# Why, exactly, is a unit root a problem?

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?

• Because a unit root will mess your inferences (tests, interval estimates) up badly. Commented Mar 2, 2023 at 16:01
• 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. Commented Mar 2, 2023 at 16:11
• 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. Commented Mar 2, 2023 at 16:19
• 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. Commented Mar 3, 2023 at 7:45
• 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 Commented Mar 3, 2023 at 13:26

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.