Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0\le t_0<t_1<\dots<t_n\le T$ 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: $l=t_0<t_1<\dots<t_n=m$. 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.


share|cite|improve this question
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. – Stefan Hansen Mar 19 '12 at 11:26
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). – Batman Aug 4 '14 at 3:11

So, as far as I understand you have that if $0\leq t_0<t_1<\ldots<t_n$ you know that $W_{t_k}-W_{t_{k-1}},k=\overline{1,n}$ are independent random variables (this I will denote by A). And now you are to prove that if $W_t-W_s$ is independent from $\mathcal F_s$(this I will denote by B). First, I will prove that B follows from A. Let $0<t_0<t_1<\ldots<t_n=s<t$. We know that $W_t-W_s$ is independent from $W_{t_k}-W_{t_{k-1}},k=\overline{1,n}$ (in the next I will denote this fact as $W_t-W_s\bigsqcup W_{t_k}-W_{t_{k-1}}$). By definition it means that $W_t-W_s\bigsqcup \sigma \{W_{t_k}-W_{t_{k-1}},k=\overline{1,n}\}=\sigma\{W_{t_k},k=\overline{1,n}\}$. Hence $W_t-W_s \bigsqcup\, \bigcup\limits_{0\leq q_1<\ldots<q_m\leq s, m \in \mathbb N}\sigma\{W_{q_k},k=\overline{1,m}\}$. The set on the left is $\pi$-class (stable under finite intersection), hence (it is well known theorem about independence of sigma-algebras generated by independent $\pi$-classes are themselves independent) $W_t-W_s\bigsqcup \sigma \{ \bigcup\limits_{0\leq q_1<\ldots<q_m\leq s, m \in \mathbb N}\sigma\{W_{q_k}, \}=\mathcal F_s$. The last fact, you can prove yourself. If you want to prove that B entails A, try to use characteristic functions. And the criteria of independence which uses c.f.. Note, that the fact you want to prove is true for much bigger class of stochastic processes. The reference here could be any normal lectures on Levy processes, for instance, ``Lectures on Levy processes'' by R.Schilling. Unfortunately, I could not find it by googling. If you want, I can send you the .pdf.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.