In this video lecture the professor is proving the theorem that

For a Brownian motion $(B_t)_{t \geq 0}$ it holds that $$P(M(t)>a) = 2 P(B(t)>a)$$ where $M_t := \sup_{s: s \leq t} B(s)$.

The proof uses the following axiom that

$$P(M(t)>a) = P (\tau_a<t)$$ where $\tau_a =$ stopping time.

My confusion is that this axiom doesn't appear correct because, all those paths which just touch $B(t)=a$, but never cross $a$, are a part of the RHS of the above equation, but for such paths the LHS doesn't hold true. Hence the RHS is greater than LHS in this case. So according to me the LHS should be equal to intersection of RHS with one more condition, which would make sure that the motion goes above a and not just touch the line $y = a$.

PS: at 34:15, the professor defines the stopping time as the first time you hit the line a. If the definition were the first time you cross the line $a$, then I think there would not be any confusion.

Where am I wrong?

  • $\begingroup$ It is clearly stated in the very first line that $a>0$ in the reference. $\endgroup$ Feb 6, 2019 at 6:36
  • $\begingroup$ Ya, just noticed that. Edited the post to reflect that. $\endgroup$ Feb 6, 2019 at 6:37
  • $\begingroup$ I though the teacher defined $\tau_a$ the usual way, which is $\inf \{t:B(t) >a\}$. I am deleting my answer. $\endgroup$ Feb 6, 2019 at 6:43
  • $\begingroup$ If that were the case then there would be no confusion, but at 34:15, he defines it as min(t) {B(t)=a} $\endgroup$ Feb 6, 2019 at 6:46

1 Answer 1


The stopping times $$\tau := \inf\{t>0; B_t>a\}$$ and $$\tilde{\tau} := \inf\{t>0; B_t=a\}$$ are equal almost surely, i.e. if a Brownian motion hits the line $y=a$ then it will immediately cross this line with probability $1$. In particular, $$\mathbb{P}(\tau<t) = \mathbb{P}(\tilde{\tau}<t) \quad \text{for all $t$.}$$ Hence, $$\mathbb{P}(M_t>a) = \mathbb{P}(\tau<t) = \mathbb{P}(\tilde{\tau}<t)$$ which means that it doesn't matter whether we use $\tau$ or $\tilde{\tau}$.

To prove that the stopping times are equal almost surely, we note first of all that $\tilde{\tau} \leq \tau$ by the very definition of the stopping times and the continuity of the sample paths of Brownian motion. If we set

$$W_t := B_{\tilde{\tau}+t} - B_{\tilde{\tau}} = B_{\tilde{\tau}+t} -B_a, \qquad t \geq 0,$$

then this defines a Brownian motion, and

\begin{align*} \tau = \inf\{s \geq \tilde{\tau}; B_{s}>a\} &= \tilde{\tau} +\inf\{t>0; B_{\tilde{\tau}+t} > a\} \\ &= \tilde{\tau}+\inf\{t>0; W_t>0\} \end{align*}

Since it is known that the stopping time $$\inf\{t>0; W_t>0\}$$ equals zero almost surely (this is, for instance, a consequence of Blumenthal's 0-1 law), we conclude that $\tau=\tilde{\tau}$ almost surely.


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