# Unit-Root Asymptotics

I am using the book "Time-series-based econometrics" by Hatanaka to learn about asymptotic theory of unit roots. However, it is quite technical, so I am also using Hamilton's "Time Series Analysis" alongside it. There is an example in Hatanaka's book about the Dickey-Fuller estimator when the data generating process is $$y_t = \alpha + \rho y_{t-1} + \varepsilon_{t}$$, where $$\varepsilon_{t} \sim N(0,1)$$, with true parameters being $$\alpha_0, \rho_0$$. Assuming that $$\rho_0 = 1$$ and $$\alpha_0 \neq 0$$, it then follows that the ordinary least squares estimator $$\hat{\rho}$$ satisfies

$$T^{3/2}(\hat{\rho} - 1) \to N(0, 12\mu_0^2)$$

I understand the basics of asymptotic theory, so I made my attempt, but I don't reach something similar to what's in the book. I do get a normal distribution, but it's not the same as the one in the book. Additionally, I ran some simulations, and everything indicates that the book is correct. Could someone please help me see where my mistake is?

$$\textbf{Attempt}$$

Let's assume the data generating process is

$$y_t = \mu + y_{t-1} + \varepsilon_t = \mu t + \sum \varepsilon_t = \mu t + \zeta_t$$

(where we assume $$y_0 = 0$$), and that the following specification is used:

$$\Delta y_t = \mu + \gamma y_{t-1} + \varepsilon_t.$$

The matrix form of the OLS estimators is given by:

$$\begin{bmatrix} \hat{\mu} \\ \hat{\gamma} \end{bmatrix} = \begin{bmatrix} T & \sum y_{t-1} \\ \sum y_{t-1} & \sum y_{t-1}^2 \end{bmatrix}^{-1} \begin{bmatrix} \sum \Delta y_t \\ \sum y_{t-1} \Delta y_t \end{bmatrix} =$$

$$= \frac{1}{T \sum y_{t-1}^2 - (\sum y_{t-1})^2} \begin{bmatrix} \sum y_{t-1}^2 & -\sum y_{t-1} \\ -\sum y_{t-1} & T \end{bmatrix} \begin{bmatrix} \sum \Delta y_t \\ \sum y_{t-1} \Delta y_t \end{bmatrix}.$$

Thus,

$$\boxed{\hat{\mu} = \dfrac{\sum y_{t-1}^2 \sum \Delta y_t - \sum y_{t-1} \sum y_{t-1} \Delta y_t}{T \sum y_{t-1}^2 - \left(\sum y_{t-1}\right)^2}.}$$

$$\boxed{\hat{\gamma} = \dfrac{T \sum y_{t-1} \Delta y_t - \sum y_{t-1} \sum \Delta y_t}{T \sum y_{t-1}^2 - \left(\sum y_{t-1}\right)^2}.}$$

In this document, we will obtain the asymptotic behavior of the estimator $$\hat{\gamma}$$. To do this, we first develop the necessary sums, remembering that from this point, we assume we know the true process.

1. $$\sum y_{t-1} \Delta y_t = \mu^2 \sum (t-1) + \mu \sum (t-1) \varepsilon_t + \mu \sum \zeta_{t-1} + \sum \zeta_{t-1} \varepsilon_t.$$
2. $$\sum y_{t-1} = \mu \sum (t-1) + \sum \zeta_{t-1}.$$
3. $$\sum \Delta y_t = T \mu + \sum \varepsilon_t.$$
4. $$\sum y_{t-1}^2 = \mu^2 \sum (t-1)^2 + 2 \mu \sum (t-1) \zeta_{t-1} + \sum \zeta_{t-1}^2.$$

With the above, the expression for $$\hat{\gamma}$$ becomes

\begin{align*} \hat{\gamma} = \dfrac{T \mu^2 \sum (t-1) + \mu T \sum (t-1) \varepsilon_t + \mu T \sum \zeta_{t-1} + T \sum \zeta_{t-1} \varepsilon_t - T \mu^2 \sum (t-1) - \mu \sum (t-1) \sum \varepsilon_t - T \mu \sum \zeta_{t-1} - \sum \zeta_{t-1} \sum \varepsilon_t}{T \mu^2 \sum (t-1)^2 + 2T \mu \sum (t-1) \zeta_{t-1} + T \sum \zeta_{t-1}^2 - \mu^2 (\sum (t-1))^2 - (\sum \zeta_{t-1})^2 - 2 \mu \sum (t-1) \sum \zeta_{t-1}}. \end{align*}

$$= \dfrac{T \mu \sum (t-1) \varepsilon_t + T \sum \zeta_{t-1} \varepsilon_t - \mu \sum (t-1) \sum \varepsilon_t - \sum \zeta_{t-1} \sum \varepsilon_t}{\frac{T^4 \mu^2}{12} - \frac{T^2 \mu^2}{12} + 2T \mu \sum (t-1) \zeta_{t-1} + T \sum \zeta_{t-1}^2 - 2 \mu \sum (t-1) \sum \zeta_{t-1}}.$$

Notice that the numerator is $$O_p(T^{5/2})$$ and the denominator is $$O(T^4)$$, which means the estimator is $$O_p(T^{-3/2})$$. Now, applying what we know about the orders of convergence of the numerator and denominator, we see what this estimator converges to:

$$T^{3/2} \hat{\gamma} = \dfrac{\mu T^{-3/2} \sum t \varepsilon_t - \left(\dfrac{\mu}{2}\right) T^{-1/2} \sum \varepsilon_t}{\dfrac{\mu^2}{12}} = \frac{12 T^{-3/2} \sum t \varepsilon_t - 6 T^{-1/2} \sum \varepsilon_t}{\mu} \stackrel{a}{=}$$

$$\stackrel{a}{=} \frac{12\sigma\left[\omega(1) - \int_0^1 \omega(r) dr\right] - 6\sigma\omega(1)}{\mu}.$$

This basically tells us that the estimator is asymptotically normal.

• I would check whether Hayashi or Hamilton have any references related to that statement about $12 \mu_{0}^2$. Also, check the appendices of Hamilton. He often has a lot of derivations in the appendices of his chapters. Commented May 27 at 3:10
• Maybe unify notation of $\alpha$ and $\mu$ Commented Jul 9 at 12:48

If indeed this is the case, Hamilton gives (implicitly, invert the matrix $$Q$$ in eq. 17.4.45 and .46 ), the asymptotic variance of the centered and scaled estimate of the slope coefficient as $$12/\mu^2$$, i.e. $$\mu^2$$ divides $$12$$, it does not multiply it.