Chapter 14

Introduction to Econometrics (3rd Updated Edition)
by
James H. Stock and Mark W. Watson
Solutions to Odd-Numbered End-of-Chapter Exercises:
Chapter 14
(This version July 20, 2014)
©2015 Pearson Education, Inc. Publishing as Addison Wesley
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.1. (a) Since the probability distribution of Yt is the same as the probability distribution
of Yt–1 (this is the definition of stationarity), the means (and all other moments)
are the same.
(b) E(Yt)
0
1E(Yt–1)
0
1E(Yt),
E(ut), but E(ut)
0 and E(Yt)
E(Yt–1). Thus E(Yt)
and solving for E(Yt) yields the result.
©2015 Pearson Education, Inc. Publishing as Addison Wesley
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.3. (a) To test for a stochastic trend (unit root) in ln(IP), the ADF statistic is the tstatistic testing the hypothesis that the coefficient on ln(IPt – 1) is zero versus the
alternative hypothesis that the coefficient on ln(IPt – 1) is less than zero. The
calculated t-statistic is t 
0.0085
0.0044
 1.93. From Table 14.4, the 10% critical value
with a time trend is 3.12. Because 1.93
3.12, the test does not reject the
null hypothesis that ln(IP) has a unit autoregressive root at the 10% significance
level. That is, the test does not reject the null hypothesis that ln(IP) contains a
stochastic trend, against the alternative that it is stationary.
(b) The ADF test supports the specification used in Exercise 14.2. The use of first
differences in Exercise 14.2 eliminates random walk trend in ln(IP).
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.5. (a)
E[(W  c) 2 ]  E{[W  W )  ( W  c)]2 }
 E[(W  W )2 ]  2 E (W  W )( W  c)  ( W  c) 2
  W2  ( W  c) 2 .
(b) Using the result in part (a), the conditional mean squared error
E[(Yt  f t 1 ) 2 | Yt 1 , Yt  2 ,...]   t2|t 1  (Yt |t 1  ft 1 ) 2
with the conditional variance  t2|t 1  E[(Yt  Yt|t 1 ) 2 ]. This equation is minimized
when the second term equals zero, or when ft 1  Yt|t 1. (An alternative is to use
the hint, and notice that the result follows immediately from exercise 2.27.)
(c) Applying Equation (2.27), we know the error ut is uncorrelated with ut – 1 if
E(ut |ut – 1)
0. From Equation (14.14) for the AR(p) process, we have
ut 1  Yt 1   0  1Yt  2   2Yt 3     pYt  p 1  f (Yt 1 , Yt  2 ,..., Yt  p 1 ),
a function of Yt – 1 and its lagged values. The assumption E (ut |Yt 1 , Yt 2 ,...)  0
means that conditional on Yt – 1 and its lagged values, or any functions of Yt – 1
and its lagged values, ut has mean zero. That is,
E (ut | ut 1 )  E[ut | f (Yt 1 , Yt  2 ,..., Yt  p  2 )]  0.
Thus ut and ut – 1 are uncorrelated. A similar argument shows that ut and ut – j are
uncorrelated for all j  1. Thus ut is serially uncorrelated.
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.7. (a) From Exercise (14.1) E(Yt)
(stationarity) and E(ut)
Also, because Yt
2.5
0.7E(Yt – 1)
0, so that E(Yt)
2.5
0.7Yt – 1
0.72)
9/(1
0.72var(Yt – 1)
0 and var(Yt)
var(ut)
(b) The 1st autocovariance is
cov(Yt , Yt1 )  cov(2.5  0.7Yt1  ut , Yt1 )
 0.7 var(Yt1 )  cov(ut , Yt1 )
 0.7 Y2
 0.7  17.647  12.353.
The 2nd autocovariance is
cov(Yt , Yt  2 )  cov[(1  0.7)2.5  0.7 2 Yt  2  ut  0.7ut 1 , Yt  2 ]
 0.7 2 var(Yt  2 )  cov(ut  0.7ut 1 , Yt  2 )
 0.7 2  Y2
 0.7 2  17.647  8.6471.
(c) The 1st autocorrelation is
cov(Yt , Yt 1 )
var(Yt ) var(Yt 1 )

0.7 Y2
 Y2
 0.7.
The 2nd autocorrelation is
corr (Yt , Yt  2 ) 
cov(Yt , Yt  2 )
var(Yt ) var(Yt  2 )

0.7 2  Y2
 Y2
 0.49.
(d) The conditional expectation of YT 1 given YT is
YT 1/T
2.5
0.7YT
2.5
0.7  102.3
74.11.
©2015 Pearson Education, Inc. Publishing as Addison Wesley
2  0.7
var(Yt – 1) (stationarity), so that
17.647.
corr (Yt , Yt 1 ) 
E(Yt – 1)
2.5/(1 0.7).
ut, var(Yt)
 co1v(Yt – 1, ut). But cov(Yt – 1, ut)
var(Yt)
E(ut), but E(Yt)
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.9. (a) E(Yt)
0
E(et)
b1E(et–1)

bqE(et–q)
0
[because E(et)
0 for all
values of t].
(b)
 e2 for all t and cov(et, ei)
where the final equality follows from var(et)
t.
(c) Yt
0
Yt–j
et
b1et–1
0
et – j
cov(Yt, Yt – j)
b2et – 2
b1et – 1 – j

bqet – q and
b2et – 2 – j

bqet – q – j and
 qk 0  qm 0 bk bm cov(et  k , et  j  m ), where b0
Notice that cov(et–k, et–j–m)
1.
0 for all terms in the sum.
(d) var(Yt )   e2 1  b12  , cov(Yt , Yt  j )   e2b1 , and cov(Yt , Yt  j )  0 for j
©2015 Pearson Education, Inc. Publishing as Addison Wesley
1.
0 for i
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 14
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14.11. Write the model as Yt
Yt
0
(1
1)Yt – 1
Yt – 1
1Yt – 2
0
1(Yt – 1
Yt – 2)
ut. Rearranging yields
ut.
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