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.

Squared loss and a conditional mean

Example

Let $X = {u, v}$ with its discrete sigma-algebra, let $Y = {0, 2, 4, 6}$ with its discrete sigma-algebra, let the action space be $A = R$ with its Borel sigma-algebra, and let $F$ be all functions from $X$ to $R$ . Define $X (x, y) = x$ and $Y (x, y) = y$ on $Ω = X \times Y$ . Give the joint law by $P (u, 0) = \frac{1}{8}, P (u, 2) = \frac{1}{4}, P (u, 4) = \frac{1}{8}, P (v, 2) = \frac{1}{4}, P (v, 6) = \frac{1}{4},$ with all other atoms having probability zero, and set $P_{X} (x) : = \sum_{y} P (x, y)$ . Use squared loss $ℓ (a, y) = (a - y)^{2}$ . For $x \in {u, v}$ define the conditional mean $m (x) : = \sum_{y} y P (x, y) / P_{X} (x)$ . Then $m (u) = 2$ and $m (v) = 4$ , and the predictor $f^{*} (x) = m (x)$ uniquely minimizes population squared risk over $F$ , with risk $3$ .

Facts & Assumptions

Given: The finite spaces, joint law, real-valued predictor class and squared loss stated in the example.

[F1]

Population risk is the expectation of the nonnegative loss under the joint law. Population and empirical risk

[F2]

On a positive-probability event in a finite probability space, conditional expectation is the normalized finite weighted sum. Conditional expectation on positive-probability events in finite spaces

[F3]

Expectation splits over a finite positive-probability partition as the probability-weighted sum of the conditional expectations. Conditional expectation on positive-probability events in finite spaces

Verification

1.1givenF2algebra

The displayed atom weights sum to one and $P_{X} (u) = P_{X} (v) = 1 / 2$ . Applying [F2] to the two events ${X = u}$ and ${X = v}$ gives conditional means $m (u) = 2$ , $m (v) = 4$ , and conditional second moments $E [Y^{2} ∣ X = u] = 6$ , $E [Y^{2} ∣ X = v] = 20$ .

2.1F1F3givenstep 1.1algebra∎

For any predictor $g$ , expansion on each event gives $E [(g (u) - Y)^{2} ∣ X = u] = (g (u) - 2)^{2} + 2$ and $E [(g (v) - Y)^{2} ∣ X = v] = (g (v) - 4)^{2} + 4$ . The partition identity in [F3] and [F1] therefore give $R_{P} (g) = \frac{1}{2} ((g (u) - 2)^{2} + 2) + \frac{1}{2} ((g (v) - 4)^{2} + 4) \geq 3$ , with equality exactly when $g (u) = 2$ and $g (v) = 4$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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

Shai Shalev-Shwartz and Shai Ben-David, Understanding Machine Learning: From Theory to Algorithms, Chapter 3 §3.2.2 (standard reference, not scraped)