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

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

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 $$B_0=0$$.