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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

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up vote 3 down vote accepted
+50

Introducing the notations $X=\Delta W (J_{2k-1,n+1})$, $Y=\Delta W(J_{2k,n+1})$ and $Z=\Delta W(J_{2l-1,m+1})$, the task is to understand why $\mathrm E(XZ)=\mathrm E(YZ)$.

One sees that $X=\Delta W(A)$, $Y=\Delta W(B)$ and $Z=\Delta W(C)$, where $A$, $B$ and $C$ are some intervals such that $A\subseteq C$ and $B\subseteq C$.

Consider for example $A=(a,a')$ and $C=(c,c')$. Then $c\leqslant a\lt a'\leqslant c'$ and $Z=X'+X+X''$, where $X'=\Delta W((c,a))$ and $X''=\Delta W((a',c'))$. The increments of $W$ are independent and centered hence $X'$, $X$ and $X''$ are three independent and centered random variables. In particular $\mathrm E(X'X)=\mathrm E(X')\mathrm E(X)=0$ and $\mathrm E(X''X)=\mathrm E(X'')\mathrm E(X)=0$. Furthermore, the variance of $\Delta W(A)$ is the length of $A$, hence $$ \mathrm E(ZX)=\mathrm E(X'X)+\mathrm E(X^2)+\mathrm E(X''X)=\mathrm E(X^2)=|A|. $$ Likewise, $B\subseteq C$ hence $\mathrm E(ZY)=\mathrm E(Y^2)=|B|$. Finally, the identity $\mathrm E(XZ)=\mathrm E(YZ)$ is a consequence of the fact that $|A|=|B|$ (and this common length is $1/2^{n+1}$ in the present case).

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