Let $x_0 = 0$ and suppose that $x_{t + 1} \mid x_t \sim N((1-\alpha) x_t, \alpha^2)$. That is, $$ x_{t + 1} = (1- \alpha) x_t + \alpha w_{t+1}, \quad \mbox{for}~t \geq 1, $$ where $w_{t}$ are independent and identically distributed standard Normal random variables. Here, $\alpha \in (0, 1)$.
What is the distribution of $x_t$ as $t \to \infty$?
What I can see is that $$ x_t = \alpha \sum_{l = 0}^{t - 1} (1-\alpha)^l w_{t - l}. $$ Hence only the "most recent" $w_t$ matter. Since each of these are independent Normals, my guess would be that $x_t \to N(0, 1)$, say in distribution, as $t \to \infty$.
Is this correct?
If $X_1 \sim N(\mu_1, \sigma_1^2)$ and $X_2 \sim N(\mu_2, \sigma_2^2)$ and they are independent, then, $X_1 + X_2 \sim N(\mu_1+\mu_2, \sigma_1^2 + \sigma_2^2)$. This is known as the renewal property for normal distributions. One way to show this is to consider the characteristic function.
Then, $x_t \sim N(0, \alpha^2 \sum_{l = 0}^{t-1}(1-\alpha)^{2l})$. Hence, $x_t$ converges to $N(0, \frac{\alpha}{2-\alpha})$ in the total variation, in particular, converges in distribution, as $t \to \infty$.