Probability help/sigma-algebras

  • Thread starter Thread starter FTaylor244
  • Start date Start date
  • Tags Tags
    Probability
FTaylor244
Messages
5
Reaction score
0
Let (S, ε, P) be a probability space and let A be an element of ε with P(A)>0. Let F={AπE :E is an element of }
-Prove that F is a sigma-algebra on A.



Not sure even where to go with this really. I know that to be a sigma-algebra has to be closed under complementation and countable unions. I'm not very good with proofs, and just a push in the right direction would help me out a ton
 
Physics news on Phys.org
I assume that F is supposed to be

F:=\left\{A\cap E: E\in\epsilon\right\}.​

Now, for F to be a sigma-algebra on A, you have to first show that F\subseteq 2^A. Then, show that the following 3 conditions are satisfied by F.


  1. \emptyset,A\in F

  • If B\in F, then it is necessary that B^c\in F

  • For every sequence of sets (B_n), where everyone of them is a member of F, it is necessary that \bigcup_{n\in\mathbb{N}}B_n\in F.

Try to prove the conditions one-by-one. Also use the fact that \epsilon is a sigma-algebra. If you have some more questions, feel free to ask.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
5
Views
3K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
13
Views
592
Back
Top