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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Weighted chi-square triangular-array limit

Statement

Assume the Axiom of Choice. For each n, let Gn,1,…,Gn,mn be independent standard normal random variables, and let wn,j≥0 satisfy 2∑j=1mnwn,j2=1. If max⁡1≤j≤mnwn,j→0, then Tn=∑j=1mnwn,j(Gn,j2−1) has mean zero and variance one for every n and converges in distribution to N(0,1).

Facts & Assumptions

Given: AC; for each n, independent standard normal coordinates Gn,j; nonnegative weights with 2∑jwn,j2=1 and max⁡jwn,j→0.

[A1]

The Axiom of Choice: AC is the hypothesis of the Lindeberg–Feller supplier.

[F1]

A normal variable has finite fourth moment: for G∼N(0,1), EG2=1 and EG4=3; hence V=G2−1 has mean 0, variance 2, and finite second moment.

[F2]

Lindeberg-Feller central limit theorem: sufficiency: an independent centered triangular array with positive row standard deviations and the Lindeberg condition has its normalized row sums converging in distribution to N(0,1).

[F3]

Dominated convergence: an integrable dominating function allows limits to pass through the expectation.

Proof

1.1givenF1algebra

By [F1], each Vn,j=Gn,j2−1 is centered with variance 2. Independence within each row gives ETn=0 and Var⁡(Tn)=2∑jwn,j2=1. Thus the row standard deviation is sn=1>0.

2.1givenF1F3step 1.1algebra

Fix δ>0 and put wn∗=max⁡jwn,j. If wn,j=0, its Lindeberg summand is zero. Otherwise wn,j∣Vn,j∣>δ implies ∣Vn,j∣>δ/wn∗. Therefore ∑jE ⁣[wn,j2Vn,j21{wn,j∣Vn,j∣>δ}]≤12E ⁣[V21{∣V∣>δ/wn∗}]. Because wn∗→0 and V2 is integrable by [F1], the right side tends to zero by [F3]. This is the Lindeberg condition for the array wn,jVn,j.

3.1A1F2step 1.1step 2.1∎

The array in step 2.1 is centered and independent within rows, has finite second moments, and has row standard deviation 1. By [F2] and the Lindeberg condition just proved, Tn⇒N(0,1).

Depends on

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