0
$\begingroup$

Let $B=(B_t)_{t\ge 0}$ be a Brownian motion and $L=(L_t)_{t\ge 0}$ be its local time in zero. Given two strictly increasing functions $\phi_1$, $\phi_2: \mathbb R_+\to\mathbb R$ such that $\phi_1(0)=\phi_2(0)=0$, let us define two stopping times by $$\tau_1=\inf\Big\{t>0: B_t\notin \big(-\phi_1(B_t),\phi_1(B_t)\big)\Big\}$$ and $\tau_2=\tau_1$ if $B_{\tau_1}\notin\big(-\phi_2(B_{\tau_1}),\phi_2(B_{\tau_1})\big)$ and $$\tau_2=\inf\Big\{t>\tau_1: B_t\notin \big(-\phi_2(B_t),\phi_2(B_t)\big)\Big\}, \mbox{ else}.$$ Could we show $$\mathbb E\big[\lambda(B_{\tau_2})\big|\mathcal{F}_{\tau_1}\big]=\mathbb E\big[\lambda(B_{\tau_2})\big|B_{\tau_1}, L_{\tau_1}\big]?$$ Many thanks for the reply!

$\endgroup$
6
  • $\begingroup$ @ CodeGolf : Hi first what is $\lambda$ standing for in your last equation ? Second why do you think the Local time is involved ? $\endgroup$
    – The Bridge
    May 6, 2015 at 13:12
  • $\begingroup$ Thanks for your reply. $\lambda$ stands for a bounded continuous function. $\endgroup$
    – CodeGolf
    May 6, 2015 at 14:37
  • $\begingroup$ It is true that $B_{\tau_2}-B_{\tau_1}$ is independent of $\mathcal{F}_{\tau_1}$, but the law of $B_{\tau_2}-B_{\tau_1}$ depends on $\mathcal{F}_{\tau_1}$ $\endgroup$
    – CodeGolf
    May 6, 2015 at 14:39
  • $\begingroup$ @ CodeGolf : so what ? $\endgroup$
    – The Bridge
    May 6, 2015 at 17:23
  • $\begingroup$ So I would like to know whether the law of $B_{\tau_2}-B_{\tau_1}$ depends uniquely on $(B_{\tau_1},L_{\tau_1})$ by the construction of $\tau_1$ and $\tau_2$ $\endgroup$
    – CodeGolf
    May 7, 2015 at 5:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.