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Solution Manual Introduction to Econometrics 4th Global Edition by James Stock

Simplify your study of econometrics with this comprehensive solution manual, designed to provide students with clear solutions to econometric models and statistical methods. Ideal for economics students.<br>

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Solution Manual Introduction to Econometrics 4th Global Edition by James Stock

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  1. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 Introduction to Econometrics (4th Edition, Global Edition) by James H. Stock and Mark W. Watson Solutions to End-of-Chapter Exercises: Chapter 2* (This version September 14, 2018) complete document is available on https://solumanu.com/ *** contact me if site not loaded

  2. 2.1. (a) Probability distribution function for Y Outcome (number of heads) Probability smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 smtb98@gmail.com 1 Email: smtb98@gmail.com Telegram: https://t.me/solumanu Y = 0 0.36 Y = 1 0.48 Y = 2 0.16   (0 0.36) (1 0.48) (2 0.16)       = ( ) E Y 0.8 (b) Y Using Key Concept 2.3: var(Y)  E(Y2)-[E(Y)]2, and     2 0.6 0.4 4 0.16 1.12       2 2 Y 2 Y E Y ( ) so that  ) [ ( )] - E Y  1.12 0.8 -  2 2 2 var( ) Y E Y ( 0.48 complete document is available on https://solumanu.com/ *** contact me if site not loaded

  3. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 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) X 2 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, Y  (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. 2 E[(Y - Y)2] (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)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, XY XY 0.210.1716 0.084 corr(X,Y)    0.4425. complete document is available on https://solumanu.com/ *** contact me if site not loaded

  4. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 3  11 2 , - 4 8   V Y we have: 2.3. For the two new random variables Xand W (a)   -  -  -  .  . ( ) ( E W E V E E (11 2 ) (4 8 )  Y X 11 2 ( ) 4 8 ( )   E Y E X 11 2 0 78 4 8 0 70    . 9 44, 9 6  . . ) (b)          .  . 2 W 2 2 X var(4 8 ) X 8 64 0 21 13 44,     . - ( 2)  - 0 6864  . . 2 V 2 2 Y var(11 2 ) Y 4 0 1716 (c)   cov(4 8 ,11 2 )  X - 8 ( 2)cov( , )   - 16 0 084  -  .  - . Y X Y 1 344 WV    W 1 344 - . .  .    - . WV corr( W V , ) 0 443 13 44 0 6864 V complete document is available on https://solumanu.com/ *** contact me if site not loaded

  5. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 4   - ) 1    3 4 4 E X ( ) 0 (1 p p p 2.4. (a) (b) E(Xk) 0k(1- p)1k p  p  (c) E X ( ) 0.53  ) [ ( )] - E X  0.53 0.28 -  2 2 var( ) X E X ( 0.25    0.25 0.5 Thus, To compute the skewness, use the formula from exercise 2.21: -    ) 3[ ( - -  )][ ( )] 2[ ( )]    3 3 2 3 E X ( ) E X ( E X E X E X  - 2 3 0.53 3 0.53 2 0.53 0.12 To compute the kurtosis, use the formula from exercise 2.21: -    ) 4[ ( )][ ( - -  )] 6[ ( )] [ (  -  )] 3[ ( )] - E X 4 4 3 2 2 4 E X ( ) E X ( E X E X E X E X 6 0.53    2 3 4 0.53 4 0.53 3 0.53 1.015 complete document is available on https://solumanu.com/ *** contact me if site not loaded

  6. 2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y = 0 smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 5 when X = 32 and Y =100 when X = 212. This implies Y  (100/180)(X -32) or Y  -17.78(5/9) X. 17.78 (5/9) 65 18.33 C,   -  Y    Using Key Concept 2.3, X = 65oF implies that   Y    and X = 5oF implies (5/9) 5 2.78 C.

  7. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 6   1, 0 078,  . 1)   . Y   0 673,  . 0 751, 2.6. The table shows thatPr( Pr( 1,  X Y Pr( 1)   . X Pr( X Pr( 0)  0, X  . Y  0 12, 0) Y Pr( X 0, Y  1)  .  0 249, 0 042,  . Pr( Pr( 0 88.  . 0) 0 207, 1)  X 0) Y (a)   0 Pr(       E Y ( ) Y 0) 1 Pr( 0 88  . Y 1) Y 0 0 12 1 0 88   .   . (b) (unemployed) (labor force) Pr( 0)  Y #  Unemployment Rate #  1 Pr(  -   - ( ) 1 0 88  - . E Y  Y 1) 1 0.12 (c) Calculate the conditional probabilities first:   . Pr( X 0, Y 0) 0 078 0 751 .     0 104  . Pr( Y 0| X 0)  Pr( X 0)  X  . Pr( X 0, Y 1) 0 673 0 751 .     0 896  . Pr( Y 1| X 0)  Pr( 0)   . Pr( X 1, Y 0) 0 042 0 249 .     0 169  . Pr( Y 0| X 1)  Pr( X 1)  X  . Pr( X 1, Y 1) 0 207 0 249 .     0 831  . Pr( Y 1| X 1)  Pr( 1) The conditional expectations are       .      E Y X ( | 1) 0 Pr( 0 0 169 1 0 831   . Y 0| X 1) 1 Pr( 0 831,  . Y 1| X 1)       .      E Y X ( | 0) 0 Pr( 0 0 104 1 0 896   . Y 0| X 0) 1 Pr( 0 896  . Y 1| X 0) .

  8. Unemployment rate for college graduates = 1 – E(Y|X=1) = 1 − 0.831 = 0.169. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 7 (d) Use the solution to part (b), Unemployment rate for non-college graduates = 1 – E(Y|X=0) = 1 − 0.896 = 0.104 (e) The probability that a randomly selected worker reported as unemployed is a college graduate is  Y  . Pr( X 1, Y 0) 0 042 0 12 .     0 35  . Pr( X 1| Y 0)  Pr( 0) The probability that this worker is a non-college graduate is   1 Pr(  -   1 0 35  - . 0 65  . Pr( X 0| Y 0) X 1| Y 0) (f) Educational achievement and employment status are not independent because they do not satisfy that, for all values of x and y, Pr(X  x|Y  y)  Pr(X  x). For example, from part (e) Pr(X  0|Y  0) = 0.65, while from the table, Pr(X = 0) = 0.751.

  9. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 8 2.7. Using obvious notation, C  M  F; thus C M F and C 2M 2F 22cov(M, F). This implies   C 50 48   $98,000 (a) per year. , so that cov(M, F) MFcorr(M, F). Thus  where the units are squared thousands of (b) corr(M, F)  cov( cov(M, F) MF  15 13 0.9   M F , ) 175.50, dollars per year. (c) C 2M 2F 22cov(M, F), so that      2 C 2 2 15 13 2(175.5) 745, and   C  thousand dollars per year. 745 27.295 (d) You need to look up the current euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose this exchange rate is e (say e = 0.75 euro per dollar or 1/e = dollars per euro); each 1 dollar is therefore valued at e euros. The mean is, therefore, e × C (in units of thousands of euros per year), and the standard deviation is e × C (in units of thousands of euros per year). The correlation is unit-free, and is unchanged.

  10. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 9 1 9  -   Y     Z 3( Y 4), 2 Y E Y ( ) 4, 2.8. With var( ) Y .     -   -  -  E 3( Y 4) 3( 4) 3(4 4) 1 1 9 0 Z Y     -      2 Z 2 Y var 3( Y 4) 9 9

  11. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 10 2.9. Probability distribution of X 0.29 0.36 0.35 1.00 Value of Y 2 4 6 8 10 0.01 0.02 0.01 0.04 Value of X 3 6 9 0.04 0.10 0.13 0.27 0.09 0.06 0.11 0.26 0.03 0.15 0.04 0.22 0.12 0.03 0.06 0.21 Probability distribution of Y (a) The probability distribution is given in the table above. ( ) 2 0.27 4 0.26 6 0.22 8 0.21 10 0.04          - Y E Y E Y (b) The conditional probability of Y|X = 6 is given in the table below         E Y 4.98        2 2 2 2 2 2 E Y ( ) 2 0.27 4 0.26 6  0.22 8 0.21 10 0.04 30.6 30.6 4.98 - 2 2 var( ) ( ) [ ( )] 25.62 Value of Y 6 0.15/0.36 2 4 8 10 0.10/0.36 0.06/0.36 0.03/0.36 0.02/0.36    4 (0.06/0.36) 6 (0.15/0.36)      E Y X ( | 6) 2 (0.01/0.36) 8 (0.03/0.36) 10 (0.02/0.36)     4.944           2 2 2 2 E Y X ( | 6) 2 (0.01/0.36) 4 (0.06/0.36) 6  (0.15/0.36)  2 2 8 (0.03/0.36) 10 (0.02/0.36) 29.667  29.667 4.944 -  var( ) Y 24.723 (c)  (3 2 0.04)   (  E XY ov( , )/(  c (3 4 0.09)   ) ( ) E X E Y   X Y X Y  (  (9 10 0.01)     E XY cov( , ) X Y corr( , ) X Y ( ) 29.4 1.3764  - -   - ) 29.4 6.18 4.98 1.3764 (5.7276 24.723)  -  0.0097  - )

  12. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 11 Y - Y Y         Y 2 Y ~ N(0,1) and Appendix Table 1, 2.10. Using the fact that if then Y N : , we have - -             Y 4 5 4 3 1 3      0 6304  . (a) Pr( Y 5) Pr 3 - -                   Y 5 2 5 4 3 4 3 4  1 Pr  -   - -   0 7734  . (b) Pr( Y 2) 1 4 - - -             2 1 2 Y 1 3 1 2 1 2        - 0 2857  . (c) Pr(2 Y 5) Pr (2) 2 - - -       1 2 1 Y 2 4 2 1        - -   (2) (1 - - - (d) Pr(1 Y 4) Pr (2) ( 1) ( 1)) 1 0 8185  .

  13. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 12 2.11. (a) 0.90 (b) 0.95 (c) 0.95  2 8 Y /8~ F . Y ~ , (d) When then  8, (e) Y  Z2, where Z ~ N(0,1), thus Pr(   0.5 -    Y 0.5) Pr( Z 0.5) 0.520

  14. (d) The tdfdistribution converges to the (0,1) distribution as df gets large. The critical value coincides as df goes to infinity. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 13 2.12. (a) 0.90 (b) 0.90 (c) 0.9108 (e) 0.05 (f) 0.01

  15. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 14    4 0        16 0 16.   2 2 Y 2 2 W E Y ( ) Var( ) Y 4; E W ( ) Var( W ) 2.13. (a) (b) Y and W are symmetric around 0; thus, skewness is equal to 0; because their mean is 0, this means that the third moment is zero. 3E(Y - Y)4 (c) The kurtosis of the normal is 3, so ; transforming both Y and W Y 4 to the standard normal yields the results. (d) First, condition on X  0, so that S W:    0) 16,     3 16   2 3 4 2 E S X ( | 0) 0; E S X ( | E S X ( | 0) 0, E S X ( | 0) . Similarly,        3 4   2 3 4 2 E S X ( | 1) 0; ( E S X | 1) 4, E S X ( | 1) 0, E S X ( | 1) From the large of iterated expectations E(S)  E(S|X  0) Pr(X  0) E(S|X 1) Pr(X 1)  0   0) Pr(X        1) Pr(   2 2 2 E S ( ) E S X ( | 0) E S X ( | X 1) 16 0.1 4 0.9   5.2 E(S3) E(S3|X  0) Pr(X  0) E(S3|X 1) Pr(X 1) 0   0) Pr(X       1) Pr(   4 4 4 E S ( ) E S X ( | 0) E S X ( | X 1) 3 16     0.9 120  2 2 0.1 3 4 (e) S E(S) 0, thus E(S - S)3 E(S3) 0 from part (d). Thus skewness = 0.    -     -   ) 120.  2 S 2 2 4 4 E S ( ) E S 4.44. ( ) 5.2, E S ( ) E S ( Similarly, Thus, and S S 2 kurtosis 120 (5.2 )

  16. 2.14. The central limit theorem suggests that when the sample size (n) is large, the smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 15 2   with Y Y n. distribution of the sample average (Y) is approximately N Y,Y   Y 21,   Y 2 2 2 50, Given 2 Y n     0 42,  .  2 Y n 50, 21 50 (a) and       - -   Y 50 51 50 0 42 .     0 5613  . Pr( Y 51) Pr 1 42 0 42 . 2 Y  n    0 14,  .  2 Y n 150, 21 150 (b) and       - - . Y 50 49 40 0 14 0 9962  .  1 Pr(  -  1 Pr  -  Pr( Y 49) Y 49) 0 14 .  - - .   (2 6724) . 1 ( 2 6724) 2 Y     0 467,  .  2 Y n 45, 21 45 (c) and 64       - - . - . 50.5 50 0 467 (1 4633) . Y 50 51 50 0 467 0 928307 0 76824  .      Pr(50.5 Y 51) Pr . 0 467 (0 7317) .   - - . 0 1605  .

  17. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 16 2.15. (a)       - - - 19.6 20 4/ - Y 20 n 20.4 20 4/    n      Pr(19.6 Y 20.4) Pr n 4/ n -    19.6 20 4/ 20.4 20 4/    Pr Z n where Z ~ N(0, 1). Thus, - -       19.6 20 4/ 20.4 20 4/   -     Pr Z Pr( 1 Z 1) 0.6827 (i) n = 25; n n - -       19.6 20 4/ 20.4 20 4/   -     Pr Z Pr( 2 Z 2) 0.954 (ii) n = 100; n n - -       19.6 20 4/ 20.4 20 4/   Pr( 5.656 -     Pr Z Z 5.656) 0.954 (iii)n = 800; n n (b)          - - c Y 20 n c -       Pr(20 c Y 20 c ) Pr 4/ - n 4/ 4/ n    c c    Pr Z . 4/ n 4/ n c As n get large, gets large, and the probability converges to 1. 4/ n (c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6.

  18. 2.16. There are several ways to do this. Here is one way. Generate n draws of Y, Y1, Y2, … , smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 17 Yn. Let Xi= 1 if Yi < 5.8, otherwise set Xi = 0. Notice that Xi is a Bernoulli random variable with X = Pr(X = 1) = Pr(Y < 5.8). Compute X. Because X converges in probability to X = Pr(X = 1) = Pr(Y < 5.8); X will be an accurate approximation if n is large.

  19. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 18 2.17. Y= 0.6 and Y 2 0.40.6  0.24 (a) (i)             - - - Y 0.6 0.64 0.6 0.24/ Y 0.6  0.64) 1 Pr  -  1 Pr  -   P Y ( 0.5773 0.28 0.24/ n n 0.24/ n             - - - Y 0.6 0.56 0.6 0.24/ Y 0.6      -  P( Y 0.56) Pr Pr 1.154 0.12 (ii) 0.24/ n n 0.24/ n b) We know Pr(−1.96 Z  1.96) = 0.95, thus we want n to satisfy   -  - - - and 0.55 0.6 Solving these inequalities yields 0.61 1.96 1.96. 0.65 0.6 0.24/ n 0.24/ n n  369.

  20. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 19  0 95,  .  0 05.  . 2.18. Pr( Pr( Y $0) Y $30,000) (a) The mean of Y is    Y  $0) 30,000 Pr(     0 Pr( Y Y $30,000) $1,500 The variance of Y is            2   -  2 Y E Y Y       -   2 (0 1,500) -  - Pr Y 0 (30,000 1,500) Pr( Y 30,000) 2 0 95 (28,500)  . 0 05  . 4 27 10  .  2 2 2 ( 1,500) 1 2   Y (4 27 10 ) .   7 $6,538.35 so the standard deviation of Y is 2 Y n      7    2 Y .  $355,833 4 27 10 120 (b) (i) ( ) E Y $1500, Y (ii) Using the central limit theorem,  3000) 1 Pr(  -  Pr( Y Y 3000)       - , - , 1,500 355,833 (2 514 1 0 9939  - . Y 3 000 1 500 355,833 1 Pr  -   - . 1 5) $0 0061  . correct to 4 decim ( al pl ace ) s .

  21. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 20 2.19. (a) l Pr(Y  yj)  Pr(X  xi,Y  yj) i1 l  Pr(Yyj|Xxi)Pr(Xxi) i1 (b) k k l E(Y)  yjPr(Y  yj)  Pr(Y  yj|X  xi)Pr(X  xi) yj j1 j1 i1       k l  yjPr(Y  yj|X  xi) Pr(X xi) j1 i1  l  E(Y|Xxi)Pr(Xxi). i1 (c) When X and Y are independent, Pr(X  xi,Y  yj) Pr(X  xi)Pr(Y  yj), so XY E[(X - X)(Y - Y)] j1 j1   E(X - X)E(Y -Y) 00  0, l k  (xi-X)(yj-Y)Pr(Xxi,Yyj) i1 l k (xi-X)(yj-Y)Pr(Xxi)Pr(Yyj)  i1    l k  (xi- X)Pr(X  xi) (yj- Y)Pr(Y  yj i1 j1 XY XY 0 cor(X,Y)   0. XY

  22. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 21 l m Pr(Y  yi)  Pr(Y  yi|X  xj, Z  zh)Pr( X  xj, Z  zh) 2.20. (a) j1 h1 (b) k E(Y)  yiPr(Y  yi)Pr(Y  yi) i1 k l m  Pr(Y  yi|X  xj, Z  zh)Pr( X  xj, Z  zh) yi i1 j1   h1   Pr( l m k  yiPr(Y  yi|X  xj, Z  zh) X  xj, Z  zh) j1 h1 i1 l m  E(Y|X  xj, Z  zh)Pr( X  xj, Z  zh) j1 h1 where the first line in the definition of the mean, the second uses (a), the third is a rearrangement, and the final line uses the definition of the conditional expectation.

  23. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 22 2. 21. (a) E(X - )3 E[(X - )2(X - )] E[X3-2X2  X2- X2 2X2- 3]  E(X3)-3E(X2) 3E(X)2- 3 E(X3)-3E(X2)E(X)3E(X)[E(X)]2-[E(X)]3  E(X3)-3E(X2)E(X)2E(X)3 (b) E(X - )4 E[(X3-3X2 3X2- 3)(X - )]  E[X4-3X3 3X22- X3- X3 3X22-3X3 4]  E(X4)-4E(X3)E(X)6E(X2)E(X)2-4E(X)E(X)3 E(X)4  E(X4)-4[E(X)][E(X3)]6[E(X)]2[E(X2)]-3[E(X)]4

  24. 2. 22. The mean and variance of R are given by smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 23     - (0.06) w (1 w )(0.04) - w      - ) [0.09 0.05 0.3]   w  2 2 2 2 w (0.09) (1 ) (0.05) 2 w (1    where 0.09 0.05 0.3 between Rs and Rb. Cov R ( , R follows from the definition of the correlation ) s b     0.044; 0.036 (a)     0.056; 0.060 (b)    (c) w = 1 maximizes ; 0.09 for this value of w. (d) The derivative of 2 with respect to w is  d dw 2   - - ) 0.05  w  (2 4 ) [0.09 0.05 0.30] -  w   2 2 2 w 0.09 2(1  - 0.0158 w 0.0023   Solving for w yields positive, so that this is the global minimum.) With (Notice that the second derivative is 0.15,  w w 23 158 0.15.   0.038. R

  25. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 24 2. 23. X and Z are two independently distributed standard normal random variables, so X Z 0,X 2Z 21,XZ 0. (a) Because of the independence between X and Z,Pr(Z  z|X  x) Pr(Z  z), and E(Z|X)  E(Z)  0. Thus E(Y|X) E(X2 Z|X) E(X2|X) E(Z|X) X20  X2. (b) E(X2)X 2 X 21, and Y E(X2 Z) E(X2) Z10 1. (c) E(XY) E(X3 ZX) E(X3) E(ZX). Using the fact that the odd moments of a standard normal random variable are all zero, we have E(X3) 0. Using the independence between X and Z, we have E(ZX) ZX 0. Thus E(XY) E(X3) E(ZX) 0. (d) cov(XY)  E[(X - X)(Y - Y)] E[(X -0)(Y -1)]  E(XY - X)  E(XY)- E(X)  0-0  0. XY XY XY 0 corr(X,Y)    0.

  26. smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 25 2) 2 22 and the result follows directly. 2.24. (a) E(Yi n (b) (Yi/) is distributed i.i.d. N(0,1), W  definition of a n (Yi/)2 , and the result follows from the i1 2 random variable. 2 2 Yi 2 EYi 2 n n E(W)  E   n. (c) i1 i1 (d) Write Y1/ Y1 V   nYi 2 n (Y / )2 n-1 i2 n-1 i2 which follows from dividing the numerator and denominator by . Y1/~ N(0,1), (Yi/)2 i2 i2 follows from the definition of the t distribution. n n (Yi/)2 ~ n-1 2, and Y1/and are independent. The result then

  27. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 26 2.25. (a) (b) (c) (d) n n (abxicyi)2  (a2b2xi 2c2yi 22abxi2acyi2bcxiyi) i1 i1 n n n n n  na2b2 c2 2ab 2ac 2bc 2 2 xi yi xi yi xiyi i1 i1 i1 i1 i1

  28. uses the definition of correlation, the second uses the fact that Yi and Yj have the same smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 27 cov(Yi,Yj) YiYj cov(Yi,Yj) YY cov(Yi,Yj) Y  r, where the first equality 2.26. (a) corr(Yi,Yj) = 2 variance (and standard deviation), the third equality uses the definition of standard deviation, and the fourth uses the correlation given in the problem. Solving for cov(Yi, Yj) from the last equality gives the desired result. Y 1 2Y11 1 2E(Y)11 2E(Y2)  Y 2Y2, so that E( Y ) = (b) 4cov(Y1,Y2) Y rY 2 2 1 4var(Y1)1 4var(Y2)2 var( Y ) = 2 2 n n n Y 1 E(Y) 1 1 Y  Y (c) , so that Yi E(Yi) n n n i1 i1 i1   n 1 n var(Y)  var Yi i1 n-1 n n 1 2 var(Yi) cov(Yi,Yj) n2 n2 i1 i1 ji1 n-1 n n 1 2 Y rY 2 2 n2 n2 i1 n(n-1) i1 ji1 Y 2 rY 2 n2 n Y rY 2  1-1  2 n n where the fourth line uses for any variable a. Y n 2 1 n0, and the result follows from (c). 0 and (d) When n is large

  29. Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ 28 2.27 (a) E(u) = E[E(u|X) ] = E[E(Y−ˆY )|X] = E[ E(Y|X) – E(Y|X) ] = 0. (b) E(uX) = E[E(uX|X) ] = E[XE(u)|X] = E[ X×0] = 0 (c) Using the hint: v = (Y -ˆY )– h(X) = u  h(X), so that E(v2) = E[u2] + E[h(X)2] – 2×E[u×h(X)]. Using an argument like that in (b), E[u×h(X)] = 0. Thus, E(v2) = E(u2) + E[h(X)2], and the result follows by recognizing that E[h(X)2] ≥ 0 because h(x)2≥ 0 for any value of x.

  30. = Pr(M=3|A=0)Pr(A=0) + Pr(M=3|A=1)Pr(A=1) smtb98@gmail.com Stock/Watson - Introduction to Econometrics 4th Edition, Global Edition - Answers to Exercises: Chapter 2 _____________________________________________________________________________________________________ smtb98@gmail.com 29 2.28 (a) Pr(M = 3) = Pr(M=3,A=0) + Pr(M=3,A=1) = 0.05×0.5 + 0.01×0.5=0.03 (b) Pr(A 0| M  3)  Pr(M  3| A 0)Pr(A 0) P(M  3) 0.050.5 0.03 0.83 (c) Pr(M = 3) = Pr(M=3,A=0) + Pr(M=3,A=1) = Pr(M=3|A=0)Pr(A=0) + Pr(M=3|A=1)Pr(A=1) = 0.05×0.25 + 0.01×0.75=0.02 Pr(A 0| M  3)  Pr(M  3| A 0)Pr(A 0) P(M  3) 0.050.25 0.02 0.63

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