(New page: Virgil is right, it is <math>\sum_{i=1}^n\frac{n}{n - i + 1}\!</math> because E[X] is 1/p More over this can be simplified using the arithmetic series <math>\sum_{i=1}^n a_i = \frac...) |
(No difference)
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Latest revision as of 15:10, 6 October 2008
Virgil is right, it is
$ \sum_{i=1}^n\frac{n}{n - i + 1}\! $
because E[X] is 1/p
More over this can be simplified using the arithmetic series
$ \sum_{i=1}^n a_i = \frac{n(a_1+a_n)}{2}\! $
