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I was wondering for a Markov process:

  1. Is it true that:

    it is a stationary process, iff its transition probability function is invariant to time-translation (i.e. it is homogeneous) and its initial distribution is invariant to time-translation?

  2. When the state space is discrete, is the limit of its transition probability function related to stationarity for the process? Is its stationary distribution related to stationarity, besides both have "stationary" in their names?

Thanks and regards!

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up vote 3 down vote accepted
  1. Stationary in stationary process means that the distribution of the process is invariant by time-shifts, that is, the distributions of $(X_n)_{n\ge0}$ and $(X_{n+1})_{n\ge0}$ coincide. Hence stationarity requires the distribution of $X_n$ to be the same for every $n\ge0$ and also the distribution of $(X_n,X_{n+1})$ to be the same for every $n\ge0$. For Markov processes this is enough to guarantee that the transition probabilities are independent on $n$ and that the distribution of $X_0$ is invariant by this transition kernel. Reciprocally, if these two properties hold, the Markov chain is stationary.

  2. The first question of this item makes no sense. Regarding the second question, if the distribution of $X_0$ is stationary in the sense that it is invariant with respect to the transition kernel, and if the transition kernel is independent on $n$, then obviously the Markov process is stationary.

Recall that for a measure $\nu$ and a transition kernel $p$, $\nu p$ is the measure defined by $$(\nu p)(x)=\displaystyle\sum_y\nu(y)p(y,x), $$ for every state $x$. The measure $\nu$ is said to be invariant by the transition kernel $p$ if $\nu p=\nu$.

Let me suggest that you read this.

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Thanks! Really nice answer! – Tim Mar 31 '11 at 21:51
Is there a typo in your formula for $vp$: $p(x,y)$ or $p(y,x)$? – Tim Apr 30 '11 at 23:51
Corrected, thanks. – Did May 1 '11 at 6:03
I have read the link, which I understand to be about the density of initial distribution and the density of a later distribution wrt the invariant distribution and represented in terms of the transition kernel. But I guess I might miss your point: is the link related to your reply or my question here? Anyway, your reply there widens my eyes and is really worth reading! – Tim May 1 '11 at 20:04
@Tim: The link mainly defines objects properly (and shows that everything flows smoothly from this). Based on your questions here I thought reading it could help you understand the notions involved. – Did May 1 '11 at 20:15
  1. Markov Processes are not always stationary. It is only stationary if $P(X_{n+1} = b| X_n = a) = P(X_2 = b| X_1 = a)$. In other words $p(x_{n+1}|x_n)$ does not depend on $n$.
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?? $ $ $ $ – Did Mar 31 '11 at 21:46
Thanks anyway! Nice to see you again here. – Tim Mar 31 '11 at 21:52

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