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I am reading the book Markov Chains and Stochastic Stability from Meyn and Tweedie. They define Markov chains on a measurable state space $(E,\Sigma)$ (Chapter 3.4) and they define it on the space $\Omega = \prod_{i \in \mathbb{N}}E, $ with an $\sigma$-algebra $\mathcal{A}$ which is the smallest $\sigma$-algebra that contains all cylinder sets with only finitly many sets different from $E$ $$A_1 \times A_2 \times \dots A_n \times E \times E \times \dots$$

Then they define the Markov chain as a family of random variables $(X_n)_{n \in \mathbb{N}}$ where for $\omega=(x_n)_{n \in \mathbb{N}}\in \Omega$ they set $$X_n(\omega)=x_n .$$

Thus, all Markov chains are defined on the same set $\Omega$ and the random variables $(X_n)$ are also always the same. Now if they talk about a certain initial distribution $\mu$ and transition kernel $p(x,A)$; then they assoicate a Markov chain to it by constructing a specific measure $\mathbb{P}_\mu$. Thus, by this definition, two Markov chains only differ on the probability measure of the probability space.

My problem is that in the book they define the term $$ \mathcal{F}_n = \sigma(X_0,\dots,X_n) \subseteq \mathcal{B}(X^{n+1})$$ and they say

which is the smallest $\sigma$-field for which the random variable $\{X_0,\dots,X_n\}$ is measurable. In many cases $\mathcal{F}_n$ will coincide with $\mathcal{B}(X^{n+1})$, although this depend in particular on the initial measure $\mu$ choosen for a particular chain.

How can $\mathcal{F}_n$ depend on the initial measure? The random variable is already defined as $X_n(\omega)=x_n$, and thus the measurability of $\{X_0,\dots,X_n\}$ depends only on $\Sigma$ and $\mathcal{A}$ where does the intial measure $\mu$ comes into play?

Update: After seeing the answers, I think it is a good thing to provide my question with an example. Lets consider the case where $E=\{1,2\}$ and $\Omega = E \times E$, then the random variables $X_0$ and $X_1$ are already defined as above, in particular $X_0$ is defined as $$ X_0 ((1,1))=X_0((1,2))=1$$ and $$X_0((2,1))=X_0((2,2))=2.$$ Now if $\mathbb{P}_\mu$ is the probability that $X_0 = 1 $, then we must have $$ \mathbb{P}_\mu[\{(1,1),(1,2)\}]=1.$$ But this is completely independent from defining $\mathcal{F}_0$ (or $\mathcal{F}_n$). In this case we always have $$\mathcal{F}_0 = \{\{(1,1),(1,2)\},\{(2,1),(2,2)\},E,\emptyset \} $$ which does not depend on $\mu$. It seems to me that in the answers one believes that $\mathbb{P}_\mu[\{(2,1),(2,2)\}]=0$ implies somehow that this set should not belong to $\mathcal{F}_0$, but I think this is not correct.

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    $\begingroup$ Working with the canonical process, the sigma field does not depend on the measure indeed : you are right. $\endgroup$ – Olivier Aug 14 at 9:21
  • $\begingroup$ This being said, one could also consider the sigma-algebra generated by the measurable sets with non-null measure if we want to force the measure into, and perhaps that's what the authors have in mind... $\endgroup$ – Olivier Aug 14 at 9:26
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If we take $\mu$ to be the distribution of $X_0$, then $\sigma(X_0)$ depends on $\mu$ already: $\sigma(X_0)$ is the smallest sigma-algebra for which $X_0$ is measurable. Assume $X_0\equiv 1$ ($\mu$ is a point measure at 1), without loss of generality. Then $\sigma(X_0)=\{\Omega,\emptyset\}$. Assume $X_0\in\{1,2\}$, and $A=X_0^{-1}(1)$, then $\sigma(X_0)$ must contain at least the elements $\{\Omega,A,A^c,\emptyset\}$.

Hence, $\mathcal{F}_n$ depends on $\mu$.

Update: I believe I now better understand the question: If the RVs are already defined, they must already be measurable functions between two measure spaces. But using the generator operator $\sigma$ delivers the smallest sigma-algebra w.r.t. which the function/RV is measurable. So, if we define

$$ X_0{:}\ (\Omega,\mathcal{B}(\Omega)) \to (\mathbb{R},\mathcal{B}(\mathbb{R})) $$

but then say (as you do) that only one state of $X_0$ occurs, a smaller sigma-algebra might suffice.

Let us take your example with $E=\{1,2\}$, $\Omega=E\times E$, but now define $X_0((a,b))=a+b$. In this most general case, $\mathcal{F}_0$ is generated by $\{\emptyset,\Omega,\{(1,1)\},\{(2,1),(1,2)\},\{(2,2)\}\}$. But if you assume that $X_0\equiv 2$ (by properly choosing $\mu$), then $\mathcal{F}_0$ is generated by the smaller familiy of sets $\{\emptyset,\Omega,\{(1,1)\},\{(2,1),(1,2),(2,2)\}\}$. This was not apparent from your example, because you treated only two possible states for $X_0$.

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    $\begingroup$ +1. Why is this downvoted? Being short and simple are plusses for an answer when it is to the point, no? $\endgroup$ – Did Jun 2 '15 at 9:16
  • $\begingroup$ I have updated my question with an example which fits to your example and shows why $\sigma(X_0)$ should not depend on $\mu$. @Did I dont know - I have neither downvoted nor upvoted anything so far. $\endgroup$ – Adam Jun 2 '15 at 9:33
  • $\begingroup$ Here, $X_n(\omega)=x_n$, this is the key assumption you miss in the second example. $\endgroup$ – Olivier Aug 14 at 9:23
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$X_0$ is the random variable variable with measure $\mu$ so $\mathcal{F_n}$ does depend on its definition.

If $\mu$ was a discrete random variable it could give very different possible paths based on the number of outcomes. If $\mu$ allowed for one only one outcome $0$ then the minimum sigma field wrt which it is measurable is $\{\phi, \{0\}, Z \ \{0\}, Z\}$ where $Z$ is the state space. In comparison, a continuous random variable might allow for many, many more outcomes still, i.e. a much large sigma algebra.

To be concrete, compare Brownian Motion with a fixed initial condition, versus normally distributed initial conditions.

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  • $\begingroup$ The measure $\mu$ only gives a weight to a specific path. I don't see how this changes $\mathcal{F}_n$. For instance, $\mathcal{F}_0$ is the smallest sigma algebra, such that for any $A\in \Sigma$ we have $X_0^{-1}(A)\in \mathcal{F}_0$ this is obviously independet of $\mu$, because $X_0$ is already defined as a projection, right? $\endgroup$ – Adam Jun 2 '15 at 8:35
  • $\begingroup$ @Adam I updated to clarify the point. $\endgroup$ – muaddib Jun 2 '15 at 8:40
  • $\begingroup$ I still do not understand your point. I have updated my question with an example to clarify why I think $\mathcal{F}_n$ is independent of $\mu$. $\endgroup$ – Adam Jun 2 '15 at 9:26
  • $\begingroup$ @Adam, I'm not sure what to say, I don't really see why you don't think $X_0$ changes with the definition of $\mu$. $\endgroup$ – muaddib Jun 2 '15 at 9:34
  • $\begingroup$ In order to assure that $X_0$ is measurable, you also need that $X_0^{-1}(\{2\}) \in \mathcal{F}_0$, right? Also note that my defined $\mathcal{F}_0$ is not maximal, since $\{(1,1),(2,2)\} \notin \mathcal{F}_0$. $\endgroup$ – Adam Jun 2 '15 at 9:37

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