Tail Value-at-Risk for Discrete Losses

Why Discrete Tail Risk Needs a Fractional Threshold

For a continuous loss distribution, a particular loss value usually has probability zero. The upper tail beyond a value-at-risk threshold can therefore be selected without deciding how much probability to take at the threshold itself. A discrete loss can have positive probability at that value. If the desired upper tail is smaller than the probability at and above the threshold, including the entire threshold atom takes too much probability; excluding it takes too little.

The discrete tail value-at-risk calculation resolves this by including only the fraction of the threshold atom needed to make the selected tail probability exactly $1-\alpha$, where $\alpha$ is the confidence level. This is the quantile-based definition of TVaR used here. It need not equal the conditional mean given $X\geq \operatorname{VaR}_\alpha(X)$ or given $X>\operatorname{VaR}_\alpha(X)$.

Definition: For a loss $X$ with distribution function $F_X(x)=P(X\leq x)$ and $0<\alpha<1$, define $\operatorname{VaR}_\alpha(X)=\inf\{x:F_X(x)\geq\alpha\}$. Write $q=\operatorname{VaR}_\alpha(X)$. For a discrete loss, the quantile-based tail value-at-risk is the mean of the upper $1-\alpha$ fraction of the loss distribution, taking only the required fraction of probability at $q$.

The quantile convention matters: $q$ is the smallest loss value at which the cumulative probability reaches or exceeds $\alpha$. For a discrete distribution with positive probability at $q$, this means $F_X(q-)<\alpha\leq F_X(q)$, where $F_X(q-)=P(X<q)$. The threshold atom has probability $P(X=q)=F_X(q)-F_X(q-)$.

The Fractional-Atom Formula

The probability of a loss strictly above $q$ is $1-F_X(q)$. The desired upper tail has probability $1-\alpha$, so the additional probability to select from the threshold atom is

F_X(q)-\alpha.

This amount is nonnegative and no greater than $P(X=q)$. It can also be expressed as a fraction of the threshold atom: define $\gamma=(F_X(q)-\alpha)/P(X=q)$. Then $0\leq\gamma\leq1$, and the tail selection consists of all outcomes above $q$, plus fraction $\gamma$ of the outcomes equal to $q$.

Theorem: If $P(X=q)>0$, where $q=\operatorname{VaR}_\alpha(X)$, then

\operatorname{TVaR}_\alpha(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\bigl(F_X(q)-\alpha\bigr)} {1-\alpha}.

Equivalently, using the fraction $\gamma$ of the threshold atom,

\operatorname{TVaR}_\alpha(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\,\gamma P(X=q)} {1-\alpha}.

Here $\mathbf{1}_{\{X>q\}}$ equals 1 when $X>q$ and 0 otherwise. If $F_X(q)=\alpha$, then $\gamma=0$: no probability from the atom at $q$ is needed. If the quantile lies strictly inside a cumulative probability jump, then $0<\gamma<1$.

Why the Formula Works

The formula follows by constructing a selected tail of exactly the required probability. Give full weight to each outcome above $q$, and weight $\gamma$ to an outcome equal to $q$. The total selected probability is

P(X>q)+\gamma P(X=q) =1-F_X(q)+F_X(q)-\alpha =1-\alpha.

The total loss contributed by that selection is $E[X\,\mathbf{1}_{\{X>q\}}]+q\gamma P(X=q)$, because every selected threshold outcome has loss $q$. Dividing this selected loss total by its selected probability $1-\alpha$ gives the mean loss in the upper tail. Since $\gamma P(X=q)=F_X(q)-\alpha$, this is the stated formula. The derivation also provides a useful check: if the selected tail probability is not $1-\alpha$, the threshold probability has been handled incorrectly.

Find the quantile.
Accumulate probabilities from the smallest loss upward until the cumulative probability first reaches $\alpha$. That loss is $q$.

Find the tail probability needed at $q$.
Calculate $F_X(q)-\alpha$. This is the probability to include from the threshold atom, not the fraction of the atom itself.

Calculate the selected loss total.
Add the expected loss contributed by outcomes strictly above $q$ to $q\bigl(F_X(q)-\alpha\bigr)$.

Divide by the tail probability.
Divide the selected loss total by $1-\alpha$, then check that the selected probability is $1-\alpha$.

Worked Calculations

Worked Example: A Threshold Atom Half Included

A hypothetical one-period policy has loss $X$, in dollars, with $P(X=0)=0.65$, $P(X=100)=0.20$, $P(X=400)=0.10$, and $P(X=1{,}200)=0.05$. Calculate $\operatorname{TVaR}_{0.90}(X)$. This is a one-period loss model; no interest or mortality assumption is involved.

Method: identify the quantile, then select exactly 10% probability from the upper tail. The cumulative distribution is 0.65 at 0, 0.85 at 100, 0.95 at 400, and 1.00 at 1,200. Since it first reaches 0.90 at 400, $q=\operatorname{VaR}_{0.90}(X)=400$.

The probability strictly above 400 is 0.05. The required tail probability is $1-0.90=0.10$, so the probability to take from the atom at 400 is $0.10-0.05=0.05$. Equivalently, $F_X(400)-\alpha=0.95-0.90=0.05$. The atom has probability 0.10, so this selects half of it.

The loss total from above the threshold is $0.05(1{,}200)=60$. The loss total from the selected part at the threshold is $0.05(400)=20$. Thus the total selected loss is $60+20=80$, and

\operatorname{TVaR}_{0.90}(X)=\frac{80}{0.10}=\$800.00.

Check: the selected probability is $0.05+0.05=0.10$, as required. The calculation avoids the common mistake of including all 10% of the threshold atom.

Worked Example: A High Confidence Level and a Small Tail

A hypothetical equipment-loss model has $P(X=0)=0.50$, $P(X=200)=0.25$, $P(X=600)=0.15$, $P(X=1{,}000)=0.08$, and $P(X=3{,}000)=0.02$, with amounts in dollars. Calculate $\operatorname{TVaR}_{0.95}(X)$. No interest or mortality assumption is used.

Method: take all losses above the 95th-percentile quantile and the required portion at the quantile. The cumulative probabilities are 0.50, 0.75, 0.90, 0.98, and 1.00 at the listed losses. Therefore $q=1{,}000$, since the cumulative probability first reaches 0.95 there.

The probability strictly above 1,000 is 0.02. The required tail probability is $1-0.95=0.05$, so the probability included at 1,000 is $0.05-0.02=0.03$. This agrees with $F_X(q)-\alpha=0.98-0.95=0.03$; it is $0.03/0.08=0.375$ of the atom at 1,000.

The loss total above 1,000 is $0.02(3{,}000)=60$, and the selected threshold loss total is $0.03(1{,}000)=30$. The selected total is $60+30=90$. Therefore

\operatorname{TVaR}_{0.95}(X)=\frac{90}{0.05}=\$1{,}800.00.

Check: the selected probability is $0.02+0.03=0.05$. The threshold fraction is 0.375, not 0.03; 0.03 is the probability mass selected there. Keeping those two quantities distinct avoids a common error.

Worked Example: Several Outcomes Above the Quantile

A hypothetical cargo-claim model has loss distribution $P(X=0)=0.40$, $P(X=50)=0.25$, $P(X=150)=0.20$, $P(X=400)=0.10$, and $P(X=900)=0.05$, in dollars. Find $\operatorname{TVaR}_{0.80}(X)$. No interest or mortality assumption is involved.

Method: sum the weighted losses at every outcome above the quantile, then add the selected threshold portion. The cumulative probabilities are 0.40 at 0, 0.65 at 50, 0.85 at 150, 0.95 at 400, and 1.00 at 900. Thus $q=150$.

The desired upper tail has probability $1-0.80=0.20$. Losses strictly above 150 have probability $0.10+0.05=0.15$, so the mass needed at 150 is 0.05. This is also $F_X(150)-\alpha=0.85-0.80=0.05$, one quarter of the 0.20 atom.

The weighted loss total strictly above the quantile is $0.10(400)+0.05(900)=40+45=85$. The selected threshold contributes $0.05(150)=7.50$. The selected loss total is $85+7.50=92.50$, giving

\operatorname{TVaR}_{0.80}(X)=\frac{92.50}{0.20}=\$462.50.

Check: probability $0.10+0.05+0.05=0.20$ is selected, and the result lies between the quantile 150 and the largest possible loss 900. This calculation avoids omitting one of the outcomes strictly above the quantile.

Common Exam Trap: Conditioning on the Wrong Event

For a discrete loss, $E[X\mid X\geq q]$ includes the entire threshold atom, while the fractional-tail definition generally includes only part of it. Conversely, $E[X\mid X>q]$ excludes the threshold atom completely. Neither conditional mean is generally the quantile-based TVaR when the quantile falls within an atom.

For the first worked example, the incorrect calculation using all outcomes at or above 400 is

E[X\mid X\geq400] =\frac{0.10(400)+0.05(1{,}200)}{0.10+0.05} =\frac{100}{0.15} =\$666.67.

That calculation selects probability 0.15, although the required tail probability is 0.10. The opposite mistake, using only $X>400$, gives $E[X\mid X>400]=1{,}200$, but selects only probability 0.05. The correction is to select 0.05 probability at 400 and all 0.05 probability at 1,200, yielding $\$800.00$.

Key takeaway: At a discrete value-at-risk threshold $q$, include all probability above $q$ and exactly $F_X(q)-\alpha$ probability at $q$. Divide the selected loss total by $1-\alpha$. Always check both the selected probability and the threshold mass used.

Check Your Understanding

For each hypothetical one-period loss distribution, calculate the requested TVaR using the fractional threshold atom. Amounts are in dollars; no interest or mortality assumption applies.

$P(X=0)=0.60$, $P(X=100)=0.25$, $P(X=500)=0.10$, and $P(X=1{,}500)=0.05$. Find $\operatorname{TVaR}_{0.90}(X)$. (Answer: $\$1{,}000.00$)
$P(X=20)=0.20$, $P(X=80)=0.40$, $P(X=200)=0.25$, and $P(X=700)=0.15$. Find $\operatorname{TVaR}_{0.80}(X)$. (Answer: $\$575.00$)
$P(X=0)=0.40$, $P(X=250)=0.30$, $P(X=800)=0.20$, $P(X=1{,}600)=0.07$, and $P(X=4{,}000)=0.03$. Find $\operatorname{TVaR}_{0.95}(X)$. (Answer: $\$3{,}040.00$)
$P(X=0)=0.50$, $P(X=100)=0.20$, $P(X=300)=0.18$, $P(X=1{,}000)=0.09$, and $P(X=5{,}000)=0.03$. Find $\operatorname{TVaR}_{0.90}(X)$. (Answer: $\$2{,}200.00$)
$P(X=100)=0.90$, $P(X=500)=0.08$, $P(X=2{,}000)=0.015$, and $P(X=10{,}000)=0.005$. Find $\operatorname{TVaR}_{0.99}(X)$. (Answer: $\$6{,}000.00$)

What You'll Learn