TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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.

Expectation factors over a finite product of mutually independent random variables

Statement

If $(X_{i})_{i \in I}$ is a finite mutually independent family of real random variables, then $E ⁣ [\prod_{i \in I} X_{i}] = \prod_{i \in I} E [X_{i}] .$ For $I = \emptyset$ , both sides equal $1$ . The converse is not asserted.

Facts & Assumptions

Given: A finite mutually independent family $(X_{i})_{i \in I}$ .

[L1]

Expectation can be summed over the finite attained values of a random variable (Expectation is the sum of each attained value times its probability).

[L2]

Mutual independence factors every finite joint attained-value probability (Pairwise and mutual independence of finite-valued random variables).

[L3]

Finite Fubini interchanges iterated sums over finite products (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1

Grouping outcomes by the joint values $(x_{i})_{i \in I}$ gives $E [\prod_{i} X_{i}] = \sum_{(x_{i})} (\prod_{i} x_{i}) P (X_{i} = x_{i} for all i)$ .

L1L3

1.2

Independence changes the last probability to $\prod_{i} P (X_{i} = x_{i})$ .

2.1

Finite Fubini factors the resulting sum as $\prod_{i} \sum_{x_{i}} x_{i} P (X_{i} = x_{i}) = \prod_{i} E [X_{i}]$ . For $I = \emptyset$ , this calculation is the empty product identity $1 = 1$ .

step 1.1step 1.2L1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 41 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Matousek and J. Vondrak, The Probabilistic Method, Lemma 1.1.9 (standard reference, not scraped)
M. Bucic, Probabilistic Method, Proposition A.15 (standard reference, not scraped)