1. 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?

2. Does there exist a continuous-time Markov process with continuous sample paths which does not have independent increments?

3. Does there exist a continuous-time Markov process for which the increments have an infinitely divisible distribution but not independent increments?

4. Does there exist a continuous-time Markov process with a semi-group/generator but which does not have independent increments?

An answer to any one of these questions would be greatly appreciated.


1. 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.

2. In volume 2 of Feller, p. 305, section IX.5 (2nd ed), Feller states that "the paths are continuous with probability 1 if and only if the process is normal". Is this really true? Clearly not every normal process has independent increments, but that counter-example is not a Markov process.

Moreover, I think I proved for one of my homework assignments that a Gaussian process is Markov if and only if it has independent increments, so that would seem to imply that the answer to 2. is no, but I'm not certain of this.

3. On p. 318 of section IX.10 in Feller volume 2, he states that "the distributions associated with continuous processes [with independent but not necessarily stationary increments] are infinitely divisible". But does the other direction mean anything? I.e. if we have a process that is Markov and infinitely divisible increments, does that say anything about the stationarity or independence of those increments?

4. Also on p. 353, X.9, Feller states that "practically all Markov processes represent limiting forms of pseudo-Poisson processes". Moreover, many Markov processes can be formed by the subordination of a Levy process (a process with infinitely divisible increments). Since infinitely divisible increments, semigroups, and independent increments are all related one way or another, this seems to suggest that there ought to be some pattern to when a Markov process does or does not have independent increments.


1 Answer 1


There are many questions here, and some of them should perhaps be split into their own posts. But to address the title question, consider the Ornstein–Uhlenbeck process, which can be defined by the SDE $dX_t = dB_t - X_t \,dt$, or in various other equivalent ways. (In Wikipedia's notation, I am using the parameters $\mu = 0$ and $\theta = \sigma = 1$.) You can picture it as a particle which is trying to perform Brownian motion, but is connected to a spring which pulls it back toward the origin.

It is a continuous Gaussian process which is strong Markov, but its increments are not independent; they are negatively correlated. (In particular, if you "proved" that a Gaussian process is Markov iff it has independent increments, then you made an error.)

  • $\begingroup$ Oh oops I thought the Ornstein-Uhlenbeck process wasn't Markov; that basically is a counter-example for all of my questions. Specifically what I proved was that a Gaussian process is Markov if and only if $\Gamma(s,u)\Gamma(t,t)=\Gamma(s,t)\Gamma(t,u)$ where $\Gamma$ denotes the covariance function -- I thought this meant that the process had independent increments, but obviously not. I'm pretty sure independent increments is a sufficient condition for that, but obviously not a necessary one given the existence of the OU-process. $\endgroup$ Apr 25, 2016 at 15:05
  • $\begingroup$ as per your suggestion I will make #1 a separate question $\endgroup$ Apr 25, 2016 at 15:07

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