Are these two sub-sigma-algebras independent?

[jimholt]

Homework Statement

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

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


2. The attempt at a solution

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

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

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

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.

 

[jimholt]

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

(Right?)