Robust Dickey–Fuller tests based on ranks for time series with additive outliers

Abstract

In this paper the unit root tests proposed by Dickey and Fuller (DF) and their rank counterpart suggested by Breitung and Gouriéroux (J Econom 81(1): 7–27, 1997) (BG) are analytically investigated under the presence of additive outlier (AO) contaminations. The results show that the limiting distribution of the former test is outlier dependent, while the latter one is outlier free. The finite sample size properties of these tests are also investigated under different scenarios of testing contaminated unit root processes. In the empirical study, the alternative DF rank test suggested in Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) (GH) is also considered. In Fotopoulos and Ahn (J Time Ser Anal 24(6): 647–662, 2003), these unit root rank tests were analytically and empirically investigated and compared to the DF test, but with outlier-free processes. Thus, the results provided in this paper complement the studies of the previous works, but in the context of time series with additive outliers. Equivalently to DF and Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) unit root tests, the BG test shows to be sensitive to AO contaminations, but with less severity. In practical situations where there would be a suspicion of additive outlier, the general conclusion is that the DF and Granger and Hallman (J Time Ser Anal 12(3): 207–224, 1991) unit root tests should be avoided, however, the BG approach can still be used.

Appendix: Proofs

1.1 Proof of Theorem 2

Let us start by proving (11). Let us first prove that the numerator of (8) is not random but is just a fixed function of T.

$$\begin{aligned}&\sum _{t=2}^T\left\{ \left( r_{T,t}-\frac{T+1}{2}\right) \sum _{j=1}^{t-1}\left( r_{T,j}-\frac{T+1}{2}\right) \right\} \nonumber \\&\quad =\sum _{1\le j<t\le T}\left( r_{T,t}-\frac{T+1}{2}\right) \left( r_{T,j}-\frac{T+1}{2}\right) \nonumber \\&\quad =\frac{\left[ \sum _{j=1}^T \left( r_{T,j}-\frac{T+1}{2}\right) \right] ^2-\sum _{j=1}^T \left( r_{T,j}-\frac{T+1}{2}\right) ^2}{2}\nonumber \\&\quad =-\frac{1}{2}\sum _{j=1}^T r_{T,j}^2+\frac{T+1}{2}\left( \sum _{j=1}^T r_{T,j}\right) -\frac{T}{2}\left( \frac{T+1}{2}\right) ^2\nonumber \\&\quad =-\frac{T(T+1)(T-1)}{24}, \end{aligned}$$

(13)

where we used that $\sum _{j=1}^T (r_{T,j}-(T+1)/2)=0$ and that $\sum _{j=1}^T r_{T,j}^2=T(T+1)(2T+1)/6.$

By (13), studying the asymptotic behavior of (11) amounts to studying the following quantity

$$\begin{aligned} \frac{1}{T^4}\sum _{t=2}^T \left\{ \sum _{j=1}^{t-1}\left( r_{T,j}-\frac{T+1}{2}\right) \right\} ^2= & {} \frac{1}{T^4}\sum _{t=2}^T \left\{ \sum _{j=1}^{t-1}\left[ \left( \sum _{k=1}^T \mathbf {1}_{\{\Delta z_k\le \Delta z_j\}}\right) -\frac{T+1}{2}\right] \right\} ^2\nonumber \\= & {} \frac{1}{T}\sum _{t=2}^T \left\{ \frac{1}{\sqrt{T}}\sum _{j=1}^{t-1}\left( F_T(\zeta _j)-\frac{T+1}{2T}\right) \right\} ^2,\nonumber \\ \end{aligned}$$

(14)

where $F_T(x)=T^{-1}\sum _{t=1}^T\mathbf {1}_{\{\zeta _t\le x\}}$, where

$$\begin{aligned} \zeta _t=\Delta z_t=\varepsilon _t+\theta (\delta _t-\delta _{t-1}). \end{aligned}$$

(15)

Let us now prove that

$$\begin{aligned} \left\{ \frac{1}{\sqrt{T}}\sum _{i=1}^{[Tu]}\left( F_T(\zeta _i)-\frac{T+1}{2T}\right) ,\quad u\in [0,1]\right\} \mathop {\Longrightarrow }\limits ^{d} \{B(u),\, u\in [0,1]\}, \text { as } T\rightarrow \infty , \end{aligned}$$

(16)

where [x] denotes the integer part of x, $\{B(u),\, u\in [0,1]\}$ denotes the Brownian bridge and $\mathop {\Longrightarrow }\limits ^{d}$ denotes here the weak convergence in the space of cadlag functions of [0, 1] denoted $\mathcal {D}([0,1])$ and equipped with the topology of uniform convergence. Splitting $\sum _{j=1}^T$ into $\sum _{j=1}^{[Tu]}$ and $\sum _{j=[Tu]+1}^T$, we get

$$\begin{aligned} \frac{1}{\sqrt{T}}\sum _{i=1}^{[Tu]}\left( F_T(\zeta _i)-\frac{T+1}{2T}\right)= & {} \frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=1}^T\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{T+1}{2T}\right) \nonumber \\= & {} \frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\left\{ \sum _{j=1}^T\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \right\} -\frac{[Tu]}{2T^{3/2}}\nonumber \\= & {} \frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\left\{ \sum _{j=1}^{[Tu]}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \right\} \nonumber \\&+\,\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\left\{ \sum _{j=[Tu]+1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \right\} -\frac{[Tu]}{2T^{3/2}}.\nonumber \\ \end{aligned}$$

(17)

Let $f(x,y)=\mathbf {1}_{\{x\le y\}}-1/2$. Since $f(y,x)=-f(x,y)$, when $x\ne y$, we get that the first term in the r.h.s of (17) is equal to

$$\begin{aligned} \frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\left\{ \sum _{j=1}^{[Tu]}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \right\} =\frac{1}{T^{3/2}} \sum _{i=1}^{[Tu]}\left( \mathbf {1}_{\{\zeta _i\le \zeta _i\}}-\frac{1}{2}\right) =\frac{[Tu]}{2T^{3/2}}. \end{aligned}$$

(18)

Thus,

$$\begin{aligned} \frac{1}{\sqrt{T}}\sum _{i=1}^{[Tu]}\left( F_T(\zeta _i)-\frac{T+1}{2T}\right) =\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\left\{ \sum _{j=[Tu]+1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \right\} . \end{aligned}$$

(19)

Let $h(x,y)=\mathbf {1}_{\{x\le y\}}$. By using the Hoeffding’s decomposition, we get that

$$\begin{aligned}&\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( h(\zeta _j,\zeta _i)-\frac{1}{2}\right) \nonumber \\&\quad =\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( F(\zeta _i)-\frac{1}{2}\right) -\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( F(\zeta _j)-\frac{1}{2}\right) \nonumber \\&\qquad +\,\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( h(\zeta _j,\zeta _i)-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) , \end{aligned}$$

(20)

F denoting the c.d.f of $(\zeta _t)$. Observe that since $\varepsilon _t$ has a continuous c.d.f it is also the case of the c.d.f of $\zeta _t$.

Let us now study the behavior of the first two terms in the r.h.s of (20):

$$\begin{aligned} \frac{T-[Tu]}{T^{3/2}}\sum _{i=1}^{[Tu]}\left( F(\zeta _i)-\frac{1}{2}\right) -\frac{[Tu]}{T^{3/2}}\sum _{j=[Tu]+1}^{T}\left( F(\zeta _j)-\frac{1}{2}\right) =X_{T,u}+Y_{T,u}. \end{aligned}$$

Observe first that $(\zeta _i)$ is a strictly stationary process since it is defined as the sum of two strictly stationary processes. It is also a 1-dependent process in the sense of Example 1 p. 167 of Billingsley (1968) and $(F(\zeta _i))$ has the same properties. By using Theorem 20.1 of Billingsley (1968), we get that, as T tends to infinity, $\{X_{T,u}\}\mathop {\Longrightarrow }\limits ^{d}\{(1-u) W(u)\}$, and that $\{Y_{T,u}\}\mathop {\Longrightarrow }\limits ^{d}\{u(W(1)-W(u))\}$, where $\{W(u)\}$ denotes the Wiener process. Since the process $(X_{T,u},Y_{T,u})$ is tight and since for fixed u and s, $\text {Cov}(X_{T,u},Y_{T,s})\rightarrow \text {Cov}((1-u) W(u),s(W(1)-W(s)))$, as $T\rightarrow \infty$, $\{(X_{T,u},Y_{T,u})\}\mathop {\Longrightarrow }\limits ^{d}\{((1-u) W(u),u(W(1)-W(u))\}$.

We deduce from this and Lemma 3 that

$$\begin{aligned}&\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( h(\zeta _j,\zeta _i)-\frac{1}{2}\right) \mathop {\longrightarrow }\limits ^{d} (1-u)W(u)-u(W(1)-W(u))\nonumber \\&\quad =W(u)-uW(1), \end{aligned}$$

(21)

which with (19) gives (16).

Let $w_i=F_T(\zeta _i)-(T+1)/(2T).$ Using similar arguments as those used in (Hamilton 1994, p. 483)

$$\begin{aligned}&\frac{1}{T}\sum _{t=2}^T \left\{ \frac{1}{\sqrt{T}}\sum _{j=1}^{t-1}\left( F_T(\zeta _j)-\frac{T+1}{2T}\right) \right\} ^2\\&\quad =\frac{1}{T} \left[ \left( \frac{w_1}{\sqrt{T}}\right) ^2+\left( \frac{w_1+w_2}{\sqrt{T}}\right) ^2+\cdots +\left( \frac{w_1+\cdots +w_{T-1}}{\sqrt{T}}\right) ^2\right] \\&\quad =\int _0^1\left( \frac{1}{\sqrt{T}}\sum _{i=1}^{[Tu]}w_i\right) ^2\mathrm {d}t \mathop {\longrightarrow }\limits ^{d} \int _0^1 B(u)^2\mathrm {d}u, \end{aligned}$$

by the continuous mapping theorem and (16), which concludes the proof of (11).

Lemma 3

Under the assumptions of Theorem 2,

$$\begin{aligned}&\sup _{u\in [0,1]}\left| \frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=[Tu]+1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) \right| =\sup _{u\in [0,1]} |Z_T(u)|\nonumber \\&\qquad =o_p(1), \text { as } T\rightarrow \infty . \end{aligned}$$

(22)

Proof of Lemma 3

Observe that

$$\begin{aligned} Z_T(u)=\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) -\frac{[Tu]}{2T^{3/2}}. \end{aligned}$$

Thus it is enough to prove (22) when $Z_T(u)$ is replaced by

$$\begin{aligned} R_T(u)=\frac{1}{T^{3/2}}\sum _{i=1}^{[Tu]}\sum _{j=1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) . \end{aligned}$$

We want to apply Lemma 5.2 of Borovkova et al. (2001) to $\{R_T(u),\; t\in [0,1]\}$. Let us first prove that for s and u such that $0\le s\le u <s+\delta \le 1$,

$$\begin{aligned}&R_T(u)-R_T(s)\\&\quad \le |R_T(s+\delta )-R_T(s)| + |W_T(s+\delta )-W_T(s)| + 2 \frac{[T(s+\delta )]-[Ts]}{\sqrt{T}}, \end{aligned}$$

where

$$\begin{aligned} W_T(u)=\frac{T-[Tu]}{T^{3/2}}\sum _{i=1}^{[Tu]}\left( F(\zeta _i)-\frac{1}{2}\right) -\frac{[Tu]}{T^{3/2}}\sum _{j=[Tu]+1}^{T}\left( F(\zeta _j)-\frac{1}{2}\right) . \end{aligned}$$

(23)

Observe that

$$\begin{aligned} R_T(u)-R_T(s)=\frac{1}{T^{3/2}}\sum _{i=[Ts]+1}^{[Tu]} \sum _{j=1}^{T}\left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) . \end{aligned}$$

(24)

Thus,

$$\begin{aligned}&(R_T(u)-R_T(s))-(R_T(s+\delta )-R_T(s))\\&\quad =-\frac{1}{T^{3/2}}\sum _{i=[Tu]+1}^{[T(s+\delta )]}\sum _{j=1}^T \left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-\frac{1}{2}\right) \\&\quad =-\frac{1}{T^{3/2}}\sum _{i=[Tu]+1}^{[T(s+\delta )]}\sum _{j=1}^T \left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) \\&\qquad +\,\frac{T}{T^{3/2}}\sum _{i=[Tu]+1}^{[T(s+\delta )]} \left( F(\zeta _i)-\frac{1}{2}\right) +\frac{[Tu]-[T(s+\delta )]}{T^{3/2}} \sum _{j=1}^T\left( F(\zeta _j)-\frac{1}{2}\right) . \end{aligned}$$

Using (23),

$$\begin{aligned}&W_T(s+\delta )-W_T(s)\\&\quad =\frac{1}{\sqrt{T}}\sum _{i=[Ts]+1}^{[T(s+\delta )]}\left( F(\zeta _i)-\frac{1}{2}\right) +\frac{[Ts]-[T(s+\delta )]}{T^{3/2}}\sum _{j=1}^T\left( F(\zeta _j)-\frac{1}{2}\right) , \end{aligned}$$

and thus, we get that

$$\begin{aligned}&(R_T(u)-R_T(s))-(R_T(s+\delta )-R_T(s))-(W_T(s+\delta )-W_T(s))\\&\quad =-\frac{1}{T^{3/2}}\sum _{i=[Tu]+1}^{[T(s+\delta )]}\sum _{j=1}^T \left( \mathbf {1}_{\{\zeta _j\le \zeta _i\}}-\frac{1}{2}\right) -\frac{1}{\sqrt{T}}\sum _{i=[Ts]+1}^{[Tu]}\left( F(\zeta _i)-\frac{1}{2}\right) \\&\qquad +\,\frac{[Tu]-[Ts]}{T^{3/2}}\sum _{j=1}^T\left( F(\zeta _j)-\frac{1}{2}\right) \le \frac{3}{2}\frac{[T(s+\delta )]-[Ts]}{\sqrt{T}}. \end{aligned}$$

Since we can follow the same line of reasoning for $R_T(s)-R_T(u)$, we get Condition (5.10) of Lemma 5.2 in Borovkova et al. (2001) with $\alpha =1/2$.

Let us now check Condition (5.9) of Lemma 5.2 in Borovkova et al. (2001). Observe that for $0\le s\le u\le 1$,

$$\begin{aligned} W_T(u)-W_T(s)= & {} \frac{T-([Tu]-[Ts])}{T^{3/2}}\sum _{i=[Ts]+1}^{[Tu]}\left( F(\zeta _i)-\frac{1}{2}\right) \\&+\,\frac{[Ts]-[Tu]}{T^{3/2}}\left\{ \sum _{i=1}^{[Ts]}\left( F(\zeta _i)-\frac{1}{2}\right) + \sum _{i=[Tu]+1}^{T}\left( F(\zeta _i)-\frac{1}{2}\right) \right\} . \end{aligned}$$

Thus, for some positive constant $C_1$

$$\begin{aligned} \mathbb {E}[(W_T(u)-W_T(s))^4]\le & {} C_1\frac{(T-([Tu]-[Ts]))^4}{T^6} \mathbb {E}\left[ \left\{ \sum _{i=[Ts]+1}^{[Tu]}\left( F(\zeta _i)-\frac{1}{2}\right) \right\} ^4\right] \\&+\,C_1\frac{([Tu]-[Ts])^4}{T^{6}}\mathbb {E}\left[ \left\{ \sum _{i=1}^{[Ts]}\left( F(\zeta _i)-\frac{1}{2}\right) \right\} ^4\right] \\&+\,C_1\frac{([Tu]-[Ts])^4}{T^{6}}\mathbb {E}\left[ \left\{ \sum _{i=[Tu]+1}^{T}\left( F(\zeta _i)-\frac{1}{2}\right) \right\} ^4\right] . \end{aligned}$$

Using that $(\zeta _i)$ is a 1-dependent process and that $\mathbb {E}[(F(\zeta _i)-1/2)]=0$, $|F(\zeta _i)-1/2|\le 1/2$ for all $i\ge 1$, there exist positive constants $C_2$ and $C_3$ such that

$$\begin{aligned}&\mathbb {E}[(W_T(u)-W_T(s))^4]\\&\quad \le C_2\frac{(T-([Tu]-[Ts]))^4}{T^6} \left\{ ([Tu]-[Ts])+([Tu]-[Ts])^2\right\} \\&\qquad +\,C_2\frac{([Tu]-[Ts])^4}{ T^{6}} \left\{ [Ts]+[Ts]^2+(T-[Tu])+(T-[Tu])^2\right\} \\&\quad \le C_3 \left\{ \frac{([Tu]-[Ts])^2}{T^{2}}+\frac{([Tu]-[Ts])}{T^{2}}\right\} , \end{aligned}$$

which is Condition (5.10) of Lemma 5.2 in Borovkova et al. (2001) with $h=1$, $g(t)=t$ and $r=4$.

Let us now check Condition (5.8) of Lemma 5.2 in Borovkova et al. (2001). Using (24), we get that for $0\le s\le u\le 1$,

$$\begin{aligned}&\mathbb {E}[(R_T(u)-R_T(s))^2]\\&\quad =\mathbb {E}\left[ \left\{ \frac{1}{T^{3/2}}\sum _{i=[Ts]+1}^{[Tu]} \sum _{j=1}^{[Ts]} A_{i,j}+\frac{1}{T^{3/2}}\sum _{i=[Ts]+1}^{[Tu]}\sum _{j=[Tu]+1}^{T} A_{i,j} +\frac{[Tu]-[Ts]}{2T^{3/2}}\right\} ^2\right] \\&\quad \le \frac{3}{T^{3}} \mathbb {E}\left[ \left\{ \sum _{i=[Ts]+1}^{[Tu]} \sum _{j=1}^{[Ts]} A_{i,j}\right\} ^2\right] +\frac{3}{T^{3}} \mathbb {E}\left[ \left\{ \sum _{i=[Ts]+1}^{[Tu]}\sum _{j=[Tu]+1}^{T} A_{i,j}\right\} ^2\right] \\&\qquad +\,\frac{3([Tu]-[Ts])^2}{2T^{3}}, \end{aligned}$$

where $A_{i,j}=\mathbf {1}_{\{\zeta _j\le \zeta _i\}}-F(\zeta _i)+F(\zeta _j)-1/2$. Using that $(\zeta _i)$ is a 1-dependent process, that $\mathbb {E}(A_{i,j})=0$ for all $i\ne j$ and that $\mathbb {E}(A_{i,j}A_{k,\ell })=0$ as soon as the distances between the different indices $i,j,k,\ell$ is larger than 1, there exist positive constants $C_4$, $C_5$ and $C_6$ such that

$$\begin{aligned} \mathbb {E}[(R_T(u)-R_T(s))^2]\le & {} C_4 \frac{([Tu]-[Ts])[Ts]}{T^{3}}+ C_4\frac{([Tu]-[Ts])(T-[Tu])}{T^{3}}\\&+\,C_5\frac{([Tu]-[Ts])^2}{T^{3}} \le C_6\frac{|u-s|}{T}\le C_6\frac{|u-s|^{1-\nu }}{T}, \end{aligned}$$

for some positive $\nu$ which gives Condition (5.8) of Lemma 5.2 in Borovkova et al. (2001) with $\beta =1$ and $\gamma =1-\nu$ and thus concludes the proof of Lemma 3. $\square$