(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Are these two sub-sigma-algebras independent?

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