# Conjectures for a CDF and its generalized inverse

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{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}

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.

• 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 $F$. Oct 9, 2021 at 21:14
• 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. Oct 9, 2021 at 21:17
• @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? Oct 9, 2021 at 23:01
• @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? Oct 9, 2021 at 23:24
• @OliverDiaz Ah I see. So you just proved the first conjecture. Thank you! Any thoughts on the second one? Oct 9, 2021 at 23:30

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 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 $$P[F(X)< q]=P[Q(q)>X]=F(Q(q)-)\leq q\leq F(Q(q)), \quad 0 which is almost like the second statement made by the OP. Equality holds when $$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$$.)