# Probability distribution on a transformed variable problem

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.

• If $X>0$ then $X^{3}>0$ so $X^{3}$ cannot have a normal distribution. – Kavi Rama Murthy Mar 23 at 23:50

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 $$p_Y(y) = \frac{ d\Phi(u)}{du} \frac{du}{dy} = 12 \frac{\sqrt{y}}{\sqrt{2 \pi}\sigma}e^{-\frac{(8y^{3/2}-\mu)^2}{2\sigma^2}}$$