TVaR at an Atom Using Fractional Tail Mass

Why a Quantile Atom Requires a Fractional Tail

For a continuous loss distribution, the probability of a loss being exactly equal to a particular VaR is typically zero. A discrete distribution can assign positive probability to that exact loss. When the confidence level falls inside that probability mass, the upper tail of size $1-\alpha$ contains all losses strictly above VaR and only part of the probability at VaR. Including the whole atom would make the selected tail too large.

TVaR is an average of the worst losses in a specified probability tail. At a discrete quantile atom, it is therefore a probability-weighted average—not an average that gives each distinct loss value the same weight. The calculation depends on the probability selected at the VaR loss as well as the losses and probabilities above it.

Definition: Let $X$ be an integrable loss, and let $0<\alpha<1$ be a confidence level. Define the quantile by $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}$. Under the quantile-integral convention, $\operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du$. For a discrete distribution, this averages the losses in the upper probability tail of size $1-\alpha$, using only the needed fraction of the probability at $\operatorname{VaR}_{\alpha}(X)$.

In this tutorial, $F_X(q)=P(X\leq q)$, and $P(X>q)=1-F_X(q)$. If $q=\operatorname{VaR}_{\alpha}(X)$ is an atom, then $F_X(q-)<\alpha\leq F_X(q)$, where $F_X(q-)=P(X<q)$. The probability at $q$ is $P(X=q)=F_X(q)-F_X(q-)$.

The Fractional-Tail Formula

To form an upper tail of size $1-\alpha$, first include every loss strictly greater than $q$. Those losses contribute probability $1-F_X(q)$. The additional probability needed at $q$ is the difference between the desired tail probability and the probability already above $q$:

(1-\alpha)-P(X>q) =(1-\alpha)-\bigl(1-F_X(q)\bigr) =F_X(q)-\alpha.

This is a probability amount, not the fraction of the atom itself. To find the fraction of the atom included, divide it by $P(X=q)$. The quantile condition ensures $0\leq F_X(q)-\alpha\leq P(X=q)$. If $F_X(q)=\alpha$, no probability at $q$ is needed; if the quantile level falls strictly inside the atom, a positive but partial amount is used.

Theorem: For a discrete loss $X$, let $q=\operatorname{VaR}_{\alpha}(X)$. Then $$ \operatorname{TVaR}_{\alpha}(X) =\frac{E\!\left[X\,\mathbf{1}_{\{X>q\}}\right]+q\bigl(F_X(q)-\alpha\bigr)}{1-\alpha}. $$ For a finite distribution with possible loss values $x_j$ and probabilities $p_j=P(X=x_j)$, the first term in the numerator is $\sum_{x_j>q}x_jp_j$. The formula applies when the expectation exists; it does not mean that the whole atom at $q$ is included.

The following derivation explains why the formula matches the definition of TVaR. As $u$ moves from $\alpha$ up to $F_X(q)$, the quantile $\operatorname{VaR}_u(X)$ equals $q$. The length of this interval is $F_X(q)-\alpha$, so its contribution to the quantile integral is $q(F_X(q)-\alpha)$. Above $F_X(q)$, the quantile takes each higher loss value $x_j$ over an interval of length $p_j$. Those contributions sum to $\sum_{x_j>q}x_jp_j$. Dividing the total integral by $1-\alpha$ gives the stated result.

An equivalent form can be useful when the losses above $q$ are summarized by excesses over $q$:

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

To verify this identity, write the formula’s numerator as $q(F_X(q)-\alpha)+\sum_{x_j>q}x_jp_j$. Add and subtract $q\sum_{x_j>q}p_j$. Since $F_X(q)+P(X>q)=1$, the terms multiplied by $q$ sum to $q(1-\alpha)$; the remaining terms sum to $\sum_{x_j>q}(x_j-q)p_j=E[(X-q)_{+}]$. This alternative still requires the correct quantile $q$ and the same confidence level.

Worked Calculations

Worked Example: A Quantile Inside a Loss Atom

A hypothetical risk portfolio has loss $X$, in dollars, with $P(X=0)=0.3000$, $P(X=40)=0.2500$, $P(X=90)=0.3000$, and $P(X=150)=0.1500$. Calculate $\operatorname{TVaR}_{0.6000}(X)$. State the method and identify the mistake avoided.

Method. Find the quantile, calculate the required probability at that atom, and then divide the selected tail’s weighted loss total by its probability.

Step 1: Check the distribution and find VaR. The probabilities sum to $0.3000+0.2500+0.3000+0.1500=1.0000$. The cumulative probabilities at losses $0,40,90,150$ are $0.3000,0.5500,0.8500,1.0000$, respectively. The first cumulative probability at least $0.6000$ is $0.8500$, so $q=\operatorname{VaR}_{0.6000}(X)=90$.

Step 2: Fill the upper tail. Its required probability is $1-0.6000=0.4000$. Losses strictly above $90$ contribute probability $0.1500$, all at loss $150$. The probability to take at $90$ is $F_X(90)-0.6000=0.8500-0.6000=0.2500$. Thus the selected tail consists of probability $0.2500$ at $90$ and $0.1500$ at $150$, for a total of $0.4000$.

Step 3: Calculate the weighted average. The selected probabilities at $90$ and $150$ are $0.2500$ and $0.1500$, so the weighted average is $\bigl(90(0.2500)+150(0.1500)\bigr)/0.4000=(22.50+22.50)/0.4000=112.50$ dollars.

Check. The selected tail has probability $0.4000$, as required, and its weighted loss total is $45.00$ dollars. Therefore $\operatorname{TVaR}_{0.6000}(X)=\$112.50$.

Common mistake avoided. Including the whole atom at $90$ gives a probability of $0.3000+0.1500=0.4500$, not the required $0.4000$. The average $\bigl(90(0.3000)+150(0.1500)\bigr)/0.4500=\$110.00$ is the average conditional on $X\geq90$, not the requested TVaR. The selected probabilities are unequal, so an unweighted average of the two loss values would also be incorrect.

Worked Example: Finding the Fraction of an Atom

A hypothetical insurer models a payment per claim $X$, in dollars, with $P(X=0)=0.2000$, $P(X=100)=0.4500$, $P(X=300)=0.2500$, and $P(X=700)=0.1000$. Calculate $\operatorname{TVaR}_{0.7500}(X)$, including the fraction of the $300$-dollar atom used.

Method. Use the cumulative distribution to locate the quantile, then compare the required upper-tail probability with the probability strictly above the quantile.

Step 1: Locate the quantile. The probabilities sum to $1.0000$. Cumulative probabilities at $0,100,300,700$ are $0.2000,0.6500,0.9000,1.0000$. Since $0.6500<0.7500\leq0.9000$, $q=\operatorname{VaR}_{0.7500}(X)=300$.

Step 2: Determine the included atom probability. The upper tail must have probability $1-0.7500=0.2500$. Losses strictly above $300$ have probability $0.1000$, leaving $0.2500-0.1000=0.1500$ to take at $300$. The atom at $300$ has probability $0.2500$, so the included fraction of that atom is $0.1500/0.2500=0.6000$.

Step 3: Calculate TVaR. The weighted loss total in the selected tail is $300(0.1500)+700(0.1000)=45.00+70.00=115.00$ dollars. Dividing by the tail probability gives $115.00/0.2500=\$460.00$.

Check and common mistake avoided. The selected probability is $0.1500+0.1000=0.2500$, as required. Taking the entire $300$-dollar atom would produce probability $0.3500$ and would not average the specified upper tail. The correct answer is $\operatorname{TVaR}_{0.7500}(X)=\$460.00$.

Worked Example: A Tail Made Mostly of the VaR Atom

A hypothetical equipment-loss model has $P(X=0)=0.1000$, $P(X=50)=0.3000$, $P(X=120)=0.3500$, $P(X=200)=0.2000$, and $P(X=500)=0.0500$, with losses measured in dollars. Find $\operatorname{TVaR}_{0.8000}(X)$.

Method. Use the fractional-tail formula. This example tests whether the probability at VaR is counted as a probability amount rather than assumed to be either wholly included or wholly excluded.

Step 1: Find VaR. The probabilities sum to $1.0000$. The cumulative probabilities at $0,50,120,200,500$ are $0.1000,0.4000,0.7500,0.9500,1.0000$. Thus $q=\operatorname{VaR}_{0.8000}(X)=200$.

Step 2: Find the tail composition. The target tail probability is $1-0.8000=0.2000$. Losses above $200$ contribute $0.0500$, so the amount taken at $200$ is $F_X(200)-0.8000=0.9500-0.8000=0.1500$. Of the $0.2000$ probability at $200$, the selected fraction is $0.1500/0.2000=0.7500$.

Step 3: Average the selected losses. The selected weighted loss total is $200(0.1500)+500(0.0500)=30.00+25.00=55.00$ dollars. Dividing by $0.2000$ gives $55.00/0.2000=\$275.00$.

Check and common mistake avoided. The tail probability is $0.1500+0.0500=0.2000$. If the full atom at $200$ were included, the probability would be $0.2500$, exceeding the required tail. Therefore $\operatorname{TVaR}_{0.8000}(X)=\$275.00$.

Interpreting the Calculation Carefully

The fractional amount $F_X(q)-\alpha$ is determined by the confidence level, not chosen to make the arithmetic convenient. Its ratio to $P(X=q)$ tells how much of the atom is represented in the tail average. This is a probability-accounting device: it does not assert that a particular fraction of an individual claim or loss is paid.

A useful check is to verify both the selected probability and the weighted-loss numerator. The selected probability must equal $1-\alpha$:

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

The TVaR must also be at least $q$, because every selected loss is at least $q$. If a calculation produces a value below VaR, check whether the tail was assembled in the wrong direction or whether the numerator and denominator use different probability amounts. At a boundary where $F_X(q)=\alpha$, no mass at $q$ is selected; all the tail probability lies strictly above $q$. At a boundary where the chosen confidence level equals a cumulative probability, using the next loss value as the quantile can lead to an incorrect tail.

This fractional-mass procedure is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. It is especially useful for discrete aggregate losses, where a reported confidence level can fall inside a probability jump. The key is to respect the quantile convention, fill exactly $1-\alpha$ probability, and weight each selected loss by its selected probability.

Check Your Understanding

For each finite distribution, find the quantile first, then use the probability-weighted fractional tail. Report monetary answers to the nearest cent.

A hypothetical loss $X$ takes values $0,20,80,200$ with probabilities $0.4000,0.3000,0.2000,0.1000$, respectively. Find $\operatorname{TVaR}_{0.6000}(X)$. (Answer: $\$95.00$)
A hypothetical claim payment takes values $10,50,150,400$ with probabilities $0.2000,0.5000,0.2000,0.1000$, respectively. Find $\operatorname{TVaR}_{0.6500}(X)$. (Answer: $\$207.14$)
A hypothetical asset loss takes values $0,100,250,600$ with probabilities $0.2500,0.3000,0.2500,0.2000$, respectively. Find $\operatorname{TVaR}_{0.7000}(X)$. (Answer: $\$483.33$)
A hypothetical portfolio loss takes values $0,40,100,250$ with probabilities $0.1000,0.4000,0.3000,0.2000$, respectively. Calculate TVaR at confidence levels $0.7500$ and $0.8000$. (Answer: $\$220.00$ at $0.7500$; $\$250.00$ at $0.8000$)
A hypothetical loss $X$ takes values $0,100,300,1000$ with probabilities $0.5000,0.2000,0.2000,0.1000$, respectively. At confidence level $0.8500$, find both the fraction of the atom at VaR included in the tail and $\operatorname{TVaR}_{0.8500}(X)$. (Answer: $0.2500$ of the atom at $300$; $\$766.67$)

What You'll Learn