DefinitionDefinition: AI-adaptedProof: Not applicablePipeline-generatedprecheck passaudited 2026-10-02

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Stationary process and canonical path shift

Definition

Let $S \in {R, C}$ with Borel sigma-algebra $E$ , let $(Ω, F, P)$ be a probability space, and let $Y = (Y_{n})_{n \geq 0}$ be a stochastic process with each $Y_{n} : Ω \to S$ a random element (Stochastic processes and their finite-dimensional distributions). The process $Y$ is strictly stationary when every finite-dimensional law is unchanged by a common nonnegative time shift: for every $r \geq 1$ , every $0 \leq n_{1} < \dots < n_{r}$ and every $m \geq 0$ ,

$L (Y_{m + n_{1}}, \dots, Y_{m + n_{r}}) = L (Y_{n_{1}}, \dots, Y_{n_{r}}) .$

The canonical path law of $Y$ is the pushforward $P_{Y} = L (Y)$ of $P$ under the coordinate map $ω \mapsto (Y_{n} (ω))_{n \geq 0}$ , a probability measure on the product space $S^{N_{0}}$ with its product sigma-algebra. The left shift is $θ (z)_{n} = z_{n + 1}$ for $z = (z_{n})_{n \geq 0} \in S^{N_{0}}$ .

A strictly stationary process is called ergodic when the canonical shift $θ$ is ergodic for $P_{Y}$ in the sense of Ergodicity relative to an invariant measure. The shift-invariance needed for that definition, namely that $θ$ is measure preserving for $P_{Y}$ , is proved below for strictly stationary $Y$ , so in that case $(S^{N_{0}}, E^{\otimes N_{0}}, P_{Y}, θ)$ is a probability measure-preserving system in the sense of Measure-preserving transformations and systems.

Facts & Assumptions

Given: A process $Y = (Y_{n})_{n \geq 0}$ with values in $S \in {R, C}$ on a probability space, its coordinate map, and the left shift $θ$ .

[F1]

The finite-dimensional laws are the pushforward laws of the tuples of coordinates, including for the single time $n_{1}$ . (Stochastic processes and their finite-dimensional distributions)

[F2]

A measurable self-map $T$ of a measure space is measure preserving when $μ (T^{- 1} E) = μ (E)$ for every measurable $E$ , and the quadruple is then a measure-preserving system. (Measure-preserving transformations and systems)

[F3]

A lambda-system containing a generating pi-system contains the sigma-algebra generated by it. (Dynkin's pi-lambda theorem)

Verification

technique · verify cylinder invariance and extend by the pi-lambda theorem

1.1F1F2

The coordinate map is measurable, since each of its coordinates is a random element, so $P_{Y}$ is a well-defined probability measure on the product sigma-algebra [F1]. A cylinder $C = {z : (z_{n_{1}}, \dots, z_{n_{r}}) \in B}$ , with $0 \leq n_{1} < \dots < n_{r}$ and measurable $B \subseteq S^{r}$ , is measurable; its preimage $θ^{- 1} C = {z : (z_{n_{1} + 1}, \dots, z_{n_{r} + 1}) \in B}$ is again a cylinder, and cylinders generate the product sigma-algebra, so $θ$ is measurable.

2.1F1F2step 1.1given

Suppose $Y$ is strictly stationary and let $C = {z : (z_{n_{1}}, \dots, z_{n_{r}}) \in B}$ be a cylinder as in step 1.1. By the definition of $P_{Y}$ as the pushforward of the coordinate map, $P_{Y} (θ^{- 1} C) = P ((Y_{n_{1} + 1}, \dots, Y_{n_{r} + 1}) \in B) .$ Strict stationarity applied to the time list $(n_{1}, \dots, n_{r})$ with shift $m = 1$ says that the joint law of $(Y_{n_{1} + 1}, \dots, Y_{n_{r} + 1})$ equals the joint law of $(Y_{n_{1}}, \dots, Y_{n_{r}})$ , and the latter gives $P_{Y} (C)$ . Hence $P_{Y} (θ^{- 1} C) = P_{Y} (C)$ for every cylinder.

3.1F2F3step 2.1∎

Let $D = {A : θ^{- 1} A is measurable and P_{Y} (θ^{- 1} A) = P_{Y} (A)}$ . Preimages commute with complements and countable unions, so $D$ is a lambda-system: it contains $S^{N_{0}}$ , is closed under complements because $P_{Y}$ has total mass one, and is closed under countable disjoint unions by countable additivity. Cylinders form a pi-system generating the product sigma-algebra and lie in $D$ by step 2.1, so [F3] gives every product-measurable set. Thus, when $Y$ is strictly stationary, $θ$ is measurable and measure preserving, and [F2] makes the quadruple a probability measure-preserving system. Constant deterministic processes are included; a nonconstant deterministic process need not be stationary, and the proof of shift invariance uses only the stated finite-dimensional invariance.

Depends on

Used by

A stationary chain need not be ergodic Counterexample
Birkhoff limit for a stationary integrable process Theorem

Dependency tree · two levels

13 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

Durrett, Probability: Theory and Examples, fifth edition, §6.2 and §5.5 (standard reference, not scraped)
Charles Walkden, Ergodic Theory lecture notes, §1 (standard reference, not scraped)