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:


$\frac{d}{dx}y^{3/2}=\frac{x}{4}>0, x>0$

so we can find the pdf of $Y^{3/2}$ with inverse:

$r(y^{3/2})^{-1}=8y^{3/2}, y>0$


$=stuff$ and then getting it back to pdf of Y with :


I got lazy typing out the normal distribution parts but that is what is in place of "stuff" and the other substitutions.

  • $\begingroup$ If $X>0$ then $X^{3}>0$ so $X^{3}$ cannot have a normal distribution. $\endgroup$ – 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 $$ p_Y(y) = \frac{ d\Phi(u)}{du} \frac{du}{dy} = 12 \frac{\sqrt{y}}{\sqrt{2 \pi}\sigma}e^{-\frac{(8y^{3/2}-\mu)^2}{2\sigma^2}} $$

  • $\begingroup$ Oh great thanks, I think the part that tricked me was that I needed to use a change of variable $\endgroup$ – glockm15 Mar 27 at 1:47

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