Yesterday I asked a question about the joint law of $(M_T, B_u)$, where $M_T = \max_{t \in [0, T]} B_t$, $B_t$ is the standard Brownian motion and $u < T$.
One of the reasons I was interested in this law is because I needed to compute $\mathbb{E} \{ \tau B_u \}$, where $\tau$ is the a.s. unique point at which the maximum $M_T$ is attained: $M_T = B_\tau$.
Knowing the joint law $(M_T, B_u)$, it is possible to find the joint law of $(\tau, B_u)$ as follows: for $u > t$, $$ \begin{aligned} \mathbb{P} \{ \tau > t, \ B_u < x \} & = \mathbb{P} \{ \sup_{s \in [0, t]} B_s < B_t + \sup_{s \in [t, T]} ( B_s - B_t ), B_t + ( B_u - B_t) < x \} \\[7pt] & = \mathbb{P} \{ M_t - B_t < M_{T-t}^*, B_t + B_{u-t}^* < x \} \\[7pt] & = \int \mathbb{1} \{ Y - y < Z, y + z < x \} \cdot \mathbb{P} \left\{ \begin{aligned} & B_t \in dy \\ & M_t \in dY \end{aligned} \right\} \cdot \mathbb{P} \left\{ \begin{aligned} & B_{u-t}^* \in dz \\ & M_{T-t}^* \in dZ \end{aligned} \right\}. \end{aligned} $$ It then remains to plug in the well-known formula for the density of $(B_t, M_t)$ and the density of $(B_{u-t}^*, B_{T-t}^*)$ found in my previous quesion.
All this is doable in principle, but in practice this is a very heavy computation. Which brings me to my new question: is there a simpler way to compute $\mathbb{E}\{ \tau B_u \}$? Without going through the hard labour of finding the joint pdf of $(\tau, B_u)$.
Maybe there is a trick I do not see?
P.S. A closely related quantity, if I made no mistake, is $\mathbb{E} \{ M_T \, B_u \, B_v \}$. Using Malliavin integration by parts formula, we obtain $$ \begin{aligned} \mathbb{E} \{ M_T \, B_u \, B_v \} & = \mathbb{E} \left\{ M_T \, B_u \, \int_0^T \mathbb{1} \{ s \in [0, v] \} dB_s \right\} \\ & = \mathbb{E} \left \{ \int_0^T D_s ( M_T \, B_u ) \, \mathbb{1} \{ s \in [ 0, v ] \} ds \right\}. \end{aligned} $$ Since $D_s M_T = \mathbb{1} \{ s \in [0, \tau] \}$ and $D_s B_u = \mathbb{1} \{ s \in [0, u] \}$, applying the chain rule we get $$ D_s ( M_T \, B_u ) = \mathbb{1} \{s \in [0, \tau]\} \, B_u + M_T \, \mathbb{1} \{ s \in [0, u] \}. $$ Therefore, $$ \begin{aligned} \mathbb{E} \{ M_T \, B_u \, B_v \} & = \mathbb{E} \left\{ B_u \int_0^T \mathbb{1} \{ s \in [0, \tau \wedge v] \} ds \right\} +\mathbb{E} \left\{ M_T \int_0^T \mathbb{1} \{ s \in [0, u \wedge v] ds \right\} \\ & = \mathbb{E} \{ ( \tau \wedge v ) B_u \} +(u \wedge v) \mathbb{E} \{ M_T \}. \end{aligned} $$ Now $\mathbb{E} \{ M_T \}$ is known and the problem of finding $\mathbb{E} \{ ( \tau \wedge v ) B_u \}$ seems similar to the problem of finding $\mathbb{E} \{ \tau B_u \}$.
Unfortunately, the problem of finding $\mathbb{E} \{ M_T \, B_u \, B_v \}$ is as hard as the problem of finding $\mathbb{E} \{ \tau B_u \}$. In my previous question I managed to come near the formula for the joint law of $(M_T, B_u, B_v)$, but applying it wouldn't be easy.