I hope this is the right forum to ask this question.

Sequential filtering frameworks perform 2-steps essentially:

Process model (forecast) : $x_{t+1} = f_t(x_t) + v_t$

Observation model (update): $z_t = h_t(x_t) + w_t$

Here, $v_t, w_t$ are uncorrelated zero-mean noise processes, and $f_t, h_t$ are functions (may be linear or non-linear).

For the particular case of Kalman filters, $f_t, h_t$ are linear and the noise processes are Gaussian.

In my application, the user learns the observation model using historical data (linear regression, for example). Since that is the case, the user might as well learn an inverse observation model, such that $$x_t = g_t(z_t) + \eta_t$$

Here, the underlying state vector is expressed as a function of observations $z_t$. Assume that $\eta_t$ is Gaussian.

With the inverse observation model, the update step is simplified. Apply $g_t$ to $z_t$ and then simply assimilate the resulting vector (call it $\hat{x}_t$) to $x_t$. The assimilation here is a weighted addition of two Gaussian distributed variables where the weights are in inverse proportion of their covariance matrices (for Kalman optimal solution).

So my question is: is there anything wrong with this approach of using an inverse observation model instead of an observation model? The advantage is that I can define the weighted addition as an operator in my data structure.


1 Answer 1


If I understand well, what you propose is correct IF what is learned at time $t$ is uncorrelated from what you already know. The (conventional) Kalman update equation is a linear combination of the old state vector and the innovation, which is orthogonal to the old state vector.

  • $\begingroup$ What I am suggesting is, if $h_t$ is an invertible function, the observation model can be re-written as ${h_t}^{-1}(z_t) = x_t + w'_t$. Now Kalman update step is just a weighted linear sum of $x_t$ and ${h_t}^{-1}(z_t)$ (let's call this $\hat{x}_t$). I don't see why things have to be uncorrelated as you pointed out. Could you please explain that a bit? $\endgroup$
    – Nik
    Commented Oct 5, 2013 at 20:10
  • $\begingroup$ This adds the condition of invertibility of $h_t$ (which was no apparent in the original question). Anyway, the problem is with your statement that $z_t$ is historical data. In your original formulation, this would imply that information that is incorporated into the state vector at moment $t-1$ via $g(z_{t-1})$ would be reused in $t$ via g(z_{t})$. $\endgroup$
    – F. Tusell
    Commented Oct 6, 2013 at 18:17
  • $\begingroup$ Additionally, $h^{-1}(z_t) \not= x(t) + w_t'$ (unless $h^{-1}()$ is linear or you make $w_t'$ depend on $x_t$). $\endgroup$
    – F. Tusell
    Commented Oct 6, 2013 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.