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Let $(\Omega ,\mathcal F,P)$ a proba space and $(\mathcal F_t)_{t\geq 0}$ a filtration. Let $(B_t)_t$ a Brownian motion adapted to $\mathcal F_t$. We know that $(B_{t+s}-B_t)_{s}$ is independent of $\mathcal F_t$.

My question are the following one:

1) I know that $B_{t+s}-B_t$ is independent of $B_t$ for all $s$. Does it has sense to say that $(B_{t+s}-B_t)_s$ is independent of $(B_t)_t$ ?

2) What does it mean that $(B_{t+s}-B_t)_s$ is independent of $\mathcal F_t$ ? We it mean that for all $s$ $B_{t+s}-B_t$ is independent of $B_t$ for all $t$ ?

user352653
  • 832
  • I think you make a confusion about indexes. Let $X_t := B_{t+s}-B_s$ for some $s>0$, the Markov property states that $X_t$ is a bm independent of $F_s := \sigma(B_u, u\leq s)$.
    1. It means that for some $t$ (or ?), the process $X_. := B_{t+.}-B_t$ is independent of the bm.
     – anonymus Jul 13 '16 at 17:57

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