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I would like to apply Kalman smoothing to a series of data sampled at irregular time points. There is a claim on Stack Exchange that "For irregular spaced time series it's easy to construct a Kalman filter", but I haven't been able to find any literature that specifically addresses this.

In my situation, I'd like to use a simple exponential covariance relationship to reflect the idea that the underlying continuous process is evolving as a linear dynamical system from which we irregularly receive samples.

So: is it simply OK to apply a Kalman filter with the "predict" step using a transition model and a process noise model whose "amplitude" depend on the amount of time that has elapsed since the last measurement?

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  • $\begingroup$ It is also not bad to take the Extended Kalman Filter in this case. It is integrable in time up to step of measurement. If the discrete Kalman filter assumptions are valid (linear, Markov, uncorrelated noise,...) then they are valid in the case of the EKF. Sometimes the EKF formulation is more useful because using appropriate numerical integrators (Runga-Kutta, Adams-Bashforth-Moulton) can be more accurate and easier to set up and use in continuous domain than in discrete (z-transform) domain. $\endgroup$ – EngrStudent Apr 15 '13 at 5:04
  • $\begingroup$ Is your question related to this one? dsp.stackexchange.com/questions/13838/… $\endgroup$ – Leo May 9 '18 at 8:57
  • $\begingroup$ @Leo not really $\endgroup$ – Dan Stowell May 10 '18 at 9:57
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Yes. In fact, this is how the Kalman Filter (KF) is also set up, at least implicitly. The assumptions in place when choosing the KF model, are that the movements and measurements compose a linear dynamical system. The transition matrix, $F_{t-1}$, (in the equation: $\hat{x}_{t|t-1} = F_tx_{t-1} + ...$, where $\hat{x}$ is the predicted state estimate) is in fact indexed by time, so irregular observations shouldn't be an issue.

For a more mathematically rigorous explanation of the KF, Max Welling has a really good tutorial that I highly recommend.

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    $\begingroup$ this doesn't answer the question at all. $\endgroup$ – Taylor Mar 25 '15 at 1:38

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