Consider $0\leq s<t$ where $t,s$ represent time index.
I want to show a Brownian motion $W(s)$ is independent of $W(t)-W(s)$.
Specifically, show that $E[W(s)(W(t)-W(s))]=0$
Writing $W(s)$ as a telescoping sum and using the definition $W(0)=0$,
You can do the same for $W(t)-W(s).$
Denote the telescoping series of $W(s)$ as A and $W(t)-W(s)$ as B.
This is $E[AB].$
But since $AB$ is simply a sum of cross product of indepdent increments and each increment is normally distributed with mean zero, $E[AB]=0$. QED.
Is this proof correct?
Is there an "easier" proof?