A vanilla Kalman Filter allows for a time varying observation matrix $H_k$. Is it allowable for $H_k$ to be a function of the system state $x_k$ in a vanilla Kalman filter?

First, am I correct that this is not allowed in the vanilla Kalman filter? If it is allowed, then I don't understand how measurements would then act to correct the parameters of $x$ that predict them via $H_k$. I'd appreciate a more explicit description of this than my own hunch though!

Second, am I correct that the Extended Kalman Filter & the Unscented Kalman Filter do explicitly allow for this situation? If I am, do these filters allow observations to correct the terms of $x$ that are predicting them?


1 Answer 1


The non-linear system model that the unscented and extended Kalman filters work with (omitting control) is the following: $$x_{k+1} = F(k, x_k, v_k) \\ y_k = H(k, x_k, n_k)$$ where $F$ is the state transition function and $H$ is the observation function. The linear Kalman filter is just the special case where $F$ and $H$ have the form: $$ F(k, x, v) = A_k x + v \\ H(k, x, n) = B_k x + n$$ for sequences of matrices $A_k$ and $B_k$. In particular, the functions $F$ and $H$ are linear.

The following is a bit of a pedantic distinction (particularly in the non-linear case) but while the expression $F(k, x_k, v_k)$ clearly depends on $x_k$, the function $F$ itself does not. In the non-linear case this is no limitation. In the linear case, if the matrix $B_k$ depended $x_k$ then the function $H$ would no longer be linear.

  • $\begingroup$ Thank you – this formalises how I thought it probably was. $\endgroup$
    – Benjohn
    Jan 21, 2016 at 12:37
  • $\begingroup$ Incidentally – I originally used "linear" Kalman filter, then edited it to be "vanilla". I think linear would be the correct term, and I should revert to use that? $\endgroup$
    – Benjohn
    Jan 21, 2016 at 12:38
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    $\begingroup$ For me, "Kalman filter" by itself unambiguously means the linear case. I was a little hesitant about using "linear Kalman filter" as it makes it sound like there is a non-linear Kalman filter where there are non-linear generalizations of the Kalman filter. I don't think there's a strong reason to change the wording. $\endgroup$ Jan 21, 2016 at 13:01

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