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Sunday, December 15th 2019
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Wednesday, December 11th 2019
Wed, Dec 11th 2019

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4Vanessa Kosoy4dIn the past I considered the learning-theoretic approach to AI theory [https://www.alignmentforum.org/posts/5bd75cc58225bf0670375575/the-learning-theoretic-ai-alignment-research-agenda] as somewhat opposed to the formal logic approach popular in MIRI [https://www.alignmentforum.org/posts/3qXE6fK47JhSfkpnB/do-sufficiently-advanced-agents-use-logic] (see also discussion [https://www.alignmentforum.org/posts/3qXE6fK47JhSfkpnB/do-sufficiently-advanced-agents-use-logic#fEKc88NbDWZavkW9o] ): * Learning theory starts from formulating natural desiderata for agents, whereas "logic-AI" usually starts from postulating a logic-based model of the agent ad hoc. * Learning theory naturally allows analyzing computational complexity whereas logic-AI often uses models that are either clearly intractable or even clearly incomputable from the onset. * Learning theory focuses on objects that are observable or finite/constructive, whereas logic-AI often considers objects that unobservable, infinite and unconstructive (which I consider to be a philosophical error). * Learning theory emphasizes induction whereas logic-AI emphasizes deduction. However, recently I noticed that quasi-Bayesian reinforcement learning [https://www.alignmentforum.org/posts/Ajcq9xWi2fmgn8RBJ/the-credit-assignment-problem#X6fFvAHkxCPmQYB6v] and Turing reinforcement learning [https://www.alignmentforum.org/posts/3qXE6fK47JhSfkpnB/do-sufficiently-advanced-agents-use-logic#fEKc88NbDWZavkW9o] have very suggestive parallels to logic-AI. TRL agents have beliefs about computations they can run on the envelope: these are essentially beliefs about mathematical facts (but, we only consider computable facts and computational complexity plays some role there). QBRL agents reason in terms of hypotheses that have logical relationships between them: the order on functions corresponds to implication, taking the minimum of two functions corresponds to logical "and", taking the concave hull of two func

Tuesday, December 10th 2019
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Monday, December 9th 2019
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Sunday, December 8th 2019
Sun, Dec 8th 2019

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3Alex Mennen7dTheorem: Fuzzy beliefs (as in https://www.alignmentforum.org/posts/Ajcq9xWi2fmgn8RBJ/the-credit-assignment-problem#X6fFvAHkxCPmQYB6v [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 ϕ:ΔX→[0,1] such that sup(ϕ)=1 (where ΔX is the space of probability distributions on X). Fuzzy beliefs are partially ordered by ϕ≤ψ⟺∀μ∈ΔX:ϕ(μ)≥ψ(μ) . 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≥y, ∃d∈D such that d≥x. A DCPO is continuous if for every y , y=sup{x∣x<<y}. Lemma: Fuzzy beliefs are a DCPO. Proof sketch: Given a directed set D , (supD)(μ)=min{d(μ)∣d∈D} is convex, and {μ∣(supD)(μ)=1}=⋂d∈D{μ∣d(μ)=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 ΔX is compact. Lemma: ϕ<<ψ iff {μ∣ψ(μ)=1} is contained in the interior of {μ∣ϕ(μ)=1} and for every μ such that ψ(μ)≠1, ϕ(μ)>ψ(μ). Proof sketch: If supD≥ψ, then ⋂d∈D{μ∣d(μ)=1}⊆{μ∣ψ(μ)=1} , so by compactness of ΔX and directedness of D, there should be d∈D such that {μ∣d(μ)=1}⊆int({μ∣ϕ(μ)=1}). Similarly, for each μ such that ψ(μ)≠1, there should be dμ∈D s

Saturday, December 7th 2019
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