Chapter 16

 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]
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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)
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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).
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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.
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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 .
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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
⎦⎥
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 16
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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
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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.
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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.
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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.
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