So a random graph on n vertices is, (semi-)formally, a probability space on the set of all graphs on n vertices. Normally, we consider pretty "vanilla" probability spaces; for example, the one defined by setting the probability of there being an edge between u and v (which we'll denote by P(E(u,v)) for the remainder of this blog post) to be some constant 0 < c < 1. (Technically, yes, we can set c=0 and c=1, but those are really boring probability spaces.) However, as is well known, I'm not normal. So I'd like to investigate whether we can even construct probability spaces, in general, that have weird properties. Here's what I want to know:
Given a sentence Q of first-order logic with two free variables; say, u and v (where the variables will represent vertices of our graph, and we'll restrict ourselves to the predicate E(w,x) which, in short, means "there is an edge between w and x"), and a constant k>1, does there exist a (nontrivial*) probability space on the set of graphs with n vertices which satisfies:
P(E(u,v)|Q(u,v)) \geq k*P(E(u,v))
for all vertices u,v?
*"Nontrivial" in this context basically means "the probability of the empty graph isn't 1," or maybe "for any k, there's some n for which the probability of choosing a graph with at least k edges is positive." I'm not really sure.
By studying these distributions, we can look at "random" graphs which nonetheless are expected to have some preassigned structure (i.e., by letting Q be "there exists w: E(u,w) and E(w,v)", we're looking at graphs that are in some sense "likelier" to contain triangles than a random graph. Alternatively, by setting k very large and constructing rather intricate FOL sentences, we might be able to get some insight into Ramsey-theoretic questions. This is definitely true if we move to SOL; I'm not not quite as sure with this weaker model.) You know, just in case you're the kind of weird person who wants some sort of "motivation" for studying a problem.
Apologies for the awkward wording of my problem; I'm just not sure how else to express it.