0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It looks like you made a copy/paste error and your question became rather garbled. $\endgroup$ – Nate Eldredge Mar 30 at 23:49
  • $\begingroup$ Thank you, I corrected it. $\endgroup$ – DCPC Mar 31 at 9:15
0
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – DCPC Mar 30 at 13:41
  • $\begingroup$ Oh, very good point. I definitely jumped the gun there. I can't simply use the normal increments argument with stopping times. $\endgroup$ – mvarble Mar 31 at 22:43
  • $\begingroup$ 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 $[-T,0]$. $\endgroup$ – mvarble Apr 1 at 7:37
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.