$B$ being standard Brownian motion, its running maximum is defined as $M_t = \sup_{0\leq s\leq t} B_s$. I am trying to follow the proof of the following result but I don't understand some of the steps and they are not explained in the manuscript.

Let $a \leq b$ and $b > 0$ be real numbers. Then $$P^0\{B_t \leq a, M_t \geq b\} = P^0\{B_t \geq 2b-a\}$$

Proof: Let $\tau = \inf\{t\geq 0: B_t = b\}$. By path continuity $\{\tau \leq t\} = \{M_t \geq b\}$. So here is my first question. Why is continuity relevant here? I understand that continuity ensures $\tau$ to be a stopping time but other than that what role does continuity play in showing $\{\tau \leq t\} = \{M_t \geq b\}$?

\begin{align}P^0\{B_t(\omega) \leq a, M_t(\omega) \geq b\} ={} & P^0\{B_t(\omega) \leq a, \tau(\omega) \leq t\}\\ ={} & E^0\left[\mathbf{1}_{\{\tau(\omega)\leq t\}}P^0\{B_t \leq a\mid \mathcal{F}_{\tau}\}\right]\\={} &E^0\left[\mathbf{1}_{\{\tau(\omega)\leq t\}}P^b\{B_{t-\tau(\omega)} \leq a\}\right]\\={} &E^0\left[\mathbf{1}_{\{\tau(\omega)\leq t\}}P^b\{B_{t-\tau(\omega)} \geq 2b-a\}\right]\\={} &P^0\{B_t(\omega) \geq 2b-a, M_t(\omega) \geq b\} = P^0\{B_t(\omega) \geq 2b-a\}\end{align}

I had first thought that going from the first line to the second one was done as follows. $$P^0\{B_t(\omega) \leq a, \tau(\omega) \leq t\} = E^0\left[E^0[\mathbf{1}_{\{\tau(\omega)\leq t\}}\mathbf{1}_{\{B_t \leq a\}}\mid \mathcal{F}_{\tau}]\right]$$ But $\{\tau(\omega)\leq t\}$ is contained in $\mathcal{F}_t$, not in $\mathcal{F}_{\tau}$. So it cannot just be moved out of the inner expectation. The third and the fourth lines come from the strong Markov property and the reflection principle, respectively. That I get. But I am lost again in the fifth line. How are the probabilities put back together? The RHS of the last line is clear to me by the way. I am talking about the LHS.

  • $\begingroup$ Could you please explain how the last equation is derived, i.e., why $$\begin{align}&E^0\left[\mathbf{1}_{\{\tau(\omega)\leq t\}}P^b\{B_{t-\tau(\omega)} \geq 2b-a\}\right]\\={} &P^0\{B_t(\omega) \geq 2b-a, M_t(\omega) \geq b\} \end{align}$$ $\endgroup$
    – Vim
    Feb 6, 2019 at 15:33
  • $\begingroup$ @Vim I am thinking about it. By the way this is not a good derivation of the distribution of $M$ . There are much clearer ones (for instance in the book of Schilling on Brownian motion) $\endgroup$
    – Calculon
    Feb 6, 2019 at 17:56
  • $\begingroup$ @Vim It is actually quite straightforward as I now realize. It is the same thing as in going from the second line to the third line but in reverse. Nonetheless, I don't like $t-\tau$ as it may well be negative (still OK though since we are doing everything on the event $t \geq \tau$ thanks to the indicator function). $\endgroup$
    – Calculon
    Feb 6, 2019 at 18:12
  • $\begingroup$ Why is there sometimes an $\omega$ after $B_t$ but sometimes without? $\endgroup$
    – inbrevi
    Jun 26, 2020 at 10:38
  • $\begingroup$ In the fourth equal you're using the reflection principle $\endgroup$
    – No-one
    Oct 27, 2022 at 11:10

1 Answer 1


The reason continuity is relevant in showing $\{ \tau \leq t\} = \{M_t \geq b\}$ is because if $B_t$ were not continuous, then it could be that $B_t$ jumps over $b$ without hitting it, in which case $\tau$ might not occur even if $M_t > b$.

As for $\{\tau \leq t\}$, this set is actually measurable with respect to $\mathcal{F}_{\tau \wedge t} = \mathcal{F}_\tau\cap \mathcal{F}_t$. In general, if $T,S$ are stopping times, then $\mathcal{F}_{T \wedge S} = \mathcal{F}_T \cap \mathcal{F}_S$ and each of the following is in $\mathcal{F}_{T \wedge S}$: $$\{T < S\}, \{S < T\}, \{T \leq S\}, \{S \leq T\}, \{T = S\}.$$ This is Lemma 2.16 in Karatzas and Shreve if you want to look it up.

The last line is the reverse of the first, note that $\{M_t \geq b\} = \{\tau \leq t\}$. $$ P^0\{B_t(\omega) \geq 2b-a,\tau(\omega)\leq t\} = E^0[E^0[1_{\tau(\omega)\leq t} 1_{B_t \geq 2b-a}\mid \mathcal{F}_\tau]=... $$

  • $\begingroup$ I will think about your answer tomorrow. But a quick remark for now. If we were to define $\tau = \inf\{t: B_t \geq b\}$, continuity would not be necessary? However, in that case we wouldn't be able to make use of the strong Markov property. Does this sound right? $\endgroup$
    – Calculon
    Aug 9, 2015 at 21:21
  • 1
    $\begingroup$ As long as $B$ is right continuous, you are correct that in the case $\tau := \inf \{t: B_t \geq b\}$ we have the equality $\{ \tau \leq t\} = \{ M_t \geq b\}$. Clearly if $M_t \geq b$ then $\tau \leq t$. If $\tau \leq t$ the for every $s > t$ we can find $t<r<s$ with $B_r \geq b$. Taking a sequence of them $r_n \to t^+$ we find $B_t \geq b$ due to the right continuity of $B$. However, $\tau$ still may not be a stopping time even if $B$ is right continuous. You would need RC + adapted + $(\mathcal{F}_t)$ universally complete. See planetmath.org/hittingtimesarestoppingtimes $\endgroup$
    – nullUser
    Aug 9, 2015 at 22:41
  • $\begingroup$ Can you elaborate a bit on the last equation in your answer? It is not quite the same as the fourth line in my post. $\endgroup$
    – Calculon
    Aug 10, 2015 at 8:38
  • $\begingroup$ @nullUser Given $\tau\leq t$, the argument in your comment shows that $B_\tau\geq b$, not $B_t\geq b$. Anyway, you can then correctly deduce that $M_t\geq b.$ $\endgroup$
    – user940
    Aug 10, 2015 at 18:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .