5
$\begingroup$

An independent-increment stochastic process must be Markov. I am now wondering about the reverse case. Why do some Markov processes fail to be independent-increment?

  1. What are some examples of Markov processes that are not independent-increment?

  2. Is it possible to have some sufficient and necessary condition for a Markov process to be independent-increment?

Thanks and regards!

1 Answers 1

5

First, observe that an independent-increment process depends on the fact that the sequence is defined on $R$. A Markov Chain can be defined in any set $S$. If $S \neq R$, you might have trouble to even define what independent-increment would be.

You can also consider a process in $R$ defined in the following way: $X_{t+1} = X_{t} + Z$, where $Z|(X_{t},X_{t-1},\ldots,X_{0}) \sim N(-X_{t},1)$. This an example of a process such that the increment depends only on the last step and, therefore, is not independent but the process is Markovian.

Regarding 2, I don't think I have much to add. You can always write $X_{t} = X_{t-1} + Z_{t}$. since $X_{t}$ is Markovian, $Z_{t} = f(X_{t},U_{t})$, where $U_{t}$ is independent of $(X_{t-1},X_{t-2}, \ldots, X_{0})$. I guess you have to prove that $f(x,u)$ is constant on $x$.

  • 0
    Do you mean creating a continuous time Markov chain with exponential waiting times? Do you know any counterexamples in continuous time with continuous sample paths? Such an example would help me to better understand the relationships between Markov processes/Levy processes/processes with infinitely divisible increments/processes with independent increments/processes with semi-groups...2016-04-21