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  1. #1
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    expectation

    how do i find the expection for p(y)=(0.5)^(y+1) for y=0,1,2,3,...

    what i did was integrate over 0 to infinity for y(0.5)^(y+1) and i got a negative number instead..
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    how do i find the expection for p(y)=(0.5)^(y+1) for y=0,1,2,3,...

    what i did was integrate over 0 to infinity for y(0.5)^(y+1) and i got a negative number instead..
    This is a discrete distribution so why are you integrating? You need to evaluate \sum_{y=0}^{+\infty} \left[y \, \left(\frac{1}{2}\right)^{y+1} \right]. The calculation is very similar to that of finding the mean of a geometric distribution (which has been done several times in this subforum - use the Search tool to find the threads).
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