# Prove that Brownian Motion absorbed at the origin is Markov

I’m trying to prove that Brownian motion absorbed at the origin is a Markov process with respect to the original filtration $\{\mathcal{F}_{t}\}$. To be more specific, let $(B_{t},\mathcal{F}_{t})_{t \geq 0}$ be a 1-d brownian motion such that $B_{0}=x,\ a.s. \mathbb{P}\$ for some $x>0.\$ Let $T_{0}=\inf\{ t \geq 0, \ B_{t}=0\}$ be the hitting time of $0$, and define the Brownian motion absorbed at origin to be the process $\{B_{t \land T_{0}}, \mathcal{F}_{t} \}$. I want to show that such a process is Markov w.r.t. $\mathcal{F_{t}}, \$ i.e. $$\mathbb{P}[B_{t \land T_{0}} \in \Gamma | \mathcal{F}_{s}]=\mathbb{P}[B_{t \land T_{0}} \in \Gamma |B_{s \land T_{0}}]$$ for every pair of $0<s<t$. And further more it has transition density given by: $$\mathbb{P}[B_{(t+s) \land T_{0}} \in dy |B_{s \land T_{0}}=z]=\mathbb{P}^{z}[B_{t} \in dy, \ T_{0} >t], \ \forall y,z >0,$$ where $\mathbb{P}^{z}$ is the probability measure under which $B$ starts from $z$ almost surely.

I tried to use the strong Markov property for Brownian motion to derive these equalities, but I’m having a hard time dealing with $T_{0}$. For example, if $0 \notin \Gamma$, then$\{B_{t \land T_{0}} \in \Gamma\}=\{B_{t} \in \Gamma, T_{0} >t\}$. But how to apply strong Markov property to compute $\mathbb{P}[B_{t} \in \Gamma, T_{0} >t\ | \mathcal{F}_{s} ]\$?

Any answer or comment is greatly appreciated, thanks!

#### Solutions Collecting From Web of "Prove that Brownian Motion absorbed at the origin is Markov"

For every $r$ let $\theta_r$ denote the time-shift by $r$. Then,
$$[B_t\in\Gamma,T_0\gt t]=[T_0\gt s,B_s+(B_t-B_s)\in\Gamma,T_0\circ\theta_s\gt t-s],$$
where $[T_0\gt s]$ and $B_s$ are $\mathcal F_s$-measurable and $B_t-B_s$ is independent of $\mathcal F_s$ and distributed like $B_{t-s}$. Hence,
$$\mathbb P^x(B_t\in\Gamma,T_0\gt t\mid\mathcal F_s)=\mathbf 1_{T_0\gt s}\cdot u_{t-s}(B_s),$$
where, for every $r$ and $y$,
$$u_r(y)=\mathbb P^y(B_r\in\Gamma, T_0\gt r).$$
Likewise,
$$[B_{t\wedge T_0}\in\Gamma]=[B_t\in\Gamma,T_0\gt t]\cup[0\in\Gamma,T_0\lt t],$$
hence
$$\mathbb P^x(B_{t\wedge T_0}\in\Gamma\mid\mathcal F_s)=\mathbf 1_{T_0\gt s}\cdot u_{t-s}(B_s)+\mathbf 1_{0\in\Gamma}\cdot(1-\mathbf 1_{T_0\gt s}\cdot v_{t-s}(B_s)).$$
where, for every $r$ and $y$,
$$v_r(y)=\mathbb P^y(T_0\gt r).$$