Questions about definition of Quantile function

Question

Let $F$ be a distribution function. For $0<p<1$, the $p$-th quantile or fractile of $F$ is defined by $$\xi_p = F^{\leftarrow}(p) = \inf\{x:F(x)\geq p\}$$

my questions are following:

1. Why we take $\inf$? If we take $\min$ then what kind of problem arise?
2. Why we can not take $\sup$? If we take $\sup$ then which portion of the definition of the $F$ we have to modify? And what kind of problem arise when we take $\sup$?
3. Suppose we want to study empirical quantile function using $\sup$ in the definition then what kind of problem arise?

Any kind of help appreciable.

Thanks in advance

Answer 1

Your definition of quantile is not entirely correct.

Definition: Let $(\Omega,\mathscr{F},\mathbb{P})$ and $X$ be a probability space and a real--valued measurable function on $(\Omega,\mathscr{F})$ respectively. For any $q\in(0,1)$, a number $z_q$ such that $$ \Pr[X<z_q]\leq q\qquad\text{and}\qquad\Pr[X\leq z_q]\geq q, $$ is called a $q$--quantile of $X$.

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Notice that in principle, there may be infinitely many $q$-quantiles. For instance, a random variable $X$ whose probability distribution $F_X(x)=P[X\leq x]$ remains constant in some interval $I$, and takes value $q$ with $0<q<1$. Every the element of the interval $I$, with the possible exception of the endpoints will be a $q$-quantile.

There are two special quantiles which are in a sense the extreme or endpoints of quantiles at any level:

Consider the functions $Q$ and $Q^+$on $(0,1)$ defined by $$ \begin{align} Q(q) &=\inf\{x\in\mathbb{R}:\Pr[X\leq x]\geq q\}\\ Q^+(q)&=\sup\{x\in\mathbb{R}:\Pr[X < x]\leq q\} \end{align} $$

It turns out that

1. $Q$ an $Q^+$ are non--decreasing left--continuous and right--continuous respectively.
2. $Q(q)$ is the smallest $q$--quantile of $X$ and $Q^+(q)$ is the largest $q$--quantile of $X$.

The quantile $Q$ is rather special in the following sense. Consider the Lebesgue space $\big((0,1),\mathscr{B}(0,1),\lambda)$, where $\lambda$ is Lebesgue's measure restricted to Borel susets ($\mathscr{B}(0,1)$) of $(0,1)$.

3. Then $\lambda\Big(Q^{-1}\big((-\infty,x]\big)\Big)=\mathbb{P}[X\leq x]$. Thus, $Q$ and $X$ have the same law (probability distribution), and for any bounded measurable function $\phi$ on $\mathbb{R}$ $$ \int_{(0,1)}\phi(Q(p))\,dp=\int_\Omega \phi(X(\omega))\mathbb{P}(d\omega)$$

In other words, $Q$ is an explicit random variable in a concrete probability space ($\big((0,1),\mathscr{B}(0,1),\lambda)$) that has distribution of $X$. This implies that if one knows how to generate uniformly distributed random variables (the function $T(t)=t$ in $(0,1)$ has the uniform distribution 0-1) then we can, in principle, generate any distribution over $(\mathbb{R},\mathscr{B}(\mathbb{R})$.

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The validity of (1) and(2) lie on the following lemmas

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Lemma 1. Suppose $H:(a,b)\rightarrow\mathbb{R}$, $-\infty\leq a<b\leq \infty$, is anon--decreasing function, and define $$\begin{align} G(x)&=&H(x-)=\sup_{y<x}H(y)\qquad (a<x<b)\\ F(x)&=&H(x+)=\inf_{x<z}H(z)\qquad (a<x<b). \end{align}$$ Then, $G$ and $F$ are non--decreasing functions such that $G\leq H\leq F$, and $$\begin{align} F(x)&=& F(x+)=G(x+)\\ G(x)&=& G(x-)=F(x-). \end{align}$$

• Here is a short proof:

For any $a<x<y<z<b$, the monotonicity of $H$ implies that $$\begin{align} G(x)\leq H(x)\leq F(x)\leq G(y)\leq H(y)\leq F(y)\leq G(z)\leq H(z)\leq F(z)\tag{1}\label{monotone_chain} \end{align} $$ Letting $y\nearrow z$ we obtain $$ G(x)\leq H(x)\leq F(x)\leq G(z-)\leq G(z)\leq F(z-)\leq G(z). $$ $$ G(z-)\leq G(z)\leq G(z-). $$ Therefore, $G(z)=G(z-)=F(z-)$. Similarly, by letting first $y\searrow x$, and then letting $z\searrow x$ in~\eqref{monotone_chain}, we obtain that $F(x)=F(x+)=G(x+)$.

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Lemma 2. Suppose $F:(a,b)\rightarrow ([0,1]$ be a probability distribution function, that is $F$ is right--continuous and non--decreasing, $\lim_{x\rightarrow a+}F(x)=0$, and $\lim_{x\rightarrow b-}F(x)=1$. Let $G(x)=F(x-)$, and define the functions $$\begin{align} Q(q)&=\inf\{x\in(a,b): F(x)\geq q\},\qquad 0<q<1\\ Q^+(q)&=\sup\{x\in(a,b): G(x)\leq q\},\qquad 0 < q<1. \end{align}$$ Then, $Q$ is non--decreasing left continuous, $Q^+$ is non--decreasing right continuous, and $$\begin{align} F(Q(q))\geq q,&\qquad G(Q^+(q))\leq q\\ Q(q)=Q^+(q-),&\qquad Q^+(q)=Q(q+) \end{align}$$

• Here is a short proof of this:

The monotonicity of $F$ implies that $(Q(q),\infty)\subset \{x:F(x)\geq q\}\subset[Q(q),\infty)$ for all $0<q<1$. If $Q(q)<z_n\searrow Q(q)$, then $F(z_n)\geq q$ and so, $F(Q(q))\geq q$ by the right--continuity of $F$. Hence $$\begin{align} F(x)\geq q\quad\text{iff}\quad Q(q)\leq x\tag{2}\label{quantile_low} \end{align}$$ Clearly $Q$ is non--decreasing. If $q_n\nearrow q$, then $Q(q_n)\nearrow x^*\leq Q(q)$. Since $q_n\leq F(Q(q_n))\leq F(x^*)\leq F(Q(q))$, $q\leq F(x^*)\leq F(Q(q))$. It follows from $\eqref{quantile_low}$ that $x^*=Q(q)$, that is $Q(q-)=Q(q)$ for all $0<q<1$.

Similarly, the monotonicity of $G$ implies that $(-\infty,Q^+(q))\subset\{x:G(x)\geq q\}\subset(-\infty,Q^+(q)]$ for all $0<q<1$. If $Q^+(q)>z_n\nearrow Q^+(q)$, then $G(z_n)\leq q$ and so, $G(Q^+(q))\leq q$ by the left--continuity of $G$. Hence $$\begin{align} G(x)\leq q\quad\text{iff}\quad x\leq Q^+(q)\tag{3}\label{quantile_upper} \end{align}$$ Clearly $Q^+$ is non--decreasing. If $q_n\searrow q$, then $Q^+(q_n)\searrow x^*\geq Q^+(q)$. Since $G(Q^+(q))\leq G(x^*)\leq G(Q^+(q_n))\leq q_n$, $G(Q^+(q))\leq G(x^*)\leq q$. It follow from $\eqref{quantile_upper}$ that $x^*=Q^+(q)$, that is $Q^+(q+)=Q^+(q)$ for all $0<q<1$.

Claim I: $Q(q)\leq Q^+(q)$.} If $x<Q(q)$, then $G(x)\leq F(x)<q$ and so, $x\leq Q^+(q)$. Hence, $Q(q)\leq Q^+(q)$.\

Claim II: $Q^+(q)\leq Q(p)$ whenever $q<p$. Indeed, if $x<Q^+(q)$ then, by $\eqref{quantile_upper}$, $G(z)\leq G(Q^+(q))\leq q$ for all $x<z<Q^+(q)$. Passing to the limit $z\searrow x$ gives $G(x)\leq G(x+)=F(x)\leq q<p$. Thus, $x<Q(p)$ by $\eqref{quantile_low}$ and so, $Q^+(q)\leq Q(p)$.

From the claims above we conclude that $$\begin{align} Q(q)&=Q(q-)\leq Q^+(q-)\leq Q(q)\\ Q^+(q)&\leq Q(q+)\leq Q^+(q+)=Q^+(q) \end{align} $$

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The proof of (3) is simple from (2). The function $F(x):=\mathbb{P}[X\leq x]$ satisfies the conditions of Lemma 2. Hence, by $\eqref{quantile_low}$

$$\{q\in(0,1): Q(q)\leq x\} = \{q\in(0,1): F(x)\leq q\}$$

and so,

$$\lambda[Q\leq x] =\lambda((0,F(x)])=F(x)$$

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When $X$ is a random variable is continuous (i.e. $\mathbb{P}[X=x]=0$ for any $x$, and its law has no atoms (i.e, there are no Borel sets $A\subset \mathbb{R}$ such that $\mathbb{P}[X\in A]>0$ and for any other Borel set $B\subset A$, either $\mathbb{P}[X\in B]=0$ or $\mathbb{P}[X\in B]=\mathbb{P}[X\in A]$), then $Q=Q^+$ and so at each level $q$ the quantiles are unique.

In this situation, the distribution function $F_X(x)=\mathbb{P}[X\leq x]$ is continuous and strictly increasing.

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Edit: another interesting property the quantile function $Q$ is discussed in this posting.

Answer 2

• First, notice that $F$ is increasing. Therefore, taking the supremum would gives $+\infty $. Normally, the minimum is defined for finite set (or well ordered set). In other context, we rather talk about infimum.

• Since $F(-\infty )=0$ and $F(\infty )=1$ and $F$ increasing, the $p-$th quantile $(p\in (0,1)$ is nothing else than the first time where $F(x)\geq p$ and you have that $F(x)\geq p$ for all $x\geq \xi_p$.