VaR for Discrete Loss Distributions

Why Discrete Losses Need Cumulative Probabilities

Many loss models allow only a finite set of possible amounts. For example, a hypothetical benefit may be paid in fixed increments, or a small portfolio may have only a few possible aggregate outcomes. In these settings, calculating value at risk (VaR) is not a matter of averaging losses or choosing an outcome by its rank alone. The probabilities attached to the outcomes determine where the specified confidence level is reached.

A reliable calculation starts by listing the distinct possible losses in increasing order, then adding their probabilities from the smallest loss upward. VaR is the first loss amount at which the accumulated probability reaches the chosen confidence level. This procedure is especially useful when a probability mass function (pmf) is given directly.

Definition: Let $X$ be a loss with cumulative distribution function $F_X(x)=P(X\le x)$. At confidence level $\alpha$, where $0<\alpha<1$, define $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\ge \alpha\}. $$ For a finite-support loss with distinct possible values $x_1<x_2<\cdots<x_n$ and probabilities $p_i=P(X=x_i)$, define $C_j=\sum_{i=1}^{j}p_i$. Then VaR is the first $x_j$ for which $C_j\ge\alpha$.

From a Probability Mass Function to VaR

The probabilities in a pmf must be nonnegative and sum to one. When the possible losses are not already ordered, sort them before accumulating probabilities. If a loss amount appears more than once in a description of the outcomes, combine the probabilities for that amount so that each distinct loss has one row.

The calculation follows directly from the definition. At each support point $x_j$, the cumulative probability is the probability that the loss is no greater than $x_j$. If that cumulative probability is still below $\alpha$, then $x_j$ does not reach the target. At the first point where it reaches or exceeds $\alpha$, every smaller support point has failed to reach the target, and that point satisfies the definition of VaR.

F_X(x_j)=P(X\le x_j)=\sum_{i=1}^{j}P(X=x_i)=C_j, \qquad \operatorname{VaR}_{\alpha}(X)=x_j \ \text{for the first }j\text{ with }C_j\ge\alpha.

This argument is the reason to accumulate probability rather than interpolate between loss amounts. A discrete distribution places probability at particular outcomes; it does not necessarily assign probability to the values between them. The VaR calculation uses the stated loss support and the cumulative probability at each support point.

Check the pmf.
Confirm each probability is nonnegative and their sum is one. Combine repeated loss amounts if needed.

Sort the losses.
Write the distinct outcomes in increasing order, keeping each probability with its outcome.

Accumulate probability.
Add each probability to the total from the preceding rows.

Identify the first qualifying row.
The VaR is the loss at the first row whose cumulative probability is at least $\alpha$.

Worked Example: Selecting a Loss Quantile From a PMF

A hypothetical equipment insurer models the annual loss on one contract as follows. The probabilities are assumed to be the complete pmf. Find VaR at confidence levels $0.80$ and $0.95$.

Annual loss, $x_i$	Probability, $p_i$	Cumulative probability, $C_i$
$0	0.4000	0.4000
$1,200	0.3000	0.7000
$5,000	0.2000	0.9000
$12,000	0.1000	1.0000

Method. The support is already in increasing order. Check the probabilities: $0.4000+0.3000+0.2000+0.1000=1.0000$. The cumulative probabilities are $0.4000$, then $0.4000+0.3000=0.7000$, then $0.7000+0.2000=0.9000$, and finally $0.9000+0.1000=1.0000$.

At confidence level $0.80$, the first two cumulative probabilities are below $0.80$, while $0.9000\ge0.80$. Thus $\operatorname{VaR}_{0.80}(X)=\$5{,}000$. At confidence level $0.95$, $0.9000<0.95$ and $1.0000\ge0.95$, so $\operatorname{VaR}_{0.95}(X)=\$12{,}000$.

Check. The cumulative probability through $1,200 is only $0.7000$, so a loss of $1,200 cannot meet either target. The selected amounts do meet their respective cumulative-probability targets. A common mistake is to use the expected loss, which here is $0(0.4000)+1{,}200(0.3000)+5{,}000(0.2000)+12{,}000(0.1000)=\$2{,}560$. That is an average, not either requested VaR.

Discrete VaR for Aggregate Losses

For a total loss formed from multiple components, first determine the distribution of the total. If the components are independent, multiply their probabilities to obtain joint outcome probabilities, then add probabilities whenever different combinations produce the same total. Only after constructing and sorting the total-loss pmf should you accumulate probabilities for VaR.

Worked Example: VaR of Two Independent Loss Components

A hypothetical repair program has two independent annual costs, $A$ and $B$, measured in dollars. Assume $P(A=0)=0.7000$, $P(A=4{,}000)=0.3000$, $P(B=0)=0.5000$, and $P(B=3{,}000)=0.5000$. Find the $0.80$ VaR of the aggregate cost $T=A+B$.

Method. Independence means each joint probability is the product of the two component probabilities. The four combinations are:

$A=0,B=0$: total $0, probability $0.7000(0.5000)=0.3500$.
$A=0,B=3{,}000$: total $3,000, probability $0.7000(0.5000)=0.3500$.
$A=4{,}000,B=0$: total $4,000, probability $0.3000(0.5000)=0.1500$.
$A=4{,}000,B=3{,}000$: total $7,000, probability $0.3000(0.5000)=0.1500$.

The totals are distinct, so no rows need to be combined. Their probabilities sum to $0.3500+0.3500+0.1500+0.1500=1.0000$. In increasing order, the cumulative probabilities are $0.3500$ at $0, $0.7000$ at $3,000, $0.8500$ at $4,000, and $1.0000$ at $7,000.

The cumulative probability is still below $0.80$ at $3,000, but is $0.8500$ at $4,000. Therefore $\operatorname{VaR}_{0.80}(T)=\$4{,}000$.

Check and trap. The pmf totals one, and $P(T\le\$4{,}000)=0.8500$, while $P(T\le\$3{,}000)=0.7000$. Multiplying the component probabilities is justified here only because independence was specified. Without independence, the marginal probabilities alone would not determine the four joint probabilities or the aggregate VaR.

Calculate VaR for the Loss Actually Being Measured

In insurance applications, the random variable may be a payment rather than the underlying gross loss. A deductible and a payment limit can turn several gross-loss outcomes into the same payment. The payment pmf must reflect that mapping: when outcomes map to one payment amount, add their probabilities before finding VaR.

Definition: For a gross loss $X\ge0$, a per-loss deductible $d\ge0$, and a maximum payment $u\ge0$, the payment is $Y^L=\min\{(X-d)^{+},u\}$, where $(z)^{+}=\max(z,0)$. VaR for the insurer's payment is calculated from the distribution of $Y^L$, not by accumulating probabilities using the gross-loss amounts.

Worked Example: Payment VaR With a Deductible and Limit

A hypothetical policy has a $2,000 deductible and an $8,000 maximum payment per loss. Its gross loss $X$ takes values $0, $5,000, $10,000, and $20,000, with probabilities $0.2000$, $0.3500$, $0.3000$, and $0.1500$, respectively. Find the insurer's payment VaR at confidence level $0.60$.

Method. Apply $Y^L=\min\{(X-2{,}000)^+,8{,}000\}$ to each gross-loss outcome:

If $X=\$0$, then $(0-2{,}000)^+=0$, so $Y^L=\$0$.
If $X=\$5{,}000$, then $(5{,}000-2{,}000)^+=\$3{,}000$, so $Y^L=\$3{,}000$.
If $X=\$10{,}000$, then $(10{,}000-2{,}000)^+=\$8{,}000$, so $Y^L=\$8{,}000$.
If $X=\$20{,}000$, then $(20{,}000-2{,}000)^+=\$18{,}000$, capped at $8,000, so $Y^L=\$8{,}000$.

The two largest gross-loss outcomes map to the same payment. Add their probabilities: $0.3000+0.1500=0.4500$. The payment distribution is therefore $0 with probability $0.2000$, $3,000 with probability $0.3500$, and $8,000 with probability $0.4500$. Its cumulative probabilities are $0.2000$, $0.5500$, and $1.0000$. Since $0.5500<0.60\le1.0000$, the insurer's payment VaR is $\operatorname{VaR}_{0.60}(Y^L)=\$8{,}000$.

Check and trap. The payment probabilities sum to $1.0000$, and the selected outcome is the first payment whose cumulative probability reaches $0.60$. A common coverage error is to cap gross loss first and then subtract the deductible. At $X=\$10{,}000$, that incorrect order gives $\min(10{,}000,8{,}000)-2{,}000=\$6{,}000$, rather than the correct payment of $8,000. The limit here caps the insurer's payment after the deductible is applied.

Common Exam Traps and a Final Check

Discrete VaR problems are usually short once the correct loss variable and pmf have been identified. Errors often arise before the final comparison with $\alpha$. Check that the probabilities belong to the loss being measured, that outcomes are ordered, and that probability—not loss amount—is accumulated.

Using an average instead of a quantile. Expected loss weights outcomes by their probabilities and averages them. VaR selects a loss support point using a cumulative-probability threshold.
Sorting probabilities rather than losses. Accumulate probabilities in increasing order of the loss outcomes, not in order from smallest to largest probability.
Ignoring duplicate outcomes. Different component combinations or gross losses can produce the same total or payment. Combine those probabilities in the resulting pmf.
Assuming independence without support. Product probabilities for combinations are valid only when the relevant components are independent.
Using the wrong loss variable. If the question concerns a payment, first transform gross losses into payments and then calculate VaR for that payment distribution.

For a finite pmf, the final selection can be checked in two ways: the cumulative probability at the selected loss must be at least $\alpha$, and the cumulative probability at the preceding support point must be below $\alpha$. Together, these checks confirm that the selected loss is the first one to reach the target.

This cumulative-probability method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. It provides a direct calculation from a discrete pmf without requiring a continuous distribution or interpolation between possible losses.

Check Your Understanding

For each problem, order the outcomes, accumulate their probabilities, and identify the requested VaR.

A hypothetical equipment fund has annual loss $0 with probability $0.2500$, $2,000 with probability $0.4000$, $6,000 with probability $0.2500$, and $10,000 with probability $0.1000$. Find $\operatorname{VaR}_{0.80}$. (Answer: $6,000)
A hypothetical benefit model has losses $0, $900, $2,500, and $8,000 with probabilities $0.6000$, $0.2000$, $0.1500$, and $0.0500$, respectively. Find $\operatorname{VaR}_{0.90}$. (Answer: $2,500)
Independent hypothetical costs $A$ and $B$ satisfy $P(A=0)=0.8000$, $P(A=5{,}000)=0.2000$, $P(B=0)=0.6000$, and $P(B=2{,}000)=0.4000$. Find $\operatorname{VaR}_{0.90}(A+B)$. (Answer: $5,000)
A hypothetical policy has a $1,000 deductible and a $4,000 maximum payment. Gross loss is $0, $2,000, $5,000, or $9,000 with probabilities $0.1000$, $0.3000$, $0.4000$, and $0.2000$, respectively. Find the insurer's payment $\operatorname{VaR}_{0.75}$. (Answer: $4,000)
A hypothetical loss $X$ takes values $100, $500, $1,000, and $3,000 with probabilities $0.1000$, $0.2000$, $0.4500$, and $0.2500$. Find $\operatorname{VaR}_{0.70}(X)$, and state the cumulative probability at the preceding support point. (Answer: VaR is $1,000; the preceding cumulative probability is 0.3000)

What You'll Learn