Why does unit root imply persistance of shocks and non-stationarity from a (stochastic) dynamic systems perspective?

Unit root (of the characteristic equation) is something in terms of the evolution operator $$A$$ for the linearized system ($$y$$ is a vectors of history):

$$y_t = A y_{t-1} + z_t$$

$$A = \begin{pmatrix} a_1 & a_2 & \dots & a_n \\ 1 & 0 & \dots & 0 \\ & \ddots & \ddots & \\ 0 & 0 & 1 & 0 \end{pmatrix}$$

What is the simple connection between roots of the characteristic equation and qualitative properties of the stochastic dynanamical system?

Does it really just work out that the characteristic equation of the difference equation is the characteristic equation of the operator $$A$$ ... and we are just looking for eigenvalues? I'm trying to remember this stuff. A reference would be helpful too.

And is it never that case that roots are greater than one?

From a time series perspective, a unit root is characterized by this equation

(1 - B) X(t) = a

where B is the backshift operator i.e. B(X(t)) = X(t-1) and a is white noise

This equation essentially states that

X(t) - X(t-1) = a or X(t) = X(t-1) + a.

So the new data is the same as the old data plus some white noise. Hence the unit root implies persistence of shocks (in your words).

It implies non-stationarity since the forecast using this equation never settles to the mean even in the long run (unlike an ARMA model).

• But is this the same as the linearized dynamic systems above? And why are roots greater than one not a problem? Commented Mar 30, 2020 at 16:14
• i.e. I am asking about the clear statement of stability of the dynamical system and how to map that to the less clear statements about "unit roo tests" and stationarity. Or is there something else besides what usually take from linear dynamical systems? Commented Mar 30, 2020 at 16:15
• If the root is inside the unit circle, it will be non-stationary. In the above equation, the coefficient of B is 1. If the coefficient (lets call it theta) is > 1, then the root will be inside the unit circle. In that case, X(t) = theta * X(t-1) + a and you can see that X(t) will keep on increasing in magnitude leading to non-stationarity. Commented Mar 30, 2020 at 16:32
• Yeah, that is why I thought. I'm trying to fitgure out why unit root is the test and not sub-unit root. This is basically just an application of linear stability theory. I'm not sure why every has re-invented the wheel. Commented Mar 30, 2020 at 17:32
• @mathtick: Unit root testing and stationarity is explained quite nicely in Hamilton's "Time Series Analysis". Commented Oct 21, 2023 at 12:24

Supposing, for simplicity, that the process started at zero at $$t=0$$, $$y_0=0$$, recursive substitution (i.e., using $$y_{t-1}=y_{t-2}+z_{t-1}$$ etc.) into $$y_t=y_{t-1}+z_t$$ yields $$y_t=\sum_{s=1}^tz_s.$$ If you now compute the partial derivative of $$y_t$$ w.r.t. $$z_{t-j}$$ (an "impulse response"), $$\partial y_t/\partial z_{t-j}$$, you get a derivative of one for any $$j$$. That is, the effect of a shock $$j$$ periods ago still fully persists in the process.

• I think this is missing the $A$ operator. You will get powers of $A$ and the largest eigenvalue eigenspaces dominates? Commented Mar 30, 2020 at 15:43
• So if it is really just a spectral decomp into the eigenspaces of A, the largest eigenvalues dominate. But why aren't eigenvalues with real part larger than one a problem? Commented Mar 30, 2020 at 15:59
• So the question is now why do they talk about unit root tests when we seem to agree that what matter is real part of eigenvalues less than one as per linear stability theory. There is some part of application setup that is missing I think. Commented Mar 30, 2020 at 17:34
• Your question changed a lot through the edits, and I am no longer sure I get what you mean. Yes, in the AR(p) case (which you seem to be referring to now in view of the VAR(1) companion form), standard unit root tests test the null of the largest root of the characteristic polynomial being one, against stability. Commented Mar 31, 2020 at 4:33
• Cool, yes that is what I was getting at. It's a linear SDE (discrete time). Do the usual stability analysis. Nothing else required. Certain things happen due to the special structure of the evolution operator. For some reason everyone talks about eigenvalues less than one when in fact a full stability analyis is required. If this is correct, the whole description of these things is very confusing. Commented Mar 31, 2020 at 9:40