Study Guide, Answer of Introduction of Econometric, 2rd
PART ONE Solutions to Exercises
Chapter 2
Review of Probability
Solutions to Exercises
1. (a) Probability distribution function for Y Outcome (number of heads) probability Y 0 Y 1 Y 2
0.25
0.50
0.25
(b) Cumulative probability distribution function for Y Outcome (number of heads) Probability (c)
Y
Y 0
0
0
Y 0.25
1
1
Y 0.75
2
Y
2
1.0
= 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 ) (0 2 0.25) (12 0.50) (2 2 0.25) 1.50
so that var(Y ) E(Y 2 ) [E (Y )]2 2. We know from Table 2.2 that Pr (Y Pr ( X 1) 0 70. So (a)
Y
1.50 (1.00)2
0.50.
0) 0 22, Pr (Y
1) 0 78, Pr ( X
0) 0 30,
E(Y ) 0 Pr (Y
0) 1 Pr (Y 1)
0 0 22 1 0 78 0 78, E( X ) 0 Pr ( X 0) 1 Pr ( X 1)
X
0 0 30 1 0 70 0 70 (b)
2 X
E[( X
X
)2 ] 0) (1 0.70)2 Pr ( X 1)
(0 0.70)2 Pr ( X
( 0 70)2 0 30 0 30 2 0 70 0 21
2 Y
E[(Y
Y
)2 ] 0) (1 0.78)2 Pr (Y 1)
(0 0.78)2 Pr (Y
( 0 78)2 0 22 0 222 0 78 0 1716
4
Stock/Watson - Introduction to Econometrics - Second Edition
(c) Table 2.2 shows Pr ( X 0, Y Pr ( X 1, Y 1) 0 63. So
XY
0) 0 15, Pr ( X
0, Y 1) 0 15, Pr ( X 1, Y
0) 0 07,
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, 0 084 0 21 0 1716
cor (X , Y )
X
XY Y
0 4425
3.
For the two new random variables W (a)
3 6 X and V
20 7Y , we have:
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)
2 W 2 V
var (3 6 X ) 62
2 X 2
36 0 21 7 56,
2 Y
var (20 7Y ) ( 7)
49 0 1716 8 4084
(c)
WV
cov (3 6 X , 20 7Y ) 6( 7) cov (X , Y )
WV W V
42 0 084
3 528
cor (W , V ) 4.
3 528 7 56 8 4084 p p p p 0.3 0.09 0.21
0 4425
(a) E ( X 3 ) 03 (1
Chapter 2
Review of Probability
Solutions to Exercises
1. (a) Probability distribution function for Y Outcome (number of heads) probability Y 0 Y 1 Y 2
0.25
0.50
0.25
(b) Cumulative probability distribution function for Y Outcome (number of heads) Probability (c)
Y
Y 0
0
0
Y 0.25
1
1
Y 0.75
2
Y
2
1.0
= 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 ) (0 2 0.25) (12 0.50) (2 2 0.25) 1.50
so that var(Y ) E(Y 2 ) [E (Y )]2 2. We know from Table 2.2 that Pr (Y Pr ( X 1) 0 70. So (a)
Y
1.50 (1.00)2
0.50.
0) 0 22, Pr (Y
1) 0 78, Pr ( X
0) 0 30,
E(Y ) 0 Pr (Y
0) 1 Pr (Y 1)
0 0 22 1 0 78 0 78, E( X ) 0 Pr ( X 0) 1 Pr ( X 1)
X
0 0 30 1 0 70 0 70 (b)
2 X
E[( X
X
)2 ] 0) (1 0.70)2 Pr ( X 1)
(0 0.70)2 Pr ( X
( 0 70)2 0 30 0 30 2 0 70 0 21
2 Y
E[(Y
Y
)2 ] 0) (1 0.78)2 Pr (Y 1)
(0 0.78)2 Pr (Y
( 0 78)2 0 22 0 222 0 78 0 1716
4
Stock/Watson - Introduction to Econometrics - Second Edition
(c) Table 2.2 shows Pr ( X 0, Y Pr ( X 1, Y 1) 0 63. So
XY
0) 0 15, Pr ( X
0, Y 1) 0 15, Pr ( X 1, Y
0) 0 07,
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, 0 084 0 21 0 1716
cor (X , Y )
X
XY Y
0 4425
3.
For the two new random variables W (a)
3 6 X and V
20 7Y , we have:
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)
2 W 2 V
var (3 6 X ) 62
2 X 2
36 0 21 7 56,
2 Y
var (20 7Y ) ( 7)
49 0 1716 8 4084
(c)
WV
cov (3 6 X , 20 7Y ) 6( 7) cov (X , Y )
WV W V
42 0 084
3 528
cor (W , V ) 4.
3 528 7 56 8 4084 p p p p 0.3 0.09 0.21
0 4425
(a) E ( X 3 ) 03 (1