If $X \sim N(0,1)$ find the distribution of $Y = X^3$

Question

I have already consulted V.K. Rohatgi, and it has an example where it takes $Y=X^a$ where $a>0$ but the domain of $X$ is positive real values.
Even the theorem for transformation of continuous random values restricts the derivative of $Y$ w.r.t. $X$ to be positive or negative for the entire domain of $X$ where as in the case when $Y = X^3$ derivative of $Y$ w.r.t. $X$ is $0$ at $X=0$.
I am unable to proceed further.

Answer 1

I 've been doing some questions from Rohtagi also. It's a very good book. And to answer your question.we are give $X\sim N(0,1),f(x)=\dfrac{1}{\sqrt{2\pi}}e^\frac{-x^{2}}{2} ;-\infty<x<\infty$

$Y=X^3$ is a monotone function and for the monotone function, you can directly apply the transformation formula given to us which is

$f(y)=f(x)\bigg|\dfrac{dx}{dy}\bigg|$

$Y=X^3 \implies X=Y^{\frac{1}{3}}$

$\dfrac{dx}{dy}=\dfrac{1}{3}y^{-\frac{2}{3}}$

$f(x)=f(y^{\frac{1}{3}})$

combining it we have $f(y)=f(y^{\frac{1}{3}})\dfrac{1}{3}y^{-\frac{2}{3}}=\dfrac{1}{\sqrt{2\pi}}e^\frac{{-y^ \frac{2}{3}}}{2}\cdot\dfrac{1}{3}y^{-\frac{2}{3}} ;-\infty<y<\infty$

Answer 2

We have

\begin{align*} P(Y \leq y) &= P(X^3 \leq y)\\ &= P(X \leq y^\frac{1}{3}) \end{align*}

Now we take the derivative in this case with $F$ being the CDF of X and f being the pdf of X.

$$F(y^\frac{1}{3})' = \frac{1}{3}f(y^\frac{1}{3})y^{-\frac{2}{3}}$$

Thus you just subsitute in the PDF of X.