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Consider $$y_t =a_1 y_{t-1}+a_2 y_{t-2} +...+a_p y_{t-p} +\varepsilon_t $$

The characteristic polynomial would be: $$(1-a_1L -a_2L^2 -...-a_pL^p) $$

Suppose that there is a unit root, say that $L=1$ is a root of the characteristic polynomial. Suppose this is a root of multiplicity 1 and all other roots are greater than 1 in absolute value. I am aware of the concept that $\Delta y_t$ is weakly stationary, but I have not seen a proof for this. I am looking for a proof.

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  • $\begingroup$ How can L be 1 that’s the lag operator no? $\endgroup$
    – 1muflon1 ♦
    Commented Feb 14, 2022 at 19:04
  • $\begingroup$ L is the lag operator. That is correct. I am talking about the characteristic polynomial in which case we consider L a variable to consider the roots of that polynomial. $\endgroup$
    – Michael Gmeiner
    Commented Feb 14, 2022 at 20:36
  • $\begingroup$ I thought in that case we are supposed to introduce new variable like let’s say lambda but in any case as long as other people won’t be confused I guess it’s fine $\endgroup$
    – 1muflon1 ♦
    Commented Feb 14, 2022 at 20:43

1 Answer 1

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It is simple to see once you factorize.

In your set up there is only one unit root so the characteristic polynomial can be factorized as:

\begin{align} y_t &= a_1 y_{t-1}+a_2 y_{t-2} +...+a_p y_{t-p} +\varepsilon_t \\ (1-a_1L -a_2L^2 -...-a_pL^p)y_t &= \varepsilon_t \\ (1-L)\phi_1(L)y_t &= \varepsilon_t \\ \phi_1(L) \Delta y_t &= \varepsilon_t \end{align}

Here, $\phi_1(L)$ is a polynomial of degree $p-1$ and as you have mentioned in your set up, it will have all the roots outside the unit circle, making $\Delta y_t$ stationary.

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