Shifted two-sided Brownian Motion

Question

Let $(B_t)_{t\in\mathbb{R}}$ be a two-sided Brownian motion, defined as $B(t) = \begin{cases} B_1(t),\quad t >0 \\ 0, \quad t = 0 \\ B_2(-t), \quad t < 0 \end{cases}$. For some $a>0$ let $T:=\inf\{t\geq 0: B_t=a\}$ be the hitting time of $a$. By the strong Markov property, the process $(B_{T+t}-B_T)_{t\geq0}$ is a standard Brownian Motion.

I know that $(B_{T+t}-B_T)_{t\in\mathbb{R}}$ is not a two-sided Brownian motion, but I cannot find a rigorous argument to prove it. I get the idea that if one goes backwards in time (from $T$ to $0$), one gets something negative, which cannot be normally distributed, but I don't manage to write it down appropriately. I am grateful for any help you might give me.

Answer 1

Perhaps this is a detailed enough usage of your argument.

Define $\tilde B$ by $\tilde B_t = B_{T+t} - B_t$. As you said, if this were a two-sided Brownian motion (say its left side was $\tilde B_2$), we have a positive probability that it is a positive number: $$ P\big(\tilde B_{-T/2} \in [0,\infty)\big) = P\big( \tilde B_2(T/2) \in [0,\infty)\big) = 1/2. $$

On the other hand, because $B_1$ is almost-surely continuous, we have that $B_1|_{[0,T/2]} < a$ almost-surely, by the intermediate value theorem and definition of our stopping time $T$. Consequently, $B_{T/2}-B_T < 0$ almost-surely. This means that $$ P\big(\tilde B_{-T/2} \in [0,\infty)\big) = P\big( B_{T/2} - B_T \in [0,\infty) \big) = 0. $$

Answer 2

Someone indicated to me a reference to the answer, so I will also post it here.

It has to do with a theorem in this paper: Path Decomposition and Continuity of Local Time for One-Dimensional Difussions, I - by David Williams, which says that on the interval $[0,T_c]$, the process $(c-X_{T_c -t})$ is equal in distribution to a Bessel Process. The argument seems quite complicated to me and I haven't studied it in great detail yet, but I thought that for completion an answer is due.