VaR for Discrete and Mixed Aggregate Losses

Why VaR Needs a Quantile Convention

Value at risk (VaR) answers a probability-threshold question: how large must a loss be for the probability of not exceeding it to reach a chosen level? The answer is straightforward when the loss distribution is continuous and its cumulative distribution function rises smoothly. For discrete or mixed losses, however, the CDF can jump over the target probability. A precise convention determines which loss value is reported at that jump.

This issue is common for aggregate losses. A portfolio may have a positive probability of no claims, claim amounts may take only specified values, or a model may combine point masses with continuous severities. In each case, the CDF—not interpolation between possible losses—determines VaR.

Definition: For a loss random variable $X$ with cumulative distribution function $F_X(x)=P(X\le x)$, and a confidence level $p$ with $0<p<1$, define the lower-quantile value at risk by $\operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}$. Thus, VaR is the smallest loss value at which the cumulative probability is at least $p$. At a quantile $q$, the CDF immediately to the left is no greater than $p$, while $F_X(q)$ is at least $p$.

The convention matters at an atom: a value $q$ with $P(X=q)>0$. The CDF includes the probability at $q$, so $F_X(q)$ can be substantially larger than its left limit $F_X(q-)=P(X<q)$. If the target $p$ lies between those two values, the lower-quantile VaR is $q$. There is no need to split the probability at the atom or interpolate to a value between possible losses.

Discrete Aggregate Losses

For a discrete aggregate loss, first determine the probability at every possible aggregate amount. This may require combining the outcomes of multiple independent losses. Then order the amounts from smallest to largest, accumulate their probabilities, and select the first amount whose cumulative probability reaches or exceeds $p$.

Worked Example: Convolving Two Discrete Covered Losses

A hypothetical portfolio has two independent covered losses, measured in thousands of dollars. The first loss $X_1$ is 0 with probability 0.60 and 4 with probability 0.40. The second loss $X_2$ is 0 with probability 0.50, 3 with probability 0.30, and 8 with probability 0.20. Find $\operatorname{VaR}_{0.80}(X)$ for the aggregate loss $X=X_1+X_2$. Use the lower-quantile convention.

Method. Combine every possible pair of outcomes. Independence means that the probability of a pair is the product of the two individual probabilities. The common mistake to avoid is choosing a loss based on an unaccumulated probability, rather than checking the CDF in increasing order.

Step 1: Calculate aggregate probabilities. The six outcome pairs give:

$X=0+0=0$, with probability $0.60(0.50)=0.30$.
$X=0+3=3$, with probability $0.60(0.30)=0.18$.
$X=4+0=4$, with probability $0.40(0.50)=0.20$.
$X=4+3=7$, with probability $0.40(0.30)=0.12$.
$X=0+8=8$, with probability $0.60(0.20)=0.12$.
$X=4+8=12$, with probability $0.40(0.20)=0.08$.

The probabilities sum to $0.30+0.18+0.20+0.12+0.12+0.08=1.00$, as required.

Step 2: Accumulate from the smallest aggregate loss.

Aggregate loss $x$, in thousands	$P(X=x)$	$F_X(x)$
0	0.30	0.30
3	0.18	0.48
4	0.20	0.68
7	0.12	0.80
8	0.12	0.92
12	0.08	1.00

Step 3: Apply the quantile convention. The CDF is $0.68$ at 4 and $0.80$ at 7. The smallest aggregate loss with CDF at least $0.80$ is therefore 7. In thousands of dollars, $\operatorname{VaR}_{0.80}(X)=7$, or $7,000. As a check, $F_X(7)=0.80$, while the CDF immediately below 7 is $F_X(7-)=0.68<0.80$.

In general, if the ordered possible loss values are $x_1<x_2<\dots$, calculate $F_X(x_j)=\sum_{k\le j}P(X=x_k)$. The answer is the first $x_j$ for which this sum is at least $p$. If the CDF equals $p$ exactly at a possible loss, that loss qualifies; do not move to the next value.

Mixed Losses: An Atom and a Continuous Component

A mixed distribution contains at least one atom and also a continuous component. In aggregate-loss models, one common source of an atom at zero is a positive probability of having no claims. Positive losses may then vary continuously. To calculate VaR, write the CDF piecewise, keeping discrete probability masses and continuous probabilities distinct.

Worked Example: VaR with a No-Claim Atom

A hypothetical risk has at most one claim. The claim count $N$ is Bernoulli with $P(N=1)=0.25$ and $P(N=0)=0.75$. Given $N=1$, the claim severity $Y$, in thousands of dollars, is exponential with mean 10; its survival function is $P(Y>y)=e^{-y/10}$ for $y\ge0$. The aggregate loss is $X=NY$. Find $\operatorname{VaR}_{0.90}(X)$.

Method. Separate the no-claim probability from the conditional severity distribution. The mistake to avoid is using the severity's 90th percentile directly: the aggregate has probability 0.75 at zero, so its CDF is not the same as the severity CDF.

Step 1: Write the aggregate CDF. For $x\ge0$, the event $X\le x$ occurs either when $N=0$, or when $N=1$ and $Y\le x$. Therefore,

F_X(x)=0.75+0.25P(Y\le x) =0.75+0.25(1-e^{-x/10}) =1-0.25e^{-x/10}.

At zero, $F_X(0)=0.75$, which records the atom from the no-claim outcome. Since $0.90>0.75$, the 90th percentile is positive, and the CDF formula for $x\ge0$ can be set equal to 0.90.

Step 2: Solve for the quantile.

1-0.25e^{-x/10}=0.90 \quad\Longrightarrow\quad e^{-x/10}=\frac{0.10}{0.25}=0.40 \quad\Longrightarrow\quad x=-10\ln(0.40)=9.162907\ldots

The result is in thousands of dollars. Rounding only at the final step to the nearest cent gives $\operatorname{VaR}_{0.90}(X)=\mathbf{\$9,162.91}$. Check: $F_X(9.162907\ldots)=1-0.25(0.40)=0.90$. Because the CDF is continuous and strictly increasing for positive $x$, no smaller positive value reaches the target. Also, $F_X(0)=0.75<0.90$, so zero cannot be the answer.

The same logic identifies VaR levels within the atom. For this model, every $p$ with $0<p\le0.75$ has lower-quantile VaR equal to zero, because $F_X(0)=0.75$. For $p>0.75$, the quantile is positive and can be found from the positive-loss part of the CDF.

When the Target Falls Inside a Jump

A mixed distribution can have an atom at a positive loss as well as at zero. If the target probability falls inside the jump at that positive loss, the VaR is the loss at the jump. The CDF immediately below that loss is not enough: the probability at the atom must be included when evaluating the CDF at the loss itself.

Worked Example: A Positive-Loss Atom Crosses the Target

A hypothetical aggregate loss $X$, in thousands of dollars, has this mixture distribution: it is zero with probability 0.50; it is a point loss of 5 with probability 0.20; and with probability 0.30 it is uniformly distributed over losses from 0 to 10. Find $\operatorname{VaR}_{0.80}(X)$ and $\operatorname{VaR}_{0.90}(X)$.

Method. Build the CDF on either side of the atom at 5, then apply the lower-quantile convention separately at each confidence level. The mistake to avoid is ignoring the point mass when passing 5.

Step 1: Find the CDF below 5. For $0\le x<5$, the zero-loss outcome contributes 0.50. The uniform component contributes $0.30(x/10)=0.03x$, and the point loss at 5 contributes nothing yet. Thus $F_X(x)=0.50+0.03x$. Its left limit at 5 is $F_X(5-)=0.50+0.03(5)=0.65$.

Step 2: Include the atom at 5. At $x=5$, the continuous component has cumulative probability $0.30(5/10)=0.15$, and the point mass at 5 contributes 0.20. Hence $F_X(5)=0.50+0.15+0.20=0.85$. The CDF jumps from 0.65 to 0.85.

Since $0.65<0.80\le0.85$, the smallest loss with CDF at least 0.80 is 5. Therefore $\operatorname{VaR}_{0.80}(X)=\mathbf{\$5,000}$.

Step 3: Find the 90th percentile above 5. For $5<x\le10$, the CDF includes the zero outcome, the point mass at 5, and the continuous probability accumulated to $x$: $F_X(x)=0.50+0.20+0.30(x/10)=0.70+0.03x$. Solving $0.70+0.03x=0.90$ gives $0.03x=0.20$ and $x=6.666666\ldots$. Rounding the dollar amount to the nearest cent gives $\operatorname{VaR}_{0.90}(X)=\mathbf{\$6,666.67}$.

Check. At $x=6.666666\ldots$, $F_X(x)=0.70+0.03(6.666666\ldots)=0.90$. The CDF is increasing continuously between 5 and 10, so every smaller value in that interval has CDF below 0.90. Values below 5 also have CDF below 0.90. The answer is therefore the first crossing.

Justifying the First-Crossing Rule

The lower-quantile definition is not an arbitrary convention. It captures the smallest loss threshold that meets a probability requirement. Suppose a decision requires a threshold $q$ such that $P(X\le q)\ge p$. The set of thresholds satisfying this condition is $\{x:F_X(x)\ge p\}$. Its infimum is, by definition, $\operatorname{VaR}_p(X)$. Thus no smaller threshold can meet the requirement, while the chosen quantile reaches the target under the stated convention.

For a discrete loss with ordered values, the cumulative sum stays fixed between possible losses and jumps at each value. The first-crossing rule selects the first jump that reaches $p$. For a mixed loss, a jump may do that, or the continuous part may rise to $p$ between jumps. In either case, the same CDF definition applies; no separate quantile rule is needed.

Key takeaway: State the quantile convention, calculate or identify the CDF, and select the smallest loss $x$ for which $F_X(x)\ge p$. At an atom, compare both $F_X(x-)$ and $F_X(x)$; if the target lies between them, VaR is the atom's loss value.

Common Exam Traps

Using a strict inequality. The convention here uses $P(X\le x)$, not $P(X<x)$. Omitting the probability at a jump can incorrectly move VaR to a larger loss.
Interpolating across a jump. If $F_X(q-)<p\le F_X(q)$, the quantile is $q$. A fractional loss between $q$ and the preceding possible loss is not justified by the discrete distribution.
Ignoring the no-claim atom. For a compound loss, the mass at zero affects the aggregate CDF. A severity quantile alone does not generally give an aggregate-loss quantile.
Using the wrong aggregate probabilities. When independent losses are combined, multiply probabilities for outcome pairs and then add probabilities for pairs producing the same total.
Failing to check the answer. A proposed $q$ should satisfy $F_X(q)\ge p$, with no smaller loss meeting that condition. For an atom, explicitly check the left limit and the CDF at the atom.

These procedures are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The central discipline is to define the aggregate-loss distribution and its quantile convention before calculating a risk threshold.

Check Your Understanding

For each problem, use the lower-quantile convention and check the CDF crossing, especially where there is a jump.

Two independent hypothetical losses, in thousands of dollars, have distributions $P(X_1=0)=0.70$, $P(X_1=2)=0.30$, $P(X_2=0)=0.40$, and $P(X_2=5)=0.60$. Find the 80th percentile of $X_1+X_2$. (Answer: $5,000)
A hypothetical aggregate loss is 0 with probability 0.80 and 12 with probability 0.20. Find its VaR at confidence levels 0.80 and 0.8001. (Answer: $0 and $12,000, respectively)
Two independent losses, in thousands of dollars, have distributions $P(X_1=0)=0.50$, $P(X_1=4)=0.50$, and $P(X_2=0)=0.20$, $P(X_2=3)=0.50$, $P(X_2=6)=0.30$. Find the 70th percentile of their aggregate loss. (Answer: $7,000)
An aggregate loss is zero with probability 0.60; with probability 0.10 it is exactly 8 thousand dollars; and with probability 0.30 it is uniform from 0 to 8 thousand dollars. Find its 75th percentile. (Answer: $4,000)
An aggregate loss, in thousands of dollars, is zero with probability 0.25, exactly 6 with probability 0.30, and uniform from 0 to 12 with probability 0.45. Find its 70th percentile and its 90th percentile. (Answer: $6,000 and $9,333.33, respectively)

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