7
$\begingroup$

Suppose that we have observations $x_1,\dots,x_n$ for some process.

We want to fit an AR(k) model to these observations.

I do not understand why the naïve OLS approach to estimate our AR(k) coefficients would be inappropriate when the process has a unit root. From some simulations it seems to recover the coefficients no matter the location of the roots of the lag polynomial.

If estimating the AR coefficients is the not problem, then what is? A unit root means that uncertainty around your forecast grows with time, but that's not a problem from a mathematical point of view, it just something to keep in mind when interpreting the forecasted values.

Basically I'm asking which part of the mathematical analysis breaks down in the presence of a unit root?

Improve this question
$\endgroup$
27
  • 2
    $\begingroup$ Because a unit root will mess your inferences (tests, interval estimates) up badly. $\endgroup$
    – Alexis
    Commented Mar 2, 2023 at 16:01
  • 1
    $\begingroup$ Assuming the underlying process is an AR(k) process, you can fit the coefficients and, to the extend possible, give meaningful answers to these questions though. For example, fitting the model one sees a unit root, concludes no process mean exists, but can still say how the process mean evolves over time, in the mathematically precise and rigorous way. $\endgroup$
    – user2520938
    Commented Mar 2, 2023 at 16:11
  • 1
    $\begingroup$ My understanding is: unit root is not a problem. If it presents, it just requires statistical models/tools that are essentially different from traditional ARMA models that are designed for stationary time series. $\endgroup$
    – Zhanxiong
    Commented Mar 2, 2023 at 16:19
  • 2
    $\begingroup$ Arguably the most serious issue (among the ones I list, at least) is that when you investigate relationships between multiple unit root variables, you run into potential spurious regression, i.e. plims of regression coefficients that do not tend to zero and diverging t-ratios even if there is no relationship. stats.stackexchange.com/questions/188218/… Cointegration then becomes a relevant concept to make such studies meaningful. $\endgroup$
    – Christoph Hanck
    Commented Mar 3, 2023 at 7:45
  • 2
    $\begingroup$ Yes, as far as autoregressions are concerned that is the case. Things look different for OLS when regressing unrelated unit root variables onto each other, see my 2nd to last comment $\endgroup$
    – Christoph Hanck
    Commented Mar 3, 2023 at 13:26

2 Answers 2

Reset to default
7
$\begingroup$

Adapting this from Bauwens & Lubrano (1999), the part of the statistical procedure that "breaks down" in the presence of unit roots is asymptotic normality of the (OLS) estimator. For a model as simple as $$ y_{t} = \rho y_{t-1} + \epsilon_t $$ the asymptotic distribution of $\hat{\rho}_{OLS}$ is $\sqrt{T}(\hat{\rho}_{OLS}-\rho) \to N(0,1-\rho^2)$ if $|\rho| <1$, but $$ T(\hat{\rho}_{OLS}-\rho) \to \frac{1}{2} \frac{w(1)^{2} - 1}{\int_{0}^{1} w(r)^{2} \mathrm{d}r} \quad \quad \text{if } \rho =1.$$ where $w(\cdot)$ is a Wiener process. So in the presence of a unit root, the OLS estimator converges much faster (i.e., it is superconsistent) but to a random quantity instead of a constant. As a practical matter, any hypothesis test involving $\rho$ will require special tables.

Improve this answer
$\endgroup$
1
  • $\begingroup$ thanks also for the reference to that book, looks interesting $\endgroup$
    – user2520938
    Commented Mar 4, 2023 at 8:17
-2
$\begingroup$

The process is very different in the presence of a unit root. Many processes increase or decrease exponentially. But if there is a unit root, the process is a random walk, which is different.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ But I'm looking for a concrete example of why it matters? A specific example of something that breaks/does not work, from a mathematical point of view, in the presence of a unit root. $\endgroup$
    – user2520938
    Commented Mar 2, 2023 at 16:30
  • $\begingroup$ E.g. regression question give misleading answers, but that's not a problem from a mathematical pov, just from an interpretation pov. Another way of phrasing my question would be why time series text books usually assume the stationarity throughout. It seems that a lot of the analysis with regards to AR models in most books work in presence of unit roots without problem. $\endgroup$
    – user2520938
    Commented Mar 2, 2023 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.