Probability Spaces

  • #1

Homework Statement


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



Homework 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


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:

Answers and Replies

  • #2
EnumaElish
Science Advisor
Homework Helper
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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.
 
  • #3
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.

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}
 
  • #4
EnumaElish
Science Advisor
Homework Helper
2,322
124
What is the difference between 3 and the sigma-field property?
 
  • #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
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.
 
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