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Math Help - Independent random variables and their PDF

  1. #1
    Member courteous's Avatar
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    Question Independent random variables and their PDF

    Let f_{X,Y}(x,y)=C(x^2+y^2+1)^{-2}. Set C so that the two vector components will be independent.
    f_{X,Y}(x,y)=f_X(x)*f_Y(y)

    I can get f_X(x) (and similarly f_Y(y)) as \int_{-\infty}^\infty f_{X,Y}(x,y)dy, but I can't integrate it. I suspect this isn't the right way to proceed.

    The solution given is: "There is no such C."
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  2. #2
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    Quote Originally Posted by courteous View Post
    f_{X,Y}(x,y)=f_X(x)*f_Y(y)

    I can get f_X(x) (and similarly f_Y(y)) as \int_{-\infty}^\infty f_{X,Y}(x,y)dy, but I can't integrate it. I suspect this isn't the right way to proceed.

    The solution given is: "There is no such C."
    In each case the integration to get the marginal pdf's will involve the inverse tan function ....
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  3. #3
    Moo
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    Hello,

    There is a unique C such that f_{X,Y} is a pdf (it will be the inverse of the integral of the function). We don't care about the value, we can just keep it as C. With C being different from 0, would it be possible to write the joint pdf as a product in the form g(x)*h(y) ? (which would mean that X and Y are independent)
    The answer is obviously no...

    Hence the answer "there is no such C"
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  4. #4
    Member courteous's Avatar
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    Quote Originally Posted by Moo View Post
    There is a unique C such that f_{X,Y} is a pdf (it will be the inverse of the integral of the function).
    I don't understand what you mean by "inverse of the integral" ... isn't the "function" already p.d.f.?

    Quote Originally Posted by Moo View Post
    With C being different from 0, would it be possible to write the joint pdf as a product in the form g(x)*h(y) ? (which would mean that X and Y are independent)
    The answer is obviously no...
    Not obvious to me. Why is C\neq 0 \Rightarrow X, Y \text{ are independent } not feasible?


    All I know is that f_{X,Y}(x,y)=f_X(x)*f_Y(y) and that \int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)dxdy=1 ... which is not a lot.
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