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Singular prior covariance with positive innovation variance

Example

Let $Z \sim N (0, 1)$ and take $X_{0} = (Z, Z)^{T}$ , so the initial covariance is $P_{0 ∣ 0} = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}),$ which is singular; $I_{2}$ below denotes the $2 \times 2$ identity matrix. Use $F_{1} = I_{2}$ , $Q_{1} = 0$ , $H_{1} = (1 0)$ , and $R_{1} = 1$ , with measurement noise independent of $Z$ . At $y_{1} = 2$ , the innovation variance is $2 > 0$ , and the selected filtering-kernel value is $N_{2} ⁣ ((\begin{array}{c} 1 \\ 1 \end{array}), (\begin{array}{cc} 1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2 \end{array})) .$ 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 $P_{0 ∣ 0}$ is positive semidefinite because $u^{T} P_{0 ∣ 0} u = (u_{1} + u_{2})^{2} \geq 0$ ; it is singular because $P_{0 ∣ 0} (1, - 1)^{T} = 0$ . Since $E Z = 0$ and $Var (Z) = 1$ , the Gaussian realization $X_{0} = (Z, Z)^{T}$ has covariance $(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}) = P_{0 ∣ 0}$ . With $Q_{1} = 0$ and $F_{1} = I_{2}$ , prediction gives $m_{1 ∣ 0} = (0, 0)^{T}$ and $P_{1 ∣ 0} = P_{0 ∣ 0}$ .

2.1A1F1step 1.1algebra∎

Since $H_{1} = (1 0)$ and $R_{1} = 1$ , $S_{1} = H_{1} P_{1 ∣ 0} H_{1}^{T} + R_{1} = 2 > 0$ . Hence $K_{1} = P_{1 ∣ 0} H_{1}^{T} S_{1}^{- 1} = (1 / 2, 1 / 2)^{T}$ and $m_{1 ∣ 1} = (1, 1)^{T}$ . The covariance update is $P_{1 ∣ 1} = P_{1 ∣ 0} - K_{1} S_{1} K_{1}^{T} = (\begin{array}{cc} 1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2 \end{array})$ , which is still singular and positive semidefinite. By [F1] this is the exact filtering-kernel covariance at $y_{1} = 2$ , despite the singular prior covariance.

Depends on

Used by

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Sources

Simo Särkkä, Bayesian Filtering and Smoothing, Chapter 4, §4.3 and Appendix A.1 (standard reference, not scraped)
Kody Law, Andrew M. Stuart, and Konstantinos Zygalakis, Data Assimilation: A Mathematical Introduction, Chapter 4, §4.1 (standard reference, not scraped)