Checking Quantile Conventions and Confidence Levels

Why Quantile and Tail Conventions Need an Audit

A reported risk measure can look plausible while using the wrong confidence level, the wrong tail probability, or the wrong share of probability at a discrete quantile. These errors are especially easy to make when a loss distribution has an atom: the confidence level may fall partway through the probability accumulated at one loss amount. A reliable calculation therefore checks the cumulative distribution on both sides of the reported quantile and verifies that the TVaR tail contains exactly the probability required by the confidence level.

Throughout this tutorial, losses are represented by a random variable $X$, and higher values are worse. The confidence level is $c$, while the probability in the upper tail is $1-c$. The convention for TVaR used here is the upper-tail expected shortfall, defined by averaging quantiles above confidence level $c$. This convention handles atoms by including only the necessary fraction of the mass at the quantile.

Definition: For $0<c<1$, the lower $c$-quantile, or value-at-risk at confidence level $c$, is $$ \operatorname{VaR}_c(X)=q_c=\inf\{x:F_X(x)\geq c\}, $$ where $F_X(x)=P(X\leq x)$. Write $F_X(q_c-)=P(X<q_c)$ for the probability strictly below the quantile. At a finite lower quantile, $$ F_X(q_c-)\leq c\leq F_X(q_c). $$ The strict inequality $F_X(q_c-)<c$ holds when the quantile is at a jump that crosses $c$, but need not hold for every distribution.

The restriction $0<c<1$ matters. At $c=0$, the set in the infimum definition includes every real number, so this definition would not generally produce a finite quantile. The standard confidence levels used for VaR and TVaR lie strictly between zero and one.

The inequalities in the definition give a direct audit. If a reported value is $q$, compute $P(X<q)$ and $P(X\leq q)$. A valid lower quantile at level $c$ must have the first probability no greater than $c$, and the second no less than $c$. For a discrete distribution with a jump crossing $c$, the first probability is strictly less than $c$. For a continuous distribution, it can equal $c$.

TVaR and the Probability Assigned at the Quantile

Define TVaR as the average of the upper $1-c$ fraction of the loss distribution, using quantiles to specify exactly what “fraction” means. If $X$ is integrable and the quantiles are finite over the relevant range, this definition is

\operatorname{TVaR}_c(X)=\frac{1}{1-c}\int_c^1 \operatorname{VaR}_u(X)\,du.

When $q=\operatorname{VaR}_c(X)$, the same definition can be audited with the tail formula below. The quantity $F_X(q)-c$ is the probability mass at $q$ that must be included in the upper tail. If $F_X(q)=c$, none of the mass at $q$ is needed; if $F_X(q)>c$, only part of the mass at $q$ is included.

Theorem: For an integrable loss $X$, a finite lower quantile $q=\operatorname{VaR}_c(X)$, and $0<c<1$, $$ \operatorname{TVaR}_c(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+q\bigl(F_X(q)-c\bigr)}{1-c}. $$ The numerator uses all probability strictly above $q$, plus the amount $F_X(q)-c$ of probability assigned to the value $q$.

Here $\mathbf{1}_{\{X>q\}}$ is one when $X>q$ and zero otherwise. To see why the fractional mass is $F_X(q)-c$, note that the mass strictly above $q$ is $1-F_X(q)$. The required upper-tail probability is $1-c$. The difference is

(1-c)-\bigl(1-F_X(q)\bigr)=F_X(q)-c.

This difference is nonnegative because $F_X(q)\geq c$. It is no greater than the atom at $q$, since $F_X(q-)\leq c$. Thus the selected mass at $q$ is available and the selected total probability is exactly $1-c$.

A second form is useful as a check. Since $E[(X-q)_+]=E[X\mathbf{1}_{\{X>q\}}]-qP(X>q)$, rearranging gives

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

This identity also shows that the tail average is based on the excess above the quantile, added to the quantile itself. The formula remains valid when losses can be negative, provided the stated integrability condition holds.

Justification from the Quantile Integral

The fractional-mass formula follows directly from the quantile definition of TVaR. Let $Q(u)=\operatorname{VaR}_u(X)$. As $u$ ranges from zero to one, $Q(u)$ assigns values according to the distribution of $X$: an atom at $q$ occupies the interval of probability levels from $F_X(q-)$ to $F_X(q)$. Since $F_X(q-)\leq c\leq F_X(q)$, the part of that interval above $c$ has length $F_X(q)-c$.

Split the quantile integral at the quantile’s cumulative probability.
The levels from $c$ to $F_X(q)$ have quantile $q$, contributing $q(F_X(q)-c)$.

Account for levels above $F_X(q)$.
Those levels correspond to values strictly greater than $q$. Their integral is $E[X\mathbf{1}_{\{X>q\}}]$, because they include the full probability of each value above $q$.

Divide by the selected tail probability.
The total length of the integration interval is $1-c$. Dividing the two contributions by $1-c$ gives the stated TVaR formula.

For a continuous distribution at $q$, often $F_X(q)=c$, so the quantile contributes no separate atom term. For a discrete distribution where $F_X(q-)<c<F_X(q)$, the term is positive and corresponds to only part of the probability at $q$.

Worked Example: A Quantile Inside a Discrete Atom

A hypothetical portfolio loss $X$, measured in thousands of dollars, is $0$ with probability $0.80$, $100$ with probability $0.15$, and $300$ with probability $0.05$. Find $\operatorname{VaR}_{0.90}(X)$ and $\operatorname{TVaR}_{0.90}(X)$. State how the upper-tail probability is assembled.

Method. Calculate the cumulative probabilities at and immediately below each candidate loss, then use the fractional-mass formula. This avoids treating the entire atom at the quantile as part of the tail.

The cumulative probabilities are $F_X(0)=0.80$ and $F_X(100)=0.80+0.15=0.95$. Since $0.80<0.90\leq0.95$, the lower $0.90$-quantile is $q=100$ thousand dollars. In particular, $P(X<100)=0.80$, while $P(X\leq100)=0.95$.

The required upper-tail probability is $1-0.90=0.10$. Losses strictly above $100$ contribute probability $0.05$, so the amount of probability needed at $100$ is $0.10-0.05=0.05$. This also equals $F_X(100)-0.90=0.95-0.90=0.05$.

The numerator of TVaR, in thousands of dollars, is $300(0.05)+100(0.05)=15+5=20$. Therefore $\operatorname{TVaR}_{0.90}(X)=20/0.10=200$ thousand dollars.

Check. The selected tail contains probability $0.05+0.05=0.10$, and its average loss is $200$ thousand dollars. The result is consistent with the selected half of the probability at $100$ and all of the probability at $300$.

Boundary Checks and Continuous Distributions

A frequent audit error is to insist that $P(X<q)<c$ for every lower quantile. The correct general condition is $P(X<q)\leq c$. Equality can occur, for example, when a continuous distribution reaches the confidence level exactly at $q$. In a discrete distribution with a jump crossing $c$, the strict version does apply.

Worked Example: Equality Below a Continuous Quantile

Suppose a hypothetical loss is uniformly distributed on the interval from $0$ to $200$ thousand dollars. Find the $0.75$-quantile and its TVaR. Explain why the probability strictly below the quantile is not less than the confidence level.

Method. For a uniform distribution on $[0,200]$, $F_X(x)=x/200$ for $0\leq x\leq200$. Solve $F_X(q)=0.75$: $q/200=0.75$, so $q=150$ thousand dollars.

The cumulative probability at $q$ is $F_X(150)=150/200=0.75$. There is no atom at 150, so $P(X<150)=0.75$ as well. Thus $P(X<q)=c$, satisfying the general non-strict inequality; it does not satisfy a strict inequality.

The upper tail is the uniform interval from $150$ to $200$, whose average loss is $(150+200)/2=175$ thousand dollars. Equivalently, the quantile function is $\operatorname{VaR}_u(X)=200u$, and

\operatorname{TVaR}_{0.75}(X) =\frac{1}{0.25}\int_{0.75}^1 200u\,du =4\left[100u^2\right]_{0.75}^1 =4(100-56.25) =175.

Check. The selected tail has probability $0.25$, and its mean is $175$ thousand dollars, greater than its lower endpoint $150$ and less than its upper endpoint $200$.

Conditional Tail Averages Are Not Always TVaR

For a discrete loss, neither $E[X\mid X>q]$ nor $E[X\mid X\geq q]$ necessarily gives the quantile-integral TVaR. The first conditions on all losses strictly above $q$, which may have probability less than $1-c$. The second includes the entire atom at $q$, which may make the selected probability greater than $1-c$. TVaR instead includes precisely the fraction at $q$ needed to make the tail probability equal to $1-c$.

Worked Example: Checking the Tail Mass at a Second Confidence Level

A hypothetical aggregate loss, in thousands of dollars, is $0$ with probability $0.60$, $40$ with probability $0.20$, $100$ with probability $0.15$, and $250$ with probability $0.05$. Calculate $\operatorname{VaR}_{0.70}(X)$ and $\operatorname{TVaR}_{0.70}(X)$, and show why conditioning on $X>q$ gives a different answer.

Method. Locate the first cumulative probability at least $0.70$, then assemble the upper tail to probability $0.30$. This explicitly avoids replacing the required fractional atom by either all or none of that atom.

The cumulative probabilities are $F_X(0)=0.60$, $F_X(40)=0.80$, $F_X(100)=0.95$, and $F_X(250)=1.00$. Since $0.60<0.70\leq0.80$, $q=\operatorname{VaR}_{0.70}(X)=40$ thousand dollars. Here $P(X<q)=0.60$ and $F_X(q)=0.80$.

The required tail probability is $1-0.70=0.30$. Losses greater than $40$ have probability $0.15+0.05=0.20$, so the tail must also include $0.30-0.20=0.10$ of probability at $40$. This matches $F_X(40)-0.70=0.80-0.70=0.10$.

The contribution from losses above $40$ is $100(0.15)+250(0.05)=15+12.5=27.5$. The selected mass at $40$ contributes $40(0.10)=4$. Thus the tail’s total loss amount is $27.5+4=31.5$, and

\operatorname{TVaR}_{0.70}(X)=\frac{31.5}{0.30}=105 \quad\text{thousand dollars}.

Conditioning only on $X>40$ gives $27.5/0.20=137.5$ thousand dollars, but that conditions on a tail probability of $0.20$, not the required $0.30$. Conditioning on $X\geq40$ gives $(8+27.5)/0.40=88.75$ thousand dollars, but includes probability $0.40$, not $0.30$. Neither is the specified TVaR.

Check. The selected probability is $0.10+0.15+0.05=0.30$, and the selected loss amount is $4+15+12.5=31.5$. Their ratio is $105$ thousand dollars.

A Practical Audit Sequence

When reviewing a reported VaR or TVaR, use the same sequence regardless of whether the distribution is discrete, continuous, or mixed:

Identify the confidence convention. Confirm that $c$ is the cumulative probability level and that the corresponding upper-tail probability is $1-c$.
Verify the reported quantile. Check $F_X(q-)\leq c\leq F_X(q)$. If the distribution has a jump crossing $c$, check the stricter left-side inequality $F_X(q-)<c\leq F_X(q)$.
Build the tail probability. Add $P(X>q)=1-F_X(q)$ to the required mass $F_X(q)-c$ at $q$. The result must be $1-c$.
Build the tail loss amount. Add $E[X\mathbf{1}_{\{X>q\}}]$ and $q(F_X(q)-c)$, then divide by $1-c$.
Check units and plausibility. Keep the loss units consistent, and for a distribution with nonnegative losses, confirm that the result reflects an average of exactly the selected upper-tail probability.

The confidence level and tail probability are complements, not interchangeable labels. For example, a $0.95$ confidence level means an upper tail of $0.05$, not an upper tail of $0.95$. A result can be numerically plausible and still be wrong if it averages the wrong fraction of the distribution.

Key takeaway: A sound quantile audit verifies both sides of the CDF at $q$. A sound TVaR audit then uses all losses above $q$, adds exactly $F_X(q)-c$ probability at $q$, and divides by $1-c$.

Check Your Understanding

For each problem, use the lower-quantile convention and the upper-tail TVaR definition given above. Loss amounts are in dollars unless stated otherwise.

A loss is $-20$ with probability $0.40$, $30$ with probability $0.40$, and $90$ with probability $0.20$. Find $\operatorname{VaR}_{0.60}$ and $\operatorname{TVaR}_{0.60}$. (Answer: VaR $=30$; TVaR $=60$.)
A loss is uniform on $[0,500]$. Find the $0.80$-quantile and TVaR at confidence level $0.80$. (Answer: VaR $=400$; TVaR $=450$.)
A loss is $0$ with probability $0.50$, $80$ with probability $0.30$, and $200$ with probability $0.20$. At confidence level $0.65$, how much probability at the quantile is included, and what is TVaR? (Answer: VaR $=80$; included mass at $80$ $=0.15$; TVaR $=148.5714$.)
A loss is $-10$ with probability $0.20$, $50$ with probability $0.30$, and $100$ with probability $0.50$. At confidence level $0.50$, find VaR and TVaR, and state whether the fractional mass at VaR is positive. (Answer: VaR $=50$; TVaR $=100$; fractional mass at VaR $=0$.)
A loss is $0$ with probability $0.70$, $100$ with probability $0.20$, and $400$ with probability $0.10$. At confidence level $0.90$, calculate TVaR and identify the upper-tail probability. (Answer: VaR $=100$; upper-tail probability $=0.10$; TVaR $=400$.)

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