Let $B$ be a standard Brownian motion on $[0, T]$ and let $\tau$ be the (a.s. unique) moment at which it attains its maximum value: $M = \sup_{t \in [0, T]} B(t) = B (\tau)$.
There is a well-known formula for the joint density of $(M, B(T))$, but I cannot find a joint density of $(M, B(u))$ for $u < T$. Is it possible? I found a very similar question here, but the given answer is incomplete.
Also, I am looking for the distribution of $(M, B(u), B(v))$ with some $0<u<v<T$. If the previous question is doable, maybe the joint pdf of this vector may also be found?
Finally, I am trying to compute $\mathbb{E} \{ \tau B(u) \}$ for some fixed $u$. If I knew the joint law from above, I'd be able to find this easily. But maybe there is a chance to calculate this without knowing the joint pdf?
It seems that Thomas Kojar's suggestion does help to solve the problem. Let $0 < u < v < T$. Then, since $$ M_T = \max \{ M_u, B_u + \max_{t \in [u, T]} ( B_t - B_u ) \} = \max \{ M_u, B_u + M_{T-u}^* \}, $$ by law of total probability we obtain $$ \mathbb{P} \{ M_T < x, B_u < \alpha \} = \int_{-\infty}^\alpha \mathbb{P} \{ M_{T-u}^* < x - z \} \cdot \mathbb{P} \{ M_u < x, \ B_u \in dz \}. $$ By $$ \mathbb{P} \{ M_u \in d\xi, \ B_u \in dz \} = \frac{2 ( 2\xi - z)}{u^{3/2} \sqrt{2 \pi}} \exp \left( -\frac{( 2 \xi - z )^2}{2u} \right) $$ and $$ \mathbb{P} \{ M_T < x \} = \sqrt{\frac{2}{\pi T}} \int_0^x e^{-q^2/2T} dq $$ we obtain $$ \mathbb{P} \{ M_T < x, B_u < \alpha \} = \sqrt{\frac{2}{\pi (T-u)}} \int_{-\infty}^\alpha \int_0^{x-z} \exp \left( -\frac{q^2}{2(T-u)} \right) dq \int_0^x \frac{2 ( 2\xi - z)}{u^{3/2} \sqrt{2 \pi}} \exp \left( -\frac{( 2 \xi - z )^2}{2u} \right) \, d\xi \, dz. $$ Simplifying, we obtain $$ \mathbb{P} \{ M_T < x, B_u < \alpha \} = \frac{2}{\pi u^{3/2} \sqrt{T-u}} \int_{-\infty}^\alpha \int_0^x \int_0^x (2\xi-z) \exp \left( -\frac{(q+z)^2}{2(T-u)} -\frac{( 2 \xi - z )^2}{2u} \right) \, dq \, d\xi \, dz. $$
Okay, now I probably see how to apply this procedure to finding the law of $(M_T, B_u, B_v)$: $$ \mathbb{P} \{ M_T < x, \ B_u < y, \ B_v < z \} = \mathbb{P} \left\{ \begin{gathered} M_u < x, \ B_u + M_{v-u}^* < x, B_u + B_{v-u}^* + M_{T-v}^{**} < x, \\ B_u < \alpha, \ B_u + B_{v-u}^* < \beta \end{gathered} \right\}, $$ where $$ \begin{aligned} B_t^* & = B_{t+u} - B_u, \quad t \in [0, v-u], & M_t^* & = \max_{t \in [0, v - u]} B_t^*, \\ B_t^{**} & = B_{t+v} - B_v, \quad t \in [0, T - v], & M_{T-v}^{**} & = \max_{t \in [ 0, T - v ]} B_t^{**}. \end{aligned} $$ Most importantly, random variables with different numbers of stars (0, 1 and 2) are independent of each other. This motivates us to condition in $(B_{v-u}^*, M_{v-u}^*)$ (one star) $$ \mathbb{P} \{ M_T < x, \ B_u < y, \ B_v < z \} = \int \mathbb{P} \left\{ \begin{gathered} M_u < x, \ B_u + \xi < x, B_u + z + M_{T-v}^{**} < x, \\ B_u < \alpha, \ B_u + z < \beta \end{gathered} \right\} \cdot \mathbb{P} \{ M_{v-u}^* \in d \xi, \ B_{v-u}^* \in d z \}. $$ and then in the zero-star vector $(B_u, M_u)$: $$ \mathbb{P} \{ M_T < x, \ B_u < y, \ B_v < z \} = \int \mathbb{1} \left\{ \begin{gathered} \theta < x, \ \eta + \xi < x, \\ \eta < \alpha, \ \eta + z < \beta \end{gathered} \right\} \cdot \mathbb{P} \{ M_{T-v}^{**} < x - z - \eta \} \cdot \mathbb{P} \{ M_{v-u}^* \in d \xi, \ B_{v-u}^* \in d z \} \cdot \mathbb{P} \{ M_{u} \in d \theta, \ B_u \in d \eta \}. $$