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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.)

syeh_106
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  • What do you mean by $F_X(X)$? – MAN-MADE Jul 28 '17 at 03:42
  • See the comment by binkyhorse in the linked question – leonbloy Jul 28 '17 at 03:45
  • @MANMAID $F_X: \mathbb R \to [0,1]$ is the probability distribution function of random variable $X$. – syeh_106 Jul 28 '17 at 03:45
  • @syeh_106 I know that. But you wrote $F_X(X)$. This is a random variable. So I am asking what is $F_X(X)$? – MAN-MADE Jul 28 '17 at 03:50
  • @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. – MAN-MADE Jul 28 '17 at 03:56
  • @MANMAID My comment was directed to the OP. $F_X(X)$ is well defined, it's just the usual $F_X(x)$ evaluated at the (random) value $x=X$ – leonbloy Jul 28 '17 at 04:03
  • @MANMAID leonbloy's comment above is exactly what I meant. $F_X(X)$ is a function of the random variable $X$, so is itself a (derived) random variable. – syeh_106 Jul 28 '17 at 05:42

1 Answers1

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Your solution looks fine to me.

leonbloy
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