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Considering an Auto-Regressive Moving Average (ARMA) model, \begin{equation*} y_k = \phi_0 + \sum_{j=1}^{p} \phi_j y_{k-j} + \sum_{l=1}^{q} \theta_l \varepsilon_{k-l}+ \varepsilon_k, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} where the noise term $\varepsilon_k$ follows the Normal distribution, such that $\varepsilon_k\sim\mathcal{N}(0,\sigma^2_{\varepsilon})$.

If we split ARMA process $\{y_k\}_{k=1}^n$ into two parts: \begin{equation*} x_k = \phi_0 + \sum_{j=1}^{r} \phi_j y_{k-j} + \sum_{l=1}^{s} \theta_l \varepsilon_{k-l}, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} and \begin{equation*} z_k = \sum_{j=r+1}^{p} \phi_j y_{k-j} + \sum_{l=s+1}^{q} \theta_l \varepsilon_{k-l} + \varepsilon_k, \qquad \text{for}\quad k=1,\cdots,n \end{equation*} where $1<r<p$ and $1<s<q$, so that $y_k=x_k+z_k$.

If ARMA process $\{y_k\}_{k=1}^n$ is wide-sense stationary, can I say that both sequences $\{x_k\}_{k=1}^n$ and $\{z_k\}_{k=1}^n$ are stationary? How to prove it? Many thanks!!

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Not an answer, but maybe related:

Brockwell and Davis (Introduction to Time Series and Forecasting, 2016), Proposition 2.2.1 says the following:

Let $Z_t$ be a stationary time series with expectation zero and acf $\gamma_Z$. If $\sum_{j=-\infty}^{\infty}|\psi_j|<\infty$, then the series $$ Y_t=\sum_{j=-\infty}^{\infty}\psi_j Z_{t-j} $$ is stationary with expectation $0$ and acf $$ \gamma_Y(h)=\sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\psi_j\psi_k\gamma_Z(h+k-j). $$

If, however, one were to just delete some lags, the statement could not be proven.

E.g., having a look at the "stationarity triangle" stated here, reveal that $$ Y_t=1.1Y_{t-1}-0.2Y_{t-2}+\epsilon_t $$ is stationary (in the sense of admitting a causal representation), while omitting the second lag and just keeping the explosive first lag $1.1Y_{t-1}$ clearly isn't.

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  • $\begingroup$ given that ARMA process $\{y_t\}_{t=1}^n$ is stationary, why $z_t = 1.1 y_{t-1}$ will be explosive? $\endgroup$
    – Stephen Ge
    Jun 5, 2020 at 7:00
  • $\begingroup$ OK, I may not have understood your notation correctly. I was basically reading it as removing the first lag, to get $Y_t=1.1Y_{t-1}+\epsilon_t$ $\endgroup$ Jun 5, 2020 at 7:07
  • $\begingroup$ I probably think the same thing, if I were you :). I feel that both parts will be stationary, but I don’t know how to prove it. $\endgroup$
    – Stephen Ge
    Jun 5, 2020 at 7:12

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