Find the density $f_Y$ of $Y.$ Let $Y=X^2$. Find the denisity $f_Y$ of $Y$.
This is my approach $\rightarrow Y=X^2 \rightarrow Y=\sqrt{X}$ then 
$F_Y(Y)=P(Y \leq X)= P(Y \leq X^2) \rightarrow P(\sqrt{Y} \leq X)$.
Or am I supposed to use this formula for density:
$$\frac{1}{2\sqrt{2\pi}\sigma}e^{-\left(\frac{x-\mu}{2\sigma}\right)^2}?$$
 A: Hint : when $X$ is standard, you can extend it to the non-standard
We define the cdf of $Y$ as $$F_Y(y)=P(Y \leq y)$$
First of all, $F_Y(y)= 0$ if $y<0$, because $Y=X^2$ is non-negative.
We assume $y \geq 0$, and we have
 $$F_Y(y)=P(Y \leq y)=P(X^2 \leq y)=P(\sqrt{X^2} \leq \sqrt{y})=P(|X| \leq \sqrt{y})$$
This can be rewritten as $$P(|X| \leq \sqrt{y})=P(-\sqrt{y}\leq X\leq \sqrt{y})=P( X\leq \sqrt{y})-P( X\leq -\sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$$
Using the symmetry of $X$, we have 
$F_X(-\sqrt{y})=1-F_X(\sqrt{y})$
Thus,
 $$F_Y(y)=2F_X(\sqrt{y})-1$$
To get the density $f_Y$, we derive $F_y$ wrt $y$,
$$f_Y(y)=\frac{f_X(\sqrt{y})}{\sqrt{y}}$$
where $f_X$ is the density function of $X$
A: You found the accumulate distribution, but not the density. For all $y$, we have $F_Y(y)=\mathbb{P}(Y \leq y)=\mathbb{P}(X^2 \leq y)$. If $y<0$, then $\mathbb{P}(X^2 \leq y)=0$. If $y \geq 0$, $\mathbb{P}(X^2 \leq y)=\mathbb{P}(\left |{X}\right | \leq \sqrt[]{y})=F_X(\sqrt[]{y})-F_X(-\sqrt[]{y})$. Then, we derive, and we obtain $f_Y(y)=\frac{{\partial F_Y(y)}}{{\partial y}}=\frac{f_X(\sqrt[]{y})}{2\sqrt[]{y}}+\frac{f_X(-\sqrt[]{y})}{2\sqrt[]{y}}$ for all $y\geq 0$.
