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

[Eclair_de_XII]

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.

 

[andrewkirk]

"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 $F_k$.

 

[Eclair_de_XII]

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.

 

[andrewkirk]

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.

 

[Eclair_de_XII]

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...

 

[Eclair_de_XII]

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."