Introduction
For a discrete loss, a confidence level may fall partway through a jump in cumulative probability. The value-at-risk is the loss amount at that jump, but the probability of a loss strictly above it need not equal the nominal tail probability. Correctly locating the jump and distinguishing “greater than” from “greater than or equal to” are essential when interpreting a capital threshold.
This tutorial focuses on finite or countable discrete loss distributions. It does not require interest discounting or a mortality model: the loss amount and its probability distribution are the relevant inputs. The methods apply to hypothetical insurance losses, operational losses, or other discrete loss amounts.
Finding the Quantile at a Probability Jump
A discrete distribution places positive probability at one or more individual loss amounts. Such a point is called an atom. As the loss threshold increases through an atom, the cumulative probability jumps by the probability assigned to that loss. The value-at-risk is the loss amount where the cumulative probability first reaches or passes the chosen confidence level; it is not an interpolation within the jump.
To calculate it reliably, list the possible loss amounts in increasing order, attach their probabilities, and form cumulative probabilities from the smallest loss upward. Stop at the first cumulative probability that is at least \(\alpha\). A cumulative probability equal to \(\alpha\) qualifies: “at least” includes equality.
Write the distinct possible values of \(X\) from smallest to largest, with their probabilities.
At each loss value, add its probability to the probabilities at all smaller loss values.
The first loss value whose cumulative probability is at least \(\alpha\) is \(\operatorname{VaR}_{\alpha}(X)\).
For a finite discrete distribution, let \(q=\operatorname{VaR}_{\alpha}(X)\). The cumulative probability immediately before \(q\) is less than \(\alpha\), while the cumulative probability at \(q\) is at least \(\alpha\):
This crossing condition explains both how to locate the quantile and why equality matters. If the cumulative probability at a loss amount is exactly \(\alpha\), that loss amount is already a qualifying threshold. The next possible loss amount is not the lower quantile.
Worked Example: Locating a Quantile Inside a Jump
A hypothetical equipment policy has annual loss \(X\) with the following distribution. Find \(\operatorname{VaR}_{0.80}(X)\), then calculate the probabilities of a loss strictly above and at least as large as that amount. No interest discounting or mortality assumption is involved; the distribution is specified directly.
| Loss \(X\) | \(P(X=x)\) | Cumulative probability \(F_X(x)\) |
|---|---|---|
| $0 | 0.4000 | 0.4000 |
| $2,000 | 0.3500 | 0.7500 |
| $9,000 | 0.2000 | 0.9500 |
| $30,000 | 0.0500 | 1.0000 |
Method: cumulative-probability crossing. At $0, the cumulative probability is \(0.4000\). At $2,000, it is \(0.4000+0.3500=0.7500\). At $9,000, it is \(0.7500+0.2000=0.9500\). The first cumulative probability at least \(0.8000\) occurs at $9,000, so \(\operatorname{VaR}_{0.80}(X)=\$9{,}000\).
The strict exceedance probability is \(P(X>\$9{,}000)=P(X=\$30{,}000)=0.0500\). The non-strict exceedance probability includes the mass at $9,000: \(P(X\ge\$9{,}000)=0.2000+0.0500=0.2500\). As a check, \(P(X\le\$9{,}000)=0.9500\), so the strict exceedance probability is \(1-0.9500=0.0500\). The common mistake to avoid is reporting \(0.2500\) as the probability strictly above value-at-risk; that calculation includes equality at the threshold.
Strict and Non-Strict Exceedance
The distinction between strict and non-strict exceedance is especially important at an atom. For any threshold \(q\), the probability of a loss strictly above it is one minus the probability at or below it. The probability of a loss at least as large as it is one minus the probability strictly below it:
Equivalently, \(P(X\ge q)\) includes the probability mass at \(q\), while \(P(X>q)\) excludes it. For a discrete distribution, these two tail probabilities can differ substantially. At a value-at-risk \(q\), the crossing condition gives the following bounds:
The first inequality follows because \(F_X(q)\ge\alpha\). The second follows because \(P(X<q)<\alpha\). Thus, at a discrete quantile, the probability strictly above value-at-risk is no greater than the nominal tail probability \(1-\alpha\), whereas the probability at least as large is greater than that nominal tail probability. The mass at \(q\) accounts for the difference.
Why the Quantile Rule Works
The lower-quantile definition selects the smallest threshold whose cumulative probability covers at least the specified confidence level. In a finite discrete model, the ordered loss values divide the cumulative probability into successive steps. Before the first qualifying step, every available loss threshold has cumulative probability below \(\alpha\); at the first qualifying step, the cumulative probability reaches or exceeds \(\alpha\).
More formally, write the ordered loss values as \(x_1<\cdots<x_n\), and define \(C_j=\sum_{k=1}^{j}P(X=x_k)\), with \(C_0=0\). Let \(j\) be the smallest index such that \(C_j\ge\alpha\). By minimality, \(C_{j-1}<\alpha\), and by the choice of \(j\), \(C_j\ge\alpha\). Since \(P(X<x_j)=C_{j-1}\) and \(P(X\le x_j)=C_j\), it follows that
This proves that \(x_j\) is the first loss amount at which the cumulative probability reaches the target. It also shows why linear interpolation between two loss amounts is not part of the discrete quantile calculation: the distribution assigns probability to the stated loss values, not to interpolated values between them.
Worked Example: When Cumulative Probability Equals the Confidence Level
A hypothetical small-business coverage has annual loss \(X\), measured in thousands of dollars, with this distribution. Find \(\operatorname{VaR}_{0.50}(X)\) and both exceedance probabilities at that threshold. No interest or mortality assumptions apply; the listed values and probabilities define the loss model.
| Loss \(X\) (thousands of dollars) | \(P(X=x)\) | Cumulative probability |
|---|---|---|
| 0 | 0.2500 | 0.2500 |
| 5 | 0.2500 | 0.5000 |
| 12 | 0.3000 | 0.8000 |
| 40 | 0.2000 | 1.0000 |
Method: apply the “at least” rule at an exact boundary. At a loss of 0, the cumulative probability is \(0.2500\), which is below \(0.5000\). At a loss of 5 thousand dollars, the cumulative probability is \(0.2500+0.2500=0.5000\). Since equality meets the criterion, \(\operatorname{VaR}_{0.50}(X)=5\) thousand dollars, or $5,000.
Strictly above $5,000, the possible losses are $12,000 and $40,000. Therefore, \(P(X>\$5{,}000)=0.3000+0.2000=0.5000\). At least $5,000 includes the mass at $5,000 as well: \(P(X\ge\$5{,}000)=0.2500+0.3000+0.2000=0.7500\). The check is \(P(X\le\$5{,}000)=0.5000\), giving strict exceedance \(1-0.5000=0.5000\). The common mistake is to treat an exact cumulative probability of \(0.5000\) as insufficient and move to the next loss value. The definition requires at least the confidence level, not strictly more than it.
Worked Example: How a Small Confidence-Level Change Can Change Value-at-Risk
A hypothetical portfolio has an annual aggregate loss \(X\), in thousands of dollars, with the distribution below. Calculate value-at-risk at confidence levels \(0.90\), \(0.95\), and \(0.951\). Also calculate the strict and non-strict exceedance probabilities at the \(0.95\) value-at-risk. No discounting or mortality assumption applies; the probabilities describe the aggregate loss distribution directly.
| Loss \(X\) (thousands of dollars) | \(P(X=x)\) | Cumulative probability |
|---|---|---|
| 0 | 0.8800 | 0.8800 |
| 10 | 0.0700 | 0.9500 |
| 40 | 0.0400 | 0.9900 |
| 150 | 0.0100 | 1.0000 |
Method: compare each confidence level with the cumulative steps. At \(0.90\), the cumulative probability at 0 is \(0.8800<0.90\), while at 10 it is \(0.9500\ge0.90\). Thus, \(\operatorname{VaR}_{0.90}(X)=10\) thousand dollars.
At \(0.95\), the cumulative probability at 10 is exactly \(0.9500\), so \(\operatorname{VaR}_{0.95}(X)=10\) thousand dollars. At \(0.951\), that cumulative probability is too small: \(0.9500<0.951\). The next cumulative probability is \(0.9900\), at a loss of 40 thousand dollars, so \(\operatorname{VaR}_{0.951}(X)=40\) thousand dollars.
At the \(0.95\) value-at-risk of $10,000, strict exceedance is \(P(X>\$10{,}000)=0.0400+0.0100=0.0500\). Non-strict exceedance is \(P(X\ge\$10{,}000)=0.0700+0.0400+0.0100=0.1200\). The check is \(P(X\le\$10{,}000)=0.9500\), so strict exceedance is \(1-0.9500=0.0500\). The common mistake is to assume that a tiny increase in confidence must produce a tiny increase in value-at-risk. Here, moving from \(0.950\) to \(0.951\) crosses the probability jump at 10 and raises the quantile to 40 thousand dollars.
Exam Traps and Interpretation
The main source of errors is treating a discrete quantile as though the cumulative distribution rose smoothly. In a continuous distribution, probability at a single point is zero; in a discrete distribution, the mass at the quantile can be positive. That atom affects the distinction between strict and non-strict exceedance and can make value-at-risk stay fixed over a range of confidence levels before jumping to the next possible loss.
- Stopping at the wrong cumulative probability. Use the first cumulative probability that is at least \(\alpha\). Do not require it to be strictly greater.
- Interpolating between losses. For the discrete model specified, select a possible loss amount. Do not place the quantile between two support values because the target probability lies between their cumulative probabilities.
- Omitting the atom from a non-strict tail. \(P(X\ge q)\) includes \(P(X=q)\); \(P(X>q)\) does not.
- Claiming the strict tail equals \(1-\alpha\). For a discrete loss, it can be smaller because the cumulative probability can jump past \(\alpha\).
- Ignoring units. If loss values are stated in thousands of dollars, report the result in thousands or explicitly convert to dollars.
A useful final check is to calculate both \(F_X(q)\) and the cumulative probability immediately below \(q\). Confirm that the first is at least \(\alpha\) and the second is below \(\alpha\). Then use \(1-F_X(q)\) for strict exceedance and \(1-P(X<q)\) for non-strict exceedance.
Check Your Understanding
For each hypothetical discrete loss model, use the lower-quantile convention. Losses are in dollars unless stated otherwise. No interest or mortality assumptions are needed.
- A device-repair plan has loss values $0, $100, and $400 with probabilities 0.6000, 0.2500, and 0.1500. Find \(\operatorname{VaR}_{0.80}(X)\), \(P(X>\operatorname{VaR}_{0.80})\), and \(P(X\ge\operatorname{VaR}_{0.80})\). (Answer: VaR = $100; strict exceedance = 0.1500; non-strict exceedance = 0.4000.)
- A hypothetical loss distribution, in thousands of dollars, has values 0, 5, 12, and 40 with probabilities 0.2500, 0.2500, 0.3000, and 0.2000. Find \(\operatorname{VaR}_{0.50}(X)\). (Answer: 5 thousand dollars.)
- A hypothetical contract has losses $0, $2, $10, and $50 with probabilities 0.7000, 0.1000, 0.1500, and 0.0500. Find the 0.90 value-at-risk and both strict and non-strict exceedance probabilities at that threshold. (Answer: VaR = $10; strict exceedance = 0.0500; non-strict exceedance = 0.2000.)
- A hypothetical portfolio has losses, in thousands of dollars, 0, 20, 80, and 200 with probabilities 0.9000, 0.0600, 0.0300, and 0.0100. Find the 0.975 value-at-risk and both exceedance probabilities at that threshold. (Answer: VaR = 80 thousand dollars; strict exceedance = 0.0100; non-strict exceedance = 0.0400.)
- A discrete loss takes values 0, 3, 7, and 25 with probabilities 0.4000, 0.3500, 0.2000, and 0.0500. State the full range of confidence levels \(\alpha\) for which the value-at-risk is 7, then find both exceedance probabilities at confidence level 0.90. (Answer: \(0.7500<\alpha\le0.9500\); at 0.90, VaR = 7, strict exceedance = 0.0500, and non-strict exceedance = 0.2500.)