TITLE: Find the probability density function of $Y=X^2$
QUESTION [15 upvotes]: 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.

REPLY [14 votes]: 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
$$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$