two-to-one transformation

Jan 2010
133
7
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
 
May 2009
959
362
\(\displaystyle G(y) = \frac{2}{5} \int_{-\sqrt{y}}^{\sqrt{y}} |x| \ dx = -\frac{2}{5} \int_{-\sqrt{y}}^{0} x \ dx + \frac{2}{5} \int_{0}^{\sqrt{y}} x \ dx = \frac{2}{5} y \) 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 \(\displaystyle 1 < y \le 4 \)