# Proof that a Unit Root process is Difference Stationary

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.

• How can L be 1 that’s the lag operator no?
– 1muflon1
Commented Feb 14, 2022 at 19:04
• 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. Commented Feb 14, 2022 at 20:36
• 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
Commented Feb 14, 2022 at 20:43

\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.