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    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.
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    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.
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    (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.
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    (Original post by President Snow)
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    (Original post by atsruser)
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    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?
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    (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.
 
 
 
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