What is the distribution of $Y = \varphi(X)$ where $\varphi(x) = F_X(x)$?

Question

Let $X$ be a random variable (absolutely) continuous with $F_X$ distribution; which is strictly increasing and with inverse $F_X^{-1}(u)$, for every $u \in [0,1]$. What is the distribution of $Y = \varphi(X)$ where $\varphi(x) =F_X(x)$?

My try:

$$y = \varphi(x)$$

$$P(Y\leq y)$$ $$=P(\varphi(x)\leq y)$$ $$=P(X\leq \varphi^{-1}(y)) = y$$

So then:

$\qquad$ $F_Y(y) = \begin{cases}y \qquad if \qquad0\leq y \leq 1, \\ 1 \qquad if \qquad 1 \leq y \end{cases}$

I´m not sure if this attept is correct.

Answer 1

Suppose $y\in [0,1]$

\begin{align}P(Y\leq y) &=P(\varphi(\color{red}{X})\leq y)\\ &=P(X\leq \varphi^{-1}(y)) \\&= F_X(\varphi^{-1}(y))\\ &=\varphi(\varphi^{-1}(y)) \\&=y\end{align}

So then:

$\qquad$ $F_Y(y) = \begin{cases} 0 \qquad if \qquad y\le 0\\y \qquad if \qquad0\leq y \leq 1, \\ 1 \qquad if \qquad 1 \leq y \end{cases}$

The density function is

$\qquad$ $f_Y(y) = \begin{cases} 0 \qquad if \qquad y\le 0\\1 \qquad if \qquad0\leq y \leq 1, \\ 0 \qquad if \qquad 1 \leq y \end{cases}$

It is the uniform distribution.