What is the probability distribution of this AR(1) function? I'm preparing the exam for "stochastic models" and I encountered this exercise which is giving me a lot of problems:
Let $X_t \sim AR(1)$, with
$$X_t=-0.8X_{t-1}+ \epsilon_t,   ~~~~~~~~~~\epsilon_t \sim WN(0,4)$$

1) Compute the autocovariance and the autocorrelation functions of $X_t$ at lags $h \ge 0$
2) Assuming that $\epsilon_t \sim GWN(0,4)$, what is the probability distribution of $X_t$ for all $t$?

1) No problem for this point but I want to show you how I did it:
Autocovariance function:
$$\gamma(1)=\operatorname{Cov}(X_t,X_{t-1})~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=\operatorname{Cov}(-0.8X_{t-1}+\epsilon_t, X_{t-1})$$
$$~~~~~~~~~~~~~~~~~~~=\operatorname{Cov}(-0.8X_{t-1}, X_{t-1})+\operatorname{Cov}(\epsilon_t, X_{t-1})$$
$$=-0.8 \cdot \gamma(0)~~~~~~~~~~~~~~~~~~~~~~~~~~$$
Similarly,
$$\gamma(2)=\operatorname{Cov}(X_t,X_{t-2})~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$=\operatorname{Cov}(-0.8X_{t-1}+\epsilon_t, X_{t-2})$$
$$~~~~~~~~~~~~~~~~~~~=\operatorname{Cov}(-0.8X_{t-1}, X_{t-2})+\operatorname{Cov}(\epsilon_t, X_{t-2})$$
$$=(-0.8) \cdot \gamma(1)~~~~~~~~~~~~~~~~~~~~~~~$$
$$=(-0.8)^2 \cdot \gamma(0)~~~~~~~~~~~~~~~~~~~~~$$
$$\vdots$$
$$\gamma(h)=\gamma\left(0\right)\cdot\phi^{\left|h\right|}$$

Autocorrelation function:
$$\rho = \frac{\gamma(h)}{\gamma(0)}$$
which 
$$\gamma(0)=Var(X_t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=Var(-0.8X_{t-1})+Var(\epsilon_t)+2\operatorname{Cov}(X_{t-1},\epsilon_t)$$
$$=(-0.8)^2 Var(X_t) + 4 + 0$$
$$=11.\bar{11}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

Point 2 is my main problem because I don't know where to begin in order to solve it. One sure thing is that the Gaussian White Noise with zero-mean has a normal distribution. Then? How can I use the data that I have in order to find this probability distribution?
Any help would be appreciated
 A: Let $L$ denote the lag operator, i.e. $LX_{t}=X_{t-1}$ and $L\epsilon_t =\epsilon_{t-1}$. The given equation can be written as
$$
X_t = -.8LX_t +\epsilon_t
$$ or equivalently
$$
(1+.8L)X_t = \epsilon_t.
$$ By inverting $1+.8L$, we get
$$
X_t =\alpha+(1+.8L)^{-1}\epsilon_t =\alpha+\sum_{j=0}^\infty (-.8L)^j \epsilon_t=\alpha+\sum_{j=0}^\infty (-.8)^j \epsilon_{t-j}.
$$ Now, we claim that $\sum_{j=0}^\infty (-.8)^j \epsilon_{t-j}$ is normally distributed with mean $0$ and variance equal to $4\sum_{j=0}^\infty (.64)^j=\frac{100}{9}.$ The mean-square convergence of the series can be easily seen from the fact that
$$
\Bbb E\left[\left|\sum_{n\le j\le m}(-.8)^j \epsilon_{t-j}\right|^2\right]=4\sum_{n\le j\le m}(.64)^j \to 0
$$as $n,m\to\infty$. By examining its characteristic function, we get
$$\begin{eqnarray}
\Bbb E\left[e^{it\sum\limits_{j=0}^\infty (-.8)^j \epsilon_{t-j}}\right]&=&\prod_{j=0}^\infty \Bbb E\left[e^{it (-.8)^j \epsilon_{t-j}}\right]\\
&=&\prod_{j=0}^\infty \exp\left[-\frac{t^2 \text{Var}\left[(-.8)^j \epsilon_{t-j}\right]}{2}\right]\\&=&\prod_{j=0}^\infty \exp\left[-\frac{4t^2\cdot(.64)^j}{2}\right]\\
&=&\exp\left[-\frac{4t^2\cdot\sum_{j=0}^\infty(.64)^j}{2}\right]=\exp\left[-\frac{t^2\cdot\frac{100}{9}}{2}\right].
\end{eqnarray}$$ Since the characteristic function of $\mathcal{N}(\mu,\sigma^2)$ is $\exp\left(it\mu -\frac{\sigma^2t^2}{2}\right)$, it follows
$$
X_t \sim_d \mathcal{N}\left(\alpha,\left(\frac{10}{3}\right)^2\right)
$$ for some $\alpha$. ($\alpha$ is not determined.)
