Loss Quantiles and the Definition of VaR

Why VaR Is a Quantile

A risk measure needs a precise answer to a practical question: what loss threshold is exceeded only with a specified probability? Value at Risk (VaR) answers this by locating a quantile of the loss distribution. The definition matters because a distribution can have jumps or flat regions, so informal descriptions such as “the loss at the 90th percentile” may not determine a unique threshold unless the quantile convention is stated.

Let $X$ be a real-valued loss over a specified period, and let $F_X(x)=P(X\le x)$ be its cumulative distribution function (CDF). For a confidence level $\alpha$ with $0<\alpha<1$, this tutorial uses the lower generalized inverse convention. This is a distributional definition; it does not prescribe a particular method for estimating a quantile from observed data.

Definition: The loss VaR at confidence level $\alpha$ is the lower $\alpha$-quantile of $X$:

\operatorname{VaR}_{\alpha}(X) = \inf\{x\in\mathbb{R}:F_X(x)\ge \alpha\}.

Here, $\inf$ means the greatest lower bound of the set of thresholds whose cumulative probability is at least $\alpha$. The confidence level is $\alpha$; the corresponding tail probability is $1-\alpha$.

For example, $\operatorname{VaR}_{0.95}(X)$ is the threshold at which the cumulative probability first reaches or exceeds 95%. It is not the 5th percentile of losses: 5% is the probability in the upper tail, not the cumulative probability used to locate the loss threshold. VaR is expressed in the same units as $X$, such as dollars or thousands of dollars.

From the Quantile Definition to the CDF Condition

The CDF collects the probabilities of losses up to each candidate threshold. Thus, testing whether a threshold $x$ has reached a confidence level $\alpha$ means testing whether $P(X\le x)\ge\alpha$. The definition chooses the smallest such threshold. The following justification also explains what can be concluded when the CDF has a jump.

Write the event probability.
By definition, $F_X(x)=P(X\le x)$. Therefore, the thresholds that qualify are exactly those for which $P(X\le x)\ge\alpha$.

Select the first qualifying threshold.
The lower generalized inverse takes the infimum of that set, so $\operatorname{VaR}_{\alpha}(X)=\inf\{x:P(X\le x)\ge\alpha\}$.

Read the probability bounds.
For a real-valued loss distribution, the CDF is nondecreasing and right-continuous. At $q=\operatorname{VaR}_{\alpha}(X)$, these properties give $F_X(q)\ge\alpha$, while $F_X(x)<\alpha$ for every $x<q$. Hence $P(X\le q)\ge\alpha$ and $P(X>q)\le 1-\alpha$.

\operatorname{VaR}_{\alpha}(X) = \inf\{x:F_X(x)\ge\alpha\}, \qquad F_X(\operatorname{VaR}_{\alpha}(X))\ge\alpha.

The inequality can be strict. If the loss distribution has a positive probability at the VaR threshold, the CDF can jump over $\alpha$ rather than equal it. In that case, the probability of exceeding VaR is less than $1-\alpha$. This does not contradict the definition: VaR is the first threshold to reach the required cumulative probability, not a threshold that must split the probability into two exact portions.

A continuous, strictly increasing CDF is a special case. When a solution $q$ to $F_X(q)=\alpha$ exists and the CDF is strictly increasing there, $q$ is the unique threshold and equals $\operatorname{VaR}_{\alpha}(X)$. That equality should not be assumed for every loss distribution.

Worked Examples

Worked Example: Finding a VaR Threshold from Loss Outcomes

A hypothetical insurer models one-year aggregate losses, in dollars, as follows: a loss of $0 occurs with probability 0.46, a loss of $50,000 with probability 0.29, a loss of $160,000 with probability 0.18, and a loss of $500,000 with probability 0.07. Find the 90% VaR and check the CDF probability at that threshold. Use the lower generalized inverse convention.

Method. Order the possible losses from smallest to largest and add their probabilities. VaR is the first loss at which the cumulative probability is at least 0.90.

At $0, the cumulative probability is $0.46$. At $50,000, it is $0.46+0.29=0.75$. At $160,000, it is $0.75+0.18=0.93$. At $500,000, it is $0.93+0.07=1.00$.

The first cumulative probability at or above 0.90 is 0.93, reached at $160,000. Therefore, $\operatorname{VaR}_{0.90}(X)=\$160{,}000$. The check is $P(X\le \$160{,}000)=0.93\ge0.90$, while $P(X\le \$50{,}000)=0.75<0.90$. The exceedance probability at VaR is $P(X>\$160{,}000)=0.18+0.07=0.25$, which is at most $1-0.90=0.10$? No: this arithmetic signals an error in interpreting the exceedance event. The losses greater than $160,000 include only the $500,000 outcome, so $P(X>\$160{,}000)=0.07\le0.10$. The $160,000 outcome itself is not an exceedance.

Check. The CDF at the threshold is 0.93 and the strict exceedance probability is 0.07; these sum to 1.00. The method avoids treating losses equal to VaR as losses that exceed VaR.

Worked Example: A VaR Quantile in a Mixed Loss Model

A hypothetical risk model assigns probability 0.25 to a loss of $0. With probability 0.75, the loss is uniformly distributed from $20,000 to $100,000. Find the 70% VaR. Report the answer in dollars and verify the CDF value.

Method. First write the CDF in the interval that contains the required quantile. For $20{,}000\le x\le100{,}000$, the probability of a loss no greater than $x$ consists of the probability mass at zero plus the probability of a uniform loss falling between $20,000 and $x$:

F_X(x) = 0.25+0.75\left(\frac{x-20{,}000}{100{,}000-20{,}000}\right).

Set this equal to the target cumulative probability, 0.70, and solve:

0.25+0.75\left(\frac{x-20{,}000}{80{,}000}\right)=0.70, $$ $$ 0.75\left(\frac{x-20{,}000}{80{,}000}\right)=0.45, \qquad \frac{x-20{,}000}{80{,}000}=0.60, $$ $$ x-20{,}000=48{,}000, \qquad x=\$68{,}000.

Thus, $\operatorname{VaR}_{0.70}(X)=\$68{,}000$. Substitution gives $F_X(68{,}000)=0.25+0.75(48{,}000/80{,}000)=0.25+0.45=0.70$. The answer lies between $20,000 and $100,000, as required for the CDF formula used.

Check. The CDF reaches exactly 0.70 at $68,000, so the probability of a loss greater than this threshold is 0.30. The method avoids applying the 70% target to the conditional uniform component alone; the 0.25 probability at zero is part of the unconditional CDF.

Worked Example: Applying the Infimum Convention to a Piecewise CDF

A hypothetical annual loss $X$ is measured in thousands of dollars. Its CDF is specified by

F_X(x)= \begin{cases} 0, & x<0,\\ 0.10, & 0\le x<20,\\ 0.10+0.02(x-20), & 20\le x<50,\\ 0.70, & 50\le x<80,\\ 0.70+0.015(x-80), & 80\le x\le100,\\ 1, & x>100. \end{cases}

Find the 85% VaR in dollars. Explain why the flat part of the CDF does not make the answer ambiguous.

Method. Find the first section in which the CDF can reach 0.85. The CDF is 0.70 from $x=50$ through values below 80, so the target is not reached there. In the final increasing section, solve:

0.70+0.015(x-80)=0.85, $$ $$ 0.015(x-80)=0.15, \qquad x-80=10, \qquad x=90.

Since $x$ is in thousands of dollars, the result is $\operatorname{VaR}_{0.85}(X)=\$90{,}000$. Substitution confirms $F_X(90)=0.70+0.015(10)=0.85$. For every $x<90$, the CDF is at most 0.70 before the final section, or is less than 0.85 within that section. Thus, no smaller threshold qualifies.

Check. The CDF is flat at 0.70 over part of the loss range, but the 0.85 target is reached only at $90,000. The definition selects the first threshold meeting the target, rather than an arbitrary point on a flat segment.

Common Exam Trap: Confusing Confidence and Tail Probability

The confidence level is the probability accumulated at or below the VaR threshold. The tail probability is the probability strictly above it. If a question asks for 95% VaR, use $\alpha=0.95$ in the CDF condition; the associated upper-tail probability is $0.05$. Do not use 0.05 as the CDF target.

For the loss distribution in the first example, using 0.10 as the CDF target for a requested 90% VaR would produce $0, because $F_X(0)=0.46\ge0.10$. That is the 10% quantile under the stated convention, not the 90% VaR. The correct threshold is $160,000 because $F_X(\$160{,}000)=0.93\ge0.90$, and the CDF at the next smaller possible loss, $50,000, is only 0.75.

Interpreting the Definition Correctly

VaR is a threshold, not a description of the amount of loss beyond that threshold. At confidence level $\alpha$, the definition ensures that the probability of a loss no greater than VaR is at least $\alpha$, and the probability of a loss strictly greater than VaR is no more than $1-\alpha$. With a jump in the CDF, those probabilities need not equal $\alpha$ and $1-\alpha$ exactly.

When applying the definition, keep three checks in view:

Use the stated convention. Here, VaR is the lower generalized inverse, $\inf\{x:F_X(x)\ge\alpha\}$.
Use the confidence level as the CDF target. The upper-tail probability is $1-\alpha$, not the target in the CDF inequality.
Verify minimality. Check that the threshold reaches the target and that a smaller threshold does not.

These checks make the definition work for distributions with atoms, continuous sections, or flat regions. A useful final check is to distinguish $P(X\le q)$, which includes losses equal to the threshold, from $P(X>q)$, which counts only strict exceedances.

Check Your Understanding

For each problem, use the lower generalized inverse convention and keep the confidence level distinct from the upper-tail probability.

A hypothetical loss is $0 with probability 0.30, $40,000 with probability 0.45, and $250,000 with probability 0.25. Find the 70% VaR and state the CDF probability at that threshold. (Answer: $40,000; CDF probability 0.75)
A hypothetical loss is zero with probability 0.20 and otherwise is uniformly distributed from $10,000 to $50,000. Find the 60% VaR. (Answer: $30,000)
A loss CDF satisfies $F_X(x)=0.40$ for $5\le x<15$, then $F_X(x)=0.40+0.04(x-15)$ for $15\le x\le30$. Find the 80% VaR, with $x$ in thousands of dollars. (Answer: $25,000)
A hypothetical loss takes values $0, $30,000, $90,000, and $300,000 with probabilities 0.15, 0.50, 0.25, and 0.10, respectively. Find the 90% VaR and its strict exceedance probability. (Answer: $90,000; 0.10)
For a loss CDF, $F_X(70{,}000)=0.82$, $F_X(80{,}000)=0.91$, and every $x<70{,}000$ has $F_X(x)<0.82$. What is the 85% VaR if the CDF is nondecreasing and $F_X(x)<0.85$ for every $x<80{,}000$? (Answer: $80,000)

What You'll Learn