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.

Conditioning on a finite partition

Example

Assume AC for the conditional-class convention. Let $(A_{i})_{i = 1}^{m}$ be a finite measurable partition of a probability space, $G = σ (A_{1}, \dots, A_{m})$ , and real $X \in L^{1} (P)$ . A version has value $c_{i} = P (A_{i})^{- 1} \int_{A_{i}} X d P$ on each positive-mass cell, and zero on each zero-mass cell.

Facts & Assumptions

Given: A probability space, a finite measurable partition $(A_{i})$ , $G = σ (A_{i})$ , and real integrable X; AC is the conditional-class convention.

[F1]

A G-measurable integrable function with the defining event integrals represents the conditional class. (Conditional expectation as an ae class)

[F2]

The version is unique as a class. (Conditional expectation is unique almost surely)

[F3]

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

Verification

technique · direct

1.1

Let $T = \sum_{i} c_{i} 1_{A_{i}}$ with the displayed zero convention. It is G-measurable and $E ∣ T ∣ = \sum_{i : P (A_{i}) > 0} ∣ \int_{A_{i}} X ∣ \leq \sum_{i} \int_{A_{i}} ∣ X ∣ = E ∣ X ∣ < \infty$ . Each positive-mass cell has $\int_{A_{i}} T = c_{i} P (A_{i}) = \int_{A_{i}} X$ ; on a null cell both integrals are zero. Every G-event is a union of cells (these unions form a sigma-algebra), so adding proves every defining identity. Hence [F1]–[F2] identify T as the conditional mean.

F1F2

2.1

For a numerical instance take four atoms of weight $1 / 4$ , permitted by [F3], partitioned into $A_{1} = {1, 2}$ and $A_{2} = {3, 4}$ . For X with values $(0, 2, 4, 6)$ , the cell integrals are $1 / 2$ and $5 / 2$ and their masses are $1 / 2$ , so T has values $(1, 1, 5, 5)$ . Its mean is 3, matching $(0 + 2 + 4 + 6) / 4 = 3$ . For $X = 1_{B}$ the same positive-cell calculation is $c_{i} = P (B \cap A_{i}) / P (A_{i})$ , the finite conditional-probability formula.

step 1.1F3∎

Source notes

Durrett Example 4.1.5, printed p.208; van der Vaart Example 1.7, printed p.3. Zero-mass cells and a numerical four-atom calculation are included.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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)