Solutions Manual for Introduction to Econometrics, 3rd Edition By James Stock, Mark Watson

For Instructors Solutions to End-of-Chapter Exercises

Chapter 2

Review of Probability 2.1.

(a) Probability distribution function for Y Outcome (number of heads) Probability

Y=0 0.25

Y=1 0.50

Y=2 0.25

(b) Cumulative probability distribution function for Y Outcome (number of heads) Probability

Y<0 0

0≤Y<1 0.25

1≤Y<2 0.75

Y≥2 1.0

d Fq, ∞. (c) µY = E (Y ) = (0 × 0.25) + (1 × 0.50) + (2 × 0.25) = 1.00 . F →

Y ) E (Y 2 ) − [ E (Y )]2 , = Using Key Concept 2.3: var( and

(ui | X i ) so that

var(Y ) = E (Y 2 ) − [ E (Y )]2 = 1.50 − (1.00) 2 = 0.50.

2.2.

We know from Table 2.2 that Pr (Y = 0) = 0.22, Pr (Y = 1) = 0.78, Pr ( X = 0) = 0.30, Pr ( X = 1) = 0.70. So (a) µY = E (Y ) = 0 × Pr (Y = 0) + 1 × Pr (Y = 1) = 0 × 0.22 + 1 × 0.78 = 0.78, µX = E( X ) = 0 × Pr ( X = 0) + 1 × Pr ( X = 1) = 0 × 0.30 + 1 × 0.70 = 0.70.

σ X2 E[( X − µ X ) 2 ] (b) = =− (0 0.70) 2 × Pr ( X =+ 0) (1 − 0.70) 2 × Pr ( X = 1) = (−0.70) 2 × 0.30 + 0.302 × 0.70 = 0.21, = σ Y2 E[(Y − µY ) 2 ] = (0 − 0.78) 2 × Pr (Y = 0) + (1 − 0.78) 2 × Pr (Y = 1) = (−0.78) 2 × 0.22 + 0.222 × 0.78 = 0.1716.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

(c) σ XY =cov (X , Y ) =E[( X − µ X )(Y − µY )] = (0 − 0.70)(0 − 0.78) Pr( X == 0, Y 0)

+ (0 − 0.70)(1 − 0.78) Pr ( X = 0, Y = 1) + (1 − 0.70)(0 − 0.78) Pr ( X =1, Y = 0) + (1 − 0.70)(1 − 0.78) Pr ( X =1, Y =1) = (−0.70) × (−0.78) × 0.15 + (−0.70) × 0.22 × 0.15 + 0.30 × (−0.78) × 0.07 + 0.30 × 0.22 × 0.63 = 0.084,

σ 0.084 = corr (X , Y ) = XY = 0.4425. σ XσY 0.21 × 0.1716 2.3.

For the two new random variables W = 3 + 6 X and V= 20 − 7Y , we have: (a)

E (V ) = E (20 − 7Y ) = 20 − 7 E (Y ) = 20 − 7 × 0.78 = 14.54, E (W ) = E (3 + 6 X ) = 3 + 6 E ( X ) = 3 + 6 × 0.70 = 7.2.

(b) σ W2 = var (3 + 6 X ) = 62 ⋅ σ X2 = 36 × 0.21 = 7.56, σ V2 = var (20 − 7Y ) = (−7) 2 ⋅ σ Y2 = 49 × 0.1716 = 8.4084. (c) σ WV =cov (3 + 6 X , 20 − 7Y ) =6 × (−7)cov (X , Y ) =−42 × 0.084 =−3.528

σ WV −3.528 = =−0.4425. σWσV 7.56 × 8.4084

corr (W , V ) =

2.4.

(a) E ( X 3 ) = 03 × (1 − p ) + 13 × p = p (b) E ( X k ) = 0k × (1 − p ) + 1k × p = p (c) E ( X ) = 0.3 , and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus σ =

= σ var ( X ) =E ( X 2 ) − [ E ( X )]2 =0.3 − 0.09 =0.21

0.21 = 0.46.

= 0.21 0.46. To compute the skewness, use

the formula from exercise 2.21: 3 E ( X − µ )= E ( X 3 ) − 3[ E ( X 2 )][ E ( X )] + 2[ E ( X )]3

=0.3 − 3 × 0.32 + 2 × 0.33 =0.084

[(1 − 0.3)3 × 0.3] + [(0 − 0.3)3 × 0.7] = 0.084 Alternatively, E ( X − µ )3 = 0.084/0.463 = 0.87.

Thus, skewness =E ( X − µ )3/σ 3 = To compute the kurtosis, use the formula from exercise 2.21: 4 E ( X − µ )= E ( X 4 ) − 4[ E ( X )][ E ( X 3 )] + 6[ E ( X )]2 [ E ( X 2 )] − 3[ E ( X )]4

= 0.3 − 4 × 0.32 + 6 × 0.33 − 3 × 0.34 = 0.0777

[(1 − 0.3) 4 × 0.3] + [(0 − 0.3) 4 × 0.7] = 0.0777 Alternatively, E ( X − µ ) 4 = 4 4 4 /σ 0.0777/0.46 = 1.76 Thus, kurtosis is E ( X − µ )=

©2011 Pearson Education, Inc. Publishing as Addison Wesley

3

4

2.5.

Stock/Watson • Introduction to Econometrics, Third Edition

Let X denote temperature in °F and Y denote temperature in °C. Recall that Y = 0 when X = 32 and Y =100 when X = 212; this implies Y = (100/180) × ( X − 32) or Y = −17.78 + (5/9) × X. Using Key o −17.78 + (5/9) × 70 = 21.11°C, and σX = 7oF implies Concept 2.3, µX = 70 F implies that µY =

σ Y= (5/9) × 7= 3.89°C.

2.6.

The table shows that Pr ( X = 0, Y = 0) = 0.037, Pr ( X = 0, Y = 1) = 0.622, Pr ( X = 1, Y = 0) = 0.009, Pr ( X = 1, Y = 1) = 0.332, Pr ( X = 0) = 0.659, Pr ( X = 1) = 0.341, Pr (Y = 0) = 0.046, Pr (Y = 1) = 0.954. (a) E (Y ) = µY = 0 × Pr(Y = 0) + 1 × Pr (Y = 1) = 0 × 0.046 + 1 × 0.954 = 0.954. # (unemployed) (b) Unemployment Rate = # (labor force) = Pr (Y = 0) = 1 − Pr(Y = 1) = 1 − E (Y ) = 1 − 0.954 = 0.046.

(c) Calculate the conditional probabilities first: Pr = ( X 0,= Y 0) 0.037 = = 0.056, Pr ( X 0) 0.659 = Pr = ( X 0,= Y 1) 0.622 Pr (Y = 1| X = 0) = = = 0.944, Pr ( X 0) 0.659 = Pr (Y = 0| X = 0) =

Pr (= X 1,= Y 0) 0.009 = = 0.026, Pr ( X 1) 0.341 = Pr (= X 1,= Y 1) 0.332 Pr (Y = 1| X = 1) = 0 974. = =.

Pr ( X 1) 0.341 = Pr (Y = 0| X = 1) =

The conditional expectations are

E (Y |X = 1) = 0 × Pr (Y = 0| X = 1) + 1 × Pr (Y = 1| X = 1) = 0 × 0.026 + 1 × 0.974 = 0.974, E (Y |X = 0) = 0 × Pr (Y = 0| X = 0) + 1 × Pr (Y == 1|X 0) = 0 × 0.056 + 1 × 0.944 = 0.944. (d) Use the solution to part (b), Unemployment rate for college graduates = 1 − E(Y|X = 1) = 1 − 0.974 = 0.026 Unemployment rate for non-college graduates = 1 − E(Y|X = 0) = 1 − 0.944 = 0.056 (e) The probability that a randomly selected worker who is reported being unemployed is a college graduate is Pr (= X 1,= Y 0) 0.009 Pr ( X = 1|Y = 0) = 0 196. = =.

Pr (Y 0) 0.046 =

The probability that this worker is a non-college graduate is Pr ( X = 0|Y = 0) = 1 − Pr ( X = 1|Y = 0) = 1 − 0.196 = 0.804.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

5

(f) Educational achievement and employment status are not independent because they do not satisfy that, for all values of x and y, Pr ( X= x |Y= y )= Pr ( X= x).

For example, from part (e) Pr ( X= 0|Y= 0) = 0.804, while from the table Pr(X = 0) = 0.659.

2.7.

µ M + µ F and σ C2 = σ M2 + σ F2 + 2cov ( M , F ). This = M + F ; thus µ= Using obvious notation, C C implies (a) µC = 40 + 45 = $85,000 per year. cov( M , F )

, so that cov ( M , F ) = σ M σ F corr ( M , F ). Thus cov ( M , F ) = σ Mσ F 12 × 18 × 0.80 = 172.80, where the units are squared thousands of dollars per year.

(b) corr ( M , F ) =

(c) σ C2 = σ M2 + σ F2 + 2cov ( M , F ), so that σ C2= 122 + 182 + 2 × 172.80= 813.60, and = σC

= 813.60 28.524 thousand dollars per year.

(d) First you need to look up the current Euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose that this exchange rate is e (say e = 0.80 Euros per dollar); each 1 dollar is therefore with e Euros. The mean is therefore e × µC (in units of thousands of Euros per year), and the standard deviation is e × σC (in units of thousands of Euros per year). The correlation is unit-free, and is unchanged.

2.8.

Z (Y ) 4. With= = = µY E= (Y ) = 1, σ Y2 var 1

1 2

(Y − 1),

1



1

µ Z = E  (Y − 1) = ( µY − 1)= (1 − 1)= 0, 2 2  2 1 2

 

1 4

1 4

σ Z2 =var  (Y − 1)  = σ Y2 = × 4 =1.

Value of Y

2.9.

Value of X

1 5 8 Probability distribution of Y

14 0.02 0.17 0.02 0.21

22 0.05 0.15 0.03 0.23

30 0.10 0.05 0.15 0.30

40 0.03 0.02 0.10 0.15

65 0.01 0.01 0.09 0.11

Probability Distribution of X 0.21 0.40 0.39 1.00

(a) The probability distribution is given in the table above.

E (Y ) = 14 × 0.21 + 22 × 0.23 + 30 × 0.30 + 40 × 0.15 + 65 × 0.11 = 30.15 E (Y 2 ) = 142 × 0.21 + 222 × 0.23 + 302 × 0.30 + 402 × 0.15 + 652 × 0.11 = 1127.23 var(Y ) =E (Y 2 ) − [ E (Y )]2 =218.21

σ Y = 14.77

©2011 Pearson Education, Inc. Publishing as Addison Wesley

6

Stock/Watson • Introduction to Econometrics, Third Edition

(b) The conditional probability of Y|X = 8 is given in the table below Value of Y 14 0.02/0.39

22 0.03/0.39

30 0.15/0.39

40 0.10/0.39

65 0.09/0.39

E (Y | X = 8) = 14 × (0.02/0.39) + 22 × (0.03/0.39) + 30 × (0.15/0.39) 39.21 + 40 × (0.10/0.39) + 65 × (0.09/0.39) = E (Y 2 | X =8) =142 × (0.02/0.39) + 222 × (0.03/0.39) + 302 × (0.15/0.39) 1778.7 + 402 × (0.10/0.39) + 652 × (0.09/0.39) =

var(Y ) = 1778.7 − 39.212 = 241.65 σ Y | X =8 = 15.54 (c) E ( XY ) = (1 × 14 × 0.02) + (1 × 22 : 0.05) +  + (8 × 65 × 0.09) = 171.7 cov( X , Y ) = E ( XY ) − E ( X ) E (Y ) = 171.7 − 5.33 × 30.15 = 11.0

Y ) cov( X , Y )/(σ X σ= = 0.286 corr( X ,= Y ) 11.0 / (2.60 × 14.77) 2.10.

Using the fact that if Y  N  µY , σ Y2  then

Y − µY

σY

~ N (0, 1) and Appendix Table 1, we have

 Y −1 3 −1 (a) Pr (Y ≤ 3) = ≤ Φ (1) = Pr  0.8413. = 2   2

Y −3 0−3 ≤ (b) Pr(Y > 0) =1 − Pr(Y ≤ 0) =1 − Pr   3   3 = 1 − Φ (−1) = Φ (1) = 0.8413.

 40 − 50 Y − 50 52 − 50  Y ≤ 52) Pr  ≤ ≤ (c) Pr (40 ≤=  5 5   5 = Φ (0.4) − Φ (−2) = Φ (0.4) − [1 − Φ (2)] = 0.6554 − 1 + 0.9772 = 0.6326. 6−5 Y −5 8−5 (d) Pr (6 ≤ Y= ≤ 8) Pr  ≤ ≤  2 2   2 = Φ (2.1213) − Φ (0.7071) = 0.9831 − 0.7602 = 0.2229.

2.11.

(a) 0.90 (b) 0.05 (c) 0.05 (d) When Y ~ χ102 , then Y /10 ~ F10,∞ . (e) Y = Z 2 , where Z ~ N (0,1), thus Pr (Y ≤ 1) = Pr (−1 ≤ Z ≤ 1) = 0.32.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

2.12.

(a) 0.05 (b) 0.950 (c) 0.953 (d) The tdf distribution and N(0, 1) are approximately the same when df is large. (e) 0.10 (f) 0.01

2.13.

(a) E (Y 2 ) = Var (Y ) + µY2 =1 + 0 =1; E (W 2 ) = Var (W ) + µW2 =100 + 0 =100. (b) Y and W are symmetric around 0, thus skewness is equal to 0; because their mean is zero, this means that the third moment is zero.

3 E (Y − µY ) 4 /σ Y4 ; solving yields E(Y 4 ) = 3; a similar (c) The kurtosis of the normal is 3, so= calculation yields the results for W. (d) First, condition on X = 0, so that S = W : E ( S | X= 0)= 0; E ( S 2 | X= 0)= 100, E ( S 3 |X= 0)= 0, E ( S 4 | X= 0)= 3 × 1002.

Similarly,

E ( S | X= 1)= 0; E ( S 2 | X= 1)= 1, E ( S 3 | X= 1)= 0, E ( S 4 | X= 1)= 3.

From the law of iterated expectations E (S ) = E (S | X = 0) × Pr (X = 0) + E ( S | X = 1) × Pr( X = 1) = 0 E ( S 2 ) = E ( S 2 | X = 0) × Pr (X = 0) + E ( S 2 | X = 1) × Pr( X = 1) = 100 × 0.01 + 1 × 0.99 = 1.99 E (S 3 ) = E (S 3 | X = 0) × Pr (X = 0) + E ( S 3 | X = 1) × Pr( X = 1) = 0 E (S 4 ) = E (S 4 | X = 0) × Pr (X = 0) + E ( S 4 | X = 1) × Pr( X = 1) = 3 × 1002 × 0.01 + 3 × 1 × 0.99 =302.97

µ S E= ( S ) 0, thus E ( S − µ S )3 = E ( S 3 ) = 0 from part (d). Thus skewness = 0. Similarly, (e) =

σ S2 = E ( S − µ S ) 2 = E ( S 2 ) =1.99, and E ( S − µ S ) 4 = E ( S 4 ) = 302.97. Thus, = kurtosis 302.97 = / (1.992 ) 76.5 2.14.

The central limit theorem suggests that when the sample size (n) is large, the distribution of the 2

σ 2 43.0, sample average (Y ) is approximately N  µY , σ Y2  with σ Y2 = nY . Given µY = 100, σ Y=

(a) n = 100, σ Y2 =

σ Y2 n

=

43 100

= 0.43, and

 Y − 100 101 − 100  Pr (Y ≤ 101) = Pr  ≤  ≈ Φ (1.525) = 0.9364.

0.43   0.43

(b) n = 165, σ Y2 =

σ Y2 n

=

43 165

= 0.2606, and

 Y − 100 98 − 100  Pr (Y > 98) = 1 − Pr (Y ≤ 98) = 1 − Pr  ≤  0.2606   0.2606 1.000 (rounded to four decimal places). ≈ 1 − Φ (−3.9178) = Φ (3.9178) =

©2011 Pearson Education, Inc. Publishing as Addison Wesley

7

8

Stock/Watson • Introduction to Econometrics, Third Edition

(c) n = 64, σ Y2 =

σ Y2

64

43 = 0.6719, and 64

=

 101 − 100 Y − 100 103 − 100  Pr (101 ≤= Y ≤ 103) Pr  ≤ ≤  0.6719 0.6719   0.6719 ≈ Φ (3.6599) − Φ (1.2200) = 0.9999 − 0.8888 = 0.1111.

2.15.

 9.6 − 10 Y − 10 10.4 − 10  = ≤ ≤ Y ≤ 10.4) Pr  ≤ (a) Pr (9.6  4/n 4/n   4/n 10.4 − 10   9.6 − 10 = Pr  ≤Z≤  4/n   4/n where Z ~ N(0, 1). Thus, 10.4 − 10   9.6 − 10 (i) n = 20; Pr  ≤Z≤ Pr (−0.89 ≤ Z ≤ 0.89) = 0.63 = 4/n   4/n 10.4 − 10   9.6 − 10 (ii) n = 100; Pr  ≤Z≤ Pr(−2.00 ≤ Z ≤ 2.00) = 0.954 = 4/n   4/n 10.4 − 10   9.6 − 10 ≤Z≤ Pr(−6.32 ≤ Z ≤ 6.32) = 1.000 = 4/n   4/n

(iii) n = 1000; Pr 

 −c Y − 10 c  c ≤ Y ≤ 10 + c) Pr  ≤ ≤ (b) Pr (10 − =  4/n 4/n   4/n c   −c = Pr  ≤Z≤ .

4/n   4/n c gets large, and the probability converges to 1.

As n get large 4/n (c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6.

2.16.

There are several ways to do this. Here is one way. Generate n draws of Y, Y1, Y2, … Yn. Let Xi = 1 if Yi < 3.6, otherwise set Xi = 0. Notice that Xi is a Bernoulli random variables with µX = Pr(X = 1) = Pr(Y < 3.6). Compute X . Because X converges in probability to µX = Pr(X = 1) = Pr(Y < 3.6), X will be an accurate approximation if n is large.

2.17.

µY = 0.4 and σ Y2 = 0.4 × 0.6 = 0.24  Y − 0.4 0.43 − 0.4   Y − 0.4  (a) (i) P( Y ≥ 0.43) = Pr  ≥ = ≥ 0.6124 =  Pr   0.27 0.24/n   0.24/n  0.24/n 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

2.18.

 Y − 0.4 0.37 − 0.4   Y − 0.4  (ii) P( Y ≤ 0.37) = Pr  ≤= = ≤ −1.22  0.11  Pr  0.24/n   0.24/n  0.24/n  0.41 − 0.40 (b) We know Pr(−1.96 ≤ Z ≤ 1.96) = 0.95, thus we want n to satisfy = 0.41 > −1.96 24 / n 0.39 − 0.40 and < −1.96. Solving these inequalities yields n ≥ 9220.

24 / n Pr (Y = $0) = 0.95, Pr (Y = $20000) = 0.05.

(a) The mean of Y is

µY = 0 × Pr (Y = $0) + 20,000 × Pr (Y = $20000) = $1000.

The variance of Y is

σ Y2 = E  (Y − µY ) 2  0) + (20000 − 1000) 2 × Pr (Y = 20000) = (0 − 1000) 2 × Pr(Y = = (−1000) 2 × 0.95 + 190002 × 0.05 = 1.9 × 107, 1

so the standard deviation of Y is σ Y = (1.9 × 107 ) 2 = $4359.

$1000, σ Y2 = ) µ= (b) (i) E (Y= Y

σ Y2

=

n (ii) Using the central limit theorem,

1.9 × 107 = 1.9 × 105.

100

Pr (Y > 2000) = 1 − Pr (Y ≤ 2000)  Y − 1000 2, 000 − 1, 000  1 − Pr  = ≤  5 1.9 × 105   1.9 × 10 ≈ 1 − Φ (2.2942) = 1 − 0.9891 = 0.0109.

2.19.

l

X x ,= Y y ) ∑ Pr (=

(a) Pr (= Y y= j) =

i

i =1

j

l

(Y y= | X x )Pr= (X x ) ∑ Pr= j

i

i

i =1

(b) E (Y = )

k

k

l

Y y= Y y |= X x ) Pr (= X x) ) ∑ y ∑ Pr (= ∑ y Pr (=

=j 1

j

j

j =j 1 =i 1

 k 

 

 i =1  j =1

 

l

j

i

i

= ∑  ∑ yj Pr= xi )  Pr ( X = xi ) (Y y= j |X l

= ∑ E (Y | X= xi )Pr ( X= xi ). i =1

(c) When X and Y are independent, Pr (X = xi , = Y yj= ) Pr (X = xi )Pr (= Y yj ),

so

©2011 Pearson Education, Inc. Publishing as Addison Wesley

9

10

Stock/Watson • Introduction to Econometrics, Third Edition

σ XY =E[( X − µ X )(Y − µY )] l

k

= xi , Y = yj ) ∑ ∑ ( xi −µX )( y j −µY ) Pr ( X = =i 1 =j 1 l

k

= ∑ ∑( xi − µ X )( y j − µY ) Pr ( X = xi ) Pr (Y = y j ) =i 1 =j 1

  l  k = − = x µ X x yj  ( ) Pr ( ) X i   ∑ ( y j − µY ) Pr (Y = ∑ i =  i 1=  j 1  = E ( X − µ X ) E (Y − µY ) = 0 × 0 = 0, = corr(X ,Y)

2.20.

l

=

0 = 0.

σ XσY

m

(Y = , Z z ) Pr= (X x= ,Z z ) y | X x= ∑∑ Pr=

(a) Pr (Y = yi ) =

E (Y ) (b)=

σ σ XσY

XY =

i

h 1 =j 1 =

j

h

j

h

k

y Pr (Y y= ) Pr (Y y ) ∑= i =1

i

i

k

l

i

m

y | X x= (Y = , Z z ) Pr= (X x= ,Z z ) ∑ y ∑∑ Pr=

=i 1

i

i

=j 1 = h 1

j

h

j

h

=

 k  = yi | X x= zh )  Pr= zh ) (Y = (X x= ∑∑ j, Z j, Z  ∑ yi Pr =j 1 = h 1 =i 1 

=

Y | X x= , Z z ) Pr= (X x= ,Z z ) ∑∑ E (=

l

m

l

m

=j 1 = h 1

j

h

j

h

where the first line in the definition of the mean, the second uses (a), the third is a rearrangement, and the final line uses the definition of the conditional expectation.

2.21.

(a) E ( X − µ )3= E[( X − µ ) 2 ( X − µ )]= E[ X 3 − 2 X 2 µ + X µ 2 − X 2 µ + 2 X µ 2 − µ 3 ] 3 E ( X 3 ) − 3E ( X 2 ) E ( X ) = E ( X 3 ) − 3E ( X 2 ) µ + 3E ( X ) µ 2 − µ=

+ 3E ( X )[ E ( X )]2 − [ E ( X )]3 E ( X 3 ) − 3E ( X 2 ) E ( X ) + 2 E ( X )3 =

(b) E ( X − µ ) 4= E[( X 3 − 3 X 2 µ + 3 X µ 2 − µ 3 )( X − µ )] =E[ X 4 − 3 X 3 µ + 3 X 2 µ 2 − X µ 3 − X 3 µ + 3 X 2 µ 2 − 3 X µ 3 + µ 4 ] E ( X 4 ) − 4 E ( X 3 ) E ( X ) + 6 E ( X 2 ) E ( X ) 2 − 4 E ( X ) E ( X )3 + E ( X ) 4 = E ( X 4 ) − 4[ E ( X )][ E ( X 3 )] + 6[ E ( X )]2 [ E ( X 2 )] − 3[ E ( X )]4 =

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

2.22.

11

The mean and variance of R are given by µ = w × 0.08 + (1 − w) × 0.05

σ 2= w2 × 0.07 2 + (1 − w) 2 × 0.042 + 2 × w × (1 − w) × [0.07 × 0.04 × 0.25] Cov ( Rs , Rb ) follows from the definition of the correlation between Rs where 0.07 × 0.04 × 0.25 = and Rb. (a) µ 0.065; σ 0.044 = = (b) µ 0.0725; = = σ 0.056 (c) w = 1 maximizes µ ; σ = 0.07 for this value of w. (d) The derivative of σ2 with respect to w is

dσ 2 = 2 w × 0.07 2 − 2(1 − w) × 0.042 + (2 − 4 w) × [0.07 × 0.04 × 0.25] dw = 0.0102 w − 0.0018 Solving for w yields w 18 / 102 0.18. (Notice that the second derivative is positive, so that = = = = w 0.18, σ R 0.038. this is the global minimum.) With 2.23.

X and Z are two independently distributed standard normal random variables, so 2 2 = 0. µ= µ= 0, σ= σ= 1, σ XZ X Z X Z

(a) Because of the independence between X and Z , Pr ( Z= z| X= x= ) Pr ( Z= z ), and 2 2 2 2 E ( Z= |X ) E= ( Z ) 0. Thus E (Y | X ) = E ( X + Z| X ) = E ( X | X ) + E ( Z |X ) = X + 0 = X .

(b) E ( X 2 ) = σ X2 + µ X2 = 1, and µY =E ( X 2 + Z ) =E ( X 2 ) + µ Z =1 + 0 =1. (c) E ( XY ) = E ( X 3 + ZX ) = E ( X 3 ) + E ( ZX ). Using the fact that the odd moments of a standard normal random variable are all zero, we have E ( X 3 ) = 0. Using the independence between 3

ZX ) µ= 0. Thus E ( XY ) = E ( X ) + E ( ZX ) = 0.

X and Z , we have E (= Z µX (d)

cov (XY ) = E[( X − µ X )(Y − µY )] = E[( X − 0)(Y − 1)] = E ( XY − X = ) E ( XY ) − E ( X ) = 0 − 0 = 0. = corr (X ,Y)

2.24.

σ σ XσY

XY =

0 = 0.

σ XσY

(a) E (Yi 2 ) = σ 2 + µ 2 = σ 2 and the result follows directly. (b) (Yi/σ) is distributed i.i.d. N(0,1), W = ∑ i =1 (Yi /σ ) 2 , and the result follows from the definition n

of a χ n2 random variable. n n Yi 2 Yi 2 (c) = E (W ) E= = E n. ∑σ 2 ∑ σ2 =i 1 = i 1

(d) Write

©2011 Pearson Education, Inc. Publishing as Addison Wesley

12

Stock/Watson • Introduction to Econometrics, Third Edition

Y1 = n 2

= V

Y1 /σ

∑ i 2= ∑in 2 (Y /σ ) 2 Yi = n −1 n −1

which follows from dividing the numerator and denominator by σ. Y1/σ ~ N(0,1), n n 2 ∑ i =2 (Yi /σ )2 ~ χ n−1 , and Y1/σ and ∑ i =2 (Yi /σ )2 are independent. The result then follows from the definition of the t distribution.

2.25.

(a)

n

n

∑ axi = (ax1 + ax2 + ax3 +  + axn )= a( x1 + x2 + x3 +  + xn )= a∑ xi i 1 =i 1

(b)

n

∑ (x + y ) = (x + y + x + y +  x + y ) i =1

i

1

i

1

2

2

n

n

= ( x1 + x2 +  xn ) + ( y1 + y2 +  yn ) n

n

∑ xi + ∑ yi

=

=i 1 =i 1

(c)

n

∑ a = (a + a + a +  + a) = na i =1

n 2 i i =i 1 =i 1

(d)

n

∑ (a + bx + cy )= ∑ (a + b x + c y + 2abx + 2acy + 2bcx y ) 2

2

n

2 i

2

2 i

n

i

i

n

i

i

n

n

= na 2 + b 2 ∑ xi2 + c 2 ∑ yi2 + 2ab∑ xi + 2ac ∑ yi + 2bc ∑ xi yi =i 1 =i 1

2.26.

=i 1 =i 1 =i 1

Yi , Y j )/σ Yi σ Y j cov(= Yi , Y j )/σ Y σ Y cov( Yi , Y j )/σ Y2 ρ , where the first equality = (a) corr(Yi,Yj) = cov(= uses the definition of correlation, the second uses the fact that Yi and Yj have the same variance (and standard deviation), the third equality uses the definition of standard deviation, and the fourth uses the correlation given in the problem. Solving for cov(Yi, Yj) from the last equality gives the desired result. (b)

(c) Y =

1 1 1 1 Y1 + Y2 , so that E( Y ) = E (Y )1 + E (Y2 ) = µY 2 2 2 2 σ Y2 ρσ Y2 1 1 2 var(Y ) = var(Y1 ) + var(Y2 ) + cov(Y1 , Y2 ) = + 4 4 4 2 2

= Y

1 n 1 n 1 n = = E ( Y ) E ( Y ) µY µY Y , so that ∑ i n= ∑ ∑i n i =1 n i 1= = i 1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

1 n  var(Y ) = var  ∑ Yi   n i =1  1 n 2 n −1 n = 2 ∑ var(Yi ) + 2 ∑ ∑ cov(Yi , Y j ) n i= 1 n i = 1 j = i +1 1 n 2 2 n −1 n ρσ Y2 ∑σ Y + n2 ∑ ∑ n2 i = 1 i = 1 j = i +1

=

σ Y2

n(n − 1) ρσ Y2 2 n n 2 σ  1 = Y + 1 −  ρσ Y2 n  n

=

n −1

+

n

where the fourth line uses ∑ ∑ a= a (1 + 2 + 3 +  + n − 1)= i = 1 j = i +1

(d) When n is large 2.27

σ

2 Y

n

≈ 0 and

an(n − 1) for any variable a.

2

1 ≈ 0 , and the result follows from (c). n

(a) E(W) = E[E(W|Z)] = E[E(X − X )|Z] = E[E(X|Z) − E(X|Z)] = 0. (b) E(WZ) = E[E(WZ|Z)] = E[ZE(W)|Z] = E[ Z × 0] = 0 (c) Using the hint: V = W − h(Z), so that E(V2) = E(W2) + E[h(Z)2] − 2 × E[W × h(Z)]. Using an argument like that in (b), E[W × h(Z)] = 0. Thus, E(V2) = E(W2) + E[h(Z)2], and the result follows by recognizing that E[h(Z)2] ≥ 0 because h(z)2 ≤ 0 for any value of z.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

13

Chapter 3

Review of Statistics

3.1.

The central limit theorem suggests that when the sample size ( n ) is large, the distribution of the sample average ( Y ) is approximately N  µY , σ Y2  with σ Y2 = 2 43.0, we have σ Y=

(a) n = 100, σ Y2 =

σ Y2 n

=

σ Y2 n

. Given a population µY = 100,

43 = 0.43, and 100

 Y − 100 101 − 100  < Pr (Y < 101) = Pr   ≈ Φ (1.525) = 0.9364.

0.43   0.43

(b) n = 64, σ Y2 =

σ Y2 n

43 = 0.6719, and 64

=

 101 − 100 Y − 100 103 − 100  Pr(101 <= Y < 103) Pr  < <  0.6719 0.6719   0.6719 ≈ Φ (3.6599) − Φ (1.2200) = 0.9999 − 0.8888 = 0.1111. (c) n = 165, σ Y2 =

σ Y2 n

=

43 = 0.2606, and 165

 Y − 100 98 − 100  Pr (Y > 98) = 1 − Pr (Y ≤ 98) = 1 − Pr  ≤  0.2606   0.2606 1.0000 (rounded to four decimal places). ≈ 1 − Φ (−3.9178) = Φ (3.9178) = 3.2.

Each random draw Yi from the Bernoulli distribution takes a value of either zero or one with 1)= p and Pr (Yi = 0) = 1 − p. The random variable Yi has mean probability Pr (Y= i

0 × Pr(Y = 0) + 1 × Pr(Y = 1) = E (Yi ) = p, and variance

var( = Yi ) E[(Yi − µY ) 2 ] =(0 − p ) 2 × Pr(Yi =0) + (1 − p ) 2 × Pr(Yi =1) = p 2 (1 − p ) + (1 − p ) 2 p = p (1 − p ).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

15

(a) The fraction of successes is

pˆ=

n # (success) # (Yi = 1) ∑i =1 Yi = = = Y. n n n

 ∑in=1 Yi  1 n 1 n ˆ (b) E ( p )= E  (Yi )= = E  ∑ ∑ p= p.  =  ni1  n  n i 1=  ∑ n Yi  1 n 1 n p (1 − p ) (c) var(= . pˆ ) var  i =1 = var(= Yi ) p (1 −= p)  2 ∑ 2 ∑ n n n n = i 1 = i 1  

The second equality uses the fact that Y1 , …, Yn are i.i.d. draws and cov(Yi , Y j ) = 0, for i ≠ j.

3.3.

Denote each voter’s preference by Y . Y = 1 if the voter prefers the incumbent and Y = 0 if the voter prefers the challenger. Y is a Bernoulli random variable with probability Pr (Y= 1)= p and Pr (Y = 0) = 1 − p. From the solution to Exercise 3.2, Y has mean p and variance p (1 − p ). (a) pˆ =

215 = 0.5375.

400

pˆ (1 − pˆ ) 0.5375 × (1 − 0.5375)  pˆ ) = 6.2148 × 10−4. The (b) The estimated variance of p̂ is var( = = n 400 1 2 standard error is SE ( pˆ ) = (var( pˆ )) = 0.0249. (c) The computed t-statistic is pˆ − µ p ,0 0.5375 − 0.5 t act = = = 1.506.

SE( pˆ ) 0.0249

Because of the large sample size (n = 400), we can use Equation (3.14) in the text to get the p-value for the test H 0 : p =0.5 vs. H1 : p ≠ 0.5 :

p-value = 2Φ (−|t act |) = 2Φ (−1.506) = 2 × 0.066 = 0.132. (d) Using Equation (3.17) in the text, the p-value for the test H 0 : p =0.5 vs. H1 : p > 0.5 is

p-value = 1 − Φ (t act ) = 1 − Φ (1.506) = 1 − 0.934 = 0.066. (e) Part (c) is a two-sided test and the p-value is the area in the tails of the standard normal distribution outside ± (calculated t-statistic). Part (d) is a one-sided test and the p-value is the area under the standard normal distribution to the right of the calculated t-statistic. (f) For the test H 0 : p =0.5 vs. H1 : p > 0.5, we cannot reject the null hypothesis at the 5% significance level. The p-value 0.066 is larger than 0.05. Equivalently the calculated t-statistic 1.506 is less than the critical value 1.64 for a one-sided test with a 5% significance level. The test suggests that the survey did not contain statistically significant evidence that the incumbent was ahead of the challenger at the time of the survey.

3.4.

Using Key Concept 3.7 in the text (a) 95% confidence interval for p is

©2011 Pearson Education, Inc. Publishing as Addison Wesley

16

Stock/Watson • Introduction to Econometrics, Third Edition

pˆ ± 1.96 SE ( pˆ= ) 0.5375 ± 1.96 × 0.0249= (0.4887,0.5863). (b) 99% confidence interval for p is

pˆ ± 2.57 SE ( pˆ = ) 0.5375 ± 2.57 × 0.0249= (0.4735,0.6015). (c) Mechanically, the interval in (b) is wider because of a larger critical value (2.57 versus 1.96).

Substantively, a 99% confidence interval is wider than a 95% confidence because a 99% confidence interval must contain the true value of p in 99% of all possible samples, while a 95% confidence interval must contain the true value of p in only 95% of all possible samples. (d) Since 0.50 lies inside the 95% confidence interval for p, we cannot reject the null hypothesis at a 5% significance level.

3.5.

(a) (i) The size is given by Pr(| pˆ − 0.5| > .02), where the probability is computed assuming that p = 0.5.

Pr(|pˆ − 0.5| > 0.02) =1 − Pr(−0.02 ≤ pˆ − 0.5 ≤ .02) −0.02 pˆ − 0.5 0.02   = ≤ ≤ 1 − Pr   0.5 × 0.5/1055 0.5 × 0.5/1055   0.5 × 0.5/1055 pˆ − 0.5   = ≤ 1.30  1 − Pr  −1.30 ≤ 0.5 × 0.5/1055   = 0.19

where the final equality using the central limit theorem approximation. (ii) The power is given by Pr(| pˆ − 0.5| > 0.02), where the probability is computed assuming that p = 0.53.

Pr(|pˆ − 0.5| > 0.02) =1 − Pr(−0.02 ≤ pˆ − 0.5 ≤ .02) 0.02 pˆ − 0.5 −0.02   1 − Pr  = ≤ ≤  0.53 × 0.47/1055 0.53 × 0.47/1055   0.53 × 0.47/1055 pˆ − 0.53 −0.05 −0.01   1 − Pr  = ≤ ≤  0.53 × 0.47/1055 0.53 × 0.47/1055   0.53 × 0.47/1055 pˆ − 0.53   1 − Pr  −3.25 ≤ = ≤ −0.65  .53 × 0.47/1055   = 0.74

where the final equality using the central limit theorem approximation.

0.54 − 0.50 (b) (i) t = = 2.61, and Pr(|t= | > 2.61) 0.01, so that the null is rejected at the (0.54 × 0.46) / 1055 5% level. (ii) Pr(t > 2.61) = .004, so that the null is rejected at the 5% level.

(iii) 0.54 ± 1.96 (0.54 × 0.46) / 1055 = 0.54 ± 0.03, or 0.51 to 0.57. (iv) 0.54 ± 2.58 (0.54 × 0.46) / 1055 = 0.54 ± 0.04, or 0.50 to 0.58. (v) 0.54 ± 0.67 (0.54 × 0.46) / 1055 = 0.54 ± 0.01, or 0.53 to 0.55. ©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

17

(c) (i) The probability is 0.95 is any single survey, there are 20 independent surveys, so the probability if 0.9520 = 0.36 (ii) 95% of the 20 confidence intervals or 19. (d) The relevant equation is 1.96 × SE( pˆ ) < .01 or 1.96 × p (1 − p ) / n < .01. Thus n must be 1.962 p (1 − p ) , so that the answer depends on the value of p. Note that the 0.012 largest value that p(1 − p) can take on is 0.25 (that is, p = 0.5 makes p(1 − p) as large as 1.962 × 0.25 possible). Thus if n > = 9604, then the margin of error is less than 0.01 for all 0.012 values of p.

chosen so that n >

3.6.

(a) No. Because the p-value is less than 0.05 (= 5%), µ = 5 is rejected at the 5% level and is therefore not contained in the 95% confidence interval. (b) No. This would require calculation of the t-statistic for µ = 6, which requires Y and SE (Y ).

Only the p-value for test that µ = 5 is given in the problem.

3.7.

The null hypothesis is that the survey is a random draw from a population with p = 0.11. The tpˆ − 0.11 pˆ ) pˆ (1 − pˆ )/n. (An alternative formula for SE( p̂ ) is statistic is t = , where SE(= SE( pˆ ) 0.11 × (1 − 0.11) / n, which is valid under the null hypothesis that p = 0.11). The value of the tstatistic is 2.71, which has a p-value of that is less than 0.01. Thus the null hypothesis p = 0.11 (the survey is unbiased) can be rejected at the 1% level.

3.8

 123  1110 ± 1.96   or 1110 ± 7.62.  1000 

3.9.

Denote the life of a light bulb from the new process by Y . The mean of Y is µ and the standard deviation of Y is σ Y = 200 hours. Y is the sample mean with a sample size n = 100. The σY 200 standard deviation of the sampling distribution of Y is σ= = = 20 hours. The Y n 100 hypothesis test is H 0 : µ = 2000 vs. H1 : µ > 2000 . The manager will accept the alternative hypothesis if Y > 2100 hours. (a) The size of a test is the probability of erroneously rejecting a null hypothesis when it is valid.

The size of the manager’s test is

size = Pr(Y > 2100| µ == 2000) 1 − Pr(Y ≤ 2100| µ = 2000)  Y − 2000 2100 − 2000  = 1 − Pr  ≤ |µ = 2000  20  20  −7 = 1 − Φ (5) = 1 − 0.999999713 = 2.87 × 10 ,

©2011 Pearson Education, Inc. Publishing as Addison Wesley

18

Stock/Watson • Introduction to Econometrics, Third Edition

2000) means the probability that the sample mean is greater than 2100 where Pr(Y > 2100|µ = hours when the new process has a mean of 2000 hours. (b) The power of a test is the probability of correctly rejecting a null hypothesis when it is invalid.

We calculate first the probability of the manager erroneously accepting the null hypothesis when it is invalid:  Y − 2150 2100 − 2150  |µ = 2150  ≤ 20  20 

Pr(Y ≤ 2100|µ = 2150) = Pr  β=

=Φ (−2.5) =1 − Φ (2.5) =1 − 0.9938 =0.0062.

The power of the manager’s testing is 1 − β = 1 − 0.0062 = 0.9938. (c) For a test with 5%, the rejection region for the null hypothesis contains those values of the t-statistic exceeding 1.645.

Y act t=

act

− 2000 > 1.645 ⇒ Y act > 2000 + 1.645 × 20 = 2032.9.

20

The manager should believe the inventor’s claim if the sample mean life of the new product is greater than 2032.9 hours if she wants the size of the test to be 5%.

3.10.

(a) New Jersey sample size n1 = 100, sample average Y1 = 58, and sample standard deviation s1 = 58. The standard error of Y1 is SE( Y1 ) = s1 / n1 = 8/ 100 = 0.8. The 95% confidence interval for the mean score of all New Jersey third graders is

µ1 = Y1 ± 1.96SE(Y1 ) = 58 ± 1.96 × 0.8 = (56.432, 59.568). (b) Iowa sample size n2 = 200, sample average Y2 = 62, sample standard deviation s2 = 11. The s2 s2 64 121 standard error of Y1 − Y2 is SE (Y1 − Y2 ) = 1 + 2 = 1.1158. The 90% + = 100 200 n1 n2

confidence interval for the difference in mean score between the two states is

µ1 − µ2= (Y1 − Y2 ) ± 1.64SE(Y1 − Y2 ) = (58 − 62) ± 1.64 × 1.1158 = (−5.8299, − 2.1701).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

19

(c) The hypothesis tests for the difference in mean scores is

H 0 : µ1 − µ2= 0 vs. H1 : µ1 − µ2 ≠ 0.

From part (b) the standard error of the difference in the two sample means is SE (Y1 − Y2 ) = 1.1158. The t-statistic for testing the null hypothesis is Y −Y 58 − 62 =−3.5849. t act = 1 2 = SE(Y1 − Y2 ) 1.1158

Use Equation (3.14) in the text to compute the p-value:

p − value = 2Φ (−|t act |) = 2Φ (−3.5849) = 2 × 0.00017 = 0.00034.

Because of the extremely low p-value, we can reject the null hypothesis with a very high degree of confidence. That is, the population means for Iowa and New Jersey students are different.

3.11.

Assume that n is an even number. Then Y is constructed by applying a weight of 1/2 to the n/2 “odd” observations and a weight of 3/2 to the remaining n/2 observations.  1  1 3 1 3 E (Y1 ) + E (Y2 ) +  E (Yn −1 ) + E (Yn )    n  2 2 2 2 

E (Y )=

11 n 3 n   ⋅ ⋅ µY + ⋅ ⋅ µY =  µY n2 2 2 2 

=

1 1 9 1 9  var(Y1 ) + var(Y2 ) +  var(Yn −1 ) + var(Yn )  2  n 4 4 4 4 

) var(Y=

σ2 1 1 n 9 n  = 2  ⋅ ⋅ σ Y2 + ⋅ ⋅ σ Y2  =1.25 Y . n 4 2 4 2 n  3.12.

Sample size for men n1 = 100, sample average Y1 =3100 sample standard deviation s1 = 200.

Sample size for women n2 = 64, sample average Y 2 = 2900, sample standard deviation s2 = 320.

The standard error of Y 1 − Y 2 is SE(Y 1 − Y 2) =

s12 s22 2002 3202 + = + =44.721. n1 n2 100 64

(a) The hypothesis test for the difference in mean monthly salaries is

H 0 : µ1 − µ2= 0 vs. H1 : µ1 − µ2 ≠ 0.

The t-statistic for testing the null hypothesis is − 3100 − 2900 t act =Y 1 Y 2 = = 4.4722.

SE(Y 1 − Y 2) 44.721

Use Equation (3.14) in the text to get the p-value:

p-value = 2Φ (−|t act |) = 2Φ (−4.4722) = 2 × (3.8744 × 10−6 ) = 7.7488 × 10−6.

The extremely low level of p-value implies that the difference in the monthly salaries for men and women is statistically significant. We can reject the null hypothesis with a high degree of confidence.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

20

Stock/Watson • Introduction to Econometrics, Third Edition

(b) From part (a), there is overwhelming statistical evidence that mean earnings for men differ from mean earnings for women, and a related calculation shows overwhelming evidence that mean earning for men are greater that mean earnings for women. However, by itself, this does not imply gender discrimination by the firm. Gender discrimination means that two workers, identical in every way but gender, are paid different wages. The data description suggests that some care has been taken to make sure that workers with similar jobs are being compared.

But, it is also important to control for characteristics of the workers that may affect their productivity (education, years of experience, etc.). If these characteristics are systematically different between men and women, then they may be responsible for the difference in mean wages. (If this is true, it raises an interesting and important question of why women tend to have less education or less experience than men, but that is a question about something other than gender discrimination by this firm.) Since these characteristics are not controlled for in the statistical analysis, it is premature to reach a conclusion about gender discrimination.

3.13

(a) Sample size n = 420, sample average Y = 646.2 sample standard deviation sY= 19.5. The s 19.5 standard error of Y is SE (Y ) = Y = = 0.9515. The 95% confidence interval for the 420 n mean test score in the population is

µ = Y ± 1.96SE(Y ) = 646.2 ± 1.96 × 0.9515 = (644.34, 648.06). (b) The data are: sample size for small classes n1 = 238, sample average Y =1 657.4, sample standard deviation s1= 19.4; sample size for large classes n2 = 182, sample average 650.0, sample standard deviation s2= 17.9. The standard error of Y1 − Y2 is Y= 2 s2 s2 19.42 17.92 1.8281. The hypothesis tests for higher average SE (Y1 − Y2 ) = 1 + 2 = + = 238 182 n1 n2

scores in smaller classes is

H 0 : µ1 − µ2= 0 vs. H1 : µ1 − µ2 > 0.

The t-statistic is − 657.4 − 650.0 t act =Y 1 Y 2 = = 4.0479.

SE(Y 1 − Y 2) 1.8281

The p-value for the one-sided test is:

p-value = 1 − Φ (t act ) = 1 − Φ (4.0479) = 1 − 0.999974147 = 2.5853 × 10−5.

With the small p-value, the null hypothesis can be rejected with a high degree of confidence.

There is statistically significant evidence that the districts with smaller classes have higher average test scores.

3.14.

We have the following relations: 1 in = 0.0254 m (or 1 m= 39.37 in), 1 lb = 0.4536 kg (or 1 kg = 2.2046 lb). The summary statistics in the using metric measurements are

X = 70.5 × 0.0254 = 1.79 m; Y = 158 × 0.4536 = 71.669 kg ; s X = 1.8 × 0.0254 = 0.0457 m; sY = 14.2 × 0.4536 = 6.4411 kg ; s XY = 21.73 × 0.0254 × 0.4536 = 0.2504 m × kg , and rXY = 0.85.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

3.15.

21

From the textbook equation (2.46), we know that E( Y ) = µY and from (2.47) we know that

σ Y2

. In this problem, because Ya and Yb are Bernoulli random variables, pˆ a = Ya , pˆ b = n Yb , σ Ya2 = pa(1–pa) and σ Yb2 = pb(1–pb). The answers to (a) follow from this. For part (b), note that var( pˆ a – pˆ b ) = var( pˆ a ) + var( pˆ b ) – 2cov( pˆ a , pˆ b ). But, they are independent (and thus have

var( Y ) =

cov(pˆ a ,pˆ b ) = 0 because pˆ a and pˆ b are independent (they depend on data chosen from independent samples). Thus var( pˆ a – pˆ b ) = var( pˆ a ) + var( pˆ b ). For part (c), use equation 3.21 from the text (replacing Y with p̂ and using the result in (b) to compute the SE). For (d), apply the formula in (c) to obtain 95% CI is (.859 – .374) ± 1.96 3.16.

0.859(1 − 0.859) 0.374(1 − 0.374) or 0.485 ± 0.017. + 5801 4249

(a) The 95% confidence interval if Y ± 1.96 SE(Y ) or 1013 ± 1.96 ×

108 453

or 1013 ± 9.95.

(b) The confidence interval in (a) does not include µ = 1000, so the null hypothesis that µ = 1000 (Florida students have the same average performance as students in the U.S.) can be rejected at the 5% level.) (c) (i) The 95% confidence interval is Yprep − YNon − prep ± 1.96 SE (Yprep − YNon − prep ) where SE(Yprep − YNon − prep ) =

s 2prep n prep

+

2 snon − prep

=

nnon − prep

952 1082 + = 6.61; the 95% confidence interval is 503 453

(1019 − 1013) ± 12.96 or 6 ± 12.96.

(ii) No. The 95% confidence interval includes µ prep − µnon − prep = 0. (d) (i) Let X denote the change in the test score. The 95% confidence interval for µX is 60 X ± 1.96 SE ( X ), where SE( X ) = = 2.82; thus, the confidence interval is 9 ± 5.52.

453 (ii) Yes. The 95% confidence interval does not include µX = 0. (iii) Randomly select n students who have taken the test only one time. Randomly select one half of these students and have them take the prep course. Administer the test again to all of the n students. Compare the gain in performance of the prep-course second-time test takers to the non-prep-course second-time test takers.

3.17.

(a) The 95% confidence interval is Ym , 2008 − Ym , 1992 ± 1.96 SE(Ym , 2008 − Ym , 1992 ) where SE(Ym , 2008 − Ym , 1992 ) =

sm2 ,2008 nm ,2008

+

sm2 ,1992 nm ,1992

=

11.782 10.17 2 + = 0.37; the 95% confidence 1838 1594

interval is (24.98 − 23.27) ± 0.73 or 1.71 ± 0.73. (b) The 95% confidence interval is Yw, 2008 − Yw, 1992 ± 1.96 SE(Yw, 2008 − Yw, 1992 ) where SE(Yw, 2008 − Yw, 1992 ) =

sw2 ,2008 nw,2008

+

sw2 ,1992 nw,1992

=

9.662 7.782 + = 0.31; the 95% confidence interval 1871 1368

is (20.87 − 20.05) ± 0.60 or 0.82 ± 0.60. ©2011 Pearson Education, Inc. Publishing as Addison Wesley

22

Stock/Watson • Introduction to Econometrics, Third Edition

(c) The 95% confidence interval is (Ym , 2004 − Ym , 1992 ) − (Yw, 2004 − Yw, 1992 ) ± 1.96 SE[(Ym , 2008 − Ym , 1992 ) − (Yw, 2008 − Yw, 1992 )], where SE[(Ym , 2008 − Ym , 1992 ) − (Yw, 2008 − Yw, 1992 )] =

sm2 ,2008 nm ,2008

+

sm2 ,1992 nm ,1993

+

sw2 ,2008 nw,2008

+

sw2 ,1992

11.782 10.17 2 9.662 7.782 = + = 0.48. The 95% + + nw,1992 1838 1594 1871 1368

confidence interval is (24.98-23.27) − (20.87−20.05) ± 1.96 × 0.48 or 0.89 ± 0.95.

3.18.

Y1 , …, Yn are i.i.d. with mean µY and variance σ Y2 . The covariance cov (Y j , Yi ) = 0, j ≠ i. The 2 sampling distribution of the sample average Y has mean µY and variance var (Y= ) σ= Y

σ Y2 n

.

2 (a) E[(Yi − Y )= ] E{[(Yi − µY ) − (Y − µY )]2 }

= E[(Yi − µY ) 2 − 2(Yi − µY )(Y − µY ) + (Y − µY ) 2 ] = E[(Yi − µY ) 2 ] − 2 E[(Yi − µY )(Y − µY )] + E[(Y − µY ) 2 ] = var(Yi ) − 2cov(Yi , Y ) + var(Y ).  

(b) cov(Y , Y )= E[(Y − µY )(Yi − µY )]= E  

∑ nj =1 Y j

 

n

 

 

 

 

− µY  (Yi − µY ) 

  n   ∑ j =1 (Y j − µY )   1 1  = E   − = Y µ E[(YI − µY )2 + ∑ E[(Y j − µY )(Yi − µY )] ( )  i Y    n   n n j ≠i    

σ Y2 1 2 1 = . σ Y + ∑ cov(Y j , Yi ) = n n j ≠i n 1 n 1 n  1 n 2 2 = − = − E  sY2  E  ( Y Y ) E [( Y Y ) ] (c)= ∑ i ∑[var (Yi ) − 2cov(Yi , Y ) + var(Y )] i  n −1 ∑ n −1 i 1 = i 1 =  n − 1 i 1= 

σ Y2 σ Y2  1 n  2 1 n  n − 1 2  + σ Y  σ Y2 . = ∑ ∑ σ Y − 2 × =    n − 1 i 1= n n  n −1 i 1  n  =  =

3.19.

(a) No. E (Yi 2 ) = σ Y2 + µY2 and E (YiY j ) = µY2 for i ≠ j. Thus 2

1 n 1 n 1 1 n  2 E (Y ) ==+ E  ∑ Yi  E Y E (YiY j ) = µY2 + σ Y2 ( ) i 2 ∑ 2 ∑∑ n =i 1 j ≠ i n  n=i 1  n =i 1 2

(b) Yes. If Y gets arbitrarily close to µY with probability approaching 1 as n gets large, then Y 2 gets arbitrarily close to µY2 with probability approaching 1 as n gets large. (As it turns out, this is an example of the “continuous mapping theorem” discussed in Chapter 17.)

3.20.

Using analysis like that in equation (3.29)

= s XY =

1 n ∑ ( X i − X )(Yi − Y ) n − 1 i =1 n 1 n   n  ( X i − µ X )(Yi − µY )  −  ∑  ( X − µ X )(Y − µY )  n − 1  n i =1   n −1 ©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

p

23

p

because X → µ X and Y → µY the final term converges in probability to zero.

( X i − µ x )(Yi − µY ). Note Wi is iid with mean σXY and second moment Let Wi =

E[( X i − µ X ) 2 (Yi − µY ) 2 ]. But E[( X i − µ X ) 2 (Yi − µY ) 2 ] ≤ E ( X i − µ X ) 4 E (Yi − µY ) 4 from the

Cauchy-Schwartz inequality. Because X and Y have finite fourth moments, the second moment of

σ XY . Thus, s XY →σ XY (because Wi is finite, so that it has finite variance. Thus 1n ∑ in=1 Wi → E (Wi ) = p

the term n−n 1 → 1 ).

3.21.

Set nm = nw = n, and use equation (3.19) write the squared SE of Ym − Yw as

1 1 ∑ (Y − Y ) ∑ (Ywi − Yw ) 2 (n − 1) (n − 1) [ SE (Y − Y )] = + n n n n 2 =i 1 = mi m i 1 2 m w

n n 2 =i 1 = mi m i 1

=

∑ (Y − Y ) + ∑ (Ywi − Yw ) 2 . n(n − 1)

Similarly, using equation (3.23)

 1  1 ∑ (Y − Y ) + ∑ (Ywi − Yw ) 2   2(n − 1) (n − 1)  [ SE pooled (Y − Y )] =  2n n n 2 =i 1 = mi m i 1 2 m w

n n 2 =i 1 = mi m i 1

=

∑ (Y − Y ) + ∑ (Ywi − Yw ) 2 . n(n − 1)

©2011 Pearson Education, Inc. Publishing as Addison Wesley

p

Chapter 4

Linear Regression with One Regressor 4.1.

(a) The predicted average test score is

 TestScore = 520.4 − 5.82 × 22= 392.36 (b) The predicted change in the classroom average test score is  ∆ TestScore = (−5.82 × 19) − (−5.82 × 23) = 23.28

(c) Using the formula for β̂ 0 in Equation (4.8), we know the sample average of the test scores across the 100 classrooms is

TestScore= βˆ 0 + βˆ 1 × CS= 520.4 − 5.82 × 21.4= 395.85. (d) Use the formula for the standard error of the regression (SER) in Equation (4.19) to get the sum of squared residuals:

SSR = (n − 2) SER 2 = (100 − 2) × 11.52 = 12961.

Use the formula for R 2 in Equation (4.16) to get the total sum of squares:

= TSS

SSR 12961 = = 13044.

1 − R 2 1 − 0.082

= The sample variance is sY2 = TSS n−1 4.2.

= 131.8. Thus, standard deviation is sY=

13044 99

sY2= 11.5.

The sample size n = 200. The estimated regression equation is  Weight = (2.15) − 99.41 + (0.31) 3.94 Height ,

R 2 = 0.81, SER = 10.2.

(a) Substituting Height = 70, 65, and 74 inches into the equation, the predicted weights are 176.39, 156.69, and 192.15 pounds.  (b) ∆ Weight =3.94 × ∆Height =3.94 × 1.5 =5.91. (c) We have the following relations: 1 in = 2.54 cm and 1 lb = 0.4536 kg . Suppose the regression equation in centimeter-kilogram units is

 = γˆ + γˆ Height .

Weight 0 1

=0.7036 kg per cm. −99.41 × 0.4536 = −45.092 kg ; γˆ1 =3.94 × 0.24536 The coefficients are γˆ 0 = .54 The R 2 is unit free, so it remains at R 2 = 0.81 . The standard error of the regression is SER = 10.2 × 0.4536 = 4.6267 kg .

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

4.3.

25

(a) The coefficient 9.6 shows the marginal effect of Age on AWE; that is, AWE is expected to increase by $9.6 for each additional year of age. 696.7 is the intercept of the regression line. It determines the overall level of the line. (b) SER is in the same units as the dependent variable (Y, or AWE in this example). Thus SER is measured in dollars per week. (c) R2 is unit free. (d) (i) 696.7 + 9.6 × 25 = $936.7; (ii) 696.7 + 9.6 × 45 = $1,128.7 (e) No. The oldest worker in the sample is 65 years old. 99 years is far outside the range of the sample data. (f) No. The distribution of earning is positively skewed and has kurtosis larger than the normal. (g) βˆ = Y − βˆ X , so that Y= βˆ + βˆ X . Thus the sample mean of AWE is 696.7 + 9.6 × 41.6 = 0

$1,096.06.

4.4.

1

0

1

(a) ( R − R f )= β ( Rm − R f ) + u , so that var ( R − R f ) = β 2 × var( Rm − R f ) + var(u ) + 2 β × cov(u , Rm − R f ). But cov(u , Rm − R f ) = 0, thus var( R − R f ) =β 2 × var( Rm − R f ) + var(u ). With

β > 1, var(R − Rf) > var(Rm − Rf), follows because var(u) ≥ 0. (b) Yes. Using the expression in (a) var ( R − R f ) − var ( Rm − R f ) = ( β 2 − 1) × var ( Rm − R f ) + var(u ), which will be positive if var(u ) > (1 − β 2 ) × var ( Rm − R f ). (c) Rm − R f = 7.3% − 3.5%= 3.8%. Thus, the predicted returns are

Rˆ = R f + βˆ ( Rm − R f ) = 3.5% + βˆ × 3.8% Wal-Mart: 3.5% + 0.3×3.8% = 4.6% Kellogg: 3.5% + 0.5 × 3.8% = 5.4% Waste Management: 3.5% + 0.6 × 3.8% = 5.8% Verizon: 3.5% + 0.6 × 3.8% = 5.8% Microsoft: 3.5% + 1.0 × 3.8% = 7.3% Best Buy: 3.5% + 1.3 × 3.8% = 8.4% Bank of America: 3.5% + 2.4 × 3.8% = 11.9% 4.5.

(a) ui represents factors other than time that influence the student’s performance on the exam including amount of time studying, aptitude for the material, and so forth. Some students will have studied more than average, other less; some students will have higher than average aptitude for the subject, others lower, and so forth. (b) Because of random assignment ui is independent of Xi. Since ui represents deviations from average E(ui) = 0. Because u and X are independent E(ui|Xi) = E(ui) = 0. (c) (2) is satisfied if this year’s class is typical of other classes, that is, students in this year’s class can be viewed as random draws from the population of students that enroll in the class.

(3) is satisfied because 0 ≤ Yi ≤ 100 and Xi can take on only two values (90 and 120). = 70.6; 49 + 0.24 × 120 = 77.8; 49 + 0.24 × 150 = 85.0 (d) (i) 49 + 0.24 × 90 (ii) 0.24 × 10 = 2.4.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

26

4.6.

Stock/Watson • Introduction to Econometrics, Third Edition

Using E (ui |X i ) = 0, we have

E (Yi |X i ) =E ( β 0 + β1 X i + ui |X i ) =β 0 + β1 E ( X i |X i ) + E (ui |X i ) =β 0 + β1 X i .

4.7.

The expectation of β̂ 0 is obtained by taking expectations of both sides of Equation (4.8):    1 n  E ( βˆ0 ) = E (Y − βˆ1 X ) = E   β 0 + β1 X + ∑ ui  − βˆ1 X  n i =1     n 1 = β 0 + E ( β1 − βˆ1 ) X + ∑ E (ui ) n i =1 = β0

where the third equality in the above equation has used the facts that E(ui) = 0 and E[( β̂1 −β1) X ] = E[(E( β̂ −β1)| X ) X ] = because E[( β − βˆ ) | X ] = 0 (see text equation (4.31).)

1

4.8.

1

1

The only change is that the mean of β̂ 0 is now β0 + 2. An easy way to see this is this is to write the regression model as

Yi = ( β 0 + 2) + β1 X i + (ui − 2).

The new regression error is (ui − 2) and the new intercept is (β0 + 2). All of the assumptions of Key Concept 4.3 hold for this regression model.

4.9.

ˆ βˆ= Y . Thus ESS = 0 and R2 = 0. (a) With= βˆ1 0,= βˆ0 Y , and Y= i 0 ˆ βˆ + βˆ X , so that Yˆ = Y for all i, (b) If R2 = 0, then ESS = 0, so that Yˆi = Y for all i. But Y= i 0 1 i i ˆ which implies that β = 0, or that Xi is constant for all i. If Xi is constant for all i, then 1

0 and β̂1 is undefined (see equation (4.7)). ∑ (X − X ) = n

i −1

4.10.

i

2

(a) E(ui|X = 0) = 0 and E(ui|X = 1) = 0. (Xi, ui) are i.i.d. so that (Xi, Yi) are i.i.d. (because Yi is a function of Xi and ui). Xi is bounded and so has finite fourth moment; the fourth moment is non-zero because Pr(Xi = 0) and Pr(Xi = 1) are both non-zero so that Xi has finite, non-zero kurtosis. Following calculations like those exercise 2.13, ui also has nonzero finite fourth moment. (b) var( X i ) = 0.2 × (1 − 0.2) = 0.16 and µ X = 0.2. Also

var[( X i − µ X )ui ] =E[( X i − µ X )ui ]2 = E[( X i − µ X )ui |X i =× 0]2 Pr( X i = 0) + E[( X i − µ X )ui |X i =× 1]2 Pr( X i = 1) where the first equality follows because E[(Xi − µX)ui] = 0, and the second equality follows from the law of iterated expectations.

E[( X i − µ X )ui |X i = 0]2 = 0.22 × 1, and E[( X i − µ X )ui |X i = 1]2 = (1 − 0.2) 2 × 4.

Putting these results together = σ β2ˆ 1

1 (0.22 × 1 × 0.8) + ((1 − 0.2) 2 × 4 × 0.2) 1 = 21.25 n 0.162 n

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

4.11.

27

(a) The least squares objective function is ∑ i =1 (Yi − b1 X i ) 2. Differentiating with respect to b1 n

n

yields

∂ ∑ (Yi − b1 X i ) 2 i =1

∂b1

= −2∑ i =1 X i (Yi − b1 X i ). Setting this zero, and solving for the least n

n

∑X Y

i i

squares estimator yields βˆ = i =1 1

n

∑ X i2

.

i =1

n

∑ X (Y − 4)

(b) Following the same steps in (a) yields βˆ = i =1 1

i

i

n

∑ X i2

.

i =1

4.12.

(a) Write ESS =

n

∑ (Yˆi − Y )2 =

n

∑ (βˆ0 + βˆ1 X i − Y )2 =

=i 1 =i 1

n

∑[βˆ ( X − X )]

=i 1

1

i

2

2

 ∑ ( X i − X )(Yi − Y )  ) . = βˆ ∑ ( X i − X= ∑in=1 ( X i − X ) 2 i =1 2 1

n

2

n i =1

This implies 2

 ∑in=1 ( X i − X )(Yi − Y )  ESS 2 = = R 2 ∑in 1 (Y= ∑in 1 = ( X i − X ) 2 ∑in 1 (Yi − Y ) 2 = i −Y ) n   1 n −1 ∑ i =1 ( X i − X )(Yi − Y )  =  n1−1 ∑in 1 = ( X i − X ) 2 n1−1 ∑in 1 (Yi − Y ) 2  =

2

2

 s XY  2 = =  rXY s s  X Y (b) This follows from part (a) because rXY = rYX. n n 1 ( X i − X )(Yi − Y ) ∑ ( X i − X )(Yi − Y ) ∑ s XY s s (n − 1) i =1 i =1 (c) Because rXY = = , rXY Y = XY = = βˆ1 n n 2 s X sY 1 s X sX ∑ ( X i − X )2 ∑ ( X i − X )2 (n − 1) i 1 =i 1 =

4.13.

The answer follows the derivations in Appendix 4.3 in “Large-Sample Normal Distribution of the OLS Estimator.” In particular, the expression for νi is now νi = (Xi − µX)κui, so that var(νi) = κ3var[(Xi − µX)ui], and the term κ2 carry through the rest of the calculations.

4.14.

Because βˆ0= Y − βˆ1 X , Y= βˆ0 + β1 X . The sample regression line is = y βˆ0 + β1 x , so that the sample regression line passes through ( X , Y ).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 5

Regression with a Single Regressor: Hypothesis Tests and Confidence Intervals

5.1

(a) The 95% confidence interval for β1 is {−5.82 ± 1.96 × 2.21}, that is −10.152 ≤ β1 ≤ −1.4884. (b) Calculate the t-statistic:

βˆ − 0 −5.82 t act = 1 = =−2.6335.

SE( βˆ 1) 2.21 0 vs. H1 : β1 ≠ 0 is The p-value for the test H 0 : β1 =

p-value = 2Φ (−|t act |) = 2Φ (−2.6335) = 2 × 0.0042 = 0.0084.

The p-value is less than 0.01, so we can reject the null hypothesis at the 5% significance level, and also at the 1% significance level. (c) The t-statistic is βˆ 1 − (−5.6) 0.22 t act = = = 0.10 SE ( βˆ 1) 2.21 The p-value for the test H 0 : β1 = −5.6 vs. H1 : β1 ≠ −5.6 is

p-value = 2Φ (−|t act |) = 2Φ (−0.10) = 0.92 The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significance level. Because β1 = −5.6 is not rejected at the 5% level, this value is contained in the 95% confidence interval. (d) The 99% confidence interval for β 0 is {520.4 ± 2.58 × 20.4}, that is, 467.7 ≤ β 0 ≤ 573.0.

5.2.

(a) The estimated gender gap equals $2.12/hour.

0 vs. H1 : β1 ≠ 0. The t-statistic is (b) The null and alternative hypotheses are H 0 : β1 = = t act

βˆ 1 − 0 2.12 = = 5.89, SE ( βˆ 1) 0.36

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

29

and the p-value for the test is

p-value = 2Φ (−|t act |) = 2Φ (−5.89) = 2 × 0.0000 = 0.000 (to four decimal places) The p-value is less than 0.01, so we can reject the null hypothesis that there is no gender gap at a 1% significance level. (c) The 95% confidence interval for the gender gap β1 is {2.12 ± 1.96 × 0.36}, that is, 1.41 ≤ β1 ≤ 2.83. ˆ $12.52/hour. The sample average wage of men is (d) The sample average wage of women is β= 0 = $14.64/hour. βˆ + = βˆ $12.52 + $2.12 0

1

(e) The binary variable regression model relating wages to gender can be written as either

Wage = β 0 + β1Male + ui , or

Wage = γ 0 + γ 1 Female + vi .

In the first regression equation, Male equals 1 for men and 0 for women; β 0 is the population mean of wages for women and β 0 + β1 is the population mean of wages for men. In the second regression equation, Female equals 1 for women and 0 for men; γ 0 is the population mean of wages for men and γ 0 + γ 1 is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations:

γ 0 = β 0 + β1 , γ 0 + γ 1 = β0 .

Given the coefficient estimates β̂ 0 and β̂ 1 , we have

γˆ 0 = βˆ 0 + βˆ 1 = 14.64,

γˆ1 =βˆ 0 − γˆ 0 =− βˆ 1 =−2.12.

Due to the relationship among coefficient estimates, for each individual observation, the OLS residual is the same under the two regression equations: uˆ i = vˆ i. Thus the sum of squared residuals, SSR = ∑ i =1 uˆ i2, is the same under the two regressions. This implies that both n

1/ 2

SSR  SSR  2 are unchanged.

SER =   and R = 1 − TSS  n −1 In summary, in regressing Wages on Female, we will get  = 14.64 − 2.12 Female, Wages

R 2 =0.06, SER =4.2.

5.3.

The 99% confidence interval is 1.5 × {3.94 ± 2.58 × 0.31) or 4.71 lbs ≤ WeightGain ≤ 7.11 lbs.

5.4.

(a) − 5.38 + 1.76 × 16 − $22.78 per hour (b) The wage is expected to increase by 1.76 × 2 = $3.52 per hour.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

30

Stock/Watson • Introduction to Econometrics, Third Edition

(c) The increase in wages for college education is β1 × 4. Thus, the counselor’s assertion is that 1.76 − 2.50 = − 9.25, which has = β1 10/4 = 2.50. The t-statistic for this null hypothesis is t act 0.08 a p-value of 0.00. Thus, the counselor’s assertion can be rejected at the 1% significance level.

A 95% confidence for β1 × 4 is 4 × (1.76 ± 1.96 × 0.08) or $6.41 ≤ Gain ≤ $7.67.

5.5

(a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5 of the standard deviation in test scores, a moderate increase.

13.9 t act = 5.56, which has a p-value of 0.00. Thus the null hypothesis is (b) The t-statistic is = 2.5 rejected at the 5% (and 1%) level. (c) 13.9 ± 2.58 × 2.5 = 13.9 ± 6.45.

5.6.

5.7.

5.8.

(a) The question asks whether the variability in test scores in large classes is the same as the variability in small classes. It is hard to say. On the one hand, teachers in small classes might able so spend more time bringing all of the students along, reducing the poor performance of particularly unprepared students. On the other hand, most of the variability in test scores might be beyond the control of the teacher. (b) The formula in 5.3 is valid for heteroskesdasticity or homoskedasticity; thus inferences are valid in either case.

3.2 = 2.13 with a p-value of 0.03; since the p-value is less than 0.05, the null 1.5 hypothesis is rejected at the 5% level. (b) 3.2 ± 1.96 × 1.5 = 3.2 ± 2.94 (c) Yes. If Y and X are independent, then β1 = 0; but this null hypothesis was rejected at the 5% level in part (a). (d) β1 would be rejected at the 5% level in 5% of the samples; 95% of the confidence intervals would contain the value β1 = 0. (a) The t-statistic is

(a) 43.2 ± 2.05 × 10.2 or 43.2 ± 20.91, where 2.05 is the 5% two-sided critical value from the t28 distribution. − 55 (b) The t-statistic is= = t act 61.5 0.88, which is less (in absolute value) than the critical value of 7.4 20.5. Thus, the null hypothesis is not rejected at the 5% level. (c) The one sided 5% critical value is 1.70; tact is less than this critical value, so that the null hypothesis is not rejected at the 5% level.

5.9.

1

(a) β = n

(Y1 + Y2 +  + Yn ) so that it is linear function of Y1, Y2, …, Yn. X

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

31

(b) E(Yi|X1, …, Xn) = βXi, thus  1    n (Y1 + Y2 +  + Yn )   E ( β |X 1 ,  , X n ) = E    | ( X 1 , , X n )  X       1 β ( X1 +  + X n ) = n= β .

X

5.10.

Let n0 denote the number of observation with X = 0 and n1 denote the number of observations with n 1 n n n n 2 X = 1; note that ∑ i =1 X i = n1 ; X = 1 ; ( X i − X )= ∑ i 1 X i2 − nX=2 ∑ 1 X iYi = Y1 ; ∑ i 1 = n n1 i == n2 n n n1  n  nn n1 − 1 = n1 1 − 1  = 1 0 ; n1Y1 + n0Y0 = Y , so that= Y Y1 + 0 Y0 ∑ i =1 i n n n n n  From the least squares formula n

n

n

X )(Yi − Y ) ∑i 1 = X i (Yi − Y ) ∑i 1 X iYi − Yn1 ∑i 1 ( X i − = = βˆ1 = = = 2 2 n n ( ) ( X X n1n0 / n ∑ − ∑ i i 1 Xi − X ) =i 1 = =

n n n n  (Y1 − Y ) =  Y − 1 Y1 − 0 Y0  =Y1 − Y0 n0 n0  n n 

n  n n + n0 n and βˆ0 =Y − βˆ1 X = 0 Y0 + 1 Y1  − (Y1 − Y0 ) 1 = 1 Y0 =Y0 n  n n  n

5.11.

s Using the results from 5.10, βˆ0 = Ym and βˆ= Yw − Ym . From Chapter 3, SE (Ym ) = m and 1 nm SE (Yw − Ym ) =

sm2 sw2 + . Plugging in the numbers βˆ0 = 523.1 and SE ( βˆ0 ) = 6.22; βˆ1 = −38.0 nm nw

and SE ( βˆ1 ) = 7.65.

5.12.

Equation (4.22) gives var (H u ) µX σ β2ˆ =i i 2 , where H i = 1− Xi. 2 E ( Xi )

n  E ( H i2 ) 

0

Using the facts that E (ui | X i ) = 0 and var (ui | X i ) = σ u2 (homoskedasticity), we have       

µx

   i i   

µ

x E ( H i ui ) = E ui − Xu = E (ui ) − E [ X i E (ui | X i )] 2 E ( Xi ) E ( X i2 )

=0 −

µx

E ( X i2 )

× 0 =0,

and

©2011 Pearson Education, Inc. Publishing as Addison Wesley

32

Stock/Watson • Introduction to Econometrics, Third Edition         

2

E [( H i u= E ui − i) ]

µX E  X i2 

   2  i   

 2     i i      

Xu

2



  µ   = E u − 2  2  X u +   X 2   X i2ui2   E  X i   E  X i    

µX

2 i i

2 i  

 

E  X i2 

 

i

i

i   

2

  2  2    2   E  X i E  ui | X i     E  X i   

 µ µ   X = E u − 2 X E X E  u 2| X  +   

2

 µ   2  µ X2  2 X 2 − µ Xσ +  E 1  X i σ u =   σ .

2  E  X i2   u E  X i2   E  X i        = ( H i ui ) E[( H i ui ) 2 ], so Because E ( H i ui ) = 0, var = σ − 2 µX 2 u

2 u

 µ X2  2 1 − σ . var (H i ui ) = E[( H i ui ) 2 ] =  E ( X i2 )  u  

Also          

µX

E(H ) = E 1− X E ( X i2 ) 2 i

2     i       

      

2



  µ  2  X    E 1− 2 X X = + i i  2 E ( X i2 )  E ( X i )    2

µX

 µ  µ X2 .

2 X E X 1 + = −   i E ( X i2 )  E ( X i2 )  E ( X i2 )

µ X2 1− 2 =

( )

Thus

 µ X2  2 1 − σ  E ( X i2 )  u H u var ( ) σ u2   i i = = = σ β2ˆ 2 0  nE ( H 2 )2     2 µ X2  µ i X  n 1 −   n1 −   E ( X i2 )   E ( X i2 )      E ( X i2 )σ u2 E ( X i2 )σ u2 = = . nσ X2 n[ E ( X i2 − µ X2 )]

5.13.

(a) Yes, this follows from the assumptions in KC 4.3. (b) Yes, this follows from the assumptions in KC 4.3 and conditional homoskedasticity (c) They would be unchanged for the reasons specified in the answers to those questions. (d) (a) is unchanged; (b) is no longer true as the errors are not conditionally homosckesdastic.

5.14.

(a) From Exercise (4.11), βˆ = ∑ aiYi where ai =

Xi

∑ j=1 X 2j n

. Since the weights depend only on X i

but not on Yi , β̂ is a linear function of Y.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

33

∑ n X E (u | X ,  , X n ) (b) E ( βˆ | X 1 ,  , X n ) = β + i =1 i n i 12 β since E (ui | X 1 , , X n ) = 0 = ∑i =1 X j

∑in=1 X i 2 var (ui | X 1 ,  , X n ) σ2 = (c) var ( βˆ | X 1 ,  , X n ) = 2 ∑in=1 X 2j  ∑in=1 X 2j  (d) This follows the proof in the appendix.

5.15.

Because the samples are independent, βˆm ,1 and βˆw,1 are independent. Thus var ( βˆm ,1 − βˆw,1 ) = var ( βˆ ) + var( βˆ ). Var ( βˆ ) is consistently estimated as [SE( βˆ )]2 and Var (βˆ ) is m ,1

w ,1

m ,1

m ,1

w ,1

consistently estimated as [SE( βˆw,1 )]2 , so that var( βˆm ,1 − βˆw,1 ) is consistently estimated by [SE( βˆ )]2 + [SE( βˆ )]2 , and the result follows by noting the SE is the square root of the m ,1

w ,1

estimated variance.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 6

Linear Regression with Multiple Regressors

6.1.

By equation (6.15) in the text, we know

n −1 R2 = 1− (1 − R 2 ). n − k −1 Thus, that values of R 2 are 0.175, 0.189, and 0.193 for columns (1)–(3).

6.3.

(a) On average, a worker earns $0.29/hour more for each year he ages. (b) Sally’s earnings prediction is 4.40 + 5.48 × 1 − 2.62 × 1 + 0.29 × 29 = 15.67 dollars per hour.

Betsy’s earnings prediction is 4.40 + 5.48 × 1 − 2.62 × 1 + 0.29 × 34 = 17.12 dollars per hour.

The difference is 1.45

6.4.

(a) Workers in the Northeast earn $0.69 more per hour than workers in the West, on average, controlling for other variables in the regression. Workers in the Northeast earn $0.60 more per hour than workers in the West, on average, controlling for other variables in the regression.

Workers in the South earn $0.27 less than workers in the West. (b) The regressor West is omitted to avoid perfect multicollinearity. If West is included, then the intercept can be written as a perfect linear function of the four regional regressors. (c) The expected difference in earnings between Juanita and Jennifer is −0.27 − 0.6 =− 0.87.

6.5.

(a) $23,400 (recall that Price is measured in $1000s). (b) In this case ∆BDR = 1 and ∆Hsize = 100. The resulting expected change in price is 23.4 + 0.156 × 100 = 39.0 thousand dollars or $39,000. (c) The loss is $48,800. (d) From the text R 2 = 1 − n −n −k1−1 (1 − R 2 ), so R 2 = 1 − n −n −k1−1 (1 − R 2 ), thus, R2 = 0.727.

6.6.

(a) There are other important determinants of a country’s crime rate, including demographic characteristics of the population. (b) Suppose that the crime rate is positively affected by the fraction of young males in the population, and that counties with high crime rates tend to hire more police. In this case, the size of the police force is likely to be positively correlated with the fraction of young males in the population leading to a positive value for the omitted variable bias so that βˆ1 > β1 .

6.7.

(a) The proposed research in assessing the presence of gender bias in setting wages is too limited.

There might be some potentially important determinants of salaries: type of engineer, amount of work experience of the employee, and education level. The gender with the lower wages could reflect the type of engineer among the gender, the amount of work experience of the employee, or the education level of the employee. The research plan could be improved with the collection of additional data as indicated and an appropriate statistical technique for analyzing the data would be a multiple regression in which the dependent variable is wages and the independent variables would include a dummy variable for gender, dummy variables ©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

35

for type of engineer, work experience (time units), and education level (highest grade level completed). The potential importance of the suggested omitted variables makes a “difference in means” test inappropriate for assessing the presence of gender bias in setting wages. (b) The description suggests that the research goes a long way towards controlling for potential omitted variable bias. Yet, there still may be problems. Omitted from the analysis are characteristics associated with behavior that led to incarceration (excessive drug or alcohol use, gang activity, and so forth), that might be correlated with future earnings. Ideally, data on these variables should be included in the analysis as additional control variables.

6.8.

Omitted from the analysis are reasons why the survey respondents slept more or less than average.

People with certain chronic illnesses might sleep more than 8 hours per night. People with other illnesses might sleep less than 5 hours. This study says nothing about the causal effect of sleep on mortality.

6.9.

For omitted variable bias to occur, two conditions must be true: X1 (the included regressor) is correlated with the omitted variable, and the omitted variable is a determinant of the dependent variable. Since X1 and X2 are uncorrelated, the estimator of β1 does not suffer from omitted variable bias.

6.10.

(a) σ β2ˆ = 1

 σ u2 1 1   n 1 − ρ X2 1 , X 2  σ X2 1

Assume X1 and X2 are uncorrelated: ρ X2 1 X 2 = 0

1  1 4 400 1 − 0  6 1 4 1 = ⋅ = = 0.00167 400 6 600

σ β2ˆ = 1

(b) With ρ X1 , X 2 = 0.5

σ β2ˆ = 1

=

1  1 4 400 1 − 0.52  6 1  1 4 .0022 = 400  0.75  6

(c) The statement correctly says that the larger is the correlation between X1 and X2 the larger is the variance of βˆ1 , however the recommendation “it is best to leave X2 out of the regression” is incorrect. If X2 is a determinant of Y, then leaving X2 out of the regression will lead to omitted variable bias in βˆ1.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

36

6.11.

Stock/Watson • Introduction to Econometrics, Third Edition

(a)

∑ (Y − b X − b X )

(b)

∂ ∑ (Yi − b1 X 1i − b2 X 2i ) 2 = −2∑ X 1i (Yi − b1 X 1i − b2 X 2i ) ∂b1

i

1

1i

2

2i

2

∂ ∑ (Yi − b1 X 1i − b2 X 2i ) 2 = −2∑ X 2i (Yi − b1 X 1i − b2 X 2i ) ∂b2

(c) From (b), β̂1 satisfies

0 ∑ X (Y − βˆ X − βˆ X ) = 1i

βˆ1 =

or

1

i

1i

1

2i

∑ X 1iYi − βˆ2 ∑ X 1i X 2i ∑ X 12i

and the result follows immediately. (d) Following analysis as in (c)

βˆ2 =

∑ X 2iYi − βˆ1 ∑ X 1i X 2i ∑ X 22i

and substituting this into the expression for β̂1 in (c) yields 1i 2 i ∑ X 1iY ∑ 2 i i X1 ∑ ∑ X 1i X 2i 2 2i ∑ βˆ1 = . ∑ X 12i

X Y − βˆ

X X

Solving for β̂1 yields:

βˆ1 =

∑ X 22i ∑ X 1iYi − ∑ X 1i X 2i ∑ X 2iYi ∑ X 12i ∑ X 22i − (∑ X 1i X 2i ) 2

(e) The least squares objective function is ∑ (Yi − b0 − b1 X 1i − b2 X 2i ) 2 and the partial derivative with respect to b0 is

∂ ∑(Yi − b0 − b1 X 1i − b2 X 2i ) 2 = −2∑ (Yi − b0 − b1 X 1i − b2 X 2i ). ∂b0

Setting this to zero and solving for β̂ 0 yields: βˆ0 =− Y βˆ1 X 1 − βˆ2 X 2 . (f) Substituting βˆ0 =− Y βˆ1 X 1 − βˆ2 X 2 into the least squares objective function yields

) ∑ ( (Y − Y ) − b ( X − X ) − b ( X − X ) ) , which is identical to ∑ (Y − βˆ − b X − b X = i

0

1

1i

2

2i

2

2

i

1

1i

1

2

2i

2

the least squares objective function in part (a), except that all variables have been replaced with deviations from sample means. The result then follows as in (c).

Notice that the estimator for β1 is identical to the OLS estimator from the regression of Y onto X1, omitting X2. Said differently, when ∑ ( X 1i − X 1 )( X 2i − X 2 ) = 0 , the estimated coefficient on X1 in the OLS regression of Y onto both X1 and X2 is the same as estimated coefficient in the OLS regression of Y onto X1.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 7

Hypothesis Tests and Confidence Intervals in Multiple Regression

7.1 and 7.2 Regressor

(1)

(2)

(3)

College (X1)

5.46** (0.21)

5.48** (0.21)

5.44** (0.21)

Female (X2)

− 2.64** (0.20)

− 2.62** (0.20) 0.29** (0.04)

− 2.62** (0.20)

Age (X3)

0.69* (0.30) 0.60* (0.28)

Ntheast (X4) Midwest (X5) South (X6) Intercept

0.29** (0.04)

− 0.27 (0.26) 12.69** (0.14)

4.40** (1.05)

3.75** (1.06)

(a) The t-statistic is 5.46/0.21 = 26.0, which exceeds 1.96 in absolute value. Thus, the coefficient is statistically significant at the 5% level. The 95% confidence interval is 5.46 ± 1.96 × 0.21. (b) t-statistic is − 2.64/0.20 = −13.2, and 13.2 > 1.96, so the coefficient is statistically significant at the 5% level. The 95% confidence interval is −2.64 ± 1.96 × 0.20.

7.3.

(a) Yes, age is an important determinant of earnings. Using a t-test, the t-statistic is 0.29 / 0.04 = 7.25, with a p-value of 4.2 × 10−13, implying that the coefficient on age is statistically significant at the 1% level. The 95% confidence interval is 0.29 ± 1.96 × 0.04. (b) ∆Age × [0.29 ± 1.96 × 0.04] = 5 × [0.29 ± 1.96 × 0.04] = 1.45 ± 1.96 × 0.20 = $1.06 to $1.84

7.4.

(a) The F-statistic testing the coefficients on the regional regressors are zero is 6.10. The 1% critical value (from the F3, ∞ distribution) is 3.78. Because 6.10 > 3.78, the regional effects are significant at the 1% level. (b) The expected difference between Juanita and Molly is (X6,Juanita − X6,Molly) × β6 = β6. Thus a 95% confidence interval is −0.27 ± 1.96 × 0.26.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

38

Stock/Watson - Introduction to Econometrics - Second Edition

(c) The expected difference between Juanita and Jennifer is (X5,Juanita − X5,Jennifer) × β5 + (X6,Juanita − X6,Jennifer) × β6 = −β5 + β6. A 95% confidence interval could be constructed using the general methods discussed in Section 7.3. In this case, an easy way to do this is to omit Midwest from the regression and replace it with X5 = West. In this new regression the coefficient on South measures the difference in wages between the South and the Midwest, and a 95% confidence interval can be computed directly.

7.5.

The t-statistic for the difference in the college coefficients is t =

βˆcollege,1998 − βˆcollege,1992 . Because SE( βˆ − βˆ ) college,1998

college, 1992

β̂ college,1998 and β̂ college,1992 are computed from independent samples, they are independent, which − βˆ ) = , βˆ ) = 0 Thus, var( βˆ means that cov( βˆ college,1998

var( βˆ

college,1998

) + var( βˆ

college,1992

college,1998

college,1998

) . This implies that SE( βˆ

college,1998

college,1992

− βˆcollege,1992 ) = (0.212 + 0.202 ) 2 .

1

5.48 − 5.29 = 0.6552. There is no significant change since the calculated t-statistic 0.212 + 0.202 is less than 1.96, the 5% critical value.

= Thus, t act

7.6.

In isolation, these results do imply gender discrimination. Gender discrimination means that two workers, identical in every way but gender, are paid different wages. Thus, it is also important to control for characteristics of the workers that may affect their productivity (education, years of experience, etc.) If these characteristics are systematically different between men and women, then they may be responsible for the difference in mean wages. (If this were true, it would raise an interesting and important question of why women tend to have less education or less experience than men, but that is a question about something other than gender discrimination in the U.S. labor market.) These are potentially important omitted variables in the regression that will lead to bias in the OLS coefficient estimator for Female. Since these characteristics were not controlled for in the statistical analysis, it is premature to reach a conclusion about gender discrimination.

7.7.

2.61 0.186 < 1.96. Therefore, the coefficient on BDR is not (a) The t-statistic is 0.485 /= statistically significantly different from zero. (b) The coefficient on BDR measures the partial effect of the number of bedrooms holding house size (Hsize) constant. Yet, the typical 5-bedroom house is much larger than the typical 2-bedroom house. Thus, the results in (a) says little about the conventional wisdom. (c) The 99% confidence interval for effect of lot size on price is 2000 × [0.002 ± 2.58 × 0.00048] or 1.52 to 6.48 (in thousands of dollars). (d) Choosing the scale of the variables should be done to make the regression results easy to read and to interpret. If the lot size were measured in thousands of square feet, the estimate coefficient would be 2 instead of 0.002. (e) The 10% critical value from the F2,∞ distribution is 2.30. Because 0.08 < 2.30, the coefficients are not jointly significant at the 10% level.

7.8.

(a) Using the expressions for R2 and R 2, algebra shows that

n −1 n − k −1 R2 = 1− (1 − R 2 ), so R 2 = 1− (1 − R 2 ). n − k −1 n −1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

420 − 1 − 1 Column 1: R 2 = 1− (1 − 0.049) = 0.051 420 − 1 420 − 2 − 1 Column 2: R 2 = 1− (1 − 0.424) = 0.427 420 − 1 420 − 3 − 1 Column 3: R 2 = 1− (1 − 0.773) = 0.775 420 − 1 420 − 3 − 1 Column 4: R 2 = 1− (1 − 0.626) = 0.629 420 − 1 420 − 4 − 1 Column 5: R 2 = 1− (1 − 0.773) = 0.775 420 − 1 (b) H 0 : β= β= 0 3 4 H1 : β 3 ≠, β 4 ≠ 0 2 β 0 + β1 X 1 + β 2 X 2 + β3 X 3 + β 4 X 4 , Runrestricted = 0.775 Unrestricted regression (Column 5): Y = 2 = β 0 + β1 X 1 + β 2 X 2 , Rrestricted 0.427 Restricted regression (Column 2): Y =

FHomoskedasticityOnly = =

2 2 ( Runrestricted )/q − Rrestricted , n 420, kunrestricted q 2 = = 4,= 2 (1 − Runrestricted )/(n − kunrestricted − 1)

(0.775 − 0.427)/2 0.348/2 0.174 = = = 322.22 (1 − 0.775)/(420 − 4 − 1) (0.225)/415 0.00054

5% Critical value form F2,00 = 4.61; FHomoskedasticityOnly = F2,00 so Ho is rejected at the 5% level. (c) t3 = −13.921 and t4 = 0.814, q = 2; |t3| > c (Where c = 2.807, the 1% Benferroni critical value from Table 7.3). Thus the null hypothesis is rejected at the 1% level. (d) − 1.01 ± 2.58 × 0.27 7.9.

(a) Estimate Yi = β 0 + γ X 1i + β 2 ( X 1i + X 2i ) + ui

and test whether γ = 0. (b) Estimate β 0 + γ X 1i + β 2 ( X 2i − aX 1i ) + ui Yi =

and test whether γ = 0. (c) Estimate Yi − X 1i =β 0 + γ X 1i + β 2 ( X 2i − X 1i ) + ui

and test whether γ = 0.

7.10.

SSRrestricted − SSRunrestricted SSR 2 2 1− , Runrestricted − Rrestricted = and Because R 2 = TSS TSS SSRunrestricted 2 1 − Runrestricted = . Thus TSS

©2011 Pearson Education, Inc. Publishing as Addison Wesley

39

40

Stock/Watson - Introduction to Econometrics - Second Edition 2 2 ( Runrestricted )/q − Rrestricted = 2 (1 − Runrestricted )/(n − kunrestricted − 1)

F = = 7.11.

SSRrestricted − SSRunrestricted TSS

SSRunrestricted TSS

/q

/(n − kunrestricted − 1)

( SSRrestricted − SSRunrestricted )/q SSRunrestricted / (n − kunrestricted − 1)

(a) Treatment (assignment to small classes) was not randomly assigned in the population (the continuing and newly-enrolled students) because of the difference in the proportion of treated continuing and newly-enrolled students. Thus, the treatment indicator X1 is correlated with X2.

If newly-enrolled students perform systematically differently on standardized tests than continuing students (perhaps because of adjustment to a new school), then this becomes part of the error term u in (a). This leads to correlation between X1 and u, so that E(u|Xl) ≠ 0.

Because E(u|Xl) ≠ 0, the β̂1 is biased and inconsistent. (b) Because treatment was randomly assigned conditional on enrollment status (continuing or newly-enrolled), E(u | X1, X2) will not depend on X1. This means that the assumption of conditional mean independence is satisfied, and β̂1 is unbiased and consistent. However, because X2 was not randomly assigned (newly-enrolled students may, on average, have attributes other than being newly enrolled that affect test scores), E(u | X1, X2) may depend of X2, so that β̂ 2 may be biased and inconsistent.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 8

Nonlinear Regression Functions

8.1.

198 − 196 = 1.0204%. The approximation 196 is 100 × [ln (198) − ln (196)] = 1.0152%.

(a) The percentage increase in sales is 100 ×

205 − 196 = 4.5918% and the 196 approximation is 100 × [ln (205) − ln (196)] = 4.4895%. When Sales2010 = 250, the percentage 250 − 196 = 27.551% and the approximation is 100 × [ln (250) − ln (196)] = increase is 100 × 196 500 − 196 24.335%. When Sales2010 = 500, the percentage increase is 100 × = 155.1% and the 196 approximation is 100 × [ln (500) − ln (196)] = 93.649%. (c) The approximation works well when the change is small. The quality of the approximation deteriorates as the percentage change increases.

(b) When Sales2010 = 205, the percentage increase is 100 ×

8.2.

(a) According to the regression results in column (1), the house price is expected to increase by 21% (= 100% × 0.00042 × 500 ) with an additional 500 square feet and other factors held constant. The 95% confidence interval for the percentage change is 100% × 500 × (0.00042 ± 1.96 × 0.000038) = [17.276% to 24.724%]. (b) Because the regressions in columns (1) and (2) have the same dependent variable, R 2 can be used to compare the fit of these two regressions. The log-log regression in column (2) has the higher R 2 , so it is better so use ln(Size) to explain house prices. (c) The house price is expected to increase by 7.1% (= 100% × 0.071 × 1). The 95% confidence interval for this effect is 100% × (0.071 ± 1.96 × 0.034) = [0.436% to 13.764%]. (d) The house price is expected to increase by 0.36% (100% × 0.0036 × 1 = 0.36%) with an additional bedroom while other factors are held constant. The effect is not statistically 0.0036 = | t | = 0.09730 < 1.96. Note that this coefficient significant at a 5% significance level: 0.037 measures the effect of an additional bedroom holding the size of the house constant. Thus, it measures the effect of converting existing space (from, say a family room) into a bedroom. (e) The quadratic term ln(Size)2 is not important. The coefficient estimate is not statistically 0.0078 significant at a 5% significance level: = | t | = 0.05571 < 1.96.

0.14 (f) The house price is expected to increase by 7.1% (= 100% × 0.071 × 1) when a swimming pool is added to a house without a view and other factors are held constant. The house price is expected to increase by 7.32% (= 100% × (0.071 × 1 + 0.0022 × 1)) when a swimming pool is added to a house with a view and other factors are held constant. The difference in the expected percentage change in price is 0.22%. The difference is not statistically significant at 0.0022 a 5% significance level: = |t | = 0.022 < 1.96.

0.10 ©2011 Pearson Education, Inc. Publishing as Addison Wesley

42

Stock/Watson - Introduction to Econometrics - Second Edition

8.3.

(a) The regression functions for hypothetical values of the regression coefficients that are consistent with the educator’s statement are: β1 > 0 and β 2 < 0. When TestScore is plotted against STR the regression will show three horizontal segments. The first segment will be for values of STR < 20; the next segment for 20 ≤ STR ≤ 25; the final segment for STR > 25. The first segment will be higher than the second, and the second segment will be higher than the third. (b) It happens because of perfect multicollinearity. With all three class size binary variables included in the regression, it is impossible to compute the OLS estimates because the intercept is a perfect linear function of the three class size regressors.

8.4.

(a) With 2 years of experience, the man’s expected AHE is

AHE ln( ) (0.1032 × 16) − (0.451 × 0) + (0.0134 × 0 × 16) + (0.0134 × 2) − 0.000211 × 22 ) = −(0.095 × 0) − (0.092 × 0) − (0.023 × 1) + 1.503 = 3.159 With 3 years of experience, the man’s expected AHE is

AHE = ln( ) (0.1032 × 16) − (0.451 × 0) + (0.0134 × 0 × 16) + (0.0134 × 3) − 0.000211 × 32 ) −(0.095 × 0) − (0.092 × 0) − (0.023 × 1) + 1.503 = 3.172 Difference = 3.172 − 3.159 = 0.013 (or 1.3%) (b) With 10 years of experience, the man’s expected AHE is AHE ln( = ) (0.1032 × 16) − (0.451 × 0) + (0.0134 × 0 × 16) + (0.0134 × 10) − 0.000211 × 102 )

−(0.095 × 0) − (0.092 × 0) − (0.023 × 1) + 1.503 = 3.253 With 11 years of experience, the man’s expected AHE is AHE ln( = ) (0.1032 × 16) − (0.451 × 0) + (0.0134 × 0 × 16) + (0.0134 × 11) − 0.000211 × 112 )

−(0.095 × 0) − (0.092 × 0) − (0.023 × 1) + 1.503 = 3.263 Difference 3.263 − 0.010 (or 1.0%) (c) The regression in nonlinear in experience (it includes Potential experience2). (d) Let β1 denote the coefficient on Potential Experience and β2 denote the coefficient on (Potential Experience)2. Then, following the discussion in the paragraphs just about equation (8.7) in the text, the expected change in part (a) is given by β1 + 5β2 and the expected change in (b) is given by β1 + 21β2. The difference between these, say (b) − (a), is 16β2. Because the estimated value of β2 is significant at the 5% level (the t-statistic for β̂ 2 is −0.000211/0.000023 = − 9.2), the difference between the effects in (a) and (b) (=16β2) is significant at the 5% level. (e) No. This would affect the level of ln(AHE), but not the change associated with another year of experience. (f) Include interaction terms Female × Potential experience and Female × (Potential experience)2.

8.5.

(a) (1) The demand for older journals is less elastic than for younger journals because the interaction term between the log of journal age and price per citation is positive. (2) There is a linear relationship between log price and log of quantity follows because the estimated coefficients on log price squared and log price cubed are both insignificant. (3) The demand is greater for journals with more characters follows from the positive and statistically significant coefficient estimate on the log of characters.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

43

(b) (i) The effect of ln(Price per citation) is given by [− 0.899 + 0.141 × ln(Age)] × ln(Price per citation). Using Age = 80, the elasticity is [− 0.899 + 0.141 × ln(80)] = − 0.28. (ii) As described in equation (8.8) and the footnote on page 261, the standard error can be found by dividing 0.28, the absolute value of the estimate, by the square root of the F-statistic testing βln(Price per citation) + ln(80) × βln(Age)×ln(Price per citation) = 0.  Characters  (c) ln =  ln(Characters ) − ln(a ) for any constant a. Thus, estimated parameter on a   Characters will not change and the constant (intercept) will change.

8.6.

(a) (i) There are several ways to do this. Here is one. Create an indicator variable, say DV1, that equals one if %Eligible is greater than 20% and less than 50%. Create another indicator, say DV2, that equals one if %Eligible is greater than 50%. Run the regression: TestScore = β 0 + β1 % Eligible + β 2 DV 1× % Eligible + β3 DV 2 × % Eligible + other regressors

The coefficient β1 shows the marginal effect of %Eligible on TestScores for values of %Eligible < 20%, β1 + β2 shows the marginal effect for values of %Eligible between 20% and 50% and β1 + β3 shows the marginal effect for values of %Eligible greater than 50%. (ii) The linear model implies that β2 = β3 = 0, which can be tested using an F-test. (b) (i) There are several ways to do this, perhaps the easiest is to include an interaction term STR × ln(Income) to the regression in column (7). (ii) Estimate the regression in part (b.i) and test the null hypothesis that the coefficient on the interaction term is equal to zero.

8.7.

(a) (i) ln(Earnings) for females are, on average, 0.44 lower for men than for women. (ii) The error term has a standard deviation of 2.65 (measured in log-points). (iii) Yes. However the regression does not control for many factors (size of firm, industry, profitability, experience and so forth). (iv) No. In isolation, these results do not imply gender discrimination. Gender discrimination means that two workers, identical in every way but gender, are paid different wages. Thus, it is also important to control for characteristics of the workers that may affect their productivity (education, years of experience, etc.) If these characteristics are systematically different between men and women, then they may be responsible for the difference in mean wages. (If this were true, it would raise an interesting and important question of why women tend to have less education or less experience than men, but that is a question about something other than gender discrimination in top corporate jobs.) These are potentially important omitted variables in the regression that will lead to bias in the OLS coefficient estimator for Female. Since these characteristics were not controlled for in the statistical analysis, it is premature to reach a conclusion about gender discrimination. (b) (i) If MarketValue increases by 1%, earnings increase by 0.37% (ii) Female is correlated with the two new included variables and at least one of the variables is important for explaining ln(Earnings). Thus the regression in part (a) suffered from omitted variable bias. (c) Forgetting about the effect or Return, whose effects seems small and statistically insignificant, the omitted variable bias formula (see equation (6.1)) suggests that Female is negatively correlated with ln(MarketValue).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

44

Stock/Watson - Introduction to Econometrics - Second Edition

8.8

(a) and (b)

(c)

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

(d)

(e)

©2011 Pearson Education, Inc. Publishing as Addison Wesley

45

46

Stock/Watson - Introduction to Econometrics - Second Edition

8.9.

Note that

Y =β 0 + β1 X + β 2 X 2 =β 0 + ( β1 + 21β 2 ) X + β 2 ( X 2 − 21X ).

Define a new independent variable = Z X 2 − 21X , and estimate Y = β 0 + γ X + β 2 Z + ui .

The confidence interval is γˆ ± 1.96 × SE ( γˆ ) .

8.10.

(a) ∆ Y= f ( X 1 + ∆ X 1 , X 2 ) − f ( X 1 , X 2 )= β1∆ X 1 + β 3 ∆ X 1 × X 2 , so

∆Y = β1 + β 3 X 2 . ∆X 1

(b) ∆ Y= f ( X 1 , X 2 + ∆ X 2 ) − f ( X 1 , X 2 )= β 2 ∆ X 2 + β 3 X 1 × ∆ X 2 , so

∆Y = β 2 + β3 X 1. ∆X 2

(c) = ∆Y f ( X 1 + ∆ X 1 , X 2 + ∆ X 2 ) − f ( X 1 , X 2 ) = β 0 + β1 ( X 1 + ∆ X 1 ) + β 2 ( X 2 + ∆ X 2 ) + β 3 ( X 1 + ∆ X 1 )( X 2 + ∆ X 2 ) − ( β 0 + β1 X 1 + β 2 X 2 + β 3 X 1 X 2 ) = ( β1 + β 3 X 2 )∆ X 1 + ( β 2 + β 3 X 1 )∆ X 2 + β 3 ∆ X 1∆ X 2 .

8.11.

Linear model: E(Y | X) = β0 + β1X, so that

β1

β1 X X = E (Y | X ) β 0 + β1 X

(

dE (Y | X ) = β1 and the elasticity is dX

)

β 0 + β1 ln( X ) e β0 + β1 ln( X ) + u | X e= E (eu | X ) ce β0 + β1 ln( X ) , where c = Log-Log Model: E(Y | X) = E=

u

E(e | X), which does not depend on X because u and X are assumed to be independent. dE (Y | X ) β1 β0 + β1 ln( X ) E (Y | X ) β1 ce Thus= = , and the elasticity is β1. dX X X 8.12.

(a) Because of random assignment within the group of returning students E(X1i | ui) = 0 in “γ-regression,” so that γˆ1 is an unbiased estimator of γ1. (b) Because of random assignment within the group of returning students E(X1i | ui) = 0 in “δ- regression,” so that δˆ1 is an unbiased estimator of δ1. (c) Write E(ui | X1i, X2i) = E(ui | X2i) = λ0 + λ1X2i, where linearity is assumed for the conditional expected value. Thus,

E (Y |X 1 , X 2 ) = β 0 + β1 X 1 + β 2 X 2 + β 3 X 1 X 2 + E (ui | X 1i , X 2i ) = (β 0 + λ 0 ) + β1 X 1 + (β 2 + λ1 ) X 2 + β 3 X 1 X 2

.

Using this expression, E(Y| X1 = 1, X2 = 0) − E(Y| X1 = 0, X2 = 0) = β1, which is equal to γ1from (a). Also, E(Y| X1 = 1, X2 = 1) − E(Y| X1 = 0, X2 = 1) = β1 + β3, which is equal to δ1 from (b). Together, these results imply that β3 = δ1 − γ1. (d) Defining vi = ui − E(ui | X1i, X2i) = ui − E(ui | X2i), then Yi = (β0 + λ0) + β1X1 + (β2 + λ1) X2 + β3X1X2 + vi,

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

47

where E(vi | X1i, X2i) = 0. Thus, applying OLS to the equation will yield a biased estimate of the constant term [ E ( βˆ0 ) = β0 + λ0], an unbiased estimate of β1[ E ( βˆ1 ) = β1], a biased estimate of β2 [ E ( βˆ ) = β2 + λ1], and an unbiased estimate of β3 [ E ( βˆ ) = β3].

2

3

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 9

Assessing Studies Based on Multiple Regression

9.1.

As explained in the text, potential threats to external validity arise from differences between the population and setting studied and the population and setting of interest. The statistical results based on New York in the 1970s are likely to apply to Boston in the 1970s but not to Los Angeles in the 1970s. In 1970, New York and Boston had large and widely used public transportation systems. Attitudes about smoking were roughly the same in New York and Boston in the 1970s. In contrast, Los Angeles had a considerably smaller public transportation system in 1970. Most residents of Los Angeles relied on their cars to commute to work, school, and so forth. The results from New York in the 1970s are unlikely to apply to New York in 2010. Attitudes towards smoking changed significantly from 1970 to 2010.

9.2.

Yi + wi , or Y=i Yi − wi . Substituting the 2nd (a) When Yi is measured with error, we have Y= i equation into the regression model Yi = β 0 + β1 X i + ui gives Yi − wi = β 0 + β1 X i + ui , or Yi = β 0 + β1 X i + ui + wi . Thus v= ui + wi . i (b) (1) The error term vi has conditional mean zero given Xi: E (vi |X i ) = E (ui + wi |X i ) = E (ui |X i ) + E ( wi |X i ) = 0 + 0 = 0.

Yi + wi is i.i.d since both Yi and wi are i.i.d. and mutually independent; Xi and Yj (i ≠ j ) (2) Y= i are independent since Xi is independent of both Yj and wj. Thus, ( X i , Yi ), i = 1, , n are i.i.d. draws from their joint distribution.

(3) v= ui + wi has a finite fourth moment because both ui and wi have finite fourth moments i and are mutually independent. So (Xi, vi) have nonzero finite fourth moments. (c) The OLS estimators are consistent because the least squares assumptions hold. (d) Because of the validity of the least squares assumptions, we can construct the confidence intervals in the usual way. (e) The answer here is the economists’ “On the one hand, and on the other hand.” On the one hand, the statement is true: i.i.d. measurement error in X means that the OLS estimators are inconsistent and inferences based on OLS are invalid. OLS estimators are consistent and OLS inference is valid when Y has i.i.d. measurement error. On the other hand, even if the measurement error in Y is i.i.d. and independent of Yi and Xi, it increases the variance of the 2 σ u2 + σ w2 ), and this will increase the variance of the OLS estimators. regression error (σ= v Also, measurement error that is not i.i.d. may change these results, although this would need to be studied on a case-by-case basis.

9.3.

The key is that the selected sample contains only employed women. Consider two women, Beth and Julie. Beth has no children; Julie has one child. Beth and Julie are otherwise identical. Both can earn $25,000 per year in the labor market. Each must compare the $25,000 benefit to the costs of working. For Beth, the cost of working is forgone leisure. For Julie, it is forgone leisure and the costs (pecuniary and other) of child care. If Beth is just on the margin between working in the

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

49

labor market or not, then Julie, who has a higher opportunity cost, will decide not to work in the labor market. Instead, Julie will work in “home production,” caring for children, and so forth.

Thus, on average, women with children who decide to work are women who earn higher wages in the labor market.

9.4.

Estimated Effect of a 10% Increase in Average Income State

βln(Income)

Calif.

11.57 (1.81) 16.53 (3.15)

Mass.

Std. Dev. of Scores

In Points

19.1

In Std. Dev.

1.157 (0.18) 1.65 (0.31)

15.1

0.06 (0.001) 0.11 (0.021)

The income effect in Massachusetts is roughly twice as large as the effect in California.

9.5. = (a) Q

γ 1β 0 − γ 0 β1 γ 1u − β1v . + γ 1 − β1 γ 1 − β1

and P = (b) E (Q) =

β0 − γ 0 u − v . + γ 1 − β1 γ 1 − β1

γ 1β 0 − γ 0 β1 β −γ0 , E ( P) = 0 γ 1 − β1 γ 1 − β1 2

 1  2 2 2 2 (c) var(Q)  =  (γ 1 σ u + β1 σ v ), var( P ) γ β − 1   1 2

 1  2 2 =   (σ u + σ v ),  γ 1 − β1  and 2

 1  2 2 = cov( P, Q)   (γ 1σ u + β1σ V ) γ β − 1   1

γ 1σ u2 + β1σ v2 p cov(Q, P ) (d) (i) βˆ1 = , → var( P ) σ u2 + σ v2

p βˆ0 → E (Q) − E ( P )

cov( P, Q) var( P )

σ (γ − β ) (ii) βˆ1 − β1 → u 2 1 2 1 > 0, using the fact that γ1 > 0 (supply curves slope up) and β1 < 0 σu + σv (demand curves slope down). p

2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

50

Stock/Watson - Introduction to Econometrics - Second Edition

9.6.

(a) The parameter estimates do not change. Nor does the R2. The sum of squared residuals from the 100 observation regression is SER200= (100 − 2) × 15.12= 22, 344.98, and the sum of squared residuals from the 200 observation regression is twice this value: SSR200 = 2 × 22, 344.98. Thus,

= the SER from the 200 observation regression is SER200

1 = SSR200 15.02. The standard 200 − 2

errors for the regression coefficients are now computed using equation (5.4) where 2 2 200 2 ˆ ∑i200 =1 ( X i − X ) ui and ∑ i =1 ( X i − X ) are twice their value from the 100 observation regression. Thus the standard errors for the 200 observation regression are the standard errors 100 − 2 in the 100 observation regression multiplied by = 0.704. In summary, the results for 200 − 2 the 200 observation regression are Yˆ = 32.1 + 66.8 X, SER = 15.02, R 2 = 0.81

(10.63) (8.59) (b) The observations are not i.i.d.: half of the observations are identical to the other half, so that the observations are not independent.

9.7.

(a) True. Correlation between regressors and error terms means that the OLS estimator is inconsistent. (b) True.

9.8.

Not directly. Test scores in California and Massachusetts are for different tests and have different means and variances. However, converting (9.5) into units for Massachusetts yields the implied regression to TestScore(MA units) = 740.9 − 1.80 × STR, which is similar to the regression using Massachusetts data shown in Column 1 of Table 9.2. After this adjustment the regression could be somewhat useful; however, the regression in Column 1 of Table 9.2 has a low R2, suggesting that it will not provide an accurate forecast of test scores.

9.9.

Both regressions suffer from omitted variable bias so that they will not provide reliable estimates of the causal effect of income on test scores. However, the nonlinear regression in (8.18) fits the data well, so that it could be used for forecasting.

9.10.

There are several reasons for concern. Here are a few.

Internal consistency: omitted variable bias as explained in the last paragraph of the box.

Internal consistency: sample selection may be a problem as the regressions were estimated using a sample of full-time workers. (See exercise 9.3 for a related problem.)

External consistency: Returns to education may change over time because of the relative demands and supplies of skilled and unskilled workers in the economy. To the extent that this is important, the results shown in the box (based on 2008 data) may not accurately estimate today’s returns to education.

9.11.

Again, there are reasons for concern. Here are a few.

Internal consistency: To the extent that price is affected by demand, there may be simultaneous equation bias.

External consistency: The internet and introduction of “E-journals” may induce important changes in the market for academic journals so that the results for 2000 may not be relevant for today’s market.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

9.12.

51

(a) See the answer to part (c) of exercise 2.27. (b) Because X = E(X|Z), then E(X| X ) = X . Thus E(w| X ) = E[(X − X | X )] = E(X| X ) − E( X | X ) = X − X = 0. (c) Xi = X + wi, so that Yi = β0 + β1( X + wi) + ui = β0 + β1 X + vi, where vi = ui + β1 wi. Because E(u| Z) = 0 and X depends only on Z (that is, X = E(X|Z)), then E(u| X ) = 0. Together with the result in (b), this implies that E(vi | X ) = 0. You can then verify the other assumptions in Key Concept (4.3), and the result follows from the consistency of the OLS estimator under these assumptions.

9.13.

300 ∑ ( X i − X )(Yi − Y ) . Because all of the Xi’s are used (although some are used for the (a) βˆ1 = i =1 300 ∑ i =1 ( X i − X )2

Y β1 ( X i − X ) + ui − u . Using wrong values of Yj), X = X , and ∑ i =1 ( X i − X ) 2 . Also, Yi −= n

these expressions:

0.8 n n n ( X i − X )2 ( X i − X )( X i − X ) ∑ i = ( X i − X )(ui − u ) ∑ ∑ 1 0.8 n +1 1 i= i= ˆ β1 = β1 n + β1 + n n ( X i − X )2 X i − X )2 ∑ i 1= ∑ i 1 (= ∑ i 1 ( X i − X )2 =

1 0.8 n 1 n 1 n  ( X i − X )2 ( X i − X )( X i − X ) ( X i − X )(ui − u ) ∑ ∑ ∑ i= 1 0.8 n +1 1 i= i= β1 n = + β1 n +n 1 n 1 n 1 n ( X i − X )2 ( X i − X )2 ( X i − X )2 ∑ ∑ ∑ = i =1 i 1= i 1 n n n where n = 300, and the last equality uses an ordering of the observations so that the first 240 observations (= 0.8 × n) correspond to the correctly measured observations ( X i = Xi).

As is done elsewhere in the book, we interpret n = 300 as a large sample, so we use the approximation of n tending to infinity. The solution provided here thus shows that these expressions are approximately true for n large and hold in the limit that n tends to infinity.

Each of the averages in the expression for β̂1 have the following probability limits: p 1 n 2 X X σ X2 , − → ( ) ∑ i i =1 n p 1 0.8 n 2 X − X → ( ) 0.8σ X2 , ∑ i i =1 n p 1 n  ( X i − X )(ui − u ) → 0 , and ∑ i =1 n p 1 n  − − → X X X X ( )( ) 0, ∑ i i n =i 0.8 n +1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

52

Stock/Watson - Introduction to Econometrics - Second Edition

where the last result follows because X i ≠ Xi for the scrambled observations and Xj is p

independent of Xi for i ≠ j. Taken together, these results imply βˆ1 → 0.8β1 . p

p

(b) Because βˆ1 → 0.8β1 , βˆ1 / 0.8 → β1 , so a consistent estimator of β1 is the OLS estimator divided by 0.8. (c) Yes, the estimator based on the first 240 observations is better than the adjusted estimator from part (b). Equation (4.21) in Key Concept 4.4 (page 129) implies that the estimator based on the first 240 observations has a variance that is var( βˆ1 (240obs )) =

1 var [ ( X i − µ X )ui ] .

2 240 [ var( X i )]

From part (a), the OLS estimator based on all of the observations has two sources of sampling 300 ( X i − X )(ui − u ) ∑ i =1 which is the usual source that comes from the error. The first is 300 ∑ i =1 ( X i − X )2

∑ omitted factors (u). The second is β 1

300 i = 241

( X i − X )( X i − X )

∑

300

( X i − X )2 i =1

, which is the source that comes

from scrambling the data. These two terms are uncorrelated in large samples, and their respective large-sample variances are:  ∑ 300 ( X i − X )(ui − u )  1 var [ ( X i − µ X )ui ] = var  i =1 300 2  ∑ i =1 ( X i − X )2  300 [ var( X i )] 

and

 ∑ 300 ( X i − X )( X i − X )  0.2  = β12 var  β1 i = 241 300 .

2   300 − ( X X ) ∑ i i =1   Thus  βˆ (300obs )  1  1 var [ ( X i − µ X )ui ] 0.2  = + β12 var  1    2   0.64  300 0.8 300  X var( ) [ ]   i  

which is larger than the variance of the estimator that only uses the first 240 observations.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 10

Regression with Panel Data

10.1.

(a) With a $1 increase in the beer tax, the expected number of lives that would be saved is 0.45 per 10,000 people. Since New Jersey has a population of 8.1 million, the expected number of lives saved is 0.45 × 810 = 364.5. The 95% confidence interval is (0.45 ± 1.96 × 0.22) × 810 = [15.228, 713.77]. (b) When New Jersey lowers its drinking age from 21 to 18, the expected fatality rate increases by 0.028 deaths per 10,000. The 95% confidence interval for the change in death rate is 0.028 ± 1.96 × 0.066 = [− 0.1014, 0.1574]. With a population of 8.1 million, the number of fatalities will increase by 0.028 × 810 = 22.68 with a 95% confidence interval [− 0.1014, 0.1574] × 810 = [−82.134, 127.49]. (c) When real income per capita in New Jersey increases by 1%, the expected fatality rate increases by 1.81 deaths per 10,000. The 90% confidence interval for the change in death rate is 1.81 ± 1.64 × 0.47 = [1.04, 2.58]. With a population of 8.1 million, the number of fatalities will increase by 1.81 × 810 = 1466.1 with a 90% confidence interval [1.04, 2.58] × 810 = [840, 2092]. (d) The low p-value (or high F-statistic) associated with the F-test on the assumption that time effects are zero suggests that the time effects should be included in the regression. (e) Define a binary variable west which equals 1 for the western states and 0 for the other states.

Include the interaction term between the binary variable west and the unemployment rate, west × (unemployment rate), in the regression equation corresponding to column (4). Suppose the coefficient associated with unemployment rate is β and the coefficient associated with west × (unemployment rate) is γ. Then β captures the effect of the unemployment rate in the eastern states, and β + β captures the effect of the unemployment rate in the western states. The difference in the effect of the unemployment rate in the western and eastern states is β. Using the coefficient estimate (γˆ ) and the standard error SE(γˆ ), you can calculate the t-statistic to test whether γ is statistically significant at a given significance level.

10.2.

(a) For each observation, there is one and only one binary regressor equal to one. That is, D1i + D 2i + D3i = 1= X 0, it . (b) For each observation, there is one and only one binary regressor that equals 1. That is, D1i + D 2i +  + Dni =1 =X 0, it . (c) The inclusion of all the binary regressors and the “constant” regressor causes perfect multicollinearity. The constant regressor is a perfect linear function of the n binary regressors.

OLS estimators cannot be computed in this case. Your computer program should print out a message to this effect. (Different programs print different messages for this problem. Why not try this, and see what your program says?)

©2011 Pearson Education, Inc. Publishing as Addison Wesley

54

Stock/Watson - Introduction to Econometrics - Second Edition

10.3.

The five potential threats to the internal validity of a regression study are: omitted variables, misspecification of the functional form, imprecise measurement of the independent variables, sample selection, and simultaneous causality. You should think about these threats one-by-one.

Are there important omitted variables that affect traffic fatalities and that may be correlated with the other variables included in the regression? The most obvious candidates are the safety of roads, weather, and so forth. These variables are essentially constant over the sample period, so their effect is captured by the state fixed effects. You may think of something that we missed. Since most of the variables are binary variables, the largest functional form choice involves the Beer Tax variable. A linear specification is used in the text, which seems generally consistent with the data in Figure 8.2. To check the reliability of the linear specification, it would be useful to consider a log specification or a quadratic. Measurement error does not appear to a problem, as variables like traffic fatalities and taxes are accurately measured. Similarly, sample selection is a not a problem because data were used from all of the states. Simultaneous causality could be a potential problem.

That is, states with high fatality rates might decide to increase taxes to reduce consumption. Expert knowledge is required to determine if this is a problem.

10.4.

(a) slope = β1, intercept = β0 (b) slope = β1, intercept = β0 (c) slope = β1, intercept = β0 + γ3 (d) slope = β1, intercept = β0 + γ3

10.5.

Let D2i = 1 if i = 2 and 0 otherwise; D3i = 1 if i = 3 and 0 otherwise … Dni = 1 if i = n and 0 otherwise. Let B2t = 1 if t = 2 and 0 otherwise; B3t = 1 if t = 3 and 0 otherwise … BTt = 1 if t = T and 0 otherwise. Let β0 = α1 + λ1; γi = α i − α1 and δt = λ t − β1.

10.6.

E (vit ) E= ( X it uit ) E[ X it E (uit | X i1 , X i 2 ,..., X 0 from assumption vit = X it uit . First note that = = iT , α i )] = v v ) E= (v v ) E ( X X u u ) . The assumptions in Key Concept 10.3 do not 1. Thus, cov( it is

it is

it

is it is

imply that this term is zero. That is, Key Concept 10.3 allows the errors for individual i to be correlated across time periods.

10.7.

(a) Average snow fall does not vary over time, and thus will be perfectly collinear with the state fixed effect. (b) Snowit does vary with time, and so this method can be used along with state fixed effects.

10.8.

There are several ways. Here is one: let Yit = β0 + β1X1, it + β2t + γ2D2i + ⋅⋅⋅ + γnDni + δ2(D2i × t) + ⋅⋅⋅ + δn(Dni × t) + uit, where D2i = 1 if i = 2 and 0 otherwise and so forth. The coefficient λi = β2 + δi.

10.9.

(a) αˆ i=

1 T

∑Tt =1 Yit which has variance

small. αˆ i is not consistent.

σ u2 T

. Because T is not growing, the variance is not getting

(b) The average in (a) is computed over T observations. In this case T is small (T = 4), so the normal approximation from the CLT is not likely to be very good.

10.10. No, one of the regressors is Yit −1 . This depends on Yit −1 . This means that assumption (1) is violated.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

10.11 Using the hint, equation (10.22) can be written as n

βˆ1DM =

1

1



∑  4 ( X − X )(Y − Y ) + 4 ( X − X )(Y − Y )  i2

i =1

i1

i2

1

i1

1

i2

i1

i2



∑  4 ( X − X ) + 4 ( X − X )  n

2

i2

i =1

i1

2

i2

i1

∑ ( X − X )(Y − Y ) = = βˆ ∑ (X − X ) n

i =1

i2

i1

i2

n

i =1

i1

2

i2

BA 1

i1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

i1

55

Chapter 11

Regression with a Binary Dependent Variable

11.1.

(a) The t-statistic for the coefficient on Experience is 0.031/0.009 = 3.44, which is significant at the 1% level. (b) zMatthew= 0.712 + 0.031 × 10 = 1.022; Φ (1.022) = 0.847 (c) zChristopher = 0.712 + 0.031 ×= 0 0.712; Φ (0.712) = 0.762 (d) z Jed= 0.712 + 0.031 × 80 = 3.192; Φ (3.192) = 0.999, this is unlikely to be accurate because the sample did not include anyone with more that 40 years of driving experience.

11.2.

(a) The t-statistic for the coefficient on Experience is t = 0.040/0.016 = 2.5, which is significant at the 5% level.

1 1 = = = Prob 0.811 Matthew − (1.059 + 0.040 × 10) 1+ e 1 + e −1.459 1 1 Prob= = = 0.742 Christopher − (1.059 + 0 .040 × 0) 1+ e 1 + e −1.059 (b)

The shape of the regression functions are similar, but the logit regression lies below the probit regression for experience in the range of 0 = 60 years.

11.3.

(a) The t-statistic for the coefficient on Experience is t = 0.006/0.002 = 3, which is significant a the 1% level.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

(b)

57

ProbMatther = 0.774 + 0.006 × 10 = 0.836 ProbChristopher = 0.774 + 0.006 × 0 = 0.774

The probabilities are similar except when experience in large ( > 40 years). In this case the LPM model produces nonsensical results (probabilities greater than 1.0).

11.4.

(a) Group Men Women

Probit

Logit

Φ(1.282 − 0.333) = 0.829

1 1+ e

− (2.197 − 0.622)

Φ(1.282) = 0.900

1 1+ e

− (2.197)

= 0.829 = 0.900

LPM 0.829 0.900

(b) Because there is only regressor and it is binary (Male), estimates for each model show the fraction on males and females passing the test. Thus, the results are identical for all models.

11.5.

(a) Φ(0.806 + 0.041 × 10 × 0.174 × 1 − 0.015 × 1 × 10) = 0.814 (b) Φ(0.806 + 0.041 × 2 − 0.174 × 0 − 0.015 × 0 × 2) = 0.813 (c) The t-stat on the interaction term is −0.015/0.019 = −0.79, which is not significant at the 10% level.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

58

Stock/Watson - Introduction to Econometrics - Second Edition

11.6.

(a) For a black applicant having a P/I ratio of 0.35, the probability that the application will be denied is Φ(− 2.26 + 2.74 × 0.35 + 0.71) = Φ(− 0.59) = 27.76%. (b) With the P/I ratio reduced to 0.30, the probability of being denied is Φ(− 2.26 + 2.74 × 0.30 + 0.71) = Φ(− 0.73) = 23.27%. The difference in denial probabilities compared to (a) is 4.4 percentage points lower. (c) For a white applicant having a P/I ratio of 0.35, the probability that the application will be denied is Φ(−2.26 + 2.74 × 0.35) = 9.7%. If the P/I ratio is reduced to 0.30, the probability of being denied is Φ(−2.26 + 2.74 × 0.30) = 7.5%. The difference in denial probabilities is 2.2 percentage points lower. (d) From the results in parts (a)–(c), we can see that the marginal effect of the P/I ratio on the probability of mortgage denial depends on race. In the probit regression functional form, the marginal effect depends on the level of probability which in turn depends on the race of the applicant. The coefficient on black is statistically significant at the 1% level.

11.7.

(a) For a black applicant having a P/I ratio of 0.35, the probability that the application will be 1 denied is F (−4.13 + 5.37 × 0.35 +1.27) = = 27.28%.

1 + e0.9805 (b) With the P/I ratio reduced to 0.30, the probability of being denied is 1 F (−4.13 + 5.37 × 0.30 + 1.27) = = 22.29% . The difference in denial probabilities 1 + e1.249 compared to (a) is 4.99 percentage points lower. (c) For a white applicant having a P/I ratio of 0.35, the probability that the application will be 1 = = 9.53%. If the P/I ratio is reduced to 0.30, the denied is F (−4.13 + 5.37 × 0.35) 1 + e 2.2505 1 = = 7.45%. The difference in probability of being denied is F (−4.13 + 5.37 × 0.30) 1 + e 2.519 denial probabilities is 2.08 percentage points lower. (d) From the results in parts (a)–(c), we can see that the marginal effect of the P/I ratio on the probability of mortgage denial depends on race. In the logit regression functional form, the marginal effect depends on the level of probability which in turn depends on the race of the applicant. The coefficient on black is statistically significant at the 1% level. The logit and probit results are similar.

11.8.

(a) Since Yi is binary variable, we know E(Yi |Xi) = 1 × Pr(Yi = 1|Xi) + 0 × Pr(Yi = 0|Xi) = Pr(Yi = 1|Xi) = β0 + β1Xi. Thus

E (ui | X i ) = E[Yi − ( β 0 + β1 X i )| X i ] = E (Yi|X i ) − ( β 0 + β1 X= 0 i) (b) Using Equation (2.7), we have var(Yi | X i ) = Pr(Yi = 1| X i )[1 − Pr(Yi + 1| X i )] = ( β 0 + β1 X i )[1 − ( β 0 + β1 X i )].

Thus var(ui | X i )= var[Yi − ( β 0 + β1 X i )i | X i ] = var(Yi | X i ) = ( β 0 + β1 X i )[1 − ( β 0 + β1 X i )].

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

59

(c) var(ui |Xi) depends on the value of Xi, so ui is heteroskedastic. (d) The probability that Yi = 1 conditional on Xi is pi = β0 + β1Xi. The conditional probability = Yi yi |= X i ) piyi (1 − pi )1− yi . Assuming that (Xi, Yi) are distribution for the ith observation is Pr( i.i.d., i = 1,…, n, the joint probability distribution of Y1,…, Yn conditional on the X′s is = = Pr(Y1 y1= ,  , Yn yn | X 1 ,  , Xn) =

n

= Pr(Y y |X ) ∏ n

i

i

∏ p (1 − p ) i =1

=

i

i =1

yi i

1− yi

i

n

∏ (β + β X ) [1 − (β + β X )] i =1

0

1

i

yi

0

1

i

1− yi

.

The likelihood function is the above joint probability distribution treated as a function of the unknown coefficients (β0 and β1).

11.9.

(a) The coefficient on black is 0.084, indicating an estimated denial probability that is 8.4 percentage points higher for the black applicant. (b) The 95% confidence interval is 0.084 ± 1.96 × 0.023 = [3.89%, 12.91%]. (c) The answer in (a) will be biased if there are omitted variables which are race-related and have impacts on mortgage denial. Such variables would have to be related with race and also be related with the probability of default on the mortgage (which in turn would lead to denial of the mortgage application). Standard measures of default probability (past credit history and employment variables) are included in the regressions shown in Table 9.2, so these omitted variables are unlikely to bias the answer in (a). Other variables such as education, marital status, and occupation may also be related the probability of default, and these variables are omitted from the regression in column. Adding these variables (see columns (4)–(6)) have little effect on the estimated effect of black on the probability of mortgage denial.

11.10. (a) Let n1 = #(Y = 1), the number of observations on the random variable Y which equals 1; and n2 = #(Y = 2). Then #(Y = 3) = n − n1 − n2. The joint probability distribution of Y1,…, Yn is Pr(Y1 = y1 , , Yn = yn ) =

n

∏ Pr(Y = y ) = p q (1 − p − q) i =1

i

i

n1

n2

n − n1 − n2

.

The likelihood function is the above joint probability distribution treated as a function of the unknown coefficients (p and q). (b) The MLEs of p and q maximize the likelihood function. Let’s use the log-likelihood function = L ln[Pr( = Y1 y1= , , Yn yn )] = n1 ln p + n2 ln q + (n − n1 − n2 ) ln(1 − p − q ).

Using calculus, the partial derivatives of L are ∂L n1 n − n1 − n2 = − , and ∂p p 1 − p − q ∂L n2 n − n1 − n2 = − . ∂q q 1− p − q

©2011 Pearson Education, Inc. Publishing as Addison Wesley

60

Stock/Watson - Introduction to Econometrics - Second Edition

Setting these two equations equal to zero and solving the resulting equations yield the MLE of p and q:

= pˆ

n1 n2 = , qˆ . n n

11.11. (a) This is a censored or truncated regression model (note the dependent variable might be zero). (b) This is an ordered response model. (c) This is the discrete choice (or multiple choice) model. (d) This is a model with count data.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 12

Instrumental Variables Regression

12.1.

(a) The change in the regressor, ln( Pi ,cigarettes ) − ln( Pi ,cigarettes ), from a $0.50 per pack increase in the 1995 1985 retail price is ln(8.00) − ln(7.50) = 0.0645. The expected percentage change in cigarette demand is −0.94 × 0.0645 × 100% = − 6.07%. The 95% confidence interval is (− 0.94 ± 1.96 × 0.21) × 0.0645 × 100% = [− 8.72%, −3.41%]. (b) With a 2% reduction in income, the expected percentage change in cigarette demand is 0.53 × (−0.02) × 100% = −1.06%. (c) The regression in column (1) will not provide a reliable answer to the question in (b) when recessions last less than 1 year. The regression in column (1) studies the long-run price and income elasticity. Cigarettes are addictive. The response of demand to an income decrease will be smaller in the short run than in the long run. (d) The instrumental variable would be too weak (irrelevant) if the F-statistic in column (1) was 3.6 instead of 33.6, and we cannot rely on the standard methods for statistical inference. Thus the regression would not provide a reliable answer to the question posed in (a).

12.2.

(a) When there is only one X, we only need to check that the instrument enters the first stage population regression. Since the instrument is Z = X, the regression of X onto Z will have a coefficient of 1.0 on Z, so that the instrument enters the first stage population regression. Key Concept 4.3 implies corr(Xi, ui) = 0, and this implies corr(Zi, ui) = 0. Thus, the instrument is exogenous. (b) Condition 1 is satisfied because there are no W’s. Key Concept 4.3 implies that condition 2 is satisfied because (Xi, Zi, Yi) are i.i.d. draws from their joint distribution. Condition 3 is also satisfied by applying assumption 3 in Key Concept 4.3. Condition 4 is satisfied because of conclusion in part (a). s (c) The TSLS estimator is βˆ1TSLS = ZY using Equation (10.4) in the text. Since Zi = Xi, we have sZX sZY s XY = = βˆ1OLS .

2 sZX sX

= βˆ1TSLS

12.3.

(a) The estimator σˆ a2 =n −1 2 ∑in=1 (Yi − βˆ0TSLS − βˆ1TSLS Xˆ i ) 2 is not consistent. Write this as σˆ a2 = 1 Y − βˆ TSLS − βˆ TSLS X . Replacing βˆ TSLS with β1, ∑ n (uˆ − βˆ TSLS ( Xˆ − X )) 2 , where uˆ = n−2

i =1

1

i

i

i

i

i

0

1

i

1

as suggested in the question, write this as σˆ a2 ≈ 1n ∑in 1 = (ui − β1 ( Xˆ i − X i )) 2 = 1n ∑in 1 ui2 + = 1 ∑ n [ β 2 ( Xˆ − X ) 2 + 2u β ( Xˆ − X )]. The first term on the right hand side of the equation n

i =1

1

i

i

i

1

i

i

converges to σˆ , but the second term converges to something that is non-zero. Thus σˆ a2 is not consistent.

2 u

©2011 Pearson Education, Inc. Publishing as Addison Wesley

62

Stock/Watson - Introduction to Econometrics - Second Edition

(b) The estimator σˆ b2 =n −1 2 Σin=1 (Yi − βˆ0TSLS − βˆ1TSLS X i ) 2 is consistent. Using the same notation as in (a), we can write σˆ b2 ≈ 1n Σin=1ui2 , and this estimator converges in probability to σ u2 .

12.4.

ˆ πˆ + πˆ Z and Using Xˆ= πˆ 0 + πˆ1Z i , we have X= 0 1 i = s XY ˆ

n

n

∑ ( Xˆ i − Xˆ )(Yi − Y=) πˆ1 ∑ (Zi − Z )(Yi − Y=) πˆ1sZY ,

=i 1 =i 1

s X2ˆ=

n

n

∑ ( Xˆ i − Xˆ )2= πˆ12 ∑ (Zi − Z )2= πˆ12 sZ2 .

=i 1 =i 1

Using the formula for the OLS estimator in Key Concept 4.2, we have s sZ

πˆ1 = ZX2 .

Thus the TSLS estimator

βˆ1TSLS = 12.5.

s XY sZY πˆ1sZY ˆ = = = 2 2 2 s Xˆ πˆ1 sZ πˆ1sZ2

sZY sZY . = 2 × sZ sZX s2

sZX Z

(a) Instrument relevance. Z i does not enter the population regression for X i (b) Z is not a valid instrument. X̂ will be perfectly collinear with W. (Alternatively, the first stage regression suffers from perfect multicollinearity.) (c) W is perfectly collinear with the constant term. (d) Z is not a valid instrument because it is correlated with the error term.

12.6.

Use R 2 to compute the homoskedastic-only F statistic as 0.05 R2 / k FHomoskedasitcOnly = = = 5.16 with 100 observations in which case we 2 1 − R / (T − k − 1) 0.95 / 98

(

)

conclude that the instrument may be week. With 500 observations the FHomoskedasitcOnly = 26.2 so the instrument is not weak.

12.7.

(a) Under the null hypothesis of instrument exogeneity, the J statistic is distributed as a χ12 random variable, with a 1% critical value of 6.63. Thus the statistic is significant, and instrument exogeneity E(ui|Z1i, Z2i) = 0 is rejected. (b) The J test suggests that E(ui |Z1i, Z2i) ≠ 0, but doesn’t provide evidence about whether the problem is with Z1 or Z2 or both.

12.8.

(a) Solving for P yields P =

−σ γ 0 − β 0 uid − uis ; thus cov( P, u s ) = u + β1 β1 β1

2

s

(b) Because cov(P,u) ≠ 0, the OLS estimator is inconsistent (see (6.1)). (c) We need a instrumental variable, something that is correlated with P but uncorrelated with us.

In this case Q can serve as the instrument, because demand is completely inelastic (so that Q is not affected by shifts in supply). γ0 can be estimated by OLS (equivalently as the sample mean of Qi).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

12.9.

63

(a) There are other factors that could affect both the choice to serve in the military and annual earnings. One example could be education, although this could be included in the regression as a control variable. Another variable is “ability” which is difficult to measure, and thus difficult to control for in the regression. (b) The draft was determined by a national lottery so the choice of serving in the military was random. Because it was randomly selected, the lottery number is uncorrelated with individual characteristics that may affect earning and hence the instrument is exogenous. Because it affected the probability of serving in the military, the lottery number is relevant.

12.10. = βˆTSLS

cov( Z i , Yi ) cov( Z i , β 0 + β1 X i + β 2Wi + ui ) β1cov( Z i , X i ) + β 2 cov( Z i ,Wi ) = = cov( Z i , X i ) cov( Z i , X i ) cov( Z i , X i )

(a) If cov( Z i ,Wi ) = 0 the IV estimator is consistent. (b) If cov( Z i ,Wi ) ≠ 0 the IV estimator is not consistent.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 13 Experiments and Quasi-Experiments 13.1.

For students in kindergarten, the estimated small class treatment effect relative to being in a regular class is an increase of 13.90 points on the test with a standard error 2.45. The 95% confidence interval is 13.90 ± 1.96 × 2.45 = [9.098, 18.702].

For students in grade 1, the estimated small class treatment effect relative to being in a regular class is an increase of 29.78 points on the test with a standard error 2.83. The 95% confidence interval is 29.78 ± 1.96 × 2.83 = [24.233, 35.327].

For students in grade 2, the estimated small class treatment effect relative to being in a regular class is an increase of 19.39 points on the test with a standard error 2.71. The 95% confidence interval is 19.39 ± 1.96 × 2.71 = [14.078, 24.702].

For students in grade 3, the estimated small class treatment effect relative to being in a regular class is an increase of 15.59 points on the test with a standard error 2.40. The 95% confidence interval is 15.59 ± 1.96 × 2.40 = [10.886, 20.294].

13.2.

(a) On average, a student in class A (the “small class”) is expected to score higher than a student in class B (the “regular class”) by 15.89 points with a standard error 2.16. The 95% confidence interval for the predicted difference in average test scores is 15.89 ± 1.96 × 2.16 = [11.656, 20.124]. (b) On average, a student in class A taught by a teacher with 5 years of experience is expected to score lower than a student in class B taught by a teacher with 10 years of experience by 0.66 × 5 = 3.3 points. The standard error for the score difference is 0.17 × 5 = 0.85. The 95% confidence interval for the predicted lower score for students in classroom A is 3.3 ± 1.96 × 0.85 = [1.634, 4.966]. (c) The expected difference in average test scores in 15.89 + 0.66 × (−5) = 12.59. Because of random assignment, the estimators of the small class effect and the teacher experience effect are uncorrelated. Thus, the standard error for the difference in average test scores is 1 [2.162 + (−5) 2 × 0.17 2 ] 2 =2.3212. The 95% confidence interval for the predicted difference in average test scores in classrooms A and B is 12.59 ± 1.96 × 2.3212 = [8.0404, 17.140]. (d) The intercept is not included in the regression to avoid the perfect multicollinearity problem that exists among the intercept and school indicator variables.

13.3.

(a) The estimated average treatment effect is X TreatmentGroup − X Control = 1241 − 1201 = 40 points.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

65

(b) There would be nonrandom assignment if men (or women) had different probabilities of being assigned to the treatment and control groups. Let pMen denote the probability that a male is assigned to the treatment group. Random assignment means pMen = 0.5. Testing this null pˆ Men − 0.5 0.55 − 0.50 = = 1.00, hypothesis results in= a t-statistic of tMen 1 1 0.55(1 − 0.55) pˆ Men (1 − pˆ Men ) n Men 100 so that the null of random assignment cannot be rejected at the 10% level. A similar result is found for women.

13.4.

(a) (i) Xit = 0, Gi = 1, Dt = 0 (ii) Xit = 1, Gi = 1, Dt = 1 (iii) Xit = 0, Gi = 0, Dt = 0 (iv) Xit = 0, Gi = 0, Dt = 1 (b) (i) β0 + β2 (ii) β0 + β1 + β2 + β3 (iii) β0 (iv) β0 + β3 (c) β1 (d) “New Jersey after − New Jersey before” = β1 + β3, where β3 denotes the time effect associated with changes in the economy between 1991 and 1993. “1993 New Jersey − 1993 Pennsylvania” = β1 + β2, where β2 denotes the average difference in employment between NJ and PA.

13.5.

(a) This is an example of attrition, which poses a threat to internal validity. After the male athletes leave the experiment, the remaining subjects are representative of a population that excludes male athletes. If the average causal effect for this population is the same as the average causal effect for the population that includes the male athletes, then the attrition does not affect the internal validity of the experiment. On the other hand, if the average causal effect for male athletes differs from the rest of population, internal validity has been compromised. (b) This is an example of partial compliance which is a threat to internal validity. The local area network is a failure to follow treatment protocol, and this leads to bias in the OLS estimator of the average causal effect. (c) This poses no threat to internal validity. As stated, the study is focused on the effect of dorm room Internet connections. The treatment is making the connections available in the room; the treatment is not the use of the Internet. Thus, the art majors received the treatment (although they chose not to use the Internet). (d) As in part (b) this is an example of partial compliance. Failure to follow treatment protocol leads to bias in the OLS estimator.

13.6.

The treatment effect is modeled using the fixed effects specification Yit = α i + β1 X it + uit . (a) αi is an individual-specific intercept. The random effect in the regression has variance

var(α i + uit= ) var(α i ) + var(uit ) + 2cov(α i , uit ) = σ α2 + σ u2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

66

Stock/Watson - Introduction to Econometrics - Second Edition

which is homoskedastic. The differences estimator is constructed using data from time period t = 2. Using Equation (5.27), it is straightforward to see that the variance for the differences estimator n var( βˆ1differences ) →

var(α i + ui 2 ) σ α2 + σ u2 = . var( X i 2 ) var( X i 2 )

(b) The regression equation using the differences-in-differences estimator is ∆Yi = β1∆X i + vi

with ∆Yi = Yi2 − Yi1, ∆Xi = Xi2 − Xi1, and vi = ui2 − ui1. If the ith individual is in the treatment group at time t = 2, then ∆Xi = Xi2 − Xi1 = 1 − 0 = 1 = Xi2. If the ith individual is in the control group at time t = 2, then ∆Xi = Xi2 − Xi1 = 0 − 0 = 0 = Xi2. Thus ∆Xi is a binary treatment variable and ∆Xi = Xi2, which in turn implies var(∆Xi) = var(Xi2). The variance for the new error term is 2 var(ui 2 − ui1= ) var(ui 2 ) + var(ui1 ) − 2cov(ui 2 , ui1= ) 2σ u2 , σ= v

which is homoskedastic. Using Equation (5.27), it is straightforward to see that the variance for the differences-in-differences estimator n var( βˆ1diffs −in − diffs ) →

(

)

σ v2

2σ u2 = . var(∆X i ) var( X i 2 )

(

)

(c) When σ α2 > σ u2 , we’ll have var βˆ1differences > var βˆ1diffs −in −diffs and the differences-indifferences estimator is more efficient then the differences estimator. Thus, if there is considerable large variance in the individual-specific fixed effects, it is better to use the differences-in-differences estimator.

13.7.

From the population regression Yit =α i + β1 X it + β 2 ( Dt × Wi ) + β 0 Dt + vit ,

we have = β1 ( X i 2 − X i1 ) + β 2 [( D2 − D1 ) × Wi ] + β 0 ( D2 − D1 ) + (vi 2 − vi1 ).

Yi 2 − Y i1

By defining ∆Yi = Yi2 − Yi1, ∆Xi = Xi2 − Xi1 (a binary treatment variable) and ui = vi2 − vi1, and using D1 = 0 and D2 = 1, we can rewrite this equation as ∆Yi = β 0 + β1 X i + β 2Wi + ui ,

which is Equation (13.5) in the case of a single W regressor.

13.8.

The regression model is Yit = β 0 + β1 X it + β 2Gi + β 3 Bt + uit ,

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

67

Using the results in Section 8.3 Y control,before = βˆ0 = βˆ + βˆ Y control,after 0

3

0

1

= βˆ0 + βˆ2 Y Y treatment,after = βˆ + βˆ + βˆ + βˆ treatment,before

2

3

Thus = βˆ diffs-in-diffs (Y treatment,after − Y treatment,before ) − (Y control,after − Y control,before ) = ( βˆ + βˆ ) − ( βˆ ) = βˆ 1

13.9.

3

3

1

The covariance between β1i X i and Xi is

E{[ β1i X i − E ( β1i X i )][ X i − E ( X i )]} cov( β1i X i , X i ) = = E{β1i X i2 − E ( β1i X i ) X i − β1i X i E ( X i ) + E ( β1i X i ) E ( X i )} = E ( β1i X i2 ) − E ( β1i X i ) E ( X i ) Because Xi is randomly assigned, Xi is distributed independently of β1i. The independence means

= E ( β1i X i ) E= ( β1i ) E ( X i ) and E ( β1i X i2 ) E ( β1i ) E ( X i2 ).

Thus cov( β1i X i , X i ) can be further simplified:

cov( β1i X i , X i ) E ( β1i )[ E ( X i2 ) − E 2 ( X i )] = = E ( β1i )σ X2 .

So cov( β1i X i , X i ) E ( β1i )σ X2 E ( β1i ). = = 2 2

σX

σX

13.10. (a) This is achieved by adding and subtracting β0 + β1Xi to the right hand side of the equation and rearranging terms. (b) E[ui |Xi] = E[β0i − β0 |Xi] + E[(β1i − β1)Xi |Xi] |Xi] + E[vi |Xi] = 0. (c) (1) is shown in (b). (2) follows from the assumption that (vi, Xi, β0i, β1i) are i.i.d. random variables. (d) Yes, the assumptions in KC 4.3 are satisfied. (e) If β1i and Xi are positively correlated, cov(β1i, Xi) = E[(β1i − β1)(Xi − βX)] = E[(β1i − β1)Xi ] > 0, where the first equality follows because β1 = E(β1i) and the inequality follows because the covariance and correlation have the same sign. Note 0 < E[(β1i − β1)Xi] = E{E[(β1i − β1)Xi |Xi]} by the law of iterated expectations, so it must the case that E[(β1i − β1)Xi |Xi] > 0 for some values of Xi. Thus assumption (1) is violated. This induces positive correlation between the regressors and error term, leading the inconsistency in OLS. Thus, the methods in Chapter 4 are not appropriate.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

68

Stock/Watson - Introduction to Econometrics - Second Edition

13.11. Following the notation used in Chapter 13, let π1i denote the coefficient on state sales tax in the “first stage” IV regression, and let −β1i denote cigarette demand elasticity. (In both cases, suppose that income has been controlled for in the analysis.) From (13.11) p

βˆ TSLS →

cov( β1i , π 1i ) cov( β1i , π 1i ) E ( β1i π 1i ) Average Treatment Effect + , E ( β1i ) + = = E (π 1i ) E (π 1i ) E (π 1i )

where the first equality uses the uses properties of covariances (equation (2.34)), and the second equality uses the definition of the average treatment effect. Evidently, the local average treatment effect will deviate from the average treatment effect when cov( β1i , π 1i ) ≠ 0. As discussed in Section 13.6, this covariance is zero when β1i or π1i are constant. This seems likely. But, for the sake of argument, suppose that they are not constant; that is, suppose the demand elasticity differs from state to state (β1i is not constant) as does the effect of sales taxes on cigarette prices (π1i is not constant). Are β1i and π1i related? Microeconomics suggests that they might be. Recall from your microeconomics class that the lower is the demand elasticity, the larger fraction of a sales tax is passed along to consumers in terms of higher prices. This suggests that β1i and π1i are positively related, so that cov( β1i , π 1i ) > 0. Because E(π1i) > 0, this suggests that the local average treatment effect is greater than the average treatment effect when β1i varies from state to state.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 14 Introduction to Time Series Regression and Forecasting 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) = β0 + β1E(Yt–1) + E(ut), but E(ut) = 0 and E(Yt) = E(Yt–1). Thus E(Yt) = β0 + β1E(Yt), and solving for E(Yt) yields the result.

14.2.

(a) The statement is correct. The monthly percentage change in IP is

IPt − IPt −1 × 100 which can IPt −1

 IP  be approximated by [ln( IPt ) − ln( IPt −1 )] × 100 = 100 × ln  t  when the change is small.  IPt −1   IP  Converting this into an annual (12 month) change yields 1200 × ln  t  .  IPt −1  (b) The values of Y from the table are Date

2000:7

2000:8

2000:9

2000:10

2000:11

2000:12

IP Y

147.595

148.650 8.55

148.973 2.60

148.660 −2.52

148.206 −3.67

146.300 −7.36

The forecasted value of Yt in January 2001 is

1.377 + [0.318 × (−7.36)] + [0.123 × (−3.67)] Yˆ= t |t −1 + [0.068 × (−2.52)] + [0.001 × (2.60)] = −1.58.

−0.054 = −1.0189 with an absolute value less than 1.96, so the 0.053 coefficient is not statistically significant at the 5% level. (d) For the QLR test, there are 5 coefficients (including the constant) that are being allowed to break. Compared to the critical values for q = 5 in Table 14.5, the QLR statistic 3.45 is larger than the 10% critical value (3.26), but less than the 5% critical value (3.66). Thus the hypothesis that these coefficients are stable is rejected at the 10% significance level, but not at the 5% significance level. (c) The t-statistic on Yt–12 is t =

©2011 Pearson Education, Inc. Publishing as Addison Wesley

70

Stock/Watson - Introduction to Econometrics - Second Edition

(e) There are 41 × 12 = 492 number of observations on the dependent variable. The BIC and AIC ln T  SSR ( p )  are calculated from the formulas = and BIC( p ) ln   + ( p + 1) T  T  2  SSR ( p )  = AIC ( p ) ln   + ( p + 1) .

T  T  AR Order ( p)

1

2

3

4

5

6

SSR (p)  SSR ( p )  ln    T 

29175 4.0826

28538 4.0605

28393 4.0554

28391 4.0553

28378 4.0549

28317 4.0527

ln T T 2 ( p + 1) T BIC AIC

0.0252

0.0378

0.0504

0.0630

0.0756

0.0882

0.0081

0.0122

0.0163

0.0203

0.0244

0.0285

4.1078 4.0907

4.0983 4.0727

4.1058 4.0717

4.1183 4.0757

4.1305 4.0793

4.1409 4.0812

( p + 1)

The BIC is smallest when p = 2. Thus the BIC estimate of the lag length is 2. The AIC is smallest when p = 3. Thus the AIC estimate of the lag length is 3.

14.3.

(a) To test for a stochastic trend (unit root) in ln(IP), the ADF statistic is the t-statistic testing the hypothesis that the coefficient on ln(IPt – 1) is zero versus the alternative hypothesis that the −0.018 = −2.5714. coefficient on ln(IPt – 1) is less than zero. The calculated t-statistic is t = 0.007 From Table 14.4, the 10% critical value with a time trend is −3.12. Because −2.5714 > −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).

14.4.

(a) The critical value for the F-test is 2.372 at a 5% significance level. Since the Grangercausality F-statistic 2.35 is less than the critical value, we cannot reject the null hypothesis that interest rates have no predictive content for IP growth at the 5% level. The Grangercausality statistic is significant at the 10% level. (b) The Granger-causality F-statistic of 2.87 is larger than the 5% critical value, so we conclude at the 5% significance level that IP growth helps to predict future interest rates.

14.5.

(a) E[(W − c) 2 ]= E{[W − µW ) + ( µW − c)]2 } = E[(W − µW ) 2 ] + 2 E (W − µW )( µW − c) + ( µW − c) 2 =σ W2 + ( µW − c) 2 .

(b) Using the result in part (a), the conditional mean squared error E[(Yt − f t −1 ) 2 |Yt −1 , Yt − 2 ,...] =σ t2|t −1 + (Yt |t −1 − f t −1 ) 2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

71

2 with the conditional variance σ= E[(Yt − Yt |t −1 ) 2 ]. This equation is minimized when the t |t −1

second term equals zero, or when f t −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 − β 0 − β1Yt − 2 − β 2Yt −3 −  − β 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.

14.6.

This exercise requires a Monte Carlo simulation on spurious regression. The answer to (a) will depend on the particular “draw” from your simulation, but your answers should be similar to the ones that we found. (b) When we did these simulations, the 5%, 50% and 95% quantiles of the R2 were 0.00, 0.19, and 0.73. The 5%, 50% and 95% quantiles of the t-statistic were −12.9, −0.02 and 13.01. Your simulations should yield similar values. In 76% of the draws the absolute value of the tstatistic exceeded 1.96. (c) When we did these simulations with T = 50, the 5%, 50% and 95% quantiles of the R2 were 0.00, 0.16 and 0.68. The 5%, 50% and 95% quantiles of the t-statistic were −8.3, −0.20 and 7.8. Your simulations should yield similar values. In 66% of the draws the absolute value of the t-statistic exceeded 1.96.

When we did these simulations with T = 200, the 5%, 50% and 95% quantiles of the R2 were 0.00, 0.17, and 0.68. The 5%, 50% and 95% quantiles of the t-statistic were −16.8, −0.76 and 17.24. Your simulations should yield similar values. In 83% of the draws the absolute value of the t-statistic exceeded 1.96.

The quantiles of the R2 do not seem to change as the sample size changes. However the distribution of the t-statistic becomes more dispersed. In the limit as T grows large, the fraction of the t-statistics that exceed 1.96 in absolute values seems to approach 1.0. (You t − statistic might find it interesting that has a well-behaved limiting distribution. This is T consistent with the Monte Carlo presented in this problem.)

14.7.

(a) From Exercise (14.1) E(Yt) = 2.5 + 0.7E(Yt – 1) + E(ut), but E(Yt) = E(Yt – 1) (stationarity) and E(ut) = 0, so that E(Yt) = 2.5/(1 − 0.7). Also, because Yt = 2.5 + 0.7Yt – 1 + ut, var(Yt) = 0.72var(Yt – 1) + var(ut) + 2 × 0.7 × cov(Yt – 1, ut). But cov(Yt – 1, ut) = 0 and var(Yt) = var(Yt – 1) (stationarity), so that var(Yt) = 9/(1 − 0.72) = 17.647.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

72

Stock/Watson - Introduction to Econometrics - Second Edition

(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 ) = 0.7σ Y2 = 0.7 × 17.647 = 12.353.

The 2nd autocovariance is cov(Yt , Yt −= cov[(1 + 0.7)2.5 + 0.72 Yt − 2 + ut + 0.7ut −1 , Yt − 2 ] 2) = 0.72 var(Yt − 2 ) + cov(ut + 0.7ut −1 , Yt − 2 ) = 0.72 σ Y2 0.72 × 17.647 = 8.6471. =

(c) The 1st autocorrelation is

corr (Yt = , Yt −1 )

cov(Yt , Yt −1 ) 0.7σ Y2 = = 0.7. σ Y2 var(Yt ) var(Yt −1 )

The 2nd autocorrelation is

corr (Y= t , Yt − 2 )

cov(Yt , Yt − 2 ) 0.7 2 σ Y2 = = 0.49. σ Y2 var(Yt ) var(Yt − 2 )

(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.

14.8.

Because E(ut |Febt, Mart,… Dect) = 0, E(Yt | Febt, Mart,… Dect) = β0 + β1Febt + β2Mart + ⋅⋅⋅ + β11Dect. For observations in January, all of the regressors are equal to zero, thus µJan = β0. For observations in February, Febt = 1 and the other regressors equal 0, so that µFeb = β0 + β1. Similar calculations apply for the other months.

14.9.

(a) E(Yt) = β0 + E(et) + b1E(et–1) + ⋅⋅⋅ + bqE(et–q) = β0 [because E(et) = 0 for all values of t].

var(et ) + b12 var(et −1 ) +  + bq2 var(et − q ) (b) var(Y= t) +2b1 cov(et , et −1 ) +  + 2bq −1bq cov(et − q +1 , et − q ) = σ e2 (1 + b12 +  + bq2 ) where the final equality follows from var(et) = σ e2 for all t and cov(et, ei) = 0 for i≠ t. (c) Yt = β0 + et + b1et–1 + b2et – 2 + ⋅⋅⋅ + bqet – q and Yt–j = β0 + et – j + b1et – 1 – j + b2et – 2 – j + ⋅⋅⋅ q q + bqet – q – j and cov(Yt, Yt – j) = ∑ = k 0 ∑ = m 0 bk bm cov(et − k , et − j − m ), where b0 = 1. Notice that cov(et–k, et–j–m) = 0 for all terms in the sum.

(

)

Yt ) σ e2 1 + b12 , cov(Yt , Yt − j ) = σ e2b1 , and cov(Yt , Yt − j ) = 0 for j > 1. (d) var(= 14.10. A few things to note: first, computing the QLR using 25% trimming will result in a statistic that is at least as large as just choosing one date (the usual F statistic) and a statistic that can be no larger than the QLR with 15% trimming (because with the test with 15% trimming chooses that maximum over a larger number of statistics). Thus the 25%-trimming critical values will be larger than the

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

73

critical values for the F statistic and smaller than the critical values for the 15%-trimming QLR statistic. (a) The F statistic is larger than the 5% CV for 15% trimming (3.66), so it must be larger than the critical value for 25% trimming (which must be less than 3.66), so the null is rejected. (b) The F statistic is smaller than the 5% critical value from the F5,∞ distribution (2.21), so that it must be smaller than the critical value with 25% (which must be greater than 2.21), so the null is not rejected. (c) This is the intermediate case. Critical values for the 25% trimming would have to be computed.

14.11. Write the model as Yt − Yt – 1 = β0 + β1(Yt – 1 − Yt – 2) + ut. Rearranging yields Yt = β0 + (1 + β1)Yt – 1 − β1Yt – 2 + ut.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 15

Estimation of Dynamic Causal Effects

15.1.

(a) See the table below. βi is the dynamic multiplier. With the 25% oil price jump, the predicted effect on output growth for the ith quarter is 25βi percentage points.

Period ahead (i)

Dynamic multiplier (βi)

Predicted effect on output growth (25βi)

95% confidence interval 25 × [βi ± 1.96SE (βi)]

0 1 2 3 4 5 6 7 8

− 0.055 − 0.026 − 0.031 − 0.109 − 0.128 0.008 0.025

−1.375 − 0.65 − 0.775 −2.725 −3.2 0.2 0.625

− 0.019 0.067

− 0.475 1.675

[−4.021, 1.271] [−3.443, 2.143] [−3.127, 1.577] [−4.783, −0.667] [−5.797, −0.603] [−1.025, 1.425] [−1.727, 2.977] [−2.386, 1.436] [−0.015, 0.149]

(b) The 95% confidence interval for the predicted effect on output growth for the ith quarter from the 25% oil price jump is 25 × [βi ± 1.96SE (βi)] percentage points. The confidence interval is reported in the table in (a). (c) The predicted cumulative change in GDP growth over eight quarters is 25 × (−0.055 − 0.026 − 0.031 − 0.109 − 0.128 + 0.008 + 0.025 − 0.019) = −8.375%. (d) The 1% critical value for the F-test is 2.407. Since the HAC F-statistic 3.49 is larger than the critical value, we reject the null hypothesis that all the coefficients are zero at the 1% level.

15.2.

(a) See the table below. βi is the dynamic multiplier. With the 25% oil price jump, the predicted change in interest rates for the ith quarter is 25βi.

Period ahead (i) 0 1 2 3 4 5 6 7 8

Dynamic multiplier (βi)

0.062

Predicted change in interest rates (25βi)

95% confidence interval 25 × [βi ± 1.96SE (βi)]

1.55

0.048

1.2

− 0.014 − 0.086 − 0.000 0.023

− 0.35 −2.15 0 0.575

− 0.010 − 0.100 −0.014

− 0.25 −2.5 −0.35

©2011 Pearson Education, Inc. Publishing as Addison Wesley

[−0.655, 3.755] [− 0.466, 2.866] [−1.722, 1.022] [−10.431, 6.131] [−2.842, 2.842] [−2.61, 3.76] [−2.553, 2.053] [−4.362, −0.638] [−1.575, 0.875]

Solutions to End-of-Chapter Exercises

75

(b) The 95% confidence interval for the predicted change in interest rates for the ith quarter from the 25% oil price jump is 25 × [βi ± 1.96SE (βi)]. The confidence interval is reported in the table in (a). (c) The effect of this change in oil prices on the level of interest rates in period t + 8 is the price change implied by the cumulative multiplier: 25 × (0.062 + 0.048 − 0.014 − 0.086 − 0.000 + 0.023 − 0.010 − 0.100 − 0.014) = −2.275. (d) The 1% critical value for the F-test is 2.407. Since the HAC F-statistic 4.25 is larger than the critical value, we reject the null hypothesis that all the coefficients are zero at the 1% level.

15.3.

The dynamic causal effects are for experiment A. The regression in exercise 15.1 does not control for interest rates, so that interest rates are assumed to evolve in their “normal pattern” given changes in oil prices.

15.4.

When oil prices are strictly exogenous, there are two methods to improve upon the estimates.

The first method is to use OLS to estimate the coefficients in an ADL model, and to calculate the dynamic multipliers from the estimated ADL coefficients. The second method is to use generalized least squares (GLS) to estimate the coefficients of the distributed lag model.

15.5.

Substituting

∆ X t + X t −1 Xt = = ∆ X t + ∆ X t −1 + X t − 2 =  = ∆ X t + ∆ X t −1 +  + ∆ X t − p +1 + X t − p into Equation (15.4), we have

Yt = β 0 + β1 X t + β 2 X t −1 + β 3 X t − 2 +  + β r +1 X t − r + ut = β 0 + β1 (∆ X t + ∆ X t −1 +  + ∆ X t − r +1 + X t − r ) + β 2 (∆ X t −1 +  + ∆ X t − r +1 + X t − r ) +  + β r (∆ X t − r +1 + X t − r ) + β r +1 X t − r + ut =β 0 + β1∆ X t + ( β1 + β 2 )∆ X t −1 + ( β1 + β 2 + β3 )∆ X t − 2 +  + ( β1 + β 2 +  + β r )∆ X t − r +1 + ( β1 + β 2 +  + β r + β r +1 ) X t − r + ut .

Comparing the above equation to Equation (15.7), we see δ0 = β0, δ1 = β1, δ2 = β1 + β2, δ3 = β1 + β2+ β3,…, and δr + 1 = β1 + β2 +⋅⋅⋅+ βr + βr + 1.

15.6.

(a) Write

var(u= φ12 var(ut −1 ) + var(ut ) + 2φ1 cov(ut −1 , ut ) t) = φ12 var(ut ) + σ u2 where the second equality follows from stationarity (so that var(ut) = var(ut–1) and cov(ut–1, ut ) = 0.

The result follows by solving for var(ut). The result of var(Xt) is similar.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

76

Stock/Watson - Introduction to Econometrics - Second Edition

(b) Write = ut , ut −1 ) φ1 var(ut −1 ) + cov(ut −1 , ut ) cov( = φ1 var(ut )

showing the result for j = 1. For j = 2, write = cov(ut , ut − 2 ) φ1 cov(ut −1 , ut − 2 ) + cov(ut − 2 , ut ) = φ1 cov(ut , ut −1 ) = φ12 var(ut )

and similarly for other values of j. The result for X is similar. cov(ut , ut − j ) cov(ut , ut − j ) = from stationarity. The result then follows from (c) cor(ut, ut−j) = var(ut ) var(ut ) var(ut − j ) (b). The result for X is similar. (d) vt = (Xt − µX)ut

E[( X t − µ X ) 2 ut2 ] = E[( X t − µ X ) 2 ]E[ut2 ] = σ X2 σ u2 , where the second equality follows (i) E (vt2 ) = because Xt and ut are independent. (ii) cov(vtvt – j) = E[(Xt − µX)( Xt – j − µX)utut – j] = E[(Xt − µX)( Xt – j − µX)]E[utut – j] = 1 + γ 1φ1 . γ 1j var( X t )φ1j var(ut ), so that cor(vt, vt – j) = γ 1jφ1j . Thus f∞ = 1 + 2 ∑i∞=1 (γ 1φ1 )i = 1 − γ 1φ 15.7.

∑i∞ 0 φ1i ut −i Write ut == (a) Because E (ui | Xt ) = 0 for all i and t, E(ui |Xt) = 0 for all i and t, so that Xt is strictly exogenous. (b) Because E (ut − j | ut +1 ) = 0 for j ≥ 0, Xt is exogenous. However E(ut+1 | ut +1 ) = ut +1 so that Xt is not strictly exogenous.

15.8.

(a) Because Xt is exogenous, OLS is consistent. (b) The GLS estimator regresses Yt − φ1Yt – 1 onto Xt − φ1Xt – 1. The error term in this regression is ut . Because Xt = ut +1 , Xt − φ1 Xt–1 = ut +1 − φ1 ut , which is correlated with the regression error. Thus the GLS estimator will be inconsistent. σ 2φ cov( X t − φ1 X t −1 , ut ) φ1 p (c) βˆ1 → β1 + = β1 − 2 u 1 2 = β1 − var( X t − φ1 X t −1 ) (1 + φ12 ) σ u (1 + φ1 )

15.9.

(a) This follows from the material around equation (3.2). (b) Quasi-differencing the equation yields Yt – φ1Yt – 1 = (1 − φ1)β0 + ut , and the GLS estimator of (1 − φ1)β0 is the mean of Yt – φ1Yt – 1= = T1−1 ∑Tt 2 (Yt − φ1Yt −1 ) . Dividing by (1−φ1) yields the GLS estimator of β0. (c) This is a rearrangement of the result in (b). (d) Write βˆ =1 ∑T Y =1 (Y + Y ) + T −1 1 ∑T −1 Y , so that βˆ − βˆ GLS =

0 1 = t 1= t T t 2 T T T T −1

1 1 1−φ T −1

t

0

(YT − Y1 ) and the variance is seen to be proportional to

0

1 .

T2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

1 T

(YT + Y1 ) − T1 T1−1 ∑Tt =−21 Yt −

Solutions to End-of-Chapter Exercises

Multipliers

15.10.

Lag

Multiplier

Cumulative Multiplier

0 1 2 3 4 5

2 0 0 0 0 0

2 2 2 2 2 2

The long-run cumulative multiplier = 2.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

77

Chapter 16

Additional Topics in Time Series Regression

16.1.

Yt follows a stationary AR(1) model, Yt = (Yt ) µY E= β 0 + β1Yt −1 + ut . The mean of Yt is= and E (ut |Yt ) = 0.

β0 , 1 − β1

(a) The h-period ahead forecast of Yt , Yt + h|t = E (Yt + h |Yt , Yt −1 ,), is Yt + h|t = E (Yt + h |Yt , Yt −1 ,) = E ( β 0 + β1Yt + h −1 + ut |Yt , Yt −1 ,) = 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 =  = (1 + β1 +  + 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 Y = ∑ δ [µ + β (Y − µ )] i

t + i |t =i 0=i 0 ∞

i

Y

i 1

t

Y

∞

= µY ∑ δ i + (Yt − µY )∑ ( β1δ )i

=i 0=i 0

= 16.2.

µY Yt − µY . + 1 − δ 1 − β1δ

(a) Because R1t follows a random walk (R1t = R1t–1 + ut), the i-period ahead forecast of R1t is R1= R1t += R1t +=  = R1t . Thus t + i |t i −1|t i − 2|t

1 k 1 k R1t + i|t + et = ∑ ∑ R1t + et = R1t + et . k i 1= k i1 = Rkt =

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

79

(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 ∆R1= 0.5∆ + ut , ∆ R1t is stationary but R1t is not stationary. R1t =1.5 R1t −1 − 0.5 R1t − 2 + t ut , an AR(2) process with a unit autoregressive root. That is, R1t is I (1) . The i-period ahead forecast of ∆R1t is 0.5∆R1t + i −1|t = 0.52 ∆R1t + i − 2|t = 0.5i ∆R1t . ∆ R1t + i|t =  =

The i-period ahead forecast of R1t is

= R1t + i t R1t + i −1|t + ∆R1t + i|t = R1t + i − 2|t + ∆R1t + i −1|t + ∆R1t + i|t =  = R1t + ∆R1t +1|t +  + ∆R1t + i|t = R1t + (0.5 +  + 0.5i )∆R1t 0.5(1 − 0.5i ) = ∆R1t .

R1t + 1 − 0.5 Thus 1 k 1 k 1 [ R1t + (1 − 0.5i )∆R1t ] + et R e = + ∑ ∑ t t +i t k k =i 1 =i 1 = R1t + φ∆R1t + et .

Rkt =

k i 1 where φ =∑ φ R1t + et . Thus Rkt and R1t are cointegrated. The i =1 (1 − 0.5 ). Thus Rkt − R1t =∆ k cointegrating coefficient is 1. (d) When R1t = 0.5R1t – 1 + ut, R1t is stationary and does not have a stochastic trend. k i R1t + i|t = 0.5i R1t , so that, = Rkt θ R1t + et , where θ = 1k ∑i =1 0.5 . Since R1t and et are I (0), then

Rkt is I (0).

16.3.

2 1.0 + 0.5ut2−1. ut follows the ARCH process with mean E (ut) = 0 and variance σ= t

(a) For the specified ARCH process, ut has the conditional mean E (ut |ut −1 ) = 0 and the conditional variance.

2 var (ut | ut −= 1.0 + 0.5ut2−1.

1 ) σ= t

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 + 0.5var (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 + = ut ) 1.0 = / 0.5 2.

0.5var(ut) which implies var(

©2011 Pearson Education, Inc. Publishing as Addison Wesley

80

Stock/Watson - Introduction to Econometrics - Second Edition

(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 ≤ u= Pr  ≤ ≤  t ≤ 3)  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  ≤ t ≤ Pr (−3 ≤= ut ≤ 3) Pr    1.732 σ t 1.732  = Φ(1.732) − Φ(−1.732) = 0.9584 − 0.0416 = 0.9168.

16.4.

Yt follows an AR(p) model Yt = β 0 + 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 ( β 0 + β1Yt + h −1 +  + β pYt + h − p + ut + h |Yt , Yt −1 ,) = β 0 + β1 E (Yt + h −1|Yt , Yt −1 ,) +  + β p E (Yt + h − p |Yt , Yt −1 ,) + E (ut + h |Yt , Yt −1 ,) = β 0 + β1Yt + h −1|t +  + β pYt + h − p|t .

16.5.

Because Yt= Yt − Yt −1 + Yt −1= Yt −1 + ∆Yt , T

T 2 t =t 1 =t 1

T T 2 2 t t −1 =t 1 =t 1

T

Y ∑ (Y + ∆= Y ) ∑ Y + ∑ (∆Y ) + 2∑ Y ∆Y . ∑= t −1

t

2

=t 1

t −1

t

So T T 1 T 1 1 T  Yt −1∆Yt = ×  ∑ Yt 2 − ∑ Yt 2−1 − ∑ (∆Yt ) 2  . ∑ T t1 T= 2  t 1 =t 1 =t 1 = 

T 2 2 Note that ∑Tt = 1 Yt − ∑ t = 1 Yt −1 =

Y because Y = 0. Thus: ( ∑ Y + Y ) − (Y + ∑ Y ) = Y − Y = T −1 2 t= 1 t

2 T

2 0

T −1 2 t= 1 t

2 T

2 0

T 1 T 1 1  Yt −1∆Yt = × YT2 − ∑ (∆Yt ) 2  ∑ T t 1= T 2 = t 1  2  1  YT  1 T 2 = − ∆ Y ( ) .   ∑ t 2  T  T t =1 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

2 T

0

Solutions to End-of-Chapter Exercises

16.6.

81

(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 1

∑ Y ∆Y Y ∆Y ∑ YX ∑= T 16.7. βˆ = . Following the hint, the numerator is the same = 1 X Y ∆ ) ( ∑ ∑ ∑ ( ∆Y ) T

T

t t t 1 t =t 1 = T T 2 t t 1 =t 1 =

t +1

2

t +1

T

t +1

t =1 t

T

T

2

t +1

t =1

2

d σu 1 expression as (16.21) (shifted forward in time 1 period), so that= ( χ12 − 1). The ∑Tt 1 Yt ∆Yt +1 → T 2

p

2 2 2 T 1 denominator is T1 ∑Tt = 1 ( ∆Yt +1 ) =T ∑ t = 1 ut +1 → σ u by the law of large numbers. The result follows directly.

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 + γ 11 X t − 2 ) + γ 11 ( β 21Yt − 2 + γ 21 X t − 2 ) = [ β11β11 + γ 11β 21 ]Yt − 2 + [ β11γ 11 + γ 11γ 21 ] X t − 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.

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 +  + α p ut2− p ) = α 0 + α1 E ( ut2−1 ) + α 2 E ( ut2− 2 ) +  + α p E ( ut2− p ) = α 0 + α1 E ( ut2 ) + α 2 E ( ut2 ) +  + α p E ( ut2 )

©2011 Pearson Education, Inc. Publishing as Addison Wesley

82

Stock/Watson - Introduction to Econometrics - Second Edition

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.

16.10. Write ∆Yt = θ∆Xt + ∆v1t and ∆Xt = v2t; also v1t–1 = Yt–1 − θXt–1. Thus ∆Yt = −(Yt−1 − θXt–1) + u1t and ∆Xt = u2t, with u1t = v1t + θv2t and u2t = v2t.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 17

The Theory of Linear Regression with One Regressor

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

1

i

i

2

βˆ1RLS , the restricted least squares estimator of β1, minimizes the sum of squared errors. That is, βˆ1RLS satisfies the first order condition for the minimization which requires the differential

of the sum of squared errors with respect to b1 equals zero: n

0. ∑ 2(Y − b X )(− X ) = i

i =1

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: X (β X + u ) ∑ X i ui . = β + ∑ X ∑ X i2

n n n RLS 1 i = i 1= i i i 1 i i =i 1 1 1 2 2 n n n i i 1 i =i 1 =i 1 =

βˆ =

∑ XY ∑ = ∑ X

Thus

 ∑ n X i ui   ∑in 1 X i E (ui | X 1, , X n )  βˆ1RLS ) = β1 + E  i n1 = β β1 , E (= E = +   = 1 X i2  ∑in 1 X i2 =  ∑i 1 =   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 , , X n ) =0 ∑in=1 X i2

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.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

84

Stock/Watson - Introduction to Econometrics - Second Edition

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 . = n 2 1 ∑in 1 = X i2 = n ∑i 1 X i

βˆ1RLS= − β1

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 i ui ) E= [ X i E (ui | X i )] 0. By assumption 2, vi is i.i.d. By 1 assumption 3, var(vi) is finite. Let v = ∑in=1 X i ui , then σ v2 = σ v2 /n. Using the central limit n

theorem, the sample average = v /σ v

1

n

∑ v → N (0, 1)

σ v 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

−β ) n ( βˆ RLS= 1

u

∑in=1 X i ui d  var( X i ui )  → N  0, . n 2 1 E( X 2 )   n ∑ i =1 X i

1 n

(c) βˆ1RLS is a linear estimator: ∑in=1 X iYi Xi n where . βˆ1RLS = = aY , = ai ∑ 2 n n i =1 i i ∑i 1 = Xi ∑i 1 X i2 =

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

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

85

βˆ1RLS is conditionally unbiased because   ∑n X u , X n E  β1 + i =n1 i 2 i | X 1 , , X n  E ( βˆ1RLS | X 1 ,= ∑i =1 X i   n ∑ X u  β1 + E  i =n1 i 2 i | X 1 ,, X n  = β1. =  ∑i =1 X i 

The final equality used the fact that

∑ X u  ∑ X E (u | X 1 , , X n ) E = | X , , X  = 0 ∑ X i2  ∑ X 

n n =i 1 = i i i 1 i i n 1 n n 2 =i 1 = i i 1

because the observations are i.i.d. and E (ui |Xi) = 0. (d) The conditional variance of βˆ RLS , given X1,…, Xn, is 1

  ∑n X u var( βˆ1RLS | X1, , X n ) var  β1 + i =n1 i 2 i | X1 ,, X n  = ∑i =1 X i   2 n ∑ X var(u | X ,, X n ) = i =1 i n i 2 12 (∑i =1 X i ) = =

∑in=1 X i2σ u2 (∑in=1 X i2 )2

σ u2

∑in=1 X i2

.

(e) The conditional variance of the OLS estimator β̂1 is var( βˆ1 |X1 ,, X n ) =

σ u2

∑in=1 ( X i − X )2

.

Since n

n n 2 2 i i =i 1 =i 1

n 2 i =i 1

n 2 2 i =i 1

∑ ( X − X ) = ∑ X − 2 X ∑ X + nX = ∑ X − nX < ∑ X ,

=i 1

2 i

the OLS estimator has a larger conditional variance: var( β1 |X 1 , , X n ) > var( βˆ1RLS |X 1 , , X n ).

The restricted least squares estimator βˆ RLS is more efficient.

1

(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

©2011 Pearson Education, Inc. Publishing as Addison Wesley

86

Stock/Watson - Introduction to Econometrics - Second Edition

Using the conditional mean and conditional variance of βˆ1RLS derived in parts (c) and (d) respectively, the sampling distribution of βˆ RLS , conditional on X1,…, Xn, is 1



βˆ1RLS ~ N  β1 , 

∑

 .

X 

σ u2

n i =1

2 i

(g) The estimator ∑ Y ∑ ( β X + ui ) = ∑n u = = β1 + ni 1 i ∑ X ∑ X = ∑i 1 X i

n n =i 1 = i i 1 1 i 1 n n =i 1 = i i 1 i

β=

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 ∑ var(ui | X 1 , , X n ) = i =1 (∑in=1 X i ) 2 =

nσ u2 . (∑in=1 X i ) 2

The difference in the conditional variance of β1 and βˆ1RLS is nσ u2 σ u2 var( β1 | X 1 , , X n ) − var( βˆ1RLS | X 1 , , X n ) = . − (∑in 1 = X i ) 2 ∑in 1 X i2 =

In order to prove var( β1 | X 1 , , X n ) ≥ var( βˆ1RLS | X 1 , , X n ), we need to show 1 n ≥ n 2 2 n ( ) X ∑ ∑ i i 1 Xi =i 1 =

or equivalently 2

n  n  n∑ X i2 ≥  ∑ X i  . =i 1 = i1 

This inequality comes directly by applying the Cauchy-Schwartz inequality 2

n n  n  2 2  ∑ (ai ⋅ bi )  ≤ ∑ ai ⋅ ∑ 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 i1  = i1 =i 1 =i 1

That is nΣin 1 = X i2 ≥ (Σ nx 1 X i ) 2, or var( β1 | X 1 , , X n ) ≥ var( βˆ1RLS | X 1 , , X n ). = Note: because β1 is linear and conditionally unbiased, the result var( β | X , , X ) ≥ var( βˆ RLS | X , , X ) follows directly from the Gauss-Markov theorem.

1

1

n

1

1

n

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

17.2.

87

The sample covariance is 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 = s XY

n 1  n ∑ ( X i − µ X )(Yi − µY ) − ∑ ( X − µ X )(Yi − µY ) 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 )  − n − 1 ( X − µ X )(Y − µY ) n − 1  n i =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 is easy to p p µY so µ X and Y → see the second term converges in probability to zero because X → p ( 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

1 n p σ XY .

E[( X i − µ X )(Yi − µY )] = ( X i − µ X )(Yi − µY ) → ∑ n i =1 Also, n−n 1 → 1, so the first term for sXY converges in probability to σXY. Combining results on the p

two terms for sXY , we have s XY → σ XY .

©2011 Pearson Education, Inc. Publishing as Addison Wesley

88

Stock/Watson - Introduction to Econometrics - Second Edition

17.3.

(a) Using Equation (17.19), we have ∑in=1 ( X i − X )ui n 2 n ∑ i =1 ( X i − X )

1

n ( βˆ1 − β1 ) = n n1 1

= nn

∑in=1[( X i − µ X ) − ( X − µ X )]ui n 2 1 n ∑ i =1 ( X i − X )

∑ ( X − µ )u

(X − µ )

∑ ui

n n 1 1 = i 1= i X i X i 1 n n

=

∑ (X − X)

−

∑ ( X i − X )2

n n 2 1 1 = i 1= i i 1 n n

v

∑

(X − µ )

∑ ui

n n 1 1 = i 1= i X i 1 n n

=

∑ (X − X)

− 2

∑ ( X i − X )2

n n 1 1 = i 1= i i 1 n n

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 ) =

1 n

σu

∑in=1 ui d → N (0, 1).

σu

p p The law of large numbers implies X → µ X 2 , or X − µ X → 0. By the consistency 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,

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 d → N (0, 1).

σv

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

89

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

∑ ( X t − X )2 n i =1

d → N (0, 1),

σ X2

or equivalently 1 n

∑in=1 vi

 var(vi )  d .

N  0, → 2  ∑ (Xi − X )  [var( X i )]  1 n

2

n i =1

Thus (X − µ )

n n 1 1 = i 1= i X i 1 i n n 1 1 n 2 n 2 1 1 = i 1= i i 1 i n n

= n ( βˆ − β )

∑ v

∑ (X − X )

−

∑ u

∑ (X − X )

 var(vi )  d → N  0, 2   [var( X i )] 

since the second term for 17.4.

n ( βˆ1 − β1 ) converges in probability to zero as shown in part (b).

) an S n where a= (a) Write ( βˆ1 − β1= n

1 n

and S= n

d S where n ( Bˆ1 − β1 ). Now, an → 0 and S n →

d 0 × S . Thus Pr (|βˆ1 − β1| > δ ) → 0 for S is distributed N (0, a2). By Slutsky’s theorem an S n → p any δ > 0, so that βˆ − β → 0 and βˆ is consistent.

1

(b) We have (i)

2 u 2 u

s

1

1

p

→ 1 and (ii) g ( x) = x is a continuous function; thus from the continuous

σ mapping theorem

su2 su p = → 1.

2

σu

σu

17.5.

Because E(W 4) = [E(W2)]2 + var(W2), [E(W2)]2 ≤ E (W 4) < ∞. Thus E(W2) < ∞.

17.6.

Using the law of iterated expectations, we have E= ( βˆ1 ) E[ E ( βˆ1 | X 1 , ,= X n)] E= ( β1 ) β1 .

©2011 Pearson Education, Inc. Publishing as Addison Wesley

90

Stock/Watson - Introduction to Econometrics - Second Edition

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 (ui | X i ) f (u j | X j ). f (Xi ) f (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 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 ,  X n ) = f (ui u j | X i , X j )

and the result then follows from part (b).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

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 = β 0 X 0i + β1 X 1i + ui where

= Yi

Yi = , X 0i θ 0 + θ1| X i |

= X 1i

Xi = , and u θ 0 + θ1 | X i |

1 θ 0 + θ1 | X i | ui

θ 0 + θ1| X i |

.

(c) Using equations (17.2) and (17.19), we know the OLS estimator, βˆ1 , is ∑ ( X − X )(Y − Y ) ∑ ( X i − X ) ui . = β + ∑ (X − X) ∑ ( X i − X )2

n n =i 1 = i i i 1 1 1 2 n n =i 1 = i i 1

βˆ=

As a weighted average of normally distributed variables ui , βˆ1 is normally distributed with mean E ( βˆ ) = β . The conditional variance of βˆ , given X1,…, Xn, is 1

1

1

  ∑ n ( X − X ) ui = X n ) var  β1 + i =n1 i var ( βˆ1 | X1 ,..., | X1 ,..., X n  2 ∑i =1 ( X i − X )   n 2 ∑ ( X − X ) var (ui | X1 ,..., 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 (θ 0 + 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



∑in=1 ( X i − X ) 2 (θ 0 + 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

©2011 Pearson Education, Inc. Publishing as Addison Wesley

91

92

Stock/Watson - Introduction to Econometrics - Second Edition

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

0, ∑ 2(Y − b X − b X )(− X ) = 0

i

i =1

0i

1

1i

0i

n

0. ∑ 2(Y − b X − b X )(− X ) = i

i =1

0

0i

1

1i

1i

Solving for b1 gives the WLS estimator

βˆ1WLS =

−Q01S0 + Q00 S1 Q00Q11 − Q012

∑ X X , Q = ∑ X X , Q = ∑ X X , S = ∑ X 0iYi , where Q = and S = ∑ n X Y . Substituting Yi = β 0 X 0i + β1 X 0i + ui yields

n n n n 00 0i 0i 01 0 i 1i 11 1i 1i 0 = i 1 = i 1 = i 1= i 1 1

i =1

1i

βˆ1WLS= β1 +

−Q01Z 0 + Q00 Z1 Q00Q11 − Q012

∑ X u , and Z = ∑ X 1i ui or where Z =

n n = 0 0i i 1 i 1= i 1

∑ n (Q00 X 1i − Q01 X 0i )ui .

Q00Q11 − Q012

βˆ1WLS − β1 =i =1

From this we see that the distribution of βˆ1WLS | X 1 ,... X n is N ( β1 , σ β2ˆWLS ), where 1

σ β2ˆ

WLS 1

=

σ u2 ∑in=1 (Q00 X 1i − Q01 X 0i ) 2 (Q00Q11 − Q012 ) 2

=

Q002 Q11 + Q012 Q00 − 2Q00Q012 (Q00Q11 − Q012 ) 2

=

Q00 Q00Q11 − Q012

where the first equality uses the fact that the observations are independent, the second uses σ u2 = 1, the definition of Q00, Q11, and Q01, and the third is an algebraic simplification.

17.9.

We need to prove

1 n p [( X i − X ) 2 uˆi2 − ( X i − µ X ) 2 ui2 ] → 0. ∑ n i =1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

93

Using the identity X = µ X + ( X − µ X ), 1 n 1 n 1 n [( X i − X ) 2 uˆi2 − ( X i − µ X ) 2 ui2 ] =( X − µ X ) 2 ∑ uˆi2 − 2( X − µ X ) ∑ ( X i − µ X )uˆi2 ∑ ni1 n i 1= ni1 = = n 1 + ∑ ( X i − µ X ) 2 (uˆi2 − ui2 ). n i =1

The definition of uˆi implies

uˆi2 =ui2 + ( βˆ0 − β 0 ) 2 + ( βˆ1 − β1 ) 2 X i2 − 2ui ( βˆ0 − β 0 ) − 2u ( βˆ − β ) X + 2( βˆ − β )( βˆ − β ) X . i

1

1

0

i

0

1

1

i

Substituting this into the expression for 1n Σin=1[( X i − X ) 2 uˆi2 − ( X i − µ X ) 2 ui2 ] yields a series of terms p

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, a = ( X − µ ), a = ( βˆ − β ) and so forth. The result then follows from n

X

n

1

1

Slutksy’s theorem if 1n Σin=1 X ir uis → d where d is a finite constant. Let wi = X ir uis and note that wi p

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.

17.10. Using (17.43) with W= θˆ − θ implies

Pr(|θˆ − θ | ≥ δ ) ≤

E[(θˆ − θ ) 2 ]

δ2

p Since E[(θˆ − θ ) 2 ] → 0, Pr(| θˆ − θ | > δ ) → 0, so that θˆ − θ → 0.

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

2 2    x − µ 2  x − µ X  y − µY   y − µY   1  x − µ X   1 X  × exp   − 2 ρ XY   +  +   .

2  −2(1 − ρ XY σ X  σ Y   σ Y   2  σ X   )  σ X      

©2011 Pearson Education, Inc. Publishing as Addison Wesley

94

Stock/Watson - Introduction to Econometrics - Second Edition

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. E (e u ) 17.12. (a)=

∞

∫σ

−∞

1 u

 u2   u2 σ 2  ∞ σ2  1 exp  − 2= exp  − 2 + u − u  du + u  du exp  u  ∫ 2  2π  2  −∞ σ u 2π  2σ u   2σ u

 1 σ 2  ∞  σ u2  1 2 2 σ exp  − = exp u du = exp  u  ∫ − ( )    u 2  2  −∞ σ u 2π  2   2σ u 

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).

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

σi squares estimator is the OLS estimator of Yi/σi onto 1/σi and Xi/σi , where = Write this regression as

θ 0 + θ1 X i2 .

Yi / σ i =β 0 (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).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 18

The Theory of Multiple Regression

18.1.

(a) The regression in the matrix form is Y = Xβ + U with

 TestScore1    TestScore 2  Y = X = ,       TestScore n   U1    U2  , = U = β      U n 

1 Income1  1 Income 2    1 Income n

Income12   Income 22     Income 2n 

 β0     β1  . β   2

(b) The null hypothesis is H0: Rβ = r versus H1: Rβ ≠ r, with R = (0 0 1) and r = 0.

The heteroskedasticity-robust F-statistic testing the null hypothesis is −1

F =(Rβˆ − r )′  RΣˆ βˆ R ′ (Rβˆ − r )/q   With q = 1. Under the null hypothesis, d F→ Fq ,∞ .

We reject the null hypothesis if the calculated F-statistic is larger than the critical value of the Fq ,∞ distribution at a given significance level.

18.2.

(a) The sample size n = 20. We write the regression in the matrix from: Y = Xβ + U

©2011 Pearson Education, Inc. Publishing as Addison Wesley

96

Stock/Watson - Introduction to Econometrics - Second Edition

with

 Y1    Y Y= 2   ,    Yn 

1 X 1,1  1 X 1, 2  X=    1 X 1, n

 u1    u2  , = U = β      un 

X 2,1   X 2, 2     X 2, n 

β 0     β1  β   2

The OLS estimator the coefficient vector is βˆ = ( X′X) −1 X′Y.

with  = ∑in 1 = ∑in 1 X 2i  n X 1i  n  X 1i X 12i X′= X = ∑in 1 = ∑in 1 X 1i X 2i  ,  ∑i 1 =  ∑in 1 = X 1i ∑in 1 X 1i X= ∑in 1 X 22i  = 2i 

and  ∑in=1 Yi    X′Y=  ∑in=1 X 1iYi  .  ∑in=1 X 2iYi   

Note n

∑ X =nX =20 × 7.24 =144.8, i =1

1i

1

n

∑ X =nX =20 × 4.00 =80.0, i =1

2i

2

n

∑ Y =nY =20 × 6.39 =127.8. i =1

i

By the definition of sample variance

1 n 1 n 2 n 2 (Yi −= Y )2 Yi − Y , ∑ ∑ n − 1 i 1= n −1 i 1 n −1 = = sY2

we know n

∑ Y =(n − 1) s + nY . i =1

i

2

2 Y

2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

Thus using the sample means and sample variances, we can get n

∑ X =(n − 1) s + nX 2 1i

i =1

2 X1

2 1

= (20 − 1) × 0.80 + 20 × 7.242 = 1063.6, and n

∑X i =1

2 2, i

= (n − 1) s X2 2 + nX 22 = (20 − 1) × 2.40 + 20 × 4.002 = 365.6.

By the definition of sample covariance

1 n 1 n n ( ) ( ) X − X = Y − Y X iYi − XY , ∑ ∑ i i n − 1 i 1= n −1 i 1 n −1 = = s XY

we know n

(n − 1) s ∑X Y = i i

i =1

XY

+ nXY .

Thus using the sample means and sample covariances, we can get n

(n − 1) s ∑X Y = 1i i

i =1

X 1Y

+ nX 1Y

= (20 − 1) × 0.22 + 20 × 7.24 × 6.39= 929.45, n

(n − 1) s ∑X Y = 2i i

i =1

X 2Y

+ nX 2Y

= (20 − 1) × 0.32 + 20 × 4.00 × 6.39= 517.28,

and n

(n − 1) s ∑X X = i =1

1i

2i

X1 X 2

+ nX 1 X 2

= (20 − 1) × 0.28 + 20 × 7.24 × 4.00= 584.52.

Therefore we have 144.8 80.0   20  127.8      144.8 1063.6 584.52  , = X′Y = X′Y  929.45  .  80.0 584.52 365.6   517.28     

The inverse of matrix X′X is  3.5373 −0.4631 −0.0337    ( X′ X ) =  −0.4631 0.0684 −0.0080  .  −0.0337 −0.0080 0.0229    −1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

97

98

Stock/Watson - Introduction to Econometrics - Second Edition

The OLS estimator of the coefficient vector is βˆ = ( X′Y) −1 X′Y  3.5373 −0.4631 −0.0337   127.8   4.2063        =  −0.4631 0.0684 −0.0080   929.45  =  0.2520  .  −0.0337 −0.0080 0.0229   517.28   0.1033       

That is, βˆ0 = 4.2063, βˆ1 = 0.2520, and βˆ2 = 0.1033.

With the number of slope coefficients k = 2, the squared standard error of the regression sû2 is

su2ˆ =

n 1 1 ˆ ′U ˆ. uˆi U = ∑ n − k − 1 i =1 n − k −1

ˆ =Y−Y ˆ = Y − Xβˆ , so The OLS residuals U ˆ ′U ˆ =(Y − Xβˆ )′ (Y − Xβˆ ) =Y′Y − 2βˆ 'X′Y + βˆ 'X′ Xβˆ .

U

We have n

Y′Y = ∑ Yi 2 =(n − 1) sY2 + nY 2 i =1

= (20 − 1) × 0.26 + 20 × 6.392 = 821.58,  4.2063 ′  127.8     βˆ 'X′Y = = 0.2520   929.45  825.22,  0.1033   517.28    

and

144.8 80.0   4.2063   4.2063 ′  20    ˆβ 'X′ Xβˆ  =  0.2520  144.8 1063.6 584.52   0.2520  832.23.  0.1033   80.0 584.52 365.6   0.1033      Therefore the sum of squared residuals n

ˆ ′Y + βˆ ' X′ Xβˆ ˆ ′U ˆ = SSR = U Y′Y − 2βX ∑ uˆi2 = i =1

= 821.58 − 2 × 825.22 + 832.23 = 3.37.

The squared standard error of the regression sû2 is

= su2ˆ

1 1 ˆ= ˆ ′U U × 3.37 = 0.1982.

20 − 2 − 1 n − k −1

With the total sum of squares n

TSS = ∑ (Yi − Y ) 2 = (n − 1) sY2 = (20 − 1) × 0.26 = 4.94, i =1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

the R2 of the regression is

SSR 3.37 R2 = 1− = 1− = 0.3178.

TSS 4.94 (b) When all six assumptions in Key Concept 16.1 hold, we can use the homoskedasticity-only  of the covariance matrix of βˆ , conditional on X, which is estimator ∑ β̂

−0.4631 −0.0337   3.5373   = ( X′ X) s = −0.4631 0.0684 −0.0080  × 0.1982 ∑ βˆ    −0.0337 −0.0080 0.0229    −1 2 uˆ

−0.09179 −0.0067   0.7011   = −0.0016  .  −0.09179 0.0136  −0.0067 −0.0016 0.0045    The homoskedasticity-only standard error of β̂1 is 1  ( βˆ ) 0.0136 2 SE = = 0.1166.

1

The t-statistic testing the hypothesis β1 = 0 has a tn–k–1 = t17 distribution under the null hypothesis. The value of the t-statistic is

= t

βˆ1 0.2520 = = 2.1612,  ( βˆ ) 0.1166 SE 1

and the 5% two-sided critical value is 2.11. Thus we can reject the null hypothesis β1 = 0 at the 5% significance level.

18.3.

(a) Var= (Q) E[(Q − µQ ) 2 ] =E[(Q − µQ )(Q − µQ )′] E[(c′W − c′µ W )(c′W − c′µ W )′] = =c′E[(W − µ W ) ( W − µ W )′]c ′ var( W ) c c′Σ w c = c=

where the second equality uses the fact that Q is a scalar and the third equality uses the fact that µQ = c′µw. (b) Because the covariance matrix ∑ W is positive definite, we have c′ ∑ wc > 0 for every nonzero vector from the definition. Thus, var(Q) > 0. Both the vector c and the matrix ∑ W are finite, so var(Q) = c′ ∑ wc is also finite. Thus, 0 < var(Q) < ∞.

18.4.

(a) The regression in the matrix form is Y = Xβ + U

©2011 Pearson Education, Inc. Publishing as Addison Wesley

99

100

Stock/Watson - Introduction to Econometrics - Second Edition

with  Y1    Y2  , X Y = =      Yn 

1 X 1    1 X 2  ,      1 X n 

 u1    u2  β  U = , β  0 .    β1     un 

(b) Because X i′ = (1 X i ), assumptions 1–3 in Key Concept 18.1 follow directly from assumptions 1–3 in Key Concept 4.3. Assumption 4 in Key Concept 18.1 is satisfied since observations Xi (i = 1, 2,... n) are not constant and there is no perfect multicollinearity among the two vectors of the matrix X. (c) Matrix multiplication of X′X and X′Y yields

 n ∑in=1 X i  X′ X =  n .

X i ∑in 1 X i2  =  ∑i 1 =  ∑in=1 Yi   nY  ′ = .

X Y =  ∑ n X Y   ∑ n X Y   i =1 i i   i =1 i i  The inverse of X′X is  n ∑in=1 X i  −1 ′ ( X X) =  n  X i ∑in 1 X i2  =  ∑i 1 =

−1

 ∑i 1 = X i − ∑i 1 X i  1 = =  n n 2 2  n  n ∑i 1 = X i − (∑i 1 X i )  − ∑i =1 X i n =  n

=

2

n

 ∑in=1 X i2 / n − X  1  . ∑in=1 ( X i − X ) 2  − X 1 

The estimator for the coefficient vector is βˆ = ( X′ X)−1 X′Y   ∑in=1 X i2 / n − X   nY 1    n 2  n ∑i =1 ( X i − X )  − X 1   ∑i =1 X iYi   Y ∑in 1 = X i2 − X ∑in 1 X iYi  1 = = n   ∑i =1 ( X i − X ) 2  ∑in=1 X iYi − nXY 

=

Therefore we have ∑in 1 = X iYi − nXY ∑in 1 ( X i − X )(Yi − Y ) = = βˆ = , ∑in 1 = ( X i − X )2 ∑in 1 ( X i − X ) 2 =

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

and n n 2 =i 1 = i i 1 0 n 2 i =1 i

βˆ =

Y∑

X − X ∑ X iYi ∑ (X − X )

n n 2 =i 1 = i i 1 n 2 i =1 i

=

Y ∑ (X − X + X ) − X ∑ ∑ (X − X )

X iYi

n n 2 2 =i 1 = i i 1 n 2 i =1 i

=

Y ∑ ( X − X ) + nX Y − X ∑ ∑ (X − X )

X iYi

 ∑ n X Y − nXY  = Y −  i =n1 i i X 2   ∑i =1 ( X i − X )  = Y − βˆ X .

1

We get the same expressions for β̂ 0 and β̂1 as given in Key Concept 4.2. (d) The large-sample covariance matrix of βˆ , conditional on X, converges to

1 Σβˆ = Q -1X Σ v Q -1X n with QX = E ( Xi X′i ) = and Σ v E= ( Vi Vi′ ) E ( Xi ui ui′Xi ). The column vector Xi for the ith observation is

 1  Xi =   ,  Xi  so we have  1   1 = Xi X′i =  (1 X i )   Xi   Xi  ui  = Vi X=  , i ui  X i ui 

Xi  , X i2 

and

 ui2  ui  ′ Vi Vi = =  ( ui X i ui )  2  X i ui   X i ui

X i ui2  .

X i2ui2 

Taking expectations, we get

 1 ( Xi X′i )  = Q X E=  µX

µX 

, E ( X i2 ) 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

101

102

Stock/Watson - Introduction to Econometrics - Second Edition

and

Σ v = E (Vi Vi′ )  E (ui2 ) E ( X i ui2 )  = 2 2 2   E ( X i ui ) E ( X i ui )  cov( X i ui , ui )   var(ui ) = . var( X i ui )   cov( X i ui , ui ) In the above equation, the third equality has used the fact that E (ui | X i ) = 0 so

= E (ui ) E= [ E (ui | X i )] 0, = E ( X i ui ) E= [ X i E (ui | X i )] 0, E (ui2 ) =var(ui ) + [ E (ui )]2 =var(ui ) + [ E (ui )]2 var(ui ), E ( X i2ui2 ) =var( X i ui ) + [ E ( X i ui )]2 =var( X i ui ), E ( X i2ui2 ) =cov( X i ui , ui ) + E ( X i ui ) E (ui ) =cov( X i ui , ui ).

The inverse of QX is −1

µX   E ( X i2 ) − µ x   1 1 Q = =  .  2  E ( X i2 ) − µ X2  − µ X 1   µX E( X i )  −1 X

We now can calculate the large-sample covariance matrix of βˆ , conditional on X, from ∑= βˆ

1 −1 Q X ∑ v Q −X1 n 1 = 2 n[ E ( X i ) − µ X2 ]2 cov( X i ui , ui )   E ( X i2 ) − µ X   E ( X i2 ) − µ X   var(ui ) ×  .  var( X i ui )   − µ X 1   cov( X i ui , ui ) 1   −µ X

The (1, 1) element of ∑ β̂ is 1 {[ E ( X i2 )]2 var(ui ) − 2 E ( X i2 ) µ X cov( X i ui , ui ) + µ X2 var( X i ui )} n[ EX ) − µ X2 ]2 2 i

1 var[ E ( X i2 ) ui − µ X X i ui ] n[ E ( X ) − µ X2 ]2

=

2 i

  [ E ( X i2 )]2 µX var ui − X i ui  2 2 2 2 n[ E ( X i ) − µ X ] E( X i )  

=

   µX var 1 − X i  ui  2  µ X2   E ( X i )   n 1 −  2  E( X i )  var( H i ui ) = , (the same as the expression for σ β20 given in Key Concept 4.4) n( E ( H i2 )]2

=

1

2

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

by defining Hi = 1 −

µX

E ( X i2 )

Xi.

The denominator in the last equality for the (1, 1) element of ∑ β̂ has used the facts that 2

  µ X2 2µ X µX H = X = + X i2 − Xi , 1 1 −  i 2 2 2 E (Xi ) E ( X i2 )  E( X i )  2 i

so

µ2 µ X2 2µ X µX = E ( H i2 ) = 1 + 2 X 2 2 E ( X i2 ) − 1− .

2 E( X i ) E ( X i2 ) [ E ( X i )] 18.5.

PX = X (X′X)−1X′, MX = In − PX. (a) PX is idempotent because PXPX = X(X′X)−1 X′X(X′X)−1 X′ = X(X′X)−1X′ = PX. MX is idempotent because M X M X =(I n − PX ) (I n − PX ) =I n − PX − PX + PX PX =I n − 2PX + PX =I n − PX =M X

PXMX = 0nxn because PX M X = PX (I n PX ) = PX − PX PX = PX − PX = 0n × n

(b) Because βˆ = ( X′X) −1 X′Y, we have ˆ X( X′ X) −1 X′= ˆ Xβ Y Y PX Y = =

which is Equation (18.27). The residual vector is ˆ =Y−Y ˆ = Y − P Y = (I − P )Y = M Y. U X n X X

We know that MXX is orthogonal to the columns of X: MXX = (In − PX) X = X − PXX = X −X (X′X)−1 X′X =X − X = 0 so the residual vector can be further written as ˆ= M Y= M ( Xβ + U= U ) M X Xβ + M X U= M X U X X

which is Equation (18.28).

18.6.

The matrix form for Equation of (10.14) is  Xβ   +U  = Y

©2011 Pearson Education, Inc. Publishing as Addison Wesley

103

104

Stock/Watson - Introduction to Econometrics - Second Edition

with  Y11 − Y1     Y12 − Y1        Y1T − Y1  Y −Y   21 2   Y22 − Y2   , X   = Y =     Y2T − Y2        Yn1 − Yn     Yn 2 − Yn       Y − Y  n  nT

 X 11 − X 1     X 12 − X 1        X 1T − X 1  X −X  2  21   X 22 − X 2   ,     X 2T − X 2        X n1 − X n     X n2 − X n       X −X  n  nT

 u11 − u1     u12 − u1        u1T − u1  u −u   21 2   u22 − u2    U=     u 2T − u 2        un1 − un     un 2 − un        unT − un 

β = β1 The OLS “de-meaning” fixed effects estimator is  ′X  ) −1 X  ′Y . β DM = ( X 1

Rewrite Equation (10.11) using n fixed effects as Y= X it β1 + D1i γ 1 + D 2i γ 2 +  + Dni γ n + uit . it

In matrix form this is YnT ×1 = X nT ×1β1×1 + WnT × nγ n ×1 + U nT ×1

with the subscripts denoting the size of the matrices. The matrices for variables and coefficients are  Y11     Y12        Y1T  Y   21   Y22  , X Y = =     Y2T        Yn1     Yn 2        YnT 

 X 11     X 12        X 1T  X   21   X 22   = , W     X 2T        X n1     X n2        X nT 

 D11   D11     D11  D1  2  D12     D12     D1n   D1n     D1n

D 21  Dn1   D 21  Dn1       D 21  Dn1  D 22  Dn2   D 22  Dn2      , D 22  Dn2       D 2n  Dnn   D 2n  Dnn       D 2n  Dnn 

 u11     u12        u1T  u   21   u22  U=  ,    u 2T        un1     un 2        unT 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

 γ1    γ = β β= γ  2 .

1,      γ n 

Using the expression for βˆ given in the question, we have the estimator

βˆ1BV= βˆ= ( X′M W X) −1 X′M W Y = ( (M W X)′(M W X) ) (M W X)′(M W Y). −1

where the second equality uses the fact that MW is idempotent. Using the definition of W,  X1   X1     X1  0   0 PW X =     0     0   0     0 

0 0  0 X2 X2  X2  0 0  0

 0    0       0   0    0      0       Xn    Xn       X n 

and  X 11 − X 1   X 12 − X 1     X 1T − X 1  0  0   MWX =    0     0  0      0 

0 0  0 X 21 − X 2 X 22 − X 2  X 2T − X 2  0 0  0

  0   0       0    0   0        0      X n1 − X n    X n2 − X n       X nT − X n 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

105

106

Stock/Watson - Introduction to Econometrics - Second Edition

 . A similar calculation shows M Y = Y . Thus so that M W X = X W βˆ1BV = 18.7.

 ′X ) X  ′Y  βˆ .

X (= −1

1

DN

(a) We write the regression model, Yi = β1Xi + β2Wi + ui, in the matrix form as Y = Xβ + Wγ + U with  Y1    Y2  = Y = , X      Yn 

 X1    2  X= , W       Xn 

 W1    2 W = , U       Wn 

 u1     u2  ,      un 

= β β= γ β2 .

1,

The OLS estimator is  βˆ1   X′ X X′ W  −1  X′Y   =  βˆ   W′X W′W   W′Y   2 −1

 β   X′ X X′ W   X′ U  =  1 +     β 2   W ′ X W ′ W   W ′U   β1   1 X′ X =   +  1n  β 2   n W′ X

1 n 1 n

−1

X′ W   1n X′U     W′ W   1n W′U 

X i2 1n ∑ in 1 X iWi   β1   1n ∑ in 1 = =  = +    β 2   1n ∑in 1 = Wi X i 1n ∑ in 1 Wi 2  =

−1

 1n ∑in=1 X i ui  1 n   ∑ Wu  1 = i i i n 

p p By the law of large numbers 1n ∑in 1 = X i2 → E ( X 2 ); 1n ∑in 1 Wi 2 → E (W 2 ); =

1 n

p ∑in=1 X iWi →

p E ( XW ) = 0 (because X and W are independent with means of zero); 1n ∑in=1 X i ui → 0 E ( Xu ) = p (because X and u are independent with means of zero); 1n ∑in=1 X i ui → E ( Xu ) = 0 Thus −1  βˆ1  p  β1   E ( X 2 ) 0   0   →  +     βˆ  E (W 2 )   E (Wu )   β2   0  2

β1   = E (Wu )  .  β 2 + E (W 2 )    p (b) From the answer to (a) βˆ2 → β2 +

E (Wu ) ≠ β 2 if E(Wu) is nonzero.

E (W 2 )

(c) Consider the population linear regression ui onto Wi: ui = λWi + ai

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

107

where λ = E(Wu)/E(W2). In this population regression, by construction, E(aW) = 0. Using this equation for ui rewrite the equation to be estimated as

Yi = X i β1 + Wi β 2 + ui = X i β1 + Wi ( β 2 + λ ) + ai = X i β1 + Wiθ + ai where = θ β 2 + λ . A calculation like that used in part (a) can be used to show that −1 n 1  n ( βˆ1 − β1= )   1n ∑in 1 = X i2 1n ∑in 1 X iWi   n ∑i =1 X i ai     =1 n n 2  n 1 1  n ( βˆ −=   n ∑i 1 =  W X ∑ W ∑ W a θ ) i i i i 1 n   n i =1 u i  2   −1

 E( X 2 ) 0   S1  →     E (W 2 )   S 2   0 d

where S1 is distributed N (0, σ a2 E ( X 2 )). Thus by Slutsky’s theorem

 σ a2  d n ( βˆ1 − β1 ) → N  0, 2   E( X )  Now consider the regression that omits W, which can be written as: = Yi X i β1 + d i

where di = Wiθ + ai. Calculations like those used above imply that

 σ d2  d . n βˆ1r − β1 → N  0, 2   E( X ) 

(

)

2 σ a2 + θ 2 E (W 2 ), the asymptotic variance of βˆ1r is never smaller than the asymptotic Since σ= d variance of βˆ .

1

18.8.

= ui 0.5u i −1 + ui for i = 2, 3,…, n with the random (a) The regression errors satisfy u1 = u1 and

variables ui (i = 1, 2,  , n) being i.i.d. with mean 0 and variance 1. For i > 1, continuing substituting ui – j = 0.5ui – j – 1 + ui − j ( j = 1, 2,…, i − 2) and u1 = u1 into the expression ui = 0.5ui – 1 + ui yields = ui 0.5ui −1 + ui = 0.5(0.5ui − 2 + ui −1 ) + ui = 0.52 (0.5ui −3 + ui − 2 ) + 0.5ui −1 + ui = 0.53 (0.5ui − 4 + ui −3 ) + 0.52 ui − 2 + 0.5ui −1 + ui =  =0.5i −1 u1 + 0.5i − 2 u2 + 0.5i −3 u3 +  + 0.52 ui − 2 + 0.5ui −1 + ui i

= ∑ 0.5i − j u j . j =1

©2011 Pearson Education, Inc. Publishing as Addison Wesley

108

Stock/Watson - Introduction to Econometrics - Second Edition

Though we get the expression ui = ∑ij =1 0.5i − j u j for i > 1, it is apparent that it also holds for i = 1. Thus we can get mean and variance of random variables ui (i = 1, 2,…, n): E (ui ) = ui ) σ i2 var( = =

i

0.5i − j E (u j ) 0, ∑= j =1 i

1 − (0.52 )i .

1 − 0.52

i

j u j ) ∑ (0.52 )i −= ×1 ∑ (0.5i − j )2 var(=

=j 1 =j 1

In calculating the variance, the second equality has used the fact that ui is i.i.d.

Since ui = ∑ij =1 0.5i − j u j we know for k > 0,

= ui + k

i+k

i

i+k

0.5 u 0.5 ∑ 0.5 u + ∑ 0.5 ∑= i+k − j

j= 1

j

k

i− j

j= 1

j

i+k − j

j = i +1

u j

i+k

= 0.5k ui + ∑ 0.5i + k − j u j . j = i +1

Because ui is i.i.d., the covariance between random variables ui and ui + k is i+k   = cov(ui , ui + k ) cov  ui , 0.5k ui + ∑ 0.5i + k − j u j  j = i +1   k 2 = 0.5 σ i .

Similarly we can get

cov(ui , ui − k ) = 0.5k σ i2− k .

The column vector U for the regression error is  u1    u U = 2 .      un 

It is straightforward to get

 E (u12 ) E (u1u2 )  E (u2u1 ) E (u22 ) E (UU′) =       E (un u1 ) E (un u2 )

 E (u1un )    E (u2un )  .      E (un2 ) 

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

109

( )

Because E(ui) = 0, we have E ui2 = var(ui) and E(uiuj) = cov(ui, uj). Substituting in the results on variances and covariances, we have  σ 12  2  0.5σ 1  0.52 σ 12 = Ω E= (UU′)  3 2  0.5 σ 1    n −1 2  0.5 σ 1 

0.5σ 12

σ 22 0.5σ 22 0.52 σ 22

0.52 σ 12 0.5σ 22

σ 32 0.5σ 32

 0.5

n−2

0.53 σ 12 0.52 σ 22 0.5σ 32

σ 42



σ

2 2

0.5

n −3



σ

2 3

0.5

n−4

σ 42

     

0.5n −1σ 12   0.5n − 2 σ 22  0.5n −3 σ 32   0.5n − 4 σ 42     σ n2 

1 − (0.52 )i .

1 − 0.52 (b) The original regression model is

with σ i2 =

Yi = β 0 + β1 X i + ui .

Lagging each side of the regression equation and subtracting 0.5 times this lag from each side gives 0.5β 0 + β1 ( X i − 0.5 X i −1 ) + ui − 0.5ui −1 Yi − 0.5Yi= −1

for i = 2,…, n with ui − 0.5ui −1 = ui . Also Y1 = β 0 + β1 X 1 + u1

with u1 = u1 . Thus we can define a pair of new variables

(Yi , X 1i , X 2i ) = (Yi − 0.5Yi −1 , 0.5, X i − 0.5 X i −1 ), for i = 2,…, n and (Y1 , X 11 , X 21 ) = (Y1 ,1, X 1i ), and estimate the regression equation

Yi = β 0 X 1i + β1 X 2i + ui using data for i = 1,…, n. The regression error ui is i.i.d. and distributed independently of X i , thus the new regression model can be estimated directly by the OLS. 18.9.

(a) βˆ = ( X′M W X) −1 X′M W Y

= ( X′M W X) −1 X′M W ( Xβ + Wγ + U) = β + ( X′M W X) −1 X′M W U.

The last equality has used the orthogonality MWW = 0. Thus −1 ′ = βˆ − β ( X′M = (n −1X′M W X) −1 (n −1X′M W U).

W X) X M W U

©2011 Pearson Education, Inc. Publishing as Addison Wesley

110

Stock/Watson - Introduction to Econometrics - Second Edition

(b) Using MW = In − PW and PW = W(W′W)−1W′ we can get

′M W X n −1X′(I n − PW )X n −1X= = n −1X′X − n −1X′PW X = n −1X′X − (n −1X′W )(n −1W′W ) −1 (n −1W′X).

First consider n −1X′X=

1 n

∑in=1 Xi X′i . The (j, l) element of this matrix is 1n ∑in=1 X ji Xli . By

Assumption (ii), Xi is i.i.d., so XjiXli is i.i.d. By Assumption (iii) each element of Xi has four moments, so by the Cauchy-Schwarz inequality XjiXli has two moments: E ( X 2ji X li2 ) ≤ E ( X 4ji ) ⋅ E ( X li4 ) < ∞.

Because XjiXli is i.i.d. with two moments, 1n ∑in=1 X ji X li obeys the law of large numbers, so

1 n ∑ X ji X li →p E ( X ji X li ) . n i =1 This is true for all the elements of n−1 X′X, so

1 n n −1X′X = ∑ XX. . ∑ Xi X′i →p E (Xi X′i ) = n i =1 Applying the same reasoning and using Assumption (ii) that (Xi, Wi, Yi) are i.i.d. and Assumption (iii) that (Xi, Wi, ui) have four moments, we have 1 n p Wi Wi′ → E ( Wi Wi′ ) = ∑ WW , n −1W′W = ∑ n i =1 1 n p ∑ XW , n −1X′W = Xi Wi′ → E ( Xi Wi′ ) = ∑ n i =1

and

1 n p Wi X′i → E ( Wi X′i ) = n −1W′X = ∑ WX . ∑ n i =1 From Assumption (iii) we know Σ XX , Σ WW , Σ XW , and Σ WX are all finite non-zero, Slutsky’s theorem implies

n −1X′= M W X n −1X′X − (n −1X′W ) (n −1W′W ) −1 (n −1W′X) p

→ Σ XX − Σ XW Σ -1WW Σ WX which is finite and invertible.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

111

(c) The conditional expectation  E (u1| X, W )    E (u2 | X, W )  = E (U|X, W ) =       E (un | X, W )   W1′δ   ′  W2δ  = =       Wn′δ 

 E (u1| X1 , W1 )     E (u2 | X 2 , W2 )        E (un | X n , Wn ) 

 W1′   ′ W2  = δ Wδ .       Wn′ 

The second equality used Assumption (ii) that ( Xi , Wi , Yi ) are i.i.d., and the third equality applied the conditional mean independence assumption (i). (d) In the limit p X′M W E (U|X, W ) = X′M W Wδ = 0k1 ×1 n −1X′M W U → E ( X′M W U|X, W) =

because M W W = 0. (e) n −1X′M W X converges in probability to a finite invertible matrix, and n −1X′M W U converges in probability to a zero vector. Applying Slutsky’s theorem, p 0. βˆ − β (n −1X′M W X)-1 (n −1X′M W U) → =

This implies p βˆ → β.

18.10. (a) Using the hint: Cq = λq, so that 0 = Cq − λq = CCq − λq = λCq − λq = λ2q − λq = λ(1 − λ)q, and the result follows by inspection. (b) The trace of a matrix is equal to sum of its eigenvalues. The rank of a matrix is equal to the number of non-zero eigenvalues. Thus, the result follows from (a). (c) Because C is symmetric with non-negative eigenvalues, C is positive semidefinite, and the result follows. I 18.11. (a) Using the hint C = [Q1 Q2]  r 0

0   Q1 '  , where Q′Q = I. The result follows with A=Q1.

0  Q 2 '

(b) W = A′V ∼ N(A′0, A′InA) and the result follows immediately. (c) V′CV = V′AA′V = (A′V)′(A′V) = W’W and the result follows from (b).

18.12. (a) and (b) These mimic the steps using TSLS.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

112

Stock/Watson - Introduction to Econometrics - Second Edition

18.13. (a) This follows from the definition of the Lagrangian. (b) The first order conditions are (*) X′(Y−X β ) + R′λ = 0 and (**) R β − r = 0 Solving (*) yields (***) β = βˆ + (X′X)–1R′λ.

Multiplying by R and using (**) yields r = R βˆ +R(X′X)–1R′λ, so that

λ = −[R(X′X)–1R′]−1(R βˆ − r).

Substituting this into (***) yields the result. (c) Using the result in (b), Y − X β = (Y − X βˆ ) + X(X′X)–1R′[ R(X′X)–1R′]–1(R βˆ − r), so that (Y − X β )′(Y − X β ) = (Y − X βˆ )′(Y − X βˆ ) + (R βˆ − r) ′[R(X′X)–1R′]–1(R βˆ − r) + 2(Y − X βˆ )′ X(X′X)–1R′[R(X′X)–1R′]–1(R βˆ − r).

But (Y − X βˆ )′ X = 0, so the last term vanishes, and the result follows. (d) The result in (c) shows that (R βˆ − r)′[R(X′X)–1R′]–1(R βˆ − r) = SSRRestricted − SSRUnrestricted.

Also su2 = SSRUnrestricted /(n − kUnrestricted – 1), and the result follows immediately.

18.14. (a) βˆ ′(X′X) βˆ = Y′X(X′X)–1X′Y = Y′X1HX1′Y, where H is the upper k1 × k1 block of (X′X)–1.

Also R βˆ = HX1′Y and R(X′X)–1R′ = H. Thus (Rβˆ ) '[R( X' X ) −1 R ]−1(Rβˆ ) = Y′X1HX1′Y. (b) (i) Write the second stage regression as= Y Xˆ β + U, where X̂ and the fitted values from the ˆ ˆ = 0 where Uˆ= Y − Xˆ βˆ because OLS residual are first stage regression. Note that UX orthogonal to the regressors. Now Uˆ TSLS =Y − Xβˆ =Uˆ − ( X − Xˆ )βˆ =Uˆ − Vˆ βˆ , where V̂ is the residual from the first stage regression. But, since W is a regressor in the first stage ˆ′ = 0 Thus Uˆ TSLS 'W = ˆ ′ − βˆ 'Vˆ 'W = regression, VW UW 0.

(ii) βˆ ′(X′X) βˆ = (Rβˆ ) '  R ( X ' X) −1 R  (Rβˆ ) = SSRRest − SSRUnrest for the regression in KC −1

12.6, and the result follows directly.

18.15. (a) This follows from exercise (18.6).

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

113

 X  β + u , so that Y (b) = i i i −1

 n    n  βˆ − β =  ∑ Xi ' Xi  ∑ Xi ' u i =  i 1=  i1 −1

 n    n = ∑X i ' X i  ∑ X i ' M ' Mu i =  i 1=  i1 −1

 n    n = ∑X i ' Xi  ∑ Xi ' M ' u i =  i 1=  i1 −1

 n    n  = ∑X i ' X i  ∑ Xi ' u i =  i 1=  i1 1 (c) Qˆ X= = ∑in 1 = (T −1 ∑Tt 1 ( X it − X i ) 2 ), where (T −1 ∑Tt=1 ( X it − X i ) 2 ) are i.i.d. with mean Q X and n finite variance (because Xit has finite fourth moments). The result then follows from the law of large numbers. (d) This follows the Central limit theorem. (e) This follows from Slutsky’s theorem.

(f) ηi2 are i.i.d., and the result follows from the law of large numbers.  ' uˆ =  'X  . Then (g) Let ηˆi = T −1/ 2 X ηi − T −1/ 2 ( βˆ − β ) X i i i i  ' uˆ =+  'X  ) 2 − 2T −1/ 2 ( βˆ − β )η X   ηˆi2 = T −1/ 2 X ηi2 T −1 ( βˆ − β ) 2 ( X i i i i i i ' Xi

1 2  'X  ) 2 − 2T −1/ 2 ( βˆ − β ) 1 ∑ n η X   ∑ ηˆ − ∑ η = β ) 2 1n ∑in 1 ( X T −1 ( βˆ −= = i i i i 1 i i ' Xi n n

n n 2 1 = i 1= i i 1 n

and

p p Because ( βˆ − β ) → 0 , the result follows from (a) 1n ∑in=1 ( X i ' X i ) 2 → E[( X i ' X i ) 2 ] and (b) p ∑in=1 ηi X i ' X i → E (ηi X i ' X i ). Both (a) and (b) follow from the law of large numbers; both (a) and (b) are averages of i.i.d. random variables. Completing the proof requires verifying  'X  ) 2 has two finite moments and η X   that ( X i i i i ' X i has two finite moments. These in turn follow from 8-moment assumptions for (Xit, uit) and the Cauchy-Schwartz inequality.

Alternatively, a “strong” law of large numbers can be used to show the result with finite fourth moments.

1 n

18.16 (a) Using analysis like that Appendices 4.2 and 4.3

1

∑X u Xu ∑X Y = βˆ = β+∑ = β+n 1 ∑X ∑X ∑X i i 2 i

i i 2 i

n

i i 2 i

Consistency follows by analyzing the averages in the expression: p 1 X u → ( X i ui ) 0, ∑ i i E= n

p 1 2 X → ∑ i E ( X i2 ) > 0 , n

©2011 Pearson Education, Inc. Publishing as Addison Wesley

114

Stock/Watson - Introduction to Econometrics - Second Edition

and Slutsky’s theorem.

To show unbiasness, note that

  ∑ X i ui  ∑ X i ui   E = E E X ,..., X      1 n  ∑X2    ∑ X i2  i     

 ∑ X i E ( ui X 1 ,..., X n )   ∑ X i E ( ui X i )  = E = E =    0    ∑ X i2 ∑ X i2     (b) (i) The result follows from the expression given in (a) and the definition of Ii. (ii) – (iv) As in (a), consistency follows by analyzing the sample averages in

1 ∑ Ii X i ui I X Y I X u ∑i ii= ∑i ii= n + + βˆ = β β .

1 2 ∑ Ii X i2 ∑ Ii X i2 ∑ Ii X i n Specifically, we have

p p 1 1 I i X i ui → E ( I i X i ui ) and ∑ I i X i2 → E ( I i X i2 ). ∑ n n

In parts (ii)-(iv), E ( I i X i2 ) > 0.

In part (ii) E(IiXiui) = E(E(IiXiui | Xi, ui)) = pE(Xiui) = pE(E(Xiui | Xi)) = 0 In part (iii) E(IiXiui) = E(E(IiXiui | Xi, ui)) = E(p(Xi)Xiui) = pE(E(p(Xi)Xiui | Xi)) = 0.

In part (iv) E(IiXiui) = E(E(IiXiui | Xi, ui)) = E(p(ui,Xi)Xiui) = pE(E(p(ui,Xi)Xiui | Xi)) ≠ 0.

So that β̂ is consistent in (ii) and (iii), but is not consistent in (iv).

The unbiasness analysis proceeds as in part (a), but it is necessary to condition on both Ii and Xi when carrying out the law of iterated expectations. A key result is that E(ui | Xi, Ii) = 0 in (ii) and (iii), but not in (iv). This yields unbiasness in (ii) and (iii), but not in (iv). To see that E(ui | Xi, Ii) = 0 in (ii) and (iii), notice that the conditional distribution of u given X, I can be written as = f (u | X , I )

f (u , I | X ) = f (I | X )

f ( I | X , u ) f (u | X ) = f (I | X )

f ( I | X ) f (u | X ) = f (u | X ) f (I | X )

where the first equality is the definition of the conditional density, the second is Bayes rule, and the third equality follows because Pr(I = 1 | X,u) does not depend on u in (ii) and (iii).

This implies that E(ui | Xi, Ii) = E(ui | Xi) = 0. Unbiasness then follows in (ii) and (iii) using an argument analogous that in part (a). (c) In this example, Ii = 1 if Yi ≥ 0, or Xi + ui ≥ 0, or ui ≥ −Xi. Notice that E(IiXiui|Xi) = XiE(ui|ui ≥ −Xi). A calculation based on the normal distribution shows that

φ (− X ) E (u | u ≥ − X ) = 1 − Φ(− X )

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

where Φ is the standard normal CDF and φ is standard normal density.

Because E(u|u ≥ −X) ≠ 0, the OLS estimator is biased and inconsistent.

18.17 The results follow from the hints and matrix multiplication and addition.

©2011 Pearson Education, Inc. Publishing as Addison Wesley

115