Monotonicity and Jump Behavior of Quantiles

How the Shape of a CDF Appears in Its Quantiles

A cumulative distribution function records how probability accumulates as the value $x$ increases. A quantile reverses that question: at what value has a specified probability level been reached? The reversal is not always an ordinary inverse. A CDF can jump at an atom or remain flat across a gap in the distribution’s support, so its quantile may have plateaus or jumps of its own.

Throughout, $F(x)=P(X\le x)$ is the CDF of a real-valued random variable $X$. It is nondecreasing, right-continuous, and satisfies $F(x)\to 0$ as $x\to-\infty$ and $F(x)\to 1$ as $x\to+\infty$. We use the generalized-inverse convention from the earlier material on quantile functions.

Definition: For a probability level $u\in(0,1)$, the quantile function is $$ Q(u)=\inf\{x\in\mathbb{R}:F(x)\ge u\}. $$ The infimum is finite for each $u\in(0,1)$. This definition selects the first value at which the CDF reaches or exceeds $u$.

The word “first” matters. If $F$ is flat at level $u$, there may be many values $x$ with $F(x)=u$; the quantile chooses the left edge of that level set, not an arbitrary point in it. The convention also determines which side of a jump the quantile takes at a probability level exactly equal to the CDF value.

Quantiles Are Nondecreasing

A higher probability level cannot be reached at a smaller value of $x$. This basic fact holds whether the distribution is discrete, continuous, singular, or mixed.

Theorem: If $0<u_1<u_2<1$, then $Q(u_1)\le Q(u_2)$. Thus $Q$ is nondecreasing on $(0,1)$. It need not be strictly increasing.

Proof. Define $A_u=\{x:F(x)\ge u\}$. If $u_1<u_2$, then every $x$ satisfying $F(x)\ge u_2$ also satisfies $F(x)\ge u_1$, so $A_{u_2}\subseteq A_{u_1}$. The infimum of the larger set cannot exceed the infimum of its subset. Therefore

Q(u_1)=\inf A_{u_1}\le \inf A_{u_2}=Q(u_2).

The sets are nonempty and bounded below for $u\in(0,1)$, by the CDF limits at infinity. Hence the infima used here are finite. $\square$

Nondecreasing does not mean strictly increasing. If a CDF jumps over a range of probability levels at one value of $x$, all those levels have the same quantile. Conversely, if a CDF stays at one probability level over a range of $x$-values, the quantile can jump when the probability level moves just above that plateau.

Worked Example: Two Atoms Create Quantile Plateaus

Let $P(X=0)=0.7$ and $P(X=2)=0.3$. The CDF is zero for $x<0$, equals $0.7$ for $0\le x<2$, and equals $1$ for $x\ge2$. Apply the generalized-inverse definition.

For $0<u\le0.7$, the first value at which $F(x)\ge u$ is $x=0$. For $0.7<u<1$, the CDF at $0$ is too small, and the first value reaching the required level is $x=2$. Thus

Q(u)= \begin{cases} 0, & 0<u\le0.7,\\ 2, & 0.7<u<1. \end{cases}

The atom at zero makes the quantile constant across a probability interval of width $0.7$; the atom at two makes it constant across the remaining interval, apart from the excluded endpoint $u=1$. Meanwhile, the CDF is flat at level $0.7$ between zero and two, and the quantile rises from zero to two just after level $0.7$. This example shows both kinds of shape correspondence.

What CDF Jumps and Flat Portions Do

At a point $x_0$, a CDF has a jump when $F(x_0)>F(x_0-)$, where $F(x_0-)=\lim_{x\uparrow x_0}F(x)$. The jump size is $P(X=x_0)$, the probability of an atom at $x_0$. Every level $u$ with $F(x_0-)<u\le F(x_0)$ has quantile $Q(u)=x_0$: below $x_0$, the CDF is less than $u$, and at $x_0$ it has reached at least $u$. Thus a CDF jump produces a flat portion of the quantile function. The lower endpoint of that probability interval is excluded in this general statement; at that exact level the quantile may be smaller.

A flat CDF portion has a different effect. Suppose $F(x)=c$ throughout $[a,b]$, with $0<c<1$, and $F(x)<c$ for $x<a$. Then $Q(c)=a$. For levels just above $c$, the quantile must lie beyond $b$, because no value in $[a,b]$ has CDF greater than $c$. The quantile therefore has a right-hand jump at $c$ if the values needed to exceed $c$ begin strictly to the right of $a$.

Worked Example: A Gap in Support Produces a Quantile Jump

Suppose $X$ has probability $1/2$ uniformly distributed on $[0,1]$ and probability $1/2$ uniformly distributed on $[3,4]$. Its CDF rises from zero to $1/2$ on $[0,1]$, stays at $1/2$ from $1$ through $3$, and then rises to one on $[3,4]$. The gap $(1,3)$ has no probability.

At level $u=1/2$, the first value reaching that level is $1$, so $Q(1/2)=1$. At levels just above $1/2$, the CDF must rise in the second interval, and the corresponding quantiles are just above $3$. More precisely, on the two probability ranges the quantile is

Q(u)= \begin{cases} 2u, & 0<u\le1/2,\\ 2u+2, & 1/2<u<1. \end{cases}

For example, $Q(0.4)=0.8$, $Q(0.5)=1$, and $Q(0.6)=3.2$. The jump at $u=1/2$ reflects the CDF plateau across the gap: it is not caused by an atom at a single value.

More generally, the right-hand limit of the quantile at a level $u\in(0,1)$ identifies where the CDF first becomes strictly greater than $u$. The quantile itself identifies where the CDF first reaches at least $u$. This distinction gives a precise description of a possible jump.

Theorem: For $u\in(0,1)$, $Q$ is left-continuous, and its right-hand limit is $$ \lim_{v\downarrow u,\ v>u}Q(v)=\inf\{x:F(x)>u\}. $$ In particular, the quantile has a right-hand jump at $u$ exactly when this infimum is greater than $Q(u)$.

Proof. First, $Q$ is nondecreasing by the preceding theorem. It is also attained in the sense that $F(Q(u))\ge u$. To see this, write $q=Q(u)$. By the definition of an infimum, for every positive integer $n$ there is an $x_n$ with $q\le x_n<q+1/n$ and $F(x_n)\ge u$. Right-continuity of $F$ gives $F(q)\ge u$ as $x_n\to q$ from the right (if some $x_n=q$, the conclusion is immediate).

Now let $u_n\uparrow u$, with $u_n<u$. Monotonicity implies that $Q(u_n)$ is nondecreasing and bounded above by $Q(u)$, so it has a limit $l\le Q(u)$. If $y<Q(u)$, then $F(y)<u$; otherwise $y$ would belong to $\{x:F(x)\ge u\}$, contradicting the definition of $Q(u)$. For all sufficiently large $n$, $u_n>F(y)$. Since $F(x)\le F(y)$ for every $x\le y$, no such $x$ can reach $u_n$, and therefore $Q(u_n)>y$. This holds for every $y<Q(u)$, so $l\ge Q(u)$. Hence $l=Q(u)$, proving left-continuity.

For the right-hand limit, monotonicity ensures that $\lim_{v\downarrow u,\ v>u}Q(v)$ exists. It equals $\inf_{v>u}Q(v)$, with $v\in(u,1)$. If $F(x)>u$, choose $v$ with $u<v\le F(x)$. Then $Q(v)\le x$, so $\inf_{v>u}Q(v)\le\inf\{x:F(x)>u\}$. Conversely, $F(Q(v))\ge v>u$, so $Q(v)$ belongs to $\{x:F(x)>u\}$. Thus $\inf\{x:F(x)>u\}\le Q(v)$ for every $v>u$, giving the reverse inequality after taking the infimum. This proves the formula and the stated jump criterion. $\square$

Endpoint Limits and Distribution Bounds

The behavior near probability levels zero and one describes how far the distribution extends. Define the extended-real endpoints

\alpha=\inf\{x:F(x)>0\}, \qquad \beta=\inf\{x:F(x)=1\},

where the infimum of an empty set is $+\infty$. For a real-valued random variable, the first set is nonempty; $\alpha$ may be $-\infty$, and $\beta$ may be $+\infty$. These are the lower and upper essential bounds of the distribution.

Result: With limits understood in the extended real line, $$ \lim_{u\downarrow0}Q(u)=\alpha, \qquad \lim_{u\uparrow1}Q(u)=\beta. $$

For the lower limit, if $x<\alpha$, then $F(x)=0$, so $Q(u)>x$ for every $u>0$. This gives $\lim_{u\downarrow0}Q(u)\ge\alpha$. If $F(x)>0$, then for every $u\le F(x)$, $Q(u)\le x$. Taking $u\downarrow0$ and then the infimum over such $x$ gives $\lim_{u\downarrow0}Q(u)\le\alpha$. For the upper limit, if $F(x)=1$, then $Q(u)\le x$ for all $u<1$, giving $\lim_{u\uparrow1}Q(u)\le\beta$. If $F(x)<1$, then $Q(u)>x$ whenever $u>F(x)$. Taking $u\uparrow1$ and then the supremum over such $x$ gives the reverse inequality. Together these establish the result.

Worked Example: Endpoint Limits for an Exponential Distribution

Let $X$ have an exponential distribution with rate $1$, so $F(x)=0$ for $x<0$ and $F(x)=1-e^{-x}$ for $x\ge0$. The distribution has lower essential bound $\alpha=0$, and it never reaches one at a finite $x$, so $\beta=+\infty$.

Solving $1-e^{-x}=u$ gives $Q(u)=-\log(1-u)$. As $u\downarrow0$, $1-u\to1$, so $Q(u)\to0$. As $u\uparrow1$, $1-u\downarrow0$, so $-\log(1-u)\to+\infty$. These limits agree with the distribution’s lower endpoint and unbounded upper tail.

Common Mistakes and Why the Shape Matters

Assuming the quantile is strictly increasing. An atom makes the CDF jump, and all probability levels passed by that jump share the same quantile.
Reversing the two shape relationships. A CDF jump creates a quantile plateau; a flat CDF portion can create a quantile jump. They are different features.
Using the wrong value at a plateau level. If $F=c$ across an interval, $Q(c)$ is the leftmost point at which $F\ge c$, not an arbitrary point of the interval.
Assuming finite endpoint limits. A distribution can have an unbounded lower or upper tail, in which case the corresponding quantile endpoint limit is infinite.

These relationships are useful when interpreting a distribution or sketching its quantile function. A flat stretch of quantiles signals probability concentrated at a value, while a quantile jump can signal a gap in the support. Endpoint limits distinguish bounded distributions from those extending without bound. Keeping the chosen generalized-inverse convention in view also prevents errors at exact probability levels where the CDF jumps or remains flat.

Check Your Understanding

Use the definition $Q(u)=\inf\{x:F(x)\ge u\}$ to answer the following.

Why must $Q(u_1)\le Q(u_2)$ whenever $0<u_1<u_2<1$?
If a CDF jumps from $0.25$ to $0.6$ at $x_0$, what can be said about $Q(u)$ for $0.25<u\le0.6$?
A CDF is constant at $0.4$ throughout an interval $[a,b]$, and is less than $0.4$ for $x<a$. What is $Q(0.4)$, and why can quantiles just above $0.4$ lie to the right of $b$?
What one-sided continuity property does this generalized-inverse convention give $Q$?
If a distribution has a finite lower essential bound and an unbounded upper tail, what are the limits of $Q(u)$ as $u\downarrow0$ and $u\uparrow1$?