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Variance bound for a nested scramble of a (0,m,d)-net

Statement

Assume Countable Choice. Let a1,…,an be a (0,m,d)-net in base b≥max⁡(d,2), where n=bm, and let X1,…,Xn be its nested uniform scramble. For real Borel f∈L2([0,1)d), with σ2=Var⁡(f(U)) for U∼Unif⁡([0,1)d), Var⁡(1n∑i=1nf(Xi))≤(bb−1)min⁡(m,d−1)σ2n≤e σ2n.

Facts & Assumptions

Given: Countable Choice, the stated net and scramble, and a real Borel square-integrable f.

[F1]

ANOVA variance identity for nestedly scrambled point sets: The scrambled-rule variance is n−1∑u,κΓu,κσu,κ2(f).

[F2]

Sharp gain bound for scrambled (0,m,d)-nets: For this net, every gain is at most (b/(b−1))min⁡(m,d−1).

[F3]

Orthogonal base-b multiresolution and variance decomposition: The component variances sum to Var⁡(f(U)).

[F4]

Moments, variance, and covariance on a probability space: σ2 is finite and equals the centered second moment.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is inherited through [F1] and [F3].

Proof

technique · direct

Given: The assumptions and notation in the Statement.

1.1A1F1F2

By [F1], the variance is n−1∑u,κΓu,κσu,κ2(f). Each component variance is nonnegative, and [F2] bounds each gain above by (b/(b−1))min⁡(m,d−1). Therefore the sum is at most n−1(b/(b−1))min⁡(m,d−1)∑u,κσu,κ2(f).

2.1A1F3F4step 1.1algebra∎

By [F3] the component sum in step 1.1 is σ2. Also log⁡((b/(b−1))min⁡(m,d−1))=min⁡(m,d−1)log⁡(1+1/(b−1))≤(d−1)/(b−1)≤1, because log⁡(1+x)≤x for x≥0 and b≥d. Exponentiating gives the second inequality with e.

Depends on

Used by

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Sources