Quantiles And Inverse Distribution Functions

Why Quantiles Matter

A mean describes an average loss, but it does not identify a threshold that a specified share of losses will not exceed. A quantile answers that threshold question. For example, a 90th percentile loss is a value at or below which at least 90% of the probability lies. Quantiles are useful for describing severity distributions and interpreting loss thresholds, whether the model is continuous, discrete, or a mixture of the two.

The calculation depends on the shape of the cumulative distribution function (CDF). For a continuous distribution that rises strictly, finding a quantile is usually an inverse-function calculation. For a discrete distribution, the CDF jumps, so the quantile is the loss at the first jump that reaches the target probability. The same definition handles both cases without interpolation.

Definition: Let $X$ be a loss with CDF $F_X(x)=P(X\le x)$. For $0<p<1$, the $p$-quantile of $X$ is the generalized inverse $$ Q_X(p)=\inf\{x:F_X(x)\ge p\}. $$ The $p$-quantile is also called the $100p$th percentile. The definition selects the smallest loss threshold at which the cumulative probability is at least $p$.

The definition is important when there is a jump or a flat portion in the CDF. A discrete loss may have no possible value between two possible amounts, and a mixed distribution may have both continuous probability and atoms. The generalized inverse still gives a precise answer. In contrast, simply solving $F_X(x)=p$ can fail when the CDF jumps over $p$, or when it equals $p$ across an interval.

Checking a Quantile at a Jump

For a proposed quantile $q=Q_X(p)$, the defining conditions can be checked using the CDF immediately below $q$ and at $q$. Write $F_X(q-)=P(X<q)$ for the left limit of the CDF. For ordinary real-valued loss distributions and $0<p<1$, a generalized-inverse quantile satisfies

F_X(q-)\le p\le F_X(q).

At a jump, the first inequality can be strict and the second can also be strict. In a continuous distribution whose CDF is strictly increasing, both sides equal $p$. These inequalities provide a quick check, but the definition remains the safest way to identify the correct threshold: find the first $x$ for which $F_X(x)\ge p$.

State the target probability.
Convert a percentile such as the 80th percentile to $p=0.80$.

Identify the CDF behavior.
Determine whether the CDF is continuous and strictly increasing, has jumps, or has both continuous and discrete parts.

Find the first threshold reaching $p$.
For an invertible continuous CDF, solve $F_X(x)=p$. For a discrete CDF, accumulate the probabilities and select the first loss with cumulative probability at least $p$.

Verify the result.
Check that the CDF at the proposed value is at least $p$, and that values immediately below it have CDF less than $p$, where applicable.

Continuous Quantiles: Inverting the CDF

If $F_X$ is continuous and strictly increasing over the relevant range, the generalized inverse is the ordinary inverse: $Q_X(p)$ is the unique $x$ satisfying $F_X(x)=p$. This equality is convenient, but it is not a universal quantile rule. It relies on the CDF being continuous and strictly increasing at the target. For example, a CDF with a jump can pass from a value below $p$ to a value above $p$ without ever equaling $p$.

Worked Example: Inverting a Continuous Severity CDF

A hypothetical loss $X$, measured in dollars, has the following CDF:

F_X(x)= \begin{cases} 0, & x<0,\\ (x/12{,}000)^2, & 0\le x\le 12{,}000,\\ 1, & x>12{,}000. \end{cases}

Find the 64th percentile and the probability that a loss exceeds it.

Method. On the interval from 0 to $12,000, the CDF is continuous and strictly increasing. Therefore, set the CDF equal to $p=0.64$ and solve for $x$:

\left(\frac{x}{12{,}000}\right)^2=0.64.

Taking the nonnegative square root gives $x/12{,}000=0.8$, so

x=12{,}000(0.8)=\$9{,}600.

Thus $Q_X(0.64)=\$9{,}600$. Check the CDF value:

F_X(9{,}600)=(9{,}600/12{,}000)^2=(0.8)^2=0.64.

The requested exceedance probability is

P(X>9{,}600)=1-F_X(9{,}600)=1-0.64=0.36.

The answer is a 64th-percentile loss of $9,600, with 0.36 probability of exceeding it. The common mistake to avoid is using an inverse-CDF equation without checking the CDF’s behavior. Here inversion is justified by continuity and strict increase; it would not be the correct procedure for a CDF that jumps over 0.64.

Discrete Quantiles: Locate the First Cumulative Probability

For a discrete loss, list the possible loss amounts in increasing order and add their probabilities cumulatively. The quantile is the first loss amount for which the cumulative probability is at least the target. A target probability does not need to equal one of the cumulative probabilities.

Worked Example: A Percentile Within a CDF Jump

A hypothetical claim loss has the distribution $P(X=\$0)=0.20$, $P(X=\$1{,}000)=0.50$, and $P(X=\$5{,}000)=0.30$. Find the 65th percentile.

Method. Order the possible losses and calculate the cumulative probabilities:

F_X(0)=0.20,\qquad F_X(1{,}000)=0.20+0.50=0.70,\qquad F_X(5{,}000)=0.70+0.30=1.

At $0, the CDF is $0.20$, which is less than $0.65$. At $1,000, it is $0.70$, which is at least $0.65$. Therefore, the first loss amount reaching the target is

Q_X(0.65)=\$1{,}000.

The check at the selected value is $F_X(1{,}000-)=0.20\le 0.65\le 0.70=F_X(1{,}000)$. The quantile is $1,000, even though $F_X(1{,}000)$ is not exactly 0.65.

A common mistake is to interpolate between loss amounts or CDF values. For instance, interpolating linearly between $(0,0.20)$ and $(1{,}000,0.70)$ would give

0+\frac{0.65-0.20}{0.70-0.20}(1{,}000-0)=\$900.

But $900 is not a possible loss in this model, and $F_X(900)=0.20$, which is below 0.65. The generalized inverse does not interpolate through a jump: it selects $1,000 as soon as the CDF first reaches or exceeds 0.65. The incorrect $900 calculation treats the jump as if probability were spread continuously between the two loss amounts.

Quantiles in a Distribution With Several Jumps

The CDF may jump at each possible loss amount. The selected value depends on which jump first reaches the target, not on how close the target is to the next cumulative probability. It is also useful to distinguish the probability of a loss strictly greater than the quantile from the probability of a loss at least as large as the quantile. When there is an atom at the quantile, these probabilities differ.

Worked Example: A Discrete Quantile and Its Tail Probabilities

A hypothetical equipment-loss model has possible losses of $0, $2,000, $4,000, and $9,000, with respective probabilities $0.12$, $0.28$, $0.35$, and $0.25$. Find the 70th percentile, then calculate the probability of a loss strictly greater than that percentile and the probability of a loss at least as large as it.

Method. First check that the probabilities sum to 1:

$$ 0.12+0.28+0.35+0.25=1.00. $$

Now accumulate them in order of loss:

F_X(0)=0.12,\quad F_X(2{,}000)=0.12+0.28=0.40,

F_X(4{,}000)=0.40+0.35=0.75,\quad F_X(9{,}000)=0.75+0.25=1.00.

The first cumulative probability at least $0.70$ is $0.75$, at $4,000. Thus $Q_X(0.70)=\$4{,}000$. The strict exceedance probability excludes the probability mass at $4,000:

$$ P(X>4{,}000)=P(X=9{,}000)=0.25. $$

The probability of a loss at least as large as $4,000 includes the mass at $4,000:

P(X\ge 4{,}000)=P(X=4{,}000)+P(X=9{,}000)=0.35+0.25=0.60.

As a check, $F_X(4{,}000-)=0.40\le 0.70\le 0.75=F_X(4{,}000)$. The common mistake is to use $1-F_X(q)$ for $P(X\ge q)$. In this example, $1-F_X(4{,}000)=0.25$ is $P(X>4{,}000)$; the probability of $X\ge4{,}000$ is $0.60$.

Why the Generalized Inverse Works

A CDF is nondecreasing: if $x_1<x_2$, then $F_X(x_1)\le F_X(x_2)$, because the event $\{X\le x_1\}$ is contained in $\{X\le x_2\}$. The set of thresholds at which the CDF is at least $p$ therefore begins at the first point where the cumulative probability reaches the target. The infimum in the definition selects that boundary, rather than an arbitrary point farther to the right.

More formally, let $q=\inf\{x:F_X(x)\ge p\}$. For any $x<q$, the definition of $q$ implies $F_X(x)<p$. For a real-valued loss CDF, right-continuity ensures that the threshold $q$ itself has $F_X(q)\ge p$. The left limit consequently satisfies $F_X(q-)\le p$, while the CDF at $q$ satisfies $F_X(q)\ge p$. Thus the quantile is exactly the boundary between thresholds that have not yet accumulated probability $p$ and thresholds that have.

Key takeaway: A quantile is the smallest threshold whose CDF is at least the target probability. Use ordinary inversion only when the CDF is continuous and strictly increasing; for jumps, select the first loss amount whose cumulative probability reaches the target.

Common Quantile Traps

Interpolating across a jump. A discrete CDF assigns probability at particular loss amounts. Linear interpolation can produce an amount with a CDF still below the target.
Requiring equality with the target. For a discrete distribution, $F_X(q)$ is often strictly greater than $p$. The correct condition is $F_X(q)\ge p$.
Selecting the next loss instead of the first qualifying one. Once the cumulative probability reaches the target, later loss amounts cannot be the generalized-inverse quantile.
Confusing strict and inclusive tails. $1-F_X(q)=P(X>q)$, whereas $P(X\ge q)=1-F_X(q-)$. The distinction matters when $P(X=q)>0$.

A reliable exam check is to state the cumulative probability just below the proposed quantile and at the quantile. If the former is below $p$ and the latter is at least $p$, the defining threshold condition is met. This method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the generalized inverse definition. The final answers allow you to check both the selected threshold and any requested probability.

A hypothetical continuous loss has $F_X(x)=(x/10{,}000)^2$ for $0\le x\le10{,}000$, with CDF 0 below 0 and 1 above $10,000. Find its 49th percentile and the probability of exceeding that value. (Answer: $Q_X(0.49)=\$7{,}000$; exceedance probability $=0.5100$.)
A discrete loss is $0 with probability 0.15, $1,000 with probability 0.25, and $3,000 with probability 0.60. Find the 40th percentile. (Answer: $Q_X(0.40)=\$1{,}000$, since the cumulative probabilities are 0.15, 0.40, and 1.00.)
A loss is $500 with probability 0.20, $1,500 with probability 0.30, and $4,000 with probability 0.50. Find the 75th percentile and give the CDF immediately below and at the quantile. (Answer: $Q_X(0.75)=\$4{,}000$; $F_X(4{,}000-)=0.50$, $F_X(4{,}000)=1.00$.)
A discrete loss is $0, $2, $5, or $8 thousand with respective probabilities 0.08, 0.22, 0.45, and 0.25. Find the 53rd percentile, $P(X>Q_X(0.53))$, and $P(X\ge Q_X(0.53))$. (Answer: $Q_X(0.53)=\$5{,}000$; $P(X>Q)=0.25$; $P(X\ge Q)=0.70$.)
A mixed loss has probability 0.30 at $0, probability 0.20 at $2,000, and remaining probability 0.50 spread continuously from $2,000 to $8,000 with conditional CDF $(x-2{,}000)/6{,}000$ on that interval. Find $Q_X(0.25)$ and $Q_X(0.65)$. (Answer: $Q_X(0.25)=\$0$; $Q_X(0.65)=\$3{,}800$.)

What You'll Learn