Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(E,\mathcal E)$ be a measurable space
- $(X_n)_{n\in\mathbb N_0}$ be an $(E,\mathcal E)$-valued stationary process on $(E,\mathcal E)$
- $\kappa$ be a Markov kernel on $(E,\mathcal E)$ with $$\operatorname P\left[X_1\in B\mid X_0\right]=\kappa(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal E\tag1$$
- $\mathcal L\left(X_{n_1},\ldots,X_{n_k}\right)$ denote the joint distribution of $X_{n_1},\ldots,X_{n_k}$ for $k\in\mathbb N$ and $n_1<\cdots<n_k$
Let $n\in\mathbb N_0$. I want to show that$^1$ $$\mathcal L(X_0,X_n)=\mathcal L(X_0)\otimes\kappa^n.\tag2$$
By $(1)$, $$\mathcal L(X_0,X_1)=\mathcal L(X_0)\otimes\kappa\tag3.$$ By stationarity, $$\mathcal L(X_{n-1},X_n)=\mathcal L(X_0,X_1)=\mathcal L(X_{n-1})\otimes\kappa\;\;\;\text{for all }n\in\mathbb N\tag4.$$ We conclude from $(4)$ that $$\operatorname P\left[X_n\in B\mid X_{n-1}\right]=\kappa(X_{n-1},B)\;\;\;\text{almost surely}\tag5.$$
Can we show the desired claim from $(5)$ or do we need to assume that $(X_n)_{n\in\mathbb N_0}$ is Markov, i.e. $$\operatorname P\left[X_n\in B\mid X_0,\ldots,X_m\right]=\operatorname P\left[X_n\in B\mid X_m\right]\;\;\;\text{almost surely for all }B\in\mathcal E\tag6$$ for all $m,n\in\mathbb N_0$ with $m\le n$?
Maybe I'm missing something, but if we don't need to assume that $(X_n)_{n\in\mathbb N_0}$ is Markov, it seems like the claim implies the Markov property and hence every stationary process is Markov. Is this actually true? (It's clear that every i.i.d. process is both Markov and stationary, but not every stationary process is i.i.d.)
EDIT: Please take note my related question: Are we able to show that $\text P[X_2\in B_2\mid X_0]=\int\text P[X_2\in B_2\mid X_1=x_1]\text P[X_1\in{\rm d}x_1\mid X_0]$?
Let $\kappa_1\otimes\kappa_2$ and $\kappa_1\kappa_2$ denote the product and composition of transition kernels $kappa_1$ and $\kappa_2$, respectively. Clearly, $\kappa^n:=\kappa^{n-1}\kappa$ and $\kappa^0$ is the Dirac kernel.
Presumably, it would follow by induction on $n$ as follows:
Assume $\kappa^m(X_0,B) = P[X_m \in B | X_0]$ for almost all $B \in \mathcal{E}$. Then
$$\begin{align*}\kappa^{m+1}(X_0,B) &= (\kappa \circ \kappa^m)(X_0,B) \\ &= \int_E P[X_1\in B | X_0=y] P[X_m \in dy | X_0] \\ &= \int_E P[X_{m+1} \in B|X_m=y]P[X_m\in dy|X_0] \\ &= P[X_{m+1} \in B | X_0]\end{align*}$$