Clarification about finding CDF from PDF The question I have is:
$$f_x(x) = \begin{cases}4x^3 &  0<x<1\\ 0 & otherwise\end{cases}$$
Where I have to find the CDF function of $F_y(y)$ = $y=-log(x)$.
My steps were:
Consider $P(X \le x) = 12x^2$ so:
$P(Y \le y) = P(-log(x) \le y ) = P(X \le e^{-y})$
So the CDF of Y is: $F_Y(y) = 12(e^{-y})^2$.
I'm just a bit confused on the last bit of the CDF steps, do I sub it back into the $P(X \le x) = 12x^2$ or do I do the following where I sub $P(X \le e^{-y})$ into $Y = -log(X) = -log(e^{-y})$ to get $F_Y(y) = -y$

$-\log X \leq y$ is same as $X \geq e^{-y}$ (and not $X \leq e^{-y}$)  since  you are multiplying by $-1$.
$P(X \geq e^{-y})=1-e^{-4y}$ so $f_Y(y)=4e^{-4y}$ for $y \geq 0$.
