Let $Y_t = M_t - W_t$ where $M_t$ is the running maximum of brownian motion and $W_t$ is brownian motion. I want to show that $P^0[Y_{t+s} \in dy| Y_t = x] = p(s,x,y)+p(s,x,-y)$ where $p$ is the transition density of brownian motion.
This problem is in Karatzas and Shreve (pg 124) and actually the solution is there as well. However, the solution is not easy to understand. One part of the solution reads as follows:
For $b>m \geq w$,$b \geq a$, $m \geq 0$
$P^0[W_{t+s}\in da,M_{t+s}\in db|W_t=w,M_t=m]$
$=P^0[W_{t+s} \in da, max_{0 \leq u \leq s} W_{t+u} \in db | W_t = w, M_t = m]$
$= P^w[W_s \in da, M_s \in db]$
How is this step justified? Don't we need $(W,M)$ to be a Markov family for this? How can we show that this is a Markov family?
Thank you very much.
A simple way to prove that $(W,M)$ is a Markov process is to note that, for every $t\geqslant0$ and $s\geqslant0$, $$W_{t+s}=W_t+\bar W_s,$$ where $\bar W=(\bar W_s)_{s\geqslant0}$ is a Brownian motion starting from $0$ and independent of the sigma-algebra $\mathcal W_t=\sigma(W_u\,;\,0\leqslant u\leqslant t)$. Furthermore, considering $\bar M$ the maximum process of $\bar W$, one has $$ M_{t+s}=\max\{M_t,W_t+\bar M_s\}, $$ for every $s\geqslant0$, hence the distribution of $(W_{t+s},M_{t+s})_{s\geqslant0}$ conditionally on $\mathcal W_t$ depends on $(W_t,M_t)$ only, as desired.