VaR for Discrete Losses

Why Discrete VaR Requires Care

A discrete loss distribution places positive probability on one or more specific loss amounts. Its cumulative distribution function therefore jumps at those amounts. When calculating Value at Risk, that jump can take the cumulative probability from below the target level to above it in one step. The VaR is the loss amount where the target is first reached; the probability at or below that amount can exceed the target.

It is important to distinguish two events: a loss being strictly below VaR, and a loss being at or below VaR. In general, the first probability is at most the target, while the second is at least the target. In particular, with the usual lower-quantile definition, the probability at or below VaR cannot be less than the target. It is the probability strictly below VaR that may be less than the target.

Definition: Let $X$ be a real-valued loss with cumulative distribution function $F_X(x)=P(X\le x)$. For a confidence level $p$, where $0<p<1$, define $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}. $$ The VaR is the smallest loss threshold at which the cumulative probability reaches or exceeds $p$.

For $q=\operatorname{VaR}_p(X)$, the general quantile inequalities are

P(X<q)\le p\le P(X\le q).

The left inequality can be an equality, even when $X$ is discrete. For a finite-support distribution, however, the probability strictly below VaR is strictly less than $p$: the cumulative probability immediately before the VaR loss has not yet reached the target. The distinction matters when a problem asks for the probability of losses below the VaR rather than losses at or below it.

Finding VaR from a Discrete Loss Table

For a finite set of possible losses, arrange the loss amounts from smallest to largest and add their probabilities cumulatively. Stop at the first loss amount for which the cumulative probability is at least $p$. That amount is $\operatorname{VaR}_p(X)$. Do not interpolate between possible losses: the definition selects a loss threshold, not a point between support values.

Worked Example: Locating a Quantile at a Probability Boundary

A hypothetical insurer models one-year aggregate losses $X$, in thousands of dollars, as follows. Find the 80% VaR, and calculate the probabilities of a loss strictly below and at or below that VaR.

Loss $x$, in thousands	$P(X=x)$	$F_X(x)$
0	0.55	0.55
40	0.25	0.80
180	0.15	0.95
600	0.05	1.00

Method. Compute the cumulative probabilities in ascending loss order and select the first loss whose cumulative probability is at least $p=0.80$.

At loss 0, the cumulative probability is $0.55<0.80$. At loss 40, it is $0.55+0.25=0.80$, so the first threshold reaching the target is 40 thousand dollars:

\operatorname{VaR}_{0.80}(X)=40\text{ thousand dollars}.

The probability strictly below this threshold is $P(X<40)=0.55$. The probability at or below it is $P(X\le40)=0.55+0.25=0.80$. The calculation is checked by confirming $0.55<0.80\le0.80$. Here the cumulative probability lands exactly on the target; it does not have to overshoot it.

Common mistake avoided: Treating “80% VaR” as a guarantee that exactly 20% of losses are strictly greater than VaR, or that the probability strictly below VaR is 80%. In this example $P(X>40)=0.20$, but $P(X<40)=0.55$, because there is probability mass at the VaR itself.

Why the Cumulative Probability Can Jump Past the Target

An atom is a positive probability at a single loss amount. If the probability immediately below a loss is less than $p$, and adding the probability at that loss takes the cumulative probability to at least $p$, that loss is VaR. The portion of the atom needed to reach the target does not change the loss threshold: the entire atom is included in $P(X\le\operatorname{VaR}_p(X))$.

Order the possible losses.
Write the support values as $x_1<x_2<\cdots<x_m$, with probabilities $p_j=P(X=x_j)$.

Form cumulative probabilities.
For each $j$, calculate $C_j=\sum_{k=1}^{j}p_k$, and set $C_0=0$.

Select the first cumulative probability meeting the target.
Choose the smallest $j$ such that $C_j\ge p$. Then $x_j$ is the VaR.

Check both sides of the threshold.
For this finite-support case, verify $C_{j-1}<p\le C_j$. The jump at $x_j$ is $p_j=C_j-C_{j-1}$.

This procedure follows directly from the definition. For any threshold below $x_j$, the cumulative probability is no greater than $C_{j-1}$, which is less than $p$. At $x_j$, the cumulative probability is $C_j$, which is at least $p$. Thus $x_j$ is the first threshold reaching the target.

F_X(x_j-)=C_{j-1}<p\le C_j=F_X(x_j).

Here $F_X(x_j-)=P(X<x_j)$, the cumulative probability immediately before the jump. The amount by which the cumulative probability at VaR exceeds the target is $C_j-p$, which can be zero or positive. A large atom can make this overshoot substantial.

Worked Example: Quantifying the Jump at VaR

A hypothetical commercial property portfolio has annual loss $X$, measured in thousands of dollars, with this discrete scenario distribution. Determine the 90% VaR, the probability strictly below it, the probability at or below it, and the cumulative-probability overshoot.

Loss $x$, in thousands	$P(X=x)$	Cumulative probability
0	0.60	0.60
75	0.22	0.82
220	0.13	0.95
1,000	0.05	1.00

Method. Compare the cumulative probabilities to $p=0.90$, taking the first loss amount whose cumulative probability reaches or exceeds 0.90. The probabilities sum to $0.60+0.22+0.13+0.05=1.00$.

The cumulative probability at 75 is $0.82<0.90$. At 220 it is $0.82+0.13=0.95\ge0.90$. Therefore,

\operatorname{VaR}_{0.90}(X)=220\text{ thousand dollars}.

The probability strictly below VaR is $P(X<220)=0.60+0.22=0.82$. The probability at or below VaR is $P(X\le220)=0.60+0.22+0.13=0.95$. The cumulative probability overshoots the target by $0.95-0.90=0.05$, or 5 percentage points. The check is $0.82<0.90\le0.95$.

Common mistake avoided: Reporting 90% as the probability at or below 220. That would discard part of the probability mass at 220, which the event $X\le220$ includes in full. The actual probability is 95%.

When the Probability Strictly Below VaR Can Equal the Target

For finite support, the cumulative probability immediately before the first qualifying loss must be strictly less than $p$. It is not safe to state that strict inequality for every discrete distribution, though. With countably many loss values accumulating toward a point that has no probability mass, cumulative probabilities can approach $p$ from below without any one value below that point having cumulative probability at least $p$.

The definition still gives $P(X<q)\le p\le P(X\le q)$. What can fail is the stronger strict claim $P(X<q)<p$. The following example shows why.

Worked Example: A Countable Distribution with Equality Below VaR

Let a hypothetical loss $X$ take the values $-1/n$ for each positive integer $n$, with probability $0.6\cdot2^{-n}$, and take the value 1 with probability 0.4. Find the 60% VaR and compare $P(X<\operatorname{VaR}_{0.60}(X))$ with $P(X\le\operatorname{VaR}_{0.60}(X))$.

Step 1: Check the probabilities. The geometric series satisfies $\sum_{n=1}^{\infty}2^{-n}=1$. Thus the total probability on the negative values is $0.6$, and the complete distribution sums to $0.6+0.4=1$.

Step 2: Check thresholds below zero. For any fixed $x<0$, only finitely many values $-1/n$ are at or below $x$. Their probabilities sum to less than the full negative-value probability 0.6. Therefore $F_X(x)<0.60$ for every $x<0$.

Step 3: Evaluate the cumulative probability at zero. There is no probability mass at zero, but every negative value is less than zero. Consequently $F_X(0)=P(X\le0)=0.60$. Since every threshold below zero has cumulative probability less than 0.60 and zero reaches 0.60, the infimum in the VaR definition is zero:

\operatorname{VaR}_{0.60}(X)=0.

It follows that $P(X<0)=0.60$ and $P(X\le0)=0.60$. Thus $P(X<\operatorname{VaR}_{0.60}(X))=p$, not a value strictly less than $p$. The general inequalities hold with equality on both sides. This example also shows that a quantile need not itself be a loss value with positive probability.

Common mistake avoided: Assuming that every quantile of a discrete distribution must be an atom, and that the cumulative probability strictly below it must be less than $p$. Those statements hold for finite support, but not for every countably supported distribution.

Common Calculation Traps

For a discrete loss model, keep the event in the question explicit. $P(X<q)$, $P(X=q)$, and $P(X\le q)$ are generally different. For a finite-support distribution with VaR at $x_j$, they are $C_{j-1}$, $p_j$, and $C_j$, respectively. The relationship $C_j=C_{j-1}+p_j$ is a useful check.

A related trap is choosing the largest loss whose cumulative probability is below $p$. That amount is not VaR: its cumulative probability has not reached the target. VaR is the next support value, the first threshold for which $F_X(x)\ge p$. At an exact cumulative boundary, use “at least” as written in the definition; do not move to the next loss merely because the probability equals $p$.

Key takeaway: For $q=\operatorname{VaR}_p(X)$, the general statement is $P(X<q)\le p\le P(X\le q)$. With finite support, the left inequality is strict because VaR is the first support value whose cumulative probability reaches $p$. An atom at VaR can make the cumulative probability jump above $p$.

Check Your Understanding

For each distribution, use the lower-quantile definition and check the probability immediately below and at or below the stated VaR threshold.

A hypothetical loss is 0 with probability 0.40, 25 with probability 0.35, and 90 with probability 0.25. Find $\operatorname{VaR}_{0.70}(X)$, $P(X<\operatorname{VaR}_{0.70}(X))$, and $P(X\le\operatorname{VaR}_{0.70}(X))$. (Answer: VaR $=25$; the probabilities are 0.40 and 0.75, respectively.)
A discrete loss takes values 10, 30, 80, and 200 with probabilities 0.10, 0.40, 0.30, and 0.20, respectively. Find the VaR at levels 0.80 and 0.81. (Answer: $\operatorname{VaR}_{0.80}=80$ and $\operatorname{VaR}_{0.81}=200$.)
Let $P(X=-1/n)=0.4\cdot2^{-n}$ for each positive integer $n$, and $P(X=2)=0.6$. Find the 40% VaR and both probabilities strictly below and at or below it. (Answer: VaR $=0$; both probabilities equal 0.40.)
A hypothetical portfolio loss is 0 with probability 0.65, 50 with probability 0.10, 300 with probability 0.20, and 900 with probability 0.05. Find the 90% VaR and the probability of a loss strictly greater than it. (Answer: VaR $=300$; $P(X>300)=0.05$.)
A finite-support loss has values 0, 100, and 400 with probabilities 0.20, 0.50, and 0.30. For which confidence levels $p$ does its VaR equal 100? Be precise about the endpoints. (Answer: $0.20<p\le0.70$.)

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