Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-09-23 (gpt-6-sol)
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A nonidentical Bernoulli weak law

Example

Assume countable choice and dependent choice. Let independent Xk be Bernoulli(pk) with pk=1/4 for odd k and pk=3/4 for even k. Then for Sn=∑k=1nXk, E∣Sn−ESnn∣2=316n,Sn/n⟶1/2in probability. Thus a common distribution is not required.

Facts & Assumptions

[F1]

Chebyshev weak law for uncorrelated arrays: For each n≥1, let Xn,1,…,Xn,rn be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where rn≥0 is finite. Set Sn=∑k=1rnXn,k and let bn>0 be deterministic. If vn:=bn−2∑k=1rnVar⁡(Xn,k)⟶0, then (Sn−ESn)/bn→0 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.

[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)n∈N there is a unique probability measure μ on CN such that, for every finite F, its F-coordinate marginal is ⨂n∈Fμn.

[F4]

A Bernoulli(p) variable has mean p and variance p(1−p); a binomial(n,p) variable has mean np and variance np(1−p): If X is Bernoulli(p), then E[X]=p and Var⁡(X)=p(1−p). If S is binomial(n,p), then E[S]=np,Var⁡(S)=np(1−p). These formulas include p=0, p=1, and n=0.

[F5]

For n≥1, if X0,…,Xn−1 are independent real random variables and gi:R→R are Borel measurable with each gi(Xi) integrable, then ∏i<ngi(Xi) is integrable and E[∏i<ngi(Xi)]=∏i<nE[gi(Xi)] in R (Expectations factor over finite products of independent random variables).

Verification

Given: The construction and assumptions above.

1.1F3F2F4givenF5

Under countable choice and dependent choice, take the countable product of the two-point Bernoulli probability spaces with the prescribed pk (shift the product index by one). Its coordinates are independent with the required laws. Each has mean pk and variance 3/16. Independence gives zero mixed centered moments, hence zero off-diagonal covariances.

2.1F1step 1.1algebra∎

The row weak law with the first n entries and normalizer n gives centered second moment n(3/16)/n2=3/(16n) and convergence in probability to zero. Meanwhile ESn/n=1/2 for even n and 1/2−1/(4n) for odd n, including n=1. For any ε>0 the deterministic error is eventually below ε/2, so the probability of ∣Sn/n−1/2∣>ε is at most the probability that the centered average exceeds ε/2 in absolute value, which tends to zero.

Depends on

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Sources