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?

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    $\begingroup$ So you know BM has continuous paths a.s. What do you know about the distribution of BM? For instance, would you be able to show that $\mathbb{P}(\max(B_{t_1},\ldots, B_{t_n}) \geq a) = \mathbb{P}(\max(\tilde{B}_{t_1},\ldots, \tilde{B}_{t_n}) \geq a)$? $\endgroup$ Jan 24, 2021 at 21:21
  • $\begingroup$ Thanks for the tip! :) I added a suggestion for a proof $\mathbb{P}(\max(B_s,B_t)\geq a) = \mathbb{P}(\max(\tilde{B}_s,\tilde{B}_t)\geq a)$. I think I see how to complete it if I have the finite case $\endgroup$
    – OgvRubin
    Jan 24, 2021 at 21:51
  • $\begingroup$ Thanks for the help @BrianMoehring I think I managed to complete the proof $\endgroup$
    – OgvRubin
    Jan 25, 2021 at 9:38
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    $\begingroup$ Well, I'd say that $\mathbb{P}(\tau_a \le t) = \mathbb{P}(\sup_{0 \le s \le t} B_s \ge a) = 2\mathbb{P}(B_t \ge a)$ (reflection principle). The computation with $\tilde \tau_a$ would be the same. But then since $B$ and $\tilde B$ have the same law (don't they ?), it shows that $\tau_a$ and $\tilde \tau_a$ also have the same law and thus the same expectation. I'm not 100% sure that my argument is correct but at the same time I don't see why it wouldn't be. $\endgroup$ Jan 25, 2021 at 22:04
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    $\begingroup$ @StratosFair Curious, how would you define the law/distribution of the process $B$ itself? It is certainly true $B$ and $\tilde{B}$ have the same finite-dimensional distributions, which is what I'd loosely call its law (and what I was talking about when I mentioned the distribution of BM in my comment), but I'm not sure if I've ever seen the term defined precisely. In any case, I agree there are many ways to prove it - I was just hinting toward a proof from first principles. $\endgroup$ Jan 26, 2021 at 0:00

1 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.

  • $\begingroup$ I'm not sure that $\{\max(B_s,B_t)\geq a\} = \{(B_s\geq a)\cup \Big((B_t\geq a)\cap (B_s<a)\}$. I would have rather written that $\{\max(B_s,B_t)\geq a\} = \{\Big((B_s\geq a)\cap (B_t<a)\Big) \cup \Big((B_t\geq a)\cap (B_s<a)\Big) \cup \Big((B_t\geq a)\cap (B_s \geq a)\Big)\}$. You could then split the probability as the sum of three probabilities. For the first two ones, you proceed as you already did, and for the last one you can use that $\mathbb{P}((B_t\geq a)\cap (B_s \geq a)) = \mathbb{P}((B_t - B_s\geq 0)\cap (B_s \geq a)) $ to conclude with the independence of increments. $\endgroup$ Jan 25, 2021 at 23:39
  • $\begingroup$ If $\max(B_s,B_t)(\omega)\geq a$ then either $B_s(\omega)\geq a$ or $B_s(\omega)<a$ in which case $B_t(\omega)\geq a$. I believe equality holds here? $\endgroup$
    – OgvRubin
    Jan 26, 2021 at 8:30
  • $\begingroup$ Indeed it sounds correct, and maybe it is, but, using words, it seems "more correct" to say that "either $B_s(\omega) \ge a$ OR $B_t(\omega) \ge a$". And then the statement"$B_t(\omega) \ge a$" can also be written as "either ($B_t(\omega) \ge a$ AND $B_s(\omega) < a$) OR ($B_t(\omega) \ge a$ AND $B_s(\omega) \ge a$)". This is why your statement makes me a bit uneasy, it feels like it's missing a term. $\endgroup$ Jan 26, 2021 at 11:56
  • $\begingroup$ Suppose that $\omega$ belongs to $\{\max (B_s,B_t)\geq a\}$ then either $B_s(\omega)\geq a$ or $B_t(\omega)\geq a$. In the first case $\omega\in (B_s\geq a)$, in the second: $B_t(\omega)\geq a$ we can either have $B_s(\omega)\geq a$ or $B_s(\omega)<a$. In the first situation $\omega$ again lies in the set $(B_s\geq a)$ in the second $\omega\in (B_t\geq a)\cap (B_s<a)$. This shows that $\{\max(B_s,B_t)\geq a\}\subseteq RHS$. For the reverse inclusion note that if $\omega \in RHS$ then either $B_s(\omega)\geq a$ or $B_t(\omega)\geq a$. In either case $\omega \in LHS$ $\endgroup$
    – OgvRubin
    Jan 26, 2021 at 13:07
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    $\begingroup$ Yup, after writing it down it is indeed quite obvious that it is true, sorry for wasting your time with this ! $\endgroup$ Jan 26, 2021 at 17:00

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