Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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.

Feller converse to Lindeberg-Feller

Statement

Assume AC. Let a centered row-wise independent triangular array satisfy ∑kEXn,k2=1 and max⁡kVar⁡(Xn,k)→0. If ∑kXn,k⇒N(0,1), then it satisfies the Lindeberg condition.

Facts & Assumptions

[F1]

The scalar centered exponential increment has modulus at most u^2; its proof also gives 1-cos(u)<=u^2/2. Second-order characteristic-function expansion.

[F2]

Near-one products differ from the exponential of their summed increments by o(1). Products of near-one characteristic factors.

[F3]

The independent row sum has product characteristic function. Characteristic functions under affine maps and independent sums.

[F4]

The assumed weak convergence gives pointwise characteristic-function convergence. Levy continuity theorem forward direction.

[F5]

Under AC N(0,1) has transform exp(-t^2/2). Characteristic function of a normal law.

[F7]

Exponential addition and real extension identify the modulus as exp(real part). exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential.

[F8]

The real nonnegative exponential series gives exp(d)>=1+d for d>=0. The complex exponential by its power series.

[F9]

Centering, real parts and finite sums commute with integration. The Lebesgue integral is linear on L1(μ).

[F10]

Complex expectation is bounded by the expectation of its absolute value. The modulus of an integral is bounded by the integral of the modulus.

[F11]

For total row variance one, Lindeberg is exactly convergence of the summed tail second moments. Total row variance and the Lindeberg condition.

Proof

Given: Assume AC. Let a centered row-wise independent triangular array satisfy ∑kEXn,k2=1 and max⁡kVar⁡(Xn,k)→0. If ∑kXn,k⇒N(0,1), then it satisfies the Lindeberg condition.

1.1F1F2F3F4F5F9F10

Fix real t and put vn,k=EXn,k2 and wn,k=E(eitXn,k−1). Centering, [F1], [F9] and [F10] give ∣wn,k∣≤t2vn,k. Thus ∑k∣wn,k∣≤t2, max⁡k∣wn,k∣→0, and ∑k∣wn,k∣2≤t4max⁡kvn,k→0. By [F2]–[F5] and the assumed normal limit, exp⁡(∑kwn,k)→e−t2/2. This inference uses the forward continuity theorem only, not Lindeberg sufficiency.

2.1step 1.1F1F6F7F8F9

Define the nonnegative function dt(x)=t2x2/2−(1−cos⁡(tx)). Nonnegativity follows from the cosine Taylor bound in [F1]; also dt(x)≤t2x2/2 because cos⁡(tx)≤1. Hence Dn(t):=∑kEdt(Xn,k) is finite and nonnegative. Total variance one and [F9] give Re⁡∑kwn,k=−t2/2+Dn(t). Exponential addition and Euler modulus show ∣exp⁡(∑kwn,k)∣=e−t2/2+Dn(t). Step 1.1 therefore implies eDn(t)→1. Since the nonnegative series gives 0≤Dn(t)≤eDn(t)−1, we obtain Dn(t)→0. No complex logarithm or subsequence of measures is needed.

3.1step 2.1F6F11∎

Now fix any ε>0 and take the single frequency t=4/ε. On ∣x∣>ε we have t2x2>16, and 1−cos⁡(tx)≤2 yields dt(x)≥t2x2/2−2≥t2x2/4. On the complementary set d_t is nonnegative. Integrating and summing gives ∑kE[Xn,k21{∣Xn,k∣>ε}]≤4Dn(t)/t2→0. This is precisely [F11], for every positive epsilon. AC is inherited from the target normal-law construction; the proof uses no Helly selection, uniqueness inversion or backward application of sufficiency. Zero entries and t=0 in the earlier steps are harmless, but the final chosen t is nonzero.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

59 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