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
On
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 $W^k$
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, $\langle X,Y\rangle=0$.