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I have a question about the continuous time Markov chain. In the Poisson process we have independent and stationary increments. Do we have this in a continuous time Markov chain that is time-homogeneous?

I know that a Poisson process is a time-homogeneous continuous time Markov chain, so some of these has both independent and stationary increments. But do all of them have it?

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No. For a counterexample, consider a continuous time Markov chain $X_t$ whose state-space is $\{0,1\}$ and both states are recurrent. Let $t_3>t_2>t_1$. If the increment $X_{t_2}-X_{t_1}$ equals $1$, this clearly implies that $X_{t_2} = 1$, hence increment $X_{t_3}-X_{t_2}$ is either $0$ or $-1$. On the other hand, if the increment $X_{t_2}-X_{t_1}$ is $-1$, $X_{t_2} = -1$, hence increment $X_{t_3}-X_{t_2}$ is either $0$ or $1$. Thus, as the recurring states assumption also implies that $P(\textrm{increment}=0)<1$, the increments $X_{t_3}-X_{t_2}$ and $X_{t_2}-X_{t_1}$ are not independent.

Furthermore, the state-space of a continuous time Markov chain could be something where subtraction (and hence) increments are not even defined.

Concerning stationarity of increments

The question is about independent and stationary increments, and the previous counterexample showed that the increments can be not independent, thus not (stationary and independent). In addition, the increments can be not stationary, too. Let $X$ be a Poisson process. Let $Y_t=X_t^2$ for all $t$. Now, $Y$ is also a time-homogenous continuous-time Markov chain, but the expected increment of $Y$ increases over time, thus the increments are not stationary.

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  • $\begingroup$ Thank you very much! I updated with another quesiton also, if you know the answer to this I would appreciate your help! $\endgroup$ – user119615 Apr 30 '14 at 13:08
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    $\begingroup$ @user119615 Please don't do that. If new question, then new post. $\endgroup$ – Did Apr 30 '14 at 13:08
  • $\begingroup$ Ok, I will remove it then. $\endgroup$ – user119615 Apr 30 '14 at 13:11
  • $\begingroup$ Sorry but I just have a quick follow-up. When it comes to stationary increments, I see as you said that they not even have to be defined because of the statespace need not be numbers. But if they are numbers how do we see that the increments need not be stationary? Stationary increments are defined as what happens in the increment only depends on the length of the increment, so could it be that what happened in an increment also depend on the state it starts in? $\endgroup$ – user119615 Apr 30 '14 at 13:46
  • $\begingroup$ @user19615 Ok, I edited my answer to contain also a counterexample for the stationarity of increments. $\endgroup$ – Juho Kokkala Apr 30 '14 at 15:24

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