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Math Help - Sum of Two Independent Random Variables (uniform)

  1. #1
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    Sum of Two Independent Random Variables (uniform)

    Suppose X and Y are Uniform(-1, 1) such that X and Y are independent and identically distributed. What is the density of Z = X + Y?

    Here is what I have done so far (I am new to this forum, so, my formatting is very bad). I know that
    fX(x) = fY(x) = 1/2 if -1<x<1 and 0 otherwise

    The density of Z will be given by
    fZ(z)= integral fX(z-y)fY(y)dy bounds of integration from -infinity to infinity

    fY(y) = integral 1/2 if -1<y<1 and 0 otherwise
    So,

    fZ(z) = (1/2)fX(z-y)dy (bounds of integration -1 to 1)

    The integrand = 1/2 if -1<z-y<1 or z-1<y<z+1
    or 0 otherwise

    This is where I get stuck and not sure how to proceed. Thanks
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    MHF Contributor chisigma's Avatar
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    If x has PDF f(x) and y has PDF g(y), the z=x+y has PDF...

    \displaystyle \varphi(z)= \int_{- \infty}^{+ \infty} f(\xi)\ g(z-\xi)\ d \xi (1)

    If f(*) and g(*) are 'rectangular' then \varphi(*) is 'triangular'...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor matheagle's Avatar
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    This is just like the sum of two U(0,1).
    It's easy to solve it via geometry.
    Draw a square (-1,1) by (-1,1)
    Obtain F_Z(z)=P(X+Y\le z) by figuring the area "under" x+y<z for -2<z<2.
    The third dimension is just 1/4, since the density is constant over that square.
    Last edited by matheagle; February 4th 2011 at 06:22 AM.
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  4. #4
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    Quote Originally Posted by matheagle View Post
    This is just like the sum of two U(0,1).
    It's easy to solve it via geometry.
    Draw a square (-1,1) by (-1,1)
    Obtain F_Z(z)=P(X+Y\le z) by figuring the area "under" x+y<z for -2<z<2.
    The third demsion is just 1/4, since the density is constant over that square.
    Indeed. @OP: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf page 8.
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