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The stationary mean asymptotically dominates an AMS law

Statement

For an AMS law P with stationary mean Pbar, proves P(T^{-n}G) tends to zero for every event G with Pbar(G)=0.

Facts & Assumptions

Given: A probability measure $P$ on a measurable space $(Ω, F)$ , a measurable map $T$ , and an AMS stationary mean $\overset{ˉ}{P}$ as in Asymptotically mean stationary sources.

[F1]

Asymptotically mean stationary sources: $\overset{ˉ}{P} (G) = \lim_{N} N^{- 1} \sum_{j < N} P (T^{- j} G)$ for every measurable $G$ , and $\overset{ˉ}{P}$ is stationary.

[F2]

Fatou's lemma: for nonnegative measurable functions $f_{n}$ , $\int {lim inf}_{n} f_{n} d P \leq {lim inf}_{n} \int f_{n} d P$ .

Proof

Given: The AMS probability system and [F1]–[F2].

1.1F1given

If $B$ is strictly invariant, then $P (T^{- j} B) = P (B)$ for every $j$ , so [F1] gives $\overset{ˉ}{P} (B) = P (B)$ . Thus the two laws agree on strictly invariant events.

2.1F1step 1.1algebra

Fix measurable $G$ with $\overset{ˉ}{P} (G) = 0$ and set $B = {lim sup}_{n} T^{- n} G = ⋂_{N} ⋃_{n \geq N} T^{- n} G$ . Stationarity gives $\overset{ˉ}{P} (T^{- n} G) = 0$ for each $n$ , hence $\overset{ˉ}{P} (B) = 0$ by countable subadditivity. Also $T^{- 1} B = B$ because removing the first term of a limsup does not change it. Step 1.1 yields $P (B) = 0$ .

3.1F2step 2.1algebra∎

Apply [F2] to $f_{n} = 1_{(T^{- n} G)^{c}}$ . Since ${lim inf}_{n} (T^{- n} G)^{c} = B^{c}$ and $P (B^{c}) = 1$ by step 2.1, Fatou gives $1 = P (B^{c}) = \int {lim inf}_{n} f_{n} d P \leq {lim inf}_{n} (1 - P (T^{- n} G)) = 1 - {lim sup}_{n} P (T^{- n} G) .$ Thus ${lim sup}_{n} P (T^{- n} G) \leq 0$ , and nonnegativity implies $P (T^{- n} G) \to 0$ . This is asymptotic domination of $P$ by its stationary mean.

Depends on

Used by

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Sources

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