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Thread: Probability question involving sequences of mutually exclusive events.

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

    Let $\displaystyle \{A_1, A_2, A_3, ...\}$ be a sequence of events of a sample space $\displaystyle S$. Find a sequence $\displaystyle \{B_1, B_2, B_3, ...\}$ of mutually exclusive events such that for all $\displaystyle n \geq 1$, $\displaystyle \bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i$.
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    Quote Originally Posted by Zennie View Post
    Let $\displaystyle \{A_1, A_2, A_3, ...\}$ be a sequence of events of a sample space $\displaystyle S$. Find a sequence $\displaystyle \{B_1, B_2, B_3, ...\}$ of mutually exclusive events such that for all $\displaystyle n \geq 1$, $\displaystyle \bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i$.
    What about $\displaystyle B_i := A_i\backslash \bigcup_{j<i}A_j$?
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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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    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 $\displaystyle B_i$ of mutually exclusive events (i.e. disjoint sets) that has the same union as the sequence $\displaystyle A_i$.
    So let's start by setting $\displaystyle B_1 := A_1$. Now for $\displaystyle B_2$ we can take $\displaystyle A_2$ but in order to keep it disjoint from $\displaystyle B_1$ we take away $\displaystyle A_1$. - So $\displaystyle B_1$ and $\displaystyle B_2 := A_2\backslash A_1$ are disjoint, agreed?

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

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

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

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

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

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

    $\displaystyle \bigcup_i B_i \subseteq \bigcup_i A_i$

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