Quantile Conventions at Probability Atoms

Why Quantile Conventions Matter

For a discrete loss, the cumulative distribution function jumps at each possible loss amount. A confidence level can fall inside one of these jumps, or it can equal the cumulative probability at the top of a jump. The distinction matters: two reasonable inverse-CDF conventions agree in the first case but can select different loss amounts in the second.

For VaR calculations, the convention must be stated before the calculation. The standard lower-quantile definition uses the smallest loss amount whose cumulative probability is at least the confidence level. An alternative strict inverse uses the smallest loss amount whose cumulative probability is greater than the confidence level. At a cumulative-probability boundary, the strict version moves to the next loss amount.

Definition: Let $X$ be a loss with cumulative distribution function $F_X(x)=P(X\le x)$, and let $0<p<1$. The lower quantile is $$ Q_-(p)=\inf\{x:F_X(x)\ge p\}. $$ The strict inverse quantile is $$ Q_+(p)=\inf\{x:F_X(x)>p\}. $$ Under the lower-quantile convention, $\operatorname{VaR}_p(X)=Q_-(p)$. The strict inverse is an alternative convention, not the default definition used here for VaR. For a finite discrete loss distribution and $0<p<1$, both quantiles are finite.

Reading a Discrete Cumulative Distribution

Suppose $X$ can take ordered values $x_1<x_2<\cdots<x_n$, with probabilities $r_j=P(X=x_j)$. Define the cumulative probabilities $C_j=P(X\le x_j)=\sum_{k=1}^j r_k$, and set $C_0=0$. The lower quantile selects the first $x_j$ for which $C_j\ge p$; the strict inverse selects the first $x_j$ for which $C_j>p$.

If $C_{j-1}<p<C_j$, both conventions select $x_j$. The confidence level lies strictly inside the jump, so $x_j$ is the first amount at which the cumulative probability reaches or exceeds $p$, and also the first amount at which it exceeds $p$. If instead $p=C_j$, the lower quantile selects $x_j$, while the strict inverse selects the next loss amount $x_{j+1}$, provided $p<1$.

This comparison is about the choice of inverse, not a change to the underlying loss distribution. The probabilities at the loss amounts remain the same. In particular, a cumulative probability equal to the confidence level is not a reason to round the confidence level or to move to another convention without saying so.

Why Equality Produces a Difference

At a loss value $x_j$, the distribution has accumulated exactly $C_j$ probability. If $p=C_j$, that amount already meets the lower-quantile requirement $F_X(x)\ge p$. It does not meet the strict requirement $F_X(x)>p$. The strict condition first becomes true at a larger loss value, where the cumulative probability has increased.

F_X(x_j)=C_j=p \quad\Longrightarrow\quad Q_-(p)=x_j,\qquad Q_+(p)=x_{j+1}

This result assumes a discrete distribution with a next larger support point $x_{j+1}$ and $p<1$. It does not say that every pair of inverse conventions gives different answers at every atom. They differ at the confidence levels that equal a cumulative probability; when $p$ falls strictly between consecutive cumulative probabilities, they agree.

Order the loss amounts.
List the possible losses from smallest to largest, keeping each probability attached to its loss.

Build cumulative probabilities.
At each loss $x_j$, calculate $C_j=P(X\le x_j)$ by adding the probabilities through that point.

Compare with the confidence level.
For $Q_-(p)$, find the first $C_j\ge p$. For $Q_+(p)$, find the first $C_j>p$.

Check for equality.
If a cumulative probability equals $p$, the lower quantile is that loss amount and the strict inverse advances to the next loss amount.

Worked Examples

Worked Example: Confidence Level at a Cumulative Boundary

A hypothetical equipment insurer models the annual claim loss $X$ as follows. Find both inverse quantiles at $p=0.75$, and state the VaR under the lower-quantile convention.

Loss $x$ (dollars)	$P(X=x)$	$F_X(x)$
0	0.20	0.20
40	0.30	0.50
100	0.25	0.75
300	0.25	1.00

Method. Add the probabilities in loss order, then apply each inequality separately. The cumulative probability at 100 dollars is $0.20+0.30+0.25=0.75$, exactly the stated confidence level.

For the lower quantile, the first cumulative probability at least $0.75$ is $F_X(100)=0.75$. Therefore $Q_-(0.75)=100$ dollars. For the strict inverse, 100 dollars does not qualify because $F_X(100)$ is not greater than $0.75$. At 300 dollars, $F_X(300)=1.00>0.75$, so $Q_+(0.75)=300$ dollars.

Thus the lower-quantile VaR is 100 dollars, while the strict inverse gives 300 dollars. The result checks against the table: $P(X\le100)=0.75$, whereas $P(X\le40)=0.50<0.75$. Common mistake avoided: selecting 300 dollars for the standard VaR merely because the cumulative probability at 100 dollars equals, rather than exceeds, the confidence level.

Worked Example: Confidence Level Inside a Probability Jump

A hypothetical warranty portfolio has annual loss amounts of 0, 50, 120, and 200 dollars, with respective probabilities $0.10, 0.40, 0.30,$ and $0.20$. Find $Q_-(0.65)$ and $Q_+(0.65)$. Explain why equality does not create a difference in this case.

Method. Compute cumulative probabilities in ascending loss order:

At 0 dollars, $F_X(0)=0.10$. At 50 dollars, $F_X(50)=0.10+0.40=0.50$. At 120 dollars, $F_X(120)=0.50+0.30=0.80$. At 200 dollars, $F_X(200)=1.00$. The probabilities sum to $1.00$, as required.

The confidence level $0.65$ lies strictly between $0.50$ and $0.80$. Therefore 120 dollars is the first loss amount whose cumulative probability is at least $0.65$, and it is also the first amount whose cumulative probability is greater than $0.65$. Hence $Q_-(0.65)=Q_+(0.65)=120$ dollars.

The check is $P(X\le50)=0.50<0.65$ and $P(X\le120)=0.80>0.65$. Common mistake avoided: assuming that any discrete probability mass automatically makes the two conventions differ. The difference requires the confidence level to equal a cumulative probability.

Worked Example: Finding the Boundary Before Applying the Convention

For a hypothetical group-benefit portfolio, annual loss $X$ has the following distribution: 10 dollars with probability $0.15$, 30 dollars with probability $0.35$, 80 dollars with probability $0.20$, and 150 dollars with probability $0.30$. Find both quantiles at confidence level $0.50$, then find both at confidence level $0.60$.

Method. First calculate every cumulative probability:

$F_X(10)=0.15$; $F_X(30)=0.15+0.35=0.50$; $F_X(80)=0.50+0.20=0.70$; and $F_X(150)=0.70+0.30=1.00$. The probabilities sum to $1.00$.

At $p=0.50$, the lower quantile is 30 dollars because $F_X(30)=0.50\ge0.50$, and no smaller loss has cumulative probability at least $0.50$. The strict inverse must move on: $F_X(30)=0.50$ is not greater than $0.50$, while $F_X(80)=0.70>0.50$. Thus $Q_-(0.50)=30$ dollars and $Q_+(0.50)=80$ dollars.

At $p=0.60$, the level is strictly between $F_X(30)=0.50$ and $F_X(80)=0.70$. Both conventions therefore give 80 dollars. The check is that losses up to 30 dollars have probability $0.50<0.60$, and losses up to 80 dollars have probability $0.70>0.60$. Common mistake avoided: using the answer at one confidence level for another without checking whether the new level is exactly a cumulative probability.

Interpreting the Quantile Conditions

The lower quantile can also be checked using probabilities immediately below and at the selected loss. If $q=Q_-(p)$ is a support point, then

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

The strict inequality on the left reflects that $q$ is the first loss amount at which the cumulative probability reaches $p$. When $P(X\le q)=p$, the lower quantile is exactly $q$; it does not need to pass beyond $q$. Under the strict inverse, the corresponding condition is $P(X<q)\le p<P(X\le q)$, so equality at the cumulative probability before the next support point can shift the answer upward.

These checks are useful because they make the convention visible. Reporting only a number can hide whether the calculation used $\ge$ or $>$. In a discrete problem, write down the cumulative probability at the selected loss and compare it explicitly with the stated confidence level.

Common Exam Traps

Replacing “at least” by “greater than.” The lower-quantile VaR condition is $F_X(x)\ge p$. If equality holds, that loss qualifies.
Ignoring an exact boundary. A confidence level such as $0.75$ may match a cumulative probability exactly. Test equality before deciding whether both conventions agree.
Advancing one row automatically. Move to the next loss amount under the strict inverse only when the current cumulative probability fails the strict $>$ condition.
Confusing a probability with a loss amount. Cumulative probabilities identify which loss qualifies; they are not themselves VaR values.

For a discrete loss model, a reliable response states the convention, displays the cumulative probabilities, and applies the relevant inequality. This is especially important when a problem gives a confidence level that coincides with one of those cumulative probabilities. The distinction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use $Q_-(p)=\inf\{x:F_X(x)\ge p\}$ and $Q_+(p)=\inf\{x:F_X(x)>p\}$. All amounts are hypothetical losses in dollars.

A loss is 0 with probability 0.25, 20 with probability 0.25, and 60 with probability 0.50. Find both quantiles at $p=0.50$. (Answer: $Q_-=20$, $Q_+=60$.)
A loss is 5 with probability 0.10, 25 with probability 0.20, 45 with probability 0.40, and 90 with probability 0.30. Find both quantiles at $p=0.30$. (Answer: $Q_-=25$, $Q_+=45$.)
A loss is 0 with probability 0.40, 40 with probability 0.15, 100 with probability 0.25, and 160 with probability 0.20. Find both quantiles at $p=0.55$. (Answer: $Q_-=40$, $Q_+=100$.)
A loss is 10 with probability 0.20, 30 with probability 0.30, 70 with probability 0.10, and 110 with probability 0.40. Find both quantiles at $p=0.50$, and identify whether this level is a cumulative-probability boundary. (Answer: $Q_-=30$, $Q_+=70$; yes, $F_X(30)=0.50$.)
A loss is 0 with probability 0.12, 25 with probability 0.28, 75 with probability 0.35, and 200 with probability 0.25. Find both quantiles at $p=0.70$, and state whether the conventions differ. (Answer: $Q_-=75$, $Q_+=75$; they agree because $0.40<0.70<0.75$.)

What You'll Learn