Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $I\subseteq\mathbb R$
- $(\mathcal F_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$
- $(E,\mathcal E)$ be a measurable space
- $X$ be a $(E,\mathcal E)$-valued $\mathcal F$-Markov process with transition semigroup $\left(\kappa_{s,\:t}:s,t\in I\text{ with }s\le t\right)$
If $Y$ is a random variable on $(\Omega,\mathcal A,\operatorname P)$, let $$\mathfrak L(Y):=\operatorname P\circ Y^{-1}.$$
Let $n\in\mathbb N_0$ and $t_0,\ldots,t_n\in I$ with $t_0\le\cdots\le t_n$. We can show that $$\mathfrak L\left(X_{t_0},\ldots,X_{t_n}\right)=\mathfrak L\left(X_{t_0}\right)\otimes\bigotimes_{i=1}^n\kappa_{t_{i-1},\:t_i}\tag1.$$ I want to conclude $$\operatorname P\left[\left(X_{t_1},\ldots,X_{t_n}\right)\in\;\cdot\;\mid\mathcal F_{t_0}\right]=\bigotimes_{i=1}^n\kappa_{t_{i-1},\:t_i}\left(X_{t_0},\;\cdot\;\right).\tag2$$
From $(1)$, we obtain \begin{equation}\begin{split}\operatorname P\left[\left(X_{t_0},\ldots,X_{t_n}\right)\in B\times C\right]&=\int_B\operatorname P\left[X_{t_0}\in{\rm d}x_0\right]\bigotimes_{i=1}^n\kappa_{t_{i-1},\:t_i}(x_0,C)\\&=\operatorname E\left[1_{\left\{\:X_{t_0}\:\in\:B\:\right\}}\bigotimes_{i=1}^n\kappa_{t_{i-1},\:t_i}\left(X_{t_0},C\right)\right]\end{split}\tag3\end{equation} for all $B\in\mathcal E$ and $C\in\mathcal E^{\otimes n}$.
Why does $(2)$ immediately follow from $(3)$?
I guess it's some kind of conditional independence argument needed.
For a fixed measurable set $C$ and $t_0 \leq \ldots \leq t_n$ define
$$\begin{align*} U &:= 1_{\{(X_{t_1},\ldots,X_{t_n}) \in C\}} \quad \text{and} \quad V := \bigotimes_{i=1}^n \kappa_{t_{i-1},t_i}(X_{t_0},C). \end{align*}$$
Clearly, $(3)$ shows that
$$\mathbb{E}(1_{\{X_{t_0} \in B\}} U) = \mathbb{E}(1_{\{X_{t_0} \in B\}} V)$$
for any $B \in \mathcal{E}$. Since the sets $\{X_{t_0} \in B\}$, $B \in \mathcal{E}$, generate $\sigma(X_{t_0})$ and $V$ is $\sigma(X_{t_0})$-measurable, this implies that
$$\mathbb{E}(U \mid X_{t_0}) = V,$$
i.e.
$$\mathbb{P}((X_{t_1},\ldots,X_{t_n}) \in C \mid X_{t_0}) = \bigotimes_{i=1}^n \kappa_{t_{i-1},t_i}(X_{t_0},C). \tag{4}$$
Finally, we note that the Markov property gives
$$\mathbb{P}((X_{t_1},\ldots,X_{t_n}) \in C \mid \mathcal{F}_{t_0})= \mathbb{P}((X_{t_1},\ldots,X_{t_n}) \in C \mid X_{t_0}) \tag{5},$$
and combining this with $(4)$ this proves the assertion.