Let $v_k$ and $w_k$, $k \in \mathbb{N}$, be independent Gaussian processes with $v_k \sim \mathcal{N}(0,\sigma_v) $ and $w_k \sim \mathcal{N}(\mu_w,\sigma_w)$. Let $p \in (0,1)$ and consider the switched process: $$x_{k+1} = \cases{ ax_k + v_k \quad \text{ w.p. } \quad 1-p,\cr bx_k + w_k \quad \text{ w.p. } \quad p, }$$ where $a,b \in (0,1)$ and "w.p." stands for "with probability". That is, with probability $p$, $x_{k+1}$ samples from $ax_k + v_k$, and with probability $1-p$, it samples from $bx_k + w_k$, for all $k \in \mathbb{N}$.
I would like to find a closed form of the stationary distribution of $x_k$. From Monte Carlo simulations, I know it does have a stationary distribution. Moreover, for $b=0$, I can find a closed form of the distribution by writing $x_{k}$ as an infinite-dimensional Markov Chain. For $b=0$, the stationary distribution is an infinite Gaussian mixture given by $$x_{k} \sim \cases{ \mathcal{N}(\mu_w,\sigma_w) \quad \text{ w.p. } \quad p,\cr \mathcal{N}(a\mu_w,a^2\sigma_w + \sigma_v) \quad \text{ w.p. } \quad (1-p)p,\cr \mathcal{N}(a^2\mu_w,a^4\sigma_w + (a^2+1)\sigma_v) \quad \text{ w.p. } \quad (1-p)^2p,\cr \mathcal{N}(a^3\mu_w,a^6\sigma_w + (a^4+a^2+1)\sigma_v) \quad \text{ w.p. } \quad (1-p)^3p,\cr \mathcal{N}(a^4\mu_w,a^8\sigma_w + (a^6 + a^4+a^2+1)\sigma_v) \quad \text{ w.p. } \quad (1-p)^4p,\cr \qquad \qquad \qquad \qquad \qquad \qquad \vdots }$$ Now, for $b \neq 0$, I cannot write the infinite-dimensional Markov Chain as every state has an infinite number of transitions. Any hint about how to tackle this problem? Thanks!