Computing the transition probabilities for the Markov process $(B_t, \sup_{0\le s\le t} B_s). $

Question

I would like to follow up on a previous question of mine that was asked in this forum earlier (Is $(\sup_{s\le t} B_s, B_t) $ a Markov process?)

I am trying to play with Markov processes and become more familiar with certain technicalities involved in dealing with them. Right now I am wondering about the best approach to find the transition probabilities for the Markov process $(B_t, M_t:=\sup_{0\le s \le t} B_s) $ where $B=\{B_t: t \in [0, \infty)\} $ is BM on a given filtered probability space.

For univariate processes, there might be a certain amount of algebra involved, but ideally I know how to attack the problem. Here, I do know the pdf for the joint probability of $(B_t, M_t) $ which is a standard result in the theory of BM and not even particularly complicated, but I feel a bit clumsy on how to organize my work to compute the transition probabilities. Again, any help would be greatly appreciated.

Thank you Maurice

Answer 1

Suppose $s,t>0$. Consider the conditional distribution of $(B_{t+s},M_{t+s})$ given $\mathcal F_t$ (the history of the Brownian motion up to time $t$). We have that $B_{t+s}=B_t+(B_{t+s}-B_t)$ and $$ M_{t+s} = M_t\vee\left[B_t+\sup_{t\le u\le t+s}(B_u-B_t)\right]. $$ By the independent increments of Brownian motion, the conditional distribution of $(B_{t+s},M_{t+s})$ given $\mathcal F_t$ is therefore the same as that of $(B_t+\tilde B_s,M_t\wedge (B_t+\tilde M_s))$, in which the pair $(\tilde B_s,\tilde M_s)$ is independent of $B_t$ and has the same distribution as $(B_s,M_s)$ under the initial condition $B_0=0$.