Distribution function of random variable

Question

Given the distribution function of a random variable $X$, $F_X$, I would like to determine the distribution function of the random variable $Y = |X^2-1|$, that is, $Y = g(X)$ for $g(x)=|x^2-1|$. So \begin{align*} F_Y(y) &= \mathrm P\{Y \leq y\} = \mathrm P \{ |X^2-1| \leq y \} = \mathrm P \{ -y \leq X^2- 1 \leq y\}\\ &= \mathrm P\{-y+ 1 \leq X^2 \leq y + 1\} = \mathrm P\{\sqrt{1-y} \leq X \leq \sqrt{1+y}\}\\ \end{align*} assumming that $y \geq 0$ clearly since $|X^2-1|\leq y$ or, equivalently, $g^{-1}(-\infty,y] = \emptyset$ for all $y < 0$. Hence $F_Y(y) = 0$ for $y < 0$ but I don't know how to justify that $y \leq 1$ (otherwise $\sqrt{1-y}\not\in\mathbb R$). Thanks in advance.

Edit I It is clear that the previous attempted justification is absurd, so I would do the following: $$ \{ 1-y\leq X^2 \leq 1+y\} = \{1-y\leq X^2 < 0\} \sqcup \{0\leq X^2 \leq 1+y\} $$ where the first set in the union is $\emptyset$. So I'll need to discuss that if $y > 1$: $$ F_Y(y) = \mathrm P \{ 0\leq X \leq \sqrt{1+y}\}. $$ Is this right?

Answer 1

For $y\geqslant0$, let $A_y=[−y+1≤X^2≤y+1]$ then;

• If $0\leqslant y\leqslant1$, then $A_y=[-\sqrt{y+1}\leqslant X\leqslant-\sqrt{1-y}]\cup[\sqrt{1-y}\leqslant X\leqslant\sqrt{y+1}]$.
• If $y\geqslant1$, then $A_y=[-\sqrt{y+1}\leqslant X\leqslant\sqrt{y+1}]$.

This allows to deduce $F_Y$ in terms of $F_X$.