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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Truncation weak law for independent arrays

Statement

For each n1 let Xn,1,,Xn,rn be independent real random variables on one probability space, with finite rn0. Let deterministic bn>0 tend to infinity and set Yn,k=Xn,k1{Xn,kbn}. If k=1rnP(Xn,k>bn)0,bn2k=1rnVar(Yn,k)0, then k=1rnXn,kk=1rnEYn,kbn0in probability. No independence between rows is required.

Facts & Assumptions

[F1]

Zero truncation at a positive level: For a real random variable X and a deterministic level A>0, its zero truncation is X(A)=X1{XA}. The threshold event is measurable because X is measurable and [A,A] is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus X(A) is a real random variable as in def-random-element-and-real-random-variable. It equals X at both cutoff endpoints and is zero outside the interval. Since X(A)A, for every 0<p< its absolute pth moment is at most ApP(Ω)=Ap. This is not clipping to the endpoints.

[F2]

Chebyshev weak law for uncorrelated arrays: For each n1, let Xn,1,,Xn,rn be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where rn0 is finite. Set Sn=k=1rnXn,k and let bn>0 be deterministic. If vn:=bn2k=1rnVar(Xn,k)0, then (SnESn)/bn0 in L2 and in probability. More precisely, its second moment is vn, and its probability of absolute value at least ε>0 is at most vn/ε2. No independence between rows is required.

[F3]

Measurable coordinatewise functions preserve independence: Let (Xi)iI be an independent family of random elements Xi:(Ω,F,P)(Si,Σi). For each i, let gi:(Si,Σi)(Ti,Ti) be measurable. Then the family (giXi)iI is independent.

[F4]

Convergence in probability: For real random variables (Xn) and X on one probability space, write XnX in probability when, for every ε>0, P(XnX>ε)0. This is precisely def-convergence-in-measure for the probability measure.

[F5]

Finite and countable subadditivity of measures: Let μ be a measure and let (Ek)kN be measurable. Then μ(kNEk)k=0μ(Ek). For every mN one also has μ(k<mEk)k<mμ(Ek), including m=0, where both sides are 0.

[F6]

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.

Proof

Given: The objects and hypotheses of the statement.

1.1

Each truncated row is independent by measurable transformations and bounded by bn. Its distinct centered mixed moments vanish: independence factors expectations of bounded products, so the row is uncorrelated. The row weak law therefore makes bn1k(Yn,kEYn,k) tend to zero in probability.

F1F3F2givenF6
2.1

Let En=k=1rn{Xn,k>bn}. Outside En the original and truncated sums agree. Hence for ε>0 the probability of the claimed error exceeding ε is bounded by kP(Xn,k>bn) plus the corresponding centered truncated probability. Both tend to zero. For empty rows all sums and the union are zero or empty, so the argument includes them; equality at the cutoff is retained.

F5F4step 1.1givenalgebra

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