# Thread: Probability question involving sequences of mutually exclusive events.

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

2. Originally Posted by Zennie
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 $\displaystyle B_i := A_i\backslash \bigcup_{j<i}A_j$?

3. 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.

4. 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 $\displaystyle B_i$ 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 $\displaystyle \bigcup_i B_i \supseteq \backslash \bigcup_i A_i$