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Math Help - Silly mistake somewhere in integration?

  1. #1
    Junior Member
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    Dec 2010
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    Silly mistake somewhere in integration?

    Hi, I have the following problem.

    Suppose that the random variables X and Y vary in accordance with the joint pdf  f_{X, Y}(x, y) = c(x + y), 0 < x < y <1
    Find c.

    As I see it then we can say F_{X, Y}(x, y) = \int_{0}^1 \int_{0}^y \! c(x + y) \, \mathrm{d}x \, \mathrm{d}y = 1

    Then separating the variables F_{X, Y}(x, y) = \int_{0}^1 cy \int_{0}^y \! cx \, \mathrm{d}x \, \mathrm{d}y = 1

    So F_{X, Y}(x, y) = \int_{0}^1 \! cy [{c\over2} y^2] \, \mathrm{d}y = \int_{0}^1 \! {c^2\over2} y^3 \, \mathrm{d}y = 1

    So, {c^2 \over 8} = 1

    so c = \sqrt{8}

    But according to the solution manual c = 2

    Can anyone point out where I have gone wrong?
    Thanks in advance. MD.
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  2. #2
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    Re: Silly mistake somewhere in integration?

    The equation after separating the variables should be:
    F_{X, Y}(x, y) = \int_{0}^1 cy^2 \mathrm{d}y + \int_{0}^1 \int_{0}^y \! cx \, \mathrm{d}x \, \mathrm{d}y = 1
    Last edited by tupo; April 3rd 2012 at 07:56 AM.
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