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Invariant Gaussian law and uniqueness for a stable autoregression

Statement

Assume AC. Let $A \in R^{d \times d}$ and $G \in R^{d \times m}$ , with finite $d, m \geq 1$ . Suppose there are $C < \infty$ and $0 < θ < 1$ such that $∥ A^{n} ∥ \leq C θ^{n}$ for every integer $n \geq 0$ . A probability law $ν$ is invariant for this recursion if, whenever $Z_{0} \sim ν$ is independent of a fresh $ξ_{1} \sim N_{m} (0, I_{m})$ , the updated state $A Z_{0} + G ξ_{1}$ also has law $ν$ . Consider the Markov chain $Z_{n + 1} = A Z_{n} + G ξ_{n + 1},$ where each $ξ_{n + 1} \sim N_{m} (0, I_{m})$ is independent of the current state and of the earlier innovations. Then the chain has a unique invariant probability law. It is $N_{d} (0, P)$ , where the convergent positive-semidefinite series $P = \sum_{j = 0}^{\infty} A^{j} G G^{T} (A^{T})^{j}$ satisfies $P = A P A^{T} + G G^{T}$ . This is the unique finite matrix solution of that discrete Lyapunov equation.

Facts & Assumptions

Given: AC, matrices $A, G$ with the power bound above, and the stated Gaussian autoregression.

[F1]

Multivariate normal law, including singular covariance and Characteristic function of a multivariate normal law: for every positive-semidefinite covariance $P$ , $N_{d} (0, P)$ exists and has characteristic function $u \mapsto e^{- u^{T} P u / 2}$ , including singular $P$ .

[F2]

Characteristic functions under affine maps and independent sums: affine images transform characteristic functions by the corresponding linear map, and the characteristic functions of finite sums of independent vectors multiply coordinatewise through their scalar projections.

[F3]

Independent random elements and For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique: finite product probability spaces exist and their coordinate random elements are independent.

[F4]

Characteristic function of a multivariate normal law: any Borel probability law with characteristic function $u \mapsto e^{- u^{T} P u / 2}$ equals $N_{d} (0, P)$ , including singular $P$ .

[F5]

Dominated convergence: pointwise convergence dominated by one integrable function permits passage to the limit in the integral.

[A1]

The Axiom of Choice: AC is inherited through [F1]–[F5]; no choice is used in the convergent series or finite-dimensional characteristic-function iteration.

Proof

Given: The assumptions above.

1.1givenalgebra

The power bound gives $∥ A^{j} G G^{T} (A^{T})^{j} ∥ \leq C^{2} θ^{2 j} ∥ G ∥^{2}$ . The geometric series therefore converges entrywise and in matrix norm. Each summand is symmetric positive semidefinite because $x^{T} A^{j} G G^{T} (A^{T})^{j} x = ∣ G^{T} (A^{T})^{j} x ∣^{2} \geq 0$ , so $P$ is symmetric positive semidefinite. Shifting the summation index gives $A P A^{T} + G G^{T} = P$ .

2.1A1F1F2F3F4step 1.1givenalgebra

Let $Z \sim N_{d} (0, P)$ and $ξ \sim N_{m} (0, I_{m})$ be independent; [F3] supplies their finite product probability space. By [F1]–[F2], $A Z + G ξ$ has characteristic function $u \mapsto \exp [- \frac{1}{2} u^{T} (A P A^{T} + G G^{T}) u] = \exp [- \frac{1}{2} u^{T} P u]$ . The covariance identity in step 1.1 and [F4] show that $A Z + G ξ$ has the same law as $Z$ . Thus $N_{d} (0, P)$ is invariant.

3.1A1F1F2F3F4F5givenstep 1.1step 2.1algebra∎

Let $ν$ be any invariant probability law and let $φ (u) = \int e^{i u \cdot x} ν (d x)$ . Invariance and independence of a fresh innovation imply [A1, F1, F2, F3, F4, F5, given, step 1.1, step 2.1, algebra] $φ (u) = φ (A^{T} u) \exp ⁣ (- \frac{1}{2} u^{T} G G^{T} u) .$ Iterating $n$ times gives $φ (u) = φ ((A^{T})^{n} u) \exp ⁣ (- \frac{1}{2} u^{T} [\sum_{j = 0}^{n - 1} A^{j} G G^{T} (A^{T})^{j}] u) .$ The power bound gives $(A^{T})^{n} u \to 0$ , so the dominated convergence theorem [F5] in the integral defining $φ$ gives $φ ((A^{T})^{n} u) \to 1$ . Step 1.1 makes the matrix sum converge to $P$ , whence $φ (u) = e^{- u^{T} P u / 2}$ , the characteristic function of the invariant Gaussian from step 2.1. By [F4], $ν = N_{d} (0, P)$ . If $P^{'}$ is another finite matrix solution of the Lyapunov equation, $D = P^{'} - P$ satisfies $D = A D A^{T}$ , hence $D = A^{n} D (A^{T})^{n} \to 0$ by the power bound; therefore $P^{'} = P$ . This proves uniqueness of the invariant law and covariance solution.

Depends on

Used by

Exact invariant covariance and stability threshold for scalar harmonic BAOAB Theorem

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Sources

Jean-Noël Roux, Stéphane Rodts, and Gabriel Stoltz, Introduction à la physique statistique et à la physique quantique, §5.2.2 (standard reference, not scraped)
Benedict Leimkuhler and Charles Matthews, Rational Construction of Stochastic Numerical Methods for Molecular Sampling, §2 (standard reference, not scraped)