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This question is related to Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X, with the difference being that $F_X$ (the probability distribution function of random variable $X$) is an arbitrary continuous distribution function, not necessarily strictly increasing.

I think the proof is similar, but we have to take care of the possibility that $F_X$ may not be $1$-to-$1$. I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved. Thanks a lot!

The goal is to show $F_Y(y)=y$ for any $y \in [0,1]$. To do so, note that $F_Y(y)\triangleq\mathbb P(\{Y\le y\})$, and

$\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.$

Since $F_X$ is continuous, $F_X^{-1}([0,y])$ must be closed. So it follows that

$\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),$ which let's denote by $a$.

Therefore, $F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad$ (Q.E.D.)

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  • $\begingroup$ What do you mean by $F_X(X)$? $\endgroup$
    – MAN-MADE
    Commented Jul 28, 2017 at 3:42
  • $\begingroup$ See the comment by binkyhorse in the linked question $\endgroup$
    – leonbloy
    Commented Jul 28, 2017 at 3:45
  • $\begingroup$ @MANMAID $F_X: \mathbb R \to [0,1]$ is the probability distribution function of random variable $X$. $\endgroup$
    – syeh_106
    Commented Jul 28, 2017 at 3:45
  • $\begingroup$ @syeh_106 I know that. But you wrote $F_X(X)$. This is a random variable. So I am asking what is $F_X(X)$? $\endgroup$
    – MAN-MADE
    Commented Jul 28, 2017 at 3:50
  • $\begingroup$ @leonbloy I asked this question due to notation confusion. In $F_X(X)$ OP wrote $X$ twice, and yet he could not give a definition of that. $\endgroup$
    – MAN-MADE
    Commented Jul 28, 2017 at 3:56

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Your solution looks fine to me.

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