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?


1 Answer 1


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 $i \ne j$.

  • 2
    $\begingroup$ And $\mathrm{Cov}(B^i_t,B^j_s)=0$ (to answer the question asked by the OP). $\endgroup$
    – Did
    Dec 18, 2011 at 20:19
  • $\begingroup$ What about a fractional Brownian motion defined by it's covariance $$ R_{H}(t,s) := \frac{H}{2} ( {|t|}^{2{H}} + {|s|}^{2{H}} - {|t-s|}^{2{H}}) $$, there is no independance. $\endgroup$
    – Zbigniew
    Dec 18, 2011 at 20:39
  • 1
    $\begingroup$ $E[B_tB_s]$ means the expectation of the kronecker product between $B_t$ and $B_s$, so it is a matrix. If $B^i$ and $B^j$ are independent then the matrix contains a lot of zeros, if not, then it does not :) $\endgroup$
    – Martingalo
    Mar 13, 2014 at 7:57

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