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I have a doubt on this theorem:

The probability integral transform states that if ${\displaystyle X}$ is a continuous random variable with cumulative distribution function ${\displaystyle F_{X}}$, then the random variable ${\displaystyle Y=F_{X}(X)}$ has a uniform distribution on $[0, 1]$.

The theorem is longer, but I don't understand this first part. If I have a continuous $F(X)$ and I set $Y=F(X)$, then this implies that $Y\sim U(0,1)$? Why?

Thanks!

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Since $X$ is a continuous random variable, for any $y$ with $0 \lt y \lt 1$ we have an $x$ with $F_X(x)=y$,

so $F_Y(y)= P(Y\le y)= P(F_X(X)\le y)=P(X\le x)=F_X(x)=y$

meaning that $Y \sim U(0,1)$

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  • $\begingroup$ This is better than the other answer, the only thing to be mentioned is that the result doesn't depend on which $x$ you picked, if $x$ happens to be non-unique. $\endgroup$
    – Ian
    Commented May 3, 2017 at 16:35
  • $\begingroup$ @Ian: Indeed, and that becomes particularly relevant if $X$ is bounded and you then want to extend to $0 \le y \le 1$ $\endgroup$
    – Henry
    Commented May 3, 2017 at 16:37
  • $\begingroup$ Could you detail how you went from $P(F_X(X) \leq y)$ to $P(X \leq x)$? Or I guess from $P(F_X(X) \leq F_X(x))$ to $P(X \leq x)$ ? $\endgroup$
    – matthieu
    Commented Jan 30, 2022 at 1:35
  • $\begingroup$ @matthieu: In general any cdf $F_X(x)$ is a weakly increasing function of $x$ so $P(F_X(X) \leq F_X(x)) $ $=P((F_X(X) \lt F_X(x))\cup(F_X(X) = F_X(x))) $ $=P(F_X(X) \lt F_X(x)) +P(F_X(X) $ $= F_X(x)) $ $=P(X<x)+P(X=x) = P(X\le x)$. As an aside, note that in this case you have a continuous random variable so $P(X=x)=0$. $\endgroup$
    – Henry
    Commented Jan 30, 2022 at 12:01
  • $\begingroup$ Okay, so how do you go from $P(F_X(X) \lt F_X(x))$ to $P(X \lt x)$ ? Can I do that for any increasing function $f$? Like $P(f(X) \lt f(x)) = P(X \lt x)$ ? Or is it specific to $F_X$ as the distribution function of $X$? $\endgroup$
    – matthieu
    Commented Jan 31, 2022 at 20:22

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