Name of the probability distribution

Question

If $X\sim N(0,1)$, then the density function of random variable $X^3$ is as follows: $$f(y)=\frac{1}{3\sqrt{2\pi}}\left | y \right |^{-\frac{2}{3}}e^{-\frac{1}{2}\left | y \right |^{\frac{2}{3}}}$$ I'm interested in the properties of this distribution. I could derive them, but maybe it has some special name, so I could search for it specifically. So, my question, what is the name of the above distribution with the above pdf?

Answer 1

Extended comment.

Here is a simulation experiment (in R), that suggests $Var(X^3) = 15.$

 x = rnorm(10^6);  y = x^3
 var(y);  mean(abs(y) < 40)
 ## 15.02577;  0.999341
 summary(y)
 ##        Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
 ##  -1.020e+02 -3.077e-01  1.118e-08 -2.277e-04  3.101e-01  1.409e+02 

  hist(y[abs(y)< 40], prob=T, br=50, col="wheat", main="")
  curve(1/(3*sqrt(2*pi))*abs(x)^(-(2/3))*exp(-.5*abs(x)^(2/3)),
      lwd=2, n = 1001, col="blue", add=T)

Answer 2

An explicit derivation of Robert Israel's claim is fairly straightforward by induction: $$\begin{align*} \operatorname{E}[X^{2m}] &= \int_{x=-\infty}^\infty x^{2m} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \end{align*}$$ and with the choice $$u = x^{2m-1}, \quad du = (2m-1) x^{2(m-1)} \, dx, \quad dv = x e^{-x^2/2} \, dx, \quad v = -e^{-x^2/2},$$ we obtain $$\begin{align*} \operatorname{E}[X^{2m}] &= \left[ -\frac{x^{2m-1} e^{-x^2/2}}{\sqrt{2\pi}} \right]_{x=-\infty}^\infty + (2m-1)\int_{x=-\infty}^\infty x^{2(m-1)} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \\ &= (2m-1)\operatorname{E}[X^{2(m-1)}]. \end{align*}$$ Since $\operatorname{E}[X^0] = 1$, we obtain $$\operatorname{E}[X^{2m}] = \prod_{j=1}^m (2j-1) = (2m-1)!! = \frac{(2m)!}{2^m m!}.$$