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.

  • 1
    $\begingroup$ What is a pre-Brownian motion...? $\endgroup$
    – saz
    Commented May 9, 2018 at 6:54
  • $\begingroup$ "Can someone explain to me how to prove..." Sure, but what did you try yourself? $\endgroup$
    – Did
    Commented May 9, 2018 at 7:46
  • $\begingroup$ @saz My understanding is a standard Brownian motion without the assumption of continuous sample paths. $\endgroup$
    – Math1000
    Commented May 9, 2018 at 8:12

1 Answer 1


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

  • 1
    $\begingroup$ Of course $B^{a}(t)$ stands for $B(t+a)-B(a)$, not for $B(t+a)$. Otherwise the result is wrong. $\endgroup$
    – Did
    Commented May 9, 2018 at 7:46
  • $\begingroup$ @Did Very silly of me. I have edited the answer. $\endgroup$ Commented May 10, 2018 at 4:54
  • $\begingroup$ @Kavi Rama Murthy Thanks for your answer :). It really helped me. $\endgroup$
    – Crystal
    Commented May 10, 2018 at 12:12
  • $\begingroup$ @KaviRamaMurthy Thanks for this answer. But I see you have made the increments discrete but the canonical sigma algebra we are looking at has is specifically "continuous" increments, so surely this proof is not suitable because there are uncountably many random variables that generate the sigma algebra $\endgroup$
    – MinaThuma
    Commented Sep 21, 2019 at 14:11
  • $\begingroup$ @MinaThuma Proving that two sigma algebras are independent is equivalent to proving that any two finite collections of events from them are independent. I am certainly not considering a discrete process. $\endgroup$ Commented Sep 21, 2019 at 23:09

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