I have a random variable $X$ that is governed by a Folded Normal distribution.
$f_{X}(x) = \frac{2}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.
I am trying to find a transformation $h(X) = Y$ so that $Y$ is governed by a Gamma($\alpha, \beta$) distribution.
$g_{Y}(y) = \frac{1}{\Gamma (\alpha) \beta^\alpha}y^{\alpha-1} e^{-y/\beta}$.
I have started by removing the normalizing constants of both distributions and looking only at their respective kernels.
$f(x) = e^{-x^2}$ and $g(y) = y^{\alpha - 1}e^{-y}$, so I suppose that $y = x^2$. Is this correct? If so, where do I go from here to find $\alpha, \beta$ of the Gamma distribution?
Look at the kernel for each distribution. The kernel for $X$ is $e^{-x^2/2}$, and the kernel for $Y$ is $y^{a-1}e^{-by}$.
This is sort of a trial and error situation, but begin by letting $y = x^2, y \in (0, \infty)$ since they appear in both kernels.
$$y = x^2 \rightarrow x = + \sqrt{y} \rightarrow \frac{d}{dy} = \frac{1}{2\sqrt{y}}$$
$$f_{Y}(y) = f_{X}(\sqrt{y}) \bigg|{\frac{1}{2\sqrt{y}}} \bigg| = \frac{2}{\sqrt{2 \pi}}e^{-\sqrt{y}^2} \frac{1}{2\sqrt{y}} = \frac{1}{\sqrt{2\pi}}y^{-½} e^{-y/2} $$
Now try to see how this would fit into a Gamma density. Recall that $\Gamma(½) = \sqrt{\pi}$.
$$\frac{\big( \frac{1}{2} \big)^{½}}{\Gamma(½)} y^{½ -1} e^{-½ y}$$ which follows a Gamma density with $\alpha = ½, \beta = ½$.