# When are increasing functions on posets (specifically, lattices) the CDF of a probability measure?

This is perhaps a basic question, but I couldn't find a reference. Let $$P = (X,\leq)$$ be a poset. Given a probability measure $$\mu$$ on $$P$$ (with respect to the Borel $$\sigma$$-algebra generated by sets $$S_p = \{q \in X: q \leq p\}$$ for all $$p \in X$$), one can define its `CDF' $$f_\mu(p) = \mu(S_p)$$, the probability that a corresponding random element of $$X$$ is $$\leq p$$. This function is increasing with respect to the order $$\leq$$, and furthermore satisfies $$\lim_{n \to \infty} f_\mu(p_n) = 1 \text{ if S_{p_n} \nearrow X}$$ and $$\lim_{n \to \infty} f_\mu(p_n) = 0 \text{ if S_{p_n} \searrow \emptyset}.$$ The question: What conditions on either the function or the poset allow one to go backward and, given a function $$f$$ satisfying the above properties, realize it as the CDF (in the above sense) of a probability measure on our poset? I specifically want the case where $$P$$ is a lattice where each element covers a finite number of others, but may be countably infinite and have some infinite intervals.

In principle one can write down $$\mu(\{p\})$$ as an infinite linear combination of $$f_\mu(q)$$ for different $$q$$ by Möbius inversion, but given an $$f$$, it's not obvious to me when these linear combinations are nonnegative—and, for infinite posets, when they converge.

Edit: Thanks very much for the comments so far. Since from these it doesn't look like there's a nice well-known general theorem, let me say more about the specific setup I'm after and maybe this will be helpful. I take $$X = \{(\lambda_1,\ldots,\lambda_d) \in (\mathbb{Z} \cup \{-\infty\})^d: \lambda_1 \geq \ldots \geq \lambda_d\}$$ ('extended integer signatures'), and the ordering is given by the dominance order $$\lambda \geq \mu \Leftrightarrow \sum_{j=1}^i \lambda_j \geq \sum_{j=1}^i \mu_j \text{ for all }i=1,\ldots,d$$ (all sums with $$-\infty$$s in them are taken to be $$-\infty$$). This poset doesn't seem to satisfy the locally chain finite condition mentioned in the comments, but it's quite close to the poset of integer signatures $$\{(\lambda_1,\ldots,\lambda_d) \in \mathbb{Z}^d: \lambda_1 \geq \ldots \geq \lambda_d\}$$ with dominance order, which has all sorts of nice properties (e.g. all intervals are finite).

• What do you mean by "covers"? Commented May 3 at 17:50
• In the real case the function should additionally be cadlag in the order topology. You might try this condition. Commented May 3 at 17:53
• I don't think that there are any interesting non-linear orders for which these conditions will suffice. Consider e.g. the set $X$ being the set of all integral points in $[0,n]^2$ and $f(x,y)=\max(x,y)$ for $(x,y)\in X$. Commented May 3 at 17:53
• I think you probably want some further finiteness conditions on the posets you're considering, Roger. For example, the set of rational numbers $\mathbb{Q}$ with its usual total order is a countable lattice where each element covers finitely many elements (in fact, no elements at all!) but is probably not what you had in mind. A possible additional condition you might want is "locally chain finite," i.e., no interval $[x,y]$ contains an infinite chain (although it may be infinite). Commented May 3 at 17:57
• Thanks everyone, I've added more detail about the setting I'm after (I didn't put it in earlier because I figured there's probably a general theorem and didn't want to cloud the question with details, but it looks like maybe this isn't the case). @JulianHölz I think in my example one can get away with sequences rather than nets, though in more general settings I agree you're right. Commented May 3 at 18:53