# two-to-one transformation

#### rebghb

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

#### Random Variable

$$\displaystyle G(y) = \frac{2}{5} \int_{-\sqrt{y}}^{\sqrt{y}} |x| \ dx = -\frac{2}{5} \int_{-\sqrt{y}}^{0} x \ dx + \frac{2}{5} \int_{0}^{\sqrt{y}} x \ dx = \frac{2}{5} y$$ 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 $$\displaystyle 1 < y \le 4$$