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Unclear Step in Proof

  1. Oct 12, 2008 #1

    I may be missing something obvious here, but I am reading a paper on random sampling, and in one of the proofs, two consecutive steps run like this:

    = \sum_{d=0}^m \left[ {{m}\choose{d}} \left(\frac{1}{n}\right)^d \left(\frac{n-1}{n}\right)^{m-d} \right] \left( \frac{m-d}{m}\right)
    = \sum_{d=0}^{m-1} {{m}\choose{d}} \left(\frac{1}{n}\right)^d \left(\frac{n-1}{n}\right)^{m-d} \ \left( \frac{m-d}{m}\right)

    I cant see how the second step works .... the summation index is decreased by one, but nothing obvious changes inside the summation...is there an assumption or step i am missing? Any help appreciated!
  2. jcsd
  3. Oct 12, 2008 #2


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    The (m-d)/m term is in the summation right? At d=m, that term will become 0, and so d=m doesn't contribute to the sum.
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