Kalman filter innovation residual inversion

Question

I'm trying to implement a Kalman filter in a computationally efficient way. The main issue is the inversion of the innovation residual:

  S=HPH^t+R

  K=PH^t*inv(S)

My question is, can one assume that the S matrix in positive definite? This would make inverting it more computationally efficient....

perhaps I should move this to the math community – staple Jul 01 '14 at 15:34 — staple, Jul 01 '14 at 15:34

score 1 · Accepted Answer · edited May 23 '17 at 12:09

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Yes, S is symmetric positive definite as mentioned here, and also how it can be automatically recognized:

Matrix Expressions and BLAS/LAPACK; SciPy 2013 Presentation

I find that presentation fascinating!

EDIT: I have done a little research and it turns out one has to be very careful when implementing the Kalman-Filter in order to retain the symmetric positive definiteness. Keywords: "stabilized Kalman filter" and "Josepf form". See for example Linear-Optimal Estimation for Discrete-Time Systems, page 7 or Understanding and Applying Kalman Filtering, page 28.

To address bits_international's comment, see Under what circumstance will a covariance matrix be positive semi-definite rather than positive definite? or page 7 in Understanding and Applying Kalman Filtering. In short, it is positive definite unless there is a linear dependence among the components.

edited May 23 '17 at 12:09

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answered Jul 01 '14 at 15:44

Ali

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Thanks! Can you point be toward a proof (however complicated)? – staple Jul 01 '14 at 16:32
as a covariance matrix S is positive-semidefinite, I am not sure if this is definite – 4pie0 Jul 01 '14 at 16:49
@staple I am glad to hear it helped! – Ali Jul 01 '14 at 19:31

score 1 · Answer 2 · answered Mar 23 '15 at 05:33

Another way to reduce the cost of inverting S is to use serial updates. In a nutshell: Predict, then repeat the update step for each row of H. In this case, HPH', R, and therefore S are all scalar, so inverting S is trivial. If R is diagonal (as is often the case) this doesn't require any extra work. If there are off-diagonal terms, you can orthogonalize your measurements. Most of the effort there can be pre-computed.

Several square root KF formulations rely on serial updates, so look at those for more ideas.

Kalman filter innovation residual inversion

2 Answers2