• $(\Omega,\mathcal A,\operatorname P)$ 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

$$\begin{align*}\kappa^{m+1}(X_0,B) &= (\kappa \circ \kappa^m)(X_0,B) \\ &= \int_E P[X_1\in B | X_0=y] P[X_m \in dy | X_0] \\ &= \int_E P[X_{m+1} \in B|X_m=y]P[X_m\in dy|X_0] \\ &= P[X_{m+1} \in B | X_0]\end{align*}$$

  • $\begingroup$ Thank you for your answer! Please take note of my edit. So do you think the claim is true? I'm wondering whether I'm missing something, since the argumentation seems to imply that a stationary process is always Markov. $\endgroup$ – 0xbadf00d Aug 10 '19 at 19:02
  • $\begingroup$ I don't see the Markov property showing up anywhere, actually. The composition of the Markov kernels "integrates out" the intermediate random variables, so the composition itself never has a long memory. I don't have the background to definitively answer your new claim, but my gut says that there should be a stationary, non-Markov, process. $\endgroup$ – Brian Moehring Aug 10 '19 at 19:23
  • $\begingroup$ How do you obtain, for example, the last equality? If $(X_n)_{n\in\mathbb N_0}$ is Markov, then $$\operatorname P\left[X_{m+1}\in B\mid X_m,\ldots,X_0\right]=\operatorname P\left[X_{m+1}\in B\mid X_m\right]=\kappa(X_m,B)$$ almost surely for all $B\in\mathcal E$. The first equality seems to be crucial, since it allows us to write $$\text P\left[X_0\in A,X_{m+1}\in B\right]=\text E[\text P[X_0\in A,X_{m+1}\in B\mid X_0,\ldots,X_m]]=\text E[1_A(X_0)\text P[X_{m+1}\in B\mid X_0,\ldots,X_m]]=\text E[1_A(X_0)\kappa(X_m,B)]$$ for all $A,B\in\mathcal E$. $\endgroup$ – 0xbadf00d Aug 11 '19 at 4:52
  • $\begingroup$ BTW: Note that you need regular versions of the conditional distributions occurring in your integrals (such as $\kappa$ for the conditional distribution of $X_1$ given $X_0)$ in order for them to be well-defined. $\endgroup$ – 0xbadf00d Aug 11 '19 at 4:52
  • $\begingroup$ I think that you're assuming that $X_{m+1}$ is conditionally independent of $X_0$ given $X_m$ in your last equation. See this question: math.stackexchange.com/q/3319882/47771 $\endgroup$ – 0xbadf00d Aug 11 '19 at 9:44

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