Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Summable untruncated variances are not necessary

Statement refuted

It is false that almost-sure convergence of a series of independent centered square-integrable variables forces summability of their untruncated variances. Assume countable choice and dependent choice. Let X1=0 and for n2 take independent Xn with P(Xn=n)=P(Xn=n)=12n2,P(Xn=0)=11n2. Then nXn converges absolutely almost surely, although EXn=0 and Var(Xn)=1 for every n2.

Facts & Assumptions

[F1]

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.

[F2]

Coordinate random elements of a countable product are independent: Under the measure of thm-countable-product-of-probability-spaces, the coordinate maps Xn(x)=xn have laws μn and are independent.

[F3]

Assuming countable and dependent choice, countable products of arbitrary probability spaces: Assume countable choice and dependent choice. For probability spaces (En,En,μn)nN there is a unique probability measure μ on CN such that, for every finite F, its F-coordinate marginal is nFμn.

[F4]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

[F5]

Almost-sure convergence of a random series: For real random variables (Xn)n1, the series n1Xn converges almost surely if its partial sums Sn converge to a finite real limit on an event of probability one, as in def-almost-sure-convergence-of-random-variables. With S0=0 from def-partial-sums-and-sample-means, its convergence event is C=r1N1jiN{SjSi<1/r}. This is exactly the real Cauchy condition, with the indexing of thm-series-cauchy-criterion shifted by one. Measurable arithmetic makes every event in this countable expression measurable. For any fixed m, the union over N may be restricted to Nm; then each difference uses only Xm+1,Xm+2,. Thus C is in the tail sigma-algebra, without assuming independence. Under independence, cor-almost-sure-convergence-of-an-independent-series-is-a-zero-one-event gives P(C){0,1}. Set S=limnSn on C and S=0 off C. The functions 1CSn converge everywhere to S, so thm-sequential-suprema-infima-limsup-liminf-and-pointwise-limits-are-measurable and thm-arithmetic-and-lattice-operations-preserve-measurability make S measurable. For Borel sets Bn, the event {XnBn infinitely often}=mnm{XnBn} is likewise tail measurable. Changing finitely many summands adds an eventually constant finite difference to Sn; divided by deterministic cn>0 tending to infinity that difference tends to zero, so the normalized limsup is unchanged. The sign of the unnormalized limsup need not be unchanged: the all-zero sequence has limsup zero, while changing its first term to 1 makes the limsup of partial sums equal to 1.

Counterexample

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice take the countable product of these finite probability spaces. The specified masses are nonnegative and sum to one; its independent coordinates have the desired laws. For n2, direct finite expectation gives EXn=0 and EXn2=n2(1/n2)=1, hence variance 1. The first coordinate is zero.

F3F2givenalgebra
2.1

The sum n2P(Xn0)=n2n2 is finite. The first Borel–Cantelli lemma gives only finitely many nonzero terms almost surely. On that event the absolute sum is a finite sum of finite numbers, hence finite, and the original partial sums converge. But their untruncated variance sum is n21=. The example has no uniform bound on all summands.

F4F1F5step 1.1algebra

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