M-estimator based unit root tests in the ESTAR framework

Abstract

Building on work of Lucas (Econom Theory 11:331–346, 1995a) we derive the limiting distribution for M-estimator based unit root tests in the ESTAR model. This yields that the LS based unit root tests by Kapetanios et al. (J Econom 112:359–379, 2003) and Rothe and Sibbertsen (Allg Stat Arch 90:439–456, 2006) are robustified in an outlier context. We also consider an LS based heteroscedasticity robust version (White’s) of one of our unit root tests as a “quick fix” solution to the problems of outliers. Finite sample properties of the tests are examined, and in the case of additive outliers it is shown that the LS based tests are grossly over-sized whereas the size of our tests is close to the nominal size. If an ESTAR model with innovation outliers is considered, significant power gains over the LS based tests are shown by using our robust tests.

Appendices

Appendix 1

The following Lemma will facilitate the proofs of Theorem 1.

Lemma

Given Assumptions 1 and 2, $\delta =6$ , and $x_{t}=x_{t-1}+\epsilon _{t}$. Then,

• (a) $T^{-4}\sum _{t}x_{t-1}^{6}\psi (\epsilon _{t})^{2}\Rightarrow \sigma _{\psi }^{2}\bar{\sigma }_{\epsilon }^{6}\int b_{1}^{6},$

• (b) $T^{-4}\sum _{t}x_{t-1}^{6}\psi ^{\prime }(\epsilon _{t})\Rightarrow \mu _{\psi }\bar{\sigma }_{\epsilon }^{6}\int b_{1}^{6},$

• (c) $T^{-2}\sum _{t}x_{t-1}^{3}\psi (\epsilon _{t})\Rightarrow \bar{ \sigma }_{\psi }\bar{\sigma }_{\epsilon }^{3}\int b_{1}^{3}db_{2}+3/2\bar{ \sigma }_{\epsilon }^{2}(\bar{\sigma }_{\epsilon \psi }-\sigma _{\epsilon \psi })\int b_{1}^{2}.$

Proof of Lemma

The proof of (a). Write

$$\begin{aligned} T^{-4}\sum \nolimits _{t}x_{t-1}^{6}\psi (\epsilon _{t})^{2}=T^{-4}\sum \nolimits _{t}x_{t-1}^{6}(\psi (\epsilon _{t})^{2}-\sigma _{\psi }^{2})+T^{-4}\sigma _{\psi }^{2}\sum \nolimits _{t}x_{t-1}^{6}. \end{aligned}$$

Here, $T^{-4}\sigma _{\psi }^{2}\sum _{t}x_{t-1}^{6}\Rightarrow \sigma _{\psi }^{2}\bar{\sigma }_{\epsilon }^{6}\int b_{1}^{6}$ by the continuous mapping theorem, and because $\{ \psi (\epsilon _{t})^{2}-\sigma _{\psi }^{2}\}$ under the present assumptions satisfies the conditions of Theorem 3.3 in Hansen (1992) it follows that

$$\begin{aligned} \sup _{s\in [0,1]}\left| T^{-4}\sum \nolimits _{t=1}^{[Ts]}x_{t-1}^{6}(\psi (\epsilon _{t})^{2}-\sigma _{\psi }^{2})\right| \overset{p}{\rightarrow }0, \end{aligned}$$

and thus $T^{-4}\sum _{t}x_{t-1}^{6}\psi (\epsilon _{t})^{2}\Rightarrow \sigma _{\psi }^{2}\bar{\sigma }_{\epsilon }^{6}\int b_{1}^{6}$.

The proof of (b) is similar to the proof of (a) noticing that under the present assumptions $\{ \psi ^{\prime }(\epsilon _{t})-\mu _{\psi }\}$ satisfies the conditions of Theorem 3.3 in Hansen (1992).

Next, the proof of (c). Along the lines in the proof of Theorem 1 in Sandberg (2009), from which the (sufficient) moment condition $\delta =6$ is taken, it is straightforward to show that

$$\begin{aligned}&T^{-2}\sum \nolimits _{t} x_{t-1}^{3}\psi (\epsilon _{t})\Rightarrow \int B_{1}^{3}dB_{2}+3\sigma _{\epsilon \psi ,1}\int B_{1}^{2}, \end{aligned}$$

(11)

where $\sigma _{\epsilon \psi ,1}=\lim _{T\rightarrow \infty }T^{-1}\sum \nolimits _{i=1}^{T}\sum \nolimits _{j=i+1}^{\infty }\mathrm {E} \epsilon _{i}\psi (\epsilon _{j})$. The proof now follows by substituting for $\int B_{1}^{3}dB_{2}=\bar{\sigma }_{\psi }\bar{\sigma }_{\epsilon }^{3}\int b_{1}^{3}db_{2}$, $\sigma _{\epsilon \psi ,1}=(\bar{\sigma } _{\epsilon \psi }-\sigma _{\epsilon \psi })/2$, and $\int B_{1}^{2}=\bar{ \sigma }_{\epsilon }^{2}\int b_{1}^{2}$ into (11). $\square$

Proof of Theorem 1

The proof of (i). A (appropriately scaled) Taylor-series approximation of ( 5) with respect to $\hat{e}_{t}$ around $\epsilon _{t}$ yields

$$\begin{aligned}&0=T^{-2}\sum \nolimits _{t}y_{t-1}^{3}\psi (\epsilon _{t})-\left( T^{2}\hat{ \beta }_{\psi }\right) T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi ^{\prime }(\epsilon _{t})+T^{-2}R_{T}, \end{aligned}$$

(12)

where we have used that $\hat{e}_{t}-\epsilon _{t}=-\hat{\beta }_{\psi }y_{t-1}^{3}$ holds under the null hypothesis, and $R_{T}$ is a remainder term. Next, the triangle inequality, the first-order Lipschitz condition for $\psi ^{\prime }(\epsilon _{t})$, and that $\hat{e}_{t}-\epsilon _{t}=o_{p}(1)$ holds uniformly for all $t$ implies

$$\begin{aligned} \left| T^{-2}R_{T}\right| \le \kappa \sum \nolimits _{t}\left| ( \hat{e}_{t}-\epsilon _{t})T^{-4}y_{t-1}^{6}\left( T^{2}\hat{\beta }_{\psi }\right) \right| \le o_{p}(1)T^{2}\hat{\beta }_{\psi }, \end{aligned}$$

for some $\kappa \in (0,\infty )$ (independent of $\hat{e}_{t}$ and $\epsilon _{t}$). Hence, by (12) we can now write

$$\begin{aligned}&T^{2}\hat{\beta }_{\psi }=T^{-2}\sum \nolimits _{t}y_{t-1}^{3}\psi (\epsilon _{t})/\left( T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi ^{\prime }(\epsilon _{t})+o_{p}(1)\right) . \end{aligned}$$

(13)

Finally, apply parts (b) and (c) of the Lemma to (13) to complete the proof.

The proof of (ii). Write

$$\begin{aligned} t_{\psi ,\beta }&= T^{2}\hat{\beta }_{\psi }/\left( \left( T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi ^{\prime }(\hat{e}_{t})\right) ^{-1}\left( T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi (\hat{e}_{t})^{2}\right) \right. \nonumber \\&\left. \left( T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi ^{\prime }(\hat{e}_{t})\right) ^{-1}\right) ^{1/2}. \end{aligned}$$

Next, it is straightforward to show that $T^{-4}\sum \nolimits _{t}y_{t-1}^{6} \psi ^{\prime }(\hat{e}_{t})=T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi ^{\prime }(\epsilon _{t})+o_{p}(1)$ and $T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi (\hat{ e}_{t})^{2}=T^{-4}\sum \nolimits _{t}y_{t-1}^{6}\psi (\epsilon _{t})^{2}+o_{p}(1)$ since $\hat{e}_{t}-\epsilon _{t}=o_{p}(1)$. It follows now by parts (a) and (b) in the Lemma that the denominator in the above expression for $t_{\psi ,\beta }$ converges weakly to the square root of $\sigma _{\psi }^{2}\mu _{\psi }^{-2}\bar{\sigma }_{\epsilon }^{-6}\left( \int b_{1}^{6}\right) ^{-1}$. Combine this result with the convergence result for the numerator in (13) in part (i) of the proof to yield the desirable result. $\square$

Proof of Corollary 3

The proof of (i). We shall first derive an equivalent expression for $\int b_{1}^{3}db_{2}$. Thus, notice first that $\left[ b_{1}(s),b_{2}(s)\right] ^{\prime }\sim BM(\tilde{\Omega })$ where

$$\begin{aligned} \tilde{\Omega }=\left[ \begin{array}{c@{\quad }c} 1 &{} \rho \\ \rho &{} 1 \end{array} \right] . \end{aligned}$$

Next, a lower triangular decomposition of $\tilde{\Omega }$ is given by

$$\begin{aligned} \tilde{\Omega }^{1/2}=\left[ \begin{array}{c@{\quad }c} 1 &{} 0 \\ \rho &{} \sqrt{1-\rho ^{2}} \end{array} \right] , \end{aligned}$$

and we can write $\left[ b_{1}(s),b_{2}(s)\right] ^{\prime }=\tilde{\Omega } ^{1/2}\left[ w_{1}(s),w_{2}(s)\right] ^{\prime }$ where $w_{1}(s)$ and $w_{2}(s)$ are two independent standard Brownian motions on $C[0,1]$. Accordingly, substitute for the identities $b_{1}=w_{1}$ and $b_{2}=\rho w_{1}+\sqrt{1-\rho ^{2}}w_{2}$ into $\int b_{1}^{3}db_{2}$ to obtain

$$\begin{aligned} \int b_{1}^{3}db_{2}&= \rho \int w_{1}^{3}dw_{1}+\sqrt{1-\rho ^{2}}\int w_{1}^{3}dw_{2} \\&\\&= \rho \left( w_{1}^{4}(1)-6\int w_{1}^{2}\right) /4+\left( 1-\rho ^{2}\right) ^{1/2}\left( \int w_{1}^{6}\right) ^{1/2}\eta , \end{aligned}$$

where $\eta \sim N(0,1)$ and is independent of $w_{1}$. The last equality follows by using elementary results on Ito calculus and that $\int w_{1}^{3}dw_{2}$ is a scale mixture of normal distributions. Thus, part (i) of Corollary 3 can also be written as

$$\begin{aligned} \tilde{C}_{\psi ,\beta }^{N}\Rightarrow \rho \left( w_{1}^{4}(1)-6\int w_{1}^{2}\right) /\left( 4\int w_{1}^{6}\right) +\left( 1-\rho ^{2}\right) ^{1/2}\left( \int w_{1}^{6}\right) ^{-1/2}\eta . \end{aligned}$$

Lastly, noticing that $\hat{\rho }\tilde{C}_{RS}\Rightarrow \rho \left( w_{1}^{4}(1)-6\int w_{1}^{2}\right) /(4\int w_{1}^{6})$ completes the proof.

The proof of (ii) is trivial. $\square$

Appendix 2

This appendix presents the 1, 5, and 10 % percentiles of the limiting distribution of $\tilde{t}_{\psi ,\beta }$ in Corollary 2 as a function of $\rho$.¹⁷ As an estimate of $\rho$ we may use $\bar{s}_{\epsilon \psi }/(\bar{s}_{\epsilon } \bar{s}_{\psi })$ (i.e. the long-run sample correlation between $\hat{e} _{t}$ and $\psi (\hat{e}_{t})$) (Table 6).

Table 6 Percentiles of the limiting distribution of $\tilde{t} _{\psi ,\beta }$ in Corollary 2 as a function of $\rho$