3

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 ...


3

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 ...


3

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{...


1

(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 ...


Only top voted, non community-wiki answers of a minimum length are eligible