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Consider a Markov Chain $(X,\mathcal{F},\mu,T)$, where $X = (1,2,\dots,n)^\mathbb{Z}$ and $T$ is the left shift and a transition matrix $P=(p_{i,j})$ and stationary distribution $\pi$ such that $\pi P = \pi$, with the Markov measure defined as:

$ \mu(\{x \in X: x_{-k} = a_{-k},\dots,x_k = a_k\}) = \pi_{a_{-k}}p_{a_{-k}a_{-k+1}}\dots p_{a_{k-1}a_{k}}$

In all the literature I've found on the matter, it is stated that extends to the full $\sigma$-algebra by Kolmogorov's Extension theorem. My problem is, I don't understand how this extends to cylinders where only a single coordinate is specified, for eg:

$ \mu(\{x \in X: x_0=i\}) = \,?$

Is it just $\pi_i$, $\pi_i p_{ii}$, or some summation of $p_{ij}$?

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    I'm not sure what you mean, but I just want to know what $\mu(\{x \in X: x_0 = i\})$ is :(2012-12-13

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The definition is $\mu\{x\in X,x_j=a_j,-k\leqslant j\leqslant k\}:=\pi_{a_{-k}}\prod_{j=-k}^{k-1}p_{a_ja_{j+1}}.$ When $k=0$, the set of indexes of the product is empty and by convention the product is $1$, so $\mu\{x\in X,x_0=i\}=\pi_{i}$.

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    And it's seems coherent with the definition of initial measure.2012-12-13