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Math Help - Another Joint pdf question

  1. #1
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    Another Joint pdf question

    Hi,

    Iam trying to find the conditional pdf f(x│1/2) and have the joint pdf of

    f(x,y) = 3.3 - y , 0 < y < < 1

    Iam not sure how to set up the limits of integration to get f(x│y) because of the x^2.

    Any help how be apprectiated
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  2. #2
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    Quote Originally Posted by Number Cruncher 20 View Post
    Hi,

    Iam trying to find the conditional pdf f(x│1/2) and have the joint pdf of

    f(x,y) = 3.3 - y , 0 < y < < 1

    Iam not sure how to set up the limits of integration to get f(x│y) because of the x^2.

    Any help how be apprectiated
    f(x | y) = \frac{f(x, y)}{f_Y(y)}.

    The joint pdf is non-zero for values of X and Y lying in the region of the XY-plane enclosed by x = 0, x = 1, y = 0 and y = x^2. Therefore the marginal density function of Y is

    f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    f(x | y) = \frac{f(x, y)}{f_Y(y)}.

    The joint pdf is non-zero for values of X and Y lying in the region of the XY-plane enclosed by x = 0, x = 1, y = 0 and y = x^2. Therefore the marginal density function of Y is

    f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx.
    My mistake. It should be

    f_Y(y) = \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx + \int_{x=-1}^{x = -\sqrt{y}} f(x, y) \, dx = 2 \int_{x = +\sqrt{y}}^{x=1} f(x, y) \, dx (since f(x, y) is symmetric in x).

    And by now you've realised that it's 1.8 and not 3.3 (see my other mistake in the other post)
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