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Math Help - marginal distribution

  1. #1
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    marginal distribution

    I have a pdf of two random variables given as f(x,y) = \frac{K}{x} if 0<y<x  \mbox{; and ;} 0<x<2, and 0 otherwise.

    I found out the value of K by integrating the pdf equal to 1, that is:

    \int_0^2 \int_0^x \frac{K}{x} \mbox{dy} \mbox{dx} = 1

    and got k = 1/2

    then I found of the marginal pdf of x by:

    f_{X}(x) = \int_0^x \frac{1}{2x} dy = \frac{1}{2}

    but when I tried to find the marginal pdf of y by doing:

    f_{Y}(x) = \int_0^2 \frac{1}{2x} dx, I am getting infinity!!

    Is that correct or did I make a mistake while finding the marginal pdf of Y??

    Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    the lower bound on that last integral is y
    The region is 0<y<x<2.
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  3. #3
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    Thank you Matheagle!

    I did not grasp this material completely.

    So f_{Y}(y) = \int_0^x \frac{1}{2x} dx is what I need to evaluate?
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  4. #4
    MHF Contributor matheagle's Avatar
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    you cannot integrate x from 0 to itself, key word is ITSELF, 0<x<x does not make sense.

    You want {1\over 2}\int_y^2{dx\over x} where 0<y<2.

    which is a valid, but strange density, I checked.
    Thanks from downthesun01
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  5. #5
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    Quote Originally Posted by matheagle View Post
    you cannot integrate x from 0 to itself, key word is ITSELF, 0<x<x does not make sense.

    You want {1\over 2}\int_y^2{dx\over x} where 0<y<2.

    which is a valid, but strange density, I checked.
    Thanks again Matheagle..


    I got f_{Y}(x) = \frac{log(2)-log(y)}{2}

    and finally I would use this to find the condiitional pdf for X given Y by:

     \frac{f(x,y)}{f_{Y}(y)}

    Why did you call this density strange?
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  6. #6
    MHF Contributor matheagle's Avatar
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    It's just weird looking to me, it's not a multiple of anything I am used to seeing.
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