# Let $B = \{B_t : t \geq 0\}$ be a pre-Brownian motion. Show that $B^a$ is independent of $\sigma(B_r : r \leq a)$ for every $a \geq 0$.

Let $$B = \{B_t : t \geq 0\}$$ be a pre-Brownian motion.

I know the following definition of a pre-Brownian motion:

Let $$(\Omega, \mathcal{F}, \mathbb{P})$$ be a measure space and let $$I = [0, \infty)$$. A pre-Brownian motion on $$I$$ is a Gaussian process $$B = \{B_t : t \in I\}$$ satisfying

(1) $$B_0(\omega) = 0\ \forall\ \omega \in \Omega$$

(2) $$\mathbb{E}[B_t] = 0,\ \forall\ t \geq 0$$

(3) $$Cov(B_s, B_t) = min(s, t)\ \forall\ s, t \geq 0$$

I would like to prove that, for every $$a \geq 0$$, the

process $$B^a = \{B_{t+a} - B_a, t \geq 0\}$$ is independent of the natural filtration $$\sigma(B_r : 0 \leq r \leq a)$$.

In fact, this is the statement of the Markov property (for pre-Brownian motions). I have proved that $$B^a$$ is a pre-Brownian motion. I guess that independence follows from the independent increments property of $$B^a$$, i.e.

for all $$0 < t_1 < \ldots < t_n$$, the increments $$B_{t_1}, B_{t_2}-B_{t_1}, \ldots, B_{t_n} - B_{t_{n-1}}$$ are independent.

Attempt: Let $$G$$ be the Gaussian space generated by the process $$B$$ and denote by $$G_a$$ and $$\tilde{G_a}$$ the vector spaces generated by $$\{B_t\}_{0 \leq t \leq a}$$ and by $$\{B_{a+t} - B_a\}_{t \geq 0}$$. Since $$\sigma(\{G_a\})$$ and $$\sigma(\{\tilde{G_a}\})$$ are independent by the above mentioned property, it follows that $$B^a = \{B_{t+a} - B_a, t \geq 0\}$$ is independent of $$\sigma(B_r :0 \leq r \leq a)$$.

Any suggestions of improvement will be appreciated.

• What is a pre-Brownian motion...? – saz May 9 '18 at 6:54
• "Can someone explain to me how to prove..." Sure, but what did you try yourself? – Did May 9 '18 at 7:46
• @saz My understanding is a standard Brownian motion without the assumption of continuous sample paths. – Math1000 May 9 '18 at 8:12

I suppose $B^{a}$ stands for the process $B(t+a)-B(a):t\geq 0$. You have to show that $0<t_1<t_2,...,<t_n, s_1<s_2<...<s_k\leq a$ implies $(B(t_1+a)-B(a),...,B(t_n+a))-B(a)$ is independent of $(B(s_1),...,B(s_k))$. Since $B(t)$ has independent increments $(B(t_1+a)-B(a),B(t_2+a)-B(t_1+a),...,B(t_n+a)-B(t_{n-1}+a))$ and $(B(s_1),B(s_2)-B(s_1),...,B(s_k)-B(s_{k-1}))$ are independent. If two collections are independent then any measurable functions of these are independent. Apply suitable linear maps on the two vectors to complete the proof. [ The maps are of the type $(x_1,x_2,..,x_m) \to (x_1,x_1+x_2,...,x_1+...+x_m)$
• Of course $B^{a}(t)$ stands for $B(t+a)-B(a)$, not for $B(t+a)$. Otherwise the result is wrong. – Did May 9 '18 at 7:46