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Suppose we have the $(W_t)$ Brownian Motion and the filtration $F=(\mathcal{F}_t)$, where $\mathcal{F}_t:=\sigma(W_s;s\le t)$. I know that for any $n\in \mathbb{N}$ and $0\le t_0 the increments $W_{t_i}-W_{t_{i-1}}$ are independent by definition. Now let $t\ge 0$ and $h>0$. In my lecture notes is a proof which I do not understand. We want to show that $W_{t+h}-W_t$ is independent of $F$.

They say: By the independence of $W_{t_i}-W_{t_{i-1}}$, $W_{t+h}-W_t$ is independent of the family of all increments $W_l-W_m$ with $m\le l\le t$, since it gives the independence for each finite subfamily.

I guess the last sentence means, that I can split $l-m$ into a finite "partition", like this: $l=t_0. However, how we see then independence of $W_l-W_m$.?

It would be appreciated if someone could explain in a more detailed way, how we get this independence. Thanks in advance.

hulik

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    I guess you want to show that $W_{t+h}-W_t$ is independent of $\mathcal{F}_t$ and not the whole filtration. Try to put $t_0=0, t_1=m, t_2=l, t_3=t$ and $t_4=t+h$ and see if that doesn't give you what you need.2012-03-19
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    What is your definition of brownian motion? This should fall out immediately from most definitions (e.g. the ones provided in Kartzsas and Shreve's Brownian Motion and Stochastic calculus).2014-08-04

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