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Are these two sub-sigma-algebras independent?

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]\Omega=\{HH,HT,TH,TT\} [/tex] and let [tex]F [/tex] and [tex]U [/tex] be two partitions of [tex] \Omega: F=\{G,K\} [/tex] with [tex]G=\{HH,TT\} [/tex] and [tex]K= \Omega\backslash G[/tex], while [tex]U=\{V,W\} [/tex] where [tex]V=\{HH,TH\} [/tex] and [tex]W= \Omega\backslash V[/tex].

    If [tex]2^\Omega[/tex] is the [tex]\sigma[/tex]-algebra of [tex]\Omega[/tex], and [tex]A=2^F [/tex] and [tex]B=2^U [/tex] are the sub-[tex]\sigma[/tex]-algebras of [tex]F [/tex] and [tex]U [/tex] (respectively), are [tex]A [/tex] and [tex]B [/tex] independent?

    2. The attempt at a solution

    [tex]A [/tex] and [tex]B [/tex] are independent if, for any [tex]a \in A [/tex] and [tex]b \in B[/tex], [tex] \Pr(a \cap b) = \Pr(a) \Pr(b) [/tex].

    So let's say [tex]a=G[/tex] and [tex]b=V[/tex]. Then [tex] \Pr(G \cap V) = \Pr(HH) [/tex]. Now if the two [tex]\sigma[/tex]-algebras are independent, [tex] \Pr(G)\Pr(V) = \Pr(HH) [/tex].

    Since [tex] \Pr(G) = \Pr(HH) + \Pr(TT) [/tex] and [tex]\Pr(V) = \Pr(HH) + \Pr(TH)[/tex] we need to show [tex](\Pr(HH) + \Pr(TT))(\Pr(HH) + \Pr(TH))=\Pr(HH)[/tex].

    Now, permit me for not writing out all my algebra, but I'm just not seeing it...
    I want to say A and B must be independent because, e.g., if you know G happened (either 2 H's or 2 T's), you still don't know anything about the probability of V or W (an H second, or a T second). Is this right, or am I thinking about it all wrong? I am still trying to get a grasp on this whole sigma-algebra concept, so please go easy on me.
    Last edited: Aug 11, 2010
  2. jcsd
  3. Aug 11, 2010 #2
    Ok, so me thinks they are not in general independent. If we know G happened, we do indeed have information to update our probability of V (or W).

    E.g., if the probabilities over [tex]\Omega=\{HH,HT,TH,TT\} [/tex] are given by the 4-vector [tex](q,r,s,1-q-r-s)[/tex], then
    [tex] \Pr(G \cap V) = \Pr(HH) = q [/tex],
    [tex] \Pr(G) = \Pr(HH) + \Pr(TT) = 1-r-s[/tex] and
    [tex]\Pr(V) = \Pr(HH) + \Pr(TH) = q+s[/tex],
    so [tex] \Pr(G)\Pr(V) = \Pr(G \cap V) [/tex] only in the special case [tex] (1-r-s)(q+s)=q[/tex] or [tex]q=r(1-r-s)/(r+s) [/tex].

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