Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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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.
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Kolmogorov maximal inequality

Statement

Let X1,,Xn be independent centered square-integrable real random variables, n1, and Sk=j=1kXj. For every λ>0, P(max1knSkλ)Var(Sn)λ2=j=1nVar(Xj)λ2. Thus controlling the whole finite maximum costs no larger bound than controlling the final sum by Chebyshev.

Facts & Assumptions

[F1]

Disjoint groups of an independent sigma-algebra family remain independent: Let (Fi)iI be an independent family of sigma-algebras on a probability space, and let J0,,Jm1I be pairwise disjoint index sets. For each r<m, define Gr:=σ(iJrFi). Then the sigma-algebras G0,,Gm1 are independent.

[F2]

Expectations factor over finite products of independent random variables: Let n1, let X0,,Xn1 be independent real random variables on a common probability space, and let gi:RR be Borel measurable for each i<n. 1. If every gi is nonnegative, then E[i<ngi(Xi)]=i<nE[gi(Xi)] in [0,+]. 2. If every gi(Xi) is integrable, then i<ngi(Xi) is integrable and the same factorization holds in R.

[F3]

Variance and covariance identities for random variables: Let X,Y be square-integrable real random variables on one probability space. Then Var(X)=E[X2]E[X]2, Cov(X,Y)=E[XY]E[X]E[Y]. Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

[F4]

Arithmetic and lattice operations preserve measurability whenever they are defined: Let (X,A) be a measurable space and let f,g:XR be measurable. Then: 1. cf is measurable for every real scalar c; 2. max(f,g), min(f,g), f, f+, and f are measurable; 3. if f+g is pointwise defined, then f+g is measurable; 4. with the convention of rem-zero-times-infinity-convention-for-pointwise-products, the pointwise product fg is measurable.

Proof

Given: The objects and hypotheses of the statement.

1.1

Let Ak={Sj<λ (j<k), Skλ}. These measurable events are disjoint, and their union is the event in the statement. Measurability follows by finite arithmetic.

F4given
2.1

Grouping shows that Sk1Ak and SnSk are independent. Both are integrable (their squares have finite expectation), and the latter has mean zero. Factorization therefore gives E[Sk1Ak(SnSk)]=0. For k=n the tail is zero and the identity still holds.

F1F2step 1.1
3.1

Expand the square on Ak: E[Sn21Ak]=E[Sk21Ak]+E[(SnSk)21Ak]λ2P(Ak). Sum over the disjoint events. Centering gives ESn2=Var(Sn), and covariance bilinearity with factorization cancels every off-diagonal covariance. This proves the bound even when the variance is zero or n=1.

F3F2step 1.1step 2.1algebra

Depends on

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Sources