If two martingales are independent, they are orthogonal. Check that the converse does not hold.

Let \(B\) and \(W\) be two independent Brownian motions. The martingales \(W\) and \(M\) where \(M_{t}=\int_{0}^{t} W_{s} d B_{s}\) are orthogonal and not independent. Indeed, the martingales \(W\) and \(M\) satisfy \(\langle W, Mangle=0\). However, the bracket of \(M\), that is \(\langle Mangle_{t}=\int_{0}^{t} W_{s}^{2} d s\) is \(\mathbf{F}^{W}\)-adapted. One can also note that

\[\mathbb{E}\left(\exp \left(i \lambda \int_{0}^{t} W_{s} d B_{s}\right) \mid \mathcal{F}_{\infty}^{W}\right)=\exp \left(-\frac{\lambda^{2}}{2} \int_{0}^{t} W_{s}^{2} d s\right)\]

and the right-hand side is not a constant as it would be if the independence property held. The martingales \(M\) and \(N\) where \(N_{t}=\int_{0}^{t} B_{s} d W_{s}\) (or \(M\) and \(\widetilde{N}_{t}:=\int_{0}^{t} W_{s} d W_{s}\) ) are also orthogonal and not independent.