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Thread: two-to-one transformation

  1. #1
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    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
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  2. #2
    Super Member Random Variable's Avatar
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    $\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 $
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