LemmaStatement: AI-adaptedProof: 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.

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An IID sample and ghost point are exchangeable

Statement

Let $(Z, Σ)$ be a measurable space, let $P$ be a probability measure on it, and let $m \geq 1$ . If $Z_{1}, \dots, Z_{m}, Z_{*}$ are IID random elements with law $P$ , define $S = (Z_{1}, \dots, Z_{m})$ and, for $i \in {1, \dots, m}$ , let $S^{(i)}$ be $S$ with coordinate $i$ replaced by $Z_{*}$ . Then the random vectors $(S, Z_{*})$ and $(S^{(i)}, Z_{i})$ have the same distribution on the finite product measurable space $(Z^{m + 1}, Σ^{\otimes (m + 1)})$ .

Statement references

Hardt, Recht, and Singer, §2.2, proof of Theorem 2.2: full arXiv paper.

Matni, Lecture 16, §3.1, proof of Theorem 2: full lecture notes.

Facts & Assumptions

Given: A common measurable target, a probability law $P$ , and the IID random elements in the Statement.

[F1]

Identical distribution and IID families: IID means mutual independence together with equality of all coordinate laws.

[F2]

Independent sigma-algebras and independent events: for any finite selection of coordinate events, their intersection has probability equal to the product of their probabilities.

[F3]

The product sigma-algebra and its finite iterates: finite product sigma-algebras are generated recursively by coordinate measurable rectangles.

[F4]

For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: a finite product measure is uniquely determined by its values on coordinate rectangles; this follows by iterating the stated two-factor uniqueness theorem.

Proof

1.1F1F2given

Fix $i \in {1, \dots, m}$ and a measurable rectangle $B_{1} \times \dots \times B_{m + 1}$ . By [F1] and [F2], its probability under $(S, Z_{*})$ is $\prod_{j = 1}^{m + 1} P (B_{j})$ . The vector $(S^{(i)}, Z_{i})$ swaps coordinates $i$ and $m + 1$ , so the same rectangle has probability $\prod_{j = 1}^{m + 1} P (B_{j})$ under that vector as well.

2.1F3F4step 1.1given∎

Each vector is measurable: the preimage of a coordinate rectangle is a finite intersection of measurable coordinate events, and [F3] says these rectangles generate $Σ^{\otimes (m + 1)}$ . Thus both vector laws are probability measures on that product sigma-algebra. They agree on its generating coordinate rectangles by step 1.1; iterating [F4] along the recursive finite product in [F3] identifies both with the same product measure. Hence their distributions agree.

Depends on

Used by

Uniform stability implies expected generalization Theorem

Dependency tree · two levels

21 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

Moritz Hardt, Benjamin Recht, and Yoram Singer, Train faster, generalize better: Stability of stochastic gradient descent, §2.2, proof of Theorem 2.2 (standard reference, not scraped)
Nikolai Matni, ESE680 Lecture 16: Algorithmic Stability and Stochastic Gradient Descent, §3.1 (standard reference, not scraped)