I have been asked to prove that the Brownian motion absorbed at the origin is a Markov process.
Formally, let $B_t^x$ be a Brownian motion originating from $x>0$ and let $\tau^x_0 = \inf\{t>0 : B^x_t = 0\}$ be the first passage from the origin. We need to show that the process $$ W_t^x := B^x_{t \land \tau_0^x} = \begin{cases} x + B_t & \text{if } t \le \tau_0^x\\ 0 & \text{if } t > \tau_0^x \end{cases}$$ is a markov process.
Now, intuitively the claim looks fine and the reasoning should be something like:
if we sampled the process up to time $s$ then, if $s \ge \tau^x_0$, i.e. if the process already reached $0$, then of course we do not need the whole filtration to conclude that it will remain in $0$ - the $\sigma-$algebra generated by $W_s^x$ will be enough.
If instead $s < \tau_0^x$ the process is a actually a Brownian motion, so again the the $\sigma-$algebra generated by $W_s^x$ will be enough to estimate the future evolution (as we know the Brownian motion is Markov).
I have some doubts on how to formalize this reasoning:
we want to prove that $ \mathbb{P}^{s,x}(W_t \in A | \mathcal{F}_s) = \mathbb{P}^{s,x}(W_t \in A | W_s)$. I would write something like
$$\mathbb{P}^{s,x}(W_t \in A ~|~ \mathcal{F}_s) =\\ \mathbb{P}^{s,x}(W_t \in A,~ s < \tau_0^x~ |~ \mathcal{F}_s) + \mathbb{P}^{s,x}(W_t \in A,~ s\ge \tau_0^x ~|~ \mathcal{F}_s) $$
first term:
$$ \mathbb{P}^{s,x}(W_t \in A,~ s < \tau_0^x~ |~ \mathcal{F}_s) = \mathbb{P}^{s,x}(W_t \in A~ |~\mathbb{1}_{\{s < \tau_0^x\}}, \mathcal{F}_s)\cdot \mathbb{P}^{s,x}(s < \tau_0^x ~ | ~ \mathcal{F}_s) =\\ \mathbb{P}^{s,x}(W_t \in A ~|~\mathbb{1}_{\{s < \tau_0^x\}}, \mathcal{F}_s) \cdot \mathbb{1}_{\{s < \tau_0^x\}} $$
I don't continue with the proof as it's not really short. Here my only doubt is: do I need to show that, if $s < \tau_0^x$ , then $\mathcal{F}_s = \mathcal{F}_x^B$, i.e. the filtration is the same as the one generated by the underlying brownian motion? (and, in particular, that conditioning w.r.t. the brownian motion at time $s$ is the same that conditioning w.r.t. to $W_s$). If yes, how can I do it? Is it enough to say that the process before $\tau_0^x$ is exactly just a brownian motion?
second term:
$$ \mathbb{P}^{s,x}(W_t \in A,~ s \ge \tau_0^x~ |~ \mathcal{F}_s) = \mathbb{P}^{s,x}(W_t \in A~ |~\mathbb{1}_{\{s \ge \tau_0^x\}}, \mathcal{F}_s)\cdot \mathbb{P}^{s,x}(s \ge \tau_0^x ~ | ~ \mathcal{F}_s) =\\ \delta_0(A) \cdot \mathbb{1}_{\{s \ge \tau_0^x\}} = \mathbb{P}^{s,x}(W_t \in A~ |~\mathbb{1}_{\{s \ge \tau_0^x\}}, W_s)\cdot \mathbb{1}_{\{s \ge \tau_0^x\}} $$
where $\delta_0$ is the Dirac Delta centered in $0$.
In the end I should arrive to have
$$ \mathbb{P}^{s,x}(W_t \in A ~|~ \mathcal{F}_s) = \mathbb{P}^{s,x}(W_t \in A~ |~\mathbb{1}_{\{s < \tau_0^x\}}, W_s)\cdot \mathbb{1}_{\{s < \tau_0^x\}}+\mathbb{P}^{s,x}(W_t \in A~ |~\mathbb{1}_{\{s \ge \tau_0^x\}}, W_s)\cdot \mathbb{1}_{\{s \ge \tau_0^x\}} =\\ \mathbb{P}^{s,x}(W_t \in A,~ s < \tau_0^x~ |~ W_s) + \mathbb{P}^{s,x}(W_t \in A,~ s\ge \tau_0^x ~|~ W_s) = \mathbb{P}^{s,x}(W_t \in A~ |~ W_s)$$
Is this reasoning formally correct?
Sorry, I was a bit long but I hope my question is easily readable. Any help is highly appreciated.