The PDF of $X^3$ where $ X \sim \text{Normal}(0, 1)$

Question

I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)

Here's what I have tried. Let $Y = X^3$, where X is a standard normal distribution with mean 0 and variance 1. Then

$P(Y \leq y) = P(X^3 \leq y) = P(X \leq y^{1/3})$, for both $y$ positive or negative.

The above is the CDF of $X^3$; differentiating w.r.t. to $y$ should give me the PDF, which is

$f_X(y^{1/3})\frac{1}{3}y^{-2/3}$, where $f_X$ is the PDF of the standard normal.

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But THAT cannot be the answer, because it would take on negative values for $y < 0$. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes $y$ to $|y|$ and results in an integral that converges (in Mathematica).

But where can I introduce the absolute terms!? Much thanks :D

Answer 1

Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $\Phi,\,\phi$ denote the CDF and PDF of $X$. For $y\ge 0$, $$P(Y\le y)=\Phi(y^{1/3})\implies f_Y(y)=\frac{1}{3}y^{-2/3}\phi(y^{1/3}).$$Then, for arbitrary $y\in\mathbb{R}$,$$f_Y(y)=f_Y(|y|)=\frac{1}{3}|y|^{-2/3}\phi(|y|^{1/3}).$$

Answer 2

If $y \ne 0$, the $2$ in the index would make it positive,

$$f_X(y^{1/3})\frac{1}{3}y^{-2/3}=f_X(y^{1/3})\frac{1}{3}y^{-\color{red}2/3} $$