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The Kushner equation is not suitable for numerical solution, because of its nonlinearity, but it does give the quantity, a normalized measure, you ultimately want. The Zakai equation, in contrast, can be readily solved numerically (Galerkin method), and if it has a unique solution it gives the solution of the Kushner equation upon normalization. So the ...

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I'm only going to answer the special case $b\equiv 1$ and $\bar W$ independent from the signal noise because I'm not familiar enough with the case of multiplicative/correlated observation noise. However, as far as I have been able to glean from the literature, adding those generalizations is pretty straightforward, if a little technical. With this ...

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If we assume $A$ constant $$\frac{d}{dt}\mathbb{E}(X_t )=A \mathbb{E}(X_t )$$so $\mathbb{E}(X_t)=e^{tA}X_0$ and $\mathbb{E}Y_t = Y_0 + \int_0^t H_s e^{sA}X_0ds$. For the variance, we can assume $X_0=0$ and $Y_0=0$. And we have $$\frac{d}{dt}\mathbb{E}(X_tX_t^T )=A\mathbb{E}(X_tX_t^T )+\mathbb{E}(X_tX_t^T )A^T+C_tC_t^T$$so $$\mathbb{E}(X_tX_t^T )=\int_0^t e^... 2 For simplicity take E=\Bbb R and the time interval to be [0,1], and think of X=(X_t)_{0\le t\le 1} as a random element of C=C([0,1]\to\Bbb R), a Polish space. We then have a regular conditional distribution of X given \mathcal G, call it Q=Q(\omega,B), \omega\in\Omega, B\in\mathcal B(C). And the induced "marginal conditional ... 1 No. E.g., let n=m=1 and X_t=Z_t=B_t, where B is the standard Brownian motion. Take the natural filtrations, so that E(f(X_t)|\mathcal G_t)=f(B_t). Let f(x) to be something like \max(0,x). It should be easy to show that the process (f(B_t)) is not Markov. Indeed, to simplify calculations, let f(x):=1(x>0). Then$$P(f(B_3)=1|f(B_2)=0)=\frac{...

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(Answering your comment) Off the top of my head, I'd look in vol.2 of Probabilités et Potentiel (Dellacherie & Meyer) or in Limit Theorems for Stochastic Processes (Jacod & Shiryaev). Another convenient resource is the blog https://almostsure.wordpress.com of Geo. Lowther. The key is that for a bounded rc martingale $M$, the predictable projection ${}... 1 Continuity at zero is included in the RCLL property. If$X(\cdot,\omega)$is discontinuous at some$t \in (0,t_0)$, then the left and right limits at$t$(which exist) must differ. Thus there must exist an integer$n>0\$ such that $$\|\lim_{s\to t^{-}} X(s,\omega)-\lim_{s\to t^{+}} X(s,\omega)\|>1/n \,.$$ Since the limits exists, this implies that ...

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