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So I have a discrete stochastic process defined by

$x_{k+1}=Ax_k+Bw_k$

where $w_k$ is zero mean Gaussian white noise with covariance $R_w$, and where $A$ has its eigenvalues in the unit disk.

I can get the mean and covariance updates for $x$ as

$E[x_{k+1}]=AE[x_k]=A^{k+1}E[x_0]$

$E[x_{k+1}x_{k+1}^T]=AE[x_{k}x_{k}^T]A^T+BR_w B^T$

and with $A$ having its eigenvalues in the unit disk implying $A^n$ goes to $0$ as $n$ goes to $\infty$, I can see how the mean of $x$ will converge to $0$.

However, apparently the covariance of $x$ should also converge making $x$ converge to a stationary process, but I am having trouble showing this mainly because the integrated covariance seems to have a summation with an unbounded amount of terms $A^nBR_wB^T[A^T]^n$ with different $n$.

How can I show that the covariance of $x$ will also converge as $k$ goes to $\infty$?

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1 Answer 1

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For $k \in \mathbb{N}$ the covariance of $x_k$ is given by $\mathbb{E}\left[ x_k x_k^\top \right] - \mathbb{E}\left[ x_k \right] \mathbb{E}\left[ x_k \right]^\top$ where

$$\mathbb{E}\left[ x_k x_k^\top \right] = A^{k} \mathbb{E}\left[ x_0 x_0^\top \right] \left( A^k \right)^\top + \sum_{i=0}^k A^i B R_w B^\top \left( A^\top \right)^i$$

and

$$\mathbb{E}\left[ x_k \right] \mathbb{E}\left[ x_k \right]^\top = A^k \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \left( A^\top \right)^k$$

so the covariance is

$$A^{k} \left( \mathbb{E}\left[ x_0 x_0^\top \right] - \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \right) \left( A^k \right)^\top + \sum_{i=0}^k A^i B R_w B^\top \left( A^\top \right)^i$$

As you've correctly identified,

$$k \to \infty \implies A^{k} \left( \mathbb{E}\left[ x_0 x_0^\top \right] - \mathbb{E}\left[ x_0 \right] \mathbb{E}\left[ x_0 \right]^\top \right) \left( A^k \right)^\top \to 0$$

For the remaining terms, we need some functional analysis facts.

  • $\mathbb{R}^{m \times m}$ with the operator norm is a Banach space.
  • The set of PSD matrices $\mathbb{S}^m_+ \subseteq \mathbb{R}^{m \times m}$ is also a Banach space.
  • In a Banach space $X$, every absolutely convergent series converges in $X$.

Therefore,

$$\sum_{i=0}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert < \infty \implies \sum_{i=0}^{\infty} A^i B R_w B^\top \left( A^\top \right)^i \text{ converges in $\mathbb{S}_+^m$}$$

and it only remains for us to show $\sum_{i=1}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert < \infty$.

$$\sum_{i=0}^\infty \left \lVert A^i B R_w B^\top \left( A^\top \right)^i \right \rVert \leq \left \lVert B R_w B^\top \right \rVert \sum_{i=0}^\infty \lambda_{\max}\left( A \right)^{2i} < \infty$$

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