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There is an Autoregressive model of order 1 (AR(1)) that is excited by a non-linear signal as the input: $$x_t = \rho x_{t-1} + u_t \tag{1}$$ The time series $u_t$ is generated from a nonlinear map, $$u_t = f(u_{t-1},\mathbf{w}) \tag{2}$$ where $f$ is the nonlinear function. The observations are $$y_t = x_t + v_t \tag{3}$$ where $v_t$ is the measurement noise that is an Additive White Gaussian Noise.

Q1: Can I re-write the model (1) as a state space in the following way:

$$x_t = Ax_{t-1} + f(u_{t-1},\mathbf{w}) $$

$$y_t = x_t+ v_t \tag{4}$$

Is the above representation correct? If not then I shall be grateful for the correct technique to represent it.

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    $\begingroup$ Does it matter how $u_t$ is generated? Why are you applying the Unscented KF? What parameters are you interested in estimating? Your simple substitution probably isn't quite there. You will probably have to include $u$ as part of the state, and then $A$ will instead be a nonlinear operation on $x_{t-1}$ and $u_{t-1}$. $\endgroup$ – Peter K.♦ Aug 7 '15 at 17:38
  • $\begingroup$ Not sure if my answer is clear; please comment on it and I'll try to update it. $\endgroup$ – Peter K.♦ Aug 11 '15 at 0:08
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Rather than write

$$ x_t = Ax_{t-1} + f(u_{t-1},\mathbf{w}) $$

as the state update equation, I'd write:

$$ \xi = \left[ \begin{array}{c} x_t\\ u_t \end{array} \right] $$

and then $$ \xi_t = g(\xi_{t-1},A,\mathbf{w}) $$

so that $$ y_t = \left[ 1\ \ \ 0 \right] \xi_t + v_t $$

Then you could apply the non-additive noise formulation of the EKF to get your filter equations.

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  • $\begingroup$ The representation makes sense but I am unsure about the non additive part. I need to do augmented state representation so how can I do augmented state representation where the process noise $f$ and the measurement noise are augmented states? $\xi$ appears to be the augmented state for signal state and process noise = $[\mathbf{x^T_{t-1},u_t^T}]$;. Please correct me if wrong $\endgroup$ – Srishti M Aug 12 '15 at 17:02
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    $\begingroup$ Well, your output equation is already linear, so you don't need to augment the state for the measurement noise; it's just the process noise that is problematic. Yes, $\xi$ as above is an augmented state. $\endgroup$ – Peter K.♦ Aug 12 '15 at 17:48

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