I am not sure if my process to solve this particular problem is correct and not looking for particular solutions.
The Question:
Given $X^3 $~ $ N( \mu , \sigma^2), $ x>0$ $ and $Y=(\frac{X^2}{4})$, find distribution of Y
My attempt:
$Y^{3/2}=(\frac{X^2}{4})^{3/2})=\frac{X^3}{8}$
$\frac{d}{dx}y^{3/2}=\frac{x}{4}>0, x>0$
so we can find the pdf of $Y^{3/2}$ with inverse:
$r(y^{3/2})^{-1}=8y^{3/2}, y>0$
$f(y)=f_x(r(y)^{-1})\frac{d}{dy}[r(y)^{-1}]$
$=stuff$ and then getting it back to pdf of Y with :
$f(y)=stuff^{2/3}$
I got lazy typing out the normal distribution parts but that is what is in place of "stuff" and the other substitutions.
Let $W = X^3$. Then $Y = 1/4 W^{2/3}$.
Furthermore, $$ P( Y < y ) = P( \frac{1}{4} W^{2/3} < y ) = P( W < 8y^{3/2}) = \Phi( u ), $$ where $\Phi$ is CDF of standard normal and $u = (8y^{3/2} - \mu)/\sigma$.
Probability density of $Y$ is $$ p_Y(y) = \frac{ d\Phi(u)}{du} \frac{du}{dy} = 12 \frac{\sqrt{y}}{\sqrt{2 \pi}\sigma}e^{-\frac{(8y^{3/2}-\mu)^2}{2\sigma^2}} $$