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Reflected Brownian Motion is a Markov process

0

$begingroup$

Let $B=(B_t,tgeq 0)$ be a Brownian Motion and $M=(M_t, t geq 0)$ the running maximum of $B$. To be more precise, this means $M_t=sup_sleq t B_s$ for all $tgeq 0$. In Problem 6.1 c) in SchillingPartzschSolutions they proceed in the following way to show that $Y=(Y_t, tgeq 0)$ with $Y_t=M_t-B_t$ for all $tgeq 0$ 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 $mathbbR^2$
$$mathbbE[f(X,Y) | mathcalF_s]=mathbbE[f(x,Y)vert_x=Xquad a.s.$$
holds true for a $mathcalF_s$-measurable random variable $X$ and a random variable $Y$ which is independent of $mathcalF_s$. But in the proof in SchillingPartzsch we have two random variables which are independent of $mathcalF_s$.

brownian-motion conditional-expectation

share|cite|improve this question

asked Mar 18 at 10:26

Math95Math95

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $B=(B_t,tgeq 0)$ be a Brownian Motion and $M=(M_t, t geq 0)$ the running maximum of $B$. To be more precise, this means $M_t=sup_sleq t B_s$ for all $tgeq 0$. In Problem 6.1 c) in SchillingPartzschSolutions they proceed in the following way to show that $Y=(Y_t, tgeq 0)$ with $Y_t=M_t-B_t$ for all $tgeq 0$ 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 $mathbbR^2$
$$mathbbE[f(X,Y) | mathcalF_s]=mathbbE[f(x,Y)vert_x=Xquad a.s.$$
holds true for a $mathcalF_s$-measurable random variable $X$ and a random variable $Y$ which is independent of $mathcalF_s$. But in the proof in SchillingPartzsch we have two random variables which are independent of $mathcalF_s$.

brownian-motion conditional-expectation

share|cite|improve this question

asked Mar 18 at 10:26

Math95Math95

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

Let $B=(B_t,tgeq 0)$ be a Brownian Motion and $M=(M_t, t geq 0)$ the running maximum of $B$. To be more precise, this means $M_t=sup_sleq t B_s$ for all $tgeq 0$. In Problem 6.1 c) in SchillingPartzschSolutions they proceed in the following way to show that $Y=(Y_t, tgeq 0)$ with $Y_t=M_t-B_t$ for all $tgeq 0$ 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 $mathbbR^2$
$$mathbbE[f(X,Y) | mathcalF_s]=mathbbE[f(x,Y)vert_x=Xquad a.s.$$
holds true for a $mathcalF_s$-measurable random variable $X$ and a random variable $Y$ which is independent of $mathcalF_s$. But in the proof in SchillingPartzsch we have two random variables which are independent of $mathcalF_s$.

brownian-motion conditional-expectation

share|cite|improve this question

asked Mar 18 at 10:26

Math95Math95

114

$endgroup$

share|cite|improve this question

asked Mar 18 at 10:26

Math95Math95

114

share|cite|improve this question

asked Mar 18 at 10:26

Math95Math95

114

asked Mar 18 at 10:26

Math95Math95

114

asked Mar 18 at 10:26

Math95Math95

114