# Does an embedded discrete-time Markov chain preserve its properties in continuous time?

Given a discrete-time Markov chain without independent increments, is the embedding of it into a continuous time Markov chain (i.e. via the use of exponential waiting times) an example of a continuous time Markov process without independent increments?

Note: This question is forked from one I asked previously, since the Ornstein-Uhlenbeck process was an answer to every other part of that question except the part being asked again here.

Context:

As an answer to this related question, user @madprob gave an example of a discrete time Markov chain that does not have independent increments.

Specifically, let a discrete time process with state space $\mathbb{R}$ be defined as follows: $X_{n+1} = X_n + Z$, where $Z|(X_n,X_{n-1},...,X_0) \sim N(-X_n, 1)$.

In general, any Markov process in discrete time can be written as $X_{n+1}=X_n+Z_{n+1}$ where $Z_n = f(X_n,U_n)$ for some suitable $f$ and a random variable $U_n$ that is independent of $(X_{n-1},...,X_1,X_0)$ -- thus the problem of finding a discrete time Markov process that does not have independent increments reduces to the problem of choosing an appropriate $f$.

Such a counterexample presumably exists given the answer to this question.

EDIT: I suppose the process of embedding a Markov chain into continuous time is essentially subordinating a Poisson process. So this question is probably just a special case of whether or not subordinating infinitely divisible distributions preserves independence properties.

If $\{X_n:n=1,2,\ldots\}$ is a Markov chain and we define the holding times $\{T_n:n=1,2,\ldots\}$ by $T_n\sim\mathsf{Exp}(\lambda_{X_n})$, then the sequence $\{T_n\}$ only defines a Poisson process if $\lambda_{X_n}$ is equal for all $n$. Otherwise, the intensity would be a stochastic process itself, in which case the counting process $N(t):= \sum_{n=1}^\infty \mathsf 1_{(0,t]} T_n$ would be a Cox process. (The jump process defined by the state transitions does of course define a CTMC, though.)