Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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.

Conditional Hoeffding bound for bounded martingale differences

Statement

Assume AC. Let DL1 satisfy E[DG]=0 almost surely. Let A,B be finite G-measurable random variables such that ADB and BAc almost surely for a deterministic c0. Then for every λR, E[eλDG]eλ2c2/8a.s.

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Basic algebra and order properties of conditional expectation supplies conditional linearity, order, and preservation of constants.

[F2]

Conditioning a known variable and an independent variable says a G-measurable integrable variable conditions to itself.

[F3]

The two-point convexity inequality for the exponential function gives the chord bound for the exponential.

[F4]

The Axiom of Choice states AC, assumed here because F1, F2, and F5 use conditional expectations and chosen representatives.

[F5]

Taking out what is known permits a finite G-measurable factor to be taken outside conditional expectation when the input and product are integrable.

Proof

1.1

The inequalities and the deterministic width give DcADBD+c, so A,BL1. Conditional order and F2 yield A=E[AG]0E[BG]=B. Thus DBAc, making eλD bounded and its conditional expectation well-defined.

F1F2
2.1

On {A=B}, step 1.1 forces D=A=B=0, so the result is equality. On {A<B} write w=BA and θ=(DA)/w[0,1]. F3 gives eλDBDweλA+DAweλB. The ratios B/w and A/w lie in [0,1], while the G-measurable divided difference K=(eλBeλA)/w is bounded by λeλc. Thus the right side is the sum of the two displayed bounded endpoint terms and KD; conditional linearity, the known-variable rule, F5, and E[DG]=0 turn its conditional expectation into BweλA+AweλB. This justifies pulling out the potentially small-width coefficient rather than formally dividing inside a conditional expectation.

F1F2F3F5
3.1

Put u=A/w[0,1] and z=λw. The last expression is hu(z)=(1u)euz+ue(1u)z. For gu=loghu, direct differentiation gives gu(0)=gu(0)=0 and 0gu(z)=pz(1pz)14, where pz=ue(1u)z/hu(z)[0,1]. Integrating the second-derivative bound from 0 to z (with reversed limits when z<0) gives gu(z)z2/8. Hence hu(z)ez2/8eλ2c2/8.

step 2.1
4.1

Combining the two measurable cases proves the conditional inequality. Random endpoints cause no hidden selection; only their deterministic width enters the final bound. AC is used exactly as recorded in F4.

F4step 1.1step 2.1step 3.1

Depends on

Used by

Dependency tree · two levels

20 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