# When an ARMAX model is stationary? Why stationarity or invertibility is needed?

Let $y_t$ a stochastic process and $\tau_t$ presents the time duration between the $t$ and $t-1$ event.The ARMA(p,q,r) with exogenous variables is defined as:

$$y_t = \varepsilon_t + \sum_{i=1}^p \alpha_i y_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i} + \sum_{i=0}^b \eta_i \tau_{t-i}.\,$$

where $\eta_1, \ldots, \eta_b$ are the parameters of the exogenous input $\tau_t$ and $\varepsilon_t$ is a white noise.

# Question 1:

I have read that this process is stationary if and only if the roots of the AR polynomial must be outside the unit circle, but I don't understand it. I understand the proof when the model doesn't include exogenous variables but I don't do it when it includes ones.

If for example $p=q=1$ and $r=0$ the model is $$y_t = \alpha_0 + \alpha_1 y_{t-1} + \tau_t + \varepsilon_t,$$

Let assume that the $y_t$ process is mean stationary, then $$E(y_t) = \mu,$$ for all t.

Then $$E(y_t) = \alpha_0 + \alpha_1 E(y_{t-1}) +E(\tau_t) \rightarrow \mu=\alpha_0 +\alpha_1\mu +E(\tau_t)\rightarrow \mu = \frac{\alpha_0 +E(\tau_t)}{1-\alpha_1}.$$

So the process is mean stationary if and only $\alpha_1 \neq1$ and the process $\tau_t$ is mean stationary.

# Question 2:

I would like to ask why do we want a time series to be stationary / invertible. What is the intuition behind this? What are the consequences of non-stationarity. Would you please recommend any reference to clarify this theme? Also if I estimated the unknown parameters of the model without imposing restrictions to them, this would influence my results? In what sense?

• The question appears to be answered?... is there any other concern?... – Brethlosze May 28 '15 at 18:38

You arrived at $$E(y_t) = \frac{\alpha_0 +E(\tau_t)}{1-\alpha_1}$$
You want $E(y_t) = E(y_{t+k}) = \mu_y$, a constant. Obviously, this will hold if and only if
$E(\tau_t) = E(\tau_{t+k}) = \mu_{\tau}$, a constant. So mean-stationarity of $y_t$ requires mean-stationarity of $\tau_t$.