introductiontoeconometrics (3 u pdatededition, global...

©2015 Pearson Education, Ltd.

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)

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


1

2.1. (a)

(b)

Probability distribution function for Y

Outcome (number of heads) Y = 0 Y = 1 Y = 2 Probability 0.36 0.48 0.16

µY = (E Y ) = (0×0.36) (1+ ×0.48) + (2×0.16) = 0.8

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

(E Y ) 2= 0.6× ×0.4 + 0.82 2 =1.12so that var(Y ) (= E Y ) [− (E Y )]2 2 1.12= (− 0.8)2 = 0.48

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


2.3.

(b)

(c)

For the two new random variables W = 4 + 8X and V = 11 – 2Y, we have: (a) (11 2Y ) 2E(Y ) 0.78 9.44

(4 8 ) 8E(X ) 0.70 9.6− =11− =11 2− × =+ = 4 + = 4 8+ × =

E(V ) = EE(W ) = E X

2 2 2

2 2 2

64 0.21 13.44

v= ar(11 2Y ) 2)σ σ 0.1716 0.6864

σ v= ar(4 8X ) 8=+ =σ × =

− (= − 4= × =XW

V Y

cov(4 + 8X ,11− 2Y ) = 6 (× −2)cov(X ,Y ) = −12×0.084 = −1.0081.008corr( ,W V ) 0.3319

13.44 0.6864

σ

σ σ

=

σ −= = =

×

WV

WV

W V



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)× .UsX ing Key Concept 2.3, µX = 65oF implies thatσ = −17.78 (+ 5/9)×65 =18.33 C° , and JX = 5oF implies

Y

Y

J = (5/9) 5× = 2.78°C.



2.7. Using obvious notation, C M ;F thus M F C and 2 cov(MC F2 2 2 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 )

2MC F

2 2 2(c) cov(M, F),

(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.

15 × 13 × 0.9 = 175.50, where the units are squared thousands of dollars per year.

so that 22CC

2 =15 +13 + 2(175.5) = 745 and σ σ = 745 = 27.295 thousand dollars per year.



2.9.

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

Value of Y Probability Distribution of X 2 4 6 8 10

Value of X 3 0.04 0.09 0.03 0.12 0.01 0.29 6 0.10 0.06 0.15 0.03 0.02 0.36 9 0.13 0.11 0.04 0.06 0.01 0.35

Probability distribution of Y 0.27 0.26 0.22 0.21 0.04 1.00

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

(c)

2 4 6 8 10 0.10/0.36 0.06/0.36 0.15/0.36 0.03/0.36 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.

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



2.11. (a) 0.90 (b) 0.95 (c) 0.95

(d) When 8Y 2~ , then 8, /10 ~Y F .

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



2.13. (a)

(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)

(d)

Similarly,

From the law of iterated expectations

(e)

) = Var2 (Y ) + µ 4= + =02 4; ) = Var(W ) + µ2 2 =16 0+ =16Y W(E Y E(W

4

4

E Z −( µ )σ

3 = Z

Z

; by transforming both Y and W to the The kurtosis of the normal is 3,

standard normal so yields the results. First, condition on X = 0, so that S = W:

E(S | X = 0) = 0;E(S | X = 0)2 =16;E(S | X = 0)3 = 0;E(S | X = 0)4 = 3×16 2

E(S | X =1) = 0;E(S | X =12 ) = 4; E(S | X =13 ) = 0; E(S | X =14 ) = 3× 4 2

22 2

33 3

44 4 2 2

(S) E(S |(S ) E(S | 0)(S ) E(S | 0)(S ) E(S | 0)

= 0)= Pr× (X 0)= E+ (S | X 1)= Pr× (X 1)= 0=

= = Pr× (X 0)= E+ (S | X 1)= Pr× (X 1)= 16= 0.1× 4+ 0.9× 5.2=

= = Pr× (X 0)= E+ (S | X 1)= Pr(X× =1) 0=

= = Pr× (X 0)= E+ (S | X 1)= Pr× (X 1)= 3= 16× 0.1× 3+ 4× 0.×

E XE XE XE X 9 =120

µ = ) = 0,S E(S thus µ )3− = E(S )3 0,=SE(S from part (d). Thus skewness = 0. Similarly, σ 2 = − )2 5.2= ,S S(E S )µ 2 = (E S Sand E(S µ )− = E(S4 4 ) =120.

Thus, kurtosis = 120/(5.22) = 4.44.



2.15. (a) 9.6 10 10 10.4 10Pr (9.6 10.4) Pr4/ 4/ 4/

9.6 10 10.4 10Pr4/ 4/

YYn n n

Zn n

≤ ≤

≤

⎛ ⎞− − −≤ = ≤⎜ ⎟⎝ ⎠

− −⎛ ⎞= ≤⎜ ⎟⎝ ⎠

where Z ~ N(0, 1). Thus,

(i) n = 20; 9.6 10 10.4 10Pr Pr ( 0.89 0.89) 0.634/ 4/

Z Zn n

≤ ≤− −⎛ ⎞≤ = − ≤ =⎜ ⎟⎝ ⎠

(ii) n = 100; 9.6 10 10.4 10Pr Pr( 2.00 2.00) 0.9544/ 4/

Z Zn n

≤ ≤− −⎛ ⎞≤ = − ≤ =⎜ ⎟⎝ ⎠

(iii) n = 1000; 9.6 10 10.4 10Pr Pr( 6.32 6.32) 1.0004/ 4/

Z Zn n

≤ ≤− −⎛ ⎞≤ = − ≤ =⎜ ⎟⎝ ⎠

(b) 10Pr (10 10 ) Pr4/ 4/ 4/

Pr .4/ 4/

c Y cc Y cn n nc cZn n

≤ ≤

≤

⎛ ⎞− −− ≤ + = ≤⎜ ⎟⎝ ⎠

−⎛ ⎞= ≤⎜ ⎟⎝ ⎠

As n get large 4 /cn

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.



(a) (i)

2.17. YµY = 0.6 and σ 2 = 0.4 0.6× = 0.24

0.6 0.64 0.60.64)≥ = Pr0.24 0.24

0.04 0.577 0.21570.24 0.24 0.24

Y − − ≤

0.6Y − 0.6Y −Pr= ≤ Pr = ≤ =

P(Y

nn

n n n

(ii)

(b)

0.56 0.6≤ 0.56) = Pr Pr 1.157

0.24 0.24 0.24

− 0.6−Y = =≤ ≤ −

0.6−YP(Y

nn n

0.12

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

0.61= 0.65 0.60> −1.96 and− 0.65 0.60

< −− 1.96.

0.24 0.24nn

Solving these inequalities yields n ≥ 368.



2.19. (a) 1

1

Pr ( ) Pr ( , )

Pr ( | )Pr ( )

l

j i jil

j i ii

Y y X x Y y

Y y X x X x

=

=

= = = =

= = = =

∑

∑

(b) 1 1 1

11

1

( ) Pr ( ) Pr ( | ) Pr ( )

Pr ( | ) Pr ( )

( | )Pr ( )

i

i i

k k l

j j j j i ij j i

kl

j j iji

l

i

E Y y Y y y Y y X x X x

y Y y X x X x

E Y X x X x

= = =

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎜ ⎟== ⎝ ⎠

=

= = = = = =

= = =

= = = .

∑ ∑ ∑

∑∑

∑

(c) When X and Y are independent,

Pr ( , ) Pr ( )Pr ( )i j i jX x Y y X x Y y= = = = = ,

so

1 1

1 1

1 1

[( )( )]

( )( )Pr ( , )

( )( )Pr ( )Pr ( )

( )Pr ( ) ( )Pr (

( ) ( ) 0 0 0,

XY X Yl k

i X j Y i ji j

l k

i X j Y i ji j

l k

i X i j Y ji j

X Y

E X Y

x y X x Y y

x y X x Y y

x X x y Y y

E X E Y

σ µ µ

µ µ

µ µ

µ µ

µ µ

= =

= =

= =

= − −

= − − = =

= − − = =

⎛ ⎞⎛ ⎞= − = − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= − − = × =

∑∑

∑∑

∑ ∑

0( , ) 0XY

X Y X Y

cor X Y σσ σ σ σ

= = = .



3]

2.21. (a) 3 2 3 2 2 2 2 3

3 2 2 3 3 2

2 3

23 3

[( )] E[ 2E X( ) 3 ) 3 ) E X( E X( )E X() 3 )

E X(3 (E X )[E(X )] [ )]E X( E X() 3E (X )E X( ) 2 )

(E X E X ) ( X X XE(X E(X

X X 2 X ]

)

(b) 4 3 2 2 3

4 3 2 2 3 3 2 2 3 4

34 2 2 3 4

34 2 2 4

[( 3 3[ 3 3 3 3 ]

E X( ) 4 (E X ) (E X ) 6 (E X ) (E X ) 4 (E X ) (E X ) (X )) 4[ (E X )][ (E X )] 6[ (E X )] [ (E X )] 3[ (E X )]

(E X XE X X XX X

EE(X

) E X X X )( )]

X X



2.23. X and Z are two independently distributed standard normal random variables, so 2 20, 1,ZX X Z 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 Z|X ) E(X2 2|X ) E (Z|X ) X 02 X 2

(b) E X( )2 2 2 1,XX and 1 0 1Y Z E X Z( ) E X2 2( )

(c) E XY( ) E X( ZX ) E X(3 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 0.Z XE ZX( ) Thus E XY( ) E X( 3) E(ZX ) 0 .

(d) cov(( ) E XY( ) E X( )

00 00corr ( ,X Y )

X Y

XY

X Y X Y

XY ) E[(XE XY X

)(Y

0

)] E [(X 0)( Y 1)]



2.25. (a) 1 2 3 1 2 31 1

ax ax( ax a x x) x( x a x)n n

i n n ii i

ax ax

(b) 111 21

1 2 1 2

11

( ) ( )

) )

n

ii n ni

n nnn

iiii

x x y

x y

yx

y x y x y

x x( y y(

(c) 1

a na)n

ia a a ( a

(d) 2 2 2 2 2 2

1 1

2 2 2 2 2

1 1 1 1 1

( ) 2 2 2 )

22 2

n n

i i i i i i i ii i

n n n n n

i i i i i ii i i i i

a cy a b x c y abx acy bcx y

na b x c y ab x ac y bc x y

bx

(



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.

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