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Any advice?

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- Thread starter jetoso
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- #1

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Any advice?

- #2

matt grime

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http://mathworld.wolfram.com/Sigma-Algebra.html

indicates

- #3

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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.

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matt grime

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(demonstrating two sets have equal cardinality says nothing about them being equal, necessarily)

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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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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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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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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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