Markov Property and FDDs

Question

Let $X,Y$ be two discrete time $\mathbb{R}^n$-valued stochastic processes with the same finite dimensional distributions. It may be that $X,Y$ are defined on two different probability spaces. Now, if $X$ is Markovian, i.e., $P(X_{t+1}\in A \mid \sigma\{X_s, s \leq t\}) = P(X_{t+1}\in A \mid X_t)$ for every $A \in \mathbb{B}(\mathbb{R}^n)$ then is it true that $Y$ is also Markovian?

Answer 1

Yes, that's true. Recall the following characterization of conditional expectation.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and, let $\mathcal{G} \subseteq \mathcal{F}$ be a $\cap$-stable generator of $\mathcal{F}$. For any two random variables $X,Y \in L^1(\mathbb{P})$ the following two statements are equivalent:

1. $Y$ is a version of $\mathbb{E}(X \mid \mathcal{F})$
2. $Y$ is $\mathcal{F}$-measurable and $$\forall G \in \mathcal{G}: \qquad \int_G Y \, d\mathbb{P} = \int_G X \, d\mathbb{P}.$$

Now let $(X_t)_{t \in \mathbb{N}}$, $(Y_t)_{t \in \mathbb{N}}$ be two discrete-time stochastic processes which have the same finite-dimensional distributions, and assume that$(X_t)_{t \in \mathbb{N}}$ is Markovian in the sense that

$$\mathbb{P}(X_{t+1} \in A \mid \mathcal{F}_t^X) = \mathbb{P}(X_{t+1} \in A \mid X_t), \qquad t \in \mathbb{N}, A \in \mathcal{B}(\mathbb{R}^n) \tag{1}$$

for the canonical filtration $\mathcal{F}_t^X := \sigma(X_s; s \leq t)$. For fixed $t \in \mathbb{N}$ and $A \in \mathcal{B}(\mathbb{R}^n)$ there exists, by the factorization lemma, a Borel measurable mapping $h$ such that $$\mathbb{P}(X_{t+1} \in A \mid X_t) = h(X_t).$$ Since $(X_t)_t$ and $(Y_t)_t$ have the same finite-dimensional distributions, we find that $$\int_{\{Y_{t} \in B\}} h(Y_t) \, d\mathbb{P} = \int_{\{X_{t} \in B\}} h(X_t) \, d\mathbb{P} = \int_{\{X_{t} \in B\}} 1_A(X_{t+1}) \, d\mathbb{P} = \int_{\{Y_{t} \in B\}} 1_A(Y_{t+1}) \, d\mathbb{P}$$ for any Borel set $B$, and so we conclude that $$\mathbb{P}(Y_{t+1} \in A \mid Y_t) = h(Y_t). \tag{2}$$ This implies $$\begin{align*} \int_{(Y_{s_1},\ldots,Y_{s_k}) \in B_1 \times \ldots \times B_k} 1_A(Y_{t+1}) \, d\mathbb{P} &= \int_{(X_{s_1},\ldots,X_{s_k}) \in B_1 \times \ldots \times B_k} 1_A(X_{t+1}) \, d\mathbb{P} \\ &\stackrel{(1)}{=} \int_{(X_{s_1},\ldots,X_{s_k}) \in B_1 \times \ldots \times B_k} \mathbb{P}(X_{t+1} \in A \mid X_t) \, d\mathbb{P} \\ &= \int_{(X_{s_1},\ldots,X_{s_k}) \in B_1 \times \ldots \times B_k} h(X_t) \, d\mathbb{P} \\ &= \int_{(Y_{s_1},\ldots,Y_{s_k}) \in B_1 \times \ldots \times B_k} h(Y_t) \, d\mathbb{P} \\ &\stackrel{(2)}{=} \int_{(Y_{s_1},\ldots,Y_{s_k}) \in B_1 \times \ldots \times B_k} \mathbb{P}(Y_{t+1} \in A \mid Y_t) \, d\mathbb{P} \end{align*}$$ for any $k \in \mathbb{N}$, $s_1 \leq \ldots \leq s_k \leq t$ and Borel sets $B_j$. Since

$$\mathcal{G} := \left\{ \{(Y_{s_1},\ldots,Y_{s_k}) \in B_1 \times \ldots \times B_k\}; k \in \mathbb{N}, s_1 \leq \ldots \leq s_k \leq t, B_j \in \mathcal{B}(\mathbb{R}^n)\right\}$$

is a $\cap$-stable generator of the canonical filtration $\mathcal{F}_t^Y$, it follows from the characterization of conditional expectation that

$$\mathbb{P}(Y_{t+1} \in A \mid Y_t) = \mathbb{P}(Y_{t+1} \in A \mid \mathcal{F}_t^Y).$$

Remark: The same reasoning works if the processes are defined on different probability spaces.

Answer 2

Yes. Markov property is equivalent to the statement $P\{X_{t+1} \in A, B\}=\int_B P\{X_{t+1} \in A|X_t)\}$ for every set $B$ of the type $\{X_{s_{1}}^{-1} (E_1)\cap X_{s_{2}}^{-1} (E_2)\cap...\cap X_{s_{k}}^{-1} (E_k)\}$ where $k$ is a positive integer, $E_i$'s are Borel sets in $\mathbb R$, $s_i$'s are points in the indexing set. This condition depends only on finite dimensional distributions, so if it holds for $\{X_t\}$ it holds for $\{Y_t\}$ .