Find the probability density function of $Y=X^2$ Consider the random variable X with probability density function $$f(x) = \begin{cases} 
 3x^2; & \text{ if, } 0 < x < 1 \\
 0; & \text{ otherwise }
\end{cases}$$ Find the probability density function of $Y=X^2$.
This is the first question of this type I have encountered, I have started by noting that since $0<x<1$, we have that $0< x^2<1$. So $X^2$ is distributed over $(0,1)$. I'm not really sure how to progress or what method to take to actually find the pdf.
 A: One method is to integrate wrt $x$ to find the CDF and then differentiate it back wrt $y$.
Another is to use the "change of variable transformation", which involves one differentiation.
$\begin{align}
\\ f_Y(y) & = f_X{\circ}x(y)\; \Bigg|\frac{\mathrm d x(y)}{\mathrm d y}\Bigg|\tag{1: change of variable}
\\[1ex]
f_X(x) & = 3x^2\tag{2 given}
\\[1ex] y & = x^2\tag{3 given}
\\[1ex] x(y) & = \pm\surd y\tag{4 $\impliedby 3$}
\\[1ex] \frac{\mathrm d x(y)}{\mathrm d y} & = \mp\tfrac 1 2 y^{-1/2}\tag{$5\impliedby 4$}
\\[2ex] f_Y(y) & = f_X{\circ}x(y)\; \Bigg|\frac{\mathrm d x(y)}{\mathrm d y}\Bigg|\tag{1}
\\[1ex] & = \frac{3y}{|2\surd y|}\tag{$\impliedby 2,4,5$}
\\[1ex] & = \tfrac 3 2 \surd y\tag{7}
\\[1ex]\sup(f_X) = [0,1] & \implies \sup(f_Y)=[0,1]\tag{8$\impliedby 3$}
\end{align}$

This is because: $\displaystyle\frac{\operatorname d}{\operatorname dy}\int_{\sup(f_X)} f_X(x) \operatorname d x = \int_{\sup(f_Y)} f_X{\circ}x(y) \bigg|\frac{\operatorname d x(y)}{\operatorname d y}\bigg|\operatorname d y$
So if we do not need the CDF, this technique might save effort.   It is also handy when the pdf can not be readily integrated (see: Gaussian Normal Distributions).
A: Note that $$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_0^\sqrt{y}3t^2\,dt=y^{3/2}$$ hence $$f_Y(y)=\frac{3}{2}\sqrt{y}$$
This holds for $0<y<1$.
A: This theorem will be helpful
$\mathbf {Theorem:}$
Let $X$ have density function $f_X$ and let $A=\{x~:~f_X(x)>0 \}$.  If $g$ is strictly monotone on $A$, then $Y=g(X)$ has density
$$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$
