Suppose that $(X_k)_{k \in \mathbb{N}}$ is a stationary Markov chain with state space $\mathbb{R}$ and $0<\lambda<1$. Can we say something about the weak convergence of the sequence $Y_n = \sum_{k=0}^n \lambda^k X_{n-k}$?
This sequence obeys the autoregressive dynamics $Y_{n} = \lambda Y_{n-1} + X_n$ with $Y_0 = 0$. Finding the limiting distribution of $Y$ then reduces to finding a stationary distribution of the joint Markov chain $(X_{n+1}, Y_n)_{n \in \mathbb{N}}$. If $(X_k)_{k \in \mathbb{N}}$ were an i.i.d. Gaussian sequence, we had an AR(1) process and we could immediately use standard Markov chain theory to deduce that $(Y_n)_{n \in \mathbb{N}}$ converges weakly to its stationary distribution (as in Meyn & Tweedie (1993), Markov Chains and Stochastic Stability, Section 10.5.4).
However, in my case, the independence assumption is obviously dropped. In this case, one could find a weak limit by inspecting the stationary distribution of the joint process $(X_{n+1}, Y_n)_{n \in \mathbb{N}}$. Intuitively, if the sequence $X$ extended in both directions like $X=(X_k)_{k \in \mathbb{Z}}$, I would guess that the marginal law of $Y_n$ under this stationary distribution is something like the law of the infinite moving average $\sum_{k=0}^\infty \lambda^k X_{n-k}$. But in my case, $X$ starts at $k=0$ and not at $k=-\infty$ so that such a construction is not available.