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Math Help - Y=1-x^2

  1. #1
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    Y=1-x^2

    The random variable X has density given with
    p(x)={
    (x+1)/2 , -1<x<1
    0, in other cases
    }
    find the density of the random variable Y=1-X^2

    I'm stuck with this problem
    any help would be appreciated
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  2. #2
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    Re: Y=1-x^2

    Hey dejandenib.

    Are you aware of the transformation theorem?

    Transformations of Variables

    Since f(x) = 1 - x^2 has an inverse in the appropriate domain, you can use this.
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  3. #3
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    Re: Y=1-x^2

    yes, I'm aware of that. I used it, and don't know how to solve it completely.
    After I integrate it, I have: (1/4)x^2 + x/2 + C, -1<x<1,
    and because in the main formula : y=1-x^2, i have x squared, when I calculate the root, I'll have both signs, plus and minus, in front of the square root.
    I don't know how to handle these two cases in one.
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  4. #4
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    Re: Y=1-x^2

    Quote Originally Posted by dejandenib View Post
    The random variable X has density given with
    p(x)={
    (x+1)/2 , -1<x<1
    0, in other cases
    }
    find the density of the random variable Y=1-X^2

    I'm stuck with this problem
    any help would be appreciated
    Do you have the right answer?my answer comes out to be -(\frac{1}{4}+\frac{1}{4\sqrt{1-y}}) .If it matches with your answer, post it,i will post the solution.I might have done something wrong.
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  5. #5
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    Re: Y=1-x^2

    So I'm given the density (x+1)/2
    To find out the function of distribution, I integrate it
    integrate &#40;x&#43;1&#41;&#47;2 dx - Wolfram|Alpha
    what do I do next?
    I dont know what exactly is your answer you are showing me.
    I need hint on what to do next after this step.
    I know that eventually I'll use Py(x)=|Px(f[-1](x))*|f[-1](x)|
    but I'm having problems with the domain, because I have a square in the equation, and the domain of x can be negative.
    Thanx for the help so far
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  6. #6
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    Re: Y=1-x^2

    Given a r.v. X with pdf f(x) with a r.v. Y = g(X) then:

    g(y) = f(g^-1(y)) * |dg^-1(y)/dy|

    Now g(x) = 1 - X^2 so g^-1(x) = SQRT(1 - x) and dg^-1(x)/dx = -1*1/2*1/SQRT(1-x)

    This information will give you the PDF for Y.
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  7. #7
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    Re: Y=1-x^2

    Two questions:
    1.
    why is g^-1(x) = SQRT(1 - x)
    but not = + - SQRT(1-x)
    ?
    2.
    after i substitute the values in the formula for the PDF of Y, do I get:
    g(y) =[(sqrt(1-x)+1)/2 ]*[1/2*sqrt(1-x)]
    which is
    (\frac{1}{4}+\frac{1}{4\sqrt{1-x}})
    Last edited by dejandenib; January 27th 2013 at 05:36 AM.
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  8. #8
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    Re: Y=1-x^2

    Quote Originally Posted by dejandenib View Post
    Two questions:
    1.
    why is g^-1(x) = SQRT(1 - x)
    but not = + - SQRT(1-x)
    ?
    2.
    after i substitute the values in the formula for the PDF of Y, do I get:
    g(y) =[(sqrt(1-x)+1)/2 ]*[1/2*sqrt(1-x)]
    which is
    (\frac{1}{4}+\frac{1}{4\sqrt{1-x}})
    AS you can see, I got the same answer too(excuse the typo...) but with a (-) sign in front of it.Do you have the correct answer??
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  9. #9
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    Re: Y=1-x^2

    i believe the minus should be ignored because we calculate the ABSOLUTE value of the derivative.
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  10. #10
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    Re: Y=1-x^2

    can anyone explain how to handle the case with both signs, plus and minus in front of the sqrt(1-x) ?
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