I am referencing from the book Probability from Dava Khoshnevisan.

Thats my definiton of a brownian motion:

1) $W(0) =0$ and for all $t > 0$ is W(t) normal distributed with mean 0 and variance t.

2) For all $0<s<t$ is $W(t)-W(s)$ independent of $\lbrace W(u) \rbrace_{0 \leq u\leq s}$.

3) $W(t)-W(s)$ have the same distribution as $W(t-s)$

4) $t \mapsto W(t)$ sind (P-)a.s. continuous.

I don't understand a part of the proof of the wiener's theorem. It is written that if $W$ is a gaussian process with $W(0)=0$ and for all $0 \leq s \leq t$ $E[|W(t)-W(s)|^2= t-s$, than $W$ is a brownian motion.

The part where I struggle is 3) all the other parts are fine and already proofed. I know from 1) that $W(t)$ is a gaussian process with mean 0 and variance t. If $W(t)-W(s)$ is a gaussian process then it is clear that

$E[W(t)-W(s)]=0$ and $Var[W(t)-W(s)]=E[|W(t)-W(s)|^2= t-s$

and my work is done. But it isn't clear that $W(t)-W(s)$ is a gaussian process, is it?

  • $\begingroup$ "Gaussian process" means that the joint distribution of $W(t_1), W(t_2),\ldots,W(t_n)$ is multivariate normal, for each choice of $0\le t_1<t_2<\cdots<t_n$. In particular, $(W(s), W(t))$ is bivariate normal, so the linear combination $W(t)-W(s)$ is univariate normal. $\endgroup$ – John Dawkins May 25 at 16:52
  • $\begingroup$ And why are linear combination univariate normal. Do you know any proof or a book where I can read it? $\endgroup$ – Matzi May 25 at 17:36

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