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I am using the following equations for Kalman filter. The state vector is $x$ and observations are $z$. Both $x$ and $z$ are multivariate vectors of same length.

Forecast model: $x_{t+1} = 0.963 x_t + \mathcal{N}(2.19,3.34)$

Observation model: $z_t = -0.00542849 x_t + \mathcal{N}(0.4524,0.038952)$

These relations hold between state and observation vectors element by element (since the lengths are same).

The observations are confined within an interval [0, 1] whereas $x$ values may range from [0,80].

I am not sure if the noise values I have determined are correct. I just fit a linear regression between $z$ and $x$ and used the MSE to find the standard deviation of noise. Is that correct?

Somehow, the update step introduces a lot of noise. So much that I am in fact better off not updating at all.

How does one estimate the noise covariance matrices given a set of $(x_{t+1},x_t)$ and $(z_t,x_t)$ values? Is my methods of fitting a linear regression and using $\sqrt{MSE}$ as standard deviation of noise correct?

EDIT: I should mention here that the relationship between $z$ and $x$ seems linear element-by-element and that's why I am learning a regression model.

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    $\begingroup$ One good way of estimating parameters in Kalman filter in such a simple setting is to use Maximum Likelihood Estimation. The exact likelihood is an easy analytic expression. I recommend the Durbin Koopman book (amazon.com/Series-Analysis-Methods-Statistical-Science/dp/…) for many practical issues (Chapter 7 for parameter estimation). eui.eu/Personal/Canova/Articles/ch6.pdf link formula (6.21) is a little more difficult to read but the same thing. By the way it is strange that you have X values at hand. If they are observable you don't need a kalman filter. $\endgroup$
    – Cagdas Ozgenc
    Nov 18, 2013 at 14:41
  • $\begingroup$ The $x$ values are known in retrospect but not available at the time of estimation. $\endgroup$
    – Nik
    Nov 20, 2013 at 0:47
  • $\begingroup$ If X values are available for the past then the Noise is also available to you (by Z-X). When signal and noise spectra is known I suppose you can design better filters. $\endgroup$
    – Cagdas Ozgenc
    Nov 20, 2013 at 7:55

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