Quantile Conventions for Losses

Why Quantile Conventions Matter

A quantile specifies a location in a loss distribution by reference to a probability level. For a smooth, strictly increasing cumulative distribution function, common quantile conventions give the same answer. For distributions with atoms or gaps, however, the CDF can jump over a level or remain constant at it across an interval. The convention then determines which boundary is selected.

This distinction matters whenever a loss quantile is used as an input to a risk measure or capital calculation. The calculation must identify not just the probability level, but also the quantile convention. In particular, an infimum is a boundary of a set: it need not itself be a point in that set. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a loss random variable with cumulative distribution function $F_X(x)=P(X\leq x)$, and let $0<\alpha<1$. The lower $\alpha$-quantile is $$ q^-_\alpha(X)=\inf\{x\in\mathbb{R}:F_X(x)\geq\alpha\}. $$ The upper $\alpha$-quantile is $$ q^+_\alpha(X)=\inf\{x\in\mathbb{R}:F_X(x)>\alpha\}. $$

The only difference in the definitions is the inequality: the lower quantile uses $F_X(x)\geq\alpha$, while the upper quantile uses $F_X(x)>\alpha$. The upper quantile is the infimum of points where the CDF is strictly beyond the level. It is not defined as a point where the CDF must strictly exceed the level.

For $0<\alpha<1$, both sets in these definitions are nonempty and bounded below for an ordinary probability distribution on the real line: $F_X(x)$ tends to 1 as $x$ increases and to 0 as $x$ decreases. Their infima therefore identify finite boundary locations. When losses are nonnegative, the relevant boundaries will also be nonnegative.

What the Two Conventions Select

The lower quantile marks the boundary where the CDF has reached or passed the probability level. The upper quantile marks the boundary of the set where the CDF is strictly greater than that level. When the CDF rises continuously and strictly through $\alpha$, these boundaries coincide. If the CDF stays at $\alpha$ over a range, the lower convention selects the start of that range and the upper convention its end.

An atom is probability concentrated at a single loss amount, so the CDF jumps there. A gap is a range of losses to which no probability is assigned; across such a range the CDF is constant. A flat portion at level $\alpha$ can arise in a gap, or across a stretch with no probability mass for another reason. The quantile definitions remain valid in all these cases.

Worked Example: A Discrete Loss at an Exact Cumulative Level

A hypothetical equipment policy has loss $X$ with the following distribution. Find the lower and upper quantiles at $\alpha=0.50$, and then at $\alpha=0.62$.

Loss $x$, dollars	$P(X=x)$	$F_X(x)$
0	0.15	0.15
40	0.35	0.50
100	0.30	0.80
250	0.20	1.00

Step 1: Check the distribution. The probabilities sum to $0.15+0.35+0.30+0.20=1.00$. The cumulative probabilities are obtained by adding the masses in order of loss.

Step 2: Use $\alpha=0.50$. The CDF first reaches 0.50 at $x=40$, so $q^-_{0.50}(X)=40$. At every $x$ from 40 up to, but not including, 100, the CDF remains 0.50. It first becomes strictly greater than 0.50 at $x=100$, where it is 0.80. Thus $q^+_{0.50}(X)=100$.

Step 3: Use $\alpha=0.62$. At $x=40$, the CDF is 0.50, which is below 0.62. At $x=100$, it is 0.80, which is both at least 0.62 and strictly greater than 0.62. Both infima are therefore 100: $q^-_{0.62}(X)=q^+_{0.62}(X)=100$.

Check. At the first probability level, the exact equality $F_X(40)=0.50$ separates the two conventions. At the second level, the CDF jumps across 0.62 at the same loss amount, so the quantiles agree.

Why the Infimum Convention Works

The infimum makes the definitions precise even when there is no smallest point satisfying the strict inequality. The key checks follow from monotonicity and right-continuity of a CDF. Monotonicity means that $F_X(x)$ cannot decrease as $x$ increases; right-continuity means that its value at a point agrees with its limit when approached from the right.

Start with the lower quantile.
Write $q=q^-_\alpha(X)$. For every $y<q$, $F_X(y)<\alpha$; otherwise $y$ would already be in the set whose infimum is $q$.

Approach the boundary from the right.
There are points $x_n$ in the defining set with $q\leq x_n<q+1/n$. Thus $x_n$ approaches $q$ from the right and $F_X(x_n)\geq\alpha$.

Apply right-continuity.
Since $F_X(x_n)$ tends to $F_X(q)$, it follows that $F_X(q)\geq\alpha$. Together with $F_X(y)<\alpha$ for $y<q$, this verifies the lower boundary.

Repeat for the upper quantile.
For $q=q^+_\alpha(X)$, every $y<q$ has $F_X(y)\leq\alpha$. Points in the defining set can approach $q$ from the right, so right-continuity gives $F_X(q)\geq\alpha$, but not necessarily $F_X(q)>\alpha$.

The derivation explains the crucial distinction: at the upper-quantile boundary, the CDF may equal $\alpha$. The set of points with $F_X(x)>\alpha$ can begin just to the right of that boundary, making the boundary its infimum even though it is not in the set.

Boundary check: Both conventions satisfy $F_X(q)\geq\alpha$ at their quantile $q$, and $F_X(y)\leq\alpha$ for every $y<q$. For a lower quantile, the defining set additionally contains $q$. For an upper quantile, it may not.

Worked Example: A Gap Where the Upper Infimum Is Not in Its Set

A hypothetical portfolio loss has probability 0.40 spread uniformly over losses from 0 to 10, and probability 0.60 spread uniformly over losses from 20 to 30. Find both quantiles at $\alpha=0.40$.

Step 1: Write the CDF over each region. On $0\leq x\leq10$, the first component has accumulated probability $0.40(x/10)$. The second component contributes nothing below 20. Therefore $F_X(x)=0.04x$ on that interval. For $10\leq x\leq20$, the first component is fully accumulated and there is no probability in the gap, so $F_X(x)=0.40$. For $20\leq x\leq30$, the CDF is $0.40+0.60((x-20)/10)$.

Step 2: Find the lower quantile. On the first interval, $F_X(10)=0.04(10)=0.40$, and $F_X(x)<0.40$ for $x<10$. Hence $q^-_{0.40}(X)=10$.

Step 3: Find the upper quantile. The CDF equals 0.40 throughout the gap, including at $x=20$. For every $x>20$ sufficiently close to 20, the second component has begun accumulating probability, so $F_X(x)>0.40$. Consequently the set where $F_X(x)>0.40$ has infimum 20, and $q^+_{0.40}(X)=20$.

Check. At the upper quantile, $F_X(20)=0.40$, not a value strictly greater than 0.40. The quantile is still 20 because it is the infimum of points where the CDF is strictly greater than 0.40. The mixture weights $0.40+0.60=1.00$, as required.

Atoms, Gaps, and Equal Quantiles

An atom does not always make the two quantiles different. If the probability level falls strictly inside a jump, the jump crosses the level and both conventions generally select the location of that jump. They differ when the CDF reaches exactly $\alpha$ and then remains at that level before rising again. The lower quantile can be at the first point of equality, while the upper quantile is at the boundary beyond which the CDF becomes larger.

A useful way to identify this situation is to examine the set where $F_X(x)=\alpha$. If that set has a nonzero interval, the lower quantile is its left boundary and the upper quantile is its right boundary, provided the CDF eventually exceeds $\alpha$. The upper boundary itself may still have CDF equal to $\alpha$. If the set has just one point, the two conventions may coincide, as they do when a continuous, strictly increasing CDF passes through the level.

Worked Example: An Atom Followed by a Continuous Loss Range

For a hypothetical claim, $P(X=0)=0.25$. With the remaining probability 0.75, the loss is uniform on the interval from 10 to 16. Find both quantiles at $\alpha=0.25$ and $\alpha=0.625$.

Step 1: State the CDF where probability accumulates. At $x=0$, the atom makes $F_X(0)=0.25$. For $10\leq x\leq16$, the uniform component contributes $0.75(x-10)/6$, so $$ F_X(x)=0.25+0.75\frac{x-10}{6}. $$ Between 0 and 10 the CDF remains 0.25; below 0 it is 0.

Step 2: Calculate the quantiles at $\alpha=0.25$. The lower set includes $x=0$, because $F_X(0)=0.25$, and no smaller $x$ qualifies. Thus $q^-_{0.25}(X)=0$. The CDF is not strictly greater than 0.25 until losses exceed 10. The infimum of those losses is 10, so $q^+_{0.25}(X)=10$, even though $F_X(10)=0.25$.

Step 3: Calculate the quantiles at $\alpha=0.625$. Set the CDF in the uniform range equal to 0.625: $$ 0.25+0.75\frac{x-10}{6}=0.625. $$ Subtracting 0.25 gives $0.75(x-10)/6=0.375$. Dividing by 0.75 gives $(x-10)/6=0.5$, so $x-10=3$ and $x=13$. The CDF rises strictly through 0.625 at 13, so both infima equal 13.

Check. Substitution gives $F_X(13)=0.25+0.75(3/6)=0.25+0.375=0.625$. The component probabilities sum to 1, and the two conventions differ at 0.25 but agree at 0.625.

Common Exam Trap

A frequent error is to describe the upper quantile as “the first loss amount where the CDF exceeds $\alpha$.” That wording assumes there is a first such amount. In the gap example, every $x>20$ close to 20 has $F_X(x)>0.40$, but $F_X(20)=0.40$. There is no smallest loss amount strictly greater than 20 with CDF above 0.40; the infimum is 20.

The incorrect calculation is to require $F_X(q^+_\alpha)>\alpha$, then conclude that 20 cannot be the upper quantile. The correction is to apply the definition: $q^+_\alpha=\inf\{x:F_X(x)>\alpha\}$. For that example, the set contains values greater than 20 and arbitrarily close to 20, so its infimum is 20. The infimum need not belong to the set.

Key takeaway: Always state which quantile convention is being used. Calculate the CDF carefully at atoms and across flat portions, then apply the relevant infimum definition. Do not replace the upper-quantile definition with a requirement that the CDF strictly exceed $\alpha$ at the quantile itself.

Check Your Understanding

For each problem, determine the requested quantile or quantiles using the definitions above.

A hypothetical deductible-loss variable has $P(X=0)=0.10$, $P(X=50)=0.40$, and $P(X=200)=0.50$. Find the lower and upper quantiles at $\alpha=0.10$. (Answer: $q^-_{0.10}=0$, $q^+_{0.10}=50$.)
A hypothetical loss is uniform on $[0,4]$ with probability 0.30 and uniform on $[8,18]$ with probability 0.70. Find both quantiles at $\alpha=0.30$. (Answer: $q^-_{0.30}=4$, $q^+_{0.30}=8$.)
A hypothetical contract has $P(X=5)=0.20$, with the remaining probability 0.80 uniformly distributed on $[9,19]$. Find both quantiles at $\alpha=0.60$. (Answer: $q^-_{0.60}=q^+_{0.60}=14$.)
A discrete hypothetical loss has probabilities 0.25, 0.25, and 0.50 at losses 2, 7, and 12, respectively. Find the lower and upper quantiles at $\alpha=0.50$. (Answer: $q^-_{0.50}=7$, $q^+_{0.50}=12$.)
A loss has CDF $F_X(x)=x^2/2500$ for $0\leq x\leq50$, with $F_X(x)=0$ below 0 and $F_X(x)=1$ above 50. Find both quantiles at $\alpha=0.36$. (Answer: $q^-_{0.36}=q^+_{0.36}=30$.)

What You'll Learn