TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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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 · two levels

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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)