Introduction to Econometrics (3rd Updated Edition)
by
James H. Stock and Mark W. Watson
Solutions to End-of-Chapter Exercises: Chapter 17*
(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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17.1. (a) Suppose there are n observations. Let b1 be an arbitrary estimator of β1. Given
the estimator b1, the sum of squared errors for the given regression model is
n
∑ (Y − b X ) .
i =1
2
i
1
i
βˆ1RLS , the restricted least squares estimator of β1, minimizes the sum of squared
errors. That is, βˆ RLS satisfies the first order condition for the minimization which
1
requires the differential of the sum of squared errors with respect to b1 equals
zero:
n
∑ 2(Y − b X )(− X ) = 0.
i =1
i
1
i
i
Solving for b1 from the first order condition leads to the restricted least squares
estimator
βˆ1RLS =
∑in=1 X iYi
.
∑in=1 X i2
(b) We show first that βˆ1RLS is unbiased. We can represent the restricted least
squares estimator βˆ1RLS in terms of the regressors and errors:
βˆ1RLS =
∑in=1 X iYi ∑in=1 X i ( β1 X i + ui )
∑in=1 X i ui
=
=
β
+
.
1
∑in=1 X i2
∑in=1 X i2
∑in=1 X i2
Thus
⎛ ∑n X u ⎞
⎡ ∑n X E (u | X ,K , X n ) ⎤
E ( βˆ1RLS ) = β1 + E ⎜ i =n1 i 2 i ⎟ = β1 + E ⎢ i =1 i ni 1 2
⎥ = β1 ,
∑i =1 X i
⎝ ∑i =1 X i ⎠
⎣
⎦
where the second equality follows by using the law of iterated expectations, and
the third equality follows from
∑in=1 X i E (ui | X 1 ,K , X n )
=0
∑in=1 X i2
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17.1 (continued)
because the observations are i.i.d. and E(ui |Xi) = 0. (Note, E(ui |X1,…, Xn) =
E(ui |Xi) because the observations are i.i.d.
Under assumptions 1−3 of Key Concept 17.1, βˆ1RLS is asymptotically normally
distributed. The large sample normal approximation to the limiting distribution
of βˆ1RLS follows from considering
∑in=1 X i ui 1n ∑in=1 X i ui
RLS
ˆ
β1 − β1 = n 2 = 1 n 2 .
∑i =1 X i
n ∑ i =1 X i
Consider first the numerator which is the sample average of vi = Xiui. By
assumption 1 of Key Concept 17.1, vi has mean zero:
E ( X iui ) = E[ X i E (ui | X i )] = 0. By assumption 2, vi is i.i.d. By assumption 3,
var(vi) is finite. Let v = 1n ∑in=1 X iui , then σ v2 = σ v2 /n. Using the central limit
theorem, the sample average
v /σ v =
1
σv
n
∑ v → N (0, 1)
n
i =1
i
d
or
1 n
d
X i ui →
N (0, σ v2 ).
∑
n i =1
For the denominator, X i2 is i.i.d. with finite second variance (because X has a
finite fourth moment), so that by the law of large numbers
1 n 2 p
X i → E ( X 2 ).
∑
n i =1
Combining the results on the numerator and the denominator and applying
Slutsky’s theorem lead to
(continued on the next page)
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17.1 (continued)
n ( βˆ1RLS − βu ) =
1
n
1
n
∑in=1 X i ui
∑
n
i =1
X
2
i
⎛ var( X i ui ) ⎞
d
→
N ⎜ 0,
⎟.
E( X 2 ) ⎠
⎝
(c) βˆ1RLS is a linear estimator:
βˆ1RLS =
∑in=1 X iYi
n
= ∑ i =1 aiYi ,
n
2
∑i =1 X i
where ai =
Xi
.
∑ X i2
n
i =1
The weight ai (i = 1,…, n) depends on X1,…, Xn but not on Y1,…, Yn.
Thus
βˆ1RLS = β1 +
∑in=1 X i ui
.
∑in=1 X i2
βˆ1RLS is conditionally unbiased because
n
⎛
⎞
∑ i=1
Xu
RLS
ˆ
E( β1 |X 1 ,…, X n = E ⎜ β1 + n i 2 i |X 1 ,…, X n ⎟
∑ i=1 X i
⎝
⎠
n
⎛ ∑ i=1
⎞
Xu
= β1 + E ⎜ n i 2 i |X 1 ,…, X n ⎟
⎝ ∑ i=1 X i
⎠
= β1.
The final equality used the fact that
n
⎛ ∑ n X iui
⎞ ∑ i=1
X i E(ui |X 1 ,…, X n )
E ⎜ i=1
|X
,…,
X
=
=0
1
n⎟
n
2
n
∑ i=1
X i2
⎝ ∑ i=1 X i
⎠
because the observations are i.i.d. and E (ui |Xi) = 0.
(continued on the next page)
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17.1 (continued)
(d) The conditional variance of βˆ1RLS , given X1,…, Xn, is
⎛
⎞
∑ n X iui
var( βˆ1RLS |X1,…, X n ) = var ⎜ β1 + i=1
|X 1 ,…, X n ⎟
n
2
∑ i=1 X i
⎝
⎠
=
n
∑ i=1
X i2 var(ui |X 1 ,…, X n )
n
(∑ i=1
X i2 )2
n
∑ i=1
X i2σ u2
= n 2 2
(∑ i=1 X i )
=
σ u2
.
n
∑ i=1
X i2
(e) The conditional variance of the OLS estimator βˆ1 is
var( βˆ1|X 1 ,K , X n ) =
σ u2
∑in=1 ( X i − X ) 2
.
Since
n
n
n
n
n
i =1
i =1
i =1
i =1
i =1
∑ ( X i − X )2 = ∑ X i2 − 2 X ∑ X i + nX 2 = ∑ X i2 − nX 2 < ∑ X i2 ,
the OLS estimator has a larger conditional variance:
var(β1|X1 ,K , X n ) > var(βˆ1RLS |X1,K , X n ).
The restricted least squares estimator βˆ1RLS is more efficient.
(f) Under assumption 5 of Key Concept 17.1, conditional on X1,…, Xn, βˆ1RLS is
normally distributed since it is a weighted average of normally distributed
variables ui:
βˆ1RLS = β1 +
∑in=1 X i ui
.
∑in=1 X i2
(continued on the next page)
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17.1 (continued)
Using the conditional mean and conditional variance of βˆ1RLS derived in parts (c)
and (d) respectively, the sampling distribution of βˆ1RLS , conditional on X1,…, Xn,
is
⎛
βˆ1RLS ~ N ⎜ β1 ,
⎝
⎞
⎟.
X ⎠
σ u2
∑
n
i =1
2
i
(g) The estimator
n
n
n
∑ i=1
Yi ∑ i=1
( β1 X i + ui )
∑ i=1
u
!
β1 = n
=
= β1 + n i
n
∑ i=1 X i
∑ i=1 X i
∑ i=1 X i
The conditional variance is
⎛
⎞
∑n u
var( β!1 |X 1 ,…, X n) = var ⎜ β1 + ni=1 i |X 1 ,…, X n⎟
∑ i=1 X i
⎝
⎠
=
n
∑ i=1
var(ui |X 1 ,…, X n )
n
(∑ i=1
X i )2
nσ u2
= n
.
(∑ i=1 X i )2
The difference in the conditional variance of β!1 and βˆ1RLS is
var( β!1|X 1 ,…, X n ) − var( βˆ1RLS |X 1 ,…, X n ) =
nσ u2
σ u2
− n 2.
n
(∑ i=1
X i )2 ∑ i=1
Xi
In order to prove var( β!1|X 1 ,…, X n ) ≥ var( βˆ1RLS |X 1 ,…, X n ), we need to show
n
1
≥ n
2
(∑ X i )
∑i =1 X i2
n
i =1
or equivalently
(continued on the next page)
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17.1 (continued)
2
⎛ n
⎞
n∑ X ≥ ⎜ ∑ X i ⎟ .
i =1
⎝ i =1 ⎠
n
2
i
This inequality comes directly by applying the Cauchy-Schwartz inequality
2
n
n
⎡ n
⎤
2
2
(
a
⋅
b
)
≤
a
⋅
⎢∑ i i ⎥ ∑ i ∑ bi
i =1
i =1
⎣ i =1
⎦
which implies
2
2
n
n
n
⎛ n
⎞ ⎛ n
⎞
2
2
X
=
1⋅
X
≤
1
⋅
X
=
n
X i2 .
∑
∑
∑
i⎟
i
⎜⎝ ∑ i ⎟⎠ ⎜⎝ ∑
⎠
i=1
i=1
i=1
i=1
i=1
n
X i2 ≥ (Σ nx=1 X i )2, or var( β!1|X 1 ,…, X n ) ≥ var( βˆ1RLS |X 1 ,…, X n ).
That is nΣ i=1
Note: because β!1 is linear and conditionally unbiased, the result
var( β!1 |X 1 ,…, X n) ≥ var( βˆ1RLS |X 1 ,…, X n ) follows directly from the Gauss-Markov
theorem.
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.2. The sample covariance is
s XY =
1 n
∑ ( X i − X )(Yi − Y )
n − 1 i =1
=
1 n
∑{[ X i − µ X ) − ( X − µ X )][Yi − µY ) − (Y − µY )]}
n = 1 i =1
=
n
1 ⎧ n
(
X
−
µ
)(
Y
−
µ
)
−
( X − µ X )(Yi − µY )
⎨∑ i
∑
X
i
Y
n − 1 ⎩ i =1
i =1
n
n
⎫
−∑ ( X i − µ X )(Y − µY ) + ∑ ( X − µ X )(Y − µY ) ⎬
i =1
i =1
⎭
n
n ⎡1
n
⎤
=
( X i − µ X )(Yi − µY ) ⎥ −
( X − µ X )(Y − µY )
∑
⎢
n − 1 ⎣ n i =1
⎦ n −1
where the final equality follows from the definition of X and Y which implies that
Σin=1 ( X i − µ X ) = n( X − µ X ) and Σin=1 (Yi − µY ) = n(Y − µY ), and by collecting terms.
We apply the law of large numbers on sXY to check its convergence in probability. It
p
is easy to see the second term converges in probability to zero because X →
µ X and
p
p
Y →
µY so ( X − µ X )(Y − µY ) →
0 by Slutsky’s theorem. Let’s look at the first
term. Since (Xi, Yi) are i.i.d., the random sequence (Xi − µX) (Yi − µY) are i.i.d. By the
definition of covariance, we have E[( X i − µ X )(Yi − µY )] = σ XY . To apply the law of
large numbers on the first term, we need to have
var[( X i − µ X )(Yi − µY )] < ∞
which is satisfied since
var[( X i − µ X )(Yi − µY )] < E[( X i − µ X ) 2 (Yi − µY ) 2 ]
≤ E[( X i − µ X ) 4 ]E [(Yi − µY ) 4 ] < ∞.
The second inequality follows by applying the Cauchy-Schwartz inequality, and the
third inequality follows because of the finite fourth moments for (Xi, Yi). Applying
the law of large numbers, we have
(continued on the next page)
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17.2 (continued)
1 n
p
( X i − µ X )(Yi − µY ) →
E[( X i − µ X )(Yi − µY )] = σ XY .
∑
n i =1
Also,
n
n−1
→ 1, so the first term for sXY converges in probability to σXY. Combining
p
results on the two terms for sXY , we have s XY →
σ XY .
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.3. (a) Using Equation (17.19), we have
n
1
n ∑ i =1 ( X i − X )ui
ˆ
n ( β1 − β1 ) = n 1 n
2
n ∑ i =1 ( X i − X )
1
n
= n
=
=
∑in=1[( X i − µ X ) − ( X − µ X )]ui
n
2
1
n ∑ i =1 ( X i − X )
1
n
∑in=1 ( X i − µ X )ui
1
n
∑in=1 ( X i − X ) 2
1
n
∑in=1 vi
∑in=1 ( X i − X ) 2
1
n
−
−
( X − µX )
1
n
∑in=1 ui
∑in=1 ( X i − X ) 2
( X − µX )
1
n
1
n
1
n
∑in=1 ui
∑in=1 ( X i − X ) 2
by defining vi = (Xi − µX)ui.
(b) The random variables u1,…, un are i.i.d. with mean µu = 0 and variance
0 < σ u2 < ∞. By the central limit theorem,
n (u − µu )
σu
=
1
n
∑in=1 ui
σu
d
→
N (0, 1).
The law of large numbers implies X → µ X 2 , or X − µ X → 0. By the consistency
p
p
of sample variance, 1n Σin=1 ( X i − X ) 2 converges in probability to population
variance, var(Xi), which is finite and non-zero. The result then follows from
Slutsky’s theorem.
(c) The random variable vi = (Xi − µX) ui has finite variance:
var(vi ) = var[( X i − µ X ) µi ]
≤ E[( X i − µ X ) 2 ui2 ]
≤ E[( X i − µ X ) 4 ]E[(ui ) 4 ] < ∞.
The inequality follows by applying the Cauchy-Schwartz inequality, and the
second inequality follows because of the finite fourth moments for (Xi, ui). The
finite variance along with the fact that vi has mean zero (by assumption 1 of Key
Concept 15.1) and vi is i.i.d. (by assumption 2) implies that the sample average
v satisfies the requirements of the central limit theorem. Thus,
(continued on the next page)
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17.3 (continued)
v
σv
=
1
n
∑in=1 vi
σv
satisfies the central limit theorem.
(d) Applying the central limit theorem, we have
1
n
∑in=1 vi
σv
d
→
N (0, 1).
Because the sample variance is a consistent estimator of the population variance,
we have
1
n
∑in=1 ( X i − X ) 2 p
→ 1.
var( X i )
Using Slutsky’s theorem,
1
n
1
n
∑in=1 vt
σv
d
→
N (0, 1),
∑ ( X t − X )2
σ X2
n
i =1
or equivalently
∑in=1 vi
⎛
var(vi ) ⎞
d
→
N
0,
.
⎜
n
2
2 ⎟
1
⎝ [var( X i )] ⎠
n ∑ i =1 ( X i − X )
1
n
Thus
1
n
n ( βˆ1 − β1 ) =
1
n
∑in=1 vi
∑in=1 ( X i − X ) 2
−
( X − µX )
1
n
1
n
∑in=1 ui
∑in=1 ( X i − X ) 2
⎛
var(vi ) ⎞
d
→
N ⎜ 0,
2 ⎟
⎝ [var( X i )] ⎠
since the second term for
n (βˆ1 − β1 ) converges in probability to zero as shown
in part (b).
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17.4. (a) Write (βˆ1 − β1 ) = an Sn where an =
1
n
and Sn = n ( Bˆ1 − β1 ). Now,
d
an → 0 and S n →
S where S is distributed N (0, a2). By Slutsky’s theorem
p
d
an Sn →
0 × S . Thus Pr (|βˆ1 − β1| > δ ) → 0 for any δ > 0, so that βˆ1 − β1 → 0 and βˆ1
is consistent.
(b) We have (i)
su2
σ
2
u
p
→ 1 and (ii) g ( x) = x is a continuous function; thus from the
continuous mapping theorem
su2
σ
2
u
=
su
σu
p
→ 1.
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17.5. Because E(W 4) = [E(W2)]2 + var(W2), [E(W2)]2 ≤ E (W 4) < ∞. Thus E(W2) < ∞.
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17.6. Using the law of iterated expectations, we have
E( βˆ1 ) = E[E( βˆ1|X 1 ,…, X n)] = E( β1 ) = β1.
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17.7. (a) The joint probability distribution function of ui, uj, Xi, Xj is f (ui, uj, Xi, Xj). The
conditional probability distribution function of ui and Xi given uj and Xj is f (ui,
Xi |uj, Xj). Since ui, Xi, i = 1,…, n are i.i.d., f (ui, Xi |uj, Xj) = f (ui, Xi). By definition
of the conditional probability distribution function, we have
f (ui , u j , X i , X j ) = f (ui , X i | u j , X j ) f (u j , X j )
= f (ui , X i ) f (u j , X j ).
(b) The conditional probability distribution function of ui and uj given Xi and Xj
equals
f (ui , u j | X i , X j ) =
f (ui , u j , X i , X j )
f (Xi, X j )
=
f (ui , X i ) f (u j , X j )
f (Xi ) f (X j )
= f (ui | X i ) f (u j | X j ).
The first and third equalities used the definition of the conditional probability
distribution function. The second equality used the conclusion the from part (a)
and the independence between Xi and Xj. Substituting
f (ui , u j | X i , X j ) = f (ui | X i ) f (u j | X j )
into the definition of the conditional expectation, we have
E (ui u j | X i , X j ) = ∫ ∫ ui u j f (ui , u j |X i , X j ) dui du j
= ∫ ∫ ui u j f (ui | X i ) f (u j | X j )dui du j
= ∫ ui f (ui | X i )dui ∫ u j f (u j | X j )du j
= E (ui | X i ) E (u j | X j ).
(c) Let Q = (X1, X2,…, Xi – 1, Xi + 1,…, Xn), so that f (ui|X1,…, Xn) = f (ui |Xi, Q). Write
(continued on next page)
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17.7 (continued)
f (ui | X i , Q) =
f (ui , X i , Q)
f ( X i , Q)
=
f (ui , X i ) f (Q)
f ( X i ) f (Q)
=
f (ui , X i )
f (Xi )
= f (ui | X i )
where the first equality uses the definition of the conditional density, the second
uses the fact that (ui, Xi) and Q are independent, and the final equality uses the
definition of the conditional density. The result then follows directly.
(d) An argument like that used in (c) implies
f (ui u j | X i , K X n ) = f (ui u j | X i , X j )
and the result then follows from part (b).
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17.8. (a) Because the errors are heteroskedastic, the Gauss-Markov theorem does not
apply. The OLS estimator of β1 is not BLUE.
(b) We obtain the BLUE estimator of β1 from OLS in the following
Y!i = β0 X! 0i + β1 X! 1i + u!i
where
Y!i =
X! 1i =
Yi
θ 0 + θ1|X i |
Xi
θ 0 + θ1|X i |
, X! 0i =
1
θ 0 + θ1|X i |
, and u! =
ui
θ 0 + θ1|X i |
.
(c) Using equations (17.2) and (17.19), we know the OLS estimator, βˆ1 , is
∑in=1 ( X i − X )(Yi − Y )
∑in=1 ( X i − X ) ui
ˆ
β1 =
= β1 + n
.
∑in=1 ( X i − X )2
∑i =1 ( X i − X ) 2
As a weighted average of normally distributed variables ui , βˆ1 is normally
distributed with mean E ( βˆ1 ) = β1. The conditional variance of βˆ1 , given X1,…,
Xn, is
⎛
⎞
∑ n ( X − X ) ui
var ( βˆ1| X 1 ,..., X n ) = var ⎜ β1 + i =n1 i
| X 1 ,..., X n ⎟
2
∑i =1 ( X i − X )
⎝
⎠
n
2
∑ ( X − X ) var (ui | X 1 ,..., X n )
= i =1 i n
[∑i =1 ( X i − X ) 2 ]2
∑in=1 ( X i − X ) 2 var(ui | X i )
=
[∑in=1 ( X i − X ) 2 ]2
=
∑in=1 ( X i − X ) 2 (θ0 + θ1| X i |)
.
[∑in=1 ( X i − X ) 2 ]2
Thus the exact sampling distribution of the OLS estimator, βˆ1 , conditional on
X1,…, Xn, is
(continued on next page)
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17.8 (continued)
⎛
∑in=1 ( X i − X )2 (θ0 + θ1| X i |) ⎞
⎟.
[∑in=1 ( X i − X )2 ]2
⎠
βˆ1 ~ N ⎜ β1 ,
⎝
(d) The weighted least squares (WLS) estimators, βˆ0WLS and βˆ1WLS , are solutions to
n
min ∑ (Y!i − b0 X! 0i − b1 X! 1i )2 ,
b0, b1
i=1
the minimization of the sum of squared errors of the weighted regression. The
first order conditions of the minimization with respect to b0 and b1 are
n
∑ 2(Y! − b X!
i
0
0i
− b1 X! 1i )(− X! 0i ) = 0,
i=1
n
∑ 2(Y! − b X!
i
0
0i
− b1 X! 1i )(− X! 1i ) = 0.
i=1
Solving for b1 gives the WLS estimator
−Q S + Q00 S1
βˆ1WLS = 01 0
Q00Q11 − Q012
where
n
n
n
n
n
!
Q00 = ∑ i=1
X! 0i X! 0i , Q01 = ∑ i=1
X! 0i X! 1i , Q11 = ∑ i=1
X! 1i X! 1i , S0 = ∑ i=1
X! 0iY!i , and S1 = ∑ i=1
X! 1iY.
!
!
!
Substituting Yi = β0 X 0i + β1 X 0i + u!i yields
−Q Z + Q Z
βˆ1WLS = β1 + 01 0 002 1
Q00Q11 − Q01
n
n
where Z0 = ∑ i=1 X! 0iu!i , and Z1 = ∑ i=1 X! 1iu!i or
n
∑ i=1
(Q00 X! 1i − Q01 X! 0i )u!i
WLS
ˆ
β1 − β1 =
.
Q00Q11 − Q012
From this we see that the distribution of βˆ1WLS | X1 ,... X n is N ( β1 , σ β2ˆWLS ), where
1
(continued on next page)
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Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.8 (continued)
σ
2
βˆ WLS
1
n
σ u2! ∑ i=1
(Q00 X! 1i − Q01 X! 0i )2
=
(Q00Q11 − Q012 )2
=
Q002Q11 + Q012Q00 − 2Q00Q012
(Q00Q11 − Q012 )2
=
Q00
Q00Q11 − Q012
where the first equality uses the fact that the observations are independent, the
2
second uses σ u! = 1, the definition of Q00, Q11, and Q01, and the third is an
algebraic simplification.
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.9. We need to prove
1 n
p
[( X i − X ) 2 uˆi2 − ( X i − µ X ) 2 ui2 ] → 0.
∑
n i =1
Using the identity X = µ X + ( X − µ X ),
n
1 n
2 2
2 2
2 1
ˆ
[(
X
−
X
)
u
−
(
X
−
µ
)
u
]
=
(
X
−
µ
)
∑ i
∑ uˆi2
i
i
X
i
X
n i =1
n i =1
1 n
− 2( X − µ X ) ∑ ( X i − µ X )uˆi2
n i =1
1 n
+ ∑ ( X i − µ X ) 2 (uˆi2 − ui2 ).
n i =1
The definition of uˆi implies
uˆi2 = ui2 + ( βˆ0 − β0 )2 + ( βˆ1 − β1 )2 X i2 − 2ui ( βˆ0 − β0 )
− 2u ( βˆ − β ) X + 2( βˆ − β )( βˆ − β ) X .
i
1
1
i
Substituting this into the expression for
0
1
n
0
1
1
i
Σin=1[( X i − X ) 2 uˆi2 − ( X i − µ X ) 2 ui2 ] yields a
p
series of terms each of which can be written as anbn where an → 0 and
bn = 1n Σin=1 X ir uis where r and s are integers. For example,
an = ( X − µ X ), an = (βˆ1 − β1 ) and so forth. The result then follows from Slutksy’s
theorem if
1
n
p
Σin=1 X ir uis → d where d is a finite constant. Let wi = X ir uis and note that
wi is i.i.d. The law of large numbers can then be used for the desired result if
E ( wi2 ) < ∞. There are two cases that need to be addressed. In the first, both r and s
are non-zero. In this case write
E (wi2 ) = E ( X i2 r ui2 s ) < [ E ( X i4 r )][ E (ui4 s )]
and this term is finite if r and s are less than 2. Inspection of the terms shows that
this is true. In the second case, either r = 0 or s = 0. In this case the result follows
directly if the non-zero exponent (r or s) is less than 4. Inspection of the terms shows
that this is true.
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
20
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17.10. Using (17.43) with W = θˆ − θ implies
E[(θˆ − θ ) 2 ]
ˆ
Pr(|θ − θ | ≥ δ ) ≤
δ2
p
Since E[(θˆ − θ ) 2 ] → 0, Pr(| θˆ − θ | > δ ) → 0, so that θˆ − θ →
0.
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
21
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17.11. Note: in early printing of the third edition there was a typographical error in the
expression for µY|X. The correct expression is µY | X = µY + (σ XY / σ X2 )( x − µ X ) .
(a) Using the hint and equation (17.38)
fY | X = x ( y ) =
1
2
σ (1 − ρ XY
)
2
Y
⎛
⎛ ⎛ x − µ ⎞2
⎛ x − µ X ⎞⎛ y − µY
1
X
⎜⎜
× exp ⎜
− 2 ρ XY ⎜
⎟
⎟⎜
2
⎜ −2(1 − ρ XY ) ⎜ ⎝ σ X ⎠
⎝ σ X ⎠⎝ σ Y
⎝
⎝
2
2
⎞ ⎛ y − µY ⎞ ⎞ 1 ⎛ x − µ X ⎞ ⎞
⎟+⎜
⎟ ⎟+ ⎜
⎟ ⎟.
⎠ ⎝ σ Y ⎠ ⎟⎠ 2 ⎝ σ X ⎠ ⎟⎠
Simplifying yields the desired expression.
(b) The result follows by noting that fY|X=x(y) is a normal density (see equation
(17.36)) with µ = µT|X and σ2 = σ Y2|X .
(c) Let b = σXY/ σ X2 and a = µY −bµX.
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.12. (a)
⎛ u2
⎞
⎛ u2
⎛ σ u2 ⎞ ∞
σ u2 ⎞
1
exp
−
+
u
du
=
exp
exp
−
+
u
−
⎜
⎟ du
⎜ ⎟∫
2
∫−∞ σ 2π ⎜⎝ 2σ u2 ⎟⎠
2
2
σ
2
σ
2
π
⎝ ⎠ −∞ u
u
⎝
⎠
u
∞
2
2
2⎞
⎛ 1
⎛σ ⎞
⎛σ ⎞
1
= exp ⎜ u ⎟ ∫
exp ⎜ − 2 ( u − σ u2 ) ⎟ du = exp ⎜ u ⎟
⎝ 2 ⎠ −∞ σ u 2π
⎝ 2 ⎠
⎝ 2σ u
⎠
E (e u ) =
∞
1
where the final equality follows because the integrand is the density of a normal
random variable with mean and variance equal to σ u2 . Because the integrand is a
density, it integrates to 1.
(b) The result follows directly from (a).
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.13 (a) The answer is provided by equation (13.10) and the discussion following the
equation. The result was also shown in Exercise 13.10, and the approach used
in the exercise is discussed in part (b).
(b) Write the regression model as Yi = β0 + β1Xi + vi, where β0 = E(β0i), β1 = E(β1i), and vi = ui + (β0i − β0) + (β1i − β1)Xi. Notice that E(vi | Xi) = E(ui|Xi) + E(β0i − β0| Xi) + XiE(β1i − β1|Xi) = 0 because β0i and β1i are independent of Xi. Because E(vi | Xi) = 0, the OLS regression of Yi on Xi will provide consistent estimates of β0 = E(β0i) and β1 = E(β1i). Recall that the weighted least squares estimator is the OLS estimator of Yi/σi onto 1/σi and Xi/σi , where σ i = θ0 + θ1 X i2 . Write this regression as Yi / σ i = β0 (1/ σ i ) + β1 ( X i / σ i ) + vi / σ i . This regression has two regressors, 1/σi and Xi/σi. Because these regressors depend only on Xi, E(vi|Xi) = 0 implies that E(vi/σi | (1/σi), Xi/σi) = 0. Thus, weighted least squares provides a consistent estimator of β0 = E(β0i) and β1 = E(β1i). ©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.14
(a) Yi = (Yi − µ) + µ, so that Yi 2 = (Yi − µ)2 + µ2 + 2(Yi − µ)µ. The result follows after
taking the expected value of both sides of the equation.
(b) This follows from the large of large numbers because Yi is i.i.d with mean E(Yi) =
µ and finite variance.
(c) This follows from the large of large numbers because Yi 2 is i.i.d with mean E( Yi 2 )
= µ2 + σ2 (from (a)) and finite variance (because Yi has a finite fourth moment, Yi 2
has a finite second moment).
(d)
(
2
1 n
1 n 2
Y
−
Y
=
(
)
∑
∑ Y + Y 2 − 2YYi
n i=1 i
n i=1 i
)
=
1 n 2
1 n
2
Y
+
Y
−
2Y
∑
∑Y 2
n i=1 i
n i=1 i
=
1 n 2
∑Y − Y 2
n i=1 i
p
(e) This follows from (a)-(d) and Y 2 → µ 2
(f) This follows from (e) and n/(n−1) → 1.
©2015 Pearson Education, Inc.
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 17
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17.15
n
(a) Write W =
∑Z
i=1
2
i
d
where Zi ~ N(0,1). From the law of large number W/n → E( Z i2 )
= 1.
(b) The numerator is N(0,1) and the denominator converges in probability to 1. The
result follows from Slutsky’s theorem (equation (17.9)).
(c) V/m is distributed χ m2 / m and the denominator converges in probability to 1. The
result follows from Slutsky’s theorem (equation (17.9)).
©2015 Pearson Education, Inc.

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