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Suppose I have a multidimensional brownian motion $W=\{W_t\}$. Why is the following true:

$$\langle W^k,W^l\rangle_t = \delta_{k,l}t$$

where $W^k$ denotes the k-th coordinate, $\langle \cdot,\cdot\rangle$ denotes the bracket process and as usual $\delta_{k,l}$ the kronecker symbol.



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up vote 1 down vote accepted

The part which is not a definition is a consequence of the fact that two independent martingales $X$ and $Y$ are such that $\langle X,Y\rangle=0$.

This fact itself is a consequence of the following simple computation. For every $s\leqslant t$, $X_tY_t$ is $X_sY_s$ plus a sum of three terms $(X_t-X_s)Y_s+(Y_t-Y_s)X_s+(X_t-X_s)(Y_t-Y_s)$, where each term has zero conditional expectation conditionally on $\sigma(X_u,Y_u;u\leqslant s)$. Thus, $XY$ is a martingale and, in particular, $\langle X,Y\rangle=0$.

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can you give me a reference for: if $X,Y$ are two independent stochastic processes, then $\langle X,Y\rangle =0$? I didn't know this result. With it, everything is clear. – math Jun 11 '12 at 11:15
Typo: two independent martingales. – Did Jun 11 '12 at 12:23
Why is it true, that if $XY$ is a martingale, then $\langle X,Y\rangle=0$? – math Jun 11 '12 at 12:48
Well, $XY$ and $XY-\langle X,Y\rangle$ are both martingales hence $\langle X,Y\rangle$ is a martingale. But $\langle X,Y\rangle$ has finite variation since $4\langle X,Y\rangle=\langle X+Y\rangle-\langle X-Y\rangle$ is the difference of two nondecreasing processes. And a martingale with finite variation is... – Did Jun 11 '12 at 14:43
constant :) Thanks a lot! – math Jun 11 '12 at 14:56

I can't give you a rigorous proof but it's esentially due to the stochastic nature of the Brownian motion: all $W^j$ are independent, hence the Kronecker delta and $\langle W^j,W^j \rangle \propto t$ (actually $2Dt$, where $D$ is the diffussion coefficient).

$\langle W^j,W^k \rangle$ gives information about the correlation of $W^j$ and $W^k$

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