Verifying use of method to find pdf of Y Let's assume that we are given $f_{X}(x)=0.5e^{-|x|}$, with x being in the set of all real numbers and Y=$|X|^{1/3}$.  If I'm asked to find the pdf of Y, do I just follow the formula and do the following?
$f_{Y}(Y)$=$f_{x}(g^{-1}(y))$|$g^{-1}$'(y) to get something like:
$0.5e^{-|y^{1/3}|} |y^{-2/3}/3|$
Is it just a matter of following the formula or are there other things to consider?
 A: In this case it is easiest to follow @Henry'2 suggestion to
use $Z = |X| \sim Exp(1)$ with density $f_Z(z) = e^{-z},$ for $z > 1$.
Then you can use a monotone transformation to get $Y = Z^{1/3}.$
Furthermore, the supports of $Z$ and $Y$ are the same.
(BTW: The random variable $X$ has a Laplace distribution;
see your text or Wikipedia.)
Alternatively, you could (1) transform separately for the
negative and positive parts of the support of $X$ or (2)
use the so-called CDF method to get
the CDF of $Y$ and differentiate that CDF to get the density of $Y$.
Below is a simulation run with $m = 100,000$ realizations of
$X$ transformed directly to $Y$. Histograms show simulated realizations of $X$ and $Y.$ Your density for $X$ is superimposed
on the histogram at left, and the correct density for $Y$ is
shown at the right. Intuitively, $P(|X| > 8)$ is quite small
and $P(Y > 2)$ must be just as small.
 m = 10^5;  x = sample(c(-1,1), m, rep=T)*rexp(m)
 y = abs(x)^(1/3)
 mean(x);  sd(x)
 ## -1.785632e-05  # approx E(X) = 0
 ## 1.408634       # approx SD(X) = sqrt(2)
 mean(y);  sd(y)
 ## 0.893701
 ## 0.3232807


A: Hint (not a complete solution):
$\newcommand{\P}[1]{\mathbb{P}\left(#1\right)}$$$F_{Y}(y) = \P{Y \leq y} = \P{|X|^{1/3}\leq y} = \P{|X| \leq y^{3}} = \P{-y^3 < X < y^3} = \int\limits_{-y^3}^{y^3}0.5e^{-|x|}\text{ d}x\text{.}$$
It is easy  to show that $f_{X}(x)$ is symmetric about $0$. (Why?)
So, $$\int\limits_{-y^3}^{y^3}0.5e^{-|x|}\text{ d}x = 2\int\limits_{0}^{y^3}0.5e^{-|x|}\text{ d}x = \int\limits_{0}^{y^3}e^{-|x|}\text{ d}x\text{.}$$
Since $y^3 \geq 0$ (why?), $$\int\limits_{0}^{y^3}e^{-|x|}\text{ d}x = \int\limits_{0}^{y^3}e^{-x}\text{ d}x = 1 - e^{-y^3}\text{, }\quad y \geq 0\text{.}$$
Please also see the document I created on transformations.
