ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Singular state covariance with positive innovation covariance

Example

Take $T = 1$ with scalar state and observation, $X_{0} \sim N_{1} (0, 1)$ , $F_{1} = 0$ , $Q_{1} = 0$ , $H_{1} = 1$ , and $R_{1} = 1$ . Let $q_{1} \sim N_{1} (0, 0) = δ_{0}$ and let $r_{1} \sim N_{1} (0, 1)$ be independent of $X_{0}$ . Then $X_{1} = F_{1} X_{0} + q_{1} = 0$ and $Y_{1} = H_{1} X_{1} + r_{1} = r_{1}$ .

With no earlier observations, define the predicted state mean $m_{1 ∣ 0} : = E [X_{1}]$ , the state prediction covariance $P_{1 ∣ 0} : = Var (X_{1})$ , the observation prediction residual $v_{1} : = Y_{1} - H_{1} m_{1 ∣ 0}$ , and its innovation covariance $S_{1} : = Var (v_{1})$ . Then $P_{1 ∣ 0}$ is singular while $S_{1}$ is positive definite.

Facts & Assumptions

Given: The one-step scalar model in the Example.

[A1]

AC is required by the multivariate-normal and finite path-law suppliers. (The Axiom of Choice)

[F1]

The linear-Gaussian model permits positive-semidefinite initial, process-noise, and observation-noise covariance matrices. (Linear-Gaussian state-space models with independent noises)

[F2]

The finite linear-Gaussian path law exists and may have singular covariance. (The finite linear-Gaussian state-space path is jointly Gaussian)

[F3]

A law $N_{d} (m, Σ)$ has mean $m$ and covariance $Σ$ ; singular covariance is allowed. (Multivariate normal law, including singular covariance)

[F4]

Variance of a square-integrable real random variable is its centered second moment. (Moments, variance, and covariance on a probability space)

[F5]

The finite-horizon kernel model has the ordered joint law stated in its definition. (Finite-horizon state-space kernel law exists and is unique)

[F6]

A scalar normal law with zero variance is the point mass at its mean. (Standard normal and normal laws)

Verification

Given: $F_{1} = 0$ , $Q_{1} = 0$ , $H_{1} = 1$ , $R_{1} = 1$ , $X_{0} \sim N_{1} (0, 1)$ , $q_{1} \sim N_{1} (0, 0) = δ_{0}$ , and independent $r_{1} \sim N_{1} (0, 1)$ .

1.1A1F1F2F4F5F6givenalgebra

By [F6], $q_{1} \sim N_{1} (0, 0) = δ_{0}$ , hence $q_{1} = 0$ almost surely. These parameters satisfy the linear-Gaussian contract, and the finite path law exists by [F2,F5]. The recursion gives $X_{1} = 0$ , so $m_{1 ∣ 0} = 0$ and $P_{1 ∣ 0} = E [(X_{1} - m_{1 ∣ 0})^{2}] = 0$ by [F4]. The $1 \times 1$ zero matrix is singular.

2.1F3F4step 1.1algebra

Since $Y_{1} = r_{1}$ , the residual is $v_{1} = Y_{1} - H_{1} m_{1 ∣ 0} = r_{1}$ . Its mean is zero and its variance is $E [r_{1}^{2}] = 1$ by [F3,F4]. Hence $S_{1} = [1]$ is positive definite, since $u^{T} S_{1} u = u^{2} > 0$ for every nonzero $u \in R$ .

3.1step 1.1step 2.1algebra∎

Thus $P_{1 ∣ 0} = 0$ is singular while $S_{1} = [1]$ is positive definite. No inverse of the state covariance is used; the observation innovation covariance remains invertible.

Depends on

Used by

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Sources

Nicolai Meinshausen, Time Series Analysis: Notes for the course HS 2022 (standard reference, not scraped)
Dewei Wang, STAT 720 Time Series Analysis, Spring 2015 lecture notes (standard reference, not scraped)