For what values of α is the AR(1) process above a unit-root process? I have this AR(1) model : $y_t = c + \alpha y_{t-1} + \epsilon_{t}$ where $\epsilon_{t}$ is iid $(0,\sigma^{2})$.
So the question is : For what values of α is the AR(1) process above a unit-root process?
I think that I can rewrite it like follow $(1- \alpha L)y_t = c +  \epsilon_{t}$ where L is the lag operator. So to be a unit-root process, I think that $ \alpha $ must be = 1 but does $ \alpha = -1$ also leads to a unit-root process?
Thanks in advance!
 A: We would usually say that an ARMA process is a "unit-root" process if it has at least one root on the unit circle.  The sole root of the auto-regressive characteristic polynomial here is $\alpha$, and this lies on the unit circle if $|\alpha| = 1$ (i.e., if $\alpha = 1$ or $\alpha = -1$).  That is the ordinary meaning of "unit-root" in this context, so unless you are using a more restricted meaning, I'd say that $\alpha=-1$ also gives a "unit-root" process.
Irrespective of the terminology here, it is possible to investigate what this process looks like in the case where $\alpha = -1$.  In this case the model reduces down to the quasi-MA($\infty$) form:
$$\begin{align}
y_t 
&= c - y_{t-1} + \epsilon_t \\[6pt]
&= y_{t-2} + \epsilon_t - \epsilon_{t-1} \\[6pt]
&= c - y_{t-3} + \epsilon_t - \epsilon_{t-1} + \epsilon_{t-2} \\[6pt]
&= y_{t-4} + \epsilon_t - \epsilon_{t-1} + \epsilon_{t-2} - \epsilon_{t-3} \\[6pt]
&\ \ \vdots \\[6pt]
&= \lim_{k \rightarrow \infty} y_{t-2k} + \sum_{k=0}^\infty (-1)^k \epsilon_{t-k}. \\[6pt]
\end{align}$$
Now, in the special case where $\sigma=0$ this time-series is a deterministic process that oscillates between the points $a, c-a, a, c-a, ...$ for some $a \in \mathbb{R}$.  In the case where $\sigma>0$ the time-series is a sum of this deterministic part, plus a discrete-time shifted-Wiener process that can have any mean and has infinite variance.
