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Thread: if X is uniform, find density function of Y=X^2

  1. #1
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    if X is uniform, find density function of Y=X^2

    If $\displaystyle X$ has the uniform distribution $\displaystyle U(-1,3)$, find the probability density function of $\displaystyle Y=X^2$.
    Since $\displaystyle -1<x<3, 0<y<9$. Also, $\displaystyle X=\sqrt{Y},-\sqrt{Y}$. I've done some work and ended up with $\displaystyle f(x)=\frac{1}{2\sqrt{y}}$. But when I integrate this from 0 to 9, it does not give me 1. Can I get some help?
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  2. #2
    Super Member Random Variable's Avatar
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    You have to break it up.

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

    and $\displaystyle G(y) = \frac{1}{2} + \frac{1}{4} \int_{1}^{\sqrt{y}} \ dx = \frac{1}{2} + \frac{1}{4} (\sqrt{y} -1) = \frac{1}{4} + \frac{\sqrt{y}}{4} $ if $\displaystyle 1 < y \le 9$

    so $\displaystyle g(y) = \frac{1}{4\sqrt{y}} $ if $\displaystyle 0 \le y<1$

    and $\displaystyle g(y)= \frac{1}{8 \sqrt{y}} $ if $\displaystyle 1 \le y \le 9 $
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  3. #3
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    Quote Originally Posted by Random Variable View Post
    $\displaystyle G(y) = \frac{1}{2} + \frac{1}{4} \int_{1}^{\sqrt{y}} \ dx = \frac{1}{2} + \frac{1}{4} (\sqrt{y} -1) = \frac{1}{4} + \frac{\sqrt{y}}{4} $ if $\displaystyle 1 < y \le 9$
    How did you get $\displaystyle \frac{1}{2}$?
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  4. #4
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    Quote Originally Posted by dori1123 View Post
    How did you get $\displaystyle \frac{1}{2}$?
    G(1) = 1/2. Therefore G(y) = what RV said when $\displaystyle 1 < y \leq 9$.
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