Alphabeta Math
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A smooth integrand satisfies the bound while an integrable singularity does not

Example

Assume Countable Choice. In one dimension compare f1(x)=x with the integrable singular function f2(x)=x−3/4 for x>0 and f2(0)=0. The first satisfies the smooth scrambled-net variance hypotheses; the second does not even have a finite second moment.

Facts & Assumptions

Given: Countable Choice, a nested uniform scramble of a (0,m,1)-net in base 2, and the two integrands.

[F1]

Smooth-integrand variance rate for nestedly scrambled (λ,t,m,d)-nets: A continuously differentiable one-dimensional integrand has scrambled-net variance O(n−3) for fixed base and net quality.

[F2]

Randomized quasi-Monte Carlo point sets and estimators: Each scrambled point is marginally uniform when the RQMC marginal condition holds.

[F3]

Expectation of a nonnegative or integrable random variable: Integrability and second moments are Lebesgue integrals against the uniform law.

[F4]

Nested uniform scrambles preserve net counts and uniform marginals: Each scrambled point has uniform law and the net property is preserved almost surely.

[F5]

Monotone convergence for the integral: Increasing nonnegative measurable truncations have integrals increasing to the integral of their pointwise limit.

[F6]

Change of variables for expectation: For a nonnegative Borel function g and random element X, E[g(X)]=∫g dPX.

[F7]

Moments, variance, and covariance on a probability space: Variance is defined for square-integrable variables and is finite only when their second moment is finite.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is inherited from [F1].

Verification

1.1A1F1given

The function f1(x)=x has continuous derivative 1 on [0,1]. With n=2m, the hypotheses of [F1] hold, so its randomized-rule variance is O(n−3).

1.2F3F5givenalgebra

For N≥1, the nonnegative Borel truncations gN=f21[1/N,1] increase pointwise to f2 and satisfy ∫01gN(x) dx=4(1−N−1/4). Likewise, hN=f221[1/N,1] increase pointwise to f22 and satisfy ∫01hN(x) dx=2(N−1). By [F5], ∫01f2=4 while ∫01f22=+∞; f2 is not continuously differentiable at 0.

2.1F2F4F6F7step 1.2algebra∎

By [F2], I^n(f2)=n−1∑if2(Xi); [F4] gives every point uniform law. Since f2≥0, [F6] and step 1.2 give E[f2(X1)2]=+∞. The squared estimate obeys I^n(f2)2≥n−2f2(X1)2, so its second moment is infinite. By [F7] it has no finite variance, and the smooth variance theorem does not apply to this integrable singularity.

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