8
$\begingroup$

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!

$\endgroup$

1 Answer 1

6
$\begingroup$

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 $x$.

$\endgroup$
5
  • 1
    $\begingroup$ +1. Thanks (1) In the example, " the increment depends only on the last step and, therefore, is not independent". Why is $X_{t+1} - X_t$ not independent of previous increments $X_{t} - X_{t-1}$ etc? $Z$ depends on $X_t$, so I guess it is better to use $Z_t$ instead of $Z$? (2) In the last paragraph, how are $f$ and $U_t$ chosen/determined usually? Maybe some references? $\endgroup$
    – Tim
    Dec 11, 2012 at 18:55
  • 1
    $\begingroup$ Take $X_{0} = 0$. Observe that $Z_{1}=X_{1}-X_{0}=X_{1}$. Similarly, $Z_{2}=X_{2}-X_{1}$. By construction $Z_{2}$ is not independent of $Z_{1}$. Regarding 2, maybe another way around is to prove that the process is increment-independent iff $Z_{t}$ is independent of $X_{t-1}$ (Under the markov process assumption)? $\endgroup$
    – madprob
    Dec 11, 2012 at 19:04
  • 1
    $\begingroup$ The counterexample only pertains to discrete time processes, does it not? It seems like OP's question pertains only to continuous time processes, so I don't see how it helps. Markov process usually refers to a continuous time process with the continuous time version of the Markov property, and Markov chain refers to any discrete time process (with discrete or continuous state space) that has the discrete time version of the Markov property. $\endgroup$ Apr 19, 2016 at 21:23
  • 1
    $\begingroup$ The OP, 4 years ago, seemed quite satisfied (as you can see in his comment, above). You believe you can interpret his question better than himself? Anyway, if you are interested in a counterexample for a continuous problem, simply interpolate the points in my example ;) $\endgroup$
    – madprob
    Apr 20, 2016 at 7:00
  • $\begingroup$ 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... $\endgroup$ Apr 21, 2016 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.