Consider the linear discrete-time stochastic systems: \begin{equation} x_{k+1} = Ax_k + v_k, \end{equation} with time-instants $k \in \mathbb{N}$, state $x_k \in \mathbb{R}^n$, stochastic process $v_k \in \mathbb{R}^n$, and matrix $A \in \mathbb{R}^{n \times n}$. The process $v_k$ is an i.i.d. multivariate stochastic processes with $E[v_k]=\mathbf{0}$ and finite covariance matrix $R_{1} := E[v_k v_k^T] \in \mathbb{R}^{n \times n}$, $R_{1} > 0$ (positive definite). The initial state $x_1$ is a random vector with $E[x_1]=\mathbf{0}$ and finite covariance matrix $R_0 \in \mathbb{R}^{n \times n}$, $R_0 >0$. The process $v_k$, $k \in \mathbb{N}$ and the initial condition $x_1$ are mutually independent.

What I know is that if the eigenvalues of $A$ are inside the open unit disk, $x_1 \sim \mathcal{N}(\mathbf{0},R_0)$, and $v_k \sim \mathcal{N}(\mathbf{0},R_1)$, the process $x_k$ converges in distribution to $\mathcal{N}(\mathbf{0},P)$, where $P$ denotes the unique solution of $APA^T-P = R_1$.

My question is if we can conclude something about the asymptotic distribution of $x_k$ if $x_1$ and $v_k$ are not Gaussian. I know that even if they are not Gaussian the asymptotic mean is zero and the asymptotic covariance matrix is given by the same $P$. However, does having zero mean and finite covariance matrix imply the existence and uniqueness of an asymptotic distribution?

I have tried simulations with different distribution for $x_1$ and $v_k$, and as long as $E[x_1]=\mathbf{0}$ and $E[v_k]=\mathbf{0}$ with finite covariance matrices, I get unique asymptotic distributions.

Any hint or reference about this? Cheers, Carlos.