IntroductiontoEconometrics (3 U pdatedEdition, Global Edition

Introduction to Econometrics (3rd Updated Edition, Global Edition) by James H. Stock and Mark W. Watson

Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 2 (This version August 17, 2014)

©2015 Pearson Education, Ltd.

Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 2 1 _____________________________________________________________________________________________________

2.1.

(a) Probability distribution function for Y

Outcome (number of heads) Probability (b)

Y=0 0.36

µY = E (Y ) = (0 × 0.36) + (1 × 0.48) + (2 × 0.16) = 0.8 Using Key Concept 2.3: var(Y = ) E (Y 2 ) − [ E (Y )]2 , and

E (Y 2 ) = 2 × 0.6 × 0.4 + 0.82 = 1.12 so that

var(Y ) =E (Y 2 ) − [ E (Y )]2 = 1.12 − (0.8) 2 =0.48


Y=1 0.48

Y=2 0.16


2.3.

For the two new random variables W = 4 + 8X and V = 11 – 2Y, we have:

(a) E(V ) = E (11 − 2Y ) = 11 − 2E(Y ) = 11 − 2 × 0.78 = 9.44 E(W ) = E (4 + 8 X ) = 4 + 8E( X ) = 4 + 8 × 0.70 = 9.6

(b) σ W2 =ar(4 v + 8X ) = 82 σ X2 = 64 × 0.21 = 13.44 σ V2 = 0.6864 var(11 − 2Y ) = (−2) 2 σ Y2 = 4 × 0.1716 = (c)

σ WV =cov(4 + 8X ,11 − 2Y ) =6 × (−2) cov( X ,Y ) =−12 × 0.084 =−1.008 σ WV −1.008 corr(W ,V ) = = = 0.3319 σW σV 13.44 × 0.6864



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 o Concept 2.3, µX = 65 F implies that σ Y = −17.78 + (5/9) × 65 = 18.33°C, and JX = 5oF implies J Y= (5/9) × 5= 2.78°C.



2.7.

Using obvious notation, C  M  F ; thus C  M  F and  C2   M2   F2  2 cov(M, F). This implies (a) C  50 + 48 = $98,000 per year. (b) corr ( M , F ) 

cov(M , F )

 M F

, so that cov(M, F)   M  F corr ( M , F ). Thus cov ( M , F ) 

15 × 13 × 0.9 = 175.50, where the units are squared thousands of dollars per year. (c)  C2   M2   F2  2 cov(M, F), so that σ C2 = 152 +132 + 2(175.5) = 745 and σ C = 745 = 27.295 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 = 0.75 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.



2.9.

Value of X

3 6 9 Probability distribution of Y

2 0.04 0.10 0.13 0.27

4 0.09 0.06 0.11 0.26

Value of Y 6 8 0.03 0.12 0.15 0.03 0.04 0.06 0.22 0.21

10 0.01 0.02 0.01 0.04

Probability Distribution of X 0.29 0.36 0.35 1.00

(a) The probability distribution is given in the table above. E(Y) = 2(0.27) + 4(0.26) + 6(0.22) + 8(0.21) + 10(0.04) = 4.98.

E(Y2) = 22(0.27) + 42(0.26) + 62(0.22) + 82(0.21) + 102(0.04) = 30.6. Var(Y) = E(Y2) – [E(Y)] 2 = 30.6 – 4.98 = 25.62. (b) The conditional probability of Y|X = 6 is given in the table below Value of Y

2 0.10/0.36

4 0.06/0.36

6 0.15/0.36

8 0.03/0.36

10 0.02/0.36

E(Y|X = 6) = 2(0.1/0.36) + 4(0.06/0.36) + 6(0.15/0.36) + 8(0.03/0.36) + 10(0.02/0.36) = 4.944 E(Y2|X = 6) = 22(0.1/0.36) + 42(0.06/0.36) + 62(0.15/0.36) + 82(0.03/0.36) + 102(0.02/0.36) = 29.667. Var(Y) = 29.667 – 4.944 = 24.723. (c)

E(XY) = 3 × 2 × 0.04 + 3 × 4 × 0.09 + ⋅ ⋅ ⋅ + 9 × 10 × 0.01 = 29.4 Cov(X, Y) = E(XY) – E(X)E(Y) = 29.4 – 6.18 × 4.98 = –1.3764 Corr(X, Y) = cov(X, Y)/(var(X)var(Y)) = –1.3764/(5.7276 × 24.723) = –0.0097


rd

Stock/Watson - Introduction to Econometrics - 3 Updated Edition - Answers to Exercises: Chapter 2 6 _____________________________________________________________________________________________________

2.11.

(a) 0.90 (b) 0.95 (c) 0.95 (d) When Y ~ 82 , then Y /10 ~ F8,  .

(e) Y = Z 2 , where Z ~ N (0,1), thus Pr(Y ≤ .5) = Pr(−.5 ≤ Z ≤ .5) = 0.383.



2.13.

(a) E (Y 2 ) = Var (Y ) + µY2 = 4 + 0 = 4;

E(W 2 ) = Var(W ) + µW2 = 16 + 0 = 16

(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, 3 =

E (Z − µZ )4

; by transforming both Y and W to the

σ Z4

standard normal so yields the results.

(d) First, condition on X = 0, so that S = W:

E(S | X = 0) = 0; E(S

2

| X = 0) =16; E(S

3

| X = 0) = 0; E(S

3

| X =1) = 0; E(S

4

| X = 0) = 3×162

Similarly,

E(S | X = 1) = 0; E(S

2

| X =1) = 4; E(S

4

| X =1) = 3× 42

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) = 16 × 0.1 + 4 × 0.9 = 5.2 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 =Pr 1) × ( X = 1) = 3 × 162 × 0.1 + 3 × 42 × 0.9 = 120

µ S E(S = ) 0, thus E(S − µ S )3 = E(S 3 ) = 0, from part (d). Thus skewness = 0. Similarly, (e) = σ S2 = E ( S − µ S ) 2 = E ( S 2 ) =5.2, and E(S − µ S ) 4 = E(S 4 ) = 120. Thus, kurtosis = 120/(5.22) = 4.44.



2.15. (a)

⎛ 9.6 − 10 Y − 10 10.4 − 10 ⎞ ≤ ≤ 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 (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 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.



2.17.

µY = 0.6 and σ Y2 = 0.4 × 0.6 = 0.24    Y − 0.6 0.64 − 0.6   (a) (i) P(Y ≥ 0.64) ≤ = Pr   0.24 0.24    n n        Y − 0.6    0.04  Y − 0.6  0.2157 = Pr  ≤ = Pr  ≤ 0.577=  0.24  0.24  0.24      n n  n   

     Y − 0.6 0.56 − 0.6    Y − 0.6  Pr  = ≤ 0.56) Pr  ≤ = ≤= −1.157  0.12 (ii) P(Y  0.24  0.24  0.24      n n n    

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

= 0.61

0.65 − 0.60 0.65 − 0.60 > −1.96 and < −1.96. Solving these inequalities yields n ≥ 368. 0.24 0.24 n n


Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 2 10 _____________________________________________________________________________________________________ l

2.19. (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 = y j ) = ∑ y j ∑ Pr (Y = y j |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 = y j ), 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 XσY σ XσY



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


]3


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

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 )   X2   X2  1, and Y  E ( X 2  Z )  E ( X 2 )  Z  1  0  1 (c) E ( XY )  E ( X 3  ZX )  E ( X 3 )  E (ZX ). Using the fact that the odd moments of a standard normal random variable are all zero, we have E ( X 3 )  0. Using the independence between 3 X and Z , we have E ( ZX )  Z  X  0. Thus E ( XY )  E ( X )  E(ZX )  0.

(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 ) 

 XY 0   0  XY  XY



2.25.

(a)

n

 ax i 1

(b)

i

n

 (ax1  ax2  ax3    axn )  a( x1  x2  x3    xn )  a  xi i 1

n

(x  y )  (x i 1

i

i

1

 y1  x  y2   xn  yn )

 ( x1  x2   xn )  ( y1  y2   yn ) n

n

i 1

i 1

  xi   yi (c)

n

 a  (a  a  a    a)  na i 1

(d)

n

 (a  bx i 1

i

n

 cyi ) 2   (a 2  b 2 xi2  c 2 yi2  2abxi  2acyi  2bcxi yi ) i 1

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



2.27 (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.


IntroductiontoEconometrics (3 U pdatedEdition, Global Edition

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