Solutions Manual for Introduction to Econometrics 4th Edition by James Stock & Mark Watson

SOLUTION MANUAL

SOLUTION MANUAL

Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 1 _____________________________________________________________________________________________________

2.1. (a) Probability distribution function for Y Outcome (number of heads) Y=0 Probability 0.25 (b) Cumulative probability distribution function for Y Outcome (number of Y<0 0£Y<1 heads) Probability 0 0.25 (c) µY = E(Y ) = (0 × 0.25) + (1× 0.50) + (2 × 0.25) = 1.00 Using Key Concept 2.3: var(Y ) = E(Y 2 ) − [E(Y )]2 , and

E(Y 2 ) = (02 × 0.25) + (12 × 0.50) + (22 × 0.25) = 1.50 so that

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

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Y=1 0.50

Y=2 0.25

1£Y<2

Y³2

0.75

1.0

Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 2 _____________________________________________________________________________________________________

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,

µ X = E( X ) = 0 × Pr ( X = 0) + 1× Pr ( X = 1) = 0 × 0.30 + 1× 0.70 = 0.70. (b)

σ 2X = E[( X − µ X )2 ] = (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.

(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, corr (X , Y ) =

σ XY 0.084 = = 0.4425. σ Xσ Y 0.21× 0.1716

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 3 _____________________________________________________________________________________________________

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

σ W2 = var (3+ 6 X ) = 62 σ 2X = 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 = −3.52

corr (W , V ) =

σ WV −3.528 = = −0.4425. σ Wσ V 7.56 × 8.4084

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 4 _____________________________________________________________________________________________________

2.4. (a) E( X 3 ) = 03 × (1− p) + 13 × p = p (b) E( X k ) = 0 k × (1− p) + 1k × p = p (c) E( X ) = 0.3

var ( X ) = E( X 2 ) − [E( X )]2 = 0.3− 0.09 = 0.21 Thus, σ = 0.21 = 0.46.

To compute the skewness, use the formula from exercise 2.21:

E( X − µ )3 = E( X 3 ) − 3[E( X 2 )][E( X )] + 2[E( X )]3 = 0.3− 3× 0.32 + 2 × 0.33 = 0.084 Alternatively, E( X − µ )3 = [(1− 0.3)3 × 0.3] + [(0 − 0.3)3 × 0.7] = 0.084 Thus, skewness = E( X − µ )3/σ 3 = .084/0.463 = 0.87.

To compute the kurtosis, use the formula from exercise 2.21:

E( X − µ )4 = 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 Alternatively, E( X − µ )4 = [(1− 0.3)4 × 0.3] + [(0 − 0.3)4 × 0.7] = 0.0777 Thus, kurtosis is E( X − µ )4/σ 4 = .0777/0.464 = 1.76

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 5 _____________________________________________________________________________________________________

2.5.

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 Concept 2.3, µX = 70oF implies that µY = −17.78 + (5/9) × 70 = 21.11°C, and sX = 7oF implies σ Y = (5/9) × 7 = 3.89°C.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 6 _____________________________________________________________________________________________________

2.6.The table shows that Pr ( X = 0, Y = 0) = 0.026, Pr ( X = 0, Y = 1) = 0.576, Pr ( X = 1, Y = 0) = 0.009, Pr ( X = 1, Y = 1) = 0.389, Pr ( X = 0) = 0.602, Pr ( X = 1) = 0.398, Pr (Y = 0) = 0.035, Pr (Y = 1) = 0.965. (a) E(Y ) = µY = 0 × Pr(Y = 0) + 1× Pr (Y = 1) = 0 × 0.035 + 1× 0.965 = 0.965.

(b)

Unemployment Rate =

#(unemployed) #(labor force)

= Pr (Y = 0) = 1− Pr(Y = 1) = 1− E(Y ) = 1− 0.965 = 0.035.

(c) Calculate the conditional probabilities first: Pr (Y = 0|X = 0) =

Pr ( X = 0, Y = 0) 0.026 = = 0.043, Pr ( X = 0) 0.602

Pr (Y = 1|X = 0) =

Pr ( X = 0, Y = 1) 0.576 = = 0.957, Pr ( X = 0) 0.602

Pr (Y = 0|X = 1) =

Pr ( X = 1, Y = 0) 0.009 = = 0.023, Pr ( X = 1) 0.398

Pr (Y = 1|X = 1) =

Pr ( X = 1, Y = 1) 0.389 = = 0.978.

Pr ( X = 1) 0.398

The conditional expectations are E(Y|X = 1) = 0 × Pr (Y = 0|X = 1) + 1× Pr (Y = 1|X = 1) = 0 × 0.023+ 1× 0.978 = 0.978, E(Y|X = 0) = 0 × Pr (Y = 0|X = 0) + 1× Pr (Y = 1|X = 0) = 0 × 0.043+ 1× 0.957 = 0.957.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 7 _____________________________________________________________________________________________________

(d) Use the solution to part (b), Unemployment rate for college graduates = 1 – E(Y|X=1) = 1−0.978 = 0.023.

Unemployment rate for non-college graduates = 1 – E(Y|X=0) = 1−0.957 = 0.043

(e) The probability that a randomly selected worker who is reported being unemployed is a college graduate is Pr ( X = 1|Y = 0) =

Pr ( X = 1, Y = 0) 0.009 = = 0.257.

Pr (Y = 0) 0.035

The probability that this worker is a non-college graduate is Pr ( X = 0|Y = 0) = 1− Pr ( X = 1|Y = 0) = 1− 0.257 = 0.743.

(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) = 0.743, while from the table Pr(X = 0) = 0.602.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 8 _____________________________________________________________________________________________________

2.7. Using obvious notation, C = M + F; thus µC = µ M + µ F and

σ C2 = σ 2M + σ F2 + 2cov ( M, F ).

This implies

(a) µC = 40 + 45 = $85,000 per year.

M, F) , so that cov ( M, F ) = σ M σ F corr ( M, F ). Thus (b) corr ( M, F ) = cov( σ Mσ F

cov ( M, F ) = 12 × 18 × 0.80 = 172.80, where the units are squared thousands of

dollars per year.

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

σ C = 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 = 0.85 Euros per Dollar or 1/e = 1.18 Dollars per Euro); each 1 Eollar 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×sC (in units of thousands of euros per year). The correlation is unit-free, and is unchanged.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 9 _____________________________________________________________________________________________________

2.8. µY = E(Y ) = 1, σ Y2 = var (Y ) = 4. With Z = 12 (Y − 1),

⎛1 ⎞ 1 1 µ Z = E ⎜ (Y − 1)⎟ = ( µY − 1) = (1− 1) = 0, 2 ⎝2 ⎠ 2 ⎛1 ⎞ 1 1 σ Z2 = var ⎜ (Y − 1)⎟ = σ Y2 = × 4 = 1.

4 ⎝2 ⎠ 4

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 10 _____________________________________________________________________________________________________

2.9.

Value of Y 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

Probability Distribution 65 of X 0.01 0.21 0.01 0.40 0.09 0.39 0.11 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 (b) The conditional probability of Y|X = 8 is given in the table below Value of Y 14 22 30 40 0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39

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

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

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

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65 0.09/0.39

Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 11 _____________________________________________________________________________________________________

2.10. Using the fact that if Y ∼ N ⎛⎜⎝ µY , σ Y2 ⎞⎟⎠ then

Y − µY ~ N (0, 1) and Appendix Table 1, σY

we have

⎛ Y − 1 3− 1⎞ ≤ = Φ(1) = 0.8413. (a) Pr (Y ≤ 3) = Pr ⎜ 2 ⎟⎠ ⎝ 2 (b)

Pr(Y > 0) = 1− Pr(Y ≤ 0) ⎛ Y − 3 0 − 3⎞ = 1− Pr ⎜ ≤ = 1− Φ(−1) = Φ(1) = 0.8413.

3 ⎟⎠ ⎝ 3 (c) ⎛ 40 − 50 Y − 50 52 − 50 ⎞ Pr (40 ≤ Y ≤ 52) = Pr ⎜ ≤ ≤ 5 5 ⎟⎠ ⎝ 5 = Φ(0.4) − Φ(−2) = Φ(0.4) − [1− Φ(2)] = 0.6554 − 1+ 0.9772 = 0.6326.

(d) ⎛ 6 − 5 Y − 5 8 − 5⎞ Pr (6 ≤ Y ≤ 8) = Pr ⎜ ≤ ≤ ⎟ ⎝ 2 2 2 ⎠ = Φ(2.1213) − Φ(0.7071) = 0.9831− 0.7602 = 0.2229.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 12 _____________________________________________________________________________________________________

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.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 13 _____________________________________________________________________________________________________

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

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 14 _____________________________________________________________________________________________________

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.

E(Y − µY )4 (c) The kurtosis of the normal is 3, so 3 = ; solving yields E(Y 4 ) = 3; a σ Y4 similar 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 large 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

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

Similarly, 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

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 15 _____________________________________________________________________________________________________

2.14. The central limit theorem suggests that when the sample size (n) is large, the σ2 ⎛ ⎞ N ⎜⎝ µY , σ Y2 ⎟⎠ (Y ) σ Y2 = nY . distribution of the sample average is approximately with Given

µY = 100, σ Y2 = 43.0, σ2

43 = 0.43, and (a) n = 100, σ Y2 = nY = 100

⎛ Y − 100 101− 100 ⎞ Pr (Y ≤ 101) = Pr ⎜ ≤ ⎟ ≈ Φ(1.525) = 0.9364. ⎝ 0.43 0.43 ⎠

σ2

43 = 0.2606, and (b) n = 165, σ Y2 = nY = 165

⎛ Y − 100 98 − 100 ⎞ Pr (Y > 98) = 1− Pr (Y ≤ 98) = 1− Pr ⎜ ≤ ⎟ ⎝ 0.2606 0.2606 ⎠ ≈ 1− Φ(−3.9178) = Φ(3.9178) = 1.000 (rounded to four decimal places).

σ2

43 = 0.6719, and (c) n = 64, σ Y2 = 64Y = 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.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 16 _____________________________________________________________________________________________________

2.15. (a)

⎛ 9.6 − 10 Y − 10 10.4 − 10 ⎞ Pr (9.6 ≤ Y ≤ 10.4) = Pr ⎜ ≤ ≤ ⎟ ⎝ 4/n 4/n 4/n ⎠ ⎛ 9.6 − 10 10.4 − 10 ⎞ = Pr ⎜ ≤Z≤ ⎟ ⎝ 4/n 4/n ⎠ where Z ~ N(0, 1). Thus,

⎛ 9.6 − 10 10.4 − 10 ⎞ (i) n = 20; Pr ⎜ ≤Z≤ ⎟ = Pr (−0.89 ≤ Z ≤ 0.89) = 0.63 ⎝ 4/n 4/n ⎠ ⎛ 9.6 − 10 10.4 − 10 ⎞ (ii) n = 100; Pr ⎜ ≤Z≤ ⎟ = Pr(−2.00 ≤ Z ≤ 2.00) = 0.954 ⎝ 4/n 4/n ⎠ ⎛ 9.6 − 10 10.4 − 10 ⎞ (iii)n = 1000; Pr ⎜ ≤Z≤ ⎟ = Pr(−6.32 ≤ Z ≤ 6.32) = 1.000 ⎝ 4/n 4/n ⎠

(b)

⎛ −c Y − 10 c ⎞ Pr (10 − c ≤ Y ≤ 10 + c) = Pr ⎜ ≤ ≤ ⎟ ⎝ 4/n 4/n 4/n ⎠ ⎛ −c c ⎞ = Pr ⎜ ≤Z≤ ⎟. ⎝ 4/n 4/n ⎠ As n get large

c 4/ n

gets large, and the probability converges to 1.

(c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 17 _____________________________________________________________________________________________________

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 X. X variables with µX = Pr(X = 1) = Pr(Y < 3.6). Compute Because converges in X probability to µX = Pr(X = 1) = Pr(Y < 3.6), will be an accurate approximation if n is large.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 18 _____________________________________________________________________________________________________

2.17. µY = 0.4 and σ Y2 = 0.4 × 0.6 = 0.24

⎛ Y − 0.4 0.43− 0.4 ⎞ ⎛ Y − 0.4 ⎞ ≥ = Pr ≥ 0.6124 (a) (i) P( Y ³ 0.43) = Pr ⎜ ⎟ ⎜⎝ ⎟⎠ = 0.27 ⎝ 0.24/n 0.24/n ⎠ 0.24/n ⎛ Y − 0.4 0.37 − 0.4 ⎞ ⎛ Y − 0.4 ⎞ ≤ = Pr ≤ −1.22 (ii) P( Y £ 0.37) = Pr ⎜ ⎟ ⎜⎝ ⎟⎠ = 0.11 ⎝ 0.24/n 0.24/n ⎠ 0.24/n

b) We know Pr(−1.96 £ Z £ 1.96) = 0.95, thus we want n to satisfy

0.41 = 0.41−0.4 > −1.96 and 0.39−0.4 < −1.96. Solving these inequalities yields n ³ 0.24/n 0.24/n 9220.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 19 _____________________________________________________________________________________________________

2.18. 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 ⎡

2⎤

σ Y2 = E ⎢⎢ ⎛⎝⎜ Y − µY ⎞⎠⎟ ⎥⎥ ⎣

⎦

= (0 − 1000) × Pr (Y = 0 ) + (20000 − 1000)2 × Pr (Y = 20000) 2

= (−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.

σ2

= 1.9 × 105. (b) (i) E(Y ) = µY = $1000, σ Y2 = nY = 1.9×10 100 7

(ii) Using the central limit theorem,

Pr (Y > 2000) = 1− Pr (Y ≤ 2000) ⎛ Y − 1000 2,000 − 1,000 ⎞ = 1− Pr ⎜ ≤ ⎟ ⎝ 1.9 × 105 1.9 × 105 ⎠ ≈ 1− Φ(2.2942) = 1− 0.9891 = 0.0109.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 20 _____________________________________________________________________________________________________

2.19. (a) l

Pr (Y = y j ) = ∑ Pr ( X = xi , Y = y j ) i=1

l

= ∑ Pr (Y= y j|X =xi )Pr ( X =xi ) i=1

(b) k

k

l

j=1

j=1

i=1

E (Y ) = ∑ y j Pr (Y = yj ) = ∑ yj ∑ Pr (Y = yj |X = xi ) Pr ( X = xi ) l ⎛ k ⎜ ⎜ ⎜ i=1 ⎜⎝ j=1

⎞ ⎟

=∑ ∑ yj Pr (Y = yj |X = xi ) ⎟ Pr ( X = xi ) ⎟ ⎟⎠

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

σ XY = E[( X − µ X )(Y − µY )] l

k

=∑ ∑ (xi − µ X )( y j − µY ) Pr ( X =xi , Y= y j ) 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 = ⎜ ∑ (xi − µ X ) Pr ( X = xi )⎟ ⎜ ∑ ( yj − µY ) Pr (Y = yj ⎟ ⎝ i=1 ⎠ ⎝ j=1 ⎠ = E( X − µ X )E(Y − µY ) = 0 × 0 = 0,

cor (X , Y ) =

σ XY 0 = = 0. σ Xσ Y σ Xσ Y

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 21 _____________________________________________________________________________________________________ l

m

2.20. (a) Pr (Y = yi ) = ∑ ∑ Pr (Y = yi |X = xj , Z = z h ) Pr ( X = xj , Z = z h ) j=1 h=1

(b) k

E(Y ) = ∑ yi Pr (Y = yi ) Pr (Y = yi ) i=1 k

l

m

= ∑ yi ∑ ∑ Pr (Y = yi |X = xj , Z = z h ) Pr ( X = xj , Z = z h ) i=1

j=1 h=1

⎡ k ⎤ = ∑ ∑ ⎢ ∑ yi Pr (Y = yi |X = xj , Z = z h ) ⎥ Pr ( X = xj , Z = z h ) j=1 h=1 ⎣ i=1 ⎦ l

m

l

m

= ∑ ∑ E(Y|X = xj , Z = z h ) Pr ( X = xj , Z = z h ) j=1 h=1

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.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 22 _____________________________________________________________________________________________________

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

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

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 23 _____________________________________________________________________________________________________

2. 22. 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] where 0.07 × 0.04 × 0.25 = Cov (Rs , Rb ) follows from the definition of the correlation between Rs 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 s2 with respect to w is

dσ 2 = 2w × .07 2 − 2(1− w) × 0.042 + (2 − 4w) × [0.07 × 0.04 × 0.25] dw = 0.0102w − 0.0018 Solving for w yields w = 18 / 102 = 0.18. (Notice that the second derivative is positive, so that this is the global minimum.) With w = 0.18, σ R = .038.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 24 _____________________________________________________________________________________________________

2. 23. X and Z are two independently distributed standard normal random variables, so

µ X = µ Z = 0, σ 2X = σ Z2 = 1, σ XZ = 0. (a) Because of the independence between X and Z, Pr (Z = z|X = x) = Pr (Z = z), and E(Z|X ) = E(Z ) = 0. Thus

E(Y|X ) = E( X 2 + Z|X ) = E( X 2|X ) + E(Z|X ) = X 2 + 0 = X 2 .

(b) E( X 2 ) = σ 2X + µ 2X = 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 X and Z, we have E(ZX ) = µ Z µ X = 0. Thus

E( XY ) = E( X 3 ) + E(ZX ) = 0.

(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 ) =

σ XY 0 = = 0. σ Xσ Y σ Xσ Y

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 25 _____________________________________________________________________________________________________

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

definition of a χ n2 random variable. n n Y2 Y2 (c) E(W ) = E ∑ i 2 = ∑ E i 2 = n. σ i=1 σ i=1

(d) Write V=

Y1 n ∑ i=2 Yi n−1

= 2

Y1 /σ n ∑ i=2 (Y /σ )2 n−1

which follows from dividing the numerator and denominator by s. Y1/s ~ N(0,1),

∑ (Y /σ ) ~ χ n

i=2

2

i

2 n−1

, and Y1/s and ∑ i=2 (Yi /σ )2 are independent. The result then n

follows from the definition of the t distribution.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 26 _____________________________________________________________________________________________________ n

n

i=1

i=1

2.25. (a) ∑ axi = (ax1 + ax2 + ax3 +!+ axn ) = a(x1 + x2 + x3 +!+ xn ) = a∑ xi (b) n

∑ (x + y ) = (x + y + x + y +!x + y ) i

i

1

1

2

2

n

n

i=1

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

n

i=1

i=1

= ∑ xi + ∑ yi

n

(c) ∑ a = (a + a + a +!+ a) = na i=1

(d) n

n

i=1

i=1

∑ (a + bxi + cyi )2 = ∑ (a 2 + b2 xi2 + c2 yi2 + 2abxi + 2acyi + 2bcxi yi ) n

n

n

n

n

i=1

i=1

i=1

i=1

i=1

= na 2 + b2 ∑ xi2 + c 2 ∑ yi2 + 2ab∑ xi + 2ac∑ yi + 2bc∑ xi yi

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 27 _____________________________________________________________________________________________________

2.26. (a) corr(Yi,Yj) =

cov(Yi ,Y j )

σYσY i

=

cov(Yi ,Y j )

j

σ Yσ Y

=

cov(Yi ,Y j )

σ Y2

= ρ , where the first equality

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.

1 1 1 1 (b) Y = Y1 + Y2 , so that E( Y ) = E(Y )1 + E(Y2 ) = µY 2 2 2 2 var( Y ) =

(c) Y =

σ 2 ρσ Y2 1 1 2 var(Y1 ) + var(Y2 ) + cov(Y1 ,Y2 ) = Y + 4 4 4 2 2

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

⎛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 σ + 2 ∑ ∑ ρσ Y2 2 ∑ Y n i=1 n 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

an(n − 1) for any 2

variable a.

σ Y2 1 (d) When n is large ≈ 0 and ≈ 0 , and the result follows from (c). n n

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 28 _____________________________________________________________________________________________________

2.27 (a) E(u) = E[E(u|X) ] = E[E(Y− Ŷ )|X] = E[ E(Y|X) – E(Y|X) ] = 0.

(b) E(uX) = E[E(uX|X) ] = E[XE(u)|X] = E[ X×0] = 0

(c) Using the hint: v = (Y - Ŷ ) – h(X) = u - h(X) , so that E(v2) = E[u2] + E[h(X)2] – 2×E[u×h(X)]. Using an argument like that in (b), E[u×h(X)] = 0. Thus, E(v2) = E(u2) + E[h(X)2], and the result follows by recognizing that E[h(X)2] ≥ 0 because h(x)2 ≥ 0 for any value of x.

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Stock/Watson - Introduction to Econometrics 4th Edition - Answers to Exercises: Chapter 2 29 _____________________________________________________________________________________________________

2.28 (a) Pr(M = 3) = Pr(M=3,A=0) + Pr(M=3,A=1) = Pr(M=3|A=0)Pr(A=0) + Pr(M=3|A=1)Pr(A=1) = 0.05×0.5 + 0.01×0.5=0.03

(b) Pr( A = 0 | M = 3) = Pr( M = 3 | A = 0) ×

Pr( A = 0) 0.5 = 0.05 × = 0.83 P( M = 3) 0.03

(c) Pr(M = 3) = Pr(M=3,A=0) + Pr(M=3,A=1) = Pr(M=3|A=0)Pr(A=0) + Pr(M=3|A=1)Pr(A=1) = 0.05×0.25 + 0.01×0.75=0.02 Pr( A = 0 | M = 3) = Pr( M = 3 | A = 0) ×

Pr( A = 0) 0.25 = 0.05 × = 0.63 P( M = 3) 0.02

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Solutions to End-of-Chapter Exercises

(c)  XY = cov (X , Y ) = E[( X −  X )(Y − Y )] = (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, corr (X , Y ) =

2.3.

 XY 0084 = = 04425  XY 021  01716

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)  W2 = var (3 + 6 X ) = 62   X2 = 36  021 = 756,  V2 = var (20 − 7Y ) = (−7) 2   Y2 = 49  01716 = 84084 (c)  WV = cov(3 + 6 X , 20 − 7Y ) = 6  (−7)cov(X , Y ) = −42  0084 = −3528 corr (W , V ) =

2.4.

 WV −3528 = = −04425 WV 756  84084

3 3 3 (a) E ( X ) = 0  (1 − p ) + 1  p = p k k k (b) E ( X ) = 0  (1 − p ) + 1  p = p

(c) E ( X ) = 0.3 , and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus  =

0.21 = 0.46.

var ( X ) = E ( X 2 ) − [ E ( X )]2 = 0.3 − 0.09 = 0.21  = 0.21 = 0.46. To compute the skewness, use the formula from exercise 2.21: E ( X −  )3 = E ( X 3 ) − 3[ E ( X 2 )][ E ( X )] + 2[ E ( X )]3 = 0.3 − 3  0.32 + 2  0.33 = 0.084 Alternatively, E ( X −  ) = [(1 − 0.3)  0.3] + [(0 − 0.3)  0.7] = 0.084 3

3

3

3 3 3 Thus, skewness = E ( X −  ) / = 0.084/0.46 = 0.87.

To compute the kurtosis, use the formula from exercise 2.21:

E ( X −  )4 = 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 Alternatively, E ( X −  ) = [(1 − 0.3)  0.3] + [(0 − 0.3)  0.7] = 0.0777 4

4

4

4 4 4 Thus, kurtosis is E ( X −  ) / = 0.0777/0.46 = 1.76

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4

Stock/Watson • Introduction to Econometrics, Third Edition

2.5.

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 Concept 2.3, X = 70oF implies that Y = −17.78 + (5/9)  70 = 21.11C, and X = 7oF implies  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 ) = Y = 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 (Y = 0| X = 0) =

Pr ( X = 0, Y = 0) 0037 = = 0056, Pr ( X = 0) 0659

Pr (Y = 1| X = 0) =

Pr ( X = 0, Y = 1) 0622 = = 0944, Pr ( X = 0) 0659

Pr (Y = 0| X = 1) =

Pr ( X = 1, Y = 0) 0009 = = 0026, Pr ( X = 1) 0341

Pr (Y = 1| X = 1) =

Pr ( X = 1, Y = 1) 0332 = = 0974 Pr ( X = 1) 0341

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) =

Pr ( X = 1, Y = 0) 0009 = = 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

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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.

Using obvious notation, C = M + F ; thus C =  M +  F and  C =  M +  F + 2cov ( M , F ). This implies 2

2

2

(a) C = 40 + 45 = $85,000 per year. cov( M , F )

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

(b) corr ( M , F ) =

2 2 2 2 2 2 (c)  C =  M +  F + 2cov ( M , F ), so that  C = 12 + 18 + 2  172.80 = 813.60, and

 C = 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  C (in units of thousands of Euros per year), and the standard deviation is e  C (in units of thousands of Euros per year). The correlation is unit-free, and is unchanged.

2.8.

Y = E (Y ) = 1,  Y2 = var (Y ) = 4. With Z = 12 (Y − 1), 1 2

 

1 2

1 2

 Z = E  (Y − 1)  = ( Y − 1) = (1 − 1) = 0, 1 2

 

1 4

1 4

 Z2 = var  (Y − 1)  =  Y2 =  4 = 1

2.9.

Value of Y 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

 Y = 14.77

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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

22

30

40

65

0.02/0.39

0.03/0.39

0.15/0.39

0.10/0.39

0.09/0.39

E (Y | X = 8) = 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) = 39.21 E (Y 2 | X = 8) = 142  (0.02/0.39) + 222  (0.03/0.39) + 302  (0.15/0.39) + 402  (0.10/0.39) + 652  (0.09/0.39) = 1778.7

var(Y ) = 1778.7 − 39.212 = 241.65

 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

corr( X , Y ) = cov( X , Y )/( X  Y ) = 11.0 / (2.60  14.77) = 0.286 2.10.

Using the fact that if Y

N  Y ,  Y2  then

Y − Y

Y

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

 Y −1 3 −1 (a) Pr (Y  3) = Pr    =  (1) = 08413 2   2

Y −3 0−3  (b) Pr(Y  0) = 1 − Pr(Y  0) = 1 − Pr   3   3 = 1 −  (−1) = (1) = 08413

 40 − 50 Y − 50 52 − 50    (c) Pr (40  Y  52) = Pr   5 5   5 =  (04) −  (−2) =  (04) − [1 −  (2)] = 06554 − 1 + 09772 = 06326 6−5 Y −5 8−5 (d) Pr (6  Y  8) = Pr     2 2   2 =  (21213) −  (07071) = 09831 − 07602 = 02229

2.11.

(a) 0.90 (b) 0.05 (c) 0.05 (d) When Y ~ 10 , then Y /10 ~ F10, .

2

(e) Y = Z 2 , where Z ~ N (0,1), thus Pr (Y  1) = Pr (−1  Z  1) = 0.32.

©2018 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.

2 2 2 2 (a) E (Y ) = Var (Y ) + Y = 1 + 0 = 1; E (W ) = Var (W ) + W = 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. (c) The kurtosis of the normal is 3, so 3 = E (Y − Y ) / Y ; solving yields E(Y ) = 3; a similar calculation yields the results for W. (d) First, condition on X = 0, so that S = W: 4

4

4

E ( S | X = 0) = 0; E ( S 2 | X = 0) = 100, E ( S 3 |X = 0) = 0, E ( S 4 | X = 0) = 3  100 2.

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

 S2 = E ( S −  S ) 2 = E ( S 2 ) = 1.99, and E ( S −  S ) 4 = E ( S 4 ) = 302.97. Thus, kurtosis = 302.97 / (1.99 2 ) = 76.5 2.14.

The central limit theorem suggests that when the sample size (n) is large, the distribution of the 2 2 sample average (Y ) is approximately N  Y ,  Y2  with  Y2 = nY . Given Y = 100,  Y = 430, 2

43 = 043, and (a) n = 100,  Y2 = nY = 100

 Y − 100 101 − 100  Pr (Y  101) = Pr      (1525) = 09364 043   043 2

43 = 02606, and (b) n = 165,  Y2 = nY = 165

 Y − 100 98 − 100  Pr (Y  98) = 1 − Pr (Y  98) = 1 − Pr    02606   02606  1 −  (−39178) =  (39178) = 1000 (rounded to four decimal places)

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Stock/Watson • Introduction to Econometrics, Third Edition

(c) n = 64,  Y2 =

 Y2 64

=

43 = 06719, and 64

 101 − 100 Y − 100 103 − 100  Pr (101  Y  103) = Pr     06719 06719   06719   (36599) −  (12200) = 09999 − 08888 = 01111 2.15.

 9.6 − 10 Y − 10 10.4 − 10    (a) Pr (9.6  Y  10.4) = Pr   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    (b) Pr (10 − c  Y  10 + c) = Pr   4/n 4/n   4/n c   −c = Pr  Z .

4/n   4/n

As n get large

c

gets large, and the probability converges to 1.

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 X = Pr(X = 1) = Pr(Y  3.6). Compute X . Because X converges in probability to X = Pr(X = 1) = Pr(Y  3.6), X will be an accurate approximation if n is large.

2.17.

Y = 0.4 and  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  = 0.27  = Pr  0.24/n   0.24/n  0.24/n 

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Solutions to End-of-Chapter Exercises

 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 

(b) We know Pr(−1.96  Z  1.96) = 0.95, thus we want n to satisfy 0.41 = 0.39 − 0.40

 −1.96. Solving these inequalities yields n  9220.

24 / n Pr (Y = $0) = 095, Pr (Y = $20000) = 005.

and

2.18.

0.41 − 0.40

(a) The mean of Y is

Y = 0  Pr (Y = $0) + 20,000  Pr (Y = $20000) = $1000.

The variance of Y is

 Y2 = E  (Y − Y ) 2  = (0 − 1000) 2  Pr(Y = 0) + (20000 − 1000) 2  Pr (Y = 20000) = (−1000) 2  095 + 19000 2  005 = 19  107,

so the standard deviation of Y is  Y = (19  107 ) 2 = $4359 1

(b) (i) E (Y ) = Y = $1000,  Y2 =

 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 −  (22942) = 1 − 09891 = 00109

2.19.

l

(a) Pr (Y = y j ) =  Pr ( X = xi , Y = y j ) i =1 l

=  Pr (Y = y j | X = xi )Pr ( X = xi ) i =1

k

k

l

j =1

j =1

i =1

(b) E (Y ) =  y j Pr (Y = yj ) =  yj  Pr (Y = yj |X = xi ) Pr ( X = xi )  k 

 

 i =1  j =1

  

l

=   yj Pr (Y = yj |X = xi )  Pr ( X = xi ) 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

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24 / n

 −1.96

9

10

Stock/Watson • Introduction to Econometrics, Third Edition

 XY = E[( X −  X )(Y − Y )] l

k

=  ( xi −  X )( y j − Y ) Pr ( X = xi , Y = y j ) i =1 j =1 l

k

= ( xi −  X )( y j − Y ) Pr ( X = xi ) Pr (Y = y j ) i =1 j =1

  l  k =   ( xi −  X ) Pr ( X = xi )    ( yj − Y ) Pr (Y = yj   i =1   j =1  = E ( X −  X ) E (Y − Y ) = 0  0 = 0, corr(X , Y ) =

2.20.

l

 XY 0 = = 0  XY  XY

m

(a) Pr (Y = yi ) =  Pr (Y = yi | X = xj , Z = zh ) Pr ( X = xj , Z = zh ) j =1 h =1

k

(b) E (Y ) =  yi Pr (Y = yi ) Pr (Y = yi ) i =1 k

l

m

=  yi  Pr (Y = yi | X = xj , Z = zh ) Pr (X = xj , Z = zh ) i =1

j =1 h =1

 k  =    yi Pr (Y = yi | X = xj , Z = zh )  Pr (X = xj , Z = zh ) j =1 h =1  i =1  l

m

l

m

=  E (Y | X = xj , Z = zh ) Pr (X = xj , Z = zh ) j =1 h =1

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

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

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Solutions to End-of-Chapter Exercises

2.22.

11

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

 2 = w2  0.072 + (1 − w)2  0.042 + 2  w  (1 − w)  [0.07  0.04  0.25] where 0.07  0.04  0.25 = Cov ( Rs , Rb ) follows from the definition of the correlation between Rs and Rb. (a)  = 0.065;  = 0.044 (b)  = 0.0725;  = 0.056 (c) w = 1 maximizes  ;  = 0.07 for this value of w. (d) The derivative of 2 with respect to w is

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

2.23.

X and Z are two independently distributed standard normal random variables, so

 X =  Z = 0,  X2 =  Z2 = 1,  XZ = 0. (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  2 2 2 2 2 (b) E ( X ) =  X +  X = 1, and Y = E ( X + Z ) = E ( X ) +  Z = 1 + 0 = 1 3 3 (c) E ( XY ) = E ( X + ZX ) = E ( X ) + E ( ZX ). Using the fact that the odd moments of a standard

normal random variable are all zero, we have E ( X ) = 0. Using the independence between 3

X and Z , we have E ( ZX ) =  Z  X = 0. Thus E ( XY ) = E ( X ) + E ( ZX ) = 0.

3

(d)

cov (XY ) = E[( X −  X )(Y − Y )] = E[( X − 0)(Y − 1)] = E ( XY − X ) = E ( XY ) − E ( X ) = 0 − 0 = 0 corr (X , Y ) =

2.24.

 XY 0 = = 0  XY  XY

2 2 2 2 (a) E (Yi ) =  +  =  and the result follows directly.

(b) (Yi/) is distributed i.i.d. N(0,1), W =  i =1 (Yi / ) 2 , and the result follows from the definition n

of a  n random variable.

2

n n Y2 Y2 (c) E (W ) = E  i 2 =  E i 2 = n.  i =1  i =1

(d) Write

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12

Stock/Watson • Introduction to Econometrics, Third Edition

V=

Y1 in=2 Yi2 n −1

=

Y1 / in=2 (Y / ) 2 n −1

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

2.25.

(a)

 axi = (ax1 + ax2 + ax3 +

+ axn ) = a( x1 + x2 + x3 +

i =1 n

(b)

 (x + y ) = (x + y + x + y + i =1

i

i

1

1

2

= ( x1 + x2 + n

n

i =1

i =1

n

+ xn ) = a  xi i =1

xn + y n )

2

xn ) + ( y1 + y2 +

yn )

=  xi +  yi n

(c)

 a = (a + a + a +

+ a) = na

n

n

i =1

(d)

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

i =1

i

i

2

2

i =1

2 i

2

2 i

i

i

i

i

n

n

n

n

n

i =1

i =1

i =1

i =1

i =1

= na 2 + b 2  xi2 + c 2  yi2 + 2ab xi + 2ac  yi + 2bc xi yi

2.26.

(a) corr(Yi,Yj) = cov(Yi , Y j )/ Yi  Y j = cov(Yi , Y j )/ Y  Y = cov(Yi , Y j )/ Y2 =  , where the first equality 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 Y = Y1 + Y2 , so that E( Y ) = E (Y )1 + E (Y2 ) = Y 2 2 2 2  2  Y2 1 1 2 var(Y ) = var(Y1 ) + var(Y2 ) + cov(Y1 , Y2 ) = Y + 4 4 4 2 2

1 n 1 n 1 n Y E ( Y ) = E ( Y ) = Y = Y , so that i  i n n i =1 n i =1 i =1

©2018 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  Y2  Y + n2   n 2 i =1 i =1 j = i +1

=

 Y2

n(n − 1)  Y2 n n2 2  1 = Y + 1 −   Y2 n  n

=

n −1

+

n

where the fourth line uses   a = a (1 + 2 + 3 + i =1 j = i +1

(d) When n is large 2.27

 Y2 n

 0 and

+ n − 1) =

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.

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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 2 sample average ( Y ) is approximately N  Y ,  Y2  with  Y2 = Y . Given a population Y = 100, n 2  Y = 430, we have (a) n = 100,  Y2 =

 Y2 n

=

43 = 043, and 100

 Y − 100 101 − 100  Pr (Y  101) = Pr      (1.525) = 09364 043   043

(b) n = 64,  Y2 =

 Y2 n

=

43 = 06719, and 64

 101 − 100 Y − 100 103 − 100  Pr(101  Y  103) = Pr     06719 06719   06719  (36599) − (12200) = 09999 − 08888 = 01111 (c) n = 165,  Y2 =

 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 − (−39178) =  (39178) = 10000 (rounded to four decimal places) 3.2.

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

E (Yi ) = 0  Pr(Y = 0) + 1 Pr(Y = 1) = p, and variance var(Yi ) = E[(Yi − Y ) 2 ] = (0 − p ) 2  Pr(Yi = 0) + (1 − p) 2  Pr(Yi = 1) = p 2 (1 − p ) + (1 − p ) 2 p = p (1 − p )

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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  =  E (Yi ) =  p = p   n  n i =1 n i =1     in=1 Yi  1 n 1 n p (1 − p) ˆ (c) var( p ) = var  = 2  var(Yi ) = 2  p (1 − p ) =    n  n i =1 n i =1 n  

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) = = 62148  10−4. The n 400 1 2 standard error is SE ( pˆ ) = (var( pˆ )) = 00249.

(b) The estimated variance of p̂ is var( pˆ ) =

(c) The computed t-statistic is

t act =

pˆ −  p0 SE( pˆ )

=

05375 − 05 = 1506 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 = 2 (−|t act |) = 2 (−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 −  (t act ) = 1 −  (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

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16

Stock/Watson • Introduction to Econometrics, Third Edition

pˆ  1.96SE ( 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 − Pr  −1.30   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 −0.05 pˆ − 0.53 −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. ©2018 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 = 9604, then the margin of error is less than 0.01 for all possible). Thus if n  0.012 values of p.

chosen so that n 

3.6.

(a) No. Because the p-value is less than 0.05 (= 5%),  = 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  = 6, which requires Y and SE (Y ).

Only the p-value for test that  = 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 statistic is t = , where SE( pˆ ) = pˆ (1 − pˆ )/n. (An alternative formula for SE( p̂ ) is 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  and the standard deviation of Y is  Y = 200 hours. Y is the sample mean with a sample size n = 100. The  200 standard deviation of the sampling distribution of Y is  Y = Y = = 20 hours. The n 100 hypothesis test is H 0 :  = 2000 vs. H1    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|  = 2000) = 1 − Pr(Y  2100|  = 2000)  Y − 2000 2100 − 2000  = 1 − Pr   | = 2000  20  20  −7 = 1 − (5) = 1 − 0999999713 = 287  10 ,

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18

Stock/Watson • Introduction to Econometrics, Third Edition

where Pr(Y  2100| = 2000) means the probability that the sample mean is greater than 2100 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   | = 2150  20  20 

 = Pr(Y  2100| = 2150) = Pr 

= (−25) = 1 −  (25) = 1 − 09938 = 00062 The power of the manager’s testing is 1 −  = 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. t act = Y

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

1 = 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 standard error of Y1 − Y2 is SE (Y1 − Y2 ) =

s12 s22 64 121 + = + = 11158. The 90% n1 n2 100 200

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

1 − 2 = (Y1 − Y2 )  164SE(Y1 − Y2 ) = (58 − 62)  164  11158 = (−58299 − 21701)

©2018 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  1 − 2 = 0 vs H1  1 − 2  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

t act =

Y1 − Y2 58 − 62 = = −35849 SE(Y1 − Y2 ) 11158

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

p − value = 2 ( −|t act |) = 2 ( −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.

E (Y ) =

1  1 3 E (Y1 ) + E (Y2 ) +   n2 2

 1 3 E (Yn −1 ) + E (Yn )   2 2 

11 n 3 n  =    Y +   Y  = Y n2 2 2 2  1 1 9 1 9  var(Y ) = 2  var(Y1 ) + var(Y2 ) + var(Yn −1 ) + var(Yn )  n 4 4 4 4  =

3.12.

 Y2 1 1 n 2 9 n 2    +    = 1  25  Y Y   n2  4 2 4 2 n 

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 + = n1 n2

2002 3202 + = 44721.

100 64

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

H 0  1 − 2 = 0 vs H1  1 − 2  0 The t-statistic for testing the null hypothesis is t act =

Y 1 − Y 2 = 3100 − 2900 = 44722 SE(Y 1 − Y 2) 44721

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

p -value = 2 (−|t act |) = 2 (−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.

©2018 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 n 420 mean test score in the population is

 = 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

Y 2 = 6500, sample standard deviation s2 = 179. The standard error of Y1 − Y2 is SE (Y1 − Y2 ) =

s12 s22 19.42 17.92 + = + = 18281. The hypothesis tests for higher average n1 n2 238 182

scores in smaller classes is

H 0  1 − 2 = 0 vs H1  1 − 2  0 The t-statistic is t act =

Y 1 − Y 2 = 6574 − 6500 = 40479 SE(Y 1 − Y 2) 18281

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

p -value = 1 −  (t act ) = 1 −  (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 1kg = 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.

©2018 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 ) = Y and from (2.47) we know that 2 var( Y ) = Y . In this problem, because Ya and Yb are Bernoulli random variables, pˆ a = Ya , pˆ b = n 2 Yb ,  Ya = pa(1–pa) and  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 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  = 1000, so the null hypothesis that  = 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  prep − non − prep = 0. (d) (i) Let X denote the change in the test score. The 95% confidence interval for X 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 X = 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. ©2018 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 nw,1992

11.782 10.17 2 9.662 7.782 + + + = 0.48. The 95% 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 Y and variance  Y2 . The covariance cov (Y j , Yi ) = 0, j  i. The sampling distribution of the sample average Y has mean Y and variance var (Y ) =  Y2 =

 Y2 n

.

(a) E[(Yi − Y ) 2 ] = E{[(Yi − Y ) − (Y − Y )]2 } = E[(Yi − Y ) 2 − 2(Yi − Y )(Y − Y ) + (Y − Y ) 2 ] = E[(Yi − Y ) 2 ] − 2 E[(Yi − Y )(Y − Y )] + E[(Y − Y ) 2 ] = var(Yi ) − 2cov(Yi , Y ) + var(Y ).  

 nj =1 Y j

 

n

(b) cov(Y  Y ) = E[(Y − Y )(Yi − Y )] = E  

 

 

 

 

− Y  (Yi − Y ) 

  n    j =1 (Y j − Y )   1 1  = E[(Y −  )2 + = E   ( Y −  )  i Y I Y   E[(Y j − Y )(Yi − Y )]   n   n n j i    

2 1 1 =  Y2 +  cov(Y j , Yi ) = Y  n n j i n 1 n 1 n  1 n 2 2 ( Y − Y ) = E [( Y − Y ) ] = (c) E  sY2  = E   i [var (Yi ) − 2cov(Yi  Y ) + var(Y )] i  n −1  n − 1 i =1 i =1  n − 1 i =1  =

3.19.

 Y2  Y2  1 n  2 1 n  n − 1 2  +  Y  =  Y2 .     Y − 2  =   n − 1 i =1  n n  n − 1 i =1  n 

(a) No. E (Yi 2 ) =  Y2 + Y2 and E (YiY j ) = Y2 for i  j. Thus 2

1 n 1 n 1 2 1 n  2 E (Y 2 ) = E   Yi  = 2  E (Yi 2 ) + 2  E (YY Y i j ) = Y + n i =1 j i n  n i =1  n i =1 (b) Yes. If Y gets arbitrarily close to Y with probability approaching 1 as n gets large, then Y 2 2 gets arbitrarily close to Y 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 −  X )(Yi − Y )  −    ( X −  X )(Y − Y )  n − 1  n i =1   n −1 ©2018 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

23

because X →  X and Y → Y the final term converges in probability to zero. p

p

Let Wi = ( X i −  x )(Yi − Y ). Note Wi is iid with mean XY and second moment

E[( X i −  X ) 2 (Yi − Y ) 2 ]. But E[( X i −  X ) 2 (Yi − Y ) 2 ]  E ( X i −  X ) 4 E (Yi − Y ) 4 from the Cauchy-Schwartz inequality. Because X and Y have finite fourth moments, the second moment of Wi is finite, so that it has finite variance. Thus 1n  in=1 Wi → E (Wi ) =  XY . Thus, s XY → XY (because 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 in=1 (Ymi − Ym ) 2  in=1 (Ywi − Yw ) 2 (n − 1) ( n − 1) [ SE (Ym − Yw )]2 = + n n =

in=1 (Ymi − Ym ) 2 +  in=1 (Ywi − Yw ) 2 . n(n − 1)

Similarly, using equation (3.23)

 1  n 1 i =1 (Ymi − Ym ) 2 + in=1 (Ywi − Yw ) 2   2(n − 1)  (n − 1)  [ SE pooled (Ym − Yw )]2 = 2n =

in=1 (Ymi − Ym ) 2 +  in=1 (Ywi − Yw ) 2 . n(n − 1)

©2018 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 ˆ 0 in Equation (4.8), we know the sample average of the test scores across the 100 classrooms is

TestScore = ˆ 0 + ˆ 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

= 13044 = 1318. Thus, standard deviation is sY = sY2 = 115.

The sample variance is sY = TSS n−1 99 2

4.2.

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

Weight = ˆ 0 + ˆ1 Height .

= 07036 kg per cm.

The coefficients are ˆ 0 = −9941  04536 = −45092 kg ; ˆ1 = 394  024536 54 The R 2 is unit free, so it remains at R2 = 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) ˆ = Y − ˆ X , so that Y = ˆ + ˆ X . Thus the sample mean of AWE is 696.7 + 9.6  41.6 = 0

1

0

1

$1,096.06.

4.4.

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

 > 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 ) = (  2 − 1)  var ( Rm − R f ) + var(u ), which will be positive if var(u )  (1 −  2 )  var ( Rm − R f ). (c) Rm − R f = 7.3% − 3.5% = 3.8%. Thus, the predicted returns are

Rˆ = R f + ˆ ( Rm − R f ) = 3.5% + ˆ  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). (d) (i) 49 + 0.24  90 = 70.6; 49 + 0.24  120 = 77.8; 49 + 0.24  150 = 85.0 (ii) 0.24 10 = 2.4.

©2018 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 (  0 + 1 X i + ui |X i ) =  0 + 1 E ( X i |X i ) + E (ui |X i ) =  0 + 1 X i  4.7.

The expectation of ̂ 0 is obtained by taking expectations of both sides of Equation (4.8):    1 n  E ( ˆ0 ) = E (Y − ˆ1 X ) = E    0 + 1 X +  ui  − ˆ1 X  n i =1     n 1 =  0 + E ( 1 − ˆ1 ) X +  E (ui ) n i =1 = 0

where the third equality in the above equation has used the facts that E(ui) = 0 and E[( ˆ1 −1) X ] = E[(E( ˆ −1)| X ) X ] = because E[(  − ˆ ) | X ] = 0 (see text equation (4.31).)

1

1

4.8.

1

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

Yi = (  0 + 2) + 1 X i + (ui − 2).

The new regression error is (ui − 2) and the new intercept is (0 + 2). All of the assumptions of Key Concept 4.3 hold for this regression model.

4.9.

(a) With ˆ1 = 0, ˆ0 = Y , and Yˆi = ˆ0 = Y . Thus ESS = 0 and R2 = 0. (b) If R2 = 0, then ESS = 0, so that Yˆi = Y for all i. But Yˆi = ˆ0 + ˆ1 X i , so that Yˆi = Y for all i, which implies that ˆ = 0, or that Xi is constant for all i. If Xi is constant for all i, then 1

 ( X − X ) = 0 and ˆ1 is undefined (see equation (4.7)). n

i −1

4.10.

2

i

(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  X = 0.2. Also

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

E[( X i −  X )ui |X i = 0]2 = 0.2 2  1, and E[( X i −  X )ui |X i = 1]2 = (1 − 0.2) 2  4.

Putting these results together

 2ˆ = 1

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

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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 ˆ = i =1 1

n

 X i2

.

i =1

n

 X (Y − 4)

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

i

i

.

n

 X i2 i =1

4.12.

(a) Write n

n

i =1

i =1

n

ESS =  (Yˆi − Y ) 2 =  ( ˆ0 + ˆ1 X i − Y ) 2 =  [ ˆ1 ( X i − X )]2 i =1

2

  ( X i − X )(Yi − Y )  = ˆ  ( X i − X ) =  .  in=1 ( X i − X ) 2 i =1 n

2 1

2

n i =1

This implies 2

  in=1 ( X i − X )(Yi − Y )  ESS 2 R = n =  i =1 (Yi − Y ) 2  in=1 ( X i − X ) 2  in=1 (Yi − Y ) 2  = 

1 n −1 1 n −1

  n 2 1  n −1  i =1 (Yi − Y ) 

 in=1 ( X i − X )(Yi − Y )

 in=1 ( X i − X ) 2

2

2

 s  2 =  XY  = 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 s ( n − 1) i =1 s (c) Because rXY = XY , rXY Y = XY = = i =1 n = ˆ1 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 i is now i = (Xi − X)ui, so that var(i) = 3var[(Xi − X)ui], and the term 2 carry through the rest of the calculations.

4.14.

Because ˆ0 = Y − ˆ1 X , Y = ˆ0 + 1 X . The sample regression line is y = ˆ0 + 1 x , so that the sample regression line passes through ( X , Y ).

©2018 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 1 is {−582  196  221}, that is −10152  1  −14884. (b) Calculate the t-statistic: t act =

ˆ 1 − 0 −582 = = −26335 SE( ˆ 1) 221

The p-value for the test H 0  1 = 0 vs. H1  1  0 is

p -value = 2 (−|t act |) = 2 (−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 t act =

ˆ 1 − (−5.6) 022 = = 0.10 SE ( ˆ 1) 221

The p-value for the test H 0 : 1 = −5.6 vs. H1 : 1  −5.6 is

p -value = 2 ( −|t act |) = 2 ( −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 1 = −5.6 is not rejected at the 5% level, this value is contained in the 95% confidence interval. (d) The 99% confidence interval for  0 is {520.4  2.58  20.4}, that is, 467.7   0  573.0.

5.2.

(a) The estimated gender gap equals $2.12/hour. (b) The null and alternative hypotheses are H 0  1 = 0 vs. H1  1  0. The t-statistic is t act =

ˆ 1 − 0 2.12 = = 5.89 SE ( ˆ 1) 036

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Solutions to End-of-Chapter Exercises

29

and the p-value for the test is

p -value = 2 (−|t act |) = 2 ( −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 1 is {212  196  036}, that is, 141  1  2.83. (d) The sample average wage of women is ˆ0 = $1252/hour. The sample average wage of men is ˆ + ˆ = $12.52 + $2.12 = $14.64/hour.

0

1

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

Wage = 0 + 1Male + ui  or

Wage =  0 +  1 Female + vi  In the first regression equation, Male equals 1 for men and 0 for women;  0 is the population mean of wages for women and 0 + 1 is the population mean of wages for men. In the second regression equation, Female equals 1 for women and 0 for men;  0 is the population mean of wages for men and  0 +  1 is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations:

 0 = 0 + 1   0 +  1 = 0  Given the coefficient estimates ˆ 0 and ̂ 1 , we have

ˆ 0 = ˆ 0 + ˆ 1 = 14.64 ˆ1 = ˆ 0 − ˆ 0 = − ˆ 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 = vˆi. 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 Wages = 14.64 − 212 Female

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.

©2018 Pearson Education, Inc. Publishing as Addison Wesley

30

Stock/Watson • Introduction to Econometrics, Third Edition

(c) The increase in wages for college education is 1  4. Thus, the counselor’s assertion is that

1.76 − 2.50 − 9.25, which has 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 1  4 is 4  (1.76  1.96  0.08) or $6.41  Gain  $7.67.

1 = 10/4 = 2.50. The t-statistic for this null hypothesis is t act =

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 = 5.56, which has a p-value of 0.00. Thus the null hypothesis is 2.5 rejected at the 5% (and 1%) level.

(b) The t-statistic is t act =

(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 1 = 0; but this null hypothesis was rejected at the 5% level in part (a). (d) 1 would be rejected at the 5% level in 5% of the samples; 95% of the confidence intervals would contain the value 1 = 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. (b) The t-statistic is t act = 61.57.4−55 = 0.88, which is less (in absolute value) than the critical value of 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.

1

5.9.

(a)  = n

(Y1 + Y2 + X

+ Yn )

so that it is linear function of Y1, Y2, , Yn.

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Solutions to End-of-Chapter Exercises

31

(b) E(Yi|X1, , Xn) = Xi, thus  1   n (Y1 + Y2 + + Yn )  , X n ) = E   | ( X1, X     1  ( X1 + + X n ) =n = .

X

E (  |X 1 ,

5.10.

  , X n )  

Let n0 denote the number of observation with X = 0 and n1 denote the number of observations with n 1 n n n n X = 1; note that  i =1 X i = n1 ; X = 1 ;  X iYi = Y1;  i =1 ( X i − X )2 = i =1 X i2 − nX 2 = n n1 i =1 n2 n n n  n  nn n1 − 1 = n1 1 − 1  = 1 0 ; n1Y1 + n0Y0 =  i =1Yi , so that Y = 1 Y1 + 0 Y0 n n n n n  From the least squares formula

ˆ1 = =

in=1 ( X i − X )(Yi − Y ) in=1 X i (Yi − Y ) in=1 X iYi − Yn1 = n = in=1 ( X i − X ) 2 i =1 ( X i − X ) 2 n1n0 / n n n n n  (Y1 − Y ) =  Y − 1 Y1 − 0 Y0  = Y1 − Y0 n0 n0  n n 

n  n n + n0 n and ˆ0 = Y − ˆ1 X =  0 Y0 + 1 Y1  − (Y1 − Y0 ) 1 = 1 Y0 = Y0 n  n n  n

5.11.

s Using the results from 5.10, ˆ0 = Ym and ˆ1 = Yw − Ym . From Chapter 3, SE (Ym ) = m and nm SE (Yw − Ym ) =

sm2 sw2 + . Plugging in the numbers ˆ0 = 523.1 and SE ( ˆ0 ) = 6.22; ˆ1 = −38.0 nm nw

and SE ( ˆ1 ) = 7.65.

5.12.

Equation (4.22) gives

 2ˆ = 0

var (H i ui )

n  E ( H i2 ) 

2

 where H i = 1 −

X

E ( X i2 )

Xi

Using the facts that E (ui | X i ) = 0 and var (ui | X i ) =  u (homoskedasticity), we have 2

      

E ( H i ui ) = E ui − =0−

x

E(X )

x

E ( X i2 )

2 i

   i i   

X u = E (ui ) −

x

E ( X i2 )

E [ X i E (ui | X i )]

 0 = 0

and

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32

Stock/Watson • Introduction to Econometrics, Third Edition         

E [( H i ui ) ] = E ui − 2

X E  X i2 

   2  i   

   i i   

Xu

2       

2

     = E u − 2  2  X u +   X 2   X i2ui2   E  X i   E  X i    

X

2 i i

2

    2  2     X = E u − 2 X 2 E  X i E  ui2 | X i   +   E  X i E  ui | X i     2   

2 i  



E  X i 







 E  X i  





2

    2  2   X =  − 2 X 2   u2 +  E  X i  u2 = 1 −  X 2   u2 .    2 2 u

X





 E  X i   2 Because E ( H i ui ) = 0, var ( H i ui ) = E[( H i ui ) ] so E  X i 





 

E  X i  

  X2  2   var (H i ui ) = E[( H i ui ) 2 ] = 1 −  E ( X i2 )  u  

Also         

E (H ) = E 1− 2 i

X

E ( X i2 )

X

   i   

2      

      

2



    2  X   = E 1− 2 X + X i i  2 E ( X i2 )  E ( X i )    2

X

 X2 +   X  E X 2 = 1−  X2  =1− 2   i 2 2 2 E ( Xi )

 E ( X i ) 

( )

E ( Xi )

Thus   X2  2 1 −   E ( X i2 )  u var ( H u )  u2   i i  2ˆ = = = 2 0  nE ( H 2 )2     2  X2   i X   n 1 −  n 1 −   E ( X i2 )   E ( X i2 )      =

E ( X i2 ) u2

n[ E ( X i2 −  X2 )]

=

E ( X i2 )  u2 n 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), ˆ =  aiYi where ai =

Xi

 j =1 X 2j n

. Since the weights depend only on X i

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

©2011 Pearson Education, Inc. Publishing as Addison Wesley

Solutions to End-of-Chapter Exercises

in=1 X i E (ui | X 1 , in=1 X 2j

, X n)

=  since E (ui | X 1 ,

in=1 X i 2 var (ui | X 1 ,

, X n)

=

(b) E ( ˆ | X 1 ,

, X n) =  +

(c) var ( ˆ | X 1 ,

, X n) =

2

 in=1 X 2j  (d) This follows the proof in the appendix.

5.15.

, X n) = 0

2 in=1 X 2j

Because the samples are independent, ˆm ,1 and ˆw ,1 are independent. Thus var ( ˆm ,1 − ˆw,1 ) = var ( ˆ ) + var( ˆ ). Var ( ˆ ) is consistently estimated as [SE( ˆ )]2 and Var (ˆ ) is m ,1

w ,1

m ,1

m ,1

w ,1

consistently estimated as [SE( ˆw,1 )]2 , so that var( ˆm ,1 − ˆw,1 ) is consistently estimated by [SE( ˆ )]2 + [SE( ˆ )]2 , and the result follows by noting the SE is the square root of the m ,1

w ,1

estimated variance.

©2018 Pearson Education, Inc. Publishing as Addison Wesley

33

Chapter 6 Linear Regression with Multiple Regressors 6.1.

By equation (6.15) in the text, we know

R2 = 1−

n −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 ˆ1  1.

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

©2018 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 1 does not suffer from omitted variable bias.

6.10.

(a)  2ˆ = 1

  u2 1 1   n 1 −  X2 1 , X 2   X2 1

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

1  1 4 400 1 − 0  6 1 4 1 =  = = 0.00167 400 6 600

 2ˆ = 1

(b) With  X1 , X 2 = 0.5

 2ˆ = 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 ˆ1 , 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 ˆ1.

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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

2

2i

  (Yi − b1 X 1i − b2 X 2i ) 2 = −2 X 2i (Yi − b1 X 1i − b2 X 2i ) b2

(c) From (b), ˆ1 satisfies

 X (Y − ˆ X − ˆ X ) = 0 1i

ˆ1 =

or

i

1

1i

1

2i

 X 1iYi − ˆ2  X 1i X 2i  X 12i

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

ˆ2 =

 X 2iYi − ˆ1  X 1i X 2i  X 22i

and substituting this into the expression for ˆ1 in (c) yields 1i 2 i  X 1iY  2 i i X1   X 1i X 2i 2 2i  ˆ1 = .

2  X 1i

X Y − ˆ

X X

Solving for ˆ1 yields:

ˆ1 =

 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 ̂ 0 yields: ˆ0 = Y − ˆ1 X 1 − ˆ2 X 2 . (f) Substituting ˆ0 = Y − ˆ1 X 1 − ˆ2 X 2 into the least squares objective function yields

 (Y − ˆ − b X − b X ) =  ( (Y − Y ) − b ( X − X ) − b ( X − X ) ) , which is identical to 2

2

i

0

1

1i

2

2i

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 1 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.

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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)

− 2.62** (0.20)

0.29** (0.04)

0.29** (0.04)

Age (X3) Ntheast (X4)

0.69* (0.30)

Midwest (X5)

0.60* (0.28)

South (X6)

− 0.27 (0.26)

Intercept

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)  6 = 6. Thus a 95% confidence interval is −0.27  1.96  0.26.

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(c) The expected difference between Juanita and Jennifer is (X5,Juanita − X5,Jennifer)  5 + (X6,Juanita − X6,Jennifer)  6 = −5 + 6. 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 =

ˆcollege,1998 − ˆcollege,1992 SE( ˆcollege,1998 − ˆcollege, 1992 )

. Because

̂college,1998 and ̂college,1992 are computed from independent samples, they are independent, which means that cov( ˆcollege,1998 , ˆcollege,1992 ) = 0 Thus, var( ˆcollege,1998 − ˆcollege,1992 ) = 1 var( ˆcollege,1998 ) + var( ˆcollege,1998 ) . This implies that SE( ˆcollege,1998 − ˆcollege,1992 ) = (0.212 + 0.202 ) 2 .

=

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.

(a) The t-statistic is 0.485 / 2.61 = 0.186  1.96. Therefore, the coefficient on BDR is not 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

R2 = 1−

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

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Solutions to End-of-Chapter Exercises

420 − 1 − 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 Column 1: R 2 = 1 −

(b) H 0 : 3 =  4 = 0

H1 :  3  ,  4  0 Unrestricted regression (Column 5): Y =  0 + 1 X 1 +  2 X 2 +  3 X 3 +  4 X 4 , Runrestricted = 0.775 2

Restricted regression (Column 2): Y =  0 + 1 X 1 +  2 X 2 , Rrestricted = 0.427 2

FHomoskedasticityOnly = =

2 2 ( Runrestricted − Rrestricted )/q , n = 420, kunrestricted = 4, q = 2 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 =  0 +  X 1i +  2 ( X 1i + X 2i ) + ui

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

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

and test whether  = 0.

7.10.

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

Because R 2 = 1 −

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39

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Stock/Watson - Introduction to Econometrics - Second Edition

F= = 7.11.

SSRrestricted − SSRunrestricted 2 2 /q ( Runrestricted − Rrestricted )/q TSS = SSRunrestricted 2 (1 − Runrestricted )/( n − kunrestricted − 1) /( n − kunrestricted − 1) TSS

( 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 ˆ1 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 ˆ1 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 ̂ 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 increase is 100  = 27.551% and the approximation is 100  [ln (250) − ln (196)] = 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 = 0.09730  1.96. Note that this coefficient significant at a 5% significance level: | t | = 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 = 0.05571  1.96. significant at a 5% significance level: | t | = 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 = 0.022  1.96. a 5% significance level: | t | = 0.10 ©2018 Pearson Education, Inc. Publishing as Addison Wesley

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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: 1  0 and  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 ln( AHE ) = (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 ln( AHE ) = (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 ln( AHE ) = (0.1032  16) − (0.451  0) + (0.0134  0  16) + (0.0134  10) − 0.000211  10 2 ) −(0.095  0) − (0.092  0) − (0.023  1) + 1.503 = 3.253

With 11 years of experience, the man’s expected AHE is ln( AHE ) = (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 1 denote the coefficient on Potential Experience and 2 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 1 + 52 and the expected change in (b) is given by 1 + 212. The difference between these, say (b) − (a), is 162. Because the estimated value of 2 is significant at the 5% level (the t-statistic for ̂ 2 is −.000211/0.000023 = − 9.2), the difference between the effects in (a) and (b) (=162) 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 ln(Price per citation) + ln(80)  ln(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 = 0 + 1 % Eligible +  2 DV 1 % Eligible + 3 DV 2  % Eligible +

other regressors The coefficient 1 shows the marginal effect of %Eligible on TestScores for values of %Eligible  20%, 1 + 2 shows the marginal effect for values of %Eligible between 20% and 50% and 1 + 3 shows the marginal effect for values of %Eligible greater than 50%. (ii) The linear model implies that 2 = 3 = 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).

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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)

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45

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Stock/Watson - Introduction to Econometrics - Second Edition

8.9.

Note that

Y =  0 + 1 X +  2 X 2 =  0 + ( 1 + 21 2 ) X +  2 ( X 2 − 21X ).

Define a new independent variable Z = X 2 − 21X , and estimate Y =  0 +  X +  2 Z + ui 

The confidence interval is ˆ  196  SE ( ˆ ) .

8.10.

(a)  Y = f ( X 1 +  X 1 , X 2 ) − f ( X 1 , X 2 ) = 1 X 1 + 3  X 1  X 2 , so

Y = 1 +  3 X 2 . X 1

(b)  Y = f ( X 1 , X 2 +  X 2 ) − f ( X 1 , X 2 ) =  2  X 2 + 3 X 1   X 2 , so

Y =  2 + 3 X 1. X 2

(c) Y = f ( X 1 +  X 1 , X 2 +  X 2 ) − f ( X 1 , X 2 ) =  0 + 1 ( X 1 +  X 1 ) +  2 ( X 2 +  X 2 ) + 3 ( X 1 +  X 1 )( X 2 +  X 2 ) − (  0 + 1 X 1 +  2 X 2 +  3 X 1 X 2 ) = ( 1 + 3 X 2 ) X 1 + (  2 + 3 X 1 ) X 2 +  3  X 1 X 2 .

8.11.

Linear model: E(Y | X) = 0 + 1X, so that

1

1 X X = E (Y | X )  0 + 1 X

(

dE (Y | X ) = 1 and the elasticity is dX

 + 1 ln( X ) + u

Log-Log Model: E(Y | X) = E e 0

)

| X = e 0 + 1 ln( X ) E (eu | X ) = ce 0 + 1 ln( X ) , where c =

E(eu | X), which does not depend on X because u and X are assumed to be independent.

Thus 8.12.

dE (Y | X ) 1 0 + 1 ln( X ) E (Y | X ) = ce = 1 , and the elasticity is 1. dX X X

(a) Because of random assignment within the group of returning students E(X1i | ui) = 0 in “-regression,” so that ˆ1 is an unbiased estimator of 1. (b) Because of random assignment within the group of returning students E(X1i | ui) = 0 in “- regression,” so that ˆ1 is an unbiased estimator of 1. (c) Write E(ui | X1i, X2i) = E(ui | X2i) = 0 + 1X2i, where linearity is assumed for the conditional expected value. Thus, E (Y |X 1 , X 2 ) =  0 + 1 X 1 +  2 X 2 +  3 X 1 X 2 + E (ui | X 1i , X 2i ) = ( 0 +  0 ) + 1 X 1 + ( 2 + 1 ) X 2 +  3 X 1 X 2

.

Using this expression, E(Y| X1 = 1, X2 = 0) − E(Y| X1 = 0, X2 = 0) = 1, which is equal to 1from (a). Also, E(Y| X1 = 1, X2 = 1) − E(Y| X1 = 0, X2 = 1) = 1 + 3, which is equal to 1 from (b). Together, these results imply that 3 = 1 − 1. (d) Defining vi = ui − E(ui | X1i, X2i) = ui − E(ui | X2i), then Yi = (0 + 0) + 1X1 + (2 + 1) X2 + 3X1X2 + vi,

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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 ( ˆ0 ) = 0 + 0], an unbiased estimate of 1[ E ( ˆ1 ) = 1], a biased estimate of 2 [ E ( ˆ ) = 2 + 1], and an unbiased estimate of 3 [ E ( ˆ ) = 3].

2

3

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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.

(a) When Yi is measured with error, we have Yi = Yi + wi , or Yi = Yi − wi . Substituting the 2nd equation into the regression model Yi =  0 + 1 X i + ui gives Yi − wi =  0 + 1 X i + ui , or Yi =  0 + 1 X i + ui + wi . Thus vi = ui + wi . (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.

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

(3) vi = ui + wi has a finite fourth moment because both ui and wi have finite fourth moments 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 2 2 regression error ( v =  u +  w ), and this will increase the variance of the OLS estimators.

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

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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

ln(Income)

Calif.

11.57 (1.81) 16.53 (3.15)

Mass.

Std. Dev. of Scores 19.1 15.1

In Points

In Std. Dev.

1.157 (0.18) 1.65 (0.31)

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 =

 1 0 −  0 1  1u − 1v + .  1 − 1  1 − 1

and P = (b) E (Q) =

0 −  0 u − v + .  1 − 1  1 − 1  1 0 −  0 1  −0 , E ( P) = 0  1 − 1  1 − 1 2

 1  2 2 2 2 (c) var(Q) =   ( 1  u + 1  v ), var( P)  −  1   1 2

 1  2 2 =  ( u +  v ),   1 − 1 

and 2

 1  2 2 cov( P, Q) =   ( 1 u + 1 V )  −  1   1

  2 + 1 v2 p cov(Q, P) (d) (i) ˆ1 → = 1 u2 , var( P)  u +  v2

p ˆ0 → E (Q) − E ( P)

cov( P, Q) var( P)

 ( −  ) (ii) ˆ1 − 1 → u 2 1 2 1  0, using the fact that 1  0 (supply curves slope up) and 1  0 u + v (demand curves slope down). p

2

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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 2 100 observation regression is SER200 = (100 − 2)  15.1 = 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 =

1 200 − 2

SSR200 = 15.02. The standard

errors for the regression coefficients are now computed using equation (5.4) where 200 2 2 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.

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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 = 0 + 1( X + wi) + ui = 0 + 1 X + vi, where vi = ui + 1 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.

 ( X i − X )(Yi − Y ) . Because all of the Xi’s are used (although some are used for the (a) ˆ1 = i =1 300  i =1 ( X i − X )2 300

9.13.

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

these expressions:

 i =1 ˆ = 

0.8 n

1

1

( X i − X )2

 (X − X ) n

i =1

2

 +

n i = 0.8 n +1

1

i

( X i − X )( X i − X )

 (X − X ) n

i =1

i

2

 ( X − X )(u − u ) +  (X − X ) n

i =1

i

i

n

i =1

2

i

1 0.8 n 1 n 1 n ( X i − X )2 ( X i − X )( X i − X )    i =1 ( X i − X )(ui − u ) i =1 i = 0.8 n +1 = 1 n + 1 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 ˆ1 have the following probability limits: p 1 n 2 ( X − X ) →  X2 ,  i i =1 n p 1 0.8n 2 ( X − X ) → 0.8 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 =0.8n +1

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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 ˆ1 → 0.81 . p

p

(b) Because ˆ1 → 0.81 , ˆ1 / 0.8 → 1 , so a consistent estimator of 1 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( ˆ1 (240obs )) =

1 var  ( X i −  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 error. The first is which is the usual source that comes from the 300  i =1 ( X i − X )2

 omitted factors (u). The second is  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 −  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  = 12 var  1 i = 241 300 .

2   300 ( X − X )  i i =1   Thus  ˆ (300obs)  1  1 var  ( X i −  X )ui  0.2  var  1 = + 12    2   0.64  300 0.8 300  var( X )     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  and the coefficient associated with west  (unemployment rate) is . Then  captures the effect of the unemployment rate in the eastern states, and  +  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 . Using the coefficient estimate (ˆ ) and the standard error SE(ˆ ), you can calculate the t-statistic to test whether  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?)

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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 = 1, intercept = 0 (b) slope = 1, intercept = 0 (c) slope = 1, intercept = 0 + 3 (d) slope = 1, intercept = 0 + 3

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 0 = 1 + 1; i =  i − 1 and t =  t − 1.

10.6.

vit = X it uit . First note that E (vit ) = E ( X it uit ) = E[ X it E (uit | X i1 , X i 2 ,..., X iT ,  i )] = 0 from assumption 1. Thus, cov(vit vis ) = E (vit vis ) = E ( X it X is uit uis ) . The assumptions in Key Concept 10.3 do not 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 = 0 + 1X1, it + 2t + 2D2i +  + nDni + 2(D2i  t) +  + n(Dni  t) + uit, where D2i = 1 if i = 2 and 0 otherwise and so forth. The coefficient i = 2 + i.

10.9.

(a) ˆi = T1 Tt =1 Yit which has variance small. ˆ i is not consistent.

 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.

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Solutions to End-of-Chapter Exercises

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

ˆ1DM =

1



1

  4 ( X − X )(Y − Y ) + 4 ( X − X )(Y − Y )  i =1

i2

i1

i2

i1

1

i2

i1

1

i2



  4 ( X − X ) + 4 ( X − X )  n

2

i =1

i2

i1

2

i2

i1

 ( X − X )(Y − Y ) = ˆ =  (X − X ) n

i =1

i2

i1

i2

n

i =1

i1

2

i2

BA 1

i1

©2018 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;  (1.022) = 0.847 (c) zChristopher = 0.712 + 0.031 0 = 0.712; (0.712) = 0.762 (d) z Jed = 0.712 + 0.031 80 = 3.192; (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 = 0.811 1+ e 1 + e−1.459 1 1 ProbChristopher = = = 0.742 − (1.059 + 0 .040  0) 1+ e 1 + e−1.059 Prob Matthew =

− (1.059 + 0.040  10)

=

(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.

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Solutions to End-of-Chapter Exercises

57

ProbMatther = 0.774 + 0.006  10 = 0.836 ProbChristopher = 0.774 + 0.006  0 = 0.774 (b)

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

Probit

Logit

(1.282 − 0.333) = 0.829

1 1+ e

Women

− (2.197 − 0.622)

(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) (0.806 + 0.041  10  0.174  1 − 0.015  1  10) = 0.814 (b) (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.

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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 (− 2.26 + 2.74  0.35 + 0.71) = (− 0.59) = 27.76%. (b) With the P/I ratio reduced to 0.30, the probability of being denied is (− 2.26 + 2.74  0.30 + 0.71) = (− 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 (−2.26 + 2.74  0.35) = 9.7%. If the P/I ratio is reduced to 0.30, the probability of being denied is (−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 = 27.28%. denied is F (−4.13 + 5.37  0.35 +1.27) = 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 + e2.2505 1 = 7.45%. The difference in probability of being denied is F (−4.13 + 5.37  0.30) = 1 + e2.519 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) = 0 + 1Xi. Thus

E (ui | X i ) = E[Yi − (  0 + 1 X i )| X i ] = E (Yi|X i ) − (  0 + 1 X i ) = 0 (b) Using Equation (2.7), we have

var(Yi | X i ) = Pr(Yi = 1| X i )[1 − Pr(Yi + 1| X i )] = ( 0 + 1 X i )[1 − ( 0 + 1 X i )].

Thus

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

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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 = 0 + 1Xi. The conditional probability y 1− y distribution for the ith observation is Pr(Yi = yi | X i ) = pi i (1 − pi ) i . Assuming that (Xi, Yi) are 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 ,

n

, X n ) =  Pr(Yi = yi |X i ) i =1

n

=  piyi (1 − pi )1− yi i =1 n

=  (  0 + 1 X i ) yi [1 − (  0 + 1 X i )]1− yi . i =1

The likelihood function is the above joint probability distribution treated as a function of the unknown coefficients (0 and 1).

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 ,

n

, Yn = yn ) =  Pr(Yi = yi ) = p n1 q n2 (1 − p − q) n − n1 − n2 . i =1

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

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

pˆ =

n1 n , qˆ = 2 . 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.

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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% = − .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 TSLS (c) The TSLS estimator is ˆ1 = ZY using Equation (10.4) in the text. Since Zi = Xi, we have sZX

ˆ1TSLS = 12.3.

sZY s XY = 2 = ˆ1OLS . sZX sX

2 (a) The estimator ˆ a2 = n −1 2 in=1 (Yi − ˆ0TSLS − ˆ1TSLS Xˆ i ) 2 is not consistent. Write this as ˆ a = 1  n (uˆ − ˆ TSLS ( Xˆ − X )) 2 , where uˆ = Y − ˆ TSLS − ˆ TSLS X . Replacing ˆ TSLS with 1, n−2

i =1

i

1

i

i

i

i

0

1

i

1

as suggested in the question, write this as ˆ a2  1n  in=1 (ui − 1 ( Xˆ i − X i )) 2 = 1n  in=1 ui2 + 1 n

 in=1[ 12 ( Xˆ i − X i ) 2 + 2ui 1 ( Xˆ i − X i )]. The first term on the right hand side of the equation

converges to ˆ u , but the second term converges to something that is non-zero. Thus ˆ a is not consistent.

2

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(b) The estimator ˆ b2 = n −1 2 in=1 (Yi − ˆ0TSLS − ˆ1TSLS X i ) 2 is consistent. Using the same notation as in (a), we can write ˆ b2  1n in=1ui2 , and this estimator converges in probability to  u .

2

12.4.

Using Xˆ i = ˆ0 + ˆ1Zi , we have Xˆ = ˆ0 + ˆ1Z and n

n

ˆ ˆ s XY ˆ =  ( X i − X )(Yi − Y ) = ˆ1  ( Z i − Z )(Yi − Y ) = ˆ1 sZY , i =1

i =1

n

n

i =1

i =1

s X2ˆ =  ( Xˆ i − Xˆ ) 2 = ˆ12  ( Z i − Z ) 2 = ˆ12 sZ2 .

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

ˆ1 = ZX2 .

Thus the TSLS estimator

ˆ1TSLS = 12.5.

s XY ˆ s

2 Xˆ

=

ˆ1sZY s s s = ZY2 = s ZY 2 = ZY .

2 2 ˆ1 sZ ˆ1sZ  sZ sZX s ZX 2 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 R2 / k 0.05 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 1 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.

2

−  − 0 uid − uis (a) Solving for P yields P = 0 ; thus cov( P, u s ) = u + 1 1 1

2

12.8.

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). 0 can be estimated by OLS (equivalently as the sample mean of Qi).

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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.

cov( Z i , Yi ) cov( Z i ,  0 + 1 X i +  2Wi + ui ) 1cov( Z i , X i ) +  2 cov( Z i ,Wi ) 12.10. ˆTSLS = = = 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.

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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.

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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 pˆ Men (1 − pˆ Men ) 0.55(1 − 0.55) 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) 0 + 2 (ii) 0 + 1 + 2 + 3 (iii) 0 (iv) 0 + 3 (c) 1 (d) “New Jersey after − New Jersey before” = 1 + 3, where 3 denotes the time effect associated with changes in the economy between 1991 and 1993. “1993 New Jersey − 1993 Pennsylvania” = 1 + 2, where 2 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 =  i + 1 X it + uit . (a) i is an individual-specific intercept. The random effect in the regression has variance var( i + uit ) = var( i ) + var(uit ) + 2cov( i , uit ) =  2 +  u2

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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( ˆ1differences ) →

var( i + ui 2 )  2 +  u2 = . var( X i 2 ) var( X i 2 )

(b) The regression equation using the differences-in-differences estimator is Yi = 1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

 v2 = var(ui 2 − ui1 ) = var(ui 2 ) + var(ui1 ) − 2cov(ui 2 , ui1 ) = 2 u2 , which is homoskedastic. Using Equation (5.27), it is straightforward to see that the variance for the differences-in-differences estimator

n var( ˆ1diffs −in −diffs ) →

 v2 var(X i )

=

2 u2 . var( X i 2 )

(c) When  2   u2 , we’ll have var ( ˆ1differences )  var ( ˆ1diffs −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 =  i + 1 X it +  2 ( Dt  Wi ) +  0 Dt + vit ,

we have Yi 2 − Yi1 = 1 ( X i 2 − X i1 ) +  2 [( D2 − D1 )  Wi ] + 0 ( D2 − D1 ) + (vi 2 − vi1 ).

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 =  0 + 1 X i +  2Wi + ui ,

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

13.8.

The regression model is Yit =  0 + 1 X it +  2Gi + 3 Bt + uit ,

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Solutions to End-of-Chapter Exercises

67

Using the results in Section 8.3 Y control,before = ˆ0 Y control,after = ˆ + ˆ 0

Y

3

treatment,before

= ˆ + ˆ

treatment,after

= ˆ + ˆ + ˆ + ˆ

0

Y

2

0

1

2

3

Thus

ˆ diffs-in-diffs = (Y treatment,after − Y treatment,before ) − (Y control,after − Y control,before ) = ( ˆ + ˆ ) − ( ˆ ) = ˆ 1

13.9.

3

3

1

The covariance between 1i X i and Xi is cov( 1i X i , X i ) = E{[ 1i X i − E ( 1i X i )][ X i − E ( X i )]} = E{1i X i2 − E ( 1i X i ) X i − 1i X i E ( X i ) + E ( 1i X i ) E ( X i )} = E ( 1i X i2 ) − E ( 1i X i ) E ( X i )

Because Xi is randomly assigned, Xi is distributed independently of 1i. The independence means

E ( 1i X i ) = E ( 1i ) E ( X i ) and E ( 1i X i2 ) = E ( 1i ) E ( X i2 ).

Thus cov( 1i X i , X i ) can be further simplified:

cov( 1i X i , X i ) = E ( 1i )[ E ( X i2 ) − E 2 ( X i )] = E ( 1i ) X2 .

So

cov( 1i X i , X i )

 X2

=

E ( 1i ) X2

 X2

= E ( 1i ).

13.10. (a) This is achieved by adding and subtracting 0 + 1Xi to the right hand side of the equation and rearranging terms. (b) E[ui |Xi] = E[0i − 0 |Xi] + E[(1i − 1)Xi |Xi]|Xi] + E[vi |Xi] = 0. (c) (1) is shown in (b). (2) follows from the assumption that (vi, Xi, 0i, 1i) are i.i.d. random variables. (d) Yes, the assumptions in KC 4.3 are satisfied. (e) If 1i and Xi are positively correlated, cov(1i, Xi) = E[(1i − 1)(Xi − X)] = E[(1i − 1)Xi ]  0, where the first equality follows because 1 = E(1i) and the inequality follows because the covariance and correlation have the same sign. Note 0  E[(1i − 1)Xi] = E{E[(1i − 1)Xi |Xi]} by the law of iterated expectations, so it must the case that E[(1i − 1)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.

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13.11. Following the notation used in Chapter 13, let 1i denote the coefficient on state sales tax in the “first stage” IV regression, and let −1i denote cigarette demand elasticity. (In both cases, suppose that income has been controlled for in the analysis.) From (13.11) p

ˆ TSLS →

E ( 1i  1i ) cov( 1i ,  1i ) cov( 1i ,  1i ) = E ( 1i ) + = Average Treatment Effect + , E ( 1i ) E ( 1i ) E ( 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( 1i ,  1i )  0. As discussed in Section 13.6, this covariance is zero when 1i or 1i 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 (1i is not constant) as does the effect of sales taxes on cigarette prices ( 1i is not constant). Are 1i and 1i 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 1i and 1i are positively related, so that cov( 1i ,  1i )  0. Because E(1i)  0, this suggests that the local average treatment effect is greater than the average treatment effect when 1i varies from state to state.

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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) = 0 + 1E(Yt–1) + E(ut), but E(ut) = 0 and E(Yt) = E(Yt–1). Thus E(Yt) = 0 + 1E(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 Yˆt |t −1 = 1.377 + [0.318  ( −7.36)] + [0.123  ( −3.67)] + [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 =

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(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 BIC( p) = ln  and  + ( 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 − W ) + ( W − c)]2 } = E[(W − W ) 2 ] + 2 E (W − W )( W − c) + ( W − c) 2 =  W2 + ( W − c) 2 .

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

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with the conditional variance  t2|t −1 = E[(Yt − Yt |t −1 ) 2 ]. This equation is minimized when the second term equals zero, or when ft −1 = Yt |t −1. (An alternative is to use the hint, and notice that the result follows immediately from exercise 2.27.) (c) Applying Equation (2.27), we know the error ut is uncorrelated with ut – 1 if E(ut |ut – 1) = 0.

From Equation (14.14) for the AR(p) process, we have ut −1 = Yt −1 − 0 − 1Yt − 2 −  2Yt −3 −

−  pYt − p −1 = f (Yt −1 , Yt − 2 ,..., Yt − p −1 ),

a function of Yt – 1 and its lagged values. The assumption E (ut |Yt −1 , Yt − 2 ,...) = 0 means that conditional on Yt – 1 and its lagged values, or any functions of Yt – 1 and its lagged values, ut has mean zero. That is, E (ut | ut −1 ) = E[ut | f (Yt −1 , Yt − 2 ,..., Yt − p − 2 )] = 0.

Thus ut and ut – 1 are uncorrelated. A similar argument shows that ut and ut – j are uncorrelated for all j  1. Thus ut is serially uncorrelated.

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.

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(b) The 1st autocovariance is

cov(Yt , Yt −1 ) = cov(2.5 + 0.7Yt −1 + ut , Yt −1 ) = 0.7 var(Yt −1 ) + cov(ut , Yt −1 ) = 0.7 Y2 = 0.7  17.647 = 12.353.

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

(c) The 1st autocorrelation is corr (Yt , Yt −1 ) =

cov(Yt , Yt −1 )

0.7 Y2

= 0.7.

0.7 2  Y2

= 0.49.

=

var(Yt ) var(Yt −1 )

 Y2

The 2nd autocorrelation is corr (Yt , Yt − 2 ) =

cov(Yt , Yt − 2 ) var(Yt ) var(Yt − 2 )

=

 Y2

(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) = 0 + 1Febt + 2Mart +  + 11Dect. For observations in January, all of the regressors are equal to zero, thus Jan = 0. For observations in February, Febt = 1 and the other regressors equal 0, so that Feb = 0 + 1. Similar calculations apply for the other months.

14.9.

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

2 (b) var(Yt ) = var(et ) + b1 var(et −1 ) +

+2b1 cov(et , et −1 ) + =  (1 + b + 2 e

2 1

+ bq2 var(et − q )

+ 2bq −1bq cov(et − q +1 , et − q )

+b ) 2 q

where the final equality follows from var(et) =  e for all t and cov(et, ei) = 0 for i t.

2

(c) Yt = 0 + et + b1et–1 + b2et – 2 +  + bqet – q and Yt–j = 0 + et – j + b1et – 1 – j + b2et – 2 – j +  + bqet – q – j and cov(Yt, Yt – j) =  qk = 0  mq = 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.

(

)

2 2 (d) var(Yt ) =  e 1 + b1 , cov(Yt , Yt − j ) =  e2b1 , and cov(Yt , Yt − j ) = 0 for j  1.

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

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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 = 0 + 1(Yt – 1 − Yt – 2) + ut. Rearranging yields Yt = 0 + (1 + 1)Yt – 1 − 1Yt – 2 + ut.

©2018 Pearson Education, Inc. Publishing as Addison Wesley

Chapter 15 Estimation of Dynamic Causal Effects 15.1.

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

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

Dynamic multiplier (i) −0.055 −0.026 −0.031 −0.109 −0.128 0.008 0.025 −0.019 0.067

Predicted effect on output growth (25i)

95% confidence interval 25  [i  1.96SE (i)]

−1.375 −0.65 −0.775 −2.725 −3.2 0.2 0.625 −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  [i  1.96SE (i)] 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. i is the dynamic multiplier. With the 25% oil price jump, the predicted change in interest rates for the ith quarter is 25i.

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

Dynamic multiplier (i)

0.062 0.048 −0.014 −0.086 −0.000 0.023 −0.010 −0.100

Predicted change in interest rates (25i)

95% confidence interval 25  [i  1.96SE (i)]

1.55 1.2 −0.35 −2.15 0 0.575 −0.25 −2.5

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[−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]

Solutions to End-of-Chapter Exercises

8

−0.014

−0.35

75

[−1.575, 0.875]

(b) The 95% confidence interval for the predicted change in interest rates for the ith quarter from the 25% oil price jump is 25  [i  1.96SE (i)]. 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 + X t −1 =  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 =  0 + 1 X t +  2 X t −1 +  3 X t − 2 + =  0 + 1 ( X t +  X t −1 + +  2 ( X t −1 + +

+  r +1 X t − r + ut

+  X t − r +1 + X t − r )

+  X t − r +1 + X t − r )

+  r ( X t − r +1 + X t − r ) +  r +1 X t − r + ut

=  0 + 1 X t + ( 1 +  2 ) X t −1 + ( 1 +  2 +  3 ) X t − 2 +

+ ( 1 +  2 +

+ ( 1 +  2 +

+  r ) X t − r +1

+  r +  r +1 ) X t − r + ut .

Comparing the above equation to Equation (15.7), we see 0 = 0, 1 = 1, 2 = 1 + 2, 3 = 1 + 2+ 3,…, and r + 1 = 1 + 2 ++ r + r + 1.

15.6.

(a) Write

var(ut ) = 12 var(ut −1 ) + var(ut ) + 21 cov(ut −1 , ut ) = 12 var(ut ) +  u2 where the second equality follows from stationarity (so that var(ut) = var(ut–1) and cov(ut–1, ut ) = 0.

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

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(b) Write

cov(ut , ut −1 ) = 1 var(ut −1 ) + cov(ut −1 , ut ) = 1 var(ut ) showing the result for j = 1. For j = 2, write cov(ut , ut − 2 ) = 1 cov(ut −1 , ut − 2 ) + cov(ut − 2 , ut ) = 1 cov(ut , ut −1 ) = 12 var(ut )

and similarly for other values of j. The result for X is similar. cov(ut , ut − j ) cov(ut , ut − j ) = (c) cor(ut, ut−j) = from stationarity. The result then follows from var(ut ) var(ut ) var(ut − j ) (b). The result for X is similar. (d) vt = (Xt − X)ut 2 2 2 2 2 2 2 (i) E (vt ) = E[( X t −  X ) ut ] = E[( X t −  X ) ]E[ut ] =  X  u , where the second equality follows because Xt and ut are independent. (ii) cov(vtvt – j) = E[(Xt − X)( Xt – j − X)utut – j] = E[(Xt − X)( Xt – j − X)]E[utut – j] = 1 +  11 .  1j var( X t )1j var(ut ), so that cor(vt, vt – j) =  1j1j . Thus f = 1 + 2 i=1 ( 11 )i = 1 −  1

15.7.

 i Write ut =  i = 0 1 ut −i

(a) Because E (ui | 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 | ut +1 ) = 0 for j  0, Xt is exogenous. However E(ut+1 | ut +1 ) = ut +1 so that Xt is not strictly exogenous.

15.8.

(a) Because Xt is exogenous, OLS is consistent. (b) The GLS estimator regresses Yt − 1Yt – 1 onto Xt − 1Xt – 1. The error term in this regression is ut . Because Xt = ut +1 , Xt − 1 Xt–1 = ut +1 − 1 ut , which is correlated with the regression error. Thus the GLS estimator will be inconsistent. cov( X t − 1 X t −1 , ut )  2 1 p (c) ˆ1 → 1 + = 1 − 2 u 1 2 = 1 − var( X t − 1 X t −1 )  u (1 + 1 ) (1 + 12 )

15.9.

(a) This follows from the material around equation (3.2). (b) Quasi-differencing the equation yields Yt – 1Yt – 1 = (1 − 1)0 + ut , and the GLS estimator of (1 − 1)0 is the mean of Yt – 1Yt – 1 = T1−1 Tt= 2 (Yt − 1Yt −1 ) . Dividing by (1−1) yields the GLS estimator of 0. (c) This is a rearrangement of the result in (b). (d) Write ˆ = 1 T Y = 1 (Y + Y ) + T −1 1 T −1 Y , so that ˆ − ˆ GLS = 1 (Y + Y ) − 1 1 T −1 Y − 0

T

t =1

t

T

T

1

T

T −1

t =2

t

0

0

1 1 1 (YT − Y1 ) and the variance is seen to be proportional to 2 .

1− T −1 T

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T

T

1

T T −1

t =2

t

Solutions to End-of-Chapter Exercises

15.10.

Multipliers 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.

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Chapter 16 Additional Topics in Time Series Regression 16.1.

Yt follows a stationary AR(1) model, Yt = 0 + 1Yt −1 + ut . The mean of Yt is Y = E (Yt ) =

0 , 1 − 1

and E (ut |Yt ) = 0. (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 (  0 + 1Yt + h −1 + ut |Yt , Yt −1 , ) =  0 + 1Yt + h −1|t =  0 + 1 (  0 + 1Yt + h − 2|t ) = (1 + 1 )  0 + 12Yt + h − 2|t = (1 + 1 )  0 + 12 (  0 + 1Yt + h −3|t ) = (1 + 1 + 12 )  0 + 13Yt + h −3|t = = (1 + 1 + =

+ 1h −1 )  0 + 1hYt

1 − 1h  0 + 1hYt 1 − 1

= Y + 1h (Yt − Y ).

(b) Substituting the result from part (a) into Xt gives 



X t =   iYt +i |t =   i [ Y + 1i (Yt − Y )] i =0

i =0





i =0

i =0

= Y   i + (Yt − Y ) ( 1 )i Y − Y  = Y + t .

1 −  1 − 1 16.2.

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

Rkt =

1 k 1 k R1t +i|t + et =  R1t + et = R1t + et .  k i =1 k i =1

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

= 0.5i  R1t .

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 + (0.5 + = R1t +

+  R1t +i|t

+ 0.5i )  R1t

0.5(1 − 0.5i )  R1t .

1 − 0.5

Thus 1 k 1 k R 1 + e =  [ R1t + (1 − 0.5i )R1t ] + et t k i =1 t +i t k i =1 = R1t + R1t + et .

Rkt =

where  = 1k ik=1 (1 − 0.5i ). Thus Rkt − R1t = R1t + et . Thus Rkt and R1t are cointegrated. The cointegrating coefficient is 1. (d) When R1t = 0.5R1t – 1 + ut, R1t is stationary and does not have a stochastic trend.

R1t + i|t = 0.5i R1t , so that, Rkt =  R1t + et , where  = 1k ik=1 0.5i. Since R1t and et are I (0), then Rkt is I (0).

16.3.

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

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

var (ut | ut −1 ) =  t2 = 1.0 + 0.5ut2−1.

The unconditional mean of ut is E (ut) = 0, and the unconditional variance of ut is var (ut ) = var[ E (ut |ut −1 )] + E[var (ut | ut −1 )] = 0 + 1.0 + 0.5E (ut2−1 ) = 1.0 + 0.5var (ut −1 ).

The last equation has used the fact that E (ut ) = var(ut ) + E (ut )] = var(ut ), which follows because E (ut) = 0. Because of the stationarity, var(ut–1) = var(ut). Thus, var(ut) = 1.0 + 0.5var(ut) which implies var(ut ) = 1.0 / 0.5 = 2.

2

2

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(b) When ut −1 = 0.2,  t = 1.0 + 0.5  0.2 = 1.02. The standard deviation of ut is t = 1.01. Thus 2

2

 −3 ut 3  Pr ( −3  ut  3) = Pr      1.01  t 1.01  =  (2.9703) −  (−2.9703) = 0.9985 − 0.0015 = 0.9970.

When ut–1 = 2.0,  t = 1.0 + 0.5  2.0 = 3.0. The standard deviation of ut is t = 1.732. Thus 2

2

 −3 u 3  Pr ( −3  ut  3) = Pr   t    1.732  t 1.732  = (1.732) − (−1.732) = 0.9584 − 0.0416 = 0.9168.

16.4.

Yt follows an AR(p) model Yt =  0 + 1Yt −1 + E (ut + h |Yt , Yt −1 ,

+  pYt − p + ut . E (ut |Yt −1 , Yt − 2 , ) = 0 implies

) = 0 for h  1. The h-period ahead forecast of Yt is

Yt + h|t = E (Yt + h |Yt , Yt −1 , ) = E (  0 + 1Yt + h −1 +

+  pYt + h − p + ut + h |Yt , Yt −1 , )

=  0 + 1 E (Yt + h −1|Yt , Yt −1 , ) + +  p E (Yt + h − p |Yt , Yt −1 , ) + E (ut + h |Yt , Yt −1 , ) =  0 + 1Yt + h −1|t + 16.5.

+  pYt + h − p|t .

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

T

T

T

T

Y = (Y + Y ) = Y +  (Y ) + 2Y Y .

2

t =1

t

2

t −1

t =1

t

t =1

2 t −1

2

t =1

t

t =1

t −1

t

So T T 1 T 1 1 T  Yt −1Yt =   Yt 2 −  Yt 2−1 −  (Yt )2  .  T t =1 T 2  t =1 t =1 t =1 

(

) (

)

2 2 2 Note that Tt =1 Yt 2 − Tt =1 Yt 2−1 = Tt =−11 Yt 2 + YT2 − Y02 + Tt =−11 Yt 2 = YT − Y0 = YT because Y0 = 0. Thus: 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 

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Solutions to End-of-Chapter Exercises

16.6.

16.7.

81

(a) Rewrite the regression as Yt = 3.0 + 2.3Xt + 1.7(Xt+1 − Xt) + 0.2(Xt − Xt–1) + ut Thus  = 2.3, –1 = 1.7, 0 = 0.2 and 1 = 0.0. (b) Cointegration requires Xt to be I(1) and ut to be I(0). (i) No (ii) No (iii) Yes 1

 Y X =  Y Y = T  Y Y . Following the hint, the numerator is the same ˆ =  X  ( Y ) 1  ( Y ) T

T

t t =1 t T 2 t t =1

t =1 t T t =1

T

t +1

t =1 t

t +1

2

T

t +1

T

2

t +1

t =1

d u expression as (16.21) (shifted forward in time 1 period), so that T1 Tt =1 Yt Yt +1 → ( 12 − 1). The 2 2

denominator is T1 Tt =1 (Yt +1 )2 = T1 Tt =1 ut2+1 → u2 by the law of large numbers. The result follows directly. p

16.8.

(a) First note that Yt–1/t–2 = 11Yt–2 + 11Xt–2 and Xt–1/–2 = 21Yt–2 + 21Xt–2. Also Yt/t–2 = 11Yt–1/t–2 + 11Xt–1/t–2. Substituting yields

Yt / − 2 = 11 ( 11Yt − 2 +  11 X t − 2 ) +  11 (  21Yt −2 +  21 X t −2 ) = [ 1111 +  11 21 ]Yt −2 + [ 11 11 +  11 21 ] X t −2 so that 1 = 1111 + 1121 and 2 = 1111 + 1121. (b) There is no difference in iterated multistep or direct forecast if the values of 1 and 2 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 ( t2 ) = E ( 0 + 1ut2−1 ) =  0 + 1 E ( ut2−1 ) =  0 + 1 E ( ut2 ) 2

where the last line uses stationarity of u. Solving for E (ut ) gives the required result. (b) As in (a)

E (ut2 ) = E ( t2 ) = E ( 0 + 1ut2−1 +  2ut2− 2 +

+  p ut2− p )

=  0 + 1 E ( ut2−1 ) +  2 E ( ut2− 2 ) + =  0 + 1 E ( ut2 ) +  2 E ( ut2 ) +

+  p E ( ut2− p ) +  p E ( ut2 )

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so that E (ut2 ) =

0

1 − tp=1  i

.

2

(c) This follows from (b) and the restriction that E (ut ) > 0. (d) As in (a) E (ut2 ) = E ( t2 ) =  0 + 1 E ( ut2−1 ) + 1 E ( t2−1 ) =  0 + (1 + 1 ) E ( ut2−1 ) =  0 + (1 + 1 ) E ( ut2 ) =

0 1 − 1 − 1

( )

(e) This follows from (d) and the restriction that E ut2  0.

16.10. Write Yt = Xt + v1t and Xt = v2t; also v1t–1 = Yt–1 − Xt–1. Thus Yt = −(Yt−1 − Xt–1) + u1t and Xt = u2t, with u1t = v1t + v2t and u2t = v2t.

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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 1. Given the estimator b1, the sum of squared errors for the given regression model is n

 (Y − b X ) .

2

i =1

i

1

i

ˆ1RLS , the restricted least squares estimator of 1, minimizes the sum of squared errors. That is, ˆ1RLS 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

 2(Y − b X )(− X ) = 0. i =1

i

1

i

i

Solving for b1 from the first order condition leads to the restricted least squares estimator

ˆ1RLS =

in=1 X iYi . in=1 X i2

(b) We show first that ˆ1RLS is unbiased. We can represent the restricted least squares estimator ˆ1RLS in terms of the regressors and errors:

ˆ1RLS =

in=1 X iYi in=1 X i ( 1 X i + ui ) in=1 X i ui = =  + .

1 in=1 X i2 in=1 X i2 in=1 X i2

Thus  n X u    n X E (u | X , , X n )  E ( ˆ1RLS ) = 1 + E  i =n1 i 2 i  = 1 + E  i =1 i ni 1 2  = 1 ,  i =1 X i   i =1 X i   

where the second equality follows by using the law of iterated expectations, and the third equality follows from

in=1 X i E (ui | X 1 , in=1 X i2

, Xn)

=0

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.

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Under assumptions 1−3 of Key Concept 17.1, ˆ1RLS is asymptotically normally distributed. The large sample normal approximation to the limiting distribution of ˆ1RLS follows from considering

ˆ1RLS − 1 =

in=1 X i ui 1n in=1 X i ui = 1 n .

2 in=1 X i2 n  i =1 X i

Consider first the numerator which is the sample average of vi = Xiui. By assumption 1 of Key Concept 17.1, vi has mean zero: E ( X i ui ) = E[ X i E (ui | X i )] = 0. By assumption 2, vi is i.i.d. By assumption 3, var(vi) is finite. Let v = 1n  in=1 X i ui , then  v2 =  v2 /n. Using the central limit theorem, the sample average

1

v / v =

n

 v → N (0, 1) d

 v n i =1

i

or n

1

 X u → N (0,  ). n 2 v

d

i =1

i i

2

For the denominator, X i 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 n ( ˆ 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) ˆ1RLS is a linear estimator:

ˆ1RLS =

in=1 X iYi n =  i =1 aiYi , n 2 i =1 X i

where ai =

Xi .  X i2 n i =1

The weight ai (i = 1,, n) depends on X1,, Xn but not on Y1,, Yn. Thus

ˆ1RLS = 1 +

in=1 X i ui . in=1 X i2

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Solutions to End-of-Chapter Exercises

85

ˆ1RLS is conditionally unbiased because  n X u , X n = E  1 + i =n1 i 2 i | X 1 ,  i =1 X i 

E ( ˆ1RLS | X 1 ,

 , Xn  

 n X u = 1 + E  i =n1 i 2 i | X 1 ,   i =1 X i

 , X n  = 1 . 

The final equality used the fact that  n X u E  i =n1 i 2 i | X 1 ,   i =1 X i

  n X E (u | X , , X n  = i =1 i n i 12  i =1 X i 

, Xn)

=0

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

  n X u , X n ) = var  1 + i =n1 i 2 i | X1 , , X n   i =1 X i   n 2  X var(u | X , , X n ) = i =1 i n i 2 12 (i =1 X i )

var( ˆ1RLS | X1,

= =

in=1 X i2 u2 ( in=1 X i2 )2

 u2 in=1 X i2

.

(e) The conditional variance of the OLS estimator ˆ1 is

var( ˆ1 |X1 , , X n ) =

 u2 in=1 ( Xi − X )2

.

Since n

n

n

n

n

 ( X − X ) =  X − 2 X  X + nX =  X − nX   X , 2

i =1

i

i =1

2 i

2

i

i =1

i =1

2 i

2

the OLS estimator has a larger conditional variance: var( 1|X 1 , The restricted least squares estimator ˆ RLS is more efficient.

i =1

2 i

, X n )  var( ˆ1RLS |X 1 ,

1

(f) Under assumption 5 of Key Concept 17.1, conditional on X1,, Xn, ˆ1RLS is normally distributed since it is a weighted average of normally distributed variables ui:

ˆ1RLS = 1 +

in=1 X i ui . in=1 X i2

©2018 Pearson Education, Inc. Publishing as Addison Wesley

, X n ).

86

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Using the conditional mean and conditional variance of ˆ1RLS derived in parts (c) and (d) respectively, the sampling distribution of ˆ RLS , conditional on X1,, Xn, is 1



ˆ1RLS ~ N  1 , 

 .

X 

 u2 

n i =1

2 i

(g) The estimator

1 =

in=1 Yi in=1 ( 1 X i + ui ) in=1 ui = =  + 1 in=1 X i in=1 X i in=1 X i

The conditional variance is

  n u , X n) = var  1 + 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

var( 1 | X 1 ,

=

n u2 . (in=1 X i ) 2

The difference in the conditional variance of 1 and ˆ1RLS is

var( 1| X 1 ,

, X n ) − var( ˆ1RLS | X 1 ,

In order to prove var( 1 | X 1 ,

, X n) =

, X n )  var( ˆ1RLS | X 1 ,

n u2  u2 − . (in=1 X i )2 in=1 X i2 , X n ), we need to show

n 1  ( in=1 X i ) 2  in=1 X i2

or equivalently 2

n  n  n X i2    X i  . i =1  i =1 

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

n n  n  2 2 ( a  b )  a    i i   i  bi i =1 i =1  i =1 

which implies 2

2

n n n  n   n  2 2 X = 1  X  1  X = n X i2 .      i i i     i =1 i =1 i =1  i =1   i =1 

That is nin=1 X i2  ( nx =1 X i )2, or var( 1| X 1 ,

, X n )  var( ˆ1RLS | X 1 ,

, X n ).

Note: because 1 is linear and conditionally unbiased, the result var(  | X , , X )  var( ˆ RLS | X , , X ) follows directly from the Gauss-Markov theorem.

1

1

n

1

1

n

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Solutions to End-of-Chapter Exercises

17.2.

87

The sample covariance is s XY =

1 n  ( X i − X )(Yi − Y ) n − 1 i =1

=

1 n {[ X i −  X ) − ( X −  X )][Yi − Y ) − (Y − Y )]} n = 1 i =1

=

n 1  n  ( X i −  X )(Yi − Y ) −  ( X −  X )(Yi − Y ) n − 1  i =1 i =1

n n  − ( X i −  X )(Y − Y ) +  ( X −  X )(Y − Y )  i =1 i =1  n n 1 n  =  ( X i −  X )(Yi − Y )  − n − 1 ( X −  X )(Y − Y ) n − 1  n i =1

where the final equality follows from the definition of X and Y which implies that  i =1 ( X i −  X ) = n

n( X −  X ) and in=1 (Yi − Y ) = n(Y − Y ), 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 see the second term converges in probability to zero because X →  X and Y → Y so p ( X −  X )(Y − Y ) → 0 by Slutsky’s theorem. Let’s look at the first term. Since (Xi, Yi) are i.i.d., the random sequence (Xi − X) (Yi − Y) are i.i.d. By the definition of covariance, we have E[( X i −  X )(Yi − Y )] =  XY . To apply the law of large numbers on the first term, we need to have

var[( X i −  X )(Yi − Y )]  

which is satisfied since var[( X i −  X )(Yi − Y )]  E[( X i −  X ) 2 (Yi − Y ) 2 ]  E[( X i −  X ) 4 ]E [(Yi − Y ) 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 ( X i −  X )(Yi − Y ) → E[( X i −  X )(Yi − Y )] =  XY .  n i =1 Also, n−n 1 → 1, so the first term for sXY converges in probability to XY. Combining results on the two terms for sXY , we have s XY →  XY . p

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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 ( ˆ1 − 1 ) = n n1

in=1[( X i −  X ) − ( X −  X )]ui n 2 1 n  i =1 ( X i − X )

1

= nn =

1 n

in=1 ( X i −  X )ui

1 n

in=1 ( X i − X )2 in=1 vi

1 n

=1

in=1 ( X i − X )2

n

−

−

( X − X ) 1 n

in=1 ui

 in=1 ( X i − X )2

( X − X ) 1 n

1 n

1 n

in=1 ui

 in=1 ( X i − X )2

by defining vi = (Xi − X)ui. (b) The random variables u1,, un are i.i.d. with mean u = 0 and variance 0   u  . By the central limit theorem, 2

n (u − u )

u

1 n

=

in=1 ui d → N (0, 1).

u

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

p

variance, 1n in=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 − X) ui has finite variance:

var(vi ) = var[( X i −  X ) i ]  E[( X i −  X ) 2 ui2 ]  E[( X i −  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

v

=

1 n

in=1 vi

v

satisfies the central limit theorem. (d) Applying the central limit theorem, we have 1 n

in=1 vi d → N (0, 1).

v

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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

v

 ( X t − X )2 n i =1

d → N (0, 1),

 X2

or equivalently

1 n

 in=1 vi

 var(vi )  d → N  0, .

2   (Xi − X )  [var( X i )]  1 n

n i =1

2

Thus n ( ˆ1 − 1 ) = 1 n

1 n

in=1 vi

in=1 ( X i − X )

− 2

( X − X ) 1 n

1 n

in=1 ui

in=1 ( X i − X ) 2

 var(vi )  d → N  0, 2   [var( X i )] 

since the second term for 17.4.

n ( ˆ1 − 1 ) converges in probability to zero as shown in part (b).

d S where (a) Write ( ˆ1 − 1 ) = an S n where an = 1n and S n = n ( Bˆ1 − 1 ). Now, an → 0 and S n → d 0  S . Thus Pr (|ˆ1 − 1|   ) → 0 for S is distributed N (0, a2). By Slutsky’s theorem an S n → p any  > 0, so that ˆ −  → 0 and ˆ 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

mapping theorem su2



2 u

=

su

u

→ 1. p

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 ( ˆ1 ) = E[ E ( ˆ1| X 1 ,

, X n)] = E ( 1 ) = 1.

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17.7.

(a) The joint probability distribution function of ui, uj, Xi, Xj is f (ui, uj, Xi, Xj). The conditional probability distribution function of ui and Xi given uj and Xj is f (ui, Xi |uj, Xj). Since ui, Xi, i = 1,, n are i.i.d., f (ui, Xi |uj, Xj) = f (ui, Xi). By definition of the conditional probability distribution function, we have

f (ui , u j , X i , X j ) = f (ui , X i | u j , X j ) f (u j , X j ) = f (ui , X i ) f (u j , X j ). (b) The conditional probability distribution function of ui and uj given Xi and Xj equals f (ui , u j | X i , X j ) =

f (ui , u j , X i , X j ) f (Xi , X j )

f (ui , X i ) f (u j , X j )

=

f (Xi ) f (X j )

= f (ui | X i ) f (u j | X j ).

The first and third equalities used the definition of the conditional probability distribution function. The second equality used the conclusion the from part (a) and the independence between Xi and Xj. Substituting f (ui , u j | Xi , X j ) = f (ui | Xi ) 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).

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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 1 is not BLUE. (b) We obtain the BLUE estimator of 1 from OLS in the following

Yi =  0 X 0i + 1 X 1i + ui where Yi = X 1i =

Yi

 0 + 1 | X i | Xi

 0 + 1 | X i |

, X 0i =

1

 0 + 1 | X i |

, and u =

ui

 0 + 1 | X i |

.

(c) Using equations (17.2) and (17.19), we know the OLS estimator, ˆ1 , is

ˆ1 =

in=1 ( Xi − X )(Yi − Y ) in=1 ( Xi − X ) ui =  + .

1 in=1 ( Xi − X )2 in=1 ( Xi − X )2

As a weighted average of normally distributed variables ui , ˆ1 is normally distributed with mean E ( ˆ ) =  . The conditional variance of ˆ , given X1,, Xn, is 1

1

1

   n ( X − X ) ui var ( ˆ1 | X1 ,..., X n ) = var  1 + i =n1 i | X1 ,..., X n  2 i =1 ( Xi − 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 ( 0 + 1| X i |) . [ in=1 ( X i − X )2 ]2

Thus the exact sampling distribution of the OLS estimator, ˆ1 , conditional on X1,, Xn, is 

 in=1 ( X i − X ) 2 ( 0 + 1| X i |)  . [ in=1 ( X i − X ) 2 ]2 

ˆ1 ~ N  1 , 

(d) The weighted least squares (WLS) estimators, ˆ0WLS and ˆ1WLS , are solutions to n

min  (Yi − b0 X 0i − b1 X 1i ) 2 , b0, b1

i =1

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the minimization of the sum of squared errors of the weighted regression. The first order conditions of the minimization with respect to b0 and b1 are n

 2(Y − b X − b X )(− X ) = 0, i

i =1

0

0i

1

1i

0i

n

 2(Y − b X − b X )(− X ) = 0. i

i =1

0

0i

1

1i

1i

Solving for b1 gives the WLS estimator

ˆ1WLS =

−Q01S0 + Q00 S1 Q00Q11 − Q012

where Q00 =  i =1 X 0i X 0i , Q01 =  i =1 X 0i X 1i , Q11 =  i =1 X 1i X 1i , S0 =  i =1 X 0 iYi , n

n

n

n

and S1 =  in=1 X 1iY . Substituting Yi =  0 X 0i + 1 X 0 i + ui yields

−Q01Z 0 + Q00 Z1 Q00Q11 − Q012

ˆ1WLS = 1 +

where Z 0 =  i =1 X 0 i ui , and Z1 =  i =1 X 1i ui or n

n

ˆ1WLS − 1 =

in=1 (Q00 X 1i − Q01 X 0i )ui .

Q00Q11 − Q012

From this we see that the distribution of ˆ1WLS | X 1 ,... X n is N ( 1 ,  2ˆWLS ), where 1

 2ˆ

WLS 1

=

 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  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 −  X ) 2 ui2 ] → 0.  n i =1

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Solutions to End-of-Chapter Exercises

93

Using the identity X =  X + ( X −  X ), 1 n 1 n 1 n [( X i − X ) 2 uˆi2 − ( X i −  X ) 2 ui2 ] = ( X −  X ) 2  uˆi2 − 2( X −  X )  ( X i −  X )uˆi2  n i =1 n i =1 n i =1 +

1 n  ( X i −  X )2 (uˆi2 − ui2 ). n i =1

The definition of uˆi implies

uˆi2 = ui2 + ( ˆ0 −  0 ) 2 + ( ˆ1 − 1 ) 2 X i2 − 2ui ( ˆ0 −  0 ) − 2u ( ˆ −  ) X + 2( ˆ −  )( ˆ −  ) X . i

1

1

i

0

0

1

1

i

Substituting this into the expression for 1n in=1[( X i − X )2 uˆi2 − ( X i −  X )2 ui2 ] yields a series of terms each of which can be written as anbn where an → 0 and bn = 1n in=1 X ir uis where r and s are integers. For example, a = ( X −  ), a = ( ˆ −  ) and so forth. The result then follows from p

n

X

n

1

1

Slutksy’s theorem if 1n in=1 X ir uis → d where d is a finite constant. Let wi = X i ui and note that wi r

p

s

is i.i.d. The law of large numbers can then be used for the desired result if E ( wi )  . There are two cases that need to be addressed. In the first, both r and s are non-zero. In this case write 2

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 = ˆ −  implies

Pr(|ˆ −  |   ) 

E[(ˆ −  )2 ]

2

p Since E[(ˆ −  )2 ] → 0, Pr(| ˆ −  |   ) → 0, so that ˆ −  → 0.

17.11. Note: in early printing of the third edition there was a typographical error in the expression for Y|X. The correct expression is Y | X = Y + ( XY /  X2 )( x −  X ) . (a) Using the hint and equation (17.38) fY | X = x ( y ) =

1 2  (1 −  XY ) 2 Y

2 2    x −  2  x −  X  y − Y   y − Y   1  x −  X   1 X   exp   − 2  XY   +  +   .

2  −2(1 −  XY )   X   X   Y    Y   2   X       

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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  = T|X and 2 =  Y2|X . (c) Let b = XY/  X and a = Y −bX. 2

  u2   u2   u2    u2  1 1 u E ( e ) = exp − + u du = exp exp − + u −     du   2 2  17.12. (a) 2   2  −  u 2  2 u   2 u −  u 2 2  1  2    2  1 = exp  u   exp  − 2 ( u −  u2 )  du = exp  u   2  −  u 2  2   2 u 

where the final equality follows because the integrand is the density of a normal random variable 2 with mean and variance equal to  u . 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 = 0 + 1Xi + vi, where 0 = E(0i), 1 = E(1i), and vi = ui + (0i − 0) + (1i − 1)Xi. Notice that E(vi|Xi) = E(ui|Xi) + E(0i − 0|Xi) + XiE(1i − 1|Xi) = 0 because 0i and 1i are independent of Xi. Because E(vi | Xi) = 0, the OLS regression of Yi on Xi will provide consistent estimates of 0 = E(0i) and 1 = E(1i). Recall that the weighted least squares estimator is the OLS estimator of Yi/i onto 1/i and Xi/i , where  i =  0 + 1 X i2 .

Write this regression as Yi /  i =  0 (1 /  i ) + 1 ( X i /  i ) + vi /  i .

This regression has two regressors, 1/i and Xi/i. Because these regressors depend only on Xi, E(vi|Xi) = 0 implies that E(vi/i | (1/i), Xi/i) = 0. Thus, weighted least squares provides a consistent estimator of 0 = E(0i) and 1 = E(1i).

©2018 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 = X + U with

1 Income1  1 Income2 X=    1 Incomen

 TestScore1    TestScore2  Y=  ,      TestScoren   U1    U U =  2 ,     U n 

Income12   Income22    Income2n 

 0    β =  1  .    2

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

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

F = ( Rβˆ − r )  Rˆ 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 = X + U

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with

 Y1    Y Y= 2  ,    Yn 

 1 X 1,1  1 X 1, 2 X=    1 X 1, n

 u1    u U =  2 ,      un 

X 2,1   X 2, 2    X 2, n 

 0    β =  1     2

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

with  n  XX =   in=1 X 1i  in=1 X 1i 

in=1 X 1i  in=1 X 12i  in=1 X 1i X 2i

in=1 X 2i    in=1 X 1i X 2i  ,  in=1 X 22i 

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

sY2 =

1 n 1 n 2 n 2 (Yi − Y )2 = Yi − Y ,   n − 1 i =1 n − 1 i =1 n −1

we know n

Y = (n − 1) s + nY .

2

i =1

i

2 Y

2

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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

s XY =

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

we know n

 X Y = (n − 1)s i i

i =1

XY

+ nXY .

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

 X Y = (n − 1) s 1i i

i =1

X 1Y

+ nX 1Y

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

 X Y = (n − 1) s 2i i

i =1

X 2Y

+ nX 2Y

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

and n

 X X = (n − 1) s 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   XY = 144.8 1063.6 584.52  ,  80.0 584.52 365.6   

 127.8    XY =  929.45  .  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

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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, ˆ0 = 4.2063, ˆ1 = 0.2520, and ˆ2 = 0.1033.

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

su2ˆ =

n 1 1 ˆ U ˆ. uˆi = U  n − k − 1 i =1 n − k −1

ˆ =Y−Y ˆ = Y − Xˆ , 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  6.39 2 = 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 = YY − 2βX SSR =  uˆi2 = U i =1

= 821.58 − 2  825.22 + 832.23 = 3.37.

2

The squared standard error of the regression sû is

su2ˆ =

1 1 ˆ U ˆ = U  3.37 = 0.1982. n − k −1 20 − 2 − 1

With the total sum of squares n

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

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Solutions to End-of-Chapter Exercises

the R2 of the regression is

R2 = 1 −

SSR 3.37 =1− = 0.3178.

TSS 4.94

(b) When all six assumptions in Key Concept 16.1 hold, we can use the homoskedasticity-only estimator  ˆ of the covariance matrix of ˆ , conditional on X, which is

−0.4631 −0.0337   3.5373    ˆ = ( XX) s =  −0.4631 0.0684 −0.0080   0.1982  −0.0337 −0.0080 0.0229    −1 2 uˆ

−0.09179 −0.0067   0.7011   =  −0.09179 0.0136 −0.0016  .  −0.0067 −0.0016 0.0045    The homoskedasticity-only standard error of ˆ1 is 1 SE ( ˆ1 ) = 0.0136 2 = 0.1166.

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

ˆ1 0.2520 = = 2.1612, SE ( ˆ1 ) 0.1166

and the 5% two-sided critical value is 2.11. Thus we can reject the null hypothesis 1 = 0 at the 5% significance level.

18.3.

(a) Var (Q ) = E[(Q − Q ) 2 ] = E[(Q − Q )(Q − Q )] = E[(cW − c W )(cW − c W )] = cE[(W −  W ) ( W −  W )]c = c var( W ) c = cΣ w c

where the second equality uses the fact that Q is a scalar and the third equality uses the fact that Q = cw. (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 = X + U

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with  Y1    Y Y = 2 ,      Yn 

1 X 1    1 X2  X= ,     1 X n 

 u1    u U  2 ,      un 

  β =  0 .  1 

(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 X X =  n  i =1 X i

in=1 X i  . in=1 X i2 

 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 ( X X) −1 =  n  i =1 X i

in=1 X i   in=1 X i2 

−1

=

 in=1 X i2 −  in=1 X i  1   n in=1 X i2 − ( in=1 X i ) 2  − in=1 X i 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 n 2  i =1 ( X i − X )  − X 1   i =1 X iYi 

=

 Y in=1 X i2 − X  in=1 X iYi  1   in=1 ( X i − X ) 2  in=1 X iYi − nXY 

ˆ =

in=1 X iYi − nXY in=1 ( X i − X )(Yi − Y ) = , in=1 ( X i − X )2 in=1 ( X i − X )2

Therefore we have

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Solutions to End-of-Chapter Exercises

and

ˆ0 =

Y  in=1 X i2 − X  in=1 X iYi  in=1 ( X i − X ) 2

=

Y  in=1 ( X i − X + X ) 2 − X  in=1 X iYi  in=1 ( X i − X ) 2

=

Y  in=1 ( X i − X ) 2 + nX 2Y − X  in=1 X iYi  in=1 ( X i − X ) 2

  n X Y − nXY  = Y −  i =n1 i i X 2    i =1 ( X i − X )  = Y − ˆ X .

1

We get the same expressions for ̂ 0 and ˆ1 as given in Key Concept 4.2. (d) The large-sample covariance matrix of ˆ , 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  u  Vi = Xi ui =  i  ,  X i ui 

Xi  , X i2 

and

 u  Vi Vi =  i  ( ui  X i ui 

 ui2 X i ui ) =  2  X i ui

X i ui2  .

X i2ui2 

Taking expectations, we get  1 Q X = E ( Xi Xi ) =   X

X 

, E ( X i2 ) 

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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 Q =  X −1 X

X

−1

 E ( X i2 ) −  x   1 =  .  E ( X i2 )  E ( X i2 ) −  X2  −  X 1 

We now can calculate the large-sample covariance matrix of ˆ , conditional on X, from 1  ˆ = Q −X1  v Q −X1 n 1 = 2 n[ E ( X i ) −  X2 ]2 cov( X i ui , ui )   E ( X i2 ) −  X   E ( X i2 ) −  X   var(ui )   .  var( X i ui )   −  X 1   cov( X i ui , ui ) 1   − X

The (1, 1) element of  ˆ is 1 {[ E ( X i2 )]2 var(ui ) − 2 E ( X i2 )  X cov( X i ui , ui ) +  X2 var( X i ui )} n[ EX ) −  X2 ]2 2 i

=

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

=

  [ E ( X i2 )]2 X var ui − X i ui  2 2 2 2 n[ E ( X i ) −  X ] E( X i )  

2 i

   X var  1 − X i  ui  2  E ( X i )     X2  n 1 −  2  E( X i )  var( H i ui ) = , (the same as the expression for  20 given in Key Concept 4.4) n( E ( H i2 )]2

=

1

2

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Solutions to End-of-Chapter Exercises

by defining Hi = 1 −

X E ( X i2 )

Xi.

The denominator in the last equality for the (1, 1) element of  ˆ has used the facts that 2

   X2 2 X X H = 1 − X = 1 + X i2 − Xi ,  i 2 2 2 E (Xi ) E ( X i2 )  E( X i )  2 i

so

E ( H i2 ) = 1 + 18.5.

 X2 2

2 i

2

[ E ( X )]

E ( X i2 ) −

2 X  X2 X = 1 − .

2 E( X i ) E ( X i2 )

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 = Xβˆ = X( XX)−1 XY = P Y Y X

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 ( X + U) = M X + M U = M U U X X X X X

which is Equation (18.28).

18.6.

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

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with  Y11 − Y1     Y12 − Y1       Y1T − Y1  Y −Y   21 2   Y22 − Y2   , Y=    Y2T − Y2       Yn1 − Yn     Yn 2 − Yn      Y − Y  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 

 X 11 − X 1     X 12 − X 1       X 1T − X 1  X −X  2  21   X 22 − X 2  , X=    X 2T − X 2       X n1 − X n     X n2 − X n      X −X  n  nT

 = 1 The OLS “de-meaning” fixed effects estimator is

1DM = ( XX)−1 XY. Rewrite Equation (10.11) using n fixed effects as Yit = X it 1 + D1i  1 + D 2i  2 +

+ Dni  n + uit .

In matrix form this is YnT 1 = XnT 111 + WnT  n 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  Y=  ,    Y2T       Yn1     Yn 2       YnT 

 X 11     X 12       X 1T  X   21   X 22  , X=    X 2T       X n1     X n2       X nT 

 D11   D11    D11  D1  2  D12 W=   D12    D1n   D1n    D1n

D 21 D 21 D 21 D 22 D 22 D 22 D 2n D 2n D 2n

Dn1   Dn1    Dn1  Dn2   Dn2  ,  Dn2    Dnn   Dnn    Dnn 

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 u11     u12       u1T  u   21   u22  , U=    u2T       un1     un 2       unT 

Solutions to End-of-Chapter Exercises

 = 1 ,

 1      = 2 .      n 

Using the expression for ˆ given in the question, we have the estimator

ˆ1BV = ˆ = ( 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   X n1 − X n   X n2 − X n    X nT − X n  0

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

ˆ1BV = ( XX ) XY = ˆ1DN . −1

18.7.

(a) We write the regression model, Yi = 1Xi + 2Wi + ui, in the matrix form as Y = X + W + U with  Y1    Y Y= 2,      Yn 

 X1   W1      X W X= 2, W= 2,          Xn   Wn   = 1 ,  = 2 .

 u1    u U= 2,      un 

The OLS estimator is  ˆ1   X X X W  −1  XY   =  ˆ   WX WW   WY   2 −1

 1   X X X W   XU  = +      2   W  X W  W   W U   1   1 X X =   +  1n   2   n W X

1 n 1 n

 1   1  n X 2 =   +  1 n n i =1 i   2   n i =1 Wi X i

−1

X W   1n XU     W W   1n WU  1 n

in=1 X iWi   1 in=1 Wi 2  n

−1

 1n in=1 X i ui  1 n    Wu   n i =1 i i 

p p E ( X 2 ); 1n in=1 Wi 2 → E (W 2 ); By the law of large numbers 1n in=1 X i2 →

1 n

p in=1 X iWi →

p E ( Xu ) = 0 E ( XW ) = 0 (because X and W are independent with means of zero); 1n in=1 X i ui → p E ( Xu ) = 0 Thus (because X and u are independent with means of zero); 1n in=1 X i ui → −1  ˆ1  p  1   E ( X 2 ) 0   0   →  +     ˆ  E (W 2 )   E (Wu )   2   0  2

1   = E (Wu )  .  2 +  E (W 2 )   p (b) From the answer to (a) ˆ2 → 2 +

E (Wu )   2 if E(Wu) is nonzero.

E (W 2 )

(c) Consider the population linear regression ui onto Wi: ui = Wi + ai

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Solutions to End-of-Chapter Exercises

107

where  = 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 1 + Wi  2 + ui = X i 1 + Wi (  2 +  ) + ai = X i 1 + Wi + ai

where  =  2 + . A calculation like that used in part (a) can be used to show that

 n ( ˆ1 − 1 )   1n in=1 X i2  =  n ( ˆ −  )   1n in=1 Wi X i 2  

1 n

in=1 X iWi   1 in=1 Wi 2  n

−1

 1n in=1 X i ai     1 in=1 Wu ai   n 

−1

 E( X 2 ) 0   S1  →     E (W 2 )   S2   0 d

where S1 is distributed N (0,  a E ( X 2 )). Thus by Slutsky’s theorem 2

  a2  d n ( ˆ1 − 1 ) → N  0, 2   E( X ) 

Now consider the regression that omits W, which can be written as: Yi = X i 1 + di

where di = Wi + ai. Calculations like those used above imply that   d2  d n ˆ1r − 1 → N  0, .

2   E( X ) 

(

)

Since  d =  a +  E (W ), the asymptotic variance of ˆ1r is never smaller than the asymptotic variance of ˆ .

2

2

2

2

1

18.8.

(a) The regression errors satisfy u1 = u1 and ui = 0.5u i −1 + ui for i = 2, 3,, n with the random variables ui (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 + ui − j ( j = 1, 2,, i − 2) and u1 = u1 into the expression ui = 0.5ui – 1 + ui yields ui = 0.5ui −1 + ui = 0.5(0.5ui − 2 + ui −1 ) + ui = 0.52 (0.5ui −3 + ui − 2 ) + 0.5ui −1 + ui = 0.53 (0.5ui − 4 + ui −3 ) + 0.52 ui − 2 + 0.5ui −1 + ui = = 0.5i −1 u1 + 0.5i − 2 u2 + 0.5i −3 u3 +

+ 0.52 ui − 2 + 0.5ui −1 + ui

i

=  0.5i − j u j . j =1

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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): i

E (ui ) =  0.5i − j E (u j ) = 0, j =1 i

i

j =1

j =1

 i2 = var(ui ) =  (0.5i − j ) 2 var(u j ) = (0.52 )i − j  1 =

1 − (0.52 )i .

1 − 0.52

In calculating the variance, the second equality has used the fact that ui is i.i.d.

Since ui =  ij =1 0.5i − j u j we know for k > 0, i+k

i

i+k

j =1

j = i +1

ui + k =  0.5i + k − j u j = 0.5k  0.5i − j u j +  0.5i + k − j u j j =1

i+k

= 0.5k ui +  0.5i + k − j u j . j = i +1

Because ui 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  i .

Similarly we can get

cov(ui , ui − k ) = 0.5k  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 (unu2 )

E (u1un )   E (u2un )  .   E (un2 ) 

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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   12  2  0.5 1  0.52  12  = E (UU) =  3 2  0.5  1   n −1 2  0.5  1 

0.5 12

0.52  12

0.53  12

 22 0.5 22 0.52  22

0.5 22

0.52  22

 32 0.5 32

0.5 32

0.5n − 2  22

0.5n −3  32

0.5n − 4  42

 42

0.5n −1 12   0.5n − 2  22  0.5n −3  32   0.5n − 4  42     n2 

1 − (0.52 )i .

1 − 0.52 (b) The original regression model is

with  i2 =

Yi =  0 + 1 X i + ui .

Lagging each side of the regression equation and subtracting 0.5 times this lag from each side gives Yi − 0.5Yi −1 = 0.50 + 1 ( X i − 0.5 X i −1 ) + ui − 0.5ui −1

for i = 2,, n with ui − 0.5ui −1 = ui . Also Y1 =  0 + 1 X 1 + u1

with u1 = u1. Thus we can define a pair of new variables

(Yi , X 1i , X 2i ) = (Yi − 0.5Yi −1 , 0.5, X i − 0.5 X i −1 ), for i = 2,, n and (Y1 , X 11 , X 21 ) = (Y1 ,1, X 1i ), and estimate the regression equation

Yi =  0 X 1i + 1 X 2i + ui using data for i = 1,, n. The regression error ui 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) ˆ = ( XM W X) −1 XM W Y = ( XM W X) −1 XM W ( X + W + U ) =  + ( XM W X) −1 XM W U.

The last equality has used the orthogonality MWW = 0. Thus

ˆ −  = ( XM W X)−1 XM W U = (n−1XM W X)−1 (n−1XM W U).

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(b) Using MW = In − PW and PW = W(WW)−1W we can get n −1XM W X = n −1X(I n − PW )X = 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 = 1n 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 p  X ji X li → E ( X ji X li ) . n i =1 This is true for all the elements of n−1 XX, so

n −1XX =

1 n p  Xi Xi → E (Xi Xi ) =  XX. . 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 n −1WW =

1 n p Wi Wi → E ( Wi Wi ) =  WW ,  n i =1

n −1XW =

1 n p Xi Wi → E ( Xi Wi ) =  XW ,  n i =1

n −1WX =

1 n p Wi Xi → E ( Wi Xi ) =  WX .  n i =1

and

From Assumption (iii) we know  XX ,  WW ,  XW , and  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) → Σ XX − Σ XW Σ -1WW Σ WX p

which is finite and invertible.

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(c) The conditional expectation  E (u1| X, W )   E (u1| X1 , W1 )      E (u2 | X, W )   E (u2 | X 2 , W2 )  E (U|X, W ) =  =          E (un | X, W )   E (un | X n , Wn )   W1   W1       W W =  2  =  2   = W .          Wn   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 n −1XM W U → E ( XM W U|X, W) = XM W E (U|X, W ) = XM W W = 0 k1 1

because M W W = 0. −1 −1 (e) n XM W X converges in probability to a finite invertible matrix, and n XM W U converges in probability to a zero vector. Applying Slutsky’s theorem, p ˆ −  = (n−1XM W X)-1 (n−1XM W U) → 0.

This implies p ˆ → .

18.10. (a) Using the hint: Cq = q, so that 0 = Cq − q = CCq − q = Cq − q = 2q − q = (1 − )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.

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18.13. (a) This follows from the definition of the Lagrangian. (b) The first order conditions are (*) X(Y−X  ) + R = 0 and (**) R  − r = 0 Solving (*) yields (***)  = ˆ + (XX)–1R.

Multiplying by R and using (**) yields r = R ˆ +R(XX)–1R, so that

 = −[R(XX)–1R]−1(R ˆ − r).

Substituting this into (***) yields the result. (c) Using the result in (b), Y − X  = (Y − X ˆ ) + X(XX)–1R[ R(XX)–1R]–1(R ˆ − r), so that (Y − X  )(Y − X  ) = (Y − X ˆ )(Y − X ˆ ) + (R ˆ − r) [R(XX)–1R]–1(R ˆ − r) + 2(Y − X ˆ ) X(XX)–1R[R(XX)–1R]–1(R ˆ − r).

But (Y − X ˆ ) X = 0, so the last term vanishes, and the result follows. (d) The result in (c) shows that (R ˆ − r)[R(XX)–1R]–1(R ˆ − r) = SSRRestricted − SSRUnrestricted.

Also su = SSRUnrestricted /(n − kUnrestricted – 1), and the result follows immediately.

2

18.14. (a) ˆ (XX) ˆ = YX(XX)–1XY = YX1HX1Y, where H is the upper k1  k1 block of (XX)–1.

Also R ˆ = HX1Y and R(XX)–1R = H. Thus (Rˆ ) '[R( X' X )−1 R ]−1(Rˆ ) = YX1HX1Y. (b) (i) Write the second stage regression as Y = Xˆ  + U, where X̂ and the fitted values from the ˆ ˆ = 0 where Uˆ = Y − Xˆ ˆ because OLS residual are first stage regression. Note that UX orthogonal to the regressors. Now Uˆ TSLS = Y − Xˆ = Uˆ − ( X − Xˆ )ˆ = Uˆ − Vˆ ˆ , 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 = UW ˆ  − ˆ 'Vˆ 'W = 0. regression, VW −1

(ii) ˆ (XX) ˆ = (Rβˆ ) '  R ( X ' X) −1 R  (Rβˆ ) = SSRRest − SSRUnrest for the regression in KC 12.6, and the result follows directly.

18.15. (a) This follows from exercise (18.6).

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(b) Yi = Xi β + u i , so that  n  βˆ − β =   X i ' X i   i =1 

−1 n

 n  =   Xi ' Xi   i =1 

−1 n

 n  =   Xi ' Xi   i =1 

−1 n

 n  =   Xi ' Xi   i =1 

−1 n

 X 'u i =1

i

i

 X ' M ' Mu i =1

i

 X ' M 'u i =1

i

 X 'u i =1

i

i

i

i

−1 T 2 (c) Qˆ X = 1n in=1 (T −1 Tt =1 ( X it − X i ) 2 ), where (T  t =1 ( X it − X i ) ) are i.i.d. with mean Q X and 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.

2 (f) i are i.i.d., and the result follows from the law of large numbers.

(g) Let ˆi = T −1/2 Xi ' uˆ i = i − T −1/2 ( ˆ −  ) Xi ' Xi . Then

ˆi2 = T −1/2 Xi ' uˆ i = i2 + T −1 ( ˆ −  )2 ( Xi ' Xi )2 − 2T −1/2 ( ˆ −  )i Xi ' Xi 1 and 1n in=1 ˆi2 − in=1 i2 = T −1 ( ˆ −  ) 2 1n in=1 ( Xi ' Xi ) 2 − 2T −1/ 2 ( ˆ −  ) 1n in=1 i Xi ' Xi n p p Because ( ˆ −  ) → 0 , the result follows from (a) 1n in=1 ( X i ' X i )2 → E[( X i ' X i ) 2 ] and (b)

in=1 i X i ' X i → E (i 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 2 that ( X i ' X i ) has two finite moments and i Xi ' Xi 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

p

18.16 (a) Using analysis like that Appendices 4.2 and 4.3

1

X Y = + X u = + nX u ˆ = 1 X X X i i 2 i

i i

i i 2 i

n

2 i

Consistency follows by analyzing the averages in the expression: p 1 X u →  i i E ( X i ui ) = 0, n

p 1 2 X →  i E ( X i2 )  0 , n

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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 n ˆ = =  + =  + .

1 2  Ii X i2  Ii X i2  Ii X i n

 Ii X iYi

Specifically, we have

 I i X i ui

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

2

In parts (ii)-(iv), E ( I i X i ) > 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 ̂ 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 , u ) f (u | X ) f ( I | X ) f (u | X ) = = = f (u | X ) f (I | X ) f (I | 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 E (u | u  − X ) =

 (− X ) 1 − (− X )

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Solutions to End-of-Chapter Exercises

where  is the standard normal CDF and  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.

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