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I have found an article about stationarity: Variance of a stationary AR(2) model

and I have also estimated a model:

enter image description here

And I am not sure if I have understood it, is it truly stationary? And if so, could you show me how to derive it analytically?

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    $\begingroup$ Your estimates assume a stationary model. What exactly do you wish to "derive analytically"?? $\endgroup$
    – whuber
    Commented Jun 29, 2022 at 19:23
  • $\begingroup$ @whuber What I mean is, I have heard that the AR coefficients cannot be equal nor bigger than 1, and as far as I know there are some exceptions from it. The problem is, I'm not sure how to prove it while having an actual example. $\endgroup$
    – Fatafim
    Commented Jun 29, 2022 at 20:41
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    $\begingroup$ Check out relevant posts about "unit roots" of AR processes. For a fuller appreciation, one subtlety is treated at stats.stackexchange.com/questions/406204. $\endgroup$
    – whuber
    Commented Jun 29, 2022 at 20:52

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You can write the process in companion form (as a VAR(1)): $$\left[ \begin{array}{c} x_{n} \\ x_{n-1}% \end{array}% \right] =\left[ \begin{array}{c} 2.84 \\ 0% \end{array}% \right] +\left[ \begin{array}{cc} 1.36 & -0.75 \\ 1 & 0% \end{array}% \right] \left[ \begin{array}{c} x_{n-1} \\ x_{n-2}% \end{array}% \right] +\left[ \begin{array}{c} \varepsilon _{n} \\ 0% \end{array}% \right] $$ The eigenvalues of the autoregression matrix are within the unit circle. Therefore, the process is stationary.

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  • $\begingroup$ What is the autoregression matrix? The one with elements 1.36, -0.75, 1 and 0? $\endgroup$ Commented Jun 30, 2022 at 8:51

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