# Thread: two-to-one transformation

1. ## two-to-one transformation

Hello everyone!

I got this transformation problem here:
$\displaystyle \text{let} \ f(x)=\frac{2}{5}|x|, \text{if} \ -1<x<2, \text{be the pdf of a continuous random variable} \ X$. $\displaystyle \text{find the pdf of } Y=X^2$.

The solution starts by mentioning that $\displaystyle g(x)=x^2$ is a two-to-one transformation. I'm a bit lost here, both $\displaystyle f(x)$ and $\displaystyle g(x)$ are two to one? What do we do??... The answer says:

$\displaystyle h(y)= \frac{2}{5}, 0<y<1$ and $\displaystyle h(y)=\frac{1}{5}, 1<y<4$.
Any help is appreciated

2. $\displaystyle G(y) = \frac{2}{5} \int_{-\sqrt{y}}^{\sqrt{y}} |x| \ dx = -\frac{2}{5} \int_{-\sqrt{y}}^{0} x \ dx + \frac{2}{5} \int_{0}^{\sqrt{y}} x \ dx = \frac{2}{5} y$ if $\displaystyle 0 \le y \le 1$

and $\displaystyle G(y) = G(1) + \frac{2}{5} \int_{1}^{\sqrt{y}} |x| \ dx = \frac{2}{5} + \frac{2}{5} \int_{1}^{\sqrt{y}} x \ dx = \frac{2}{5} + \frac{1}{5} (y-1)$ if $\displaystyle 1 < y \le 4$

so $\displaystyle g(y) = G'(y) = \frac{2}{5}$ if $\displaystyle 0 \le y \le 1$

and $\displaystyle g(y) = \frac{1}{5}$ if $\displaystyle 1 < y \le 4$