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From my reader on interacting particle systems:

Consider a Markov process $\{ X_t : t \geq 0\}$ with three states $\{1,2,3\}$ and "coagulate" (merge) the last two, i.e., define $Y_t = 1$ if $X_t = 1$ and $Y_t = 2$ if $X_t \in \{2,3\}$. In general, $Y_t$ will not be Markov. More generally if $X_t$ is a Markov process and $f$ is a non-bijective map, then in general $f(X_t)$ is not Markov. E.g. components of a multicomponent Markov process are in general not Markov, unless the components are e.g. independent. Notice that it can be Markov, e.g., $|W_t$ is a Markov process, despite the fact that $f(x) = |x|$ is of course not bijective.

Now my question is how to show this. If you can give an example of how the first system with $Y_t$ is non-Markovian that would already be helpful.

PianoEntropy
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  • You accepted an answer which addresses only very partially the questions in your post, roughly two hours after it was posted. Care to explain this surprising behaviour? – Did May 29 '17 at 05:56
  • You're right in that it does not fully address the question. However, if someone comes up with a better answer I could still change which answer I accept. – PianoEntropy May 30 '17 at 11:37
  • Except that when you accept an answer, you are in effect signifying that the post you accept is addressing your question and preventing many users to even visit the page. The former in untrue and you might find the latter undesirable. – Did May 30 '17 at 11:48
  • Ok, I'm unaccepting the answer to open it up to more potential viewers then. – PianoEntropy May 30 '17 at 11:56
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    https://math.stackexchange.com/q/27507/ https://math.stackexchange.com/q/113091/ https://math.stackexchange.com/q/630316/ https://math.stackexchange.com/q/226543/ – Did May 30 '17 at 12:19

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The fact that it's non-Markovian follows from the fact that you lose information on what the last state actually was. So as an example, imagine a simple markov chain on 3 vertices connected as:

1--2--3.

Then $P(X_2=1|X_1=3)=0$ but $P(X_2=1|X_1=2)>0$. So $P(Y_2=1|Y_2=2)$ can either be 0 or nonzero, i.e. you can't even evaluate it!

Sometimes however it would be Markovian. As an example, let $Y_t=1$ regardless of $X_t$. Then $Y_t$ is Markovian, despite not being a bijection.

Alex R.
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  • Ok, it makes sense, but I thought that perhaps $P(Y_2|Y_1)$ could be evaluated by conditioning: $P(Y_2 = 1|Y_1 = 2) =? P(X_2 = 1|X_1 = 2)P(X_1 = 2) + P(X_2 = 1|X_1 = 2)P(X_1 = 1) = P(X_2 = 1|X_1 = 2)P(X_1 = 1) $. Would this be incorrect then? – PianoEntropy May 22 '17 at 17:11
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    Your calculation is absolutely correct but, notice that this implies you need knowledge about $X_1$ to evaluate $P(Y_2|Y_1)$. This implies it is not sufficient to just know $Y_1$, because this means that $P(Y_{t+1}=b|Y_t=a)$ actually depends on $t$ (through $P(X_t)$), whereas for a true Markov process it suffices to just know $a,b$. – Alex R. May 22 '17 at 17:27