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If we have an integer valued stochastic process, are these implications correct?

  1. independent increments $\rightarrow$ Markov property

  2. Markov property $\nrightarrow$ independent increments

  3. stationary increments $\rightarrow$ time homogenity

  4. time homogenity $\nrightarrow$ stationary increments

  5. stationary increments $\rightarrow$ independent increments $\rightarrow$ Markov Property

  6. independent increments $\nrightarrow$ stationary increments

PS: I got number 2 answered by an example in another thread, but I would like to know all on this list aswell, because it is confusing for me. You don't need to make me examples if this is too much work for you, please then just tell me if the impliations hold or not.

I made a picture to describe it, is it correct?

pic

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    $\begingroup$ Much more interesting would be you telling us why you think these implications hold. For example, why do you think that stationary increments → independent increments? $\endgroup$ – Did Apr 30 '14 at 23:44
  • $\begingroup$ Hello. I am very unsure but on that case I thought like this: Lets say we want to calculate P([x(t+s)-x(t)=j]|[x(t)-x(0)=i]). Since stationary increments means that it only depends on the length of the interval this is P([x(t+s)-x(t)=j])=P([x(s)-x(0)=j]),so the information about what happened before doesn't mean anything, hence we also have independent increments. But I am very unsure of this. $\endgroup$ – user119615 Apr 30 '14 at 23:49
  • $\begingroup$ "Stationary increments" does not mean that, please check the definitions. $\endgroup$ – Did May 1 '14 at 6:06
  • $\begingroup$ @Did The only definition of stationary increments in my book was this one: s2.postimg.org/g5rjhnpo9/image.png But I see now that it is for a counting process, and it might not be enough for a general integer valued process. I guess that general stationary increments means that it only depends on the time interval, but also the state it is in in the start of the increment. $\endgroup$ – user119615 May 1 '14 at 7:43
  • $\begingroup$ But where you are at the start of the increment, may depend on an earlier increment. So the distribution may still depend on what happened earlier. I updated the picture, is one of the two pictures in the updated version correct?: s30.postimg.org/uyqucthox/updated.png $\endgroup$ – user119615 May 1 '14 at 7:43

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