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A reverse martingale and the tail sigma-algebra

Statement

Assume AC. Let $Z_{0}, Z_{1}, \dots$ be integrable random variables, $G_{n} = σ (Z_{n}, Z_{n + 1}, \dots)$ , and $X \in L^{1}$ . The process $Y_{n} = E [X ∣ G_{n}]$ is a reverse martingale with respect to the decreasing filtration $(G_{n})$ . Moreover, $E [X ∣ G_{n}] ⟶ E [X ∣ T]$ almost surely and in $L^{1}$ , where $T = ⋂_{n} G_{n}$ is the tail $σ$ -algebra.

Facts & Assumptions

Given: The hypotheses, objects, and conventions in the Statement.

[F1]

Tail sigma-algebra of a sequence identifies the intersection as $T$ .

[F2]

Levy downward convergence of conditional expectations gives the asserted convergence.

[F3]

Kolmogorov zero-one law from martingale convergence treats the independent case.

[F4]

The Axiom of Choice is inherited from conditional expectation.

[F5]

Tower property of conditional expectation gives $E [E [X ∣ G_{n}] ∣ G_{n + 1}] = E [X ∣ G_{n + 1}]$ almost surely.

Proof

1.1

Deleting the first generator gives $G_{n + 1} \subseteq G_{n}$ , and F1 gives $⋂_{n} G_{n} = T$ . Each $Y_{n}$ is integrable and $G_{n}$ -measurable by the conditional-expectation interface in F5. The tower identity in F5 gives $E [Y_{n} ∣ G_{n + 1}] = Y_{n + 1}$ almost surely, which is the reverse-martingale identity. F2 now applies directly to $X$ , proving both modes of convergence.

F1F2F5

2.1

If the $Z_{n}$ are independent and $A \in T$ , take $X = 1_{A}$ . F3 gives $P (A) \in {0, 1}$ , so $1_{A}$ and therefore $E [1_{A} ∣ T]$ is almost surely constant. This extra conclusion is asserted only under independence. AC has exactly the inherited role in F4.

F3F4∎

Depends on

Used by

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Dependency tree · two levels

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Sources

van der Vaart, Martingales, Diffusions and Financial Mathematics, Example 2.28 and Exercise 2.31, pp. 17–18 (standard reference, not scraped)