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Create a Sigma-Algebra

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to create the smallest sigma-algebra where [tex]\Omega[/tex] is the union of disjoint sets Ai, i=1,2,...,n.

    Now, I did this for the case where [tex]\Omega[/tex] is just the union of disjoint sets A_1 and A_2, and found that the sigma-algebra would be {[tex]\Omega[/tex], {}, A_1, A_2} since A_1 is the complement of A_2 and vice versa.

    So for the A_i's, would the sigma algebra just be {[tex]\Omega[/tex], {}, A_1, A_2,...,A_n, [tex]\Omega[/tex]-A_1, [tex]\Omega[/tex]-A_2,...,[tex]\Omega[/tex]-A_n}?

    Thank you! :)
     
  2. jcsd
  3. Aug 25, 2009 #2
    Is [itex]A_1 \cup A_2[/itex] in your candidate? Should it be?
     
  4. Aug 25, 2009 #3
    I don't understand your question. But I think it might have to do with the fact I forgot to mention a condition!

    We are supposed to find the smallest sigma-algebra that includes the whole set, the empty set, and then ONLY the specific disjoint sets (and obviously their complements).

    So, the unions wouldn't be included.

    Sorry about that.

    So, do my answers work? I think they do, but I didn't know if I was missing something or not writing it correctly.
     
  5. Aug 25, 2009 #4

    Dick

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    A sigma algebra includes unions of the basis sets. That's what a sigma algebra is. Review the definition.
     
    Last edited: Aug 26, 2009
  6. Aug 26, 2009 #5
    Yes, I know that, but for the case where you only have A_1 and A_2, that union is Omega, since they are disjoint sets. I believe I mentioned that all the A's are disjoint, no?

    And then even for the case with all the A_i's, any of those unions is still already in the set. For instance:

    A_1 U [tex]\Omega[/tex] - A_1 is just [tex]\Omega[/tex] which is already part of the sigma-algebra.

    Or

    A_2 U [tex]\Omega[/tex] - A_1 is just again [tex]\Omega[/tex]-A_1 so I think it still works.

    Am I missing something yet again??
     
  7. Aug 26, 2009 #6

    Dick

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    I think the problem is to describe ALL of the sets in the sigma algebra explicitly. You've been given {A1...An} which is a set of generators for the sigma algebra. You've enlarged the set of generators to {U,{},A1...An,A1^C...An^C}, but that's still just a set of generators, not the complete sigma algebra. The complete sigma algebra has 2^n sets in it. So far you've only listed 2n+2 of them.
     
  8. Aug 26, 2009 #7
    Yeah, I understand what you are saying. I was completely forgetting about the unions of the A's themselves. But is there an easier way to write it all out? Otherwise I have to list all the unions individually, like A_1 U A_2, A_1 U A_3, and even ones like A_1 U A_2 U A_3, but with n A's, there's got to be a better way to show that all of these are also in there.
     
  9. Aug 26, 2009 #8

    Dick

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    Well, let K={k1...km} be any subset of N={1...n}. Can you think of a way to describe a corresponding element of the sigma algebra? Suppose K={} or K=N?
     
  10. Aug 26, 2009 #9
    I'm really not sure...I'm assuming it will be some kind of union where the k's are a type of index, but that's a complete guess, and I really have no clue :\
     
  11. Aug 26, 2009 #10

    Dick

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    You mean we can take the subset K={k1...km} to correspond to the set A_k1 U A_k2 U...U A_km? That would be a pretty good guess. Now if I take the sets corresponding to all subsets of N={1...n}, would I get all of the sets in the sigma algebra??
     
  12. Aug 26, 2009 #11
    Okay, let me see if I have this straight. K = {k1,k2,...,kn} is a subset of N = {1, 2,...,n} then all the various unions of the A's A_k1 U A_k2 U...U A_kn would be taken care of? I think I might have set that up wrong? I'm so sorry I'm not getting it. Thank you for all your help though!
     
  13. Aug 26, 2009 #12
    Wait a second, don't all the complements of the individual A's (A_1 or A_2 or...) cover all the possible unions??
     
  14. Aug 26, 2009 #13

    Dick

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    If K={1,2} then A=A1 U A2. If K={1,3,4} then A=A1 U A3 U A4. If K={} then A={}. If you let K range over ALL possible subsets of N={1...n} then you get ALL possible elements of the sigma algebra. Agree?
     
  15. Aug 26, 2009 #14
    Yes, sir. I think I get it now. And those K's even take care of the entire set (Omega), the empty set (as you showed), and it would take care of the individual A's and their complements. Okay, I think I'm understanding. Again, thank you very much for being patient with me and helping out. :)
     
  16. Aug 26, 2009 #15
    Another sigma-algebra question: I have to show that the union of sigma-algebras is not necessarily a sigma-algebra. I'm pretty sure it fails for the third axiom, that it is not closed under countable union, but I'm not quite sure how to show this.
     
  17. Aug 26, 2009 #16

    Dick

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    Produce a counter example. You've been looking at these finite sigma algebras for a while now, right? Take SA={{},A} where A is one set. Take SB={{},B} where B is a disjoint set from A. Is SA a sigma algebra? Is SB a sigma algebra? Is SA U SB a sigma algebra? Put some teeth behind your feeling the union axiom fails.
     
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