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Introduction to Econometrics (Global Edition) 4th Edition By James Stock, Mark Watson (Solution Manual)

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Introduction to Econometrics (Global Edition) 4e James Stock, Mark Watson (Solution Manual) Introduction to Econometrics (Global Edition) 4e James Stock, Mark Watson (Solution Manual)

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  • July 13, 2023
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©2019 Pearson Education Ltd. Introduction to Econometrics (4th Edition, Global Edition) by James H. Stock and Mark W. Watson Solutions to End‐of‐Chapter Exercises: Chapter 2* . NOTE: For Complete File, Download link at the end of this File Stock/Watson - Introduction to Ec onometrics 4th E dition, Global Edition - Answers to Exercises: Chapter 2 _______________________________________________________________ ______________________________________ ©2019 Pearson Education Ltd. 1 2.1. (a) Probability distribution function for Y Outcome (number of heads) Y = 0 Y = 1 Y = 2 Probability 0.36 0.48 0.16 (b) = ( ) (0 0.36) (1 0.48) (2 0.16) 0.8   YEY Using Key Concept 2.3: var(Y)E(Y2)[E(Y)]2, and 22 2( ) 2 0.6 0.4 4 0.16 1.12   YY EY so that 22 2var( ) ( ) [ ( )] 1.12 0.8 0.48   YE Y E Y Stock/Watson - Introduction to Ec onometrics 4th E dition, Global Edition - Answers to Exercises: Chapter 2 _______________________________________________________________ ______________________________________ ©2019 Pearson Education Ltd. 2
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) YE(Y)0Pr(Y0)1Pr (Y1)
00221078078,
XE(X)0Pr (X0)1Pr (X1)
00301070070 (b) X2E[(X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 Y2E[(YY)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) XY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)P r( X0Y1)
(1070)(0078)P r( 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,
corr( X,Y)XY
XY0084
0210171604425 Stock/Watson - Introduction to Ec onometrics 4th E dition, Global Edition - Answers to Exercises: Chapter 2 _______________________________________________________________ ______________________________________ ©2019 Pearson Education Ltd. 3
2.3. For the two new random variables 48 WX and 11 2 , VY we have: (a) ( ) (11 2 ) 11 2 ( ) 11 2 0 78 9 44,
() ( 48) 48 () 480 7 0 9 6        
     EV E Y EY
EW E X E X (b) 22 2
22 2v a r( 4 8 ) 8 6 4 02 1 1 34 4 ,
var (11 2 ) ( 2) 4 0 1716 0 6864
              WX
VYX
Y (c) cov(4 8 ,11 2 ) 8 ( 2)cov( , ) 16 0 084 1 344        WV XY X Y 1 344corr ( , ) 0 443
13 44 0 6864
  
 WV
WVWV

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