Introduction to Econometrics, 4th Edition – James H. Stock & Mark W. Watson | Solutions Manual with Empirical Exercises, Data Sets & Replication Files

SOLUTIONS MANUAL

INTRODUCTION TO ECONOMETRICS
4TH EDITION

CHAPTER NO. 01: ECONOMIC QUESTIONS AND DATA (Solutions
not Available)

CHAPTER 2. REVIEW OF PROBABILITY

SOLUTIONS TO END-OF-CHAPTER EXERCISES

2.1. (a) Probability distribution function for Y
Outcome (number of heads) Y=0 Y=1 Y=2
Probability 0.25 0.50 0.25


(b) Cumulative probability distribution function for Y
Outcome (number of Y<0 0£Y<1 1£Y<2 Y³2
heads)
Probability 0 0.25 0.75 1.0


(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.

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

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



(a)
E(V ) = E(20 − 7Y ) = 20 − 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



σ WV −3.528
corr (W , V ) = = = −0.4425.
σ Wσ V 7.56 × 8.4084

, 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




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.