I have a question concerning the strong Markov property:
For a strong Markov process $(X_u)_{u\ge 0}$, a real time $t\in \mathbb{R}$ and an optional stopping time $T$ with $t< T$
\begin{align*} \mathcal{P}((X_u)_{u\ge T}\in B \mid (X_u)_{u\le t}\in A , X_T=x )\\ = \mathcal{P}((X_u)_{u\ge T}\in B\mid X_T=x)\quad, a.s. \end{align*}
How can I show that
\begin{align*} \mathcal{P}((X_u)_{u\le t}\in A, (X_u)_{u\ge T}\in B \mid X_T=x)\\ = \mathcal{P}((X_u)_{u\le t}\in A\mid X_T=x) * \mathcal{P}((X_u)_{u\ge T}\in B\mid X_T=x)) \quad a.s.? \end{align*}
Is this even true?
Thanks and regards!
If you are familiar with $\sigma$-algebras, then you could start by demonstrating that $$ P(F\cap G|{\mathcal C}) =P(F|{\mathcal C})\cdot P(G|{\mathcal C}),\quad\forall F\in{\mathcal A},G\in{\mathcal B} $$ is equivalent to $$ P(G|{\mathcal C}\vee{\mathcal A}) =P(G|{\mathcal C}),\quad\forall G\in{\mathcal B}. $$ Here ${\mathcal A}, {\mathcal B},{\mathcal C}$ are three $\sigma$-algebras of events in your probability space, and ${\mathcal C}\vee{\mathcal A}$ is the least $\sigma$-algebra containing both ${\mathcal C}$ and ${\mathcal A}$. (The equalities in both displays above should be understood to hold almost surely.)
Having shown this, consider the special case in which ${\mathcal A}=\sigma\{X_{\min(t,T)}: t\ge 0\}$, ${\mathcal B}=\sigma\{X_{T+s}: s\ge 0\}$, and ${\mathcal C} =\sigma\{X_T\}$. In this way the strong Markov propery is seen to be equivalent to the time-symmetric conditional independence property in your second display.