ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

Simultaneous coverage for dependent interval sets

Example

Take the location experiment with $Θ = R$ and sample space $(X, A) = (R, B (R))$ : under parameter $θ$ , $X$ equals $θ + k / 10$ with probability $1 / 10$ for each $k = 1, \dots, 10$ . For both coordinates set $ψ_{i} (θ) = θ$ and report the interval $C_{i} (x) = [x - 9 / 10, x]$ , $i = 1, 2$ . Each interval has marginal coverage $9 / 10$ . The two coverage events are identical, so their simultaneous coverage is also $9 / 10$ . With $α_{1} = α_{2} = 1 / 10$ and $α = 1 / 5$ , Bonferroni guarantees at least $4 / 5$ ; the bound is conservative here because of perfect dependence.

Facts & Assumptions

Given: The location experiment and two confidence-set procedures in the Example, together with the confidence-set definition and simultaneous-coverage theorem below.

[F1]

Statistical experiments: an experiment is a measurable sample space, parameter set, and family of probability measures on that sample space.

[F2]

Confidence sets and coverage: a confidence-set procedure has a measurable membership section for each candidate parameter, and its coverage is the probability that it contains the true parameter.

[F3]

Bonferroni simultaneous coverage: if $m$ marginal confidence sets have coverage at least $1 - α_{i}$ and $\sum_{i} α_{i} \leq α$ , their joint coverage is at least $1 - α$ without an independence assumption.

[A1]

The sample space is $(X, A) = (R, B (R))$ . For each $θ \in R$ , $P_{θ}$ places mass $1 / 10$ at each of the ten distinct points $θ + k / 10$ , $1 \leq k \leq 10$ ; $ψ_{1} (θ) = ψ_{2} (θ) = θ$ and $C_{1} (x) = C_{2} (x) = [x - 9 / 10, x]$ .

Verification

technique · direct finite calculation

Given: The family of ten-point laws, targets, and interval procedures in the Example.

1.1givenA1F1F2algebra

For each candidate $η \in R$ , the membership section of either interval procedure is ${x : η \in [x - 9 / 10, x]} = [η, η + 9 / 10]$ , a Borel set, so [F2] applies. Under $P_{θ}$ , the true value $θ$ is included exactly for $k = 1, \dots, 9$ ; hence each marginal coverage is $9 / 10$ .

2.1givenA1F3step 1.1algebra∎

The two interval procedures are equal for every observation, so their membership events coincide and joint coverage is $9 / 10$ . Since $α_{1} + α_{2} = 1 / 5$ , [F3] gives the general lower bound $1 - 1 / 5 = 4 / 5$ , which this example satisfies. The equality of the two events shows that independence is not assumed.

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