2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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, 2015 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, 2015 at 9:52

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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