Stationarity of an AR(1) process

Question

Suppose we have a AR(1) process $X_t=\theta X_{t-1}+Z_t$ with $t\in\mathbb{Z}$ and $\theta\in\mathbb{R}$ and $Z_t$ white noise. I already know how to derive the fact that if $|\theta|>1$ or $|\theta|<1$ then there exists a stationary solution. Also I know how to prove that if $\theta=1$ that no stationary solutions exists.

As we can write $X_t=X_0+Z_1+Z_2+\ldots+Z_t$, then $X_t$ is a random walk and thus satisfies $\text{Var}(X_t-X_0)=t\sigma^2\rightarrow \infty$ if $t\rightarrow\infty$ but we also have $$\text{sd}(X_t-X_0)\leq \text{sd}(X_t)+\text{sd}(X_0)$$ as $\text{sd}(X)=\sqrt{\text{Var}(X)}$ but the above should be finite so we have a contradiction. My question now is how do I prove the same result for $\theta=-1$. Then we would have (say t even) $$ X_t=X_0-Z_1+Z_2-Z_3+\ldots+Z_t$$ this is not a random walk anymore so we cannot apply the above. How do you prove it otherwise?

Answer 1

Why do you claim that $$ X_t = X_0 -Z_1 + \ldots (-)^tZ_t $$ is not a random walk? Recall that, due to the symmetry of the standard normal distribution about it's zero mean, $Z_i$ and $-Z_i$ are identically distributed. So, $X_t$ is the sum of $X_0$ and $t$ independent standard normal random variables.

Also, note that the really useful fact in your first proof is not that the $X_t$ form a random walk (with Gaussian steps), but that the $Z_i$ are assumed to be iid with finite variance. So, the proof in the case of $\theta = -1$ follows an almost identical argument as your first proof, since $$ \mathbb{V}ar(X_t-X_0) = \mathbb{V}ar\left(-Z_1 +\ldots (-)^{t}Z_t \right) = \mathbb{V}ar\left(Z_1 \right) + \dots + \mathbb{V}ar\left(Z_t \right) = t\sigma^2. $$