Turn on thread page Beta
    • Community Assistant
    • Thread Starter
    Offline

    20
    ReputationRep:
    Community Assistant


    I'm not sure how to begin with these. I know the definition of a \sigma-algebra of events and the 3 conditions, but I can't seem to see how I can use it here, unless there's some other simple thing my brain keeps missing out on.
    Offline

    17
    ReputationRep:
    The question is, I think, missing a few definitions from the top, however, assuming F and G are both sigma-algebras, this is basically saying:

    (ii) Given that F and G are sigma-algebras, show that F intersect G is a sigma-algebra.
    (iii) Given that F and G are sigma-algebras, show that F union G is not necessary a sigma-algebra.

    Essentially, for (ii), show that H (the intersection of F and G) contains the empty set, is closed under complementation, and closed under countable unions and countable intersections, given that F and G satisfy these properties.

    For (iii), you must build a counterexample: find and state specific F and G which are sigma-algebras (prove this) but that F union G is not a sigma-algebra (contradict one of the three parts of the definition).

    Does this give you something to start from?

    P.S. Sorry for lack of latex.
    Offline

    11
    ReputationRep:
    (Original post by RDKGames)


    I'm not sure how to begin with these. I know the definition of a \sigma-algebra of events and the 3 conditions, but I can't seem to see how I can use it here, unless there's some other simple thing my brain keeps missing out on.
    Assuming that F and G are sigma-algebras, you need to show, for example, that:

    A \in F \cap G \Rightarrow A^c \in F \cap G

    But A \in F \cap G \Rightarrow A \in F, A \in G, \cdots

    Now apply what it means for F,G to be sigma-algebras w.r.t complements.
    • Community Assistant
    • Thread Starter
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by President Snow)
    ...
    (Original post by atsruser)
    ...
    Ah thanks, finally clicked I think. It is indeed given that F,G are \sigma-algebras - forgot to mention, and use this fact properly. Here's what I did:

    Part (ii)
    Spoiler:
    Show





    We know that \emptyset \in F and \emptyset \in G thus \emptyset \in F \cap G hence \emptyset \in H

    If A \in H then A \in F and A \in G, but then we also have A^c \in F and A^c \in G, thus A^c \in F \cap G = H

    If A_1, A_2, ... \in H then A_1, A_2, ... \in F and this implies that \displaystyle \cap_{i=1}^{\infty} A_i \in F. Similarly, we have \displaystyle \cap_{i=1}^{\infty} A_i \in G. Thus \displaystyle \cap_{i=1}^{\infty} A_i \in F\cap G = H

    This hence proves that H forms a \sigma-algebra of events.







    Part (iii)
    Spoiler:
    Show




    Say \Omega = \{ a ,b,c \} and define F= \{ \emptyset , \Omega , \{a \} , \{ b ,c \} \} and G = \{ \emptyset , \Omega , \{b \} , \{ a,c \} \} then we have F \cup G = \{ \emptyset , \Omega , \{a \}, \{b \} , \{b ,c \} , \{a,c \} \} but a \cup b \not\in F \cup G thus it is not a \sigma- algebra





    Does that seem right? I have doubts on part (iii) - can I be more general here?
    Offline

    17
    ReputationRep:
    (Original post by RDKGames)
    Ah thanks, finally clicked I think. It is indeed given that F,G are \sigma-algebras - forgot to mention, and use this fact properly. Here's what I did:

    Part (ii)
    Spoiler:
    Show



















    We know that \emptyset \in F and \emptyset \in G thus \emptyset \in F \cap G hence \emptyset \in H

    If A \in H then A \in F and A \in G, but then we also have A^c \in F and A^c \in G, thus A^c \in F \cap G = H

    If A_1, A_2, ... \in H then A_1, A_2, ... \in F and this implies that \displaystyle \cap_{i=1}^{\infty} A_i \in F. Similarly, we have \displaystyle \cap_{i=1}^{\infty} A_i \in G. Thus \displaystyle \cap_{i=1}^{\infty} A_i \in F\cap G = H

    This hence proves that H forms a \sigma-algebra of events.





















    Part (iii)
    Spoiler:
    Show


















    Say \Omega = \{ a ,b,c \} and define F= \{ \emptyset , \Omega , \{a \} , \{ b ,c \} \} and G = \{ \emptyset , \Omega , \{b \} , \{ a,c \} \} then we have F \cup G = \{ \emptyset , \Omega , \{a \}, \{b \} , \{b ,c \} , \{a,c \} \} but a \cup b \not\in F \cup G thus it is not a \sigma- algebra



















    Does that seem right? I have doubts on part (iii) - can I be more general here?
    This is good - students often struggle coming up with a counterexample for part (iii) but you got there [this question is commonly used in all Year 1 Probability courses].

    I have a few points to note:

    If A_1, A_2, ... \in H then \displaystyle \cap_{i=1}^{\infty} A_i \in H.

    There's no need to go around the houses using F and G here [although you do need to do closure under countable unions similarly].

    Also, do not write a \cup b. This is incorrect. a and b are elements of A and B, not necessarily sets. You cannot union them. What you mean is \{a\} \cup \{b\} = \{a, b\}.

    Next, spell out for the examiner that you know F and G are sigma-algebras, even if you don't prove it. Something like this:

    Say \Omega = \{ a ,b,c \} and define F= \{ \emptyset , \Omega , \{a \} , \{ b ,c \} \} and G = \{ \emptyset , \Omega , \{b \} , \{ a,c \} \}, which we note are both sigma-algebras, then we have F \cup G = \{ \emptyset , \Omega , \{a \}, \{b \} , \{b ,c \} , \{a,c \} \} but \{a\} \cup \{b\} = \{a, b\} \not\in F \cup G thus F \cup G is not a \sigma- algebra.

    You don't need to go any more general than this. It's already fairly general and generality is not required in a counterexample.

    Well done.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 4, 2017
Poll
Favourite type of bread
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.