ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

One scrambled rule versus independent full-set replicates

Example

Take $b = 2$ , $d = 1$ , and the $(0, 1, 1)$ -net with points $a_{1} = 0$ and $a_{2} = 1 / 2$ . Use a nested uniform scramble and the integrand $f (x) = x$ . Compute the exact variance of one scrambled estimate and explain what independent full-set replicates add.

Facts & Assumptions

Given: The two-point net, its nested uniform scramble, fixed target $I = \int_{0}^{1} x d x$ , and independent repetitions of the complete scrambled set.

[F1]

$(λ, t, m, d)$ -nets in base $b$ : A $(0, 1, 1)$ -net in base $2$ has one point in each half interval.

[F2]

Nested uniform scrambling of a finite point set: The permutation at each input-prefix node is shared by points with that prefix and independent across distinct nodes.

[F3]

Marginal uniformity makes an RQMC rule unbiased: Marginally uniform points give an unbiased estimator of the cube integral.

[F4]

Independent RQMC replicates estimate randomization variance: For $R \geq 2$ independent full-set randomizations, ${\hat{V}}_{R}$ estimates the variance of their replicate mean.

[F5]

Moments, variance, and covariance on a probability space: The variance of a square-integrable real variable is its centered second moment.

[F6]

Expectations factor over finite products of independent random variables: Integrable functions of independent random variables have factored product expectation.

[F7]

The Lebesgue integral is linear on $L^{1} (μ)$ : Expectation is linear on integrable real random variables.

Verification

1.1F1F2given

The first input digits of $0$ and $1 / 2$ differ, so the first permutation sends them to opposite halves. The net count from [F1] is preserved, leaving one point in each half.

2.1F2step 1.1

The two input-prefix paths remain distinct after the first digit; hence all their subsequent node permutations are independent. Each tail is a sequence of independent uniform binary digits, so conditional on the first permutation the two scrambled points are independent uniform variables in their assigned half intervals.

3.1F3F5F6F7step 2.1algebra

A uniform variable on an interval of width $1 / 2$ has variance $(1 / 2)^{2} / 12 = 1 / 48$ . Under either value of the first permutation, step 2.1 gives the same law for $Z$ : half the sum of independent uniforms on the lower and upper halves. Thus this is also its unconditional law. The single-rule estimate $Z = (X_{1} + X_{2}) / 2$ is unbiased for $I = 1 / 2$ by [F3], and [F6]–[F7] give $Var (Z) = (1 / 48 + 1 / 48) / 4 = 1 / 96$ by [F5].

4.1F4step 3.1algebra∎

One realized estimate $Z$ cannot form the centered replicate variance formula because $R (R - 1) = 0$ at $R = 1$ . With $R \geq 2$ independent full-set scrambles, [F4] gives an unbiased estimate of $Var ({\overline{Z}}_{R}) = 1 / (96 R)$ . The target integral is fixed, so this is computation-randomization uncertainty only.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Art B. Owen, Monte Carlo Book: The Quasi-Monte Carlo Parts, Chapter 17 §§17.1 and 17.5 (standard reference, not scraped)