Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Finite response likelihood ratios change measure by cancellation

Statement

Let Y be a finite nonempty response set with a fixed enumeration (y1,…,ym), and let p and q be probability mass functions on Y satisfying p(y)>0⇒q(y)>0. Define ρ(y)=p(y)/q(y) when q(y)>0 and ρ(y)=0 when q(y)=0. Then, for every real-valued function f on Y,

∑j=1mq(yj)ρ(yj)f(yj)=∑j=1mp(yj)f(yj).

In particular, ∑j=1mq(yj)ρ(yj)=1. This is a finite response identity; it asserts no change-of-measure result on an arbitrary trajectory space.

Facts & Assumptions

Given: The finite response set Y, probability masses p,q, the support inclusion, and a real-valued function f.

[F1]

Probability measures and probability spaces: a probability law is a measure of total mass one; its nonnegative singleton masses on the finite response set sum to one.

[F2]

Finite truncated response spaces, correctness predicates, and verifiers: every prompt has a finite response set and a normalized policy mass function on its fixed support.

[F3]

Finite sums and finite products, by recursion: a sum of m real terms is the recursively defined finite sum over indices 1,…,m.

Proof

Given: The objects and hypotheses in the Statement.

1.1givenF1F2algebra

For each index j, if q(yj)=0, support inclusion forces p(yj)=0 and the definition gives q(yj)ρ(yj)=p(yj)=0. If q(yj)>0, the quotient is defined and q(yj)ρ(yj)=p(yj). Thus this equality holds for every j=1,…,m.

2.1givenF1F3step 1.1algebra∎

Multiplying the pointwise equality in step 1.1 by f(yj) and summing over j using [F3] gives the stated identity. Taking f(yj)=1 gives ∑j=1mq(yj)ρ(yj)=∑j=1mp(yj)=1. Every operation is a finite sum, so no choice or sigma-finite measure hypothesis is used.

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources