# Markov processes driven by the noise

Let $\xi_n\in \Xi$ be a sequence of iid random variables with $n \in\mathbb N\cup\{0\}$, which we call a noise process. Construct a process $$Z_{n+1} = f(Z_n,\xi_n)\quad(\star)$$ with $Z_0\in E$ and $f:E\times\Xi\to E$ is a measurable function of two variables. Clearly, $Z$ is a Markov process.

On the other hand, consider a Markov process $X$ on $E$ given by its transition kernel $P(x,A)$. Is it always possible to find a noise process (on some set $\Xi$) and a measurable function $f:E\times \Xi\to E$ such that given any initial condition (maybe random) $X_0\in E$ $$\operatorname{Law}(X_1) = \operatorname{Law}(f(X_0,\xi_0))$$

Briefly speaking, if any Markov process can be presented in the form $(\star)$?

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## 1 Answer

Yes, this is the representation of a discrete time Markov process as a randomized dynamical system as found in Proposition 8.6 (page 145) from Foundations of Modern Probability (2nd edition) by Olav Kallenberg:

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.

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– Tim Nov 18 '12 at 18:35
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