AlexMennen’s Shortform

[AlexMennen]

Theorem: Fuzzy beliefs (as in https://​​www.alignmentforum.org/​​posts/​​Ajcq9xWi2fmgn8RBJ/​​the-credit-assignment-problem#X6fFvAHkxCPmQYB6v ) form a continuous DCPO. (At least I’m pretty sure this is true. I’ve only given proof sketches so far)

The relevant definitions:

A fuzzy belief over a set $X$ is a concave function $\varphi :\Delta X\to [0,1]$ such that $sup\left(\varphi \right)=1$ (where $\Delta X$ is the space of probability distributions on $X$). Fuzzy beliefs are partially ordered by $\varphi \le \psi ⟺\forall \mu \in \Delta X:\varphi \left(\mu \right)\ge \psi \left(\mu \right)$ . The inequalities reverse because we want to think of “more specific”/​”less fuzzy” beliefs as “greater”, and these are the functions with lower values; the most specific/​least fuzzy beliefs are ordinary probability distributions, which are represented as the concave hull of the function assigning 1 to that probability distribution and 0 to all others; these should be the maximal fuzzy beliefs. Note that, because of the order-reversal, the supremum of a set of functions refers to their pointwise infimum.

A DCPO (directed-complete partial order) is a partial order in which every directed subset has a supremum.

In a DCPO, define $x<<y$ to mean that for every directed set $D$ with $supD\ge y$, $\exists d\in D$ such that $d\ge x$. A DCPO is continuous if for every $y$ , $y=sup\{x\mid x<<y\}$.

Lemma: Fuzzy beliefs are a DCPO.

Proof sketch: Given a directed set $D$ , $(supD)\left(\mu \right)=min\left\{d\right(\mu )\mid d\in D\}$ is convex, and $\{\mu \mid (supD\left)\right(\mu )=1\}={\bigcap }_{d\in D}\{\mu \mid d(\mu )=1\}$. Each of the sets in that intersection are non-empty, hence so are finite intersections of them since $D$ is directed, and hence so is the whole intersection since $\Delta X$ is compact.

Lemma: $\varphi <<\psi$ iff $\{\mu \mid \psi (\mu )=1\}$ is contained in the interior of $\{\mu \mid \varphi (\mu )=1\}$ and for every $\mu$ such that $\psi \left(\mu \right)\ne 1$, $\varphi \left(\mu \right)>\psi \left(\mu \right)$.

Proof sketch: If $supD\ge \psi$, then ${\bigcap }_{d\in D}\{\mu \mid d(\mu )=1\}\subseteq \{\mu \mid \psi (\mu )=1\}$ , so by compactness of $\Delta X$ and directedness of $D$, there should be $d\in D$ such that $\{\mu \mid d(\mu )=1\}\subseteq \text{int}\left(\right\{\mu \mid \varphi \left(\mu \right)=1\left\}\right)$. Similarly, for each $\mu$ such that $\psi \left(\mu \right)\ne 1$, there should be $d_{\mu }\in D$ such that $d_{\mu }\left(\mu \right)<\varphi \left(\mu \right)$ . By compactness, there should be some finite subset of $\left\{d\right\}\cup \{d_{\mu }\mid \psi (\mu )\ne 1\}$ such that any upper bound for all of them is at least $\varphi$ .

Lemma: $\psi =sup\{\varphi \mid \{\mu \mid \psi \left(\mu \right)=1\}\subseteq \text{int}\{\mu \mid \varphi \left(\mu \right)=1\},\forall \mu \psi (\mu )\ne 1\to \varphi (\mu )>\psi (\mu \left)\right\}$ .

Proof: clear?