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Let $\mathcal{F}$ be a sigma algebra over $\Omega$ and $M$ the set of all probability measures on $\mathcal{F}$. Let $\mathcal{C}$ be some collection of pairs $(A,B)$ with $ \ A,B\in\mathcal{F}$. Now I can declare the pairs in $\mathcal{C}$ independent by defining $M(\mathcal{C})$ to be the set of probability measures which satisfy my independence constraints, so that $P\in M(\mathcal{C})$ iff $P(A\cap B)=P(A)P(B)$ for every $(A,B)\in \mathcal{C}$. Note that it's probably enough for $\mathcal{F}$ to be a field by invoking Caratheodory extension.

Question: For what kind of $\mathcal{C}$ does $M(\mathcal{C})$ reduce to a "unique" (perhaps "unique" modulo some class of sets) measure on $\mathcal{F}$? In other words, when does an overabundance of independence induce a unique measure.

The point is that $\mathcal{C}$ gives a kind of functional equation. It seems that uniqueness will not be guaranteed on sets which do not contain any independent subsets. In particular, null sets could screw things up. However, what if for every $A\in\mathcal{F}$ with $P(A)>0$, I were to guarantee the existence of a $B\in\mathcal{F}$ with $0 < P(B) < 1$ such that $A$ and $B$ are independent?

For example, suppose $\Omega$ has cardinality $n$. Then there are $n$ points on which to define $P$ and one constraint: $P(\Omega)=1$. I would then need $n-1$ equations to determine $P$. So when $\Omega$ is finite, things are much easier.

My motivation for asking this is to better understand the concept of independence.

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Hi, I am not sure to fully understand your notation, does $\mathcal{C}$ depends only one the measure $P$ ?(so we can write it \mathcal{C}(P)) If this is the case, does $M(\mathcal{C})$ is the set of all measure $Q$ such that $\mathcal{C}(Q)=\mathcal{C}(P)$ ? Regards – The Bridge Feb 2 '11 at 9:08
In "Note that it's probably enough for $\mathcal{F}$ to be a sigma field" did you mean to say "ring"? – Mark Meckes Feb 2 '11 at 14:27
@The Bridge: sorry for the confusion, corrected now – Alex R. Feb 2 '11 at 18:00
"Sigma field" and "sigma algebra" mean the same thing. I guess from your response above that in that sentence you may have meant just "algebra". – Mark Meckes Feb 2 '11 at 19:51
Alex R, note that such events $B$ need not exist. (cf. my question from a few days ago:… ) – Tom LaGatta Feb 2 '11 at 20:14

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