$B$ 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.
Best Answer
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 $\{M_t \geq b\} = \{\tau \leq t\}$. $$ P^0\{B_t(\omega) \geq 2b-a,\tau(\omega)\leq t\} = E^0[E^0[1_{\tau(\omega)\leq t} 1_{B_t \geq 2b-a}\mid \mathcal{F}_\tau]=... $$