# Shifted two-sided Brownian Motion

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.

• It looks like you made a copy/paste error and your question became rather garbled. – Nate Eldredge Mar 30 at 23:49
• Thank you, I corrected it. – DCPC Mar 31 at 9:15

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 $$P\big(\tilde B_{-T/2} \in [0,\infty)\big) = P\big( B_{T/2} - B_T \in [0,\infty) \big) = 0.$$

• I think you mean $\tilde{B}_t:=B_{T+t}-B_T$, not $\tilde{B}_t:=B_{T+t}-B_t$. And I am not sure what result implies your first claim, namely that $\tilde{B}_{-T/2}$ should be normally distributed. What you say is that if you have a Brownian motion $B$, then $B_T$ should be normally distributed, for a stopping time $T$. I don't think this is true. – DCPC Mar 30 at 13:41
• Oh, very good point. I definitely jumped the gun there. I can't simply use the normal increments argument with stopping times. – mvarble Mar 31 at 22:43
• How about considering the event $\bigcap_{q\in(0,1)\cap\mathbb Q}\bigg(\tilde B_{-qT}\in(0,\infty)\bigg)$? I think this should have probability zero by definition of the stopping time, so continuity of Brownian motion would tell us that almost every path is negative on $[-T,0]$. – mvarble Apr 1 at 7:37

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.