13 Feb 2024

Quantile Functions

For a distribution function $F:\mathbb{R}\to [0,1]$, its quatile function $q:(0,1)\to\mathbb{R}$ is defined by¹ \[ q(u):=\inf\{x:F(x)\geq u\}. \] Noting that $F$ is right continuous, it is clear that $\{x: F(x)\geq u\}$ is closed and bounded below for any $0 < u < 1$. Hence, $q(u)=\min\{x:F(x) \geq u\}$. Interestingly, there is also (see the section Properties) \[ F(x) = \sup\{u: q(u) \leq x\}.\]

The graph of a quantile function can be conveniently derived from the graph of the corresponding distribution function.

Plot the graph $(x, F(x))$ for $x\in(-\infty,+\infty)$.
Connect all discontinuity points of $F$ by adding vertical lines.
Swap the horizontal axis with the vertical axes.
Shrink vertical lines to their lowest points to avoid multiple values.

By the definition of quantile functions, in the last step we should choose the lowest point as the proper function value while removing a vertical line. This reveals that, unlike distribution functions, quantile functions are left continuous.

Examples

Example 1. The distribution function of a exponentially distributed random variable is

$$ F(x) = \begin{cases} 1 - e^{-x},&\quad x \geq 0,\\ 0,&\quad x < 0. \end{cases} $$

Its quantile function is \[ q(u) = -\ln (1-u),\quad u\in(0,1). \]

In this case,

$$ q(F(x)) = \begin{cases} x,&\quad x \geq 0,\\ 0,&\quad x < 0, \end{cases} \quad\text{and}\quad F(q(u)) = u, \quad 0 < u < 1. $$

Example 2. The distribution function of a unit mass is

$$ F(x) = \begin{cases} 1 ,&\quad x \geq 0,\\ 0,&\quad x < 0. \end{cases} $$

Its quantile function is \[ q(u) = 0,\quad u\in(0,1).\]

In this case, $q(F(x))$ is void and \[ F(q(u)) = 1, \quad 0 < u < 1.\]

Example 3. The distribution function of two equal mass is

$$ F(x) = \begin{cases} 1 ,&\quad x \geq 1,\\ 1/2, &\quad -1 \leq x < 1,\\ 0,&\quad x < -1. \end{cases} $$

Its quantile function is

$$ q(u) = \begin{cases} 1, &\quad 1/2 < u < 1 \\ -1, &\quad 0 < u \leq 1/2. \end{cases} $$

In this case,

$$ q(F(x)) = -1,\quad -1\leq x < 1\quad\text{and}\quad F(q(u)) = \begin{cases} 1,&\quad 1/2 < u < 1,\\ 1/2,&\quad 0 < u \leq 1/2. \end{cases} $$

Note that in this example, the quantile function is left continuous.

Properties

By definition, a quantile function is clearly nondecreasing due to the nondecreasing nature of distribution functions. However, a quantile function is left continuous, while distribution functions are right continuous. Indeed, as $q$ is nondecreasing, we can show that $q(u-) = q(u)$².

By definition, \[ q(F(x_0)) = \inf\{x: F(x) \geq F(x_0)\} \leq x_0, \quad\forall x_0\in\mathbb{R}.\] This inequality even holds when $F(x_0)$ equals 0 or 1¹. This inequality leads to the equation $F(x)=\sup\{u:q(u) \leq x\}$³. Thus, \[ F(q(u_0)) = \sup\{u: q(u) \leq q(u_0)\} \geq u_0,\quad\forall u_0\in(0,1).\] Similarly, the supremum can be replaced by maximum as $q$ is left continuous.

As a result, there are \[ x \geq q(u) ⇒ F(x) \geq F(q(u)) \geq u \] and \[ F(x) \geq u ⇒ x \geq q(F(x)) \geq q(u),\] which means $x \geq q(u)$ if and only if $F(x) \geq u$. It is also true that $x < q(u)$ if and only if $F(x) < u$. However, it is possible that $x \leq q(u)$ holds but $F(x) \leq u$ fails. This happens when $F(q(u)) > u$, where $x=q(u)$ but $F(x) > u$.

Assume $U$ is a uniformly distributed variable with distribution $\mathbb{P}(U \leq u)=u$ for $u\in(0,1)$. Then the random variable $X=q(U)$ has distribution \[ \mathbb{P}(X \leq x) = \mathbb{P}(q(U) \leq x) = \mathbb{P}(U \leq F(x)) = F(x).\]

Summary of properties of quantile functions.

Domain is $(0, 1)$ and range is $\mathbb{R}$.
Nondecreasing and left continuous.
$q(F(x)) \leq x$ and $F(q(u)) \geq u$
$q(u) \leq x$ if and only if $u \leq F(x)$
$q(u) > x$ if and only if $u > F(x)$
$q(U)$ has distribution function $F$

Footnotes:

Define that $F(-\infty)=0$ and $F(\infty)=1$. Then $F$ becomes a function from $\overline{\mathbb{R}}$ to $[0,1]$ and $q$ becomes a function from $[0,1]$ to $\overline{\mathbb{R}}$.

For any $ϵ > 0$, we have $F(q(u-ϵ)) \leq F(q(u-))$. Applying $F(q(u)) \geq u$ yields $u-ϵ \leq F(q(u-))$. Hence, $F(q(u-)) \geq u$ and by the definition of quantile functions $q(u-) \geq q(u)$.

As $q(F(x)) \leq x$, we know that $F(x) \in \{u: q(u) \leq x\}$, which means $F(x) \leq \sup\{u:q(u) \leq x\}$. For the converse direction, it suffices to show that if $q(u) \leq x$ then $u \leq F(x)$. However, this is true by definition.