TVaR for Discrete Losses Without a Quantile Atom

Why the Cumulative Probability at VaR Matters

For a discrete loss, the value at the confidence-level cutoff can carry positive probability. The key question for calculating Tail Value at Risk (TVaR) is not simply whether VaR has an atom. It is whether the cumulative probability at VaR is exactly the confidence level. When it is, the quantile integral covers precisely the losses strictly above VaR, so a strict upper-tail sum gives TVaR directly.

This distinction matters in practice: an atom at VaR does not automatically make the strict-tail calculation wrong. The calculation works when the cutoff lands at the end of the quantile interval for VaR. The method below is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $X$ be a loss and let $F_X(x)=P(X\leq x)$ be its cumulative distribution function. For confidence level $p$, where $0<p<1$, define the lower quantile $q=\operatorname{VaR}_p(X)$ by $$ q=\inf\{x:F_X(x)\geq p\}. $$ Define the quantile function by $Q(u)=\inf\{x:F_X(x)\geq u\}$. The quantile-integral definition of TVaR is $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 Q(u)\,du. $$

For a finite discrete loss, $Q(u)$ is constant over intervals of probability levels. If a loss value $x_i$ has probability $r_i$, it contributes $x_i r_i$ to the expected loss across its full quantile interval. When the confidence level $p$ is exactly the cumulative probability at $q$, the quantile integral from $p$ to 1 includes the full probability mass of each loss above $q$, and none of the mass at or below $q$.

The Strict Upper-Tail Formula

Suppose $X$ is a finite discrete loss, $q=\operatorname{VaR}_p(X)$, and $F_X(q)=p$. Since $P(X>q)=1-F_X(q)$, the probability strictly above $q$ is $1-p$. The quantile integral then gives

\operatorname{TVaR}_p(X) =\frac{1}{1-p}\sum_{x_i>q}x_iP(X=x_i) =\frac{E[X\,\mathbf{1}_{\{X>q\}}]}{1-p} =E[X\mid X>q].

The last equality is valid because $P(X>q)=1-p>0$. Importantly, $P(X=q)$ can still be positive. If $F_X(q)=p$, the quantile interval associated with $q$ ends at $p$, so the integral beginning at $p$ does not include any of that interval except possibly its single endpoint, which has zero length.

Find the VaR cutoff.
Accumulate probabilities in increasing order of loss until the cumulative probability first reaches or exceeds $p$. The corresponding loss is $q$.

Check the cumulative probability.
Calculate $F_X(q)$. The strict upper-tail formula applies when $F_X(q)=p$. Do not use the presence or absence of $P(X=q)$ as a substitute for this check.

Sum the strict upper tail.
For every loss value greater than $q$, multiply the loss by its probability and add the products.

Divide by the tail probability.
Divide the sum by $1-p$. Verify that the probabilities of losses strictly above $q$ add to $1-p$.

Worked Examples

Worked Example: A Boundary Cutoff with Several Upper-Tail Losses

A hypothetical insurer models a one-year claim loss $X$, in dollars, with the following finite distribution: $P(X=0)=0.25$, $P(X=20)=0.35$, $P(X=50)=0.25$, and $P(X=100)=0.15$. Calculate $\operatorname{TVaR}_{0.60}(X)$ using the strict upper tail, then verify it using the quantile integral. The probabilities sum to $0.25+0.35+0.25+0.15=1.00$.

Method and VaR. Accumulate the probabilities in increasing loss order. At loss 0, the cumulative probability is $0.25$. At loss 20, it is $0.25+0.35=0.60$. Therefore, $q=\operatorname{VaR}_{0.60}(X)=20$, and $F_X(q)=0.60=p$.

Strict-tail calculation. The losses strictly greater than 20 are 50 and 100. Their probability-weighted losses are $50(0.25)=12.50$ and $100(0.15)=15.00$. The total is $12.50+15.00=27.50$. The upper-tail probability is $0.25+0.15=0.40=1-0.60$. Thus $$ \operatorname{TVaR}_{0.60}(X)=\frac{27.50}{0.40}=\$68.75. $$

Quantile-integral check. The quantile is 50 from probability level $0.60$ through $0.85$, and 100 from $0.85$ through 1. The integral is $50(0.85-0.60)+100(1-0.85)=50(0.25)+100(0.15)=12.50+15.00=27.50$. Dividing by $1-p=0.40$ again gives $\$68.75$.

Check and common mistake. The probabilities above VaR total $0.40$, exactly the required upper-tail probability. Although $P(X=20)=0.35$, the strict-tail calculation is valid because $F_X(20)=p$. Rejecting the method just because VaR has positive probability would be the mistake.

Worked Example: A Single Loss Value in the Upper Tail

A hypothetical equipment-protection portfolio has annual loss $X$, in thousands of dollars, with $P(X=0)=0.40$, $P(X=40)=0.20$, $P(X=90)=0.15$, and $P(X=150)=0.25$. Find $\operatorname{TVaR}_{0.75}(X)$ in thousands of dollars and verify it with the quantile integral. The probabilities sum to 1.

Method and VaR. The cumulative probabilities are $0.40$ at loss 0, $0.40+0.20=0.60$ at loss 40, and $0.60+0.15=0.75$ at loss 90. Hence $q=\operatorname{VaR}_{0.75}(X)=90$ and $F_X(90)=0.75=p$.

Strict-tail calculation. Only the loss of 150 exceeds 90. Its probability-weighted loss is $150(0.25)=37.50$ thousand dollars. The strict upper-tail probability is $0.25=1-0.75$. Therefore, $$ \operatorname{TVaR}_{0.75}(X)=\frac{37.50}{0.25}=150 $$ thousand dollars.

Quantile-integral check. The quantile is 150 for probability levels from $0.75$ to 1, an interval of length $1-0.75=0.25$. The integral is $150(0.25)=37.50$ thousand dollars, and $37.50/0.25=150$ thousand dollars.

Check and common mistake. The tail probability and denominator both equal $0.25$, so the result is the loss value 150. Dividing by the probability of all losses at or above VaR would include the mass at 90 and would not compute the specified quantile integral.

Worked Example: A Positive Atom at VaR Does Not Prevent the Calculation

A hypothetical benefit plan models a yearly covered loss $X$, in dollars, as $P(X=5)=0.20$, $P(X=25)=0.30$, $P(X=60)=0.25$, and $P(X=100)=0.25$. Calculate $\operatorname{TVaR}_{0.50}(X)$, using the strict upper tail and then the quantile integral. The probabilities sum to 1.

Method and VaR. The cumulative probability is $0.20$ at 5 and $0.20+0.30=0.50$ at 25. Thus $q=\operatorname{VaR}_{0.50}(X)=25$, $F_X(25)=p=0.50$, and $P(X=25)=0.30$.

Strict-tail calculation. The losses above 25 are 60 and 100. Their weighted losses are $60(0.25)=15.00$ and $100(0.25)=25.00$, for a total of $40.00$. Their probabilities sum to $0.25+0.25=0.50=1-p$. Therefore, $$ \operatorname{TVaR}_{0.50}(X)=\frac{40.00}{0.50}=\$80.00. $$

Quantile-integral check. The quantile is 60 from level $0.50$ to $0.75$, and 100 from $0.75$ to 1. The integral is $60(0.25)+100(0.25)=15.00+25.00=40.00$. Dividing by $0.50$ gives $\$80.00$.

Check and common mistake. The strict-tail mass is exactly $0.50$, matching $1-p$, and both calculations agree. The atom at 25 is not included in the integral from $p=0.50$ onward, because its quantile interval ends at 0.50. The incorrect claim that any atom at VaR rules out the strict-tail method would lead to the wrong conclusion here.

When the Strict-Tail Calculation Is Not Enough

For a lower quantile $q=\operatorname{VaR}_p(X)$, the cumulative probability satisfies $F_X(q)\geq p$. Consequently,

P(X>q)=1-F_X(q)\leq 1-p.

The strict-tail sum divided by $1-p$ gives the quantile-integral TVaR when $F_X(q)=p$. If $F_X(q)>p$, the strict upper tail has probability less than $1-p$; the quantile integral includes some probability from the quantile interval at $q$, so summing only losses greater than $q$ omits part of the required tail. That case needs a different calculation, treated separately from the boundary case developed here.

A reliable exam check is therefore: find $q$, calculate $F_X(q)$, and compare it with $p$. In particular, $P(X=q)>0$ does not determine whether the strict-tail method works. The relevant condition is $F_X(q)=p$, which can hold even when $q$ has positive probability.

Key takeaway: For a finite discrete loss, if $q=\operatorname{VaR}_p(X)$ and $F_X(q)=p$, then the losses strictly greater than $q$ have total probability $1-p$, and $$ \operatorname{TVaR}_p(X)=\frac{\sum_{x_i>q}x_iP(X=x_i)}{1-p}. $$ Verify both the cumulative probability at VaR and the probability total above VaR. An atom at VaR alone does not invalidate the formula.

Check Your Understanding

For each finite discrete loss distribution, find the requested VaR and TVaR, using the quantile-integral definition. Give monetary answers to the nearest cent when needed.

A hypothetical property loss $X$, in dollars, has probabilities $P(X=0)=0.30$, $P(X=30)=0.20$, $P(X=70)=0.30$, and $P(X=120)=0.20$. Find $\operatorname{VaR}_{0.50}(X)$ and $\operatorname{TVaR}_{0.50}(X)$. (Answer: VaR = $30; TVaR = $90.)
A hypothetical equipment loss $X$, in thousands of dollars, takes values 10, 40, 80, and 160 with respective probabilities $0.10, 0.40, 0.25,$ and $0.25$. Find $\operatorname{VaR}_{0.75}(X)$ and $\operatorname{TVaR}_{0.75}(X)$. (Answer: VaR = 80 thousand dollars; TVaR = 160 thousand dollars.)
A hypothetical benefit loss $X$, in dollars, takes values 0, 25, 55, and 95 with respective probabilities $0.20, 0.25, 0.15,$ and $0.40$. Find $\operatorname{VaR}_{0.45}(X)$ and $\operatorname{TVaR}_{0.45}(X)$. (Answer: VaR = $25; TVaR = $84.09.)
A hypothetical repair loss $X$, in dollars, takes values 5, 15, 35, and 75 with respective probabilities $0.20, 0.40, 0.10,$ and $0.30$. Find $\operatorname{VaR}_{0.60}(X)$ and $\operatorname{TVaR}_{0.60}(X)$. (Answer: VaR = $15; TVaR = $65.)
A hypothetical claim loss $X$, in dollars, has $P(X=0)=0.20$, $P(X=40)=0.50$, and $P(X=100)=0.30$. At $p=0.50$, is the strict upper-tail sum divided by $1-p$ sufficient to calculate quantile-integral TVaR? Find VaR and TVaR. (Answer: No; VaR = $40, $F_X(40)=0.70>0.50$, and TVaR = $76.)

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