Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27
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.

Overlap control gives a union lower bound

Statement

Let B1,…,Bm be finitely many events in a probability space, put S:=∑i=1mPr⁡[Bi], and suppose that for a real number C≥0 ∑1≤i<j≤mPr⁡[Bi∩Bj] ≤ C S. Then Pr⁡[⋃i=1mBi] ≥ S1+2C. Both quantities vanish when S=0; no hypothesis is imposed on the individual probabilities beyond S<∞, and the bound is uniform over all finite families with the stated overlap ratio.

Facts & Assumptions

Given: events B1,…,Bm on a probability space, the sum S=∑iPr⁡[Bi], and a real C≥0 with ∑i<jPr⁡[Bi∩Bj]≤CS.

[L1]

For vectors u,v in a real or complex inner product space, ∣⟨u,v⟩∣≤∥u∥∥v∥; applied to the indicator functions of two events in L2 of a finite probability space this is the inequality E[∣XY∣]≤E[X2] E[Y2] for random variables, and for a nonnegative integer-valued N it gives (EN)2≤Pr⁡[N>0] E[N2], since N=0 off the event {N>0} (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

technique · direct
1.1

Let N:=∑i=1m1Bi count the events that occur. Then N≥0 is integer valued, with EN=∑iPr⁡[Bi]=S by linearity of expectation, and {N>0}=⋃iBi.

givenalgebra
2.1

Expanding the square, N2=∑i1Bi+2∑i<j1Bi∩Bj, so taking expectations and using the hypothesis gives EN2=S+2∑i<jPr⁡[Bi∩Bj]≤S+2CS=(1+2C)S, a finite bound.

step 1.1algebra
3.1

If S=0 then EN=0 with N≥0, so every Pr⁡[Bi]=0, N=0 almost surely, and both sides of the claimed inequality are zero; assume S>0 from now on. Applying [L1] to N and to the indicator of {N>0} gives S2=(EN)2≤Pr⁡[N>0] EN2≤Pr⁡[N>0] (1+2C)S. Dividing by the positive number (1+2C)S gives Pr⁡[⋃iBi]=Pr⁡[N>0]≥S/(1+2C), which is the claim.

step 1.1step 2.1L1algebra∎

Remarks

  • The hypothesis is a ratio condition, not a smallness condition on the intersections separately: the bound is useful exactly when C is uniformly bounded, and then it loses only the factor 1+2C relative to the first moment S.
  • The Arora-Barak form of the same estimate (Claim 18.34) counts elements of finite sets, makes 2C copies of each element and reduces to inclusion-exclusion, with the weaker constant 14 and a hypothesis on the diameter of the set system; the probabilistic second-moment computation above is the convention of this page, and it applies directly to the events Bj,f of Powering amplifies a small unsatisfaction gap, whose pair overlaps are controlled by Violated-edge positions have controlled collisions.
  • The constant is sharp already for two disjoint events: then C=0, S=Pr⁡[B1]+Pr⁡[B2] and the union has probability exactly S.

Depends on

Used by

Dependency tree · two levels

6 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