Orthogonal continuous local martingales ($[M,N]=0$)

Two continuous local martingales are orthogonal when their quadratic covariation is zero. With zero initial values, the martingale product identity says that their product is a local martingale. For two Brownian motions in a common filtration, orthogonality and the Lévy characterization of multidimensional Brownian motion make the pair a two-dimensional Brownian motion and imply independence. Merely having the two marginal Brownian motion laws on the same space does not imply orthogonality.

Let $M_1,\dots ,M_N$ be continuous local martingales starting at zero, with $⟨M_i,M_j⟩=0$ for $i\ne j$ and $⟨M_j{⟩}_{\infty }=\infty$ almost surely. The individual Dambis-Dubins-Schwarz theorem time changes produce independent standard Brownian motions $B_j$ with $M_j(t)=B_j(⟨M_j{⟩}_t)$. Independence is between the Brownian motions; each may still depend on its own clock.

For orthogonal continuous local martingales $B,\vartheta$ that are standard Brownian motions, the real part $X=e^B\cos \vartheta$ and imaginary part $Y=e^B\sin \vartheta$ are continuous local martingales. The Itô formula gives $dX=XdB-Yd\vartheta$ and $dY=YdB+Xd\vartheta$, since the two diagonal second-order terms cancel. Their quadratic variations equal ${\int }_0^t{e}^{2B_s}ds$ and their quadratic covariation is zero. If $C=[B,\vartheta ]$ is nonzero, the omitted drift terms are respectively $-YdC$ and $XdC$, so the orthogonality hypothesis is essential.