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I found in the book Markov Chains by Revuz the following definition of a Markov chain. In the following $(X_n)_{n \in \mathbb{N}}$ is a sequence of random variables on a probability space $(\Omega,\mathcal{F},P_0)$ and the range is a measurable space $(E,\Sigma)$, and $\mathcal{F}_n^0 =\sigma(X_m , m \leq n)$

Definition.1 $(X_n)_{n \in \mathbb{N}}$ is a Markov chain if for every $B \in \Sigma$ we have $$P_0 [ X_m \in A \mid \mathcal{F}_n^0 ] = P_0[ X_m \in A \mid X_n] $$ for all $m\geq n$

For a transition probability $P : E \times \Sigma \to [0,1]$, i.e $x \to p(x,A)$ is measurable for all $A\in \Sigma$ and $A\to p(x,A)$ is a probability measure for all $x\in E$ he defines a family of transition probabilities $(P_n)_{n \geq 0}$ by $$ P_{n} (x,A) = \int_E p(y,A) \, P_{n-1}(x,dy) $$ and $$ P_0 := P.$$ Then he defines a homogeneous Markov chain by

Definition 2. $(X_n)_{n \in \mathbb{N}}$ is called a homogeneous Markov chain with respect to transition probability $P$ if $$ P_0[X_m \in A \mid \mathcal{F}_n^o] = P_{n-m}(x,\{X_m \in A\}).$$

I would like to know if $(X_n)$ is a Markov chain with $$ P_0[X_n \in B \mid X_0 \in A] = P_0[X_{n+m} \in B \mid X_m \in A] $$ for all $n,m\geq 0$ and $A,B \in \Sigma$, does this imply that one can find a transition probability $P$ such that $(X_n)$ is a homogeneous Markov chain with respect to $P$?

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Thankfully, YES :)

You can very simply define the needed probability transition by :

\begin{equation} P : x, \mathcal{A} \mapsto P_0[X_1 \in \mathcal{A} | X_0=x] \end{equation}

You can then deduce from the definition of the family $(P_n)$ that :

\begin{equation} P_n ( x, \mathcal{A} ) = P_0[X_{n+1} \in \mathcal{A} | X_0=x] \end{equation}

!

Then verify that your Markov chain satisfies the condition to be homogeneous with respect to $(P_n)$, by using the definition of a Markov chain and the property.

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