ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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.

Randomize at a discrete likelihood-ratio boundary

Example

On $X = {0, 1}$ with its full sigma-algebra and counting measure $μ$ , let the simple null and alternative laws be $P_{0} ({0}) = P_{0} ({1}) = 1 / 2$ and $P_{1} ({0}) = 1 / 4$ , $P_{1} ({1}) = 3 / 4$ . At level $α = 1 / 4$ , the Neyman–Pearson test randomizes with probability $1 / 2$ at $x = 1$ and rejects with probability $0$ at $x = 0$ . Its size is $1 / 4$ and its power is $3 / 8$ .

Verification

Given: The two probability laws above, counting measure $μ$ , and $α = 1 / 4$ . Assume AC for the general Neyman–Pearson comparison.

[A1] The Axiom of Choice: the Neyman–Pearson theorem is used in its AC density-selection branch; the finite example's densities are explicit.

[F1] Counting measure on an arbitrary set: each singleton has $μ$ -measure $1$ .

[F6] Counting measure is a measure and Finite, sigma-finite, and semifinite measures: the counting set function is a measure; a finite measure is sigma-finite.

[F2] Likelihood-ratio tests with boundary randomization and The likelihood-ratio rule is a measurable test of exact size: the candidate is a measurable randomized test of exact size; on ${p_{0} > 0}$ it rejects above its cutoff and randomizes by $γ$ at equality.

[F3] Test size and power: size is null rejection probability and power is alternative rejection probability.

[F4] Expectation of a nonnegative or integrable random variable, Nonnegative simple measurable functions, The integral of a nonnegative simple function, and The nonnegative integral agrees with the simple integral on simple functions: expectation of this finite-valued rejection function is the sum of its values weighted by the point masses.

[F5] Neyman–Pearson lemma with singular alternative mass: no test of size at most $1 / 4$ has larger power than the constructed likelihood-ratio test.

[F7] Probability measures and probability spaces: a measure with total mass $1$ is a probability measure.

[F8] A measurable function between measurable spaces: every real-valued function on the two-point space with its full sigma-algebra is measurable.

1.1givenF1F6F7algebra

By [F1] and [F6], $μ$ is a measure with $μ (X) = 2$ , so it is finite and hence sigma-finite. Each displayed law has nonnegative masses summing to $1$ , and any disjoint family of subsets of ${0, 1}$ has at most two nonempty members, so each law is countably additive. Thus $P_{0}, P_{1}$ are probability measures by [F7].

2.1F2F4F8step 1.1algebra

Define $p_{0} (0) = p_{0} (1) = 1 / 2$ , $p_{1} (0) = 1 / 4$ , and $p_{1} (1) = 3 / 4$ . They are finite, nonnegative, and measurable by [F8]. For every $A \subseteq X$ , the simple-integral formula in [F4] gives $\int_{A} p_{i} d μ = \sum_{x \in A} p_{i} (x) = P_{i} (A)$ , so these are density representatives and $P_{i} ≪ μ$ . Hence $R (0) = 1 / 2$ and $R (1) = 3 / 2$ . With $c = 3 / 2$ , $P_{0} (R > c) = 0$ and $P_{0} (R = c) = 1 / 2$ , so the boundary equation gives $γ = (1 / 4) / (1 / 2) = 1 / 2$ . The test has $φ^{*} (1) = 1 / 2$ and $φ^{*} (0) = 0$ .

3.1F3F4step 2.1algebra

Using [F4], its size is $E_{P_{0}} φ^{*} = (1 / 2) (1 / 2) + (1 / 2) (0) = 1 / 4$ , and its power is $E_{P_{1}} φ^{*} = (3 / 4) (1 / 2) + (1 / 4) (0) = 3 / 8$ . Thus it uses the boundary atom to attain the requested size.

4.1A1F5step 3.1∎

The Neyman–Pearson lemma [F5] makes this test most powerful among all tests whose size is at most $1 / 4$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

55 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.