I was struggling with this question for a while now:
Let the process $(B(t))_{t\ge 0}$ be a standard Brownian motion and suppose $c>0$ is a constant. Let τ be a stopping time such that $B(τ)=c$ and $B(u)<c$ for all $u<τ$.
$$Z(u)=\begin{cases} B(u) &\text{if }0\le u<τ , \\2c−B(u) &\text{if }u>τ\end{cases}$$
I did two things:
-I was struggling when showing $Z(s+t)-z(s)$ is normally distributed with mean $0$ and variance $t$ I struggle when I try to work on the case where s+t>τ and s<τ in which I ended up with $2c-B(s+t)-B(s)$ where I have no idea how to prove it is normally distributed mean $0$ variance $t$
-I then tried to work on the $Z(u)$ itself for $0\le u<τ$ it was not a problem as $Z(u)=B(u)$, then for $u>\tau $ I said $2c−B(u)=2B(\tau )−B(u)$.
Let $u=x+\tau$, then $ = 2B(\tau)−B(\tau +x)$. Here I wanted to use this lemma "For any $s \ge 0$, $(B(t + s) − B(s))_{t\ge 0}$ is a Brownian motion", but the constant doesn't let me.
I would be very grateful if someone could help me.