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Math Help - Series Expansion of Uniform MGF

  1. #1
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    Series Expansion of Uniform MGF

    X has uniform dist (0,1).

    I found the MGF to be:

    \frac{e^s - 1}{s}

    Now I need to expand the MGF in powers of s up to s^2 and use this to find the mean and variance of X. I know the series expansion for e^s = \Sigma \frac{s^r}{r!}, but I'm not sure how to apply this in this situation.
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  2. #2
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    Hello,

    e^s=\sum_{n=0}^\infty \frac{s^n}{n!}=1+\sum_{n=1}^\infty \frac{s^n}{n!}

    So \frac{e^s-1}{s}=\sum_{n=1}^\infty \frac{s^{n-1}}{n!}

    by changing the indice, this is \sum_{n=0}^\infty \frac{s^n}{(n+1)!}
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  3. #3
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    EDIT: I think I understand what you did there.

    But, I'm still stuck on another part of this question:

    In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, X_{i}, i = 1,...,n, are independently distributed as U(-0.5,+0.5). What is the approximate distribution of the total error, \Sigma X_{i}, in the sum of the n numbers?

    How would I approach this question?
    Last edited by Zenter; August 9th 2009 at 04:40 AM.
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