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تسجيل مستخدم جديدRobust filtering: Correlated noise and multidimensional observation
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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 $Y={Y_s,sin[0,t]}$. 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.
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