# Reflected Brownian Motion is a Markov process

Let $$B=(B_{t},t\geq 0)$$ be a Brownian Motion and $$M=(M_{t}, t \geq 0)$$ the running maximum of $$B$$. To be more precise, this means $$M_{t}=\sup_{s\leq t} B_{s}$$ for all $$t\geq 0$$. In Problem 6.1 c) in SchillingPartzschSolutions they proceed in the following way to show that $$Y=(Y_{t}, t\geq 0)$$ with $$Y_{t}=M_{t}-B_{t}$$ for all $$t\geq 0$$ is a Markov process:

Can someone explain me the last equality in more detail? I'm aware of the fact that for all bounded and measurable functions $$f$$ on $$\mathbb{R}^{2}$$ $$\mathbb{E}[f(X,Y) | \mathcal{F}_{s}]=\mathbb{E}[f(x,Y)\vert_{x=X}\quad a.s.$$ holds true for a $$\mathcal{F}_{s}$$-measurable random variable $$X$$ and a random variable $$Y$$ which is independent of $$\mathcal{F}_{s}$$. But in the proof in SchillingPartzsch we have two random variables which are independent of $$\mathcal{F}_{s}$$.

• Just replace the random variable $Y$ by a random vector $Y=(Y_1,Y_2)$. For further details you might want to take a look at the appendix, Lemma A.3 and Corollary A.4. – saz Mar 19 at 10:45