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

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Tail events are independent of every finite initial sigma-algebra

Statement

Let $(X_{n})_{n \in N}$ be an independent sequence of random elements. For each $m \in N$ , let

$H_{m} : = σ (X_{0}, \dots, X_{m}), G_{m} : = σ (X_{n} : n \geq m + 1) .$

Then every tail event $A \in T (X_{n} : n \in N)$ is independent of every event $B \in H_{m}$ .

Facts & Assumptions

Given: An independent sequence $(X_{n})_{n \in N}$ , an index $m$ , a tail event $A$ , and an event $B \in H_{m}$ .

[L1]

The tail sigma-algebra is $T (X_{n} : n \in N) = ⋂_{r \in N} σ (X_{n} : n \geq r)$ . (Tail sigma-algebra of a sequence)

[L2]

Disjoint groups of an independent sigma-algebra family remain independent. (Disjoint groups of an independent sigma-algebra family remain independent)

Proof

technique · direct

1.1givenL2

The independent sequence $(X_{n})$ gives an independent family of sigma-algebras $(σ (X_{n}))_{n \in N}$ . Grouping the first $m + 1$ coordinates into one block and the remaining coordinates into the other, [L2] shows that $H_{m}$ and $G_{m}$ are independent sigma-algebras.

2.1L1step 1.1∎

Because $A$ lies in the tail sigma-algebra, [L1] gives $A \in σ (X_{n} : n \geq m + 1) = G_{m}$ . Step 1.1 therefore yields $P (A \cap B) = P (A) P (B) .$ So every tail event is independent of every event in the finite initial sigma-algebra.

Depends on

Used by

Kolmogorov zero-one law from martingale convergence Corollary
Kolmogorov zero-one law Theorem

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

S. R. S. Varadhan, Probability Theory, discussion before Theorem 3.15 (standard reference, not scraped)