Moreover, the second term is derived from: 1 T ) 2 1

Transcription

1 170 Moreover, the second term is derived from: 1 T T ɛt 2 σ 2 ɛ. Therefore, 1 σ 2 ɛt T y t 1 ɛ t = 1 2 ( yt σ T ) 2 1 2σ 2 ɛ 1 T T ɛt (χ2 (1) 1). (b) Next, consider y 2 t 1. T E y 2 t 1 T T = E(y 2 t 1 ) = σ 2 ɛ(t 1) = σ 2 ɛ Thus, we obtain the following result: 1 T T E 2 y 2 t 1 a fixed value. T(T 1). 2

2 171 Therefore, 1 T 2 T y 2 t 1 a distribution. 6. Summarizing the results up to now, T( ˆφ 1 φ 1 ), not T( ˆφ 1 φ 1 ), has limiting distribution in the case of φ 1 = 1. T( ˆφ 1 φ 1 ) = (1/T) y t 1 ɛ t (1/T 2 ) y 2 t 1 a distribution. The distributions of the t statistic: of ˆφ 1. ˆφ 1 1, where s φ denotes the standard error s φ = Compare t distribution with (a) (c). = Unit Root Test (, or Dickey-Fuller (DF) Test)

3 172 t Distribution T

4 173 (a) H 0 : y t = y t 1 + ɛ t H 1 : y t = φ 1 y t 1 + ɛ t for φ 1 < 1 or 1 < φ 1 T

5 174 (b) H 0 : y t = y t 1 + ɛ t H 1 : y t = α 0 + φ 1 y t 1 + ɛ t for φ 1 < 1 or 1 < φ 1 T

6 175 (c) H 0 : y t = α 0 + y t 1 + ɛ t H 1 : y t = α 0 + α 1 t + φ 1 y t 1 + ɛ t for φ 1 < 1 or 1 < φ 1 T

7 Serially Correlated Errors Consider the case where the error term is serially correlated Augmented Dickey-Fuller (ADF) Test Consider the following AR(p) model: y t = φ 1 y t 1 + φ 2 y t φ p y t p + ɛ t, ɛ t iid(0, σ 2 ɛ), which is rewritten as: φ(l)y t = ɛ t. When the above model has a unit root, we have φ(1) = 0, i.e., φ 1 + φ φ p = 1. The above AR(p) model is written as: y t = ρy t 1 + δ 1 y t 1 + δ 2 y t δ p 1 y t p+1 + ɛ t, where ρ = φ 1 + φ φ p and δ j = (φ j+1 + φ j φ p ).

8 177 The null and alternative hypotheses are: H 0 : ρ = 1 (Unit root), H 1 : ρ < 1 (Stationary). Use the t test, where we have the same asymptotic distributions. We can utilize the same tables as before. Choose p by AIC or SBIC. Use N(0, 1) to test H 0 : δ j = 0 against H 1 : δ j 0 for j = 1, 2,, p 1. Reference Kurozumi (2008) Economic Time Series Analysis and Unit Root Tests: Development and Perspective, Japan Statistical Society, Vol.38, Series J, No.1, pp Download the above paper from:

9 178 Example of ADF Test. gen time=_n. tsset time time variable: time, 1 to 516 delta: 1 unit. gen sexpend=expend-l12.expend (12 missing values generated). corrgram sexpend LAG AC PAC Q Prob>Q [Autocorrelation] [Partial Autocor]

10 179. varsoc d.sexpend, exo(l.sexpend) maxlag(25) Selection-order criteria Sample: Number of obs = lag LL LR df p FPE AIC HQIC SBIC e e e e e e e e e e e e e * e e e e e e e e e e * e+07* * * e e

11 e Endogenous: D.sexpend Exogenous: L.sexpend _cons. dfuller sexpend, lags(23) Augmented Dickey-Fuller test for unit root Number of obs = Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) MacKinnon approximate p-value for Z(t) = dfuller sexpend, lags(13) Augmented Dickey-Fuller test for unit root Number of obs = Interpolated Dickey-Fuller Test 1% Critical 5% Critical 10% Critical Statistic Value Value Value Z(t) MacKinnon approximate p-value for Z(t) =

12 Cointegration ( ) 1. For a scalar y t, when y t = y t y t 1 is a white noise (i.e., iid), we write y t I(1). 2. Definition of Cointegration: Suppose that each series in a g 1 vector y t is I(1), i.e., each series has unit root, and that a linear combination of each series (i.e, a y t for a nonzero vector a) is I(0), i.e., stationary. Then, we say that y t has a cointegration. 3. Example: Suppose that y t = (y 1,t, y 2,t ) is the following vector autoregressive process: y 1,t = φ 1 y 2,t + ɛ 1,t, y 2,t = y 2,t 1 + ɛ 2,t. Then, y 1,t = φ 1 ɛ 2,t + ɛ 1,t ɛ 1,t 1, (MA(1) process), y 2,t = ɛ 2,t,

13 182 where both y 1,t and y 2,t are I(1) processes. The linear combination y 1,t φ 1 y 2,t is I(0). In this case, we say that y t = (y 1,t, y 2,t ) is cointegrated with a = (1, φ 1 ). a = (1, φ 1 ) is called the cointegrating vector, which is not unique. Therefore, the first element of a is set to be one. 4. Suppose that y t I(1) and x t I(1). For the regression model y t = x t β + u t, OLS does not work well if we do not have the β which satisfies u t I(0). = Spurious regression ( ) 5. Suppose that y t I(1), y t is a g 1 vector and y t = y 2,t is a k 1 vector, where k = g 1. Consider the following regression model: ( y1,t y 2,t ). y 1,t = α + γ y 2,t + u t, t = 1, 2,, T.

14 183 OLSE is given by: ( ˆα ) ( T y ) 1 ( 2,t y1,t ) = ˆγ y2,t y2,t y. 2,t y1,t y 2,t Next, consider testing the null hypothesis H 0 : Rγ = r, where R is a m k matrix (m k) and r is a m 1 vector. The F statistic, denoted by F T, is given by: where F T = 1 ( m (Rˆγ r) s2 T ( 0 R ) T y ) 1 ( 2,t 0 ) 1 (Rˆγ r), y2,t y2,t y 2,t R s 2 T = 1 T g T (y 1,t ˆα ˆγ y 2,t ) 2. When we have the γ such that y 1,t γy 2,t is stationary, OLSE of γ, i.e., ˆγ, is not statistically equal to zero. When the sample size T is large enough, H 0 is rejected by the F test. 6. Phillips, P.C.B. (1986) Understanding Spurious Regressions in Econometrics, Journal of Econometrics, Vol.33, pp

15 184 Consider a g 1 vector y t whose first difference is described by: y t = Ψ(L)ɛ t = Ψ s ɛ t s, for ɛ t an i.i.d. g 1 vector with mean zero, variance E(ɛ t ɛ t ) = PP, and finite fourth moments and where {sψ s } s=0 is absolutely summable. Let k = g 1 and Λ = Ψ(1)P. ( y1,t ) ( Σ11 Σ Partition y t as y t = and ΛΛ as ΛΛ 21 = y 2,t Σ 21 Σ 22 and Σ 21 are k 1 vectors, and Σ 22 is a k k matrix. Suppose that ΛΛ is nonsingular,and define σ 2 1 = Σ 11 Σ 21 Σ 1 22 Σ 21. s=0 ), where y 1,t and Σ 11 are scalars, y 2,t Let L 22 denote the Cholesky factor of Σ 1 22, i.e., L 22 is the lower triangular matrix satisfying Σ 1 22 = L 22L 22. Then, (a) (c) hold. (a) OLSEs of α and γ in the regression model y 1,t = α + γ y 2,t + u t, denoted by ˆα T and ˆγ T, are characterized by: ( T 1/2 ˆα T ˆγ T Σ 1 22 Σ 21 ) ( σ 1 h 1 σ 1 L 22h 2 ),

16 185 where ( h1 h 2 ) ( = 1 0 W 2 (r) dr W 2 (r)dr 1 0 W 2 (r)w 2 (r) dr ) 1 ( 1 0 W 1 (r)dr ) 1 0 W 2 (r)w 1 (r)dr. W1 (r) and W 2 (r) denote scalar and g-dimensional standard Brownian motions, and W 1 (r) is independent of W 2 (r). (b) The sum of squared residuals, denoted by RSS T = T û 2 t, satisfies T 2 RSS T σ 2 1 H, where H = ( (W 1 (r))2 0 dr W 1 (r)dr ) ( h1 1 0 W 2 (r)w 1 (r)dr h 2 (c) The F T test satisfies: T 1 F T 1 m (σ 1 R h 2 r ) σ 2 1 H ( 0 R ) (σ 1 R h 2 r ), where R = RL 22 and r = r RΣ 1 22 Σ 21. ( ) W 2 (r) dr ) W 2 (r)dr 1 ( 0 R ) 0 W 2 (r)w 2 (r) dr

17 186 Summary: (a) indicates that OLSE ˆγ T is not consistent. (b) indicates that s 2 T = 1 T g (c) indicates that F T diverges. T û 2 t diverges. = Spurious regression ( )

18 Resolution for Spurious Regression: Suppose that y 1,t = α + γ y 2,t + u t is a spurious regression. (1) Estimate y 1,t = α + γ y 2,t + φy 1,t 1 + δy 2,t 1 + u t. Then, ˆγ T is T-consistent, and the t test statistic goes to the standard normal distribution under H 0 : γ = 0. (2) Estimate y 1,t = α + γ y 2,t + u t. Then, ˆα T and ˆβ T are T-consistent, and the t test and F test make sense. (3) Estimate y 1,t = α + γ y 2,t + u t by the Cochrane-Orcutt method, assuming that u t is the first-order serially correlated error. Usually, choose (2). However, there are two exceptions. (i) The true value of φ is not one, i.e., less than one.

19 188 (ii) y 1,t and y 2,t are the cointegrated processes. In these two cases, taking the first difference leads to the misspecified regression. 8. Cointegrating Vector: Suppose that each element of y t is I(1) and that a y t is I(0). a is called a cointegrating vector ( ), which is not unique. Set z t = a y t, where z t is scalar, and a and y t are g 1 vectors. For z t I(0) (i.e., stationary) T 1 T T z 2 t = T 1 (a y t ) 2 E(z 2 t ). For z t I(1) (i.e., nonstationary, i.e., a is not a cointegrating vector), T 2 T 1 (a y t ) 2 λ 2 (W(r)) 2 dr, where W(r) denotes a standard Brownian motion and λ 2 indicates variance of (1 L)z t. 0

20 189 If a is not a cointegrating vector, T 1 T z 2 t diverges. = We can obtain a consistent estimate of a cointegrating vector by minimizing T z 2 t with respect to a, where a normalization condition on a has to be imposed. The estimator of the a including the normalization condition is super-consistent (T-consistent). Stock, J.H. (1987) Asymptotic Properties of Least Squares Estimators of Cointegrating Vectors, Econometrica, Vol.55, pp Proposition: Let y 1,t be a scalar, y 2,t be a k 1 vector, and (y 1,t, y 2,t ) be a g 1 vector, where g = k + 1. Consider the following model: y 1,t = α + γ y 2,t + z t, y 2,t = u 2,t, ( z ) t = Ψ (L)ɛ t, u 2,t

21 190 ɛ t is a g 1 i.i.d. vector with E(ɛ t ) = 0 and E(ɛ t ɛ t ) = PP. ( ˆα ) ( T y ) 1 ( 2,t y1,t OLSE is given by: = ˆγ y2,t y2,t y ). 2,t y1,t y 2,t Define λ 1, which is a g 1 vector, and Λ 2, which is a k g matrix, as follows: ( λ ) Ψ 1 (1) P =. Then, we have the following results: ( ) ( T 1/2 ( ˆα α) ) 1 Λ 2 W(r)dr ( ) T(ˆγ γ) Λ 2 W(r)dr Λ 2 (W(r)) (W(r)) dr ( h1 ) λ 1 W(1) where = ( ) W(r) (dw(r)) λ 1 + E(u 2,t z t+τ). h 2 Λ 2 W(r) denotes a g-dimensional standard Brownian motion. τ=0 Λ 2 Λ 2 1 ( h1 h 2 ), 1) OLSE of the cointegrating vector is consistent even though u t is serially correlated. 2) The consistency of OLSE implies that T 1 û 2 t σ 2. 3) Because T 1 (y 1,t y 1 ) 2 goes to infinity, a coefficient of determination, R 2, goes to one.

22 Testing Cointegration Engle-Granger Test y t I(1) y 1,t = α + γ y 2,t + u t u t I(0) = Cointegration u t I(1) = Spurious Regression Estimate y 1,t = α + γ y 2,t + u t by OLS, and obtain û t. Estimate û t = ρû t 1 + δ 1 û t 1 + δ 2 û t δ p 1 û t p+1 + e t by OLS. ADF Test: H 0 : ρ = 1 (Sprious Regression) H 1 : ρ < 1 (Cointegration) = Engle-Granger Test For example, see Engle and Granger (1987), Phillips and Ouliaris (1990) and Hansen (1992).

23 192 Asymmptotic Distribution of Residual-Based ADF Test for Cointegration # of Refressors, (a) Regressors have no drift (b) Some regressors have drift excluding constant 1% 2.5% 5% 10% 1% 2.5% 5% 10% J.D. Hamilton (1994), Time Series Analysis, p.766.