3
$\begingroup$

I'm working through the derivation for the Kalman filter using orthogonal projections. It's pretty intuitive that if we want to generate the best linear estimate for an $n$-vector $x$ using some $m$-vectors $y_0, \dots, y_t$, we would take the orthogonal projection of $x$ onto the subspace $\mathbb{R}^m$ spanned by the vectors $y_i$.

In the Kalman filter case, it seems that we assume $n>m>t$. That is, a lower dimensional observations $(y)$ than states $(x)$ we are trying to estimate. Similarly, we are trying to build a recursive estimator that uses additional observations $y_i$ such that at timestep $t$ we have $t$ observation vectors and our subspace is of dimension $t$.

My question is this: what happens when $t=m=n$? This would mean, I think, that the orthogonal projection is meaningless and it wouldn't help your filter to gain new estimates... is this correct?

$\endgroup$
1
  • $\begingroup$ $t$ is the time step here, so you should technically write it as $t_k$ where $k = 1,...,N$. The time step is irrelevant to the Kalman filter, it is just a means of dynamics propagation. What is important is the dimensionality of the state and measurements. If $m = n$ then you have as many measurements as state variables so the mapping is from $\mathbb{R}^n \mapsto \mathbb{R}^n$, which is perfectly valid. Only in the case where $m < n$ can you run into problems with observability, where the filter can’t estimate your state because of your measurement model. $\endgroup$ Commented Apr 14, 2020 at 8:31

0

Your Answer

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

Browse other questions tagged or ask your own question.