- #1
rwinston
- 36
- 0
Hi
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:
<br /> = \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)<br />
<br /> = \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)<br />
I can't 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!
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:
<br /> = \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)<br />
<br /> = \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)<br />
I can't 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!
