Independent Events Inducing Probability Measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:49:46Z http://mathoverflow.net/feeds/question/54048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54048/independent-events-inducing-probability-measures Independent Events Inducing Probability Measures Alex R. 2011-02-02T01:37:45Z 2011-02-02T21:09:11Z <p>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 $\mathcal{C}$ <em>independent</em> 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.</p> <blockquote> <p>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.</p> </blockquote> <p>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 $0 &lt; P(B) &lt; 1$ such that $A$ and $B$ are independent?</p> <p>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.</p> <p>My motivation for asking this is to better understand the concept of independence. </p>