Discrete Losses And Fractional VaR-Mass Treatment

Why Discrete Losses Need Fractional VaR Mass

For a continuous loss distribution, the probability of a loss being exactly equal to VaR is zero. A discrete loss distribution can assign positive probability to that exact value. The upper tail needed for TVaR may therefore contain all losses above VaR and only part of the probability mass at VaR. Including all of that mass, or none of it, can give the wrong tail average.

Throughout this tutorial, losses are measured at one valuation date, with no discounting. Assume $X$ is an integrable loss, and fix a confidence level $0<\alpha<1$. We use the lower-quantile convention for VaR and the quantile-integral definition of TVaR. This makes the definition precise even when the loss distribution has atoms.

Definition: The loss distribution function is $F_X(x)=P(X\le x)$, and its left limit is $F_X(x-)=P(X<x)$. Define $q=\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge\alpha\}$. Under the quantile-integral convention, $\operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du$. It is the average of the worst $1-\alpha$ fraction of the loss distribution, counting only the fraction of probability at $q$ needed to complete that tail.

The Fraction of Probability at VaR

Write $P(X>q)$ for the probability strictly above VaR. The upper tail must have total probability $1-\alpha$, so the amount of probability to take from the outcome $X=q$ is $\,1-\alpha-P(X>q)$. Since $P(X>q)=1-F_X(q)$, this required amount is $F_X(q)-\alpha$.

The quantile definition gives $F_X(q-)\le\alpha\le F_X(q)$. In particular, the probability strictly below $q$ may equal $\alpha$; it need not be strictly less. The inequalities show that the required mass is valid: it is nonnegative and no larger than the atom at $q$, because $F_X(q)-\alpha\le F_X(q)-F_X(q-)=P(X=q)$.

Theorem: For an integrable loss $X$ and $0<\alpha<1$, with $q=\operatorname{VaR}_{\alpha}(X)$, \[ \operatorname{TVaR}_{\alpha}(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\big(F_X(q)-\alpha\big)}{1-\alpha}. \] The first numerator term sums the probability-weighted losses strictly above $q$; the second adds the VaR loss multiplied by the required fraction of its probability mass.

For a discrete distribution with loss values $x_j$ and probabilities $p_j$, the strict-above term is calculated as $\sum_{x_j>q}x_jp_j$. If $P(X=q)>0$, the fraction of the atom at $q$ included in the tail is $\big(F_X(q)-\alpha\big)/P(X=q)$. This is a fraction of probability, not a claim that identical outcomes have different loss amounts.

Why the Formula Holds

The quantile-integral definition averages quantiles over the interval of probability ranks from $\alpha$ to 1. At ranks up to $F_X(q)$, the quantile is $q$; thus the part of that interval assigned the value $q$ has length $F_X(q)-\alpha$. The remaining ranks, from $F_X(q)$ to 1, correspond to losses strictly above $q$. Their quantile integral is the probability-weighted total $E[X\mathbf{1}_{\{X>q\}}]$.

Adding the contribution from the VaR ranks to the contribution from ranks above VaR gives the numerator. The rank interval has total length $1-\alpha$, so dividing by $1-\alpha$ gives its average:

\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du = q\big(F_X(q)-\alpha\big)+E[X\mathbf{1}_{\{X>q\}}], \qquad \operatorname{TVaR}_{\alpha}(X) = \frac{\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du}{1-\alpha}.

This argument also explains why the same formula works at a boundary where $F_X(q-)=\alpha$. In that case, all probability strictly below $q$ is outside the tail, but $F_X(q)-\alpha=P(X=q)$, so all the mass at $q$ is included.

Applying the Formula

Worked Example: The Cumulative Probability Stops at VaR

A hypothetical portfolio has a one-year loss $X$ with the following distribution. Losses are in thousands of dollars; probabilities sum to 1. Calculate $\operatorname{TVaR}_{0.80}(X)$.

\begin{array}{c|rrrr} x & 0 & 100 & 300 & 1000\\ \hline P(X=x) & 0.50 & 0.30 & 0.15 & 0.05 \end{array}

Method. First find the 80th-percentile loss by accumulating probability from the smallest loss upward. The cumulative probabilities are $0.50$ at 0, $0.80$ at 100, $0.95$ at 300, and $1.00$ at 1000. Therefore $q=\operatorname{VaR}_{0.80}(X)=100$, and $F_X(q)-\alpha=0.80-0.80=0$.

The losses strictly above 100 contribute $300(0.15)=45$ and $1000(0.05)=50$, for a total of $45+50=95$ thousand dollars. The required mass at VaR is zero. The tail probability is $1-0.80=0.20$, so $\operatorname{TVaR}_{0.80}(X)=95/0.20=475$ thousand dollars, or $475,000.

Check. The two outcomes above VaR have total probability $0.15+0.05=0.20$. Their conditional average loss is $95/0.20=475$ thousand dollars, matching the formula. The common mistake avoided here is adding the full probability at 100 even though the upper tail already has exactly the required probability without it.

Worked Example: Taking Part of the VaR Atom

A hypothetical insurance portfolio has annual aggregate losses $X$, in thousands of dollars, with probabilities $0.40$ at 0, $0.25$ at 50, $0.20$ at 200, and $0.15$ at 800. Calculate $\operatorname{TVaR}_{0.75}(X)$.

The cumulative probability is $0.40$ at 0, $0.65$ at 50, $0.85$ at 200, and 1 at 800. Thus $q=\operatorname{VaR}_{0.75}(X)=200$. Since $F_X(q)=0.85$, the probability required from the VaR atom is $0.85-0.75=0.10$. The atom itself has probability $0.20$, so half of its probability mass is included.

The loss strictly above 200 contributes $800(0.15)=120$ thousand dollars. The included VaR mass contributes $200(0.10)=20$ thousand dollars. The total tail loss is $120+20=140$ thousand dollars, and the tail probability is $1-0.75=0.25$. Therefore $\operatorname{TVaR}_{0.75}(X)=140/0.25=560$ thousand dollars, or $560,000.

Check. The selected tail has probability $0.15+0.10=0.25$, as required. Its probability-weighted loss is 140 thousand dollars, so its average is 560 thousand dollars. The common mistake avoided is conditioning on every loss at least 200: that would use probability $0.20+0.15=0.35$ and average $(200(0.20)+800(0.15))/0.35=160/0.35\approx457.14$ thousand dollars. It includes too much of the threshold atom.

Worked Example: Probability Below VaR Equals the Confidence Level

A hypothetical risk measure has loss distribution $P(X=0)=0.50$, $P(X=100)=0.25$, and $P(X=200)=0.25$, with losses in dollars. Calculate $\operatorname{TVaR}_{0.50}(X)$.

The cumulative probabilities at 0, 100, and 200 are $0.50$, $0.75$, and 1.00. Hence $q=\operatorname{VaR}_{0.50}(X)=100$. Here $F_X(q-)=P(X<100)=0.50=\alpha$. This equality is allowed by the quantile definition.

The required amount of probability at 100 is $F_X(q)-\alpha=0.75-0.50=0.25$, the entire atom at 100. The loss strictly above 100 contributes $200(0.25)=50$ dollars; the threshold contributes $100(0.25)=25$ dollars. Divide the total $50+25=75$ by the tail probability $1-0.50=0.50$: $\operatorname{TVaR}_{0.50}(X)=75/0.50=\mathbf{\$150}$.

Check. The selected half of the distribution consists of all the 100 outcomes and all the 200 outcomes, each with probability 0.25. Their average is $(100+200)/2=150$ dollars. The common mistake avoided is asserting $F_X(q-)<\alpha$ as a strict inequality: in this example they are equal, yet the fractional-mass formula remains valid.

Common Mistakes and Practical Checks

The formula is a probability-weighted average over a tail of size $1-\alpha$. Before accepting a result, verify that the selected probabilities add to that size and that the numerator includes only the losses in that selected tail.

Including all of the VaR atom. Use only $F_X(q)-\alpha$ probability at $q$, not automatically $P(X=q)$.
Using only losses strictly above VaR. When that probability is less than $1-\alpha$, the missing probability must come from the VaR atom.
Conditioning on $X\ge q$ or $X>q$ without checking. These conditional averages generally do not equal quantile-integral TVaR when $P(X=q)>0$. They coincide only in cases where the selected conditional event has exactly the required tail probability and loss weighting.
Assuming probability below VaR is strictly less than $\alpha$. The correct statement is $F_X(q-)\le\alpha$. Equality can occur, as in the third example.
Forgetting the denominator. Divide the selected tail’s probability-weighted loss total by $1-\alpha$, not by $P(X\ge q)$ unless those probabilities happen to be equal.

Key takeaway: For a discrete loss, find $q=\operatorname{VaR}_{\alpha}(X)$, include every loss above $q$, then add probability $F_X(q)-\alpha$ at $q$. Divide the resulting probability-weighted loss by $1-\alpha$. The required amount is always between zero and $P(X=q)$, including when $F_X(q-)=\alpha$.

Check Your Understanding

For each discrete loss distribution, use the quantile-integral convention and report TVaR in the units given.

A hypothetical loss takes values 0, 100, and 500 dollars with probabilities 0.60, 0.25, and 0.15. Find $\operatorname{TVaR}_{0.80}(X)$. (Answer: $400.)
A hypothetical portfolio loss, in thousands of dollars, is 0 with probability 0.20, 50 with probability 0.30, 250 with probability 0.30, and 900 with probability 0.20. Find $\operatorname{TVaR}_{0.70}(X)$. (Answer: $683.33 thousand.)
A hypothetical loss, in dollars, is 0 with probability 0.40, 40 with probability 0.20, 120 with probability 0.20, and 600 with probability 0.20. Find $\operatorname{TVaR}_{0.60}(X)$. (Answer: $360.)
A hypothetical loss, in dollars, is 0 with probability 0.50, 80 with probability 0.20, 300 with probability 0.20, and 1000 with probability 0.10. Find $\operatorname{TVaR}_{0.50}(X)$. (Answer: $352.)
A hypothetical loss, in dollars, is 10, 30, 90, or 210, each with probability 0.25. Find $\operatorname{TVaR}_{0.625}(X)$. (Answer: $170.)

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