We define $J_{k,n}:=((k-1)2^{-n},k2^{-n}]$ for $n\in \mathbb{N}_0$ and $k=1,\dots,2^n$. Let $W$ be a Brownian Motion. Let $n\ge m$ and we assume $J_{k,n}\subset J_{l,m}$. W.l.o.g $J_{k,n}$ lies in the left half of $J_{l,m}$. Moreover we set $\Delta W([a,b])=W_b-W_a$, which is by definition normal distributed. Hence I know $\Delta W (J_{2k-1,n+1})-\Delta W(J_{2k,n+1})$ and $\Delta W(J_{2l,m+1})$ are independent, since $J_{k,n}$ lies in the left half of $J_{l,m}$. Furthermore, $(l-1)2^{-m}\le (k-1)2^{-n}\le k2^{-n}\le(2l-1)2^{-(m+1)}$. Why is the following computation true?
$E[(\Delta W (J_{2k-1,n+1})-\Delta W(J_{2k,n+1}))\Delta W(J_{2l-1,m+1})]=2^{-(n+1)}-0-2^{-(n+1)}+0=0$
They argue that all sub intervals of a dyadic partition have the same length.
EDIT:
Here is what I did so far:
$\Delta W (J_{2k-1,n+1})=W_{(2k-1)2^{-(n+1)}}-W_{(k-1)2^{-n}}$
$\Delta W(J_{2k,n+1}) = W_{k2^{-n}}-W_{(2k-1)2^{-(n+1)}}$
$\Delta W(J_{2l-1,m+1}) = W_{(2l-1)2^{-(m+1)}}-W_{(l-1)2^{-m}}$
$\Delta W (J_{2k-1,n+1})-\Delta W(J_{2k,n+1})=2W_{(2k-1)2^{-(n+1)}}-W_{(k-1)2^{-n}}-W_{k2^{-n}}$
From here, I do not know how to proceed.
hulik