# Definition of Markov Processes and transition probability

Im getting a bit confused with the terminology of continous time markov processes. Now let $\{X_t | t \geq 0 \}$ be a stochastic Process with state space $\mathbb{R}$, and $(F_t)$ be its canonical Filtration My professor says X is a markov process, if:
$\mathbb{P}(X_t \in A | F_s)=\mathbb{P}(X_t \in A |X_s)$ for all $s \leq t$.

Now i was wondering what does this imply about the transition probability?
We can define functions $\mathbb{P}(X_t \in A | X_s=x)=\mathbb{P}_{X_t|X_s=x}(A)$. I dont know what these functions are called in english, maybe conditional distribution? If we fixed $t_1 < t_2 <....< t_n$ would the markov property given above imply:
$\mathbb{P}(X_{t_n} \in A | X_{t_{n-1}},...., X_{t_1})=\mathbb{P}(X_{t_n} \in A | X_{t_{n-1}})$ as functions in $x_1,...,x_{n-1}$?

Yes, it is true that $\mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}},\dots,X_{t_{1}}\} = \mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n}}\}$. This follows from the fact that $\sigma(X_{t_{n - 1}},\dots,X_{t_{1}}) \subseteq \mathcal{F}_{t_{n - 1}}$ and $\sigma(X_{t_{n}}) \subseteq \sigma(X_{t_{n - 1}},\dots,X_{t_{1}})$. In particular, if $A_{n - 1},\dots,A_{1}$ are Borel sets, then \begin{align*} \mathbb{P}\{X_{t_{n}} \in A, X_{t_{n - 1}} \in A_{n - 1},\dots,X_{t_{1}} \in A_{1}\} &= \mathbb{E}\left(\mathbb{P}\{X_{t_{n}} \in A \, \mid \, \mathcal{F}_{t_{n -1}}\} : X_{t_{n - 1}} \in A_{n - 1},\dots,X_{t_{1}} \in A_{1}\right) \\ &= \mathbb{E}\left(\mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}}\} : X_{t_{n - 1}} \in A_{n - 1},\dots,X_{t_{1}} \in A_{1}\right). \end{align*}
Since $\mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}}\}$ is $\sigma(X_{t_{n - 1}},\dots,X_{t_{1}})$-measurable, this proves that $\mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}}, \dots, X_{t_{1}}\} = \mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}}\}$ almost surely. More equivalent definitions of the Markov property can be found in the textbook by Chung and Walsh.
• Hello, thank you for your answer. My question regarding conditional distribution was as follows: Does for the conditional distributions hold: $\mathbb{P}_{X_{t_n}|X_{t_{n-1}},...., X_{t_1}}((x_1,....,x_{n-1}),A)$=$\mathbb{P}_{X_{t_n}|X_{t_{n-1}}}((x_{n-1}),A)$ – StefanWK Jun 19 '18 at 15:21
• Yes, that's true by the definition of the conditional distribution and the fact that $\mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}},\dots,X_{t_{1}}\} = \mathbb{P}\{X_{t_{n}} \in A \, \mid \, X_{t_{n - 1}}\}$. – fourierwho Jun 19 '18 at 17:07