TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Independent random elements are characterized by finite rectangle probabilities

Statement

Let $(X_{i})_{i \in I}$ be random elements $X_{i} : (Ω, F, P) \to (S_{i}, Σ_{i})$ . Then the following are equivalent:

the family $(X_{i})_{i \in I}$ is independent;
for every natural number $n \geq 1$ , every choice of distinct indices $i_{0}, \dots, i_{n - 1} \in I$ , and every choice of measurable sets $B_{k} \in Σ_{i_{k}}$ , $P (X_{i_{0}} \in B_{0}, \dots, X_{i_{n - 1}} \in B_{n - 1}) = \prod_{k < n} P (X_{i_{k}} \in B_{k}) .$

Facts & Assumptions

Given: Random elements $X_{i} : (Ω, F, P) \to (S_{i}, Σ_{i})$ .

[L1]

A family of random elements is independent exactly when the sigma-algebras $σ (X_{i})$ are independent. (Independent random elements)

[L2]

Independent pi-systems containing the whole space generate independent sigma-algebras. (Independent pi-systems generate independent sigma-algebras)

Proof

technique · direct

1.1

If $(X_{i})_{i \in I}$ is independent, then by [L1] the sigma-algebras $σ (X_{i})$ are independent. Since $X_{i}^{- 1} (B) \in σ (X_{i})$ for every $B \in Σ_{i}$ , the displayed rectangle identity follows immediately.

1.2

Conversely, for each $i$ let $Π_{i} : = {X_{i}^{- 1} (B) : B \in Σ_{i}} \subseteq F$ . Preimages preserve finite intersections, so each $Π_{i}$ is a pi-system containing $Ω$ . The hypothesis in clause 2 says exactly that the family $(Π_{i})_{i \in I}$ is independent. Since $σ (Π_{i}) = σ (X_{i})$ by definition, [L2] implies that the sigma-algebras $σ (X_{i})$ are independent.

givenL2

2.1

Step 1.2 proves clause 2 implies clause 1, and step 1.1 proves the reverse implication. Therefore the two conditions are equivalent.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

7 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

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 2.1.8 (standard reference, not scraped)
S. R. S. Varadhan, Probability Theory, Lemma 3.1 (standard reference, not scraped)