Variance of $n$-dimensional Brownian motion (Oksendal)

Question

I apologize if this is a trivial question but I am confused by eq. $2.2.9$ in Oksendal (see image below). If $B_t$ is $n$-dimensional Brownian motion then isn't $B_t\sim\mathcal N(\mathbf x,tI_n)$ a $n$ vector? If so, what is meant by $E^x[(B_t-x)^2]$? Is it interpreted as the determinant of the covariance matrix (this would give the same solution of $nt$)?

Answer 1

It is to be understood in the sense of a scalar product, $\newcommand{\IE}{\mathbb{E}}$ $$ \IE^x[(B_t-x)^2]=\IE^x[\|B_t-x\|^2]=\sum_{k=1}^n\IE^x[(B_{k,t}-x_k)^2]=nt. $$ Or you could also call it the sum over the main diagonal of the covariance matrix.