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:

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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 $\mathcal{F}_{s}$.

  • $\begingroup$ 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. $\endgroup$ – saz Mar 19 at 10:45

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