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?