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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 $0<t_{1}<...<t_{n}$ 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 :)

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The independence is a property of the standard Brownian Motion (Wiener Process), i.e. by definition.

You can observe that:

$$ B = \begin{bmatrix} 1 & 0 & .. & 0 \\ 1 & 1 & .. & 0 \\ ... \\ 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} B_{t_1} \\ B_{t_2} - B_{t_1} \\ .. \\ B_{t_n} - B_{t_{n-1}} \end{bmatrix} = M \alpha Z $$

where Z is a standard normal multivariate distribution, again following from the properties of the process.

So $B$ is a normal random vector: $B \sim \mathcal{N}(0, \alpha^2 MM^T)$

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  • $\begingroup$ 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? $\endgroup$ – user1trill Nov 9 at 0:21

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