Diagnosing Atom and Tail-Probability Errors

Why Atoms Make Tail Calculations Tricky

For a continuous loss distribution, the probability of any single loss amount is zero. At a chosen confidence level, the VaR threshold therefore separates the lower and upper parts of the distribution without leaving a positive probability concentrated exactly at the threshold. A discrete distribution can behave differently: several outcomes may have the same loss, so a positive share of probability can sit at VaR. That probability must be handled carefully when calculating TVaR.

The central calculation issue is that the upper tail used for TVaR has a specified probability, $1-\alpha$. If losses strictly greater than VaR do not make up that entire probability, the remaining share comes from losses equal to VaR. Counting all losses equal to VaR, or counting none of them, can give the wrong result.

Definition: For a loss $X$ with distribution function $F_X(x)=P(X\leq x)$, use the quantile convention $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\},\qquad 0<\alpha<1. $$ The confidence level is $\alpha$; the corresponding upper-tail probability is $p=1-\alpha$. In this tutorial, TVaR is the average of the losses in the worst $p$ share of the distribution, including only as much probability at VaR as is needed to make the tail probability equal to $p$.

The Boundary-Mass Calculation

Let $q=\operatorname{VaR}_{\alpha}(X)$. Under the stated quantile convention, the probability at or below $q$ is at least $\alpha$, while the probability strictly below $q$ is at most $\alpha$. The probability strictly above $q$ is $1-F_X(q)$. If this is less than $1-\alpha$, the upper tail must include part of the probability at $q$.

Theorem: For an integrable loss $X$, the quantile-integral definition of TVaR is $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du. $$ When the distribution has a probability atom at $q=\operatorname{VaR}_{\alpha}(X)$, an equivalent boundary-mass formula is $$ \operatorname{TVaR}_{\alpha}(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\bigl(F_X(q)-\alpha\bigr)}{1-\alpha}. $$ Here $\mathbf{1}_{\{X>q\}}$ equals 1 when $X>q$, and 0 otherwise. The formula assumes the stated VaR convention and includes just enough probability at $q$ to fill the upper tail.

The formula follows by accounting for the tail probability and its loss total. Losses strictly above $q$ have probability $1-F_X(q)$, so their contribution to the tail loss total is $E[X\,\mathbf{1}_{\{X>q\}}]$. The tail probability still needed is $$ (1-\alpha)-\bigl(1-F_X(q)\bigr)=F_X(q)-\alpha. $$ That amount is taken at loss $q$, contributing $q(F_X(q)-\alpha)$. Adding the two contributions gives the loss total for a tail of probability $1-\alpha$; dividing by that probability gives its average loss. If $F_X(q)=\alpha$, no boundary mass is needed. If $F_X(q)>\alpha$, only part of the atom at $q$ is included.

This is a tail-average definition, not generally the conditional expectation $E[X\mid X>q]$. That conditional expectation averages only losses strictly above $q$; it excludes the boundary probability and uses a different probability denominator. The two quantities can differ substantially when $P(X=q)>0$.

Worked Calculations With Atoms

Worked Example: A Large Atom Below the VaR Threshold

A hypothetical one-year portfolio loss is $X=0$ with probability $0.96$ and $X=\$1{,}000$ with probability $0.04$. Calculate VaR and TVaR at confidence level $\alpha=0.95$. State how the atom at VaR enters the calculation.

Method. First find the smallest loss whose cumulative probability is at least 0.95. Then construct an upper tail of probability $1-\alpha=0.05$.

The cumulative probability at zero is $F_X(0)=0.96$, which is already at least 0.95. Thus $q=\operatorname{VaR}_{0.95}(X)=\$0$. The probability strictly above zero is 0.04, so the worst 0.05 share consists of all 0.04 probability at $\$1{,}000$ and another 0.01 probability at $\$0$.

The tail loss total is $0.04(\$1{,}000)+0.01(\$0)=\$40$. Therefore $$ \operatorname{TVaR}_{0.95}(X)=\frac{\$40}{0.05}=\$800. $$

Check. The tail probabilities add to $0.04+0.01=0.05$, and their loss total is $\$40$, so the tail average is $\$800$. The result is between $\$0$ and $\$1{,}000$, as expected.

Common mistake avoided. Averaging only the losses strictly above VaR gives $\$1{,}000$, but those losses have probability 0.04 rather than 0.05. The missing 0.01 share consists of losses at zero, so the required tail average is lower.

Worked Example: A Tail Probability Smaller Than the Atom at VaR

A hypothetical catastrophe loss is $\$0$ with probability 0.97 and $\$500$ with probability 0.03. Find VaR and TVaR at confidence level 0.95.

Method. Locate the quantile first, then use the boundary-mass formula. Here the confidence level is 0.95, so the upper-tail probability is $1-0.95=0.05$.

Since $F_X(0)=0.97\geq0.95$, $\operatorname{VaR}_{0.95}(X)=\$0$. Losses strictly above VaR have probability 0.03, and their loss contribution is $0.03(\$500)=\$15$. The probability needed at the boundary is $$ F_X(0)-0.95=0.97-0.95=0.02. $$ Its loss contribution is $0.02(\$0)=\$0$. Therefore $$ \operatorname{TVaR}_{0.95}(X)=\frac{\$15+\$0}{0.05}=\$300. $$

Check. The upper tail contains probability $0.03+0.02=0.05$, and its total loss is $\$15$. Dividing by 0.05 gives $\$300$, which lies between the two possible loss amounts.

Common mistake avoided. The probability 0.03 at $\$500$ is not the whole 0.05 tail. The remaining 0.02 probability comes from the atom at $\$0$. In this example it contributes no loss, but it must still be included in the tail denominator.

Worked Example: Strict Exceedance Is Not the Whole TVaR Tail

A hypothetical equipment-loss model assigns probability 0.80 to $\$0$, probability 0.15 to $\$100$, and probability 0.05 to $\$400$. Calculate VaR and TVaR at $\alpha=0.90$, and compare TVaR with the average conditional on strict exceedance of VaR.

Method. Use the cumulative probabilities to locate VaR, then split the required 0.10 tail between losses above VaR and the needed part of the boundary atom.

The cumulative probabilities are $F_X(0)=0.80$, $F_X(100)=0.95$, and $F_X(400)=1$. Since $0.80<0.90\leq0.95$, $q=\operatorname{VaR}_{0.90}(X)=\$100$. The upper-tail probability is 0.10. Losses strictly above $q$ contribute probability 0.05 and expected loss total $0.05(\$400)=\$20$. Since $F_X(q)=0.95$, the boundary probability needed is $0.95-0.90=0.05$, contributing $0.05(\$100)=\$5$. Hence $$ \operatorname{TVaR}_{0.90}(X)=\frac{\$20+\$5}{0.10}=\$250. $$

For comparison, $E[X\mid X>\$100]=\$400$, because every loss strictly above $\$100$ is $\$400$. This is not TVaR: it excludes the needed 0.05 probability at $\$100$. The TVaR tail average is $\$250$, not $\$400$.

Check. The tail probability is $0.05+0.05=0.10$; the tail loss total is $\$20+\$5=\$25$; and $\$25/0.10=\$250$.

Common mistake avoided. Replacing the TVaR tail average with $E[X\mid X>q]$ silently changes both which outcomes are included and the probability used for averaging.

Confidence Level, Tail Probability, and Event Notation

The confidence level and tail probability serve different roles. A confidence level of 0.99 means the upper tail has probability $1-0.99=0.01$, not 0.99. A common error is to use the confidence level as the denominator when averaging the worst outcomes. The denominator for TVaR is the tail probability $1-\alpha$.

Strict and non-strict events also matter. The event $X>q$ excludes losses equal to VaR; $X\geq q$ includes all of them. Neither event alone necessarily has probability $1-\alpha$. The boundary-mass formula avoids guessing which whole event has the desired probability: it uses all losses strictly above $q$, then adds precisely $F_X(q)-\alpha$ probability at $q$.

Worked Example: A One-Percent Tail at a High Confidence Level

A hypothetical risk has loss $\$0$ with probability 0.995 and loss $\$2{,}000$ with probability 0.005. Calculate VaR and TVaR at confidence level 0.99. Explain the effect of using the wrong probability denominator.

Method. Convert the confidence level to its upper-tail probability before calculating TVaR: $1-0.99=0.01$.

Since $F_X(0)=0.995\geq0.99$, $\operatorname{VaR}_{0.99}(X)=\$0$. The strict exceedance probability is 0.005, so the 0.01 tail consists of all 0.005 probability at $\$2{,}000$ and 0.005 probability at $\$0$. The tail loss total is $$ 0.005(\$2{,}000)+0.005(\$0)=\$10. $$ Therefore $$ \operatorname{TVaR}_{0.99}(X)=\frac{\$10}{0.01}=\$1{,}000. $$

Check. The selected tail probability is $0.005+0.005=0.01$, and its average loss is $\$10/0.01=\$1{,}000$.

Common mistake avoided. Dividing by 0.99 would average over the wrong probability share. TVaR at confidence 0.99 averages the worst one percent, not the worst 99 percent.

A Reliable Calculation Routine

For a discrete loss distribution, the following sequence makes the boundary allocation explicit and helps expose errors before they affect the final answer.

Write the confidence and tail probabilities.
Record $\alpha$ and calculate $p=1-\alpha$. Keep them distinct throughout.

Build the cumulative distribution.
Add probabilities in increasing order of loss until the cumulative probability first reaches or exceeds $\alpha$. That loss is $q=\operatorname{VaR}_{\alpha}(X)$ under the convention used here.

Count strict exceedances.
Find $P(X>q)=1-F_X(q)$ and $E[X\mathbf{1}_{\{X>q\}}]$. Do not treat $X>q$ and $X\geq q$ as interchangeable.

Fill the tail at the boundary.
Add probability $F_X(q)-\alpha$ at loss $q$, then divide the total tail loss by $1-\alpha$.

Check the probability and loss total.
Confirm that the selected tail probability is $1-\alpha$. The resulting TVaR should be an average of losses in that tail, not an average over the confidence portion.

This topic matters whenever a risk distribution is discrete or has a jump, including models with fixed claim amounts, limits, or many zero-loss outcomes. Atoms are not a minor technical detail: they determine how much probability at the threshold belongs in the tail average. The boundary-mass calculation makes that allocation explicit, so strict-exceedance probabilities, tail denominators, and reported risk measures can be checked separately.

Key takeaway: First identify $q=\operatorname{VaR}_{\alpha}(X)$, then form a tail of probability $1-\alpha$. Include losses strictly above $q$ and only the required share $F_X(q)-\alpha$ at $q$. Do not substitute $P(X>q)$, $P(X\geq q)$, or $\alpha$ for the tail probability without checking what each quantity represents.

Check Your Understanding

For each discrete model, use the stated quantile convention and include boundary probability as needed. Loss amounts are in dollars.

A loss is $\$0$ with probability 0.94 and $\$100$ with probability 0.06. Find VaR and TVaR at confidence level 0.95. (Answer: VaR = $\$100$; TVaR = $\$100$.)
A loss is $\$0$ with probability 0.97 and $\$400$ with probability 0.03. Find VaR and TVaR at confidence level 0.95. (Answer: VaR = $\$0$; TVaR = $\$240$.)
A loss is $\$0$ with probability 0.85, $\$80$ with probability 0.10, and $\$280$ with probability 0.05. Find VaR and TVaR at confidence level 0.90. (Answer: VaR = $\$80$; TVaR = $\$180$.)
A loss is $\$0$ with probability 0.995 and $\$1{,}000$ with probability 0.005. Find TVaR at confidence level 0.99 and state the tail probability used. (Answer: TVaR = $\$500$; tail probability = 0.01.)
A loss is uniformly distributed between $\$0$ and $\$100$. At confidence level 0.90, find VaR and TVaR. (Answer: VaR = $\$90$; TVaR = $\$95$.)

What You'll Learn