I am trying to find the distribution function for $Y = 1 - X^2$

Question

I have some $f_X(x)$ which is defined on $(\frac{-1}{2},1)$.

I know that $$F_Y(y) \begin{align} &= P(Y\le y) = P(1-X^2 \le y ) = P (X^2 \ge 1-y)\end{align}$$

this is where I get stuck or don't know how to interpret it. Should I say $$F_Y(y)= P(|X| \ge \sqrt{1-y})$$

which if true, then can I say

$$ F_Y(y) = 1 - F_X(\sqrt{1-y})$$

any help would be appreciated.

Answer 1

$P(|X| \ge \sqrt{1-y})=P(X \ge \sqrt{1-y})+P(X \le -\sqrt{1-y})=1-F_X(\sqrt{1-y}) + F_X(-\sqrt{1-y})$