Finding $Y=X^2$ for a given continuous random variable X In the problem, I'm given that the probability density function of a continuous random variable X is $$f_{X}(x) = \begin{cases}
cx^2 & -4\leq x \leq 4\\
0 & \text{else}
 \end{cases} $$
(with $c\in \mathbb{R}$) and let $Y=X^2$. My goal is to find a probability density function $F_Y(y)$. I'm not entirely sure where to begin with this. My thoughts are that $y$ should take values $0 \leq y \leq 16$, but I'm very uncertain if that's a reasonable conclusion and I'm entirely unaware of how to find the function (unless it's as simple as $(cx^2)^2$, but I can't justify why that would be the case). I'd really appreciate some guidance in figuring out how to handle this kind of problem. 
 A: $$
\int_{-4}^4 cx^2=c\frac{128}{3}=1\quad\Longrightarrow\quad c=\frac{3}{128}
$$
so we have 
$$
f_{X}(x) = \begin{cases}
\frac{3}{128} x^2 & -4\leq x \leq 4\\
0 & \text{elsewhere}
 \end{cases}
$$
and integrating
$$
F_{X}(x) = \begin{cases}
0& x\le -4\\
\frac{1}{128} (x^3 + 64) & -4\leq x \leq 4\\
1 & x\ge 1
 \end{cases}
$$
For $Y=X^2$
$$
F_{Y}(y) = \Bbb P(Y\leq y) = \Bbb{P}(X^2 \leq y) = \Bbb{P}(-\sqrt{y} \leq X \leq \sqrt{y})=F_X(\sqrt y)-F_X(-\sqrt y)
$$
$$
f_{Y}(y) =F'_{Y}(y) =\frac{1}{2\sqrt y}\Big[f_X(\sqrt y)+f_X(-\sqrt y)\Big]
$$
and then
$$
F_{X}(x) = \begin{cases}
0& y\le 0\\
\frac{1}{64} y^{3/2} & 0\leq y \leq 16\\
1 & y\ge 16
 \end{cases}
$$
and 
$$
f_{Y}(y) =F'_{Y}(y) = \begin{cases}
\frac{3}{128} \sqrt y & 0\leq y \leq 16\\
0 & \text{elsewhere}
 \end{cases}
$$
A: Have you much experience the Jacobian change of variable transformations?
When $y=x^2$ then $\frac{\mathrm d x}{\mathrm d y} = \tfrac 1{2\surd y}$, then since $X\mapsto Y$ folds the negative and positive parts of the support for $X$ together, then we have the Jacobian transformation:
$$f_Y(y)~{~=~\left(\left\lvert \tfrac 1{2\surd y}\right\rvert f_X(\surd y) +\left\lvert \tfrac 1{2\surd y}\right\rvert f_X(-\surd y)\right)\mathbf 1_{y\in(0;16]} \\~\\ \phantom{~=~ c\surd y~\mathbf 1_{y\in(0;16]}} }$$
Which simplifies greatly.

If not familiar with that, you can get there by using the fundamental law of calculus.
$$f_Y(y) ~{~=~ \dfrac{\mathrm d~~}{\mathrm d~y}\mathsf P(-\surd y \leqslant X\leqslant \surd y)\\  ~=~ \dfrac{\mathrm d~~}{\mathrm d~y}\int_{-\surd y}^{\surd y} f_X(x)\mathrm d x \\~=~ \dfrac{\mathrm d~\surd y}{\mathrm d~y}f_X(\surd y)-\dfrac{\mathrm d~(-\surd y)}{\mathrm d~y}f_X(-\surd y)}$$
