ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

$L^{2}$ best prediction by conditional expectation

Example

Assume AC. For real $X \in L^{2} (P)$ , $U = E [X ∣ G]$ and every $Z \in L^{2} (G)$ , $E [(X - Z)^{2}] = E [(X - U)^{2}] + E [(U - Z)^{2}]$ . Equality with the minimum error holds if and only if $Z = U$ almost surely.

Facts & Assumptions

Given: AC, real $X \in L^{2} (P)$ , a sub-sigma-algebra G, and any predictor $Z \in L^{2} (G)$ .

[F1]

The conditional mean is the unique $L^{2}$ minimizer and is orthogonal to every G-measurable $L^{2}$ residual. (Conditional expectation is the l2 orthogonal projection)

[F2]

Finite weights summing to one define a probability space. (Finite probability spaces are exactly finite full-power-set probability spaces)

[F3]

The event-integral characterization identifies an integrable G-measurable version. (Conditional expectation as an ae class)

Verification

technique · direct

1.1

Write $X - Z = (X - U) + (U - Z)$ . Since $U - Z \in L^{2} (G)$ , [F1] gives $E [(X - U) (U - Z)] = 0$ . Expansion gives the displayed decomposition. The last term is the squared $L^{2}$ norm of U-Z, so it is zero exactly when Z=U almost surely. This proves both directions of the minimum-error assertion.

2.1

For an instance take four equally weighted atoms by [F2], X values (0,2,4,6), and G generated by the first pair and last pair. The G-measurable U=(1,1,5,5) has on each pair the same integral as X, namely $1 / 2$ and $5 / 2$ , so it is the conditional mean. For the constant predictor Z=3 the total error is $(9 + 1 + 1 + 9) / 4 = 5$ , the residual error is $(1 + 1 + 1 + 1) / 4 = 1$ , and the prediction displacement is $(4 + 4 + 4 + 4) / 4 = 4$ . Thus the formula reads $5 = 1 + 4$ .

step 1.1F2F3∎

Source notes

Durrett Theorem 4.1.15, printed p.213; van der Vaart Lemma 1.8, printed p.3. The four-atom calculation illustrates the orthogonal error decomposition.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Durrett, Probability: Theory and Examples, 5th ed. (standard reference, not scraped)

L2L2 best prediction by conditional expectation