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?)