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Probability Spaces

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\Omega = [0,1)[/tex]

    Let G be the collection of all subsets of [tex]\Omega[/tex] of the form
    [a1,b1),[tex]\cup[/tex][a2,b2),[tex]\cup[/tex]...[tex]\cup[/tex][ar,br)
    For r any non-negative integer and 0<=a1
    and a1 <=b1 <= a2 ....

    Show that G is a field

    Show that G is not a [tex]\sigma[/tex]-field



    2. Relevant equations

    Definition:
    Let [tex]\Omega[/tex] be any set. A collection F of subsets of [tex]\Omega[/tex]
    is a field if:

    1. [tex]\emptyset[/tex][tex]\in[/tex]F
    2. given A in F, then Ac=[tex]\Omega[/tex]\A
    3. given A and B in F, A[tex]\cup[/tex]B is in F

    In addition, the collection F is a [tex]\sigma[/tex]-field if

    given A1, A2, A3.... are all in F, so is there union

    [tex]\bigcup[/tex]Ai


    3. The attempt at a solution

    To show G is a field:

    1.
    Assume empty set is not in G, then there must be an element in empty which is not in the G - a contradiction since there are no elements in the empty set.

    2.
    If we have A in G then A=[ai,bi)
    So Ac= [tex]\Omega[/tex]/A

    3. This is where I have the problem,
    Set A in G as [ai,bi)
    And B in G as [aj,bj)
    So we want to show A U B in G
    But if bi= aj
    Then we will have A U B =[ai,bj)
    Which is not in G

    To show G is not a sigma field:
    (0,1)= the union of intervals [1/n,1)
    [1/n,1) is in G for all n
    (0,1) is not in G
    So G is not a sigma field.

    So essentially the problem is with part 3 of the definition.
     
    Last edited: Jun 18, 2009
  2. jcsd
  3. Jun 18, 2009 #2

    EnumaElish

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    Can you say "If we have A in G then A=[ai,bi)"? Isn't each element of G a union of r intervals? What would A look like when r > 1? Same comment applies to both part 2 and part 3 of your answer.
     
  4. Jun 18, 2009 #3
    Thanks yes! OK so we say

    A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br)

    Ac = omega/A

    Then 3 is now easy

    A=[ai,bi)U[a(i+1),b(i+1)U...U[ar,br)
    B=[aj,bj)U[a(j+1),b(j+1)U...U[as,bs)

    so AUB = [ai,bi)U[a(i+1),b(i+1)U...U[ar,br)U[aj,bj)U[a(j+1),b(j+1)U...U[as,bs)

    AUB=[ak,bk)U[a(k+1),b(k+1)U...U[at,bt)
    where k=min{i,j} and t=max{r,s}
     
  5. Jun 18, 2009 #4

    EnumaElish

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    What is the difference between 3 and the sigma-field property?
     
  6. Jun 18, 2009 #5
    For 3 we need the union of any two intervals to be in G
    For the sigma field property we want all possible unions of all possible intervals to be in G


    Point 3 basically requires [0,x) be in G for x<=1, which we do have.
    In order for that sigma to hold we would need the interval (0,1) to be in the collection G.

    Though i'm still not sure i fully understand the difference :S
     
  7. Jun 18, 2009 #6

    EnumaElish

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  8. Jun 19, 2009 #7
    OK well I am working with Probability and random processes By Geoffrey Grimmett, David Stirzaker, the google books preview covers the section i am dealing with (1.1 and 1.2 right at the start). This uses the definitions given in the exercise.

    http://books.google.com/books?id=G3...rontcover&dq=probability+and+random+processes

    In the link you have given i can't distinguish any definition of a field or sigma field, or any treatment of an infinite omega.
     
    Last edited: Jun 19, 2009
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