TVaR at Atoms and Boundary Quantiles

Why Atoms Change the TVaR Calculation

For a continuous loss distribution, the probability of a loss being exactly equal to its VaR threshold is zero. It can then be natural to describe TVaR as the average loss in the upper tail. A discrete loss distribution is different: a positive probability may be concentrated at the threshold itself. The upper tail may need only part of that probability mass to make up the prescribed tail probability.

The quantile-integral definition makes this partial inclusion precise. It averages exactly the largest $1-\alpha$ fraction of the loss distribution, even when that fraction cuts through an atom. This tutorial is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: For an integrable loss $X$ and confidence level $0<\alpha<1$, define the quantile by $\operatorname{VaR}_u(X)=\inf\{x:F_X(x)\ge u\}$. The quantile-integral definition of TVaR is $$ \operatorname{TVaR}_\alpha(X)=\frac{1}{1-\alpha}\int_\alpha^1 \operatorname{VaR}_u(X)\,du. $$ Here $1-\alpha$ is the probability of the upper tail being averaged. This definition applies to discrete as well as continuous distributions.

For a finite discrete loss, list the possible losses in increasing order, along with their probabilities. The cumulative probability up to a loss is the sum of the probabilities at that loss and all smaller losses. The VaR at level $\alpha$ is the first loss whose cumulative probability reaches or exceeds $\alpha$.

Discrete-loss formula: Suppose $X$ takes ordered values $x_1<\cdots<x_n$, with probabilities $p_1,\ldots,p_n$. Let $C_k=\sum_{j=1}^{k}p_j$, $C_0=0$, and choose $k$ such that $C_{k-1}<\alpha\le C_k$. Then $\operatorname{VaR}_\alpha(X)=x_k$, and $$ \operatorname{TVaR}_\alpha(X)= \frac{\displaystyle\sum_{j=k+1}^{n}p_jx_j+(C_k-\alpha)x_k}{1-\alpha}. $$ The term $C_k-\alpha$ is the portion of the probability at the VaR threshold that remains in the upper $1-\alpha$ tail.

Why the Threshold Mass Is Only Partly Included

The quantile function is constant across each interval of probability associated with a loss value. For example, if a loss value $x_k$ brings the cumulative probability from $C_{k-1}$ to $C_k$, then the quantile equals $x_k$ over that probability interval. When the integration starts at $\alpha$ inside that interval, only the portion from $\alpha$ to $C_k$ contributes. Its length is $C_k-\alpha$.

Locate the threshold.
Find $k$ such that $C_{k-1}<\alpha\le C_k$. The corresponding loss $x_k$ is $\operatorname{VaR}_\alpha(X)$.

Integrate above the threshold.
For each larger loss $x_j$, the full probability interval of length $p_j$ lies above $\alpha$, contributing $p_jx_j$.

Integrate the remaining threshold mass.
The interval for $x_k$ runs from $C_{k-1}$ to $C_k$. Only the part from $\alpha$ to $C_k$ remains, contributing $(C_k-\alpha)x_k$.

Divide by the tail probability.
The total probability included is $(C_k-\alpha)+\sum_{j=k+1}^{n}p_j=1-\alpha$, so dividing the weighted loss total by $1-\alpha$ gives the average loss in the upper tail.

This derivation also explains the boundary case $\alpha=C_k$. The threshold contribution is then $(C_k-\alpha)x_k=0$: the upper tail starts just above $x_k$. Quantile values at isolated endpoints do not change the integral, so the formula remains valid at cumulative-probability boundaries.

Worked Calculations

Worked Example: A Confidence Level Inside the VaR Atom

A hypothetical insurer models the annual loss $X$ on a small portfolio as follows: $X=0$ with probability $0.5000$, $X=\$10{,}000$ with probability $0.3000$, and $X=\$50{,}000$ with probability $0.2000$. Calculate $\operatorname{TVaR}_{0.7000}(X)$. Report the final amount to the nearest cent.

Method. Use the discrete quantile-integral formula. The cumulative probabilities are $0.5000$, $0.8000$, and $1.0000$. Since $0.5000<0.7000\le0.8000$, the VaR threshold is $x_k=\$10{,}000$, with $C_k=0.8000$.

The probability at the threshold that remains in the tail is $0.8000-0.7000=0.1000$. The full probability above the threshold is $0.2000$, at a loss of $\$50{,}000$. Thus the weighted tail loss is

(0.2000)(\$50{,}000)+(0.1000)(\$10{,}000) =\$10{,}000+\$1{,}000 =\$11{,}000.

The tail probability is $1-0.7000=0.3000$. Therefore,

\operatorname{TVaR}_{0.7000}(X) =\frac{\$11{,}000}{0.3000} =\$36{,}666.666\ldots \approx \boxed{\$36{,}666.67}.

Check. The included probability is $0.2000+0.1000=0.3000$, exactly the required upper-tail probability. The common mistake is to average all losses at or above $\$10{,}000$; that would include $0.3000+0.2000=0.5000$ probability, rather than the required $0.3000$.

Worked Example: A Boundary Confidence Level

A hypothetical risk model assigns loss $X=\$5{,}000$ with probability $0.2000$, $X=\$15{,}000$ with probability $0.3000$, $X=\$25{,}000$ with probability $0.3000$, and $X=\$60{,}000$ with probability $0.2000$. Calculate $\operatorname{TVaR}_{0.5000}(X)$ to the nearest cent.

Method. First compute the cumulative probabilities: $0.2000$, $0.5000$, $0.8000$, and $1.0000$. At $\alpha=0.5000$, the generalized-inverse quantile is $\$15{,}000$, because the cumulative probability first reaches $0.5000$ at that loss. Here $C_k-\alpha=0.5000-0.5000=0$, so none of the probability at the threshold is included.

The losses strictly above $\$15{,}000$ contribute

(0.3000)(\$25{,}000)+(0.2000)(\$60{,}000) =\$7{,}500+\$12{,}000 =\$19{,}500.

The tail probability is $1-0.5000=0.5000$, so

\operatorname{TVaR}_{0.5000}(X) =\frac{\$19{,}500}{0.5000} =\boxed{\$39{,}000.00}.

Check. The probabilities above the threshold sum to $0.3000+0.2000=0.5000$, as required. The boundary trap is to include some or all of the $\$15{,}000$ atom even though the cumulative probability has already reached $\alpha$ at that loss.

Worked Example: Losses Below Zero and a Partially Included Atom

A hypothetical hedging position has annual net loss $X=-\$20{,}000$ with probability $0.1500$, $X=\$0$ with probability $0.3000$, $X=\$30{,}000$ with probability $0.3500$, and $X=\$80{,}000$ with probability $0.2000$. Calculate $\operatorname{TVaR}_{0.6500}(X)$ to the nearest cent.

Method. Include the negative loss in the ordered distribution; it is part of $F_X$, even though the upper tail in this calculation will not contain it. The cumulative probabilities are $0.1500$, $0.4500$, $0.8000$, and $1.0000$. Since $0.4500<0.6500\le0.8000$, the VaR threshold is $\$30{,}000$.

The remaining probability at the threshold is $0.8000-0.6500=0.1500$. The full mass above it is $0.2000$ at $\$80{,}000$. The weighted tail loss is

(0.2000)(\$80{,}000)+(0.1500)(\$30{,}000) =\$16{,}000+\$4{,}500 =\$20{,}500.

The upper-tail probability is $1-0.6500=0.3500$, giving

\operatorname{TVaR}_{0.6500}(X) =\frac{\$20{,}500}{0.3500} =\$58{,}571.428\ldots \approx \boxed{\$58{,}571.43}.

Check. The included probability is $0.1500+0.2000=0.3500$. The common mistake avoided here is dropping the negative outcome before identifying the quantile. The loss values and their probabilities must first define the full distribution; the quantile and tail are then determined from that distribution.

Conditional Averages and Common Exam Traps

A frequent error is to calculate $E[X\mid X\ge \operatorname{VaR}_\alpha(X)]$ and call the result TVaR. That conditional expectation includes the entire probability mass at the threshold. If the VaR atom straddles the $\alpha$ boundary, it includes too much probability. The quantile-integral definition instead takes exactly the upper $1-\alpha$ fraction, including only the necessary part of that atom.

For the first worked example, the incorrect conditional average would be

E[X\mid X\ge \$10{,}000] =\frac{(0.3000)(\$10{,}000)+(0.2000)(\$50{,}000)}{0.5000} =\frac{\$13{,}000}{0.5000} =\$26{,}000.

This is not $\operatorname{TVaR}_{0.7000}(X)=\$36{,}666.67$. The conditional average uses $0.5000$ probability, while the requested upper tail has probability $0.3000$. Conversely, averaging only losses strictly greater than VaR can also fail: it omits the required portion of the threshold atom and generally leaves too little probability in the average.

Situation	Correct treatment at the threshold	Common error
$\alpha$ lies inside the cumulative probability interval for VaR	Include probability $C_k-\alpha$ at $x_k$	Include the full atom or none of it
$\alpha=C_k$ exactly	Include zero probability at $x_k$; average the losses above it	Add a positive portion of the threshold atom
The loss distribution includes negative values	Order all loss outcomes before locating VaR	Remove negative values before calculating the quantile

Key takeaway: TVaR from the quantile integral averages exactly the upper $1-\alpha$ probability of loss. For a discrete loss, calculate the VaR atom's cumulative probability and include only $C_k-\alpha$ of its mass. At a cumulative-probability boundary, that amount is zero.

Check Your Understanding

For each hypothetical discrete loss distribution, identify the VaR threshold and use the quantile-integral definition. Report monetary answers to the nearest cent.

A loss is $\$0$ with probability $0.4000$, $\$20{,}000$ with probability $0.4000$, and $\$80{,}000$ with probability $0.2000$. Find $\operatorname{TVaR}_{0.7000}(X)$. (Answer: $\$60{,}000.00$)
A loss is $\$10{,}000$ with probability $0.2500$, $\$30{,}000$ with probability $0.5000$, and $\$90{,}000$ with probability $0.2500$. Find $\operatorname{TVaR}_{0.7500}(X)$, noting whether the confidence level is a cumulative-probability boundary. (Answer: $\$90{,}000.00$)
A net loss is $-\$20{,}000$ with probability $0.2000$, $\$0$ with probability $0.3000$, $\$40{,}000$ with probability $0.4000$, and $\$100{,}000$ with probability $0.1000$. Find $\operatorname{TVaR}_{0.6000}(X)$. (Answer: $\$55{,}000.00$)
A loss is $\$5{,}000$ with probability $0.1000$, $\$15{,}000$ with probability $0.2000$, $\$35{,}000$ with probability $0.4000$, and $\$75{,}000$ with probability $0.3000$. Find $\operatorname{TVaR}_{0.8500}(X)$ and state the threshold contribution to the numerator. (Answer: $\$75{,}000.00$; threshold contribution $\$11{,}250$)
A loss is $\$0$ with probability $0.5000$, $\$12{,}000$ with probability $0.2500$, and $\$40{,}000$ with probability $0.2500$. Calculate both $\operatorname{TVaR}_{0.6000}(X)$ and $\operatorname{TVaR}_{0.7500}(X)$, and explain why the threshold contribution differs. (Answer: $\$29{,}500.00$ and $\$40{,}000.00$; at $0.6000$, $0.1500$ of the threshold atom is included, while at $0.7500$, none is included)

Situation	Correct treatment at the threshold	Common error
\(\alpha\) lies inside the cumulative probability interval for VaR	Include probability \(C_k-\alpha\) at \(x_k\)	Include the full atom or none of it
\(\alpha=C_k\) exactly	Include zero probability at \(x_k\); average the losses above it	Add a positive portion of the threshold atom
The loss distribution includes negative values	Order all loss outcomes before locating VaR	Remove negative values before calculating the quantile

What You'll Learn