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

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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