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Is there any way to simplify this term: $$\sum_{v=1}^m v(1-\frac{1}{v})^u\binom{m}{v}p^v(1-p)^{m-v}$$ The term $(1-\frac{1}{v})^u$ is really annoying.


This expansion appear in a specific version of the urn and ball game. Suppose there are $u$ balls that will be randomly allocated to $V$ urns with $V$ being a random variable with binomial distribution $B(m,p)$.

Then the expected number of urns with at least one ball should be: $$ \sum_{v=0}^m v(1-(1-\frac{1}{v})^u) \binom{m}{v} p^v (1-p)^{m-v} $$

Break the expansion into two parts. The first part: $$ \begin{align} \sum_{v=0}^m v\binom{m}{v}p^v(1-p)^{m-v} &= mp \sum_{v=1}^m \binom{m-1}{v-1}p^{v-1}(1-p)^{m-v} \\ & = mp \end{align} $$ My question is how to deal with the second part.

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  • $\begingroup$ Are there any conditions on $u$? $\endgroup$ – Chappers Sep 18 '15 at 19:15
  • $\begingroup$ No. See my description of where the expansion comes from $\endgroup$ – empenguin Sep 18 '15 at 19:34
  • $\begingroup$ Well, if you have $u$ balls, I would hope $u$ is a non-negative integer! $\endgroup$ – Chappers Sep 18 '15 at 19:37
  • $\begingroup$ Ok. Sorry for being not rigorous. $u$, $m$ are non-negative integers. $0 \leq p \leq 1$ $\endgroup$ – empenguin Sep 18 '15 at 19:41

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