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Does anyone know the general formula for calculating $\prod_{i \in I}(1-a_i)$ where $I$ is a set of indices?
I suspect it is $1+\sum_{\emptyset \subsetneq J \subseteq I}(-1)^{|J|}\prod_{j \in J}a_j$. I don't require a proof, just a confirmation.

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Looks almost fine. You should write either $\sum_{\emptyset \ne J \subseteq I}$ or omit the "$1$", as $(-1)^{|\emptyset|}\prod_{j \in \emptyset} a_j = 1$ gives the second "$1$" now, which you don't need. – martini Sep 28 '12 at 7:20
Done the edit as suggested. – Abhishek Anand Sep 28 '12 at 7:23
Now you are right. – martini Sep 28 '12 at 7:45

1 Answer

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We have by distributivity \begin{align*} \prod_{i\in I} (1 - a_i) &= \sum_{J \subseteq I} \prod_{j \in J} (-a_j) \cdot \prod_{j \in I \setminus J} 1 \\ &= \sum_{J \subseteq I} (-1)^{|J|} \prod_{j \in J} a_j\\ &= 1 + \sum_{\emptyset \ne J \subseteq I} (-1)^{|J|} \prod_{j \in J} a_j \end{align*}

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