I am confused on the multi-dimensional Brownian motion.
$B_t$ is a standard Brownian motion based on a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \geq 0}, \mathbb{P})$ if
$$Cov(B_t,B_s)=\mathbb{E}\left[B_tB_s\right]= \min (s,t)$$
When $B_t=(B_t^1, \cdots , B_t^d)$ is taking its values in $\mathbb{R}^d$,
the above equality is valid for each $B_t^i, \, i=1, \cdots , d$;
Which is the same for all $B_t^i$ because $Cov(B_t^i,B_s^i)= \min (s,t)$ does not depend on $i$.
How can we write the variance-covariance matrix $C_{i,j}=Cov(B_t^i,B_s^j)$ in terme of $\min (s,t)$?
$Diag(C_{i,j})= \min (s,t)$. What about the other termes?
Best Answer
In a multi-dimensional Brownian motion, the components are independent; this is normally part of the definition. So $\{B_t^i, t \ge 0\}$ is independent of $\{B_t^j, t \ge 0\}$ whenever $i \ne j$. Thus $Cov(B_t^i, B_s^j) = 0$ whenever $i \ne j$.