How can I show that as $\epsilon \rightarrow 0$ $\inf\{t\geq 0: B_t\geq 1+\epsilon\}\rightarrow \inf\{t\geq 0:B_t\geq 1\}$ almost surely?

Question

Let $B$ be a a real valued Brownian motion starting from $0$ and $\epsilon >0$. I want to show that as $\epsilon \rightarrow 0$ then $\inf\{t\geq 0: B_t\geq 1+\epsilon\}\rightarrow \inf\{t\geq 0:B_t\geq 1\}$ almost surely.

Let me define $T_a=\inf\{t\geq 0:B_t\geq a\}$. My problem is that intuitively it make sense since $B$ has continuous sample paths and hence $T_{1+\epsilon}>T_1$ and thus as epsilon converges to zero we first also this stopping times should converge. But I don't know how to prove this rigorously. I thought about using the strong Markov property at $T_1$ since $T_1<T_{1+\epsilon}$ but I also don't know what this helps.

I have the following version of the strong Markov property:

Let $\tau$ be a stopping time and $\Phi$ a measurable bounded or non-negative function. Then, \begin{align} \mathbb{E}\left(1_{\tau<\infty} \Phi\left(\left(B_{\tau+t}\right)_{t\geq 0}\right)|\mathcal{F}_\tau\right)=1_{\tau<\infty} \mathbb{E}_{B_\tau}\left(\Phi\left((B_t)_{t\geq 0}\right)\right). \end{align}

Can someone help me?

Answer 1

The continuity question of hitting time $\tau_{a}:=\inf\{t\geq 0: B_{t}\geq a\}$ is an interesting and deep discussion.

First of all, we don't have right-continuity. As mentioned here Drawdown Point Processes and in the notes on stable subordinators, we only have: left-continuity with right-limits. So one often considers the continuous modification $\tau_{a^{+}}:=\lim_{t\downarrow a}\tau(t)$ to get right-continuity. The lack of right-continuity can also be seen from the filtration not being right-continuous e.g. see here Hitting time of an open set is not a stopping time for Brownian Motion.

In Drawdown Point Processes, he studies the difference $\tau_{a^{+}}-\tau_{a}>0$ and obtains a Poisson point process describing their difference. In particular as mentioned here pg.9 Subordinators, we have that

$$\tau_{a}=\int_{0}^{\infty}x\Pi_{a}(dx),$$

where $\Pi_{a}$ is a Poisson point process with intensity

$$dt\times \rho(dx)=dt\times \frac{1}{\sqrt{2\pi}}x^{-3/2}dx.$$

However, we do have convergence in probability i.e. from pg.9 Subordinators

$$P(\tau_{a+\epsilon}-\tau_{a}\geq q)=\frac{\epsilon}{\sqrt{2\pi}}\int_{q}^{\infty}x^{-3/2}e^{-\epsilon^{2}/(2x)}dx=\frac{1}{\sqrt{2\pi}}\int_{q/\epsilon^{2}}^{\infty}x^{-3/2}e^{-1/(2x)}dx\to 0\text{ as }\epsilon\to 0.$$

This also implies convergence almost surely since by continuity of measure we have

$$P(\lim_{\epsilon_{0}\to 0}\sup_{\epsilon\leq \epsilon_{0}}\tau_{a+\epsilon}-\tau_{a}\geq q) =\lim_{\epsilon_{0}\to 0}P(\tau_{a+\epsilon_{0}}-\tau_{a}\geq q).$$

And yes as you mention (and LeGall mentions in Lemma 7.21 in his textbook "Brownian Motion, Martingales, and Stochastic Calculus"), an easier way to see this is instead to just use the strong Markov property

$$\tau_{a+\epsilon_{0}}-\tau_{a}=\tau_{\epsilon_{0}}\circ \tau_{a}\stackrel{dis}{=}\tilde{\tau}_{\epsilon_{0}},$$

where now the Brownian motion $\tilde{B}$ again starts from $0$. So as he recommends we use Proposition 2.14(i) that when $q>0$

$$\sup_{0\leq s\leq q}\tilde{B}_{s}>0\text{ a.s. }$$

to get

$$P_{0}(\tau_{a+\epsilon_{0}}-\tau_{a}\geq q)=P_{0}(\tau_{\epsilon_{0}}\geq q)=P_{0}(\epsilon_{0}\geq \sup_{0\leq s\leq q}\tilde{B}_{s}>0)$$

and finally take limit $\epsilon_{0}\to 0$ to get zero

$$P(\lim_{\epsilon_{0}\to 0}\sup_{\epsilon\leq \epsilon_{0}}\tau_{a+\epsilon}-\tau_{a}\geq q)=\lim_{\epsilon_{0}\to 0}P_{0}(\epsilon_{0}\geq \sup_{0\leq s\leq q}\tilde{B}_{s}>0)=P_{0}(0\geq \sup_{0\leq s\leq q}\tilde{B}_{s}>0)=0.$$

Answer 2

Define $\phi(\omega):=\inf\{t\ge 0:\omega_t-\omega_0\ge\varepsilon\}$, so \begin{align} \phi\Bigl((B_{T_1+t})_{t\ge0}\Bigr)&=\inf\Bigl\{t\ge 0:B_{T_1+t}-B_{T_1}\ge\varepsilon\Bigr\}\\[.4em] &=\inf\Bigl\{t\ge T_1:B_t\ge1+\varepsilon\Bigr\}-T_1\\[.4em] &=T_{1+\varepsilon}-T_1. \end{align} Then, with $\tau:=T_1$ and $\Phi(\omega):=1_{\phi(\omega)>\delta}$, the strong Markov property you gave $$ \mathbb{E}\Bigl[1_{T_1<\infty} \Phi\bigl((B_{T_1+t})_{t\geq 0}\bigr)\:\Big|\:\mathcal{F}_{T_1}\Bigr]=1_{T_1<\infty}\,\mathbb{E}_{B_{T_1}}\Phi\bigl((B_t)_{t\geq 0}\bigr) $$ translates into (I am removing the $\Bbb 1_{\{T_1<\infty\}}$ since $\Bbb P(T_1<\infty)=1$) : \begin{align} \text{a.s},\qquad\Bbb P\Bigl(T_{1+\varepsilon}-T_1>\delta\:\Big|\:\mathcal F_{T_1}\Bigr)&=\Bbb E_1\Phi\bigl((B_t)_{t\geq 0}\bigr)\\[.4em] &=\Bbb P_1\Bigl(\inf\{t\ge0:B_t-1\ge\varepsilon\}>\delta\Bigr)\\[.4em] &=\Bbb P\Bigl(\inf\{t\ge0:B_t\ge\varepsilon\}>\delta\Bigr)\\[.4em] &=\Bbb P\Bigl(T_\varepsilon>\delta\Bigr) \end{align} (the third equality just comes from the invariance by translation of $B$). Taking expectation, it follows that $$\Bbb P(T_{1+\varepsilon}-T_1\ge\delta)=\Bbb P(T_\varepsilon\ge\delta).$$ You should be able to conclude from here that $T_{1+\varepsilon}-T_1\to0$ in probabilty as $\varepsilon\to0$ (and therefore also almost surely by a monotonicity argument).