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The Kalman-Bucy filter gives the best estimates for a partially observable linear system ( here I show a simplified version for exposition)

State $\theta_t $ of system: $d\theta_t = ( a_1\theta_t)dt + b_1dW_t$

Observation $S_t $ of system: $dS_t = ( A_1\theta_t)dt + B_2dZ_t$

where $dW_t$ and $dZ_t$ are independent random walks

the best estimate $m_t = E[\theta_t | \mathcal{F}^{S_t}$] is given by

$$dm_t = a_1m_tdt + \gamma_tA_1(dS_t-A_1m_tdt) \tag 1$$

where $\gamma_t = E[(\theta_t-m_t)^2]$ solves the following Riccati equation

$$\frac{d\gamma}{dt} = (a_1\gamma_t)^2 + b_1^2 - \frac{(\gamma_tA_1)^2}{B_2^2} \tag 2$$

My question: is there a way(maybe via linearizatio via Taylor or other method) to obtain a similar solution to the following non-linear problem

State $\theta_t $ of system: $d\theta_t = ( a_1\color{red}{f(\theta_t)})dt + b_1dW_t$

Observation $S_t $ of system: $dS_t = ( A_1\theta_t)dt + B_2dZ_t$

in which the evolution of $$\gamma_t = E[(\theta_t-m_t)^2] \tag 3$$
is still given by an explicit equation that can be solved analytically?

Thank you!

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I'm sure there are edge cases for which you could contrive a system for which the answer is "yes", but -- no.

Even for a linear time-invariant system, a Kalman-Bucy filter with more than one state requires solving a differential Riccati equation (from [1], where the notation is the commonly-used notation in the control systems community):

$$\dot {\mathbf P} = -\mathbf {P C^T R_c^{-1} C P + AP + PA^T + Q_c} \tag a $$

For certain restricted problems (like a 1-state integrating system) this Riccati equation can, indeed be solved symbolically. However, in general, even for "easy" LTI systems it gets solved numerically.

You can use a fairly typical development of the Extended Kalman Filter ([1], again) to derive a continuous-time Extended Kalman filter, but -- as with a discrete-time EKF -- the result is a Kalman gain that depends on the system state and, thus, that varies with time in an unexpected way. As long as you didn't mind calculating (a) online, you could make this work -- but not with a pre-calculated $\mathbf P$ or $\gamma$.


[1] "Optimal State Estimation", Dan Simon, Wiley 2006, p234.

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