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Math Help - Probability question involving sequences of mutually exclusive events.

  1. #1
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    Probability question involving sequences of mutually exclusive events.

    Let \{A_1, A_2, A_3, ...\} be a sequence of events of a sample space S. Find a sequence \{B_1, B_2, B_3, ...\} of mutually exclusive events such that for all n \geq 1, \bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i.
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    Quote Originally Posted by Zennie View Post
    Let \{A_1, A_2, A_3, ...\} be a sequence of events of a sample space S. Find a sequence \{B_1, B_2, B_3, ...\} of mutually exclusive events such that for all n \geq 1, \bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i.
    What about B_i := A_i\backslash \bigcup_{j<i}A_j?
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    Thanks, Failure. You wouldn't be able to explain how you got your answer would you? I don't exactly understand the question... which means I don't really understand your answer either.
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Zennie View Post
    Thanks, Failure. You wouldn't be able to explain how you got your answer would you? I don't exactly understand the question... which means I don't really understand your answer either.
    You want a sequence B_i of mutually exclusive events (i.e. disjoint sets) that has the same union as the sequence A_i.
    So let's start by setting B_1 := A_1. Now for B_2 we can take A_2 but in order to keep it disjoint from B_1 we take away A_1. - So B_1 and B_2 := A_2\backslash A_1 are disjoint, agreed?

    Similarly, for B_3 we can take A_3, but in order to keep it disjoint from B_{1,2} must take away their union, or, what amounts to the same thing A_1\cup A_2. We choose B_3 := A_3\backslash (A_1\cup A_2).

    So generally, we would like to define B_i := A_i\backslash \bigcup_{j<i}A_i.

    Observe that B_i \subseteq A_i, for all i, by definition. Also, if i_1<i_2 we see that

    B_{i_1}\cap B_{i_2}\subseteq A_{i_1}\cap \left(A_{i_2}\backslash \bigcup_{j<i_2}A_j\right)\subseteq A_{i_1}\cap (A_{i_2}\backslash A_{i_1})=\emptyset

    Finally, to show that \bigcup_i B_i = \bigcup_i A_i.

    First, from B_i\subseteq A_i for all i it follows immediately that

    \bigcup_i B_i \subseteq \bigcup_i A_i

    Second, suppose that x\in \left(\bigcup_i A_i\right) \backslash \left(\bigcup_i B_i\right). In that case there must exist a smallest i_0, such that x \in A_{i_0}.
    But then we have that x\in A_{i_0}\backslash \bigcup_{j<i_0}A_j=B_{i_0} and therefore x\notin \left(\bigcup_i A_i\right) \backslash \left(\bigcup_i B_i\right), a contradiction. Thus there can be no such x, hence we have that \bigcup_i B_i \supseteq \backslash \bigcup_i A_i
    Last edited by Failure; September 9th 2010 at 01:29 AM.
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