Quantiles and Percentiles from a CDF

Why Quantiles Need a CDF Definition

A percentile gives a threshold in the distribution of a random variable. For example, a 70th percentile is a value by which at least 70% of the probability has accumulated. For a continuous distribution with a strictly increasing CDF, finding that threshold often means solving $F(x)=0.70$. But a CDF can jump over 0.70, or remain level across a range of values. In those cases, that equation alone may have no solution or more than one solution.

The generalized inverse gives one consistent answer in all these cases. It is especially useful for discrete distributions, where percentiles need not divide probability exactly, and for distributions whose support has gaps.

Definition: Let $X$ be a real-valued random variable with CDF $F_X(x)=P(X\le x)$. For $0<p<1$, the $p$-quantile is $$ q_p=\inf\{x\in\mathbb{R}:F_X(x)\ge p\}. $$ The $100p$th percentile is $q_p$. Thus, the 70th percentile is $q_{0.70}$, not $q_{70}$.

The infimum in the definition selects the leftmost value at which the CDF has reached or passed the target probability. This convention is sometimes called the left quantile. Other conventions may be used for sample-data quantiles, but here the quantile is defined from the population CDF by the formula above.

How to Read a Quantile from a CDF

For a chosen probability $p$, imagine drawing a horizontal line at height $p$ on a graph of $F_X$. Find the first point, moving from left to right, where the CDF reaches or crosses that line. The corresponding $x$-coordinate is $q_p$.

Convert the percentile to a probability.
For a percentile $r$, set $p=r/100$. For example, the 35th percentile uses $p=0.35$.

Find where the CDF first reaches $p$.
For a discrete CDF, inspect its cumulative jumps. For a continuous CDF, solve $F_X(x)=p$ where it is increasing, while respecting the support.

Check the threshold.
Confirm that the CDF at the proposed quantile is at least $p$, and that values to its left have CDF below $p$.

A discrete CDF may cross the target in a jump. Then its value at the quantile can be greater than $p$. A continuous CDF may instead reach $p$ and remain flat; the generalized inverse chooses the left endpoint of that flat section. In either situation, the definition identifies a threshold, not necessarily a value satisfying a unique equation $F_X(x)=p$.

Why the Generalized Inverse Has This Threshold Property

The CDF is nondecreasing: if $x_1<x_2$, then $F_X(x_1)\le F_X(x_2)$. For a fixed $p$, consider the set $A_p=\{x:F_X(x)\ge p\}$. Once the CDF has reached $p$, it cannot drop below $p$, so every larger $x$ is also in $A_p$. The set is therefore an upper ray. The definition of $q_p$ chooses its left boundary.

More precisely, for a proper real-valued random variable and $0<p<1$, the CDF’s right-continuity ensures that this boundary belongs to $A_p$. Consequently,

F_X(q_p)\ge p \qquad\text{and}\qquad F_X(x)<p\quad\text{for every }x<q_p.

The first inequality says that at least probability $p$ is at or below the quantile. The second says that no smaller threshold has accumulated probability $p$. In terms of the left limit $F_X(q_p-)=P(X<q_p)$, these facts give $F_X(q_p-)\le p\le F_X(q_p)$. At a jump, the first inequality can be strict. When the CDF is continuous, it has no jump at the quantile, so $F_X(q_p)=p$, even if the CDF is flat immediately to the right.

Worked Example: A Percentile from a Discrete CDF

A hypothetical claim-count variable $X$ has the following distribution: $P(X=0)=0.10$, $P(X=1)=0.25$, $P(X=2)=0.40$, and $P(X=5)=0.25$. Find its 60th percentile. State the method and identify the mistake of interpolating between possible claim counts.

Method: form the cumulative probabilities and select the first value whose CDF is at least 0.60. The probabilities sum to $0.10+0.25+0.40+0.25=1.00$, as required. The CDF at each support value is:

F_X(0)=0.10,\quad F_X(1)=0.10+0.25=0.35,\quad F_X(2)=0.35+0.40=0.75,\quad F_X(5)=1.00.

The target is $p=0.60$. At $x=1$, $F_X(1)=0.35<0.60$; at $x=2$, $F_X(2)=0.75\ge0.60$. Therefore the first value reaching the target is $q_{0.60}=2$, so the 60th percentile is 2 claims.

Check: $F_X(2)=0.75\ge0.60$, while every value below 2 has CDF at most 0.35. The answer satisfies the threshold definition.

Common mistake avoided: Interpolating between 1 and 2 to try to make the CDF equal 0.60 would produce a number between possible claim counts. The CDF has a jump from 0.35 to 0.75 at 2, so there is no value in that interval where this discrete CDF equals 0.60. The generalized inverse correctly returns 2.

Continuous CDFs: Invert, Then Check the Support

When a CDF is continuous and strictly increasing over the relevant interval, its $p$-quantile is found by solving $F_X(x)=p$. The solution must lie in the interval where that CDF formula applies. If the CDF has a flat segment, the equality may hold at many points, and the generalized inverse chooses the leftmost one.

Worked Example: A Quantile at a Flat Section

Suppose $X$ has density $f_X(x)=1/4$ on $[0,2]$ and $[4,6]$, and density zero elsewhere. Find the 50th percentile and the 70th percentile. This is a continuous distribution with no probability in the gap from 2 to 4.

Method: write the CDF on each support interval and use the first-reach convention. Each interval has length 2, so the density integrates to $2(1/4)=1/2$ on each interval. The CDF is

F_X(x)= \begin{cases} 0, & x<0,\\ x/4, & 0\le x\le 2,\\ 1/2, & 2<x<4,\\ 1/2+(x-4)/4, & 4\le x\le 6,\\ 1, & x>6. \end{cases}

The 50th percentile uses $p=0.50$. On the first support interval, $x/4=0.50$ gives $x=2$. The CDF then remains at 0.50 throughout the gap. Although every $x$ between 2 and 4 has $F_X(x)=0.50$, the infimum of the values reaching 0.50 is 2. Thus $q_{0.50}=2$.

For the 70th percentile, $p=0.70$. This probability is reached on $[4,6]$, so solve

\frac12+\frac{x-4}{4}=0.70,\qquad \frac{x-4}{4}=0.20,\qquad x-4=0.80,\qquad x=4.80.

Therefore $q_{0.70}=4.80$. Check: $F_X(4.80)=0.50+0.80/4=0.70$, and the CDF is below 0.70 at every smaller value. The support check matters: using the first-interval formula $x/4$ for $p=0.70$ would give $x=2.80$, which is outside that formula’s interval and lies in the zero-density gap.

Common mistake avoided: For $p=0.50$, solving $F_X(x)=0.50$ without applying the infimum rule gives an entire interval of answers. The quantile convention chooses its left endpoint, 2, rather than an arbitrary point in the gap.

Quantiles for a Distribution with Both a Jump and a Continuous Part

A distribution need not be purely discrete or purely continuous. A point mass creates a jump in the CDF, while a continuous component can make the CDF rise smoothly elsewhere. The same generalized-inverse definition applies without modification.

Worked Example: A Quantile from a Mixed Distribution

A hypothetical waiting-time variable $X$, measured in hours, equals 1 with probability 0.30. With probability 0.70, it is uniformly distributed over $[2,6]$. Find the 20th percentile, the 30th percentile, and the 65th percentile. Be careful about the jump at 1.

Method: locate probabilities covered by the point mass first, then invert the continuous part. The mass at 1 raises the CDF from 0 to 0.30. For $2\le x\le6$, the uniform component has conditional CDF $(x-2)/4$; multiplying by its mixture probability 0.70 gives

F_X(x)=0.30+0.70\frac{x-2}{4},\qquad 2\le x\le6.

For $p=0.20$, the jump at 1 has already brought the CDF to 0.30, which is at least 0.20. No value below 1 reaches 0.20, so $q_{0.20}=1$.

For $p=0.30$, the CDF first reaches 0.30 at the point mass $x=1$. Values below 1 have CDF 0, so $q_{0.30}=1$ as well. Notice that the quantile at this exact jump level is still 1.

For $p=0.65$, the target lies above the jump, so solve on $[2,6]$:

0.30+0.70\frac{x-2}{4}=0.65,\qquad 0.70\frac{x-2}{4}=0.35,\qquad \frac{x-2}{4}=0.50,\qquad x=4.

Thus the 65th percentile is 4 hours. Check: $F_X(4)=0.30+0.70(2/4)=0.65$; at every smaller value in the continuous interval the CDF is below 0.65, and all values below 2 have CDF at most 0.30.

Common mistake avoided: Treating the 20th percentile as a point on the continuous interval ignores the 0.30 point mass. Since the CDF reaches 0.20 at the jump, the correct quantile is 1, not a value between 2 and 6.

Common Exam Traps

Using the percentile number as a probability. The 85th percentile means $p=0.85$, not $p=85$.
Requiring equality at a jump. The condition is $F_X(q_p)\ge p$, not necessarily $F_X(q_p)=p$. A jump can carry the CDF past the target.
Ignoring the leftmost-value rule on a plateau. If the CDF equals $p$ over an interval, the quantile is the interval’s leftmost point where the CDF has reached $p$.
Using a CDF formula outside its interval. A solution obtained by algebra is not valid until it is checked against the range for which that formula applies.
Confusing a probability threshold with a guarantee about every observation. The quantile condition concerns cumulative probability: at least $p$ is at or below $q_p$. It does not assert that the variable must take the value $q_p$.

Key takeaway: For $0<p<1$, define $q_p$ as the smallest threshold whose CDF is at least $p$. This first-reach rule handles continuous rises, discrete jumps, and flat sections with one consistent convention.

Check Your Understanding

For each problem, use the generalized inverse: identify the first value at which the CDF reaches or exceeds the stated probability.

A discrete variable $X$ has $P(X=-1)=0.20$, $P(X=2)=0.50$, and $P(X=5)=0.30$. Find the 60th percentile. (Answer: 2)
A continuous variable has CDF $F_X(x)=x^2/9$ for $0\le x\le3$, with $F_X(x)=0$ below 0 and $F_X(x)=1$ above 3. Find the 80th percentile, rounded to four decimal places. (Answer: $3\sqrt{0.80}=2.6833$)
A continuous variable has density 0.40 on $[0,1]$, zero on $(1,3)$, and density 0.60 on $[3,4]$, and zero elsewhere. Find the 40th and 70th percentiles. (Answer: $q_{0.40}=1$ and $q_{0.70}=3.5$)
A mixed variable equals 0 with probability 0.25 and, with probability 0.75, is uniform on $[2,6]$. Find the 25th and 75th percentiles. (Answer: $q_{0.25}=0$ and $q_{0.75}=14/3$)
A discrete variable takes values 1, 4, and 9 with probabilities 0.15, 0.55, and 0.30, respectively. Find its 15th and 70th percentiles, and state whether the CDF equals each target probability at the corresponding quantile. (Answer: $q_{0.15}=1$, with $F_X(1)=0.15$; $q_{0.70}=4$, with $F_X(4)=0.70$)

What You'll Learn