(t-1/T<r<t/T, we have r1∑Wr(-1)=∑ Wr(r t=1 Wr(r)dr The continuous mapping theorem applies to h(Wr)=Jo Wr(r)dr. It follows that h(WT)= h(W), so that T-3/2 Y-1=oo w(r)dr, as claimed (d). For a random walk, Y=(Yt-1+ut)2=Y_1+2Yt-1ut +ut, imply- ing that Yt-1Wt=(1/2)(Y-Y21-u2 and then 2t-Yi-1ut=(1 /2)Y Y)-(1/2)>h uf. Recall that Yo=0, and thus it is convenient to write 1Y1-1t=是1-∑12. From items(a) we know that T-1( (T-12∑1u)2a2(1)amnd∑1e→o2 by Lln( Kolmogorov);then, ∑=1-1l4=>lW(12-1 (e).Wefirstobservethat∑=1Y-1=[1+(u1+u2)+(1+u2+u3)+…+(u1+ u2+3+.+-1)=T-1)u1+(T-2)u2+(T-3)u3+…+{T-(T-1)ur-= (T-t) T Therefore, T-2 ti tut=T-i2tiut-T-i2t1Yt-1. By applying the con- tinuous mapping theorem to the joint convergence of items(a) and(c), we have W(1)-/w)dr t=1 The proofs of item(f)and(g)is analogous to those of(c) and(b). First rewrite 1 tYi- in term of Wr(rt-1)=T-1/Y-1/o=T-122-lus/o,where Tt-1=(t-1)/T, so that T-5/2>, t Yt-1= oT- E tWr(rt-1).Because Wr(r)is constant for(t-1/T<r<t/T, we have r∑mWr(-1)=∑/rr( r)dr forr=tm rWr(r)dr The continuous mapping theorem applies to h(wr)=o rWr(r)dr. It follows that h(Wr)=h(W), so that T-5/2 2t, tY-1=>0 rw(r)dr,as claimed
(t − 1)/T ≤ r < t/T, we have T −1X T t=1 WT (rt−1) = X T t=1 Z t/T (t−1)/T WT (r)dr = Z 1 0 WT (r)dr. The continuous mapping theorem applies to h(WT ) = R 1 0 WT (r)dr. It follows that h(WT ) =⇒ h(W), so that T −3/2 PT t=1 Yt−1 =⇒ σ R 1 0 W(r)dr, as claimed. (d). For a random walk, Y 2 t = (Yt−1 + ut) 2 = Y 2 t−1 + 2Yt−1ut + u 2 t , implying that Yt−1ut = (1/2){Y 2 t − Y 2 t−1 − u 2 t } and then PT t=1 Yt−1ut = (1/2){Y 2 T − Y 2 0 } − (1/2)PT t=1 u 2 t . Recall that Y0 = 0, and thus it is convenient to write PT t=1 Yt−1ut = 1 2 Y 2 T − 1 2 PT t=1 u 2 t . From items (a)) we know that T −1Y 2 T (= (T −1/2 PT t=1 us) 2 L−→ σ 2W2 (1) and PT t=1 u 2 t p −→ σ 2 by LLN (Kolmogorov); then, PT t=1 Yt−1ut =⇒ 1 2 σ 2 [W(1)2 − 1]. (e). We first observe that PT t=1 Yt−1 = [u1 + (u1 +u2) + (u1 +u2 +u3) +...+ (u1 + u2 +u3 +...+uT −1)] = [T −1)u1 + (T −2)u2 + (T −3)u3 +...+[T −(T −1)uT −1] = PT t=1(T − t)ut = PT t=1 Tut − PT t=1 tut , or PT t=1 tut = T PT t=1 ut − PT t=1 Yt−1. Therefore, T − 3 2 PT t=1 tut = T − 1 2 PT t=1 ut − T − 3 2 PT t=1 Yt−1. By applying the continuous mapping theorem to the joint convergence of items (a) and (c), we have T − 3 2 X T t=1 tut ⇒ σ[W(1) − Z 1 0 W(r)dr]. The proofs of item (f) and (g) is analogous to those of (c) and (b). First rewrite T −5/2 PT t=1 tYt−1 in term of WT (rt−1) ≡ T −1/2Yt−1/σ = T −1/2 Pt−1 s=1 us/σ, where rt−1 = (t − 1)/T, so that T −5/2 PT t=1 tYt−1 = σT −2 PT t=1 tWT (rt−1). Because WT (r) is constant for (t − 1)/T ≤ r < t/T, we have T −1X T t=1 (t/T)WT (rt−1) = X T t=1 Z t/T (t−1)/T rWT (r)dr for r = t/T = Z 1 0 rWT (r)dr. The continuous mapping theorem applies to h(WT ) = R 1 0 rWT (r)dr. It follows that h(WT ) =⇒ h(W), so that T −5/2 PT t=1 tYt−1 =⇒ σ R 1 0 rW(r)dr, as claimed. 11
We also write T-3∑1tY21 in term of WT(r1-1)≡T-12Y-1/o=T-12∑=1u3/o, where rt-1=(t-1)/T, so that T-2ta1ty21=02T-22ta1tWr(rt-1)2.Be- cause Wr(r)is constant for(t-1/T<r<t/T, we have r∑mW(-1)2=∑/.mwrv)bhJo rWr(r)2dr The continuous mapping theorem applies to h(Wr)=o rWr(r)2dr. It follows that h(W)→h(W), so that T-3∑1tY21=→a2lorw(r)2br. This com pletes the proofs of this lemma 3.1.1 No Constant Term or Time Trend in the Regression; True Process is a random walk We first consider the case that no constant term or time trend in the regression model, but true process is a random walk. The asymptotic distributions of OLs unit root coefficient estimator and t-ratio test statistics are in the following Let the data Yt be generated by(11)and(12); then as T-00, for the regression 1/2{[W(1)]2-1} ∫u(r) (r-1)、1/2{W(1)2-1} w()2d}2 t=/2 11/2 and st denote the OLS estimate of the distur- bance variance
We also write T −3 PT t=1 tY 2 t−1 in term of WT (rt−1) ≡ T −1/2Yt−1/σ = T −1/2 Pt−1 s=1 us/σ, where rt−1 = (t − 1)/T, so that T −3 PT t=1 tY 2 t−1 = σ 2T −2 PT t=1 tWT (rt−1) 2 . Because WT (r) is constant for (t − 1)/T ≤ r < t/T, we have T −1X T t=1 (t/T)WT (rt−1) 2 = X T t=1 Z t/T (t−1)/T rWT (r) 2 dr for = Z 1 0 rWT (r) 2 dr. The continuous mapping theorem applies to h(WT ) = R 1 0 rWT (r) 2dr. It follows that h(WT ) =⇒ h(W), so that T −3 PT t=1 tY 2 t−1 =⇒ σ 2 R 1 0 rW(r) 2dr. This completes the proofs of this lemma. 3.1.1 No Constant Term or Time Trend in the Regression; True Process Is a Random Walk We first consider the case that no constant term or time trend in the regression model, but true process is a random walk. The asymptotic distributions of OLS unit root coefficient estimator and t-ratio test statistics are in the following. Theorem 1: Let the data Yt be generated by (11) and (12); then as T → ∞, for the regression model (13), T(β˘ T − 1) ⇒ 1/2{[W(1)]2 − 1} R 1 0 [W(r)]2dr and t˘= (β˘ T − 1) σ˘β˘ T ⇒ 1/2{[W(1)]2 − 1} { R 1 0 [W(r)]2dr} 1/2 , where ˘σ 2 β˘ T = [s 2 T ÷ PT t=1 Y 2 t−1 ] 1/2 and s 2 T denote the OLS estimate of the disturbance variance: s 2 T = X T t=1 (Yt − β˘ T Yt−1) 2 /(T − 1). 12
Since the deviation of the OLs estimate from the true value is characterized by Y t-lut T (17) t=1 which is a continuous function function of Lemma 1(b) and 1(d), it follows that under the null hypothesis that 6= 1, the OLS estimator B is characterized by T 1/2{W(1)2-1} ∫(r)2dr To prove second part of this theorem, We first show the consistency of sT Notice that the population disturbance sum of squares can be writte (yr-yT-1B)'(T-yr-1B) (yr-yr-13+yr-13-yr-13)(yr-yr-1B+yr-13-y7-13) (yT-yT-1B)(yT-yT-1B)+(yT-1B-yT-1B' (yT-1B-yr-1B) where yT=Y Y2.Yr and the cross-product have vanished, since (yr-yr-1)yr-1(B-B)=0. by the OLs orthogonality condition(Xe=0). Dividing last equation by T, (1/T)(yr-yr-1)(yr-yr-13) (1/)(yr-yr-13)(yr-yr-13)+(3-)yr-yr-1)/7(-) (1/T)(yr-yr-13)(yr-y7-13) (19) 1/)∑2)-m12(-B)y-r-1)/72r1/(-B).(20) (1/T)C12)一→E(2)=a2 by lln for i.i.d. sequence,r1/(-)→0 and (yT-]T-1)/2=0Jo[w(r)]2dr from(18)and Lemma 1(b), respectively We thus have r2(3-By-y=1)rxr1--0o2/mPb-.0=0
Proof: Since the deviation of the OLS estimate from the true value is characterized by T(β˘ T − 1) = T −1 P T t=1 Yt−1ut T −2 P T t=1 Y 2 t−1 , (17) which is a continuous function function of Lemma 1(b) and 1(d), it follows that under the null hypothesis that β = 1, the OLS estimator β˘ is characterized by T(β˘ T − 1) ⇒ 1/2{[W(1)]2 − 1} R 1 0 [W(r)]2dr . (18) To prove second part of this theorem, We first show the consistency of s 2 T . Notice that the population disturbance sum of squares can be written (yT − yT −1β) ′ (yT − yT −1β) = (yT − yT −1β˘ + yT −1β˘ − yT −1β) ′ (yT − yT −1β˘ + yT −1β˘ − yT −1β) = (yT − yT −1β˘) ′ (yT − yT −1β˘) + (yT −1β˘ − yT −1β) ′ (yT −1β˘ − yT −1β), where yT = [Y1 Y2 ... YT ] ′ and the cross-product have vanished, since (yT − yT −1β˘) ′yT −1(β˘ − β) = 0, by the OLS orthogonality condition (X′e = 0). Dividing last equation by T, (1/T)(yT − yT −1β) ′ (yT − yT −1β) = (1/T)(yT − yT −1β˘) ′ (yT − yT −1β˘) + (β˘ − β) ′ [(y ′ T −1yT −1)/T](β˘ − β) or (1/T)(yT − yT −1β˘) ′ (yT − yT −1β˘) (19) = (1/T)(X T t=1 u 2 t ) − T 1/2 (β˘ − β) ′ [(y ′ T −1yT −1)/T2 ]T 1/2 (β˘ − β). (20) Now (1/T)(PT t=1 u 2 t ) p −→ E(u 2 t ) ≡ σ 2 by LLN for i.i.d. sequence, T 1/2 (β˘−β) → 0 and (y ′ T −1yT −1)/T2 =⇒ σ 2 R 1 0 [W(r)]2dr from (18) and Lemma 1(b), respectively. We thus have T 1/2 (β˘ − β) ′ [(y ′ T −1yT −1)/T2 ]T 1/2 (β˘ − β) p −→ 0 ′σ 2 Z 1 0 [W(r)]2 dr · 0 = 0. 13
Substituting these results into(20) we have (1/T)(yT-yT-1ByT-yT-1B)02 The ols disturbance's variance estimator s2=[1/(-1)(yr-yr-1B)(yr-y7-13) =[/(T-1)(1/)(yr-yr-13)yr-yr-1) 1 therefore is a consistent estimator Finally, we can express the t statistics alternatively as T(r-1) T-2=1Y21}(5)1P which is a continuous function function of Lemma 1(b)and 1(d), it follows that under the null hypothesis that B= 1, the asymptotic distribution of OLS t statistics is characterized b 1/2a2{W(1)2-1}1/2{W(1)]2 Lo2 w()dr (oa)/ 5 w()Pdry his complete the proof of this Theorem Statistical tables for the distributions of (18 )and (24) for various sample size T are reported in the section labeled Case 1 in Table B5 and B 6, respectively This finite sample result assume gaussian innovations 3.1.2 Constant Term but No Time Trend included in the Regression True process is a random walk We next consider the case that a constant term is added in the regression model but true process is a random walk. The asymptotic distributions of OLS unit
Substituting these results into (20) we have (1/T)(yT − yT −1β˘) ′ (yT − yT −1β˘) p −→ σ 2 . The OLS disturbance’s variance estimator s 2 T = [1/(T − 1)](yT − yT −1β˘) ′ (yT − yT −1β˘) (21) = [T/(T − 1)](1/T)(yT − yT −1β˘) ′ (yT − yT −1β˘) (22) p −→ 1 · σ 2 = σ 2 , (23) therefore is a consistent estimator. Finally, we can express the t statistics alternatively as t˘T = T(β˘ T − 1) ( T −2X T t=1 Y 2 t−1 )1/2 ÷ (s 2 T ) 1/2 or t˘T = T −1 PT t=1 Yt−1ut n T −2 PT t=1 Y 2 t−1 o1/2 (s 2 T ) 1/2 , which is a continuous function function of Lemma 1(b) and 1(d), it follows that under the null hypothesis that β = 1, the asymptotic distribution of OLS t statistics is characterized by t˘T L−→ 1/2σ 2{[W(1)]2 − 1} n σ 2 R 1 0 [W(r)]2dro1/2 (σ 2 ) 1/2 = 1/2{[W(1)]2 − 1} nR 1 0 [W(r)]2dro1/2 . (24) This complete the proof of this Theorem. Statistical tables for the distributions of (18) and (24) for various sample size T are reported in the section labeled Case 1 in Table B.5 and B.6, respectively. This finite sample result assume Gaussian innovations. 3.1.2 Constant Term but No Time Trend included in the Regression; True Process Is a Random Walk We next consider the case that a constant term is added in the regression model, but true process is a random walk. The asymptotic distributions of OLS unit 14
root coefficient estimator and t-ratio test statistics are in the following Theorem 2 Let the data Yt be generated by(11) and(12); then as T-00, for the regression model(14) T(r-1) 1/2{u(1)2-1}-W(1)·JW()r fl [w(r)2dr-5o w(r)dr) =(=1→1P2u2-1}-W(),W) So[w(r)dr-5ow(r)dr T∑Y-1 I and s? denote the OLS estimate of the disturbance variance s=∑(-a7-B1-1)(-2) Proof: The proof of this theorem is analogous to that of Theorem 1 and is omitted here Statistical tables for the distributions of (25) and(26)for various sample size T are reported in the section labeled Case 2 in Table B5 and B 6, respectively This finite sample result assume Gaussian innovations These statistics test the null hypothesis that 6= 1. However, a maintained assumption on which the derivation of Theorem 2 was based on is that the true value of a is zero. Thus, it might seem more natural to test for a unit root in this specification by testing the joint hypothesis that a=0 and 6=1. Dickey and Fuller(1981) derive the limiting distribution of the likelihood ratio test for the hypothesis that(a, B)=(0, 1)and used Monte Carlo to calculate the distri bution of the OLS F test of this hypothesis. Their values are reported under the
root coefficient estimator and t-ratio test statistics are in the following. Theorem 2: Let the data Yt be generated by (11) and (12); then as T → ∞, for the regression model (14), T(βˆ T − 1) ⇒ 1/2{[W(1)]2 − 1} − W(1) · R 1 0 W(r)dr R 1 0 [W(r)]2dr − hR 1 0 W(r)dri2 (25) and tˆ= (βˆ T − 1) σˆβˆ T ⇒ 1/2{[W(1)]2 − 1} − W(1) · R 1 0 W(r)dr R 1 0 [W(r)]2dr − hR 1 0 W(r)dri2 1/2 , (26) where ˆσ 2 βˆ T = s 2 T 0 1 T P P Yt−1 Yt−1 PY 2 t−1 −1 0 1 and s 2 T denote the OLS estimate of the disturbance variance: s 2 T = X T t=1 (Yt − αˆT − βˆ T Yt−1) 2 /(T − 2). Proof: The proof of this theorem is analogous to that of Theorem 1 and is omitted here. Statistical tables for the distributions of (25) and (26) for various sample size T are reported in the section labeled Case 2 in Table B.5 and B.6, respectively. This finite sample result assume Gaussian innovations. These statistics test the null hypothesis that β = 1. However, a maintained assumption on which the derivation of Theorem 2 was based on is that the true value of α is zero. Thus, it might seem more natural to test for a unit root in this specification by testing the joint hypothesis that α = 0 and β = 1. Dickey and Fuller (1981) derive the limiting distribution of the likelihood ratio test for the hypothesis that (α,β) = (0, 1) and used Monte Carlo to calculate the distribution of the OLS F test of this hypothesis. Their values are reported under the 15