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A Markov chain $(X_i)_{i\in \mathbb{N}}$ on a measurable space $(E,\Sigma)$ is (see e.g. Revuz or Meyn/Tweedie) constructed on the following probabilty space. $$ \Omega = \{ (x_l)_{l \in \mathbb{N}} \mid x_l \in E \text{ for all } l \in \mathbb{N} \}$$ and $X_i$ is definied as $$X_i ((x_l)_{l \in \mathbb{N}}) = x_i$$ and $\Sigma$ is constructed suitable. Then one can show, for any Markov kernel p and probability measure $\mu$ exists a a probability measure $\mathbb{P}_\mu$ on $\Omega$ with $$ \mathbb{P}_\mu [X_0 \in A_0, X_1 \in A_1 ,\dots X_n \in A_n] =$$ $$ \int_{A_0} \dots \int_{A_n} p(y_{n-1},A_n) \, p(y_{n-2}, dy_{n-1}) \dots p(y_0, dy_1) \, \mu(dy_0).$$ Thus by this particular construction for all Markov chains the space $(\Omega,\Sigma)$ and the random variable $(X_i)_{i \in \mathbb{N}}$ are identical and two Markov chains only differ in the associated probability measure $\mathbb{P}_\mu$.

My question: Assume two measures $\mu_1$ and $\mu_2$ and two Markov kernels $p_1$ and $p_2$ are given. Then I have two seperate probability measures $\mathbb{P}_{\mu_1}$ and $\mathbb{P}_{\mu_2}$ where the projection process lives on. Can I conclude from this that there exits a single probability space $(\hat{\Omega},\hat{\Sigma},\mathbb{P})$ together with two stochastic processes $(Z_i)$ and $(Y_i)$ with $$ \mathbb{P} [Z_0 \in A_0, Z_1 \in A_1 ,\dots Z_n \in A_n] = \mathbb{P}_{\mu_1} [X_0 \in A_0, X_1 \in A_1 ,\dots X_n \in A_n]$$

and $$ \mathbb{P} [Y_0 \in A_0, Y_1 \in A_1 ,\dots Y_n \in A_n] = \mathbb{P}_{\mu_2} [X_0 \in A_0, X_1 \in A_1 ,\dots X_n \in A_n]?$$

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  • $\begingroup$ If I'm understanding the question correctly: if you assume $(E, \Sigma)$ is standard Borel, then the answer is certainly yes, and you don't need the restriction $\nu \ll \mu$ either. If $E$ is standard Borel then so is $\Omega$, and every probability measure on a standard Borel space (e.g. the law of any $E$-valued process) is a pushforward of, say, Lebesgue measure on $[0,1]$. $\endgroup$ Sep 3 '15 at 3:59
  • $\begingroup$ @NateEldredge I have edited my question to make it more clear. I am interested to construct two stochastic processes on the same probability space, is this possible? $\endgroup$
    – Adam
    Sep 3 '15 at 9:52
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The answer is yes: both processes can be constructed simultaneously on the same probability space. This is a standard fact (true in much greater generality - look up "product measure" and "Kolmogorov extension theorem" on Wikipedia and in any graduate-level textbook on probability theory).

To add a bit of detail, the space $\hat{\mathbb{P}}$ can be constructed as the product space of $(\Omega,\mathbb{P}_{\mu_1})$ and $(\Omega,\mathbb{P}_{\mu_1})$ (with the accompanying $\sigma$-algebras, which I will omit). That is, we define $$ \hat{\Omega} = \Omega \times \Omega $$ where $\Omega = \{ (x_l)_{l \in \mathbb{N}}\ |\ x_l \in E \textrm{ for all }l\in\mathbb{N}\}$ as in the question, and we equip $\hat{\Omega}$ with the obvious $\sigma$-algebra together with the unique probability measure $\hat{\mathbb{P}}$ satisfying $$ \hat{\mathbb{P}}(A\times B) = \mathbb{P}_{\mu_1}(A) \mathbb{P}_{\mu_2}(B) $$ for all $A,B \subset \Omega$ that belong to the $\sigma$-algebra $\Omega$ comes equipped with. (The existence and uniqueness of such a measure is a standard result in measure theory.) If you now define stochastic processes $(Z_i)_{i\in\mathbb{N}}$ and $(Y_i)_{i\in\mathbb{N}}$ on $\hat{\Omega}$ by $$ Z_i( (a_l)_{l\in\mathbb{N}},(b_l)_{l\in\mathbb{N}} ) = a_i, $$ $$ Y_i( (a_l)_{l\in\mathbb{N}},(b_l)_{l\in\mathbb{N}} ) = b_i, $$ (noting that a generic element of $\hat{\Omega}$ is of the form $((a_l)_{l\in\mathbb{N}},(b_l)_{l\in\mathbb{N}})$ ), then it is easy to check that $(Z_i)_{i\in\mathbb{N}}$ is a realization of the Markov chain with distribution $\mathbb{P}_{\mu_1}$ and $(Y_i)_{i\in\mathbb{N}}$ is a realization of the Markov chain with distribution $\mathbb{P}_{\mu_2}$, in the sense mentioned in the question.

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