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A Markov-chain law is determined by its initial law and kernel
Statement
Assume Choice. Two time-homogeneous Markov chains on the same measurable state space with the same initial law and the same transition kernel have the same finite-dimensional distributions. Consequently their induced laws on the canonical path space equipped with its cylinder sigma-algebra are equal.
Facts & Assumptions
Given: Choice and two -chains with initial law .
Every finite-dimensional law of a Markov chain is the iterated integral determined by its initial law and iterated kernels. (Finite-dimensional laws of a Markov chain)
A process law on countable-coordinate cylinder space is determined by its finite-dimensional distributions. (Finite-dimensional distributions determine a process law on the cylinder sigma-algebra)
Proof
For every finite increasing time list, [F1] gives the same iterated [F1] integral for both processes because their and agree. This includes a single time, time zero, empty rectangle events, and the full rectangle. Therefore all their finite-dimensional distributions coincide.
Push both processes forward by their path maps. The two induced [F2, step 1.1] probabilities have the finite-dimensional distributions compared in step 1.1, so [F2] makes them equal on the cylinder sigma-algebra. Choice enters through [F1]'s conditional-expectation argument; [F2] adds no new selection.
Depends on
Used by
- Identical one-time marginals do not determine a Markov chain Counterexample
Dependency tree · two levels
12 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, Section 5.1 (standard reference, not scraped)