Prove that $F^{-1} \circ F \circ F^{-1} = F^{-1}$ and $F \circ F^{-1} \circ F = F$ for the quantile function

Question

I want to prove that \begin{align} F^{-1} \circ F \circ F^{-1} = F^{-1}\quad\text{and}\quad F\circ F^{-1} \circ F = F\tag{*}\label{star} \end{align} where F is real-valued non-decreasing right continuous function over $\mathbb{R}$ such that $$\lim_{x\rightarrow-\infty}F(x)=0,\quad\lim_{x\rightarrow\infty}F(x)=1,$$ and $F^{-1}$ is the the quantile function is defined as

$$ F^{-1} (p) = \inf\{ x : F(x) \geq p \}. $$

Notice that $F^{-1}(0)=-\infty$, $F^{-1}(1)=\infty$ if and only if $F(x)<1$ for all $x\in\mathbb{R}$, and $F^{-1}(p)\in\mathbb{R}$ for all $0<p<1$

From the definition of $F^{-1}$ and the right continuity of the distribution function $F$, it follows that for any $x\in\mathbb{R}$ and $0<p<1$

$$ F(x)\geq p \quad\text{if and only if}\quad Q(p)\leq x. $$

This in turn implies that $$F(Q(p))\geq p\quad\text{and}\quad Q(F(x))\leq x$$

I've been stuck on this one for a while and don't really know where to start for proving property \eqref{star}. Any help would be appreciated.

Answer 1

$\newcommand{\R}{\mathscr{R}}\newcommand{\I}{\mathscr{I}}\newcommand{\F}{\mathscr{F}}\newcommand{\G}{\mathscr{G}}$Fix a nondecreasing right continuous $F:\Bbb R\to(0,1)$ with $F(-\infty)=0$ and $F(\infty)=1$. This ensures that $F^{-1}:(0,1)\to\Bbb R$ is well defined.

Right continuity ensures that $F(F^{-1}(x))\ge x$ is true: the infimum is attained (exercise).

The abstract stuff below can be unwound to an explicit proof: fix $x\in\Bbb R$. I find $F(x)\le FF^{-1}F(x)$ because $u\le FF^{-1}(u)$ for all $u\in(0,1)$. Also, since: $F^{-1}F(x)\le x$ I can infer $FF^{-1}F(x)\le F(x)$. This forces $F(x)=FF^{-1}F(x)$. The other equality can be treated similarly.

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(This is admittedly slightly silly)

Let $\R$ and $\I$ be the posetal categories associated to $\Bbb R$ and $(0,1)$, respectively, with their usual orderings (there is a (unique) arrow $a\to b$ iff. $a\le b$).

We have a functor $\F:\R\to\I$ induced by the nondecreasing function $F$, and a functor $\G:\I\to\R$ induced by the nondecreasing function $F^{-1}$.

$\F,\G$ form a "Galois connection": $\G\dashv\F$. For any $x\in\I$ we have an arrow $x\to\F\G(x)$ which I claim is initial: if there is an arrow $x\to\F(y)$, then $x\le F(y)$ and by definition we find $F^{-1}(x)\le y$, so there is a (unique) arrow $\G(x)\to y$ trivially making the triangle: $x\to\F\G(x)\to\F(y),x\to\F(y)$ commute. This gives an adjunction.

For any Galois connection on posets it's true that $\F\circ\G\circ\F=\F$ and $\G\circ\F\circ\G=\G$. So, we are done.

Why does that identity hold?

For any $x$, using the unit of the adjunction I have an arrow $\F(x)\to\F\G\F(x)$. I also have an arrow $\F\G\F(x)\to\F(x)$ by applying $\F$ to the counit. Because we work in a posetal category, existence of arrows in both directions implies equality (and the functors are trivially equal on arrows if they are equal on objects). This is the general form of the concrete proof given in the first section of the post.

Answer 2

To prove the properties $F^{-1} \circ F \circ F^{-1} = F^{-1}$ and $F \circ F \circ F^{-1} = F$ for the quantile function, where $F$ is a non-decreasing, right continuous, $[0,1]$-valued function over $\mathbb{R}$, and the quantile function is defined as $F^{-1}(p) = \inf \{x : F(x) \geq p\}$, we can follow these steps:

Proving $F^{-1} \circ F \circ F^{-1} = F^{-1}$: Let's take an arbitrary probability value $p$. Applying $F^{-1}$ to p gives us the infimum of all $x$ such that $F(x)$ is greater than or equal to $p$. Then, applying $F$ to $F^{-1}(p)$ gives us the cumulative distribution value of the quantile obtained from the previous step. Finally, applying $F^{-1}$ again will yield the infimum of all x such that $F(x)$ is greater than or equal to the cumulative distribution value obtained in the previous step. Since $F^{-1}$ is defined as the inverse of the cumulative distribution function $F$, applying $F$ and then $F^{-1}$ will yield the original input value $p$. Therefore, $F^{-1} \circ F \circ F^{-1} = F^{-1}$.

Proving $F \circ F^{-1} \circ F = F$: Now let's take an arbitrary value $x$. Applying $F^{-1}$ to $x$ gives us the infimum of all $p$ such that $F^{-1}(p)$ is greater than or equal to $x$. Then, applying $F$ to $F^{-1}(x)$ gives us the cumulative distribution value of the quantile obtained from the previous step. Finally, applying $F$ again will yield the cumulative distribution value of the quantile value obtained in the previous step. Since $F$ and its inverse $F^{-1}$ are defined as inverse operations, applying $F$ and then $F^{-1}$ will yield the original input value $x$. Therefore, $F \circ F^{-1} \circ F = F$.

By proving both properties, we have shown that $F^{-1} \circ F \circ F^{-1} = F^{-1}$ and $F \circ F^{-1} \circ F = F$ for the quantile function.

Answer 3

Let $Q(q)=\inf\{x: F(x)\geq q\}$. This is in principle defined in $(0,1)$ and thus, the domain of $F\circ Q\circ F$ may need to be adjusted.

Notice that while $F$ is nonnegative, nondecreasing with right continuous with left limits, $Q$ is nondecreasing, left continuous with left limits.

For $Q\circ F\circ Q$ on $(0,1)$, there are two cases to consider:

• $q$ is in a jump gap interval, that is, there is $x\in\mathbb{R}$ such that $F(x-)<q\leq F(x)$. In this case, $Q(q)=x$. Hence $$Q\circ F\circ Q(q)=Q(F(x))=x=Q(q)$$

• If $q$ is not in a jump gap interval of $F$. This means that $q=F(x)$ for some point $x$ at which $F$ is contunuous. Then, there is $x'\leq x$ such that $F(x')=F(x)$ and $F(y)<F(x')$ for $y<x'$. It follows that $$Q\circ F\circ Q(q)=Q(F(x'))=Q(q)$$

Similar arguments can be used for $F\circ Q\circ F$.

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A more streamlined proof can be obtained by noticing that $$F(x)\geq q\qquad\text{iff}\qquad Q(q)\leq x$$ which follows from the right-fontinuity of $F$. This means that $$F(Q(q))\geq q\qquad\text{and}\qquad Q(F(x))\leq x$$ As both $F$ and $Q$ are monotone nondecreasing $$Q(q)\leq Q\big(F(Q(q))\big)=(Q\circ F)(Q(q))\leq Q(q)$$ and $$F(x)\geq F\big(Q(F(x))\big)=(F\circ Q)(F(x))\geq F(x)$$