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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.

cheers

math

2 Answers 2

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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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    const$a$nt :) Thanks a lot!2012-06-11
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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 $W^k$