Show that stopping time of a Brownian Motion is independent of the particular BM

Question

Let $\{B_t\}$ and $\{\tilde{B}_t\}$ be two Brownian motions starting at $0$. Define the two random variables

$$\tau_a = \inf\{t: B_t =a\}\quad \text{and}\quad \tilde{\tau}_a = \inf\{t: \tilde{B}_t=a\}$$

I want to show that $\mathbb{E}(\tau_a) = \mathbb{E}(\tilde{\tau}_a)$.

My attempt:

The functions $\max_{s\leq t}B_s(\omega)$ and $\max_{s\leq t}\tilde{B}_s$ are clearly measurable by continuity of paths and $\{\tau_a\leq t\} = \{\max_{s\leq t}B_s\geq a\}$ so these functions are measurable. We have that

$$\mathbb{E}(\tau_a) = \int_\Omega\tau_a(\omega)\mathrm{d}\mathbb{P}(\omega) = \int_\mathbb{R}\mathbb{P}(\tau_a>t)\mathrm{d}t=\int_\mathbb{R}\mathbb{P}(\max_{s\leq t}B_s\geq a)\mathrm{d}t=?=\int_\mathbb{R}\mathbb{P}(\max_{s\leq t}\tilde B_s\geq a)\mathrm{d}t = \mathbb{E}(\tilde \tau_a)$$ Now if the max wasn't there I would be done, but I don't see how to continue. I basically need that $\mathbb{P}(\max_{s\leq t}B_s\geq a) = \mathbb{P}(\max_{s\leq t}\tilde B_s\geq a)$ to conclude the argument. Is this a viable method? Could someone give a tip on how to proceed?

Answer 1

Here is an answer based on the hint by @Brian Moehring:

We first show that $\mathbb{P}(\sup_{s\leq t} B(s)\geq a) = \mathbb{P}(\sup_{s\leq t}\tilde{B}(s)\geq a)$. We do this by proving that $\mathbb{P}(\sup_{s\leq t} B(s)\geq a) \leq \mathbb{P}(\sup_{s\leq t}\tilde{B}(s)\geq a)$. Observe that if $\{s_k\}$ is a dense sequence of $[0,t]$ then $\sup_{s\leq t}B(s) = \sup_kB(s_k)$ by the continuity of paths and hence these functions are $\mathscr{A}$-measurable.

We begin by proving that $\mathbb{P}(\max(B_s,B_t)\geq a) = \mathbb{P}(\max(\tilde B_s,\tilde{B}_t)\geq a)$. To see this we let $\epsilon>0$ be arbitrary and begin by choosing $\{r_{k}\}_{-\infty}^{k=0}$ such that $r_k<r_{k+1}<r_k+\epsilon$, $r_0 = a$ and $(-\infty,a) = \bigcup_{-\infty}^{k=-1}(r_k,r_{k+1})$. Assuming that $s<t$ we obtain \begin{align*} \mathbb{P}(\max(B_s,B_t)\geq a)& = \mathbb{P}\Big[(B_s\geq a)\cup \Big((B_t\geq a)\cap (B_s<a)\Big)\Big] \\ & = \mathbb{P}(B_s\geq a)+ \mathbb{P}\Big((B_t\geq a)\cap (B_s<a)\Big) \\ & = \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t\geq a)\cap B_s\in (r_k,r_{k+1})\Big) \\ & \leq \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t-B_s\geq a-r_{k+1})\cap B_s\in (r_k,r_{k+1})\Big) \\ & = \mathbb{P}(B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((B_t-B_s\geq a-r_{k+1})\Big) \mathbb{P}\Big(B_s\in (r_k,r_{k+1})\Big) \\ & = \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t-\tilde B_s\geq a-r_{k+1})\Big) \mathbb{P}\Big(\tilde B_s\in (r_k,r_{k+1})\Big) \\ & = \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t\geq a+r_k-r_{k+1})\cap(\tilde B_s\in (r_k,r_{k+1}))\Big) \\ & \leq \mathbb{P}(\tilde B_s\geq a)+\sum_{-\infty}^{k=-1}\mathbb{P}\Big((\tilde B_t\geq a-\epsilon)\cap(\tilde B_s\in (r_k,r_{k+1}))\Big) \\ & = \mathbb{P}(\tilde{B}_s\geq a)+\mathbb{P}\Big((\tilde{B}_t\geq a-\epsilon)\cap (\tilde{B}_s<a)\Big)\\ & \xrightarrow{\epsilon \rightarrow 0} \mathbb{P}(\tilde{B}_s\geq a)+\mathbb{P}\Big((\tilde{B}_t\geq a)\cap (\tilde{B}_s<a)\Big)\\ & = \mathbb{P}(\max(\tilde B_s,\tilde B_t)\geq a). \end{align*} As mentioned before this shows that $\mathbb{P}(\max(B_s,B_t)\geq a)=\mathbb{P}(\max(\tilde B_s,\tilde B_t)\geq a)$.

Now if $\{t_1,\dotsc,t_n\}$ is a finite sequence with $t_1<t_2<\dotsc<t_n$ we observe that $B_{t_k}-B_{t_{k-1}}$ is independent from $\max_{1\leq j \leq k-1}B_{t_j}$ and hence the above argument suitably modified gives an inductive procedure of showing that \begin{equation*} \mathbb{P}\left(\max_{1\leq j \leq n}B_{t_{j}}\geq a\right) = \mathbb{P}\left(\max_{1\leq j \leq n}\tilde B_{t_{j}}\geq a\right). \end{equation*} Choosing a dense sequence of $[0,t]$ we are thus able to conclude via the continuity of paths property that \begin{equation*} \mathbb{P}\left(\max_{0\leq s\leq t}B_s\geq a\right) = \mathbb{P}\left(\max_{0\leq s\leq t}\tilde B_s\geq a\right) \end{equation*} this in fact implies that $\max_{0\leq s\leq t}B_s$ and $\max_{0\leq s\leq t}\tilde B_s$ have the same distribution. Since $\{\omega: \tau_a(\omega)\leq t\} = \{\omega: \max_{0\leq s\leq t}B_s\geq a\}$ we conclude that $\tau_a$ and $\tilde{\tau}_a$ share distribution.