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This question already has an answer here:

$(B_t)_{t \in \mathbb R_0^+}$ are random variables on $(\Omega,\mathcal A,P)$.

$\forall r \le s, t > s, B_t-B_s,B_r$ are independent (i.e. $\sigma(B_t-B_s)$ and $\sigma(B_r)$ are independent, where with $\sigma(X)$ I denote the $\sigma$-algebra produced by the variable $X$).

Can one show that $B_t-B_s$ and $\sigma(\{B_i| r \le s\})=\sigma(\bigcup_{r \le s}\sigma(B_r))$ are independent?

Does one need to use the fact that any endless sequence of random variables from $(B_t)_{t \in \mathbb R_0^+}$ is multivariate normally distributed?

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marked as duplicate by saz, user147263, Did probability-theory Jul 5 '15 at 22:58

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  • $\begingroup$ By definition of BM increments are independent i.e. $B_t-B_s$ is independent of $B_r-B_0$ ($B_0=0$) $\endgroup$ – d.k.o. Jul 5 '15 at 20:00
  • $\begingroup$ It would be good if you gave your definition of Brownian motion. In one of the common definitions, $B_{t_3}-B_{t_2}$ is independent of $B_{t_1}-B_{t_0}$ whenever $t_0<t_1\leq t_2<t_3$. So your result is the case of this where $t_0=0$. But there is also a definition involving the covariance function, and under this definition the independence of increments is a theorem. $\endgroup$ – Ian Jul 5 '15 at 21:57
  • $\begingroup$ @lan I do use the definition involving the covariance function. So $Cov(B_t,B_s)=\sigma^2 \min(s,t)$, $B_{t_0} = 0$, all tuples of $B$s are multivariate normal, $\mathbb E(B_t)= \mu t$. $\endgroup$ – Sergey Zykov Jul 6 '15 at 7:59
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To show independence you just need to show that

$\Bbb{P} (B_t- B_s \in A_0, B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) = \Bbb{P} (B_t- B_s \in A_0)\Bbb{P} ( B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k)$ for every $r_1, \ldots r_k \leq s$ and $A_1, \ldots, A_k \in \mathcal{B}$.

This is because $\sigma(\{B_r, r \leq s\})$ is generated by $E=\{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$ and you will have shown that

$\Bbb{E} (1_{A_0}(B_t- B_s) 1_E) = \Bbb{E} (1_{A_0}(B_t- B_s))\Bbb{E} ( 1_E)$

Use a monotone class theorem and you get the result for arbitrary $E\in \sigma(\{B_r, r \leq s\})$

The monotone class theorem goes like this:

Consider $\mathcal{C} = \bigg\{E \in \sigma(\{B_r, r \leq s\}); \Bbb{E} (1_{A_0}(B_t- B_s) 1_E) = \Bbb{E} (1_{A_0}(B_t- B_s))\Bbb{E} ( 1_E)\bigg\}$

Note that

1) $\Omega \in \mathcal{C}$

2) $A \in \mathcal{C}\Rightarrow A^c \in \mathcal{C}$

3) $A_1 \subset A_2 \subset \ldots $, $A_i \in \mathcal{C} \Rightarrow \cup_i A_i \in \mathcal{C} $ (this is were one usualy uses the monotone convergence theorem and that motivates the name monotone class theorem)

Every set $\mathcal{C}$ with properties 1),2),3) that contains the elementary sets $E = \{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$ contain the sigma algebra generated by all the $E$'s of the form $E=\{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$

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  • $\begingroup$ I am not familiar with monotone class theorem. So can not really appreciate your answer. $\endgroup$ – Sergey Zykov Jul 5 '15 at 21:43
  • $\begingroup$ I edited the question, maybe now it got clearer. $\endgroup$ – Conrado Costa Jul 5 '15 at 21:54
  • $\begingroup$ Is there any way to show the same without the monotone class theorem. By now in my course I am supposed to know the answer to the question, but the lectures have not covered the theorem you are using yet. So I guess there should be a way to get away without the theorem. $\endgroup$ – Sergey Zykov Jul 6 '15 at 19:46

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