A random dynamic system is defined in Wikipedia. Its definition, which is not included in this post for the sake of clarity, reminds me how similar a Markov process is to a random dynamic system just in my very superficial impression. Let

  • $T=\mathbb{R}$ or $\mathbb{Z}$ be the index set,

  • $(\Omega, \mathcal{F}, P)$ be the probability space,

  • $X$, a measurable space (or a complete separable metric space with its Borel sigma algebra, as in Wikipedia's definition for random dynamic system), be the state space.


  1. I wonder if any Markov process $f: T \times \Omega \to X$ can be seen as a random dynamic system, i.e. can induce a random dynamic system $\varphi: T \times \Omega \times X \to X$ corresponding to it?

    If no, what kinds of Markov processes can induce random dynamic systems?

  2. When it is yes, how are the random dynamic system $\varphi$ and its base flow $\vartheta: T \times \Omega \to \Omega$ constructed from the Markov process $f$?

Thanks and regards!

I finally am able to read and understand the linked question by Ilya and reply by Byron. Yes they are closely related, in that Byron pointed out a theorem that can rewrite a discrete time Markov process into a kind of "randomized dynamic system".

Let $X$ be a process on $\mathbb{Z}_+$ with values in a Borel space $S$. Then $X$ is Markov iff there exist some measurable functions $f_1,f_2,\dots:S\times[0,1]\to S$ and iid $U(0,1)$ random variables $\xi_n$ independent of $X_0$ such that $X_n=f_n(X_{n-1},\xi_n)$ almost surely for all $n\in\mathbb{N}$. Here we may choose $f_1=f_2=\cdots =f$ iff $X$ is time homogeneous.

However, the form of $f_n$'s is not exactly $\varphi$ in the definition of a random dynamic system in the Wikipedia article I linked. So how shall I see if they are equivalent?

  • $\begingroup$ should be related $\endgroup$ – Ilya May 13 '12 at 8:48
  • $\begingroup$ @Ilya:Thanks! I will take a look. $\endgroup$ – Tim May 13 '12 at 13:02
  • $\begingroup$ @Ilya: Thanks! I wonder how to see if the form given by Byron and the definition of a random dynamic system in the Wikipedia article I linked are equivalent? $\endgroup$ – Tim Nov 18 '12 at 18:32
  • $\begingroup$ I think, measure-preserving maps are related to the stationarity - more precisely, if the shift operator over the Markov process $$ \vartheta(\omega_0,\omega_1,\omega_2,\dots) = (\omega_1,\omega_2,\dots) $$ is measure $\mathsf P_\mu$ preserving, than $\mu$ is the stationary distribution of the Markov process. Not every Markov process allows for the stationary distribution - however, maybe $\vartheta$ in that case can be something different from the shift. I don't think that the linked questions is a duplicate of yours, so I vote to reopen. $\endgroup$ – Ilya Nov 18 '12 at 19:11

A different point of view:

Although the dynamics of a particle in a random walk are indeed random, the dynamics of its probability distribution certainly are not. Indeed note the probability distributions $\{\nu^{\star k}\}_{k\in\mathbb{N}}$ evolve deterministically as $\{\delta^e P^k:k\in \mathbb{N}\}$. Thus the random walk has the structure of a dynamical system $\{M_p(G),P\}$ with fixed point attractor $\{\pi\}$. The two canonical categories of dynamical systems (for which there is an existing literature of powerful methods) are topological and measure preserving dynamical systems. Unfortunately at first remove $\{M_p(G),P\}$ appears too coarse and structureless to apply any of these powerful methods. Also the mapping function $P$ is not necessarily invertible and this poses further problems. Indeed in many examples of walks exhibiting cut-off, $P$ may be seen to be singular. Hence the assumption that needs to be made on $P$ to put a structure on $\{M_p(G),P\}$ sufficient for application of dynamical systems methods to the cut-off phenomenon is overly strict. A more fundamental problem occurs in trying to put the structure of a measure preserving dynamical system on the walk in that if a meaningful (a measure $\kappa$ wouldn't be very meaningful if $\kappa(M_p(G))=\kappa(\{\pi\})$) measure is put on $M_p(G)$, the fact that $(M_p(G))P^k\underset{k\rightarrow \infty}{\rightarrow} \{\pi\}$ would imply that $P$ is in fact not measure preserving.

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