no positive excess. Above the threshold, quantiles that exceed $v_p$ describe exactly the upper losses that contribute to $(X-v_p)_+$. More formally, if $U$ is uniform on $(0,1)$, then $Q(U)$ has the distribution of $X$; hence the positive excess above $v_p$, averaged over quantile levels, is $E[(X-v_p)_+]$. Thus

\int_p^1 (Q(q)-v_p)\,dq=E[(X-v_p)_+], \qquad \operatorname{ES}_p(X)=v_p+\frac{E[(X-v_p)_+]}{1-p}.

This argument also explains why the quantile formula remains well-defined in the presence of an atom: it averages quantile levels and does not require conditioning on an event that includes either all or none of the threshold mass.

Worked Example: A Continuous Loss Model

A hypothetical portfolio has annual loss $X\sim\operatorname{Uniform}(0,120)$, in dollars, and the required confidence level is $p=0.90$. Find the VaR and compare quantile-based ES with both conditional means. The uniform distribution has no atom at its threshold.

Method. For a uniform loss on $[0,120]$, $F_X(x)=x/120$ on that interval, so its quantile function is $Q(q)=120q$. The VaR is

v_{0.90}=Q(0.90)=120(0.90)=108.

The quantile-based measure is the average of $120q$ over $0.90\le q\le1$:

\operatorname{ES}_{0.90}(X) =\frac{1}{0.10}\int_{0.90}^1 120q\,dq =\frac{60(1-0.90^2)}{0.10} =\frac{60(0.19)}{0.10} =114.

Conditional on $X>108$, the loss is uniform from 108 to 120, so $E[X\mid X>108]=(108+120)/2=114$. Since $P(X=108)=0$, conditioning on $X\ge108$ gives the same mean. The check is complete: all three definitions give 114 dollars. The common mistake this example avoids is reporting the threshold, 108 dollars, as the tail average.

What Changes When There Is an Atom at VaR?

At a jump in the cumulative distribution, the quantile definition still gives a threshold $v_p$, but the cumulative probability at that threshold can exceed $p$. Strict conditioning discards all probability at the threshold; inclusive conditioning includes all of it. The quantile-integral measure instead averages the worst $1-p$ fraction of quantile levels. As a result, it need not equal either conditional mean.

Worked Example: A Point Mass at the VaR Threshold

A hypothetical annual loss $X$, in dollars, has this distribution: $P(X=0)=0.92$, $P(X=100)=0.04$, and $P(X=1000)=0.04$. At $p=0.90$, calculate quantile-based ES, the strict conditional mean, and the inclusive conditional mean. This is a discrete model with an atom at the VaR.

Step 1: Find VaR. Since $F_X(0)=0.92\ge0.90$ and losses cannot be below zero, $v_{0.90}=0$. The upper tail represented by quantile levels from 0.90 to 1 has probability 0.10.

Step 2: Calculate quantile-based ES. Quantile levels from 0.90 to 0.92 have loss 0; levels from 0.92 to 0.96 have loss 100; and levels from 0.96 to 1 have loss 1000. Thus

\operatorname{ES}_{0.90}(X) =\frac{(0.02)(0)+(0.04)(100)+(0.04)(1000)}{0.10} =\frac{0+4+40}{0.10} =440.

Step 3: Calculate the conditional means. Strict conditioning excludes the zero loss, so $E[X\mid X>0]=(4+40)/(0.04+0.04)=44/0.08=550$. Inclusive conditioning includes all losses, including the zero mass, so $E[X\mid X\ge0]=E[X]=(0.92)(0)+(0.04)(100)+(0.04)(1000)=44$.

The three results are 440, 550, and 44 dollars, respectively. They differ because the probability at VaR is positive and the two conditional events treat that probability differently. The common mistake is to assume that “tail expectation” must mean $E[X\mid X>v_p]$; that is a specific convention, not the quantile-integral definition.

Using the Expected-Excess Form Directly

When the model supplies the expected excess above VaR, the expression $v_p+E[(X-v_p)_+]/(1-p)$ can be more efficient than integrating quantiles. This identity calculates the quantile-based measure; it does not turn a strict conditional mean into the same quantity when there is an atom at VaR.

Worked Example: Tail Average From Expected Excess

A hypothetical, integrable portfolio loss has an unspecified distribution. At the 95% confidence level its VaR is 320 dollars, and its expected excess above that threshold is $E[(X-320)_+]=18$ dollars. Find quantile-based ES. The distribution is not assumed to be continuous; the supplied expected excess is for the full loss model.

Method. Apply the tail-excess identity with $p=0.95$, $v_{0.95}=320$, and $1-p=0.05$:

\operatorname{ES}_{0.95}(X) =320+\frac{18}{0.05} =320+360 =680.

The result is 680 dollars. As a check, the quantile-based tail total per unit of probability is the threshold contribution $320$ plus an average excess of $360$, giving $680$. It exceeds VaR, as an upper-tail average should. The common mistake is to divide the expected excess by $p$, or to add the unscaled 18 dollars directly; the correct denominator is the tail probability $1-p=0.05$.

A Practical Comparison

When a source reports one of these names, check the mathematical definition and the distributional assumptions before comparing results. In particular, verify whether the source averages quantiles, conditions on strict exceedance, or conditions on inclusive exceedance.

Measure or convention	Definition	Key point
Quantile-based TVaR or ES	$\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq$	Well-defined for an integrable loss; accounts for threshold mass through quantile levels.
Strict conditional tail mean	$E[X\mid X>v_p]$	Excludes all mass at VaR; requires $P(X>v_p)>0$.
Inclusive conditional tail mean	$E[X\mid X\ge v_p]$	Includes all mass at VaR; generally differs from the quantile-based measure when there is an atom.
Continuous case with no threshold atom	All three expressions agree	Here $F_X(v_p)=p$ and $P(X>v_p)=1-p$.

The distinction is about definitions, not merely terminology. A statement that “CTE is the average loss above VaR” is incomplete unless “above” and the treatment of a loss equal to VaR are specified. For continuous models these details may not change the answer. For discrete or mixed models they can change it materially.

Common Mistakes and Interpretation Checks

Treating the names as universally interchangeable. State whether the intended measure is the quantile integral, a strict conditional mean, or an inclusive conditional mean.
Using the wrong tail probability. At confidence level $p$, the upper tail has probability $1-p$, not $p$.
Ignoring a jump at VaR. Check $P(X=v_p)$ or compare $F_X(v_p)$ with $p$. If there is an atom, conditional means need not match quantile-based ES.
Calling VaR an average loss. VaR is a quantile threshold; TVaR or ES measures an average across the specified tail.
Comparing reported values without checking conventions. Two values with the same label may have been calculated using different definitions or different assumptions about the loss distribution.

Key takeaway: Use the quantile integral to define quantile-based TVaR or expected shortfall. For a continuous loss distribution with no atom at VaR, it equals the strict and inclusive conditional tail means. With an atom at VaR, identify the convention and use its definition explicitly. This distinction is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, identify the relevant definition before calculating. Losses are measured in dollars.

A hypothetical loss is uniform on $[0,80]$. At $p=0.75$, find VaR and quantile-based ES, and state whether the strict conditional mean agrees. (Answer: VaR $=60$; ES $=70$; the strict conditional mean also equals 70.)
A hypothetical loss is uniform on $[50,250]$. At $p=0.90$, find VaR and quantile-based ES. (Answer: VaR $=230$; ES $=240$.)
A discrete hypothetical loss has probabilities $0.80$ at 0, $0.10$ at 50, and $0.10$ at 500. At $p=0.80$, find the quantile-based ES and $E[X\mid X\ge\operatorname{VaR}_{0.80}(X)]$. (Answer: quantile-based ES $=275$; inclusive conditional mean $=55$.)
A discrete hypothetical loss has probabilities $0.92$ at 0, $0.04$ at 100, and $0.04$ at 1000. At $p=0.90$, compare quantile-based ES with the strict conditional mean. (Answer: quantile-based ES $=440$; strict conditional mean $=550$.)
For an integrable hypothetical loss, $p=0.975$, $\operatorname{VaR}_{0.975}(X)=320$, and $E[(X-320)_+]=18$. Find quantile-based ES. (Answer: $1040$.)

Measure or convention	Definition	Key point
Quantile-based TVaR or ES	\(\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq\)	Well-defined for an integrable loss; accounts for threshold mass through quantile levels.
Strict conditional tail mean	\(E[X\mid X>v_p]\)	Excludes all mass at VaR; requires \(P(X>v_p)>0\).
Inclusive conditional tail mean	\(E[X\mid X\ge v_p]\)	Includes all mass at VaR; generally differs from the quantile-based measure when there is an atom.
Continuous case with no threshold atom	All three expressions agree	Here \(F_X(v_p)=p\) and \(P(X>v_p)=1-p\).

What You'll Learn