Conditional running maximum of brownian motion Hi I wish to find an integral expression for $P(\max_{u\in(s,t]}B_u\geq b\,|\,B_s<a)$ where $a<b$.
We can write,
$$
P(\max_{u\in(s,t]}B_u\geq b\,|\,B_s<a)=P(B_s+\max_{u\in(s,t]}(B_u-B_s)\geq b\,|\, B_s<a),
$$
I know that $B_s$ is measurable wrt $\{B_s<a\}$ and that $\max_{u\in(s,t]}(B_u-B_s)$ is independent of $\{B_s<a\}$ and $B_u-B_s$ is another brownian motion. How do I proceed from here?
 A: Set
$$W_u := B_{s+u}-B_s, \qquad u \geq 0,$$
then $(W_u)_{u \geq 0}$ is a Brownian motion which is independent from $(B_u)_{u \leq s}$. Moreover,
$$\mathbb{P} \left( \max_{u \in [s,t]} B_u \geq b \mid B_s < a \right) = \mathbb{P} \left( B_s + \sup_{u \leq t-s} W_u \geq b \mid B_s<a \right). \tag{$\star$}$$
Since $(W_u)_{u \geq 0}$ and $B_s$ are independent, we have
\begin{align*} \mathbb{P} \left( \left\{B_s + \sup_{u \leq t-s} W_u \geq b \right\} \cap \{B_s < a \} \right) &= \mathbb{E} \left( f(B_s) \right) \tag{1}\end{align*}
where
$$f(x) :=1_{(-\infty,a)}(x) \mathbb{P} \left( x + \sup_{u \leq t-s} W_u \geq b \right) .$$
By the reflection principle,
\begin{align*} \mathbb{P} \left( x+ \sup_{u \leq t-s} W_u \geq b \right) &= \mathbb{P} \left( \sup_{u \leq t-s} W_u \geq b-x \right) \\ &= \mathbb{P} ( |W_{t-s}| \geq b-x) \\ &= \mathbb{P}(|W_1|>(b-x)/\sqrt{t-s})\end{align*} for every $x<a$. If we denote by $\Phi$ the cumulative distribution function of the standard Gaussian distribution, then this shows that
$$f(x)=\mathbb{P} \left( x+ \sup_{u \leq t-s} W_u \geq b \right) = 2 \left[ 1- \Phi \left( \frac{b-x}{\sqrt{t-s}} \right)\right], \qquad x<a.$$
Plugging this into $(1)$, we find that
\begin{align*} \mathbb{P} \left( \left\{B_s + \sup_{u \leq t-s} W_u \geq b \right\} \cap \{B_s < a \} \right) &=2-2 \mathbb{E} \left(1_{B_s<a} \Phi \left( \frac{b-B_s}{\sqrt{t-s}} \right) \right)  \\ &= 2-2 \mathbb{E}\left( 1_{B_1<a/\sqrt{s}} \Phi \left( \frac{b- \sqrt{s} B_1}{\sqrt{t-s}} \right)\right) \\ &= 2-2 \int_{-\infty}^{a/\sqrt{s}} \Phi \left( \frac{b-\sqrt{s} x}{\sqrt{t-s}} \right) \phi(x) \, dx \end{align*} with $\phi=\Phi'$ being the density of standard Gaussian distribution. Dividing both sides by $\mathbb{P}(B_s<a) = \Phi(a/\sqrt{s})$ gives, by $(\star)$, an integral expression for the conditional expectation: $$\mathbb{P}\left( \max_{u \in [s,t]} B_u \geq b \mid B_s < a \right) = \frac{2}{\Phi(a/\sqrt{s})} \left[ 1- \int_{-\infty}^{a/\sqrt{s}} \Phi \left( \frac{b-\sqrt{s} x}{\sqrt{t-s}} \right) \phi(x) \, dx  \right].$$
