How many elements can be in an event space with two events?

[Eclair_de_XII]

Homework Statement

"Do there exist any event spaces with just six elements?"

Homework Equations

The Attempt at a Solution

Suppose $F_1$ is an event space with a non-trivial event $A$. Then $F_1=\{∅,A,A^c,Ω\}$. So $inf(|F_1|) = 4$, since if you remove any of these events, $F_1$ is no longer closed under complements.

Suuppose $F_2$ is an event space with two non-trivial events $A$ and $B$. Then $F_2=\{∅,Ω,A,A^c,B,B^c,A\cup B, A^c\cap B^c,A\cap B, A^c\cup B^c,A^c\cap B, A\cup B^c, A\cap B^c, A^c\cup B\}$, and $inf(|F_2|)=14$ because if you remove any of these elements, then $F_2$ is no longer closed under set complements, intersections, and unions.

My argument is that there exists no integer $i$ between $1$ and $2$, so there is no event space that has greater than four elements ($|F_1|$) and fewer than fourteen elements ($|F_2|$). I see a flaw, in that there might be event spaces with one event that have more than four elements, but I'm not totally sure that this is right...

 

[Stephen Tashi]

Homework Statement

"Do there exist any event spaces with just six elements?"

We need to know exactly how your text materials define an "event space". It might be better known among mathematicians as a "sigma algebra of sets".

Suuppose $F_2$ is an event space with two non-trivial events $A$ and $B$. Then $F_2=\{∅,Ω,A,A^c,B,B^c,A\cup B, A^c\cap B^c,A\cap B, A^c\cup B^c,A^c\cap B, A\cup B^c, A\cap B^c, A^c\cup B\}$, and $inf(|F_2|)=14$ because if you remove any of these elements, then $F_2$ is no longer closed under set complements, intersections, and unions.

You also must consider special cases such as $A^c = B$. Or , if $A \subset B$ we have $A \cap B = A$. So although your list has 14 distinct symbolic expressions, these may not denote 14 distinct events.

 

[Eclair_de_XII]

I'm starting to think about it, and the more restrictions I put on what events $A,B$ can be, the more elements I get in my event space.

For example, if $B⊄ A,A^c$, then I get an event space with fourteen elements (i.e. the set in the opening post).
But if I have $B\subset A$, $B \neq A$, then I have $F_3=\{∅,Ω,A,A^c,B,B^c,A\cap B^c,A^c\cup B\}$, which has eight elements.

We need to know exactly how your text materials define an "event space".

Basically, my textbook defines an event space $F$ as set of subsets of the sample space $Ω$ such that:

(1) $F\neq ∅$
(2) If $A\in F$, then $A^c \in F$
(3) If $A_i,i\in ℕ$, then $\bigcup_{i=1}^\infty A_i \in F$

Also, finite intersections, unions, and complements are in $F$, and $F$ must include $∅$ and $Ω$.

 

[Stephen Tashi]

Basically, my textbook defines an event space $F$ as set of subsets of the sample space $Ω$ such that:

(1) $F\neq ∅$
(2) If $A\in F$, then $A^c \in F$
(3) If $A_i,i\in ℕ$, then $\bigcup_{i=1}^\infty A_i \in F$

Also, finite intersections, unions, and complements are in $F$, and $F$ must include $∅$ and $Ω$.

The "event space" of your text is the same as a "sigma algebra of sets".

If the distinct sets of sigma algebra are, for example, $A,B,C$ then the universal set $\Omega$ can be partioned into disjoint subsets by looking at all combinations of the intersection of those sets or their complements. For example, $s_1 = A \cap B \cap C,\ ,s_2 = A \cap B \cap C^c,\ s_3 = A \cap B^c \cap C,...$ etc. Some of these combinations may produce the same set, but at least we know (in a finite sigma algebra) that there are a certain number of these disjoint sets $s_1,s_2,...s_n$ that partition $\Omega$. Each set in the sigma algebra can be written as a union of some of the $s_i$ and each distinct union of the $s_i$ produces a distinct set in the sigma algebra. So perhaps you should approach the problem by considering how many distinct sets are in a sigma algebra that creates a partition of $\Omega$ into $n$ sets for n = 1, 2, 3.

 

[Eclair_de_XII]

So perhaps you should approach the problem by considering how many distinct sets are in a sigma algebra that creates a partition of $\Omega$ into $n$ sets for n = 1, 2, 3.

So you want me to divide $\Omega$ into one, two, then three distinct and disjoint subsets?

 

[Stephen Tashi]

So you want me to divide $\Omega$ into one, two, then three distinct and disjoint subsets?

Yes.

For a given $\Omega$, you can't necessarily divide it into, say, 3 distinct disjoint sets. But your question concerns all possible $\Omega$'s. So you can consider the various possibilities.

 

[Eclair_de_XII]

Okay, so I started partitioning different $\Omega$ into disjoint subsets, but I'm worried that the sigma-algebra won't be closed under finite set operations anymore. For example, in the first set with $n=1$, $A\cup A^c = \Omega$, but $\Omega∉F$. Also, I noticed that if none of the $A,B,C$ are subsets of one another, the number of elements of $F$ is equal to $2^n$.

When $n=1$, $F=\{A,A^c\}$.
When $n=2$ and $A⊄B,B⊄A$, $F=\{A\cap B, A \cap B^c, B\cap A^c, A^c \cup B^c\}$
When $n=2$ and $B\subset A, B\neq A$, $F=\{A\cap B, A\cap B^c, A^c\}$
When $n=3$ and $A⊄B,A⊄C,B⊄A,B⊄C,C⊄A,C⊄B$, then $F=\{A\cap B\cap C, A\cap B\cap C^c, A\cap C \cap B^c,C\cap B\cap A^c, A^c\cap B^c\cap C,A\cap B^c\cap C^c,B\cap A^c\cap C^c, A^c\cap B^c \cap C^c\}$

 

[Stephen Tashi]

When $n=2$ and $A⊄B,B⊄A$, $F=\{A\cap B, A \cap B^c, B\cap A^c, A^c \cup B^c\}$

That would be the case $n = 4$ since the partition contains 4 distinct sets.
The distinct sets that partition $\Omega$ are $A \cap B, A \cap B^c, A^c \cap B, A^c \cap B^c$.

The sigma algebra $F$ contains more sets than the sets in the partition and, by definition, it does include $\emptyset$ and $\Omega$.

You don't have to use complements when you express sets in the sigma algebra in terms of sets in the partition. For example if $\Omega$ is partioned by $\{s_1, s_2\}$
$F = \{\Omega, \emptyset, s_1, s_2, s_1 \cup s_2 \}$

It isn't necessary to list $s_1^c$ as a distinct set since $s_1^c = s_2$
Similarly, $s_1 \cup s_2^c = s_1$.

 

[Eclair_de_XII]

It isn't necessary to list $s^c_1$ as a distinct set since $s_1^c=s_2$

By that logic would it be necessary to list $s_1\cup s_2$, since $s_1\cup s_2 = \Omega$? Anyway, can I ask the purpose of partitioning the sample space in regards to showing that the sigma-algebra cannot have exactly six elements?

 

[Stephen Tashi]

By that logic would it be necessary to list $s_1\cup s_2$, since $s_1\cup s_2 = \Omega$?

You are correct, it wouldn't be necessary.

Anyway, can I ask the purpose of partitioning the sample space in regards to showing that the sigma-algebra cannot have exactly six elements?

I think it is simpler to count the number of distinct sets that are possible when we express each set in terms of the sets in the partition. If we have n sets in the partition, each distinct combination of the union of k of those sets is a distinct set. Each distinct set in the sigma algebra can be expressed as such a union for k = 0, 1,...n (k = 0 being the case of the empty set.)

 

[Eclair_de_XII]

So do you want me to show that it's not possible to have six elements in a sigma-algebra by exhausting all possible combinations of restrictions on the events $A,B,C$ in $\Omega$?

 

[Stephen Tashi]

So do you want me to show that it's not possible to have six elements in a sigma-algebra by exhausting all possible combinations of restrictions on the events $A,B,C$ in $\Omega$?

How are you defining $A,B,C$?

I think you should consider the number of sets that partition $\Omega$, not simply the number of "generic" sets $A,B,C$. If you think of a finite event space as a venn diagraom of sets, there are a certain number of mutually exclusive areas that partition the space. All sets in the space are composed of unions of some of these areas.

In a manner of speaking, the smallest sets that can be used to partition $\Omega$ are like the "atoms" or "points" from which all other sets in the sigma algebra are composed.

 

[Eclair_de_XII]

All sets in the space are composed of unions of some of these areas.

So how do I use this fact to show that there cannot exist six distinct sets in the space composed from these unions? Should I draw a circle, partition it into three disjoint pieces, and count how many distinct sets there are (including those formed from the unions of these sets), then do the same with four disjoint pieces?

 

[Stephen Tashi]

So how do I use this fact to show that there cannot exist six distinct sets in the space composed from these unions?

Given n things, how many distinct combinations can be formed by taking 0,1,2...or n of those n things?

 

[Eclair_de_XII]

$\binom n k$ combinations, I should think, where $k\leq n$.

 

[Stephen Tashi]

$\binom n k$ combinations, I should think, where $k\leq n$.

The sum of such combinations for k = 0,1,2,..n.

 

[Eclair_de_XII]

So:

For $n=1$: $\sum_{i=1}^1 \binom 1 i = 1$
For $n=2$: $\sum_{i=1}^2 \binom 2 i = 3$
For $n=3$: $\sum_{i=1}^3 \binom 3 i = \binom 3 1 + \binom 3 2 + \binom 3 3 = 3 + 3+ 1 = 7$

Basically, I take it, the $i$'s symbolize how many disjoint sets are in the union? Okay, I think I have my proof, now.

Thank you for your help and patience, @Stephen Tashi .

 

[Stephen Tashi]

Don't forget i= 0 to get the empty set. Do $\sum_{i=0}$.