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.
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 $(x_1,x_2,..,x_m) \to (x_1,x_1+x_2,...,x_1+...+x_m)$