Ergodicity of linear dynamical systems and convergence of covariance matrices

Question

Let $z(n+1)=Bz(n)+\xi(n+1)$ be an $N$-dimensional linear dynamical system with $\left(\xi(n)\right)_{n\in\mathbb{N}}$ being i.i.d. with $\xi(n)\sim\mathcal{N}(0,\Sigma_{\xi})$.

Assumptions: a) The covariance matrix $\Sigma_{\xi}$ is full-rank (so that the support of the distribution of $\xi(n)$ is absolutely continuous w.r.t. Lebesgue); b) $\xi\left(n+1\right)$ is independent of $z(n)$ for all $n\in\mathbb{N}$; c) the dynamical system is stable, i.e., the spectral radius of the matrix $B\in\mathbb{R}^{N\times N}$ is smaller than $1$, $\rho(B)<1$; d) the process $\left(z(n)\right)_{n\in\mathbb{N}}$ is initialized at the invariant distribution.

Define $R_k\overset{\Delta}=\mathbb{E}\left[z(n+k)z(n)^{\top}\right]$ as the $k$-lag moment associated with the process $\left(z(n)\right)_{n\in\mathbb{N}}$.

Define its empirical counterpart: $\widehat{R}_k(m)=\frac{1}{m}\sum_{i=1}^m z(i+k)z(i)^{\top}$.

Question.[Strong law] Is it true that under the discussed assumptions, we have

$$\widehat{R}_k(m)\overset{m\rightarrow \infty}\longrightarrow R_k,$$

for all $k<\infty$, almost surely?

Under these assumptions, the process $\left(z(n)\right)_{n\in\mathbb{N}}$ is geometrically ergodic -- in particular, its distribution converges to the invariant distribution w.r.t. the total variation norm.

A full proof or any reference would be greatly appreciated.

Answer 1

Under your assumptions, the following facts are true: \begin{equation*} \lim_{m\to\infty}\hat{R}_k(m)=R_k, \qquad \forall k\in \mathbb{N}_+, \quad \text{a.s.}, \tag{1} \end{equation*} where \begin{equation*} \hat{R}_k(m)=\frac1m\sum_{i=1}^{m}z(i+k)z(i)^\top, \qquad \forall k,m\in \mathbb{R}_+. \end{equation*} From your Assumption d), the intial distribution of $z$ is same as the distribution of \begin{equation*} z(0)=\sum_{n=0}^\infty B^n\xi(-n). \end{equation*}
If taking $z(0)$ as initial value for linear dynamic system $z(n+1)=Bz(n)+\xi(n+1)$, then
\begin{equation*} z(i) = \sum_{n=0}^\infty B^n\xi(i-n),\quad i\ge0, \end{equation*} hence $z=\{z(i), i\ge 0\}$ is ergodic stationary Gaussian Process. For each $k\in\mathbb{N}_+$, the \begin{equation*} y(i)=z(i+k)z(i)^\top, \quad i\in \mathbb{N}_+ \end{equation*} is also an ergodic stationary matrix-valued($\mathbb{R}^{N\times N}$-valued) process and \begin{equation*} \hat{R}_k(m)=\frac{1}{m}\sum_{i=1}^{m}y(i). \end{equation*} Therefore, using ergodicity of $y=\{y(i),i\in\mathbb{N}_+\}$, (1) is true.

Remark The following statements come from J. L. Doob, Stochastic processes, John Wiley & Sons, 1953. Ch X, $\S$1, pp452--.

In brief, the ergodicity belong to mearsure-preseving transformation $T$ defined by \begin{equation*} T\xi(i)=\xi(i+1), \quad \forall i\in \mathbb{N}_+. \tag{R1} \end{equation*} From (R1), \begin{align*} Tz(i) & =z(i+1), \\ Ty(i) & =y(i+1). \end{align*} The ergodicity of $\xi=\{\xi(i),i\in \mathbb{N}_+ \}$ is the ergodicity of $T$, which also guarantee the ergodicity of $z,y$.