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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?

This Wikipedia page on difference equations looks like the links I am trying to remember.

Nikhil Gupta
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mathtick
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2 Answers2

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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.

Christoph Hanck
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  • I think this is missing the $A$ operator. You will get powers of $A$ and the largest eigenvalue eigenspaces dominates? – mathtick Mar 30 '20 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? – mathtick Mar 30 '20 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. – mathtick Mar 30 '20 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. – Christoph Hanck Mar 31 '20 at 04:33
  • 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. – mathtick Mar 31 '20 at 09:40
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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).

Nikhil Gupta
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  • But is this the same as the linearized dynamic systems above? And why are roots greater than one not a problem? – mathtick Mar 30 '20 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? – mathtick Mar 30 '20 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. – Nikhil Gupta Mar 30 '20 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. – mathtick Mar 30 '20 at 17:32
  • From a modeling perspective roots outside the unit circle are rarely (if ever) used as the waveform become unrealistic very quickly (so your model would not produce useful results). What is useful is ARIMA (with or without seasonality) type of models which have roots on the unit circle. That is why I guess the test is for the root = unit root and not root < unit circle. – Nikhil Gupta Mar 30 '20 at 18:24
  • This is in the context of "I have some data" and "do some cointegration" study. So you you don't pick the model at all, you learn the model over differencing features. I'm trying to make sense of the cointegration crowd. I don't find the literature very good or precise from a probabilitistic point of view. I'm probably missing something critical though. – mathtick Mar 30 '20 at 19:00