# Multidimensional brownian motion, vectorial normal distribution and independence

while studying Brownian motion I got really lost in something that I suppose its pretty basic but Im a little rusty, so if anyone could help would do me a big favor.

Problem:

Let $$(B_{t})_{t \geq 0}$$ be the standar brownian motion, which is the rigorous way to show that for $$0 the vector $${B}=(B_{t_{1}},...,B_{t_{n}})^{T}$$ has a normal vectorial distribution? and how could I show that $$B_{t_{1}}, (B_{t_{k+1}}-B_{t_{k}})_{k=1}^{n-1}$$ are in fact independent.

Thanks so much for the help :)

$$B = \begin{bmatrix} 1 & 0 & .. & 0 \\ 1 & 1 & .. & 0 \\ ... \\ 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} B_{t_1} \\ B_{t_2} - B_{t_1} \\ .. \\ B_{t_n} - B_{t_{n-1}} \end{bmatrix} = M \alpha Z$$
So $$B$$ is a normal random vector: $$B \sim \mathcal{N}(0, \alpha^2 MM^T)$$
• Thanks so much for your answer, but, is there a way to show that the $B_{t_{1}}, (B_{t_{k}}-B_{t_{k-1})_{k=2}^{n}$ are independent using the fact that B is a normal random vector as you showed? – user1trill Nov 9 at 0:21