<p>It is simple to see once you factorize.</p>
<p>In your set up there is only one unit root so the characteristic polynomial can be factorized as:</p>
<p><span class="math-container">\begin{align}
y_t &amp;= 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 &amp;= \varepsilon_t \\
(1-L)\phi_1(L)y_t &amp;= \varepsilon_t \\
\phi_1(L) \Delta y_t &amp;= \varepsilon_t
\end{align}</span></p>
<p>Here, <span class="math-container">$\phi_1(L)$</span> is a polynomial of degree <span class="math-container">$p-1$</span> and as you have mentioned in your set up, it will have all the roots outside the unit circle, making <span class="math-container">$\Delta y_t$</span> stationary.</p>

CC BY-SA 4.0
2022-02-15T08:28:18.987Z
2022-02-15T08:28:18.987Z
Dayne
<p>Consider <span class="math-container">$$y_t =a_1 y_{t-1}+a_2 y_{t-2} +...+a_p y_{t-p} +\varepsilon_t $$</span></p>
<p>The characteristic polynomial would be: <span class="math-container">$$(1-a_1L -a_2L^2 -...-a_pL^p) $$</span></p>
<p>Suppose that there is a unit root, say that <span class="math-container">$L=1$</span> 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 <span class="math-container">$\Delta y_t$</span> is weakly stationary, but I have not seen a proof for this. I am looking for a proof.</p>

CC BY-SA 4.0
2022-02-14T19:04:11.520Z
How can L be 1 that’s the lag operator no?
1muflon1
CC BY-SA 4.0
2022-02-14T20:36:50.727Z
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.
Michael Gmeiner
CC BY-SA 4.0
2022-02-14T20:43:22.573Z
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
1muflon1
<time-series><stationary>
Proof that a Unit Root process is Difference Stationary
Michael Gmeiner
# Proof that a Unit Root process is Difference Stationary

(Asked by: username_0 on 2022-02-14)

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.

**Tags:** time-series, stationary
> **Comments:**
> **username_1**: How can L be 1 that’s the lag operator no?  
> **username_0**: 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.  
> **username_1**: 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  

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## 1 answers

### Answer by username_2 on 2022-02-15. Score: 4 | Accepted answer

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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