SW_2e_ex_ch02[1] - PART ONE Solutions to Chapter Exercises...

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P ART O NE Solutions to Chapter Exercises
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Chapter 2 Review of Probability ± Solutions to Exercises 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 heads) Y < 0 0 Y < 1 1 Y < 2 Y 2 Probability 0 0.25 0.75 1.0 (c) = ( ) (0 0.25) (1 0.50) (2 0.25) 1.00 Y EY µ = Using Key Concept 2.3: 22 var( ) ( ) [ ( )] , YE Y E Y =− and = 2 2 ( ) (0 0.25) (1 0.50) (2 0.25) 1.50 so that 2 var( ) ( ) [ ( )] 1.50 (1.00) 0.50. Y E Y = = 2. We know from Table 2.2 that == . Pr( 0) 0 22, Y . 1) 0 78, Y . 0) 0 30, X . 0 70. X So (a) () 0 P r ( 0 ) 1P r ( 1 ) 0 0 22 1 0 78 0 78, ( ) 0 Pr( 0) 1 Pr( 1) 0 0 30 1 0 70 0 70 Y X Y Y EX X X ×= + = ×. +×. =. + . (b) [( ) ] (0 0.70) 0) (1 0.70) ( 0 70) 0 30 0 30 0 70 0 21 [( ) ] (0 0.78) 0) (1 0.78) ( 0 78) 0 22 0 22 0 78 0 1716 XX YY σµ = + =−. ×. +. ×. = . , × =+− × = ×. = . .
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4 Stock/Watson - Introduction to Econometrics - Second Edition (c) Table 2.2 shows Pr( 0, 0) 0 15, XY == = . Pr( 1) = . 1, 0) 0 07, = . 0 63. = . So cov( , ) [( )( )] (0 - 0.70)(0 - 0.78)Pr( 0, 0) (0 0 70)(1 0 78)Pr( 0 (1 0 70)(0 0 78)Pr( 1 0) (1 0 70)(1 0 78) Pr ( 1 ( 070) ( 078) 015 ( 070) 022 015 030 ( 078) 007 0 XY X Y E X Y σµ µ = +− . . = ,= . . = . . = =−. ×−. ×. +−. ×. ×. +. ×−. ×. +. 30 0 22 0 63 0 084, 0084 ( , ) 0 4425 0 21 0 1716 XY cor X Y σ σσ =. . = . . . 3. For the two new random variables 36 WX =+ and 20 7 , VY =− we have: (a) ( ) (20 7 ) 20 7 ( ) 20 7 0 78 14 54, () ( 36)36()360 7 07 2 EV E Y EY EW E X EX = = × . = . = + =+×. =. . (b) 22 2 2 var(3 6 ) 6 36 0 21 7 56, var (20 7 ) ( 7) 49 0 1716 8 4084 X Y = = × . = . = = × . = . . (c) (3 6 , 20 7 ) 6( 7) ( , ) 42 0 084 3 528 3528 ( , ) 0 4425 7 56 8 4084 WV WV WV cov X Y cov X Y cor W V = = × . = . −. = . . . 4. (a) =×−+×= 33 3 ()0( 1 )1 p p p (b) kk k p (c) = ( ) 0.3 = = var( ) ( ) [ ( )] 0.3 0.09 0.21 XE X E X Thus, 0.21 0.46. To compute the skewness, use the formula from exercise 2.21: 2 3 23 ( ) ( ) 3[ ( )][ ( )] 2[ ( )] 0.3 3 0.3 2 0.3 0.084 −= + × + × = Alternatively, −− × + × = 3 ( ) = [(1 0.3) 0.3] [(0 0.3) 0.7] 0.084 Thus, skewness 3 ( ) / .084/0.46 0.87. µσ = =
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Solutions to Exercises in Chapter 2 5 To compute the kurtosis, use the formula from exercise 2.21: 44 3 2 2 4 234 ( ) ( ) 4[ ( )][ ( )] 6[ ( )] [ ( )] 3[ ( )] 0.3 4 0.3 6 0.3 3 0.3 0.0777 EX EX EX µ −= + =− × + × × = Alternatively, −− × + × = 4 ( ) = [(1 0.3) 0.3] [(0 0.3) 0.7] 0.0777 Thus, kurtosis is 4 ( ) / = .0777/0.46 1.76 µσ 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 (100/180) ( 32) or 17.78 (5/9) . YX Y X = + × Using Key Concept 2.3, 70 F X implies that 17.78 (5/9) 70 21.11 C, Y + × = ° and 7F X σ implies (5/9) 7 3.89 C. Y = ° 6. The table shows that Pr( 0, 0) 0 045, XY == = . Pr( 0, 1) 0 709, = . 1, 0) 0 005, = . 0 241, = . 0) 0 754, X . . 0 246, X . 0) 0 050, Y . 0 950. Y (a) × = + × = =×. +×. =. . () 0 P r ( 0 ) 1P r ( 1 ) 0 0 050 1 0 950 0 950 Y EY Y Y (b) (unemployed) Unemployment Rate (labor force) 0) 0 050 1 0 950 1 ( ) # # YE Y = = . = . = . (c) Calculate the conditional probabilities first: 0, 0) 0 045 0| 0) 0 0597, 0) 0 754 X . = = = . =. 0, 0 709 1| 0) 0 9403, 0) 0 754 X . = = = . 0) 0 005 0| 0 0203, 0 246 X . = = = . 0 241 0 9797 0 246 X . = = = . . The conditional expectations are ( | 0 Pr ( 0| 1) 1 Pr ( 0 0 0203 1 0 9797 0 9797, ( | 0) 0 Pr( 0| 0) 1 Pr( 0) 0 0 0597 1 0 9403 0 9403 EYX Y X Y X Y X Y X ==× = =+ × = = = ×. +×. = = ×= = + = .
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6 Stock/Watson - Introduction to Econometrics - Second Edition (d) Use the solution to part (b), Unemployment rate for college grads 1 ( | 1) 1 0.9797 0.0203. Unemployment rate for non-college grads 1 ( | 0) 1 0.9403 0.0597.
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This note was uploaded on 11/06/2009 for the course ECON ECON111 taught by Professor Smith during the Spring '09 term at Punjab Engineering College.

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SW_2e_ex_ch02[1] - PART ONE Solutions to Chapter Exercises...

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