Transition function for absorbed Brownian motion I need an help with the following exercise. I've already seen this question Prove that Brownian Motion absorbed at the origin is Markov but I don't understand the answer. Also I would like to prove the thing differently (at least the two ways seem different to me).

So let $\tau$ be the first time the Brownian motion arrives in $0$ and define

$W^x_t = \begin{cases} x+B_t &amp;\mbox{if } t\leq \tau \\ 
0&amp; \mbox{else }. \end{cases} $  

Here we denote with $B_t$ the standard Brownian motion starting at $0$ and with $B_t^x$ the Brownian motion starting at $x$, so $B_t^x=x+B_t$.

I want to compute the transition function for the process, i.e. $p(t,x,A)$.

I've already got an hint: show that $\mathbb P(B_t^x\in A)= \mathbb P(B^x_t\in A,B_s^x&gt;0 \,\forall s \in [0,t] )+\mathbb P(B_t^x\in -A)$, for each $x&gt;0, A\subset (0+\infty).$

Now I see that if I prove the hint, I'm done, but I don't know how to prove the hint. I know that I should use reflection principle, but how?
Any help will be really appreciated.
$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s&gt;0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$
$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s&gt;0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$