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Math Help - independence

  1. #1
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    independence

    I hope this is in the right place.

    Given the joint pdf
    f (u, v) = u + v , u > 0, v < 1

    determine if U and V are independent

    i tried to find f(u) by integrating the joint pdf from 0 to infinity... but my answer comes up to be infinity?

    for v, i integrated the joint pdf from negative infinity to 1

    is this the only way to find independence? f(u, v) = f(u) * f(v)?
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  2. #2
    MHF Contributor matheagle's Avatar
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    BY inspection the RVs are dependent.
    AND you need to find the correct region where the density is nonzero.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    BY inspection the RVs are dependent.
    AND you need to find the correct region where the density is nonzero.
    Seems like they are dependent as well. But I still need to find the marginal pdf's for the other parts of the question.

    What is the correct region? I was only given u > 0 and v < 1
    so I thought u goes from 0 to infinity and v goes from negative infinity to 1
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by helpwithassgn View Post
    Seems like they are dependent as well. But I still need to find the marginal pdf's for the other parts of the question.

    What is the correct region? I was only given u > 0 and v < 1
    so I thought u goes from 0 to infinity and v goes from negative infinity to 1
    THAT is not a valid region.
    A density cannot be negative.
    After you find the correct region, maybe 0<v<1 and 0<u<1.
    The rvs are DEPENDENT since the joint density cannot factor into two functions, one of u and one of v.
    Last edited by matheagle; January 28th 2010 at 07:15 AM.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    THAT is not a valid region.
    A density cannot be negative.
    After you find the correct region, maybe 0<v<1 and 0<u<1.
    The rvs are DEPENDENT since the joint density cannot factor into two functions, once of u and one of v.
    Why can't a density be negative? If you look at a normal distribution, the pdf goes from negative infinity to positive infinity.
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  6. #6
    MHF Contributor matheagle's Avatar
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    you are confusing the region and the FUNCTION

    f(x,y)\ge 0 and u+v can be negative in YOUR region
    SO, why don't you find the correct region.
    ONCE again, I'll bet that 0<u,v<1 is the correct region.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    you are confusing the region and the FUNCTION

    f(x,y)\ge 0 and u+v can be negative in YOUR region
    SO, why don't you find the correct region.
    ONCE again, I'll bet that 0<u,v<1 is the correct region.
    I get what you're saying now.
    I think the correct region for v is 0< v < 1
    but can't u be 0 < u < infinity?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by helpwithassgn View Post
    I get what you're saying now.
    I think the correct region for v is 0< v < 1
    but can't u be 0 < u < infinity?
    because the double integral must equal ONE.
    These facts come from the definition of a density.
    Last edited by matheagle; January 27th 2010 at 10:58 PM.
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  9. #9
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    Quote Originally Posted by matheagle View Post
    because the double integral must equal ONE.
    These facts come from the defintion of a density.
    Thanks, I think I should be able to solve it now. I'll be back tomorrow if I can't solve it so let's hope I get it!
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