# Let $X$ a random variable with a strictly increasing distrubution function $F_X$. Show that $Y=F_X(X) \sim \hom(0,1)$ distrubution.

Let $X$ a random variable with a strictly increasing distrubution function $F_X$. Show that the random variable $Y=F_X(X)$ has a $\hom(0,1)$ distrubution.

Here is what I thought:

\begin{align} F_Y(y)&=P(Y\leq y) \\ &=P(F_X(X)\leq y) \\ &=P(P(X\leq X)\leq y) \end{align}

The chance that $X\leq X$ is always $1$, right ? Therefore I would say that:

$$F_Y(y)=P(1\leq y)=1_{[1,\infty)}(y)$$

But this is wrong... Why is this ?

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\begin{align} F_Y(y)&=P(Y\leq y) \\ &=P(F_X(X)\leq y) \\ &=P(X\leq F_X^{-1}(y)) \text{ inverse exists as $F_x$ is strictly increasing}\\ &=F_X(F_X^{-1}(y))\\ &=y, \ 0\le y \le 1 \end{align}
Your move from step 2 to step 3 is illogical. $P(\omega:F_X(X)(\omega)\le y)$ is the exact meaning of the second statement, which can be rewritten using inverse. Your 3rd equation does not have any such set-based notion.
I don't understand your last equality. Why must $y$ be in $[0,1]$ ? –  Kasper Mar 9 '13 at 20:00
$Y=F_X(X)$, so the random variable $Y$ can take values only in $[0,1]$. –  Bravo Mar 9 '13 at 20:02
My brain gets completely confused when I read $F_X(X)(\omega)$. So this means $P(X\leq X)(\omega)$ ? How can I look at this ? This is a probability function and function from $\Omega \to \Bbb R$ ? –  Kasper Mar 9 '13 at 20:07
@Bravo, why can't one write $P(F_X(X) \leq y) = P(P(X\leq X)\leq y)$ this is the exact meaning ? –  user111854 Apr 5 '14 at 9:04