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Math Help - Probability of a joint p.d.f.

  1. #1
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    Probability of a joint p.d.f.

    f_{X,Y}(x,y) = \{x + y , 0 \leq x \leq 1, 0 \leq y \leq 1; 0 otherwise

    Find Pr(X+Y \leq 1)

    I understand i need to do \int \int (x+y) dy dx
    But i can't work out what the limits of the integrals should be, any help would be great, thanks.
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  2. #2
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    have you thought about doing the inner integral from 0 to 1 (since that's the range of y), and the outer integral from 0 to 1 (since that's the range of x?

    Also, if you graph the pdf, you'll notice that it's the area of a circle...so why not find the probability of landing in a circle of radius 1 inscribed in a 1x1 box? This should be the area of that circle divided by the area of the square.
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  3. #3
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    Quote Originally Posted by Rapid_W View Post
    f_{X,Y}(x,y) = \{x + y , 0 \leq x \leq 1, 0 \leq y \leq 1; 0 otherwise

    Find Pr(X+Y \leq 1)

    I understand i need to do \int \int (x+y) dy dx
    But i can't work out what the limits of the integrals should be, any help would be great, thanks.
    A diagram makes things crystal clear. The support of the joint pdf is a square. You need to integrate the joint pdf over the lower region of this square that is bounded by the line y = 1 - x.

    Quote Originally Posted by financial View Post
    have you thought about doing the inner integral from 0 to 1 (since that's the range of y), and the outer integral from 0 to 1 (since that's the range of x?

    Also, if you graph the pdf, you'll notice that it's the area of a circle...so why not find the probability of landing in a circle of radius 1 inscribed in a 1x1 box? This should be the area of that circle divided by the area of the square.
    Sorry, but this is not correct.
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