# If O is an event space, show for a finite number of events--

• Eclair_de_XII
In summary, this problem can be solved by finding a k-element set in ##B_k## that is not in any other k-element set."In summary, the set of points in the event space which belong to exactly k of the A_i's in O belongs to O.

## Homework Statement

"If ##A_1,...,A_m\in O## and ##k\in ℕ##, show that the set of points in ##Ω## (the sample space) which belong to exactly ##k## of the ##A_i## belongs to ##O## (the previous exercise is the case when ##m=2## and ##k=1##)."

## Homework Equations

Event space: O
##O\neq ∅##
##\text{If} \space A\in O\space \text{then} \space Ω\cap A^c \in O##
##\text{If} \space A_1,...\in O \space \text{then} \space \bigcup_{i=1}^\infty A_i \in O##

## The Attempt at a Solution

This is how I have done the case for when ##m=2## and ##k=1##:

Since ##A,B\in O##, then it follows that since ##O## is closed under the operations of finite unions, that ##A\cup B\in O##. Moreover, since ##O## is closed under the operations of finite intersecctions, then it is true that ##A\cap B \in O##. Moreover, since ##A\cap B \in O##, then ##Ω\cap (A\cap B)^c \in O##. In turn, the symmetric difference of ##A## and ##B##, ##(A\cup B)\cap (A\cap B)^c =AΔB\in O##.

This is my attempt to solve for ##m,n\in ℕ##:

Since ##A_i\in O, i\in [1,m]\ \cap ℕ##, then it follows that ##\bigcup_{i=1}^m A_i \in O##. Moreover, ##\bigcup_{i\neq j} A_i\cap A_j\in O##. Then ##(\bigcup_{i=1}^m A_i)\cap (\bigcup_{i\neq j} A_i\cap A_j)^c\in O##. I've tried defining a set ##B_j=\{A_i:\forall x,y\in Ω, \text{If} \space x\in A_i, y\in A_n, \text{then} \space x,y\notin A_i\cap A_n, n\neq i\}##, with ##1\leq j \leq k## and writing: ##(\bigcup_{j=1}^k B_j)\cap (\bigcup_{i\neq j} A_i\cap A_j)^c\in O##, but I'm not really sure that I understand the problem.

Eclair_de_XII said:
"If ##A_1,...,A_m\in O## and ##k\in ℕ##, show that the set of points in ##Ω## (the sample space) which belong to exactly ##k## of the ##A_i## belongs to ##O## (the previous exercise is the case when ##m=2## and ##k=1##)."
Call the set described in red text ##B_k##.

Now for ##k\in\{1,...,m\}## define ##D^m_k## to be the set of all ##k##-element subsets of ##\{1,...,m\}##. Then the set of all elements that are in ##k## or more of the ##A_j## is
$$F_k\triangleq \bigcup_{S\in D^m_k} \bigcap_{j\in S} A_j$$
Confirm this to yourself before proceeding.

Then confirm that ##F_k\in O##.

Then see if you can write ##B_k## as the result of a finite sequence of sigma-algebra-preserving operations on the ##F_k##.

andrewkirk said:
Then the set of all elements that are in ##k## or more of the ##A_j## is

##F_k\triangleq \bigcup_{S\in D^m_k} \bigcap_{j\in S} A_j##

Confirm this to yourself before proceeding.

Sorry for the late reply, but I just got really scared after seeing all these notations and symbols that I don't understand, and kind of crawled into a metaphorical corner. Anyway, is ##S## the ##\sigma##-algebra ##O##? And a quick Google search tells me that ##F_k## is equal by definition to the right-hand side of the equation? I'm sorry; I just don't think that I can confirm that ##F_k \in O## without knowing what's being said, here.

The symbol ##\triangleq## means 'is defined to be'. Some people use := instead, or ##=_{def}##.

The expression means the union, for all possible selections of k distinct integers in {1,...,m}, of the intersection of the k sets in ##A_1,...,A_m## that have those integers as index numbers.

##S## is the set containing the k chosen integers. It is not the sigma algebra.

In other words:
1. choose k of the sets ##A_1,...,A_m## and take their intersection.
2. Then take a different selection of k of those sets, take their intersection, then combine that (take the union) with the set from the previous step.
3. Repeat that last step until all possible selections of k sets have been done.

andrewkirk said:
In other words:
1. choose k of the sets ##A_1,...,A_m##
• ##A_1##,...,##A_m## and take their intersection.
• Then take a different selection of k of those sets, take their intersection, then combine that (take the union) with the set from the previous step.
• Repeat that last step until all possible selections of k sets have been done.

Okay, so the event space is closed under finite intersections and unions, so ##F_k\in O##. And following in suit of the case where ##m=2## and ##k=1##...

##B_k=\bigcup_{i=1}^{m} A_i \cap (\bigcup_{S\in D^m_k} \bigcap_{j\in S} A_j)^c##

I think...

My professor went over the problem in class, and this is something else I came up with:

"Let ##I=\{1,...,m\}##. Define ##S=\{\sigma \in \mathbf{P}(I):|\sigma|=k\}##, where ##\mathbf{P}(I)## is the power set of ##I##. Then denote the k-element sets of integers from 1 to m as ##\sigma_n##, where ##n\in ℕ## and ##1\leq n\leq |S|##. Then define ##B_k=\bigcup_{n=1}^{|S|} [(\bigcap_{i\in \sigma_n} A_i)\cap(\bigcap_{j\neq i} A_j^c)]##, which is in ##B_k## since the event space is closed under finite set operations."