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.
Best Answer
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 $0<y<1$.