# Gaussian process is a Brownian motion.

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 $$0 is $$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?

• "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. – John Dawkins May 25 at 16:52
• And why are linear combination univariate normal. Do you know any proof or a book where I can read it? – Matzi May 25 at 17:36