If $F$ is a distribution function and $x_0\in F^{-1}([0,1])$, can we characterize $\left\{t\in[0,1]:F^{-1}(t)=x_0\right\}$?

Question

Let $F:\mathbb R\to[0,1]$ be a distribution function$^1$ and $$M_t:=\left\{x\in\mathbb R:F(x)\ge t\right\}$$ and $$F^{-1}(t):=\inf M_t$$ for $t\in[0,1]$.

Let $x_0\in F^{-1}([0,1])$. Can we characterize the set $S:=\left\{t\in[0,1]:F^{-1}(t)=x_0\right\}$?

I guess that $S=\left[\lim_{x\to x_0-}F(x),\lim_{x\to x_0+}F(x)\right]$, but don't know how we can prove it.

Since $F$ is right-continuous at $x_0$, $F(x_0+)=F(x_0)$. And if $t\in S$, then $t\le F(F^{-1}(t))=F(x_0)$.

Now we only need to show that if $t\in S_{x_0}$, then $t\ge t_0:=F(x_0-)$:

Since $F$ is nondecreasing, $$t_0\le F(x_0)\tag1$$ and hence $$F^{-1}(t_0)\le x_0\tag2.$$ Moreover, $t\in S_{x_0}$ implies $$x_0=F^{-1}(t)\tag3.$$ Now I've tried to get a contradiction from assuming $t<t_0$. Since $F^{-1}$ is nondecreasing, it implies that $$x_0=F^{-1}(t)\le F^{-1}(t_0)\le x_0\tag4$$ and hence $$t_0\in S_{x_0}\tag5.$$ But that doesn't seem to be a contradiction.

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$^1$ i.e. $F$ is right-continuous and nondecreasing with $F(-\infty):=\lim_{x\to-\infty}F(x)=0$ and $F(\infty):=\lim_{x\to\infty}F(x)=1$.

Answer 1

Notice that if $F$ is the cdf of the constant random variable $0$ then for $x_0 = 0$, $S = (0,1]$ so we shouldn't expect to have that $S = \left[F(x_0-), F(x_0)\right]$ in general.

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I claim that $$S = \begin{cases}\left[F(x_0-),F(x_0)\right] &\text{ if } x \text{ is a continuity point of } F \text{ or } F(x_0-) > F(y) \text{ for all } y < x_0,\\ (F(x_0-), F(x_0)] &\text{ otherwise} \end{cases}$$ where $F(x_0-) := \lim_{x \to x_0-} F(x)$.

To try to make things easier to follow, I'll break the proof down into the following 5 claims which together settle all of the cases above. Each claim has a very short proof.

Claim 1: $S \subseteq [F(x_0-), F(x_0)]$

Proof: First, notice that if $t \in S$ then $F(x_0) \geq t$ by right-continuity so that $S \subseteq (-\infty, F(x_0)]$.

Also if $t < F(x_0-)$ then there is $y < x_0$ such that $F(y) \geq t$. Hence $F^{-1}(t) \leq y < x_0$ so that $t \not \in S$. Hence $S \subseteq [F(x_0-), F(x_0)]$.

Claim 2: $(F(x_0-), F(x_0)] \subseteq S$

Proof: If $t \in (F(x_0-), F(x_0)]$ then $F(x_0) \geq t$ and for all $y < x_0$, $t > F(x_0-) \geq F(y)$. Hence $F^{-1}(t) = x_0$ so that $t \in S$. Therefore $(F(x_0-), F(x_0)] \subseteq S$.

Claim 3: If $x_0$ is a continuity point of $F$ then $S = \{F(x_0)\}$

Proof: If $x_0$ is a continuity point of $F$ then $F(x_0-) = F(x_0)$ so since we know that $S \neq \emptyset$ and $S \subseteq \{F(x_0)\}$ we must have equality.

Claim 4: If $x_0$ is not a continuity point of $F$ and $F(x_0-) > F(y)$ for all $y < x_0$ then $F(x_0-) \in S$.

Proof: If $F(x_0-) > F(y)$ for all $y < x_0$ then since $F(x_0-) \leq F(x_0)$, we have by the definition that $F^{-1}F(x_0-) = x_0$ so that $F(x_0-) \in S$. Hence, in this case $S = [F(x_0-), F(x_0)]$, as claimed.

Claim 5: If $x_0$ is not a continuity point of $F$ and there is an $x_1 < x_0$ such that $F$ is constant on $[x_1, x_0)$ then $F(x_0-) \not \in S$.

Proof: We have that that $F^{-1}(F(x_0-)) \leq x_1 < x_0$ so that $F(x_0-) \not \in S$ so that $S= (F(x_0-), F(x_0)]$ in this case, also as claimed.

Answer 2

For a fixed $x_{0}\in\mathbb{R}$ to be found is: $$S=\left\{ t\mid\inf\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t\right\} =x_{0}\right\} =\left\{ t\mid\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t\right\} =\left[x_{0},\infty\right)\right\} $$where the second equality rests on the fact that $F$ is right-continuous.

Let $t_{0}:=F\left(x_{0}\right)$ and let $t_{1}:=\lim_{x\to x_{0}-}F\left(x\right)$ so that $t_{1}\leq t_{0}$.

If $t>t_{0}$ then $x_{0}\notin\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t\right\} $ so that $t\notin S$.

If $t_{1}<t\leq t_{0}$ then $\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t\right\} =\left[x_{0},\infty\right)$ so that $t\in S$.

If some $y\in\mathbb{R}$ exists with $y<x_{0}$ and $F\left(y\right)=t_{1}$ then $\left[y,\infty\right)\subseteq\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t_{1}\right\} $ and consequently $t_{1}\notin S$.

If $F\left(y\right)<t_{1}$ for every $y\in\mathbb{R}$ with $y<x_{0}$ then $\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t_{1}\right\} \subseteq\left[y,\infty\right)$ for every $y<x_{0}$ so that $\left\{ x\in\mathbb{R}\mid F\left(x\right)\geq t_{1}\right\} \subseteq\left[x_{0},\infty\right)$ and consequently $t_{1}\in S$.

This together shows that S equals $[t_1,t_0] $ or equals $(t_1,t_0] $. In the special case where $t_0=t_1$ (a continuity point) the first set is a singleton and the second is the empty set.