TVaR with Atoms and Quantile Conventions

Why TVaR Needs a Quantile Convention

Value at risk identifies a loss threshold at a chosen confidence level. Tail value at risk (TVaR) summarizes the severity of losses in the upper tail. For a continuous loss distribution, it is natural to think of TVaR as the average loss beyond the VaR threshold. For a discrete or mixed distribution, however, the threshold can itself have positive probability. The losses strictly above it may then account for less than the desired tail probability. TVaR must include only the fraction of probability at the threshold needed to complete that tail.

Throughout, $X$ is an integrable loss, $F_X(x)=P(X\le x)$, and the confidence level $\alpha$ satisfies $0<\alpha<1$. We use the lower-quantile convention for VaR. This convention matters whenever the cumulative distribution has a jump.

Definition: The value at risk at confidence level $\alpha$ is $q_\alpha=\operatorname{VaR}_\alpha(X)=\inf\{x:F_X(x)\ge\alpha\}$. The tail value at risk under the quantile-integral convention is $$ \operatorname{TVaR}_\alpha(X)=\frac{1}{1-\alpha}\int_\alpha^1 \operatorname{VaR}_u(X)\,du. $$ For $q_\alpha$ with $F_X(q_\alpha-)\le\alpha\le F_X(q_\alpha)$, this is equivalently $$ \operatorname{TVaR}_\alpha(X)= \frac{E[X\,\mathbf{1}_{\{X>q_\alpha\}}]+q_\alpha\big(F_X(q_\alpha)-\alpha\big)}{1-\alpha}. $$

The term $F_X(q_\alpha)-\alpha$ is the probability at the threshold that is included in the tail. It is generally smaller than the entire point mass at $q_\alpha$. The denominator $1-\alpha$ is the probability of the upper tail being averaged.

Why the Threshold-Mass Formula Holds

At confidence level $\alpha$, probability $1-\alpha$ belongs in the upper tail. The probability strictly above $q_\alpha$ is $1-F_X(q_\alpha)$. The remaining amount needed is therefore

(1-\alpha)-\big(1-F_X(q_\alpha)\big)=F_X(q_\alpha)-\alpha.

All losses strictly above $q_\alpha$ are included, while exactly this remaining probability is taken at the threshold. Its contribution to the tail total is $q_\alpha(F_X(q_\alpha)-\alpha)$. Dividing the combined loss total by $1-\alpha$ gives the stated formula.

A useful equivalent form follows by writing the expected excess above the threshold as $E[(X-q_\alpha)_+]$, where $(z)_+=\max(z,0)$. Since

E[(X-q_\alpha)_+] =E[X\,\mathbf{1}_{\{X>q_\alpha\}}]-q_\alpha\big(1-F_X(q_\alpha)\big),

adding $q_\alpha(1-\alpha)$ to both sides yields

\operatorname{TVaR}_\alpha(X) =q_\alpha+\frac{E[(X-q_\alpha)_+]}{1-\alpha}.

This identity is often convenient when an expected excess or stop-loss transform is available. It does not mean that the threshold is counted with its full probability mass.

Worked Examples

Worked Example: A Discrete Loss with a VaR Atom

A hypothetical insurer models one-year aggregate loss $X$, in dollars, with the following distribution: $P(X=0)=0.50$, $P(X=100)=0.25$, $P(X=300)=0.15$, and $P(X=800)=0.10$. Calculate $\operatorname{VaR}_{0.80}(X)$, $\operatorname{TVaR}_{0.80}(X)$, and $E[X\mid X>\operatorname{VaR}_{0.80}(X)]$.

Method. Find the first loss whose cumulative probability reaches 0.80, then allocate exactly 0.20 probability to the upper tail. This avoids substituting the strict-exceedance mean for TVaR.

The cumulative probabilities are $F_X(0)=0.50$, $F_X(100)=0.75$, and $F_X(300)=0.90$. Thus the first value with cumulative probability at least 0.80 is $q_{0.80}=300$.

The probability strictly above 300 is $0.10$, all at 800. The desired tail probability is $1-0.80=0.20$, so another $0.20-0.10=0.10$ must be taken at 300. This is $0.10/0.15=2/3$ of the probability mass at the threshold.

The loss total for this selected tail is $0.10(800)+0.10(300)=80+30=110$. Therefore,

\operatorname{TVaR}_{0.80}(X)=\frac{110}{0.20}=550.

The conditional mean strictly above VaR is $E[X\mid X>300]=800$, because every loss above 300 is 800. As a check, $550$ lies between the threshold 300 and the largest loss 800. The result is $\$550$, whereas the strict-exceedance conditional mean is $\$800$.

Worked Example: A Mixed Loss with an Atom at the Threshold

A hypothetical equipment portfolio has annual loss $X$ with $P(X=0)=0.55$, $P(X=100)=0.20$, and probability 0.25 assigned to a continuous uniform distribution on $[100,500]$. The continuous component has no point mass at either endpoint. Find $\operatorname{TVaR}_{0.70}(X)$ and compare it with the conditional mean strictly above $\operatorname{VaR}_{0.70}(X)$.

Method. Locate the quantile at the jump and combine the continuous losses above it with the fractional threshold mass. This avoids treating the atom at 100 as if it were all part of the tail.

At 0 the cumulative probability is 0.55. At 100 it is $0.55+0.20=0.75$, since the continuous component places zero probability at exactly 100. Hence $q_{0.70}=100$.

The probability strictly above 100 is 0.25, and the required tail probability is $1-0.70=0.30$. The additional probability required at 100 is $F_X(100)-0.70=0.75-0.70=0.05$. The mean of the uniform component on $[100,500]$ is $(100+500)/2=300$, so its contribution to the unconditional loss total is $0.25(300)=75$. The selected threshold mass contributes $0.05(100)=5$.

Thus the tail loss total is $75+5=80$, giving

\operatorname{TVaR}_{0.70}(X)=\frac{80}{0.30}=266.666\ldots.

Conditional on $X>100$, only the continuous component can occur. Its conditional mean is 300. The check is that TVaR exceeds the threshold 100 and is less than the strict-exceedance mean 300, as expected when some of the threshold atom is included. Rounded to the nearest cent, TVaR is $\$266.67$.

Worked Example: A Tail That Uses Only Part of a Large Atom

A hypothetical annual aggregate loss model has probabilities $P(X=0)=0.20$, $P(X=100)=0.25$, $P(X=250)=0.30$, $P(X=600)=0.15$, and $P(X=1{,}000)=0.10$. Calculate $\operatorname{TVaR}_{0.60}(X)$, and find the conditional mean strictly above VaR.

Method. Use the discrete quantile and calculate the tail total from the strict upper losses plus the required part of the VaR atom. This prevents counting more than the prescribed 0.40 tail probability.

The cumulative probabilities through losses 0, 100, and 250 are 0.20, 0.45, and 0.75, respectively. Therefore $q_{0.60}=250$. The probability above 250 is $0.15+0.10=0.25$. Since the required tail probability is $1-0.60=0.40$, the tail must include another $0.40-0.25=0.15$ at 250, exactly half of its 0.30 probability mass.

The strict-upper-loss contribution is $0.15(600)+0.10(1{,}000)=90+100=190$. The threshold contribution is $0.15(250)=37.50$. The selected tail total is $190+37.50=227.50$, and

\operatorname{TVaR}_{0.60}(X)=\frac{227.50}{0.40}=568.75.

For comparison, $E[X\mid X>250]=190/0.25=760$. As a check, TVaR is above 250 and below 760 because it averages all strict exceedances together with some losses equal to 250. The TVaR is $\$568.75$.

TVaR and Strict Exceedance Are Different Quantities

The quantile-integral definition explains the distinction: at a jump in the distribution, the VaR quantile persists over a range of confidence levels. Integrating from $\alpha$ to 1 includes only the portion of that range above $\alpha$. Equivalently, the threshold-mass formula selects just enough of the atom to make the averaged tail probability exactly $1-\alpha$.

Quantity	What it averages	How it treats a VaR atom
$\operatorname{TVaR}_\alpha(X)$	The upper tail of probability $1-\alpha$	Includes only the mass $F_X(q_\alpha)-\alpha$ needed at the threshold
$E[X\mid X>q_\alpha]$	Losses strictly exceeding the threshold	Includes none of the threshold atom; its conditioning probability is $1-F_X(q_\alpha)$

If there is no atom at $q_\alpha$ and $F_X(q_\alpha)=\alpha$, then $P(X>q_\alpha)=1-\alpha$, so TVaR equals the conditional mean strictly above VaR. With an atom at the threshold, the strict-exceedance probability can be smaller than $1-\alpha$. In that case, the two quantities generally differ; if $P(X>q_\alpha)=0$, the strict conditional mean is not defined, while the quantile-integral TVaR remains defined when the loss is integrable.

A frequent exam trap is to compute the mean of all outcomes strictly above VaR and label it TVaR. In the first worked example, that incorrect calculation gives $800$. It ignores the required 0.10 probability at the VaR value of 300. The correction is to calculate $(0.10(800)+0.10(300))/0.20=550$. The difference is not rounding; it comes from using a different tail.

For applied risk comparisons, the convention should be stated when a distribution has atoms. The quantile-integral definition makes the averaged probability explicit, remains applicable to discrete and mixed losses, and is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: Find the lower quantile $q_\alpha$, include every loss strictly above it, and add only probability $F_X(q_\alpha)-\alpha$ at the threshold. Divide that loss total by $1-\alpha$; do not substitute the mean conditional on strict exceedance unless the two tail probabilities coincide.

Check Your Understanding

For each problem, use the lower-quantile convention and the quantile-integral definition of TVaR.

A hypothetical loss $X$ takes values 0, 10, and 30 with probabilities 0.40, 0.40, and 0.20. Find $\operatorname{VaR}_{0.75}(X)$ and $\operatorname{TVaR}_{0.75}(X)$. (Answer: VaR $=10$; TVaR $=26$.)
A mixed loss has probability 0.50 at 0, probability 0.25 at 40, and probability 0.25 in a continuous uniform distribution on $[40,140]$. Find $\operatorname{TVaR}_{0.60}(X)$ and $E[X\mid X>40]$. (Answer: TVaR $=71.25$; strict-exceedance mean $=90$.)
A discrete loss takes values 0, 100, 300, and 900 with probabilities 0.10, 0.20, 0.40, and 0.30. Find $\operatorname{TVaR}_{0.50}(X)$. (Answer: VaR $=300$; TVaR $=660$.)
A loss is zero with probability 0.20; with probability 0.80 it has an exponential distribution with rate $0.01$ per dollar. Find $\operatorname{TVaR}_{0.80}(X)$. Use $E[(Z-q)_+]=e^{-0.01q}/0.01$ for an exponential $Z$ and $q\ge0$. (Answer: VaR $=\ln(4)/0.01\approx \$138.63$; TVaR $\approx \$238.63$.)
A hypothetical mixed loss has probability 0.70 at 0, probability 0.10 at 100, and probability 0.20 in a continuous uniform distribution on $[100,300]$. Find $\operatorname{VaR}_{0.85}(X)$ and $\operatorname{TVaR}_{0.85}(X)$. (Answer: VaR $=\$150$; TVaR $=\$225$.)

Quantity	What it averages	How it treats a VaR atom
\(\operatorname{TVaR}_\alpha(X)\)	The upper tail of probability \(1-\alpha\)	Includes only the mass \(F_X(q_\alpha)-\alpha\) needed at the threshold
\(E[X\mid X>q_\alpha]\)	Losses strictly exceeding the threshold	Includes none of the threshold atom; its conditioning probability is \(1-F_X(q_\alpha)\)

What You'll Learn