I'm hoping for some help in nailing down a vague idea about independence. It starts with finding the expectation of a product of three RVs (or more, but I'll stick to three for now). These are not necessarily independent, but we know all the pairwise covariances. This is enough information to predict the expected value of the product of any pair, but it is not enough to predict the expected value of the product of all three, as can be shown by an example. Let A, B, and C be three Bernoulli RVs: each takes on the values 0 and 1 with equal probability. Further, assume that they are independent. Obviously $\mathbb{E}\left[ABC\right]=1/8$. Define a new variable $Z=A(1-B)+B(1-A)$, that is, $Z=1$ iff $A\ne B$. Z is independent of A, and it is independent of B. In fact, all the pairwise joint distributions of ${A,B,C}$ are identical to the pairwise joint distributions of ${A,B,Z}$. The joint distributions of the three taken together are different, of course. And, in particular, $A=B=Z=1$ never happens, and $\mathbb{E}\left[ABZ\right]=0\ne \mathbb{E}\left[ABC\right]$. This shows that knowledge of all the pairwise joint distributions of three or more RVs is not sufficient to predict the expected value of their product. 


Is there a way, for Bernoulli RVs or more generally, to define the distribution that has specified single-variable (p's for each variable) and pairwise (correlations for each pair) statistics, but has no higher-level correlations? I know that I'm being vague--what do I mean by "higher-level correlations"? But that's the problem: I can't think how to define it. The idea is something like conditional independence, but I haven't been able to make it precise. 



higher-level independence of three or more correlated RVs
Higher order correlations are called joint cumulants or semi-invariants. They can be expressed in terms of joint moments. If you want one of the cumulants to be equal to zero, that gives you an algebraic equation on moments. A joint moment of Bernoulli r.v.'s equals the probability that the r.v.'s involved are all equal to $1$. Combining all that one obtains a system of algebraic equations on the set of probability distributions on the binary cube. That needs to be solved. So, the more cumulants you want to be zero, the more equations you have, so the system may easily be overdefined.


