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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.

✓ 4 results · all verified · 4 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 4 also cleared it.

Statistical, resampling, simulation, and numerical error ledger

1 · Prerequisites

2 · Summary

Summary

For a scalar analysis, name the exact target and every successive reference target before assigning error labels. The pathwise error is the sum of the signed differences between adjacent targets, including the final floating-point difference. The mean square of a sum equals the sum of component variances plus a squared-total-bias and covariance correction; equality with the variance sum depends on this combined correction. When total bias is zero, pairwise independence is sufficient for adding variances. The companion correlated-error example shows that centering alone does not remove covariance terms. Independent replicate variance estimates computation-randomization variability for the fixed data and rule. It does not estimate data-sampling uncertainty. A replicate CLT is stated as a separate AC-dependent corollary. The dot-product theorem assumes n u<1 and normal-range operations; cancellation can leave its absolute error bound large relative to the exact result.

The companion examples calculate bootstrap finite-B error, permutation p-value resolution, Euler discretization bias versus path-to-path variance, randomized-QMC replicate variability, and dot-product roundoff.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Successive statistical and computational error targets

Definition

Let (Ω,F,P) be a probability space as in Probability measures and probability spaces, let θ∈R be a fixed target, and let m∈N∪{0}. Take r1,…,rm and r~ to be real random variables on this space, meaning measurable maps (Ω,F)→(R,B(R)) as in Random elements and real random variables. Set r0(ω):=θ for every ω∈Ω. Deterministic values are included by using a one-point probability space. The reference chain r0,r1,…,rm must put every quantity on the same scale and state which ideal analysis stage each transition represents. Define the signed stage errors pointwise by ej:=rj−rj−1(1≤j≤m) and the final numerical error efp:=r~−rm.

Each error is measurable because finite sums and differences of measurable real-valued functions are measurable (Arithmetic and lattice operations preserve measurability whenever they are defined).

When the stages are present, e1 may record data-sampling error, a later transition may record a bootstrap or permutation approximation, a discretized target may record numerical-model bias, and a finite simulation may record computation-randomization error. The labels describe the declared adjacent references; they do not by themselves assert that errors are independent, centered, unbiased, or even random. If a stage is absent, omit it or set its two adjacent references equal. If no floating-point computation is involved, set r~=rm.

For a probability statement about these errors, the required joint probability law and any conditioning convention must also be specified. For example, fixing the observed data and varying independent randomizations concerns computation error conditional on those data; it does not include variation across new data samples. A deterministic quadrature rule has no replicate-based error estimate unless an actual randomization is added. Any numerical roundoff bound must state its arithmetic and normal-range hypotheses.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Pathwise error decomposition along a reference chain

Statement

Let θ=r0,r1,…,rm,r~ be the finite scalar reference chain and returned value of Successive statistical and computational error targets, with m≥0. Then, pointwise, r~−θ=∑j=1m(rj−rj−1)+(r~−rm), and consequently ∣r~−θ∣≤∑j=1m∣rj−rj−1∣+∣r~−rm∣. The sum is empty when m=0. No independence, unbiasedness, or distributional assumption is needed.

Facts & Assumptions

Given: A finite chain θ=r0,r1,…,rm,r~ of real values, with m≥0. Prove the stated identity and bound.

[F1]

The successive signed errors are ej=rj−rj−1 for 1≤j≤m, the last numerical error is efp=r~−rm, and r0=θ. (Successive statistical and computational error targets)

Proof

1.1givenF1algebra

By [F1], the finite sum telescopes: ∑j=1m(rj−rj−1)=rm−r0; for m=0 both sides are 0.

2.1step 1.1F1algebra

Adding r~−rm to the equality in step 1.1 and using r0=θ gives r~−θ=∑j=1m(rj−rj−1)+(r~−rm).

3.1step 2.1algebra∎

Apply the triangle inequality to the finite sum in step 2.1 to obtain the absolute-error bound.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

MSE of a sum of error components includes covariance terms

Statement

Let E1,…,Em be square-integrable real random variables on one probability space, where m≥1. Write bj=EEj. Then E[(∑j=1mEj)2]=(∑j=1mbj)2+∑j=1mVar⁡(Ej)+2∑1≤j<k≤mCov⁡(Ej,Ek). The correction to the sum of component variances is exactly (∑jbj)2+2∑j<kCov⁡(Ej,Ek). Thus the mean square equals the sum of component variances exactly when this combined correction is zero. In particular, if the total bias is zero, pairwise independence suffices.

Facts & Assumptions

Given: A probability space and a finite family of square-integrable real random variables E1,…,Em, with m≥1. Prove the formula and its pairwise-independent special case.

[F1]

For a square-integrable random variable X, Var⁡(X)=E[(X−EX)2]; for square-integrable X,Y, Cov⁡(X,Y)=E[(X−EX)(Y−EY)]. (Moments, variance, and covariance on a probability space)

[F2]

The Lebesgue integral is linear on L1: finite sums of integrable functions are integrable and their integrals add. (The Lebesgue integral is linear on L1(μ))

[F3]

For independent real random variables and integrable coordinate functions, the expectation of their product factors into the product of expectations. (Expectations factor over finite products of independent random variables)

Proof

1.1givenF2algebra

Since ∣Ej∣≤1+Ej2 and EEj2<∞, each Ej is integrable. For S=∑j=1mEj, [F2] gives ES=∑j=1mbj.

2.1step 1.1algebra

Put Cj=Ej−bj. By step 1.1, S−ES=∑j=1mCj.

3.1step 2.1algebra

Expanding the finite square gives (S−ES)2=∑jCj2+2∑j<kCjCk. Each cross product is integrable because 2∣CjCk∣≤Cj2+Ck2.

4.1step 3.1F1F2

Taking expectations in step 3.1, using [F2] and [F1], yields Var⁡(S)=∑jVar⁡(Ej)+2∑j<kCov⁡(Ej,Ek).

5.1step 1.1step 4.1F1algebra

Since E(S−ES)=0, expansion of S2=((S−ES)+ES)2 gives ES2=Var⁡(S)+(ES)2; substitute step 4.1 and step 1.1 to obtain the stated formula. Subtracting ∑jVar⁡(Ej) shows that equality with the variance sum holds exactly when (∑jbj)2+2∑j<kCov⁡(Ej,Ek)=0.

6.1givenF3step 5.1algebra∎

In the pairwise-independent case, apply [F3] to the coordinate functions x↦x−bj and y↦y−bk; their means are zero, so every covariance in step 5.1 is zero. If all components are also centered, the squared-total-bias term is zero as well.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Unbiased variance estimate from independent computation replicates

Statement

Let Z1,…,ZR be IID real random variables with mean μ and finite variance τ2, where R≥2, and let Z‾R=R−1∑r=1RZr. Then V^R:=1R(R−1)∑r=1R(Zr−Z‾R)2 is unbiased for Var⁡(Z‾R)=τ2/R. The probability law may be the computation-randomization law conditional on fixed observed data; in that use, this estimate concerns computation variability only.

Facts & Assumptions

Given: IID real random variables Z1,…,ZR with finite mean μ and variance τ2, and an integer R≥2. Prove the stated unbiasedness.

[F1]

The variance of Zr is E[(Zr−EZr)2]. (Moments, variance, and covariance on a probability space)

[F2]

Expectations of products of independent integrable real random variables factor. (Expectations factor over finite products of independent random variables)

[F3]

IID means identically distributed and independent; no moment assumption is implicit in that definition. (Identical distribution and IID families)

Proof

1.1givenF1F3algebra

By [F3] and the given moment assumptions, every Zr has EZr=μ and EZr2=μ2+τ2.

1.2givenF2F3

For r≠s, independence and [F2] give E[ZrZs]=EZr EZs=μ2.

1.3givenalgebra

Expanding the finite square gives ∑r(Zr−Z‾R)2=∑rZr2−RZ‾R2, since ∑rZr=RZ‾R.

2.1step 1.1step 1.2F1algebra

Expanding Z‾R2 and using steps 1.1–1.2 gives E[Z‾R2]=R−2{R(μ2+τ2)+R(R−1)μ2}=μ2+τ2/R; hence Var⁡(Z‾R)=τ2/R.

3.1step 1.1step 1.3step 2.1algebra

Taking expectations in step 1.3 and using steps 1.1 and 2.1 gives E∑r(Zr−Z‾R)2=R(μ2+τ2)−R(μ2+τ2/R)=(R−1)τ2.

4.1step 3.1step 2.1algebra∎

Divide step 3.1 by R(R−1) to obtain EV^R=τ2/R=Var⁡(Z‾R), as required.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

CLT for independent computation replicates

Statement

Assume AC. Let (Zr)r≥1 be IID real computation-replicate outputs with mean μ and variance 0<τ2<∞. Then, as R→∞, R(Z‾R−μ)τ⇒N(0,1),Z‾R:=1R∑r=1RZr. This is a limit for the replicate-randomization law under the stated fixed-target setup; it is not a data-sampling confidence statement and does not by itself justify a finite-R Student interval.

Facts & Assumptions

Given: Assume AC. Let (Zr)r≥1 be IID real random variables with mean μ and variance 0<τ2<∞. Prove the standardized replicate-mean convergence.

[F1]

IID means the family is independent and identically distributed. (Identical distribution and IID families)

[F2]

For a real random variable with finite second moment, variance is E[(Z−EZ)2]. (Moments, variance, and covariance on a probability space)

[F3]

Under AC, IID real variables with finite positive variance satisfy the Lindeberg–Levy normalized-sum convergence. (Lindeberg-Levy iid central limit theorem)

[A1]

AC is the axiom that every family of nonempty sets has a choice function. (The Axiom of Choice)

Proof

1.1givenF1F2algebra

The normalized replicate average is R(Z‾R−μ)/τ=(∑r=1RZr−Rμ)/(τR), the normalized sum in the Lindeberg–Levy theorem.

2.1givenA1F3step 1.1∎

Under [A1], apply [F3] to the IID sequence in step 1.1 to obtain the asserted convergence in distribution.

5 · Examples, counterexamples and false statements

None yet.

Sources