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Sigma Albegra

  1. Oct 18, 2005 #1
    Let S={1,2,3,4,5,6}, F=σ(A1,A2), ie., the σ-algebra generated by A1 and A1 (the smallest σ-algebra containing A1 and A2) with A1={1,2,3,4} and A2={3,4,5,6}. Please complete the following:

    a. List all sets in F
    b. Is the random variable X(w)= 2, w=1,2,3,4; X(w)=7, w=5,6 measurable w.r.t. F?
    c. Give an example of a r.v. on S that is NOT measurable w.r.t. F.

    What I have done so far:

    a. F={ø,1,2,3,4,5,6}

    b. The definition of a measurable function: If F is a σ-algebra over X and T is a σ-algebra over Y, then a function f:X->Y is F-measurable if the preimage of every T is in F.
    The preimage of T ={1,2,3,4,5,6} which is in F.
    X={1,2,3,4,5,6} which is a σ-algebra over F.
    We can let T={2,7} which could be a σ-algebra over Y.
    Thus, X(w) is F-measurable.

    c. Y(w)= 0, w=1,2,3,4,5,6; Y(w)=1, w>=7
    This is not F-measurable since not all the preimages are in F.


    **I am not sure if I did this problem correctly. In addition, I don't really know how to explain part b and c nicely. This is the first time I have seen a σ-algebra in my life. It would be nice if someone could help me with this. Any help/suggestion is great appreciated. :smile:
     
  2. jcsd
  3. Oct 18, 2005 #2

    HallsofIvy

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    I'm no expert in sigma-algebras myself but I note:

    In (a) you are asked to List all sets in F and your answer is
    F={ø,1,2,3,4,5,6}

    While ø, the empty set, is a set, the others are not.

    In (b) you say "We can let T={2,7} which could be a σ-algebra over Y."
    No, a sigma-algebra is a collection of sets. 2 and 7 are not sets.
     
  4. Oct 18, 2005 #3
    Thank you for the reply, but I still don't know what I need to do.

    Anyone else has any idea? thank you.
     
  5. Oct 18, 2005 #4

    Hurkyl

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    Let's start simple: do you know the definition of a sigma-algebra?
     
  6. Oct 18, 2005 #5
    A sigma-albegra F over the set X is a subsets of X which is closed under contable set operations. The empty set is in F. If E is in F, then the complement of E is also in F. If E1, E2, ..., En is a sequence in F, then their contable union is also in F.

    This is the def. of sigma-algebra in my understanding.
     
  7. Oct 19, 2005 #6

    Hurkyl

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    All right.

    "F=σ(A1,A2), ie., ... the smallest σ-algebra containing A1 and A2"

    So, {1,2,3,4} and {3,4,5,6} are elements of F.

    "A sigma-albegra F over the set X is a subsets of X which is closed under contable set operations."

    So what else must be in F?
     
  8. Oct 19, 2005 #7
    so F = empty set U {1,2,3,4} U {3,4,5,6} instead of F ={empty set, 1,2,3,4,5,6} ?

    ps. I need to have this problem done by tonight since it's due tomorrow morning (US time). :smile:
     
  9. Oct 19, 2005 #8

    Hurkyl

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    No.

    You told me that F is closed under countable set operations. That means that if you perform a countable set operation on elements of F, you produce another element of F, not F itself.
     
  10. Oct 19, 2005 #9
    so, F = empty set, {1,2,3,4}, {3,4,5,6}, {1,2,3,4,5,6}?

    where under countable union, you can get: empty set, {1,2,3,4}, {3,4,5,6}, {1,2,3,4,5,6} which are all elements of F.

    Then in this case, X(w) is not measurable, because the preimage of 7 ={5,6} is not a set in F?
     
  11. Oct 20, 2005 #10

    Hurkyl

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    You meant F = {empty set, {1,2,3,4}, {3,4,5,6}, {1,2,3,4,5,6}}

    And it's still incorrect -- you're missing some things, because this set is not closed under countable set operations. (Try something other than union)
     
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