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Math Help - 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 X has the uniform distribution U(-1,3), find the probability density function of Y=X^2.
    Since -1<x<3, 0<y<9. Also, X=\sqrt{Y},-\sqrt{Y}. I've done some work and ended up with 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.

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

    and  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 1 < y \le 9

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

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