ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A binary Neyman–Pearson boundary randomization

Example

Let the null law be $P = (2 / 3, 1 / 3)$ and the alternative law be $Q = (1 / 3, 2 / 3)$ on $X = {a, b}$ . For the type-I bound $ϵ = 1 / 2$ , the optimal test rejects with probability $1 / 4$ at $a$ and probability $1$ at $b$ .

Facts & Assumptions

Given: The displayed null $P$ , alternative $Q$ , and $ϵ = 1 / 2$ .

[F1]

Randomized tests: $ϕ (x) \in [0, 1]$ is the probability of rejecting the null after observing $x$ ; $1 - ϕ (x)$ is the acceptance probability.

[F2]

Test size and power: for a simple null $P$ the type-I error is $E_{P} ϕ$ , and under the alternative $Q$ the rejection probability is the power $E_{Q} ϕ$ .

[F3]

Finite type-II testing function with explicit orientation: $β_{ϵ} (P, Q)$ is the infimum of $E_{Q} [1 - ϕ]$ over all rejection functions with $E_{P} ϕ \leq ϵ$ .

[F4]

Finite probability vectors: $P, Q$ are probability vectors on ${a, b}$ with the displayed coordinates.

[F5]

Finite probability spaces, outcome weights, events, and event probabilities and Expectation of a real random variable on a finite probability space: the displayed vectors give finite laws, with $E_{P} h = \sum_{x} p_{x} h (x)$ and $E_{Q} h = \sum_{x} q_{x} h (x)$ .

Verification

Given: $P = (2 / 3, 1 / 3)$ is the null, $Q = (1 / 3, 2 / 3)$ is the alternative, and the test function is a rejection probability.

1.1F1F2F3F4F5constructalgebra

Define $ϕ (a) = 1 / 4$ and $ϕ (b) = 1$ . By [F1], this is a valid randomized test. Its type-I error is $(2 / 3) (1 / 4) + (1 / 3) (1) = 1 / 2$ , and its type-II error is $(1 / 3) (3 / 4) + (2 / 3) (0) = 1 / 4$ by [F2, F3, F5].

2.1F1F2F3F4F5step 1.1algebra

Consider any feasible test and put $u = ϕ (a)$ , $v = ϕ (b)$ . If $u \leq 1 / 4$ , then its alternative rejection probability is $E_{Q} ϕ = u / 3 + 2 v / 3 \leq 1 / 12 + 2 / 3 = 3 / 4$ by [F5]. If $u > 1 / 4$ , the constraint $2 u / 3 + v / 3 \leq 1 / 2$ gives $v \leq 3 / 2 - 2 u$ , hence $E_{Q} ϕ \leq u / 3 + (2 / 3) (3 / 2 - 2 u) = 1 - u < 3 / 4$ by [F5]. Thus no feasible test has smaller type-II error than the candidate in step 1.1, proving $β_{1 / 2} (P, Q) = 1 / 4$ .

3.1F1F2step 1.1algebra∎

The alternative-to-null likelihood ratios are $q_{a} / p_{a} = 1 / 2$ and $q_{b} / p_{b} = 2$ . The test of step 1.1 rejects fully at the larger ratio $b$ and randomizes at the smaller ratio $a$ to exhaust the remaining type-I budget. This verifies the threshold direction for $H_{0} : P$ against $H_{1} : Q$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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.

Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 14, §§14.1–14.2 and §14.4 (standard reference, not scraped)