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Math Help - Sum of 2 independent uniform r.v.

  1. #1
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    Sum of 2 independent uniform r.v.

    If X and Y are independent r.v, both uniformly distributed on (0, 1).

    Given that f_{X+Y}(a)=\int_{-\infty}^{\infty}f_X(a-y)f_Y(y)dy

    we obtain f_{X+Y}(a)=\int_{0}^{1}f_X(a-y)dy

    For 0\leq a\leq 1, we have f_{X+Y}=\int_{0}^{a}dy=a

    For 1<a<2, we have f_{X+Y}=\int_{a-1}^{1}dy=2-a

    How do I determine the two upper and lower bounds?
    Such as for this case we are using: \int_{0}^{a} and \int_{a-1}^{1} ??

    Another question:

    Given joint density of X and Y:

    f(x,y)=\frac{1}{4}(y-x)e^{-y}
    -y<x<y, 0<y<\infty
    In order to find the density function of X, I meet the same problem:

    for x>0, f_X(x)=\frac{1}{4}\int_{x}^\infty (y-x)e^{-y} dy

    for x<0, f_X(x)=\frac{1}{4}\int_{-x}^\infty (y-x)e^{-y} dy
    Why the two different lower bounds?
    Last edited by noob mathematician; April 20th 2009 at 08:52 PM.
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    If X and Y are independent r.v, both uniformly distributed on (0, 1).

    Given that f_{X+Y}(a)=\int_{-\infty}^{\infty}f_X(a-y)f_Y(y)dy

    we obtain f_{X+Y}(a)=\int_{0}^{1}f_X(a-y)dy

    For 0\leq a\leq 1, we have f_{X+Y}=\int_{0}^{a}dy=a

    For 1<a<2, we have f_{X+Y}=\int_{a-1}^{1}dy=2-a

    How do I determine the two upper and lower bounds?
    Such as for this case we are using: \int_{0}^{a} and \int_{a-1}^{1} ??

    [snip]
    Read example 7.3 p292 here: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
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