When is a stopped Brownian motion not a markov process? Let $W_t,t\geq 0$  be a Brownian motion start from zero and $\tau$ a stopping time. For what kinds of $\tau$, $W_{t \wedge \tau}$ is not a markov process?
I have considered choices of constant stopping time, like $\tau=1$; and the hitting time, but they do not work well.
 A: Example: How about $\tau:=\inf\{t>T_1: W_t=0\}$, where $T_1:=\inf\{t>0: W_t = 1\}$. In words, $\tau$ is the time of first return to state $0$ after the first visit to  state 1.
The process $X_t:=W_{t\wedge\tau}$ is certainly not a strong Markov process because state $0$ turns into a trap after the first visit to state $1$. It's not even a simple Markov process because for a bounded measurable $f$ and $0<s<t$,
$$
E[f(X_t)\mid\mathcal F^X_s]=1_{\{\tau\le s\}} f(0)+1_{\{\tau>s\}}h(X_s,t-s),
$$
where $h(x,u):=E^x[f(W_{u\wedge T_0})]$, $T_0$ is the hitting time of $0$ by $W$, and $E^x$ is expectation relative to the Brownian motion started at $x$. The above display shows that the conditional distribution of $X_t$, given the history up to time $s$, depends not just on $X_s$ but also on the event $\{\tau\le s\}$.
A stopping time $T$ of $W$ is called a terminal time if $T(\omega)=t+T(\theta_t\omega)$ for all $\omega$ in the event $\{t<T\}$, for each $t\ge 0$. Here $\theta_t$ is the shift operator acting on the sample space of the Brownian motion by $W_u(\theta_t\omega) = W_{u+t}(\omega)$. A hitting time like $T_A:=\inf\{t>0:W_t\in A\}$ is a terminal time. If $T$ is a terminal time then the stopped process $W_{t\wedge T}$ will be a Markov process.
