Introduction to Econometrics (3rd Updated Edition) by James H. Stock and Mark W. Watson Solutions to End-of-Chapter Exercises: Chapter 16* (This version August 17, 2014) *Limited distribution: For Instructors Only. Answers to all odd-numbered questions are provided to students on the textbook website. If you find errors in the solutions, please pass them along to us at [email protected] ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 1 _____________________________________________________________________________________________________ 16.1. Yt follows a stationary AR(1) model, Yt = β0 + β1Yt −1 + ut . The mean of Yt is µY = E (Yt ) = β0 , and E (ut |Yt ) = 0. 1 − β1 (a) The h-period ahead forecast of Yt , Yt + h|t = E (Yt + h | Yt , Yt −1 ,K ), is Yt + h|t = E (Yt + h |Yt , Yt −1 ,K ) = E ( β 0 + β1Yt + h −1 + ut |Yt , Yt −1 ,K ) = β 0 + β1Yt + h −1|t = β 0 + β1 ( β 0 + β1Yt + h −2|t ) = (1 + β1 ) β 0 + β12Yt + h − 2|t = (1 + β1 ) β 0 + β12 ( β 0 + β1Yt + h −3|t ) = (1 + β1 + β12 ) β 0 + β13Yt + h −3|t =L L = (1 + β1 + L + β1h −1 ) β 0 + β1hYt = 1 − β1h β 0 + β1hYt 1 − β1 = µY + β1h (Yt − µY ). (b) Substituting the result from part (a) into Xt gives ∞ ∞ X t = ∑ δ iYt +i|t = ∑ δ i [ µY + β1i (Yt − µY )] i =0 i =0 ∞ ∞ i =0 i =0 = µY ∑ δ i + (Yt − µY )∑ ( β1δ )i = µY Yt − µY + . 1 − δ 1 − β1δ ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 2 _____________________________________________________________________________________________________ 16.2. (a) Because R1t follows a random walk (R1t = R1t–1 + ut), the i-period ahead forecast of R1t is R1t+i|t = R1t+i−1|t = R1t+i−2|t = ! = R1t. Thus Rkt = 1 k 1 k R 1 + e = R1t + et = R1t + et . ∑ t +i|t t k ∑ k i =1 i =1 (b) R1t follows a random walk and is I (1). Rkt is also I (1). Given that both Rkt and R1t are integrated of order one, and Rkt − R1t = et is integrated of order zero, we can conclude that Rkt and R1t are cointegrated. The cointegrating coefficient is 1. (c) When ΔR1t = 0.5Δ + ut , ΔR1t is stationary but R1t is not stationary. R1t = 1.5R1t −1 − 0.5R1t −2 + ut , an AR(2) process with a unit autoregressive root. That is, R1t is I (1) . The i-period ahead forecast of ΔR1t is ΔR1t+i|t = 0.5ΔR1t+i−1|t = 0.52 ΔR1t+i−2|t = ! = 0.5i ΔR1t. The i-period ahead forecast of R1t is R1t+i t = R1t+i−1|t + ΔR1t+i|t = R1t+i−2|t + ΔR1t+i−1|t + ΔR1t+i|t =… = R1t + ΔR1t+1|t +!+ ΔR1t+i|t = R1t + (0.5 +!+ 0.5i )ΔR1t = R1t + 0.5(1− 0.5i ) ΔR1t. 1− 0.5 Thus 1 k 1 k R1t +i t + et = ∑ [ R1t + (1 − 0.5i )ΔR1t ] + et ∑ k i =1 k i =1 = R1t + φΔR1t + et . Rkt = where φ = 1k ∑ik=1 (1 − 0.5i ). Thus Rkt − R1t = φΔR1t + et .Thus Rkt and R1t are cointegrated. The cointegrating coefficient is 1. (continued on next page) ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 3 _____________________________________________________________________________________________________ 16.2 (continued) (d) When R1t = 0.5R1t – 1 + ut, R1t is stationary and does not have a stochastic trend. R1t +i|t = 0.5i R1t , so that, Rkt = θ R1t + et , where θ = 1k ∑ik=1 0.5i. Since R1t and et are I (0), then Rkt is I (0). ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 4 _____________________________________________________________________________________________________ 16.3. ut follows the ARCH process with mean E (ut) = 0 and variance σ t2 = 1.0 + 0.5ut2−1. (a) For the specified ARCH process, ut has the conditional mean E (ut |ut −1 ) = 0 and the conditional variance. var (ut | ut −1 ) = σ t2 = 1.0 + 0.5ut2−1. The unconditional mean of ut is E (ut) = 0, and the unconditional variance of ut is var (ut ) = var[ E (ut | ut −1 )] + E[var (ut | ut −1 )] = 0 + 1.0 + 0.5E (ut2−1 ) = 1.0 + 0.5 var (ut −1 ). The last equation has used the fact that E (ut2 ) = var(ut ) + E (ut )]2 = var(ut ), which follows because E (ut) = 0. Because of the stationarity, var(ut–1) = var(ut). Thus, var(ut) = 1.0 + 0.5var(ut) which implies var(ut ) = 1.0 0.5 = 2. (b) When ut −1 = 0.2, σ t2 = 1.0 + 0.5 × 0.22 = 1.02. The standard deviation of ut is σt = 1.01. Thus ⎛ −3 ut 3 ⎞ Pr (−3 ≤ ut ≤ 3) = Pr ⎜ ≤ ≤ ⎟ ⎝ 1.01 σ t 1.01 ⎠ = Φ(2.9703) − Φ(−2.9703) = 0.9985 − 0.0015 = 0.9970. When ut–1 = 2.0, σ t2 = 1.0 + 0.5 × 2.02 = 3.0. The standard deviation of ut is σt = 1.732. Thus ⎛ −3 u 3 ⎞ Pr (−3 ≤ ut ≤ 3) = Pr ⎜ ≤ t ≤ ⎟ ⎝ 1.732 σ t 1.732 ⎠ = Φ(1.732) − Φ(−1.732) = 0.9584 − 0.0416 = 0.9168. ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 5 _____________________________________________________________________________________________________ 16.4. Yt follows an AR(p) model Yt = β0 + β1Yt−1 +!+ β pYt− p + ut . E(ut |Yt−1 ,Yt−2 ,…) = 0 implies E(ut+h |Yt ,Yt−1 ,…) = 0 for h ≥ 1. The h-period ahead forecast of Yt is Yt+h|t = E (Yt+h |Yt ,Yt−1 ,…) = E( β0 + β1Yt+h−1 +!+ β pYt+h− p + ut+h |Yt , Yt−1 ,…) = β0 + β1 E(Yt+h−1|Yt ,Yt−1 ,…) +! + β p E(Yt+h− p |Yt ,Yt−1 ,…) + E(ut+h |Yt ,Yt−1 ,…) = β0 + β1Yt+h−1|t +!+ β pYt+h− p|t . ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 6 _____________________________________________________________________________________________________ 16.5. Because Yt = Yt − Yt −1 + Yt −1 = Yt −1 + ΔYt , T T T T T t =1 t =1 t =1 t =1 t =1 ∑ Yt 2 =∑ (Yt −1 + ΔYt )2 = ∑ Yt −21 + ∑ (ΔYt )2 + 2∑ Yt −1ΔYt . So 1 T 1 1⎡ T 2 T 2 T ⎤ Y Δ Y = × ⎢∑ Yt − ∑ Yt −1 − ∑ (ΔYt )2 ⎥ . ∑ t −1 t T t =1 T 2 ⎣ t =1 t =1 t =1 ⎦ ( ) ( ) Note that ∑Tt =1 Yt 2 − ∑Tt =1 Yt −21 = ∑Tt =−11 Yt 2 + YT2 − Y02 + ∑Tt =−11 Yt 2 = YT2 − Y02 = YT2 because Y0 = 0. Thus: 1 T 1 1⎡ 2 T ⎤ Y Δ Y = × ⎢YT − ∑ (ΔYt ) 2 ⎥ ∑ t −1 t T t =1 T 2⎣ t =1 ⎦ = 2 ⎤ 1 ⎡⎛ YT ⎞ 1 T − ∑ (ΔYt ) 2 ⎥ . ⎢⎜ ⎟ 2 ⎣⎢⎝ T ⎠ T t =1 ⎦⎥ ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 7 _____________________________________________________________________________________________________ 16.6. (a) Rewrite the regression as Yt = 3.0 + 2.3Xt + 1.7(Xt+1 − Xt) + 0.2(Xt − Xt–1) + ut Thus θ = 2.3, δ–1 = 1.7, δ0 = 0.2 and δ1 = 0.0. (b) Cointegration requires Xt to be I(1) and ut to be I(0). (i) No (ii) No (iii) Yes ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 8 _____________________________________________________________________________________________________ 16.7. 1 T T T ∑ Y ΔY ∑ Y X ∑ Y ΔY t t t=1 t t+1 T t=1 t t+1 . βˆ = t=1 = = ∑ Tt=1 X t2 ∑ Tt=1 (ΔYt+1 )2 1 T ∑ (ΔYt+1 )2 T t=1 Following the hint, the numerator is the same expression as (16.21) (shifted forward in time 1 period), so that 1 T d ∑Tt =1 Yt ΔYt +1 → σ u2 2 ( χ12 − 1). The denominator is p ∑Tt =1 (ΔYt +1 )2 = T1 ∑Tt =1 ut2+1 →σ u2 by the law of large numbers. The result follows directly. 1 T ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 9 _____________________________________________________________________________________________________ 16.8. (a) First note that Yt–1/t–2 = β11Yt–2 + γ11Xt–2 and Xt–1/–2 = β21Yt–2 + γ21Xt–2. Also Yt/t–2 = β11Yt–1/t–2 + γ11Xt–1/t–2. Substituting yields Yt/–2 = β11(β11Yt–2 + γ11Xt–2) + γ11(β21Yt–2 + γ21Xt–2) = [β11β11 + γ11β21]Yt–2 + [β11γ11 + γ11γ21]Xt–2 so that δ1 = β11β11 + γ11β21 and δ2 = β11γ11 + γ11γ21. (b) There is no difference in iterated multistep or direct forecast if the values of δ1 and δ2 were known. (This is shown in (a).) But, these parameters must be estimated, and the implied VAR estimates of these parameters are more accurate (have lower standard errors) if the VAR model is correctly specified. ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 10 _____________________________________________________________________________________________________ 16.9. (a) From the law of iterated expectations E (ut2 ) = E (σ t2 ) = E (α 0 + α1ut2−1 ) = α 0 + α1 E ( ut2−1 ) = α 0 + α1 E ( ut2 ) where the last line uses stationarity of u. Solving for E (ut2 ) gives the required result. (b) As in (a) E (ut2 ) = E (σ t2 ) = E (α 0 + α1ut2−1 + α 2ut2− 2 + L + α put2− p ) = α 0 + α1 E ( ut2−1 ) + α 2 E ( ut2−2 ) + L + α p E (ut2− p ) = α 0 + α1 E ( ut2 ) + α 2 E ( ut2 ) + L + α p E (ut2 ) so that E (ut2 ) = α0 1 − ∑tp=1 α i (c) This follows from (b) and the restriction that E (ut2 ) > 0. (d) As in (a) E (ut2 ) = E (σ t2 ) = α 0 + α1 E ( ut2−1 ) + φ1 E (σ t2−1 ) = α 0 + (α1 + φ1 ) E ( ut2−1 ) = α 0 + (α1 + φ1 ) E ( ut2 ) = α0 1 − α1 − φ1 ( ) (e) This follows from (d) and the restriction that E ut2 > 0. ©2015 Pearson Education, Inc. Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16 11 _____________________________________________________________________________________________________ 16. 10. Write ΔYt = θΔXt + Δv1t and ΔXt = v2t; also v1t–1 = Yt–1 − θXt–1. Thus ΔYt = −(Yt−1 − θXt–1) + u1t and ΔXt = u2t, with u1t = v1t + θv2t and u2t = v2t. ©2015 Pearson Education, Inc.

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