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A null future-tail event under the stationary mean is null under the AMS law

Statement

Assume AC. If the stationary mean asymptotically dominates P, proves every future-tail event E with Pbar(E)=0 satisfies P(E)=0.

Facts & Assumptions

Given: AC; a probability measure $P$ and measurable map $T$ whose stationary mean $\overset{ˉ}{P}$ asymptotically dominates $P$ ; and the future-tail sigma-algebra $F_{\infty} = ⋂_{n \geq 0} σ (X_{n}, X_{n + 1}, \dots)$ for the coordinate process.

[A1]

The Axiom of Choice: AC is used only for the countable family of Lebesgue decompositions and tail preimages below.

[F1]

The stationary mean asymptotically dominates an AMS law: if $\overset{ˉ}{P} (G) = 0$ , then $P (T^{- n} G) \to 0$ .

[F2]

Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure: a finite positive measure decomposes into an absolutely continuous part with density and a mutually singular part.

[F3]

Mutual singularity for signed or complex measures and A positive, signed, or complex measure concentrated on a measurable set: a singular positive part is concentrated on a measurable carrier that is null for the reference measure.

[F4]

Asymptotically mean stationary sources: the stationary mean $\overset{ˉ}{P}$ is invariant under $T$ .

Proof

Given: AC, an AMS law $P$ with stationary mean $\overset{ˉ}{P}$ , and $E \in F_{\infty}$ with $\overset{ˉ}{P} (E) = 0$ .

1.1A1F2F3given

For each $n$ , define the probability measure $P_{n} (B) = P (T^{- n} B)$ . By [F2] and countable choice [A1], write $P_{n} (B) = \int_{B} f_{n} d \overset{ˉ}{P} + σ_{n} (B)$ , where $f_{n} \geq 0$ . The singularity in [F2] and the carrier convention [F3] give a measurable $N_{n}$ with $\overset{ˉ}{P} (N_{n}) = 0$ on which $σ_{n}$ is concentrated.

2.1F1step 1.1algebra

Put $N = ⋃_{n} N_{n}$ . Then $\overset{ˉ}{P} (N) = 0$ , so [F1] gives $P_{n} (N) = P (T^{- n} N) \to 0$ . For every measurable $B$ , $0 \leq P_{n} (B) - \int_{B} f_{n} d \overset{ˉ}{P} = σ_{n} (B) \leq σ_{n} (N) \leq P_{n} (N) .$ Thus the absolutely continuous approximations have error tending to zero uniformly over $B$ .

3.1A1F4step 2.1givenalgebra∎

Since $E \in F_{\infty}$ , for every $n$ choose a measurable $B_{n}$ with $E = T^{- n} B_{n}$ . Stationarity [F4] gives $\overset{ˉ}{P} (B_{n}) = \overset{ˉ}{P} (T^{- n} B_{n}) = \overset{ˉ}{P} (E) = 0$ , so $\int_{B_{n}} f_{n} d \overset{ˉ}{P} = 0$ . Step 2.1 then gives $P (E) = P_{n} (B_{n}) \leq P_{n} (N) \to 0$ . Hence $P (E) = 0$ . The tail preimages form a countable family; no uncountable simultaneous version selection is made.

Depends on

Used by

Shannon–McMillan limit for finite-alphabet AMS ergodic sources Theorem

Dependency tree · two levels

27 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

Robert M. Gray, Probability, Random Processes, and Ergodic Properties (standard reference, not scraped)