# Econometrics with SW2

There are three important requirements for this question. 1. The crucial part of this question is the explanations and list out every single symbols, formulas, algebra steps whether it is about the differentiation, the summation operator or anything as to explain to a primary school student who has not done this subject and know nothing about this topics. Which means I would need all the theories, step by step, and all the symbols in details. 2. The actual answers, the actuals answers must be incorporate with the explanations as shown as something in the screenshot sample where the explanations and the actual answers are combined together. 3. I will have many questions to follow afterward and the tutor must be very welcoming and have the right attitude to answer any concerns I have. The answers must be typed In a word document, with all the formulas or any algebra manipulation typed in the word document using the formula option.

For example, the question starts with 1. List out all the necessary symbols, formulas, theories. 2. Explain what the 1 list out . 3. The answer with all the manipulation steps with the explanations of each manipulation and each algebra. 4. Enthusiastic attitude about any queries I am having. 5. Work in the long run if this one works out well.

lOMoARcPSD|11352339

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Introduction!to!Econometrics!(3rd!Updated!Edition)!

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by!

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James!H.!Stock!and!Mark!W.!Watson!

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Solutions!to!End7of7Chapter!Exercises:!Chapter!2*!

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(This version August 17, 2014)!

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*Limited!distribution:!For!Instructors!Only.!!Answers!to!all!odd=numbered!

questions!are!provided!to!students!on!the!textbook!website.!!!If!you!find!errors!in!

the!solutions,!please!pass!them!along!to!us!at!mwatson@princeton.edu.!!

!

©2015 Pearson Education, Inc.

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lOMoARcPSD|11352339

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

1

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!

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 −1.96 and 0.39

< −1.96. Solving these inequalities yields n ≥
0.24/n
0.24/n
9220.
!
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lOMoARcPSD|11352339
Stock/Watson - Introduction to Econometrics - 3rd Updated Edition - Answers to Exercises: Chapter 2
19
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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
σ Y2 = E ⎢⎢ (Y − µY ) ⎥⎥
⎡
2⎤
⎣
⎦
= (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.
2
×10
(b) (i) E (Y ) = µY = $1000, σ Y2 = σnY = 1.9100
= 1.9 ×105.
7
(ii) Using the central limit theorem,
Pr (Y > 2000) = 1 − Pr (Y ≤ 2000)

⎛ Y − 1000 2, 000 − 1, 000 ⎞

= 1 − Pr ⎜

≤

⎟

5

1.9 ×105 ⎠

⎝ 1.9 ×10

≈ 1 − Φ (2.2942) = 1 − 0.9891 = 0.0109.

!

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lOMoARcPSD|11352339

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

20

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

⎛ k

⎜

⎜

⎜

i =1 ⎝⎜ j =1

⎞

⎟

l

=∑ ∑ yj Pr (Y = y j |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

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lOMoARcPSD|11352339

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

21

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l

m

2.20. (a) Pr (Y = yi ) = ∑∑ Pr (Y = yi | X = xj , Z = zh ) Pr (X = xj , Z = zh )

j =1 h =1

(b)

k

E (Y ) = ∑ yi Pr (Y = yi ) Pr (Y = yi )

i =1

k

l

m

i =1

j =1 h =1

= ∑ yi ∑∑ Pr (Y = yi | X = xj , Z = zh ) Pr (X = xj , Z = zh )

l

m

⎡ k

⎤

= ∑∑ ⎢ ∑ yi Pr (Y = yi | X = xj , Z = zh ) ⎥ Pr (X = xj , Z = zh )

j =1 h =1 ⎣ i =1

⎦

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.

!

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lOMoARcPSD|11352339

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

22

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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

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

(b)

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

= E[ X 4 − 3 X 3 µ + 3 X 2 µ 2 − X µ 3 − X 3 µ + 3 X 2 µ 2 − 3 X µ 3 + µ 4 ]

= E ( X 4 ) − 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 – 3rd Updated Edition – Answers to Exercises: Chapter 2

23

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2. 22. The mean and variance of R are given by

µ = w × 0.08 + (1 − w) × 0.05

σ 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) µ = 0.065; σ = 0.044

(b) µ = 0.0725; σ = 0.056

(c) w = 1 maximizes µ ; σ = 0.07 for this value of w.

(d) The derivative of σ2 with respect to w is

dσ 2

= 2w × .072 − 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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lOMoARcPSD|11352339

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

24

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

µ X = µZ = 0, σ X2 = 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 X2 + µ X2 = 1, and µY = E ( X 2 + Z ) = E ( X 2 ) + µ Z = 1 + 0 = 1.

(c) E ( XY ) = E ( X 3 + ZX ) = E ( X 3 ) + E (ZX ). Using the fact that the odd moments of

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

independence between X and Z , we have E ( ZX ) = µZ 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 XσY σ XσY

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lOMoARcPSD|11352339

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

25

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2.24. (a) E (Yi 2 ) = σ 2 + µ 2 = σ 2 and the result follows directly.

(b) (Yi/σ) is distributed i.i.d. N(0,1), W = ∑i =1 (Yi /σ )2 , and the result follows from the

n

definition of a χ n2 random variable.

n

n

Y2

Y2

(c) E (W ) = E ∑ i 2 = ∑ E i 2 = n.

i =1

σ

i =1

σ

(d) Write

V=

Y1

∑in=2 Yi2

n −1

=

Y1 /σ

∑in=2 (Y /σ )2

n −1

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

n

n

∑i=2 (Yi /σ )2 ~ χ n2−1 , and Y1/σ and ∑i=2 (Yi /σ )2 are independent. The result then

follows from the definition of the t distribution.

!

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lOMoARcPSD|11352339

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

26

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n

n

i =1

i =1

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

(b)

n

∑ (x + y ) = (x + y + x + y +L x + y )

i =1

i

i

1

1

2

2

n

n

= ( x1 + x2 + L xn ) + ( y1 + y2 + L yn )

n

n

i =1

i =1

= ∑ xi + ∑ yi

n

(c) ∑ a = (a + a + a + L + a ) = na

i =1

(d)

n

n

i =1

i =1

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

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

!

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lOMoARcPSD|11352339

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

27

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

cov(Yi , Y j )

σY σY

i

=

cov(Yi , Y j )

σ Yσ Y

j

=

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

Y

E

Y

E

Y

µY = µY

(

)

=

(

)

=

,

so

that

∑i

∑ i n∑

n i =1

n i =1

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 + L + n − 1) =

i =1 j =i +1

an(n − 1)

for any

2

variable a.

(d) When n is large

!

σ Y2

n

≈ 0 and

1

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

n

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

28

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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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