Welcome to The Nonlinear Library, where we use Text-to-Speech software to convert the best writing from the Rationalist and EA communities into audio. This is: Infra-Bayesian physicalism: proofs part I, published by Vanessa Kosoy on November 30, 2021 on The AI Alignment Forum.
This post is an appendix to "Infra-Bayesian physicalism: a formal theory of naturalized induction".
Lemma 1: prΓ×Φ(χy∈α(s∗(θ)))∈Θ occurs for all s:Γ→Γ iff, for all g:Γ×Φ→[0,1] and s:Γ→Γ, θ(λyαx.χs(y)∈αg(s(y),x))≤Θ(λyx.g(y,x))
This lemma will be implicitly used all over the place, in order to deal with the "presence in the bridge transform" condition algebraically, in terms of function expectations. The bulk of the conditions for a contribution to lie in the bridge transform is this endofunction condition (the support condition is trivial most of the time), so it's advantageous to reformulate it. First-off, we have that
prΓ×Φ(χy∈α(s∗(θ)))∈Θ iff, for all g:Γ×Φ→[0,1], we have
prΓ×Φ(χy∈α(s∗(θ)))(λyx.g(y,x))≤Θ(λyx.g(y,x)) By LF-duality for ultradistributions, proved in Less Basic Inframeasure theory. A contribution lies in an ultracontribution set iff its expectations, w.r.t. all functions, are less than or equal to the ultracontribution expectations.
Now, we just need to unpack the left-hand-side into our desired form. Start off with
prΓ×Φ(χy∈α(s∗(θ)))(λyx.g(y,x)) Apply how projections work
=χy∈α(s∗(θ))(λyαx.g(y,x)) Now, we can move the indicator function into the function we're taking the expectation again, because there's no difference between deleting all measure outside of an event, or taking the expectation of a function that's 0 outside of that event. So, we get
=s∗(θ)(λyαx.χy∈αg(y,x)) Then we use how pushforwards are defined in probability theory. =θ(λyαx.χs(y)∈αg(s(y),x)) And that's our desired form. So, we've shown that for any particular s:Γ→Γ, we have prΓ×Φ(χy∈α(s∗(θ)))∈Θ iff, for all g:Γ×Φ→[0,1], θ(λyαx.χs(y)∈αg(s(y),x))≤Θ(λyx.g(y,x)) And so, we get our desired iff statement by going from the iff for one s to the iff for all s.
Proposition 2.1: For any Γ, Φ and Θ∈□c(Γ×Φ), Br(Θ) exists and satisfies prΓ×ΦBr(Θ)=Θ. In particular, if Θ∈□(Γ×Φ) then Br(Θ)∈□(elΓ×Φ).
Proof sketch: We'll show that for any particular contribution in θ∈Θ, there's a contribution θ∗ which lies within Br(Θ) that projects down to equal θ. And then, show the other direction, that any contribution in Br(Θ) lands within Θ when you project it down. Thus, the projection of Br(Θ) must be Θ exactly.
For the first direction, given some θ∈Θ, let θ∗:=θ⋉(λy.{y}). Note that since θ∈Δc(Γ×Φ), this means that θ∗∈Δc(Γ×2Γ×Φ), so the type signatures line up. Clearly, projecting θ∗ down to Γ×Φ makes θ again. So that leaves showing that θ∗∈Br(Θ). Applying Lemma 1, an equivalent way of stating the bridge transform is that it consists precisely of all the θ′∈Δc(Γ×2Γ×Φ) s.t. for all s:Γ→Γ and g:Γ×Φ→[0,1], θ′(λyαx.χs(y)∈αg(s(y),x))≤Θ(λyx.g(y,x)) and also, supp θ′⊆elΓ×Φ.
Clearly, for our given θ∗, everything works out with the support condition, so that leaves the endofunction condition. Let s,g be arbitrary. θ∗(λyαx.χs(y)∈αg(s(y),x))=(θ⋉(λy.{y}))(λyαx.χs(y)∈αg(s(y),x)) =θ(λyx.δ{y}(λα.χs(y)∈αg(s(y),x)))=θ(λyx.χs(y)∈{y}g(s(y),x)) =θ(λyx.χs(y)=yg(s(y),x))=θ(λyx.χs(y)=yg(y,x))≤θ(λyx.g(y,x)) ≤maxθ∈Θθ(λyx.g(y,x))=Θ(λyx.g(y,x)) In order, the equalities were unpacking the semidirect product, substituting the dirac-delta in, reexpressing the condition for the indicator function, using that s(y)=y inside the indicator function, applying monotonicity of θ, using that θ∈Θ, and then packing up the definition of Θ. And that inequality has been fulfilled, so, yes, θ∗ lies within Br(Θ). Since θ was arbitrary within Θ, this shows that the projection set is as-big-or-bigger than Θ.
Now to show the reverse direction, that anything in the projection set lies within Θ. For any particular θ′∈Θ, remember, it must fulfill, for all g,s, that
θ′(λyαx.χs(y)∈αg(s(y),x))≤Θ(λyx.g(y,x)) So, in particular, we can let s:Γ→Γ be the...