Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Necessity of the truncated mean and variance conditions

Statement

Let independent real random variables (Xn)n1 have nXn convergent almost surely. For every fixed A>0, set Yn=Xn(A)=Xn1{XnA}. Then nP(Xn>A)<,nVar(Yn)<, and the real numerical series nEYn converges.

Facts & Assumptions

[F1]

Independent-copy symmetrization of random series: Given an independent sequence (Xn)n1 on (Ω,F,P), form the product probability space (Ω2,FF,PP). Write Un(ω,ω)=Xn(ω), Vn(ω,ω)=Xn(ω), and Zn=UnVn. Then (Un) and (Vn) are independent copies of the whole sequence, and the Zn are independent symmetric real random variables. Almost-sure convergence of nXn implies almost-sure convergence of nZn. If XnA almost surely for every n, with 0A<, then Zn2A almost surely, EZn=0, and Var(Zn)=2Var(Xn).

[F2]

Bounded centered convergent series have summable variances: Let (Xn)n1 be independent centered real random variables with XnC almost surely for one finite constant C0. If nXn converges almost surely, then nVar(Xn)<. The bound is two-sided and uniform in n.

[F3]

Kolmogorov convergence criterion: For independent centered square-integrable real random variables (Xn)n1, if n1Var(Xn)<, then n1Xn converges almost surely and in L2 to the same finite real random variable.

[F4]

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.

[F5]

First Borel-Cantelli lemma for events: Let (An)nN be events in a probability space. If n=0P(An)<+, then P(An i.o.)=0. No independence hypothesis is needed.

[F6]

Second Borel-Cantelli lemma under pairwise independence: Let (An)nN be pairwise independent events with n=0P(An)=+. Then P(An i.o.)=1.

[F7]

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.

[F8]

A series converges iff each of its tail series converges, and the sum splits as sN plus the N-th tail: Let (ak) be a sequence of reals with partial sums sn=k<nak, let NN, and let tj:=i<jaN+i be the partial sums of the N-th tail series kNak (def-series). Then: 1. tj=sj+NsN for every jN; 2. ak converges if and only if its N-th tail series converges, and in that case k=0ak  =  sN  +  k=Nak; 3. hence the following are equivalent: ak converges; every tail series of ak converges; some tail series of ak converges. In words: convergence of a series is a property of its terms from any index on, and changing finitely many terms changes the sum but not the fact of convergence.

Proof

Given: The objects and hypotheses of the statement.

1.1

Convergence of partial sums implies Xn0 on its probability-one event. Therefore {Xn>A} occurs only finitely often almost surely. These events are independent, by measurable transformations. If their probability sum were infinite, the second Borel–Cantelli lemma would instead make their infinitely-often event have probability one. Hence the sum is finite.

F7F6given
2.1

The Yn are independent and bounded by A. By the first Borel–Cantelli lemma, Yn=Xn eventually almost surely. Finite-change invariance, with the real-series indices shifted by one, gives almost-sure convergence of nYn.

F4F7F5F8step 1.1
3.1

Symmetrize this bounded sequence on the two-factor product. The differences Zn are independent, centered, bounded by 2A, and their series converges almost surely. The bounded-centered lemma gives nVar(Zn)<. Since Var(Zn)=2Var(Yn), the variance sum for Yn is finite.

F1F2step 2.1
4.1

The independent centered variables YnEYn now satisfy the convergence criterion. On the intersection of its probability-one event with that from the truncations, subtract the two convergent partial sums: their difference is the deterministic sequence k=1nEYk. That sequence therefore converges in R. A probability-one event is nonempty, and this argument selects only one path to establish a deterministic conclusion. All truncated expectations are finite, including when Yn=0.

F7F3step 2.1step 3.1algebra

Depends on

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Sources