Robust filtering: Correlated noise and multidimensional observation

In the late seventies, Clark [In Communication Systems and Random Process Theory (Proc. 2nd NATO Advanced Study Inst., Darlington, 1977) (1978) 721-734, Sijthoff & Noordhoff] pointed out that it would be natural for $\pi_t$, the solution of the stochastic filtering problem, to depend continuously on the observed data $Y=\{Y_s,s\in[0,t]\}$. Indeed, if the signal and the observation noise are independent one can show that, for any suitably chosen test function $f$, there exists a continuous map $\theta^f_t$, defined on the space of continuous paths $C([0,t],\mathbb{R}^d)$ endowed with the uniform convergence topology such that $\pi_t(f)=\theta^f_t(Y)$, almost surely; see, for example, Clark [In Communication Systems and Random Process Theory (Proc. 2nd NATO Advanced Study Inst., Darlington, 1977) (1978) 721-734, Sijthoff & Noordhoff], Clark and Crisan [Probab. Theory Related Fields 133 (2005) 43-56], Davis [Z. Wahrsch. Verw. Gebiete 54 (1980) 125-139], Davis [Teor. Veroyatn. Primen. 27 (1982) 160-167], Kushner [Stochastics 3 (1979) 75-83]. As shown by Davis and Spathopoulos [SIAM J. Control Optim. 25 (1987) 260-278], Davis [In Stochastic Systems: The Mathematics of Filtering and Identification and Applications, Proc. NATO Adv. Study Inst. Les Arcs, Savoie, France 1980 505-528], [In The Oxford Handbook of Nonlinear Filtering (2011) 403-424 Oxford Univ. Press], this type of robust representation is also possible when the signal and the observation noise are correlated, provided the observation process is scalar. For a general correlated noise and multidimensional observations such a representation does not exist. By using the theory of rough paths we provide a solution to this deficiency: the observation process $Y$ is "lifted" to the process $\mathbf{Y}$ that consists of $Y$ and its corresponding L\'{e}vy area process, and we show that there exists a continuous map $\theta_t^f$, defined on a suitably chosen space of H\"{o}lder continuous paths such that $\pi_t(f)=\theta_t^f(\mathbf{Y})$, almost surely.