Probability help/sigma-algebras

[FTaylor244]

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 ε }
-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. Thanks!

 

[disregardthat]

So you will need to show that:

- A is in F
- The complement of A cap E is in F
- A countable union U_i A cap E_i is in F.

All you need to know are the properties of sets and that epsilon is a sigma algebra. Try it and show what you have tried.

 

[FTaylor244]

Oh, I was under the impression I had to do it the other way around, such that F is in A, etc..maybe that's why I'm having such a hard time here.

Ok so...

A is in F because F=A intersect E, and since an intersect of A is in F, then all of A has to be in F (not sure how to write this properly)
F=AnE is the same as F=(A^c U E^c)^c (^c=complement)
not really sure how to go about the last one

 

[disregardthat]

A is in F because F=A intersect E, and since an intersect of A is in F, then all of A has to be in F (not sure how to write this properly)
F=AnE is the same as F=(A^c U E^c)^c (^c=complement)
not really sure how to go about the last one

F is a set of subsets of A, namely the set of A cap E for E in the sigma algebra. Check the definition of a sigma algebra. Try again keeping that in mind.

 

[blue_raver22]

First, we need to show that F is closed under complementation. This means that for any AπE in F, the complement AπE^c must also be in F.

Since A is an element of ε, we know that A^c is also an element of ε. Therefore, AπE^c is a subset of A, and since AπE is also a subset of A, we have AπE^c = Aπ(E^c ∩ A). Since ε is a sigma-algebra, E^c ∩ A is also an element of ε. Therefore, AπE^c is in F, and F is closed under complementation.

Next, we need to show that F is closed under countable unions. This means that for any sequence of sets {AπE1, AπE2, ...} in F, their union must also be in F.

Let B = ⋃i=1∞ AπEi. Since A is an element of ε, we know that AπEi is also an element of ε for all i. Therefore, B is a union of sets in ε, and since ε is a sigma-algebra, B is also an element of ε. This means that B is an element of F, and F is closed under countable unions.

Finally, we need to show that A is an element of F. Since A is an element of ε, we can write A as AπS, where S is the entire sample space. Since S is an element of ε, we have AπS is in F. Therefore, A is an element of F.

Overall, we have shown that F is closed under complementation, countable unions, and contains A. Therefore, F is a sigma-algebra on A.