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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Feller converse to Lindeberg-Feller

Statement

Assume AC. Let a centered row-wise independent triangular array satisfy kEXn,k2=1 and maxkVar(Xn,k)0. If kXn,kN(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)=expzexpw, 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 maxkVar(Xn,k)0. If kXn,kN(0,1), then it satisfies the Lindeberg condition.

1.1

Fix real t and put vn,k=EXn,k2 and wn,k=E(eitXn,k1). Centering, [F1], [F9] and [F10] give wn,kt2vn,k. Thus kwn,kt2, maxkwn,k0, and kwn,k2t4maxkvn,k0. By [F2]–[F5] and the assumed normal limit, exp(kwn,k)et2/2. This inference uses the forward continuity theorem only, not Lindeberg sufficiency.

F1F2F3F4F5F9F10
2.1

Define the nonnegative function dt(x)=t2x2/2(1cos(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 Rekwn,k=t2/2+Dn(t). Exponential addition and Euler modulus show exp(kwn,k)=et2/2+Dn(t). Step 1.1 therefore implies eDn(t)1. Since the nonnegative series gives 0Dn(t)eDn(t)1, we obtain Dn(t)0. No complex logarithm or subsequence of measures is needed.

step 1.1F1F6F7F8F9
3.1

Now fix any ε>0 and take the single frequency t=4/ε. On x>ε we have t2x2>16, and 1cos(tx)2 yields dt(x)t2x2/22t2x2/4. On the complementary set d_t is nonnegative. Integrating and summing gives kE[Xn,k21{Xn,k>ε}]4Dn(t)/t20. 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.

step 2.1F6F11

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