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Exact linear-Gaussian Kalman filtering — Examples

1 · Prerequisites

2 · Summary

Objective

These examples compute a scalar update and a two-dimensional update with singular state covariance. They verify that only the innovation covariance needs to be positive definite, while the exact filtered law may remain degenerate.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A one-step scalar Kalman update

Example

Take a one-step scalar model with X0∼N(0,1), transition X1=X0+q1 for q1∼N(0,1), and observation Y1=X1+r1 for r1∼N(0,1), with the three Gaussian variables independent. At the observation y1=2, the predicted law is N(0,2), the innovation variance is 3, the gain is 2/3, and the selected filtering-kernel value is N(4/3,2/3).

Facts & Assumptions

Given: AC and the stated independent-noise linear-Gaussian model.

[A1]

AC is assumed through the Gaussian model and conditional-law theorem. The Axiom of Choice.

[F1]

Under AC, the Kalman recursion with positive innovation covariance defines an everywhere filtering-kernel version of the exact conditional state law. The finite-horizon linear-Gaussian Kalman filter is exact.

[F2]

The model permits singular positive-semidefinite initial, process, and measurement covariances with mutually independent centered Gaussian noises. Linear-Gaussian state-space models with independent noises.

[F3]

Every positive-semidefinite covariance defines a multivariate-normal law under AC. Multivariate normal law, including singular covariance.

Verification

1.1A1F2F3givenalgebra

Set m0∣0=0, P0∣0=1, F1=H1=1, and Q1=R1=1. The covariances are positive semidefinite and the noises satisfy [F2]. The prediction is m1∣0=0 and P1∣0=1⋅1⋅1+1=2. Thus S1=1⋅2⋅1+1=3>0, and for y1=2 the innovation is v1=2.

2.1A1F1step 1.1algebra∎

Solving the scalar innovation system gives K1=2/3. Therefore m1∣1=0+(2/3)2=4/3 and P1∣1=2−(2/3)3(2/3)=2/3. By [F1], the selected conditional-law version at y1=2 is exactly N(4/3,2/3).

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Singular prior covariance with positive innovation variance

Example

Let Z∼N(0,1) and take X0=(Z,Z)T, so the initial covariance is P0∣0=(1111), which is singular; I2 below denotes the 2×2 identity matrix. Use F1=I2, Q1=0, H1=(1  0), and R1=1, with measurement noise independent of Z. At y1=2, the innovation variance is 2>0, and the selected filtering-kernel value is N2 ⁣((11),(1/21/21/21/2)). The filtered covariance remains singular.

Facts & Assumptions

Given: AC and the stated one-step linear-Gaussian model.

[A1]

AC is assumed through the degenerate Gaussian model and conditional-law theorem. The Axiom of Choice.

[F1]

The exact Kalman filtering theorem permits singular state and filtered covariances whenever the innovation covariance is positive definite. The finite-horizon linear-Gaussian Kalman filter is exact.

[F2]

Positive-semidefinite covariance matrices, including zero and singular ones, define multivariate-normal laws under AC. Multivariate normal law, including singular covariance.

[F3]

A positive-semidefinite operator has a positive-semidefinite square root, so the rank-one covariance is an admissible Gaussian covariance. A non-negative operator has a unique non-negative square root.

[F4]

Initial, process, and observation covariances in the linear-Gaussian model may be positive semidefinite, and the noises are mutually independent. Linear-Gaussian state-space models with independent noises.

Verification

1.1A1F2F3F4givenalgebra

The matrix P0∣0 is positive semidefinite because uTP0∣0u=(u1+u2)2≥0; it is singular because P0∣0(1,−1)T=0. Since EZ=0 and Var⁡(Z)=1, the Gaussian realization X0=(Z,Z)T has covariance (1111)=P0∣0. With Q1=0 and F1=I2, prediction gives m1∣0=(0,0)T and P1∣0=P0∣0.

2.1A1F1step 1.1algebra∎

Since H1=(1  0) and R1=1, S1=H1P1∣0H1T+R1=2>0. Hence K1=P1∣0H1TS1−1=(1/2,1/2)T and m1∣1=(1,1)T. The covariance update is P1∣1=P1∣0−K1S1K1T=(1/21/21/21/2), which is still singular and positive semidefinite. By [F1] this is the exact filtering-kernel covariance at y1=2, despite the singular prior covariance.

Sources