Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p_{ij}$?

share|cite|improve this question
Which would mean $\mu(\{x \in X: x_0 = a_0\}) = \pi_{a_0} p_{a_0a_1} p_{a_{-1}a_0}$? Or maybe $\pi_{a_0} p_{a_0a_0}$? For me it becomes a bit ambiguous for $k=0$. – BallzofFury Dec 13 '12 at 15:00
But if we consider $\sum_{i=1}^n \mu(\{x \in X: x_0 = i\}) = \sum_{i=1}^n \pi_i p_{ii} = \pi_i$ while the collection $\{x \in X: x_0 = i\}$ with $i = 1,\dots,n$ forms a partition of $X$, so it should sum to 1. – BallzofFury Dec 13 '12 at 15:35
I think you're missing my point. If that would be the measure of such a partition, we would have $1 = \mu(X) = \mu(\bigcup_{i=1}^n \{x \in X: x_0=i\}) = \sum_{i=1}^n \mu(\{x \in X: x_0 = i\}) = \sum_{i=1}^n \pi_i p_{ii} = \pi_i$, which is a contradiction. – BallzofFury Dec 13 '12 at 15:47
I'm not sure what you mean, but I just want to know what $\mu(\{x \in X: x_0 = i\})$ is :( – BallzofFury Dec 13 '12 at 16:12
up vote 1 down vote accepted

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 $\mu\{x\in X,x_0=i\}=\pi_{i}$.

share|cite|improve this answer
That makes sense, it solves the partition problem I stated above. Thanks! – BallzofFury Dec 13 '12 at 16:35
And it's seems coherent with the definition of initial measure. – Davide Giraudo Dec 13 '12 at 16:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.