Marginal distribution of an autoregressive process of order one AR(1) I'm reading "Econometric Modelling with Time Series" by V. L. Martin, A. S. Hurn and D. Harris ( https://www.researchgate.net/file.PostFileLoader.html?id=56bccdaa6225ff0de28b45a6&assetKey=AS:328005403594753@1455213991422 ).
I don't understand example 1.9 at page 11.
Let's consider an AR(1) model : $Y_t = \rho*Y_{t-1} + \upsilon_t$ $,$
$$\upsilon_t \sim i.i.d.\text{N} \Big( 0, \sigma^{2}\Big),$$ $$|\rho|<1$$
Why is the marginal distribution of $Y_1 \sim \text{N} \Big( 0, \frac{\sigma^2}{1-\rho^2} \Big)$ ?
Thanks for your help!
Here the mentioned example


This is a good technical question.  Without being given the initial distribution of $Y_0$, the rigorous proof of this well-known result actually requires advanced probability theory.  For the sake of rigor, we need to assume $t$ ranges over $t = 0, \pm 1, \pm 2, \ldots$.
Suppose $|\rho| < 1$, Example 3.1.2 in Time Series:  Theory and Methods (2nd Edition) by Peter J. Brockwell and Richard A. Davis shows that
\begin{align}
X_k := \sum_{j = 0}^k\rho^jv_{1 - j}
\end{align}
converges to $Y_1$ almost surely. Hence to determine the (marginal) distribution of $Y_1$, it suffices to determine the limiting distribution of $X_k$ as $k \to \infty$.  Clearly, being the linear combination of independent Gaussian random variables $\{v_1, v_0, \ldots, v_{1 - k}\}$, $X_k$ also follows Gaussian distribution with mean $0$ and variance $\sigma_k^2 := \sigma^2\sum_{j = 0}^k \rho^{2j} = \frac{(1 - \rho^{2(k + 1)})\sigma^2}{1 - \rho^2}$, which implies that the distribution function of $X_k$ equals to $F_{X_k}(x) = \Phi(x/\sigma_k)$, where $\Phi$ is the CDF of $N(0, 1)$.  Define $\tau^2 := \frac{\sigma^2}{1 - \rho^2} = \lim\limits_{k \to \infty}\sigma_k^2$.  It then follows by the continuity of $\Phi$ that $F_{X_k}(x) \to \Phi(x/\tau)$ as $k \to \infty$ for all $x \in \mathbb{R}$, which by definition means $X_k$ converges in distribution to $N(0, \tau^2)$.  This proves the distribution of $Y_1$ is $N(0, \tau^2)$.
