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Math Help - Joint probability function

  1. #1
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    Joint probability function

    I understand all the concepts of joint PDFs but I just can't get the right answer to this one. The function is:

    f(x,y)=3x^(1/2)e^(-2y)

    I need to find the probabilty over these two regions:

    a) P (0 ≤ x ≤ 1; 3y)
    b) P({P ({(x, y) | 2/3x ≤ 1; 0 ≤ y)
    2

    Explanations would be much appreciated. Thanks
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  2. #2
    Moo
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hello,
    Quote Originally Posted by notecardheart View Post
    I understand all the concepts of joint PDFs but I just can't get the right answer to this one. The function is:

    f(x,y)=3x^(1/2)e^(-2y)

    I need to find the probabilty over these two regions:

    a) P (0 ≤ x ≤ 1; 3y)
    b) P({P ({(x, y) | 2/3x ≤ 1; 0 ≤ y)
    2

    Explanations would be much appreciated. Thanks
    By using wise definitions of the cumulative density functions, you should understand

    P(a<X<b)=\int_a^b f_X(x) ~dx, where f_X is the pdf of X

    For a joint distribution, it's the same reasoning :
    P(a<X<b ~,~ c<Y<d)=\int_a^b \int_c^d f(x,y) ~dy dx

    So for the first one for example, it's \int_0^1 \int_3^\infty f(x,y) ~dy dx
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  3. #3
    MHF Contributor matheagle's Avatar
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    I would like to know the support of this joint density.
    The rvs seem to be independent.
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