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

1 Answer

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.)