Probability question involving sequences of mutually exclusive events.

**Zennie** · September 8th 2010, 09:19 AM

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 $n \geq 1$ , $\bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i$ .

**Failure** · September 8th 2010, 09:29 PM

Originally Posted by Zennie

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 $B_i := A_i\backslash \bigcup_{j<i}A_j$ ?

**Zennie** · September 8th 2010, 10:34 PM

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.

**Failure** · September 8th 2010, 11:19 PM

Originally Posted by Zennie

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 $\bigcup_i B_i \supseteq \backslash \bigcup_i A_i$

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