VaR for Mixed Distributions

Why Mixed Distributions Need Care

Some loss models assign positive probability to particular amounts while also allowing a continuous range of losses. For example, a hypothetical insurance payment might be zero when no covered event occurs, take one exact value for a specified benefit, or vary continuously with the size of a covered loss. In such a model, the cumulative distribution function may rise smoothly over some ranges and jump at particular values.

To find Value at Risk (VaR), locate the requested confidence level on the full cumulative distribution function, not just on its continuous part. A confidence level reached within a jump gives VaR at the jump location; a level reached during a continuous rise usually requires solving an equation.

Definition: For a loss random variable $X$ with cumulative distribution function $F_X(x)=P(X\le x)$, the VaR at confidence level $\alpha$, where $0<\alpha<1$, is the lower quantile $$ \operatorname{VaR}_\alpha(X)=\inf\{x:F_X(x)\ge\alpha\}. $$ At a point $x$, the jump size is $P(X=x)=F_X(x)-F_X(x-)$, where $F_X(x-)=P(X<x)$. A positive jump means that $X$ has a point mass at $x$.

How to Read the Mixed Cumulative Distribution Function

The cumulative distribution function records all probability at or below each possible loss. A continuous component contributes probability that accumulates across a range. A point mass contributes a jump: immediately before the point, the distribution function has accumulated $F_X(x-)$; at the point, it has accumulated $F_X(x)$. Therefore, when $F_X(x-)<\alpha\le F_X(x)$, the lower quantile is $x$.

If the confidence level falls in a continuous section, find the loss $x$ where $F_X(x)=\alpha$, while checking that no earlier value already has cumulative probability at least $\alpha$. At a boundary, use the cumulative probability at the boundary itself. In particular, equality $F_X(x)=\alpha$ is enough for $x$ to be the lower quantile if all smaller values have cumulative probability below $\alpha$.

Key result: The lower-quantile definition can be checked by locating the first point where the cumulative probability reaches or exceeds $\alpha$. At an atom this gives $F_X(x-)<\alpha\le F_X(x)$. Within an increasing continuous section, solve $F_X(x)=\alpha$ and verify that the solution is the first such point.

The atom condition follows directly from the definition. If $F_X(x-)<\alpha$, then every value below $x$ has cumulative probability less than $\alpha$. If also $F_X(x)\ge\alpha$, the set of values with cumulative probability at least $\alpha$ begins at $x$. Thus its infimum is $x$. Between jumps, the same definition asks for the first loss at which the cumulative probability reaches the target.

Write the full CDF.
Include contributions from every point mass and continuous component, and state which ranges of loss each expression covers.

Locate the confidence level.
Compare $\alpha$ with the CDF values immediately before and at each jump, and with the CDF values across continuous sections.

Solve or select the quantile.
If $\alpha$ lies within a continuous rise, solve $F_X(x)=\alpha$. If it lies in a jump, choose the jump location.

Check the result.
Confirm that cumulative probability just below the answer is less than $\alpha$, and that the CDF at the answer is at least $\alpha$.

Worked Examples

Worked Example: A Zero Payment and an Exponential Component

A hypothetical claim payment $X$, in dollars, is zero with probability $0.30$. With probability $0.70$, it is exponential with rate $0.02$ per dollar, so the conditional CDF of a positive payment is $1-e^{-0.02x}$ for $x\ge0$. Find $\operatorname{VaR}_{0.65}(X)$. Use the lower-quantile convention and round the final answer to the nearest cent.

Method and CDF. The point mass at zero gives $F_X(0)=0.30$. For $x\ge0$, the mixture CDF is

F_X(x)=0.30+0.70(1-e^{-0.02x}).

Since $0.30<0.65$, the answer is not zero. The target is reached in the positive continuous component, so set the CDF equal to the confidence level:

0.30+0.70(1-e^{-0.02x})=0.65.

Subtract $0.30$, divide by $0.70$, and isolate the exponential term: $0.70(1-e^{-0.02x})=0.35$, $1-e^{-0.02x}=0.50$, and $e^{-0.02x}=0.50$. Taking natural logarithms gives $-0.02x=\ln(0.50)$, so $x=-\ln(0.50)/0.02=34.657359\ldots$ dollars. Rounded to the nearest cent, the VaR is $34.66.

Check. At the unrounded value, $e^{-0.02x}=0.50$, so $F_X(x)=0.30+0.70(0.50)=0.65$. For any smaller positive $x$, the exponential CDF, and hence the mixture CDF, is smaller. The result is the first loss that reaches the target. The common mistake avoided here is to treat the conditional exponential distribution as the whole model and omit the zero-payment probability.

Worked Example: A Confidence Level Inside a Jump

A hypothetical portfolio loss $X$, in thousands of dollars, has probability $0.20$ of being zero. With probability $0.50$, it is uniformly distributed from 0 to 100. The remaining probability, $0.30$, is concentrated at 120. Find $\operatorname{VaR}_{0.80}(X)$.

Method and CDF. On $0\le x<100$, the zero mass and the continuous component give $F_X(x)=0.20+0.50(x/100)$. At 100, this reaches $0.20+0.50=0.70$. There is no additional probability between 100 and 120, so $F_X(x)=0.70$ for $100\le x<120$. At 120, the point mass raises the CDF by $0.30$, to 1.

F_X(120-)=0.70,\qquad F_X(120)=1.00.

The target $0.80$ is above $F_X(120-)=0.70$ and no greater than $F_X(120)=1.00$. Therefore it is reached at the jump, and $\operatorname{VaR}_{0.80}(X)=120$ thousand dollars, or $120,000.

Check. Every loss below 120 has cumulative probability at most $0.70$, which is below $0.80$, while the CDF at 120 equals 1. The common mistake avoided here is to interpolate through the gap from 100 to 120 as if probability were spread continuously across it. It is not: that interval has no probability, and the target is reached at the point mass.

Worked Example: A Continuous Quantile Above an Atom

A hypothetical payment $X$, in dollars, has the following distribution: probability $0.10$ at zero; probability $0.50$ uniformly distributed from 0 to 40; probability $0.20$ at 60; and probability $0.20$ uniformly distributed from 60 to 100. Find $\operatorname{VaR}_{0.90}(X)$, rounded to the nearest cent.

Method and CDF. The component probabilities sum to $0.10+0.50+0.20+0.20=1.00$. Just before 60, the zero mass and the first uniform component have accumulated $0.10+0.50=0.60$. The atom at 60 raises the CDF to $0.80$. For $60\le x\le100$, the upper uniform component adds $0.20(x-60)/40$, giving

F_X(x)=0.80+0.20\frac{x-60}{40},\qquad 60\le x\le100.

Because $0.80<0.90<1.00$, the target lies in this continuous section rather than in the jump. Solve: $0.80+0.20(x-60)/40=0.90$. Subtract $0.80$ to obtain $0.20(x-60)/40=0.10$. Dividing by $0.20$ gives $(x-60)/40=0.50$, so $x-60=20$ and $x=80$. Thus the VaR is $80.00.

Check. At 80, the upper uniform component contributes $0.20(20/40)=0.10$, so $F_X(80)=0.80+0.10=0.90$. Below 80 in this continuous section, the CDF is below $0.90$; below 60 it is at most $0.80$. The answer is therefore the first loss reaching the target. The common mistake avoided here is to stop at the atom at 60 even though its cumulative probability, $0.80$, is below the requested confidence level.

Boundary Checks and Common Traps

At an atom, the distinction between probability strictly below the point and probability at or below it matters. Use $F_X(x-)$ for the former and $F_X(x)$ for the latter. If the target equals $F_X(x)$, the point $x$ can be the VaR even when the CDF does not jump past the target. If the target is greater than $F_X(x-)$ but no greater than $F_X(x)$, the jump contains the target. In either case, the lower-quantile definition determines the answer.

A frequent error is to solve $F_X(x)=\alpha$ using a formula from one component without first checking where that formula applies. A candidate outside its stated range is not a valid solution for that CDF section. Another error is to assign all confidence levels near an atom to the atom: a target above the CDF at the atom may instead fall in a later continuous section. Always verify the CDF on both sides of the proposed answer.

This method is useful whenever a loss model combines an event probability, a guaranteed benefit, a deductible that can produce zero payment, or another feature that creates an exact payment amount with a continuously varying loss. It also provides a direct way to detect whether a reported quantile is plausible: the cumulative probability below it should be less than the target, and the cumulative probability at it should reach or exceed the target. This treatment of mixed CDFs is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Key takeaway: For a mixed loss distribution, find VaR from the complete CDF. Solve within a continuous rise; select the atom when the confidence level is reached in its jump; then check that no smaller loss reaches the same level.

Check Your Understanding

For each problem, use the lower-quantile convention and check where the confidence level falls in the CDF.

A hypothetical loss is zero with probability $0.40$, and otherwise has an exponential distribution with rate $0.10$ per dollar. Find its VaR at confidence level $0.40$. (Answer: $0.00)
A payment is zero with probability $0.15$ and otherwise is uniform from $0 to $50. Find its VaR at confidence level $0.66$. (Answer: $30.00)
A mixed loss distribution has $F_X(100-)=0.72$ and $F_X(100)=0.88$. Find its VaR at confidence level $0.80$. (Answer: 100)
A loss is zero with probability $0.10$, uniform from 0 to 80 with probability $0.60$, and exactly 100 with probability $0.30$. Find its VaR at confidence level $0.55$. (Answer: 60)
A loss is zero with probability $0.20$, uniform from 0 to 40 with probability $0.50$, and exactly 80 with probability $0.30$. Find its VaR at confidence level $0.70$. (Answer: 40)

What You'll Learn