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I have a pair of conjectures for the cumulative distribution function $F(x)$ for a random variable $X$ and its generalized inverse $F^{-}(u) \stackrel{\Delta}{=} \inf\{x: F(x) \ge u\}$.

  1. $F^{-}(F(X)) \sim X$.
  2. $\forall 0 < u < 1, P(F(X) \le u) = P(X \le F^{-}(u))$.

These conjectures are meant to be the counterpart of each other, with the first focuses on $F^{-}$ and the second focuses on $F$. The idea is that if $F^{-}(F(x)) \ne x$, then $x$ must be in a "flat"1 area of $F$, which suggests $P(X=x) = 0$. Additionally, if $\{F(X) \le u\} \ne \{X \le F^{-}(u)\}$, then $F^{-}$ must be a discontinuity point at $u$, which once again correspond to a "flat" area in $F$, and the same argument follows.

1 By "flat", I mean $\{x: \exists y < x, F(y) = F(x)\}$.

While the conjectures intuitively make sense, I fail to prove them rigorously. For the first proposition, I managed to prove that $F^{-}(F(x)) = F(x)$ as long as $F^{-}$ is continuous at $F(x)$, but I cannot explain why $P(\text{$F^{-}$ is discontinuous at $F(X)$} ) = 0$. For the second proposition, here is my attempt

\begin{equation}\begin{aligned} \forall 0 < u < 1, \{u_n\} \downarrow u, P(F(X) \le u) &= \lim_{n \to \infty} P(F(X) < u_n) \\ &= \lim_{n \to \infty} P(X < F^{-}(u_n)) \\ &= P(X < \lim_{n \to \infty} F^{-}(u_n)) \\ &= P(X < F^{-}(u+)) \\ &= \text{???} \\ &= P(X \le F^{-}(u)) \end{aligned}\end{equation}

How do I prove them, or is there a counter-example? In the second case, which additional conditions on $F$ do I need, e.g. continuity and/or strict monotonicity?


EDIT: Some people pointed me to Questions about definition of Quantile function. However, it doesn't seem to dis/prove my conjectures. For example, my first conjecture is $F^{-}(F(X)) \sim X$, and the answer to that question proves $F^{-}(U) \sim X$, but $F(X) \sim U$ if and only if $F$ is continuous. What if it's not? In addition, it says nothing about my second conjecture. This is why this question is not a duplicate to that one and should be reopened.

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  • $\begingroup$ The posting you quote does not assume that $F$ is continuous. It shows that the quantile function $Q(u)$ of $F$, which is defined on $((0,1),\mathscr{B}(0,1))$ has distribution $F$. $\endgroup$
    – Mittens
    Oct 9, 2021 at 21:14
  • $\begingroup$ Could you elaborate? From my understanding, $Q(u)$ is a real-valued function, not a random variable, so it does not have a probability distribution. The answer in that post seems to assume that $U$ is a uniform random variable on $(0, 1)$ and proves $Q(U) \sim F$, which makes sense to me because both $U$ and $Q(U)$ are random variables. $\endgroup$
    – nalzok
    Oct 9, 2021 at 21:17
  • $\begingroup$ @OliverDiaz I see. You have an interesting point that $Q$ can be considered as a random variable, but when saying $\lambda\Big(Q^{-1}\big((-\infty,x]\big)\Big)=\mathbb{P}[X\leq x]$, the probability measure $\lambda$ is not necessarily the same as $\mathbb{P}$. Moreover, the sample spaces of $Q$ and $X$ will likely differ. In my question, $Y = F^{-}(F(X))$ is a function of $X$ (more explicitly, $\forall \omega \in \Omega, Y(\omega) = F^{-}(F(X(\omega)))$), so it's defined in the exact same probability space as that of $X$. Does the conjecture still hold? $\endgroup$
    – nalzok
    Oct 9, 2021 at 23:01
  • $\begingroup$ @OliverDiaz Now we are getting somewhere! Could you elaborate on the $P[F(y) \ge F(X)] = P[X \le y]$ part, though? Obviously $\{x: F(y) \ge F(x)\} \supseteq \{x: x \le y\}$, but why the other way around? $\endgroup$
    – nalzok
    Oct 9, 2021 at 23:24
  • $\begingroup$ @OliverDiaz Ah I see. So you just proved the first conjecture. Thank you! Any thoughts on the second one? $\endgroup$
    – nalzok
    Oct 9, 2021 at 23:30

1 Answer 1

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Some generalities:

The posting the OP refers to, in particular this solution shows that if $$Q(q)=\inf\{x\in\mathbb{R}: F(x)\geq q\},\qquad 0<q<1$$ Then

  1. $F(Q(q))\geq q$
  2. $F(x)\geq q$ iff $Q(q)\leq x$.

In particular, if $(\Omega,\mathscr{F},\mathbb{P})=((0,1),\mathscr{B}((0,1)),\lambda)$, where $\lambda$ is Lebesgue's measure restricted to $(0,1)$, then $Q:((0,1),\mathscr{B}((0,1))\rightarrow\mathbb{R}$ is a random variable whose cumulative distribution function is $F$: $$\lambda\big(q\in(0,1):Q(u)\leq x \big)=\lambda\big(q\in(0,1):F(x)\geq q\big)=F(x)$$

Equivalently, if $\theta$ is a random variable with $0-1$ uniform distribution, then $Q(\theta)$ is a random variable with cumulative distribution $F$.

Notice that if $q=0$, $Q(q)=-\infty$, and that unless $F(x)=1$ for some $x\in\mathbb{R}$, $Q(1)=\infty$. Thus, $Q$ can be defined on $[0,1]$ as a real-extended value function.

Solution to the OP:

Observation: For any $b\in\mathbb{R}$, $I_b:=F^{-1}((-\infty,b])$ is an interval. It could be of the form $(-\infty,\beta)$ if $\beta$ is a point of discontinuity of $F$, $F(y)>b$ for all $y>\beta$, and $F(x)=b$ for all $x$ in an interval $[a,\beta)$; it is of the form $(-\infty,\beta]$ if $F(\beta)=b$ and $F(x)>F(\beta)$ for all $x>\beta$. In particular, if $b=F(y)$ for some $y\in\mathbb{R}$, then $F(\beta-)=F(y)=b$ if $I_b=(-\infty,\beta)$, or $F(\beta)=F(y)=b$ of $I_b=(-\infty,\beta]$.

Now, by (2) and the observation above, $$P[Q(F(X))\leq y]=P[F(y)\geq F(X)]=F(y)=P[X\leq y]$$ which is what the OP suggests in his first statement.

Another application of (2) shows that $$P[F(X)< q]=P[Q(q)>X]=F(Q(q)-)\leq q\leq F(Q(q)), \quad 0<q<1$$ which is almost like the second statement made by the OP. Equality holds when $Q(q)$ is a point of continuity of $F$. I leave to the OP the task of finding an example where inequality may hold (notice that such an example my be obtained by considering a distribution $F$ that is constant in intervals of the form $[x_n,x_{n+1})$ and jumps at each $x_n$.)

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