Sigma Algebra

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Main Question or Discussion Point

Let W be a sample space with a countable number of outcomes. Let A be the set of elementary events (an elementary event is a set containing only one outcome from the sample space). Prove that the sigma-algebra generated by A is the set of all subsets.


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  • #3
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A sigma algebra es defines as a set which contains the empty set, any set and its complement and infinite unios of such subsets from the sample space.
The case here is that I came up with the fact that must prove that a sigma-algebra generated by a set of elementary events must have asmany elements as 2^W, where W is the sample space where we generated the set A of elementary events and the sigma-algebra from A.

I thought it could be proven by induction.
 
  • #4
matt grime
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No, you prove it by showing that any element in the power set can be constructed using those operations on the collection of elementary events. Note a sigma algebra is closed under countable union, not infinite union.

(demonstrating two sets have equal cardinality says nothing about them being equal, necessarily)
 
  • #5
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That is true. So it is close by finite unions and by DeMorgan's Law it is also by finite intersection, right?
Then if |W| = n, and A = {set of elementary events from sample space W}, a sigma-algebra generated by A would have 2^W elements.
 
  • #6
matt grime
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You cannot argue by cardinality, it is a useless tactic, since the same argument you've just used states that the rationals and the naturals are the same subset of R.

Let X be some subset of W.

Write X as a countable union of sets that are definitely in the sigma algebra (you do get that countable sets may be infinite too?)
 
  • #7
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Thank you very much for your advice.


I think I could say this:
Suppose we have a countable set:
W = {a1, a2, ..., an}
And a set of elementary events:
F = { {a1}, {a2}, ..., {an} }

Then the sigma-algebra generated by F, say Sigma-Algebra(F)=2^W, where 2^W is the set of all subsets.

Let A belongs to 2^W, (let A be contained in W), and
A = {b1, b2, ..., bn} then A = U{bn} from n=1, infinity

Then the sigma-algebra(F) has all the possible subsets and their complement in all possible combinations, thus we can say that: Sigma-algebra(F) = 2^W



How about that?
I feel I am still missing something...
 
  • #8
matt grime
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Notationally that is wrong.

Let S be any subset of W. We must simply show that S is in B(A) the sigma algebra generated by A.

Since S =U{s}, where the union is taken over s in S, and such is a countable union, since S is a subset of a countable set, it follows that S is in the sigma algebra generated by A.

End of proof.

Your sets have terminal elements, last ones, for some reason. They need not.
 
  • #9
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Oh, I see. Thank you very much for your explanation.
 

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