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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!

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1 Answer 1

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 $x$.

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+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? –  Tim Dec 11 '12 at 18:55
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)? –  madprob Dec 11 '12 at 19:04

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