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State-space model laws — Examples

1 · Prerequisites

2 · Summary

The first example keeps the state transition fixed while changing the observation kernel, producing different joint state-observation laws. The second has a singular state prediction covariance and a positive-definite one-step observation innovation covariance. Both calculations use the specified joint model law and make no filtering or convergence claim.

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 →

Same state dynamics with distinct observation kernels

Example

Take $T = 1$ with scalar state and observation, $X_{0} \sim N_{1} (0, 1)$ , $F_{1} = 1$ , $Q_{1} = 0$ , and $R_{1} = 1$ . Let $q_{1} \sim N_{1} (0, 0)$ and $r_{1} \sim N_{1} (0, 1)$ be independent of each other and of $X_{0}$ . For $c \in {1, 2}$ , set $H_{1} = c$ and use the equations $X_{1} = X_{0} + q_{1}, Y_{1} = c X_{1} + r_{1} .$ The two models have the same initial and transition laws, but their observation kernels are $G_{1}^{(c)} (x, \cdot) = N_{1} (c x, 1)$ .

Facts & Assumptions

Given: The two one-step model specifications in the Example.

[A1]

AC is required for the stated Gaussian laws and the finite path-law supplier. (The Axiom of Choice)

[F1]

The general model law is the finite iterated product of its initial, transition and observation kernels. (General finite-horizon state-space models by kernels)

[F2]

The linear-Gaussian definition specifies Gaussian marginal laws, positive-semidefinite covariances, mutually independent noises and the displayed equations. (Linear-Gaussian state-space models with independent noises)

[F3]

The finite path has the exact kernel-product law and a multivariate normal law with its stated mean and covariance. (The finite linear-Gaussian state-space path is jointly Gaussian)

[F4]

A multivariate normal vector has its specified mean and covariance; singular covariance is allowed. (Multivariate normal law, including singular covariance)

[F5]

Independent random elements have independent generated sigma-algebras. (Independent random elements)

[F6]

Expectations of products of integrable functions of independent random variables factor. (Expectations factor over finite products of independent random variables)

[F7]

Covariance is bilinear on finite linear combinations and equals $E [X Y] - E [X] E [Y]$ for square-integrable variables. (Variance and covariance identities for random variables)

[F8]

Covariance and variance are defined for square-integrable real random variables by centered second moments. (Moments, variance, and covariance on a probability space)

[F9]

The finite-horizon kernel theorem gives a unique law with the ordered iterated-integral probabilities. (Finite-horizon state-space kernel law exists and is unique)

[F10]

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

Verification

Given: The parameters and independent Gaussian noises in the Example.

1.1A1F1F2F3F4F9F10given

By [F10], the degenerate noise satisfies $q_{1} = 0$ almost surely, so both models have $X_{1} = X_{0}$ and the same transition kernel $K_{1} (x, A) = 1_{A} (x)$ . Each set of kernels defines the state-space model. The finite-horizon theorem and the path-law theorem supply its joint law. At $x = 1$ , the two observation-kernel means are $1$ and $2$ , so the kernels themselves are distinct.

1.2F2F3F4F5F6F7F8algebra

In the model indexed by $c$ , $Y_{1} = c X_{0} + r_{1}$ , with both variables centered, $Var (X_{0}) = Var (r_{1}) = 1$ , and $Cov (X_{0}, r_{1}) = 0$ by independence and factorization of expectations. Covariance bilinearity gives $Cov (X_{0}, Y_{1}) = c Var (X_{0}) + Cov (X_{0}, r_{1}) = c$ . Thus the two joint laws have cross-covariances $1$ and $2$ .

2.1F3F4F8step 1.2contradiction∎

Both joint laws have finite second moments. If the laws were equal, the covariance of their first and third coordinates would be equal. Step 1.2 gives different values, so the joint laws are different.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.

State-space model laws — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Same state dynamics with distinct observation kernels

Example

Facts & Assumptions

Verification

Singular state covariance with positive innovation covariance

Example

Facts & Assumptions

Verification

Sources