PART ONE Solutions to Exercises 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 = = = = = 8 Stock/Watson - Introduction to Econometrics - Second Edition (c) 4050505250 Pr(4052)Pr 555 (0 4)( 2)(0 4)1(2) 0 6554 10 97720 6326 Y Y = = . = . = . + .= (d) 65585 Pr(68)Pr 222 (2 1213)(0 7071) 0 98310 76020 2229 Y Y = = . = . .= 11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When 2 10 ,Y then 10, /10 .YF (e) 2, YZ= where N(0,1),Z thus Pr(1)Pr( 11)0.32.YZ= = 12. (a) 0.05 (b) 0.950 (c) 0.953 (d) The tdf distribution and N(0, 1) are approximately the same when df is large. (e) 0.10 (f) 0.01 13. (a) 2222 ()Var( )101;()Var()1000100. YW E YYE WW=+= +=+=+= (b) Y and W are symmetric around 0, thus skewness is equal to 0; because their mean is zero, this means that the third moment is zero. (c) The kurtosis of the normal is 3, so 4 $ () 3; Y Y E Y = solving yields 4 E()3;Y= a similar calculation yields the results for W. (d) First, condition on 0,X = so that :SW= 2342 . ( |0)0;(|0)100,(|0)0,(|0)3 100E SXE S XE S XE S X= Similarly, 234 ( |1)0;(|1)1,(|1)0,(|1)3.E SXE S XE S XE S X= From the large of iterated expectations ( )( |0)Pr(X0)( |1)Pr(1)0E SE S XE S XX=+= 222 ()(|0) Pr(X0)(|1) Pr(1)1000.01 1 0.991.99E SE S XE S XX=+=+ = 333 ()(|0)Pr(X0)(|1)Pr(1)0E SE S XE S XX=+= 4442 ()(|0) Pr(X0)(|1) Pr(1)3 1000.013 1 0.99302.97E SE S XE S XX=+= + = Solutions to Exercises in Chapter 2 9 (e) ( )0, S E S= thus 33 ()()0 S E SE S=from part d. Thus skewness = 0. Similarly, 222 ()()1.99, SS E SE S= and 44 ()()302.97. S E SE S= Thus, 2 kurtosis302.97/(1.99 )76.5= 14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average ( )Y is approximately 2 , YY N with 2 2 . Y nY = Given 100, Y = 2 43 0, Y =. (a) 100,n = 2 2 43 100 0 43, Y nY = . and 100101 100 Pr(101)Pr(1 525)0 9364 0 430 43 Y Y = .= (b) 165,n = 2 2 43 165 0 2606, Y nY = . and 10098 100 Pr(98)1Pr(98)1Pr 0 26060 2606 1( 3 9178)(3 9178)1 000 (rounded to four decimal places) Y YY = = .= .= (c) 64,n = 2 2 43 6464 0 6719, Y Y = . and 101 100100103 100 Pr(101103)Pr 0 67190 67190 6719 (3 6599)(1 2200)0 99990 88880 1111 Y Y = . .= . .= 15. (a) 9.6 101010.4 10 Pr(9.610.4)Pr 4/4/4/ 9.6 1010.4 10 Pr 4/4/ Y Y nnn Z nn = = where Z N(0, 1). Thus, (i) n = 20; 9.6 1010.4 10 PrPr( 0.890.89)0.63 4/4/ ZZ nn = (ii) n = 100; 9.6 1010.4 10 PrPr( 2.002.00)0.954 4/4/ ZZ nn = (iii) n = 1000; 9.6 1010.4 10 PrPr( 6.326.32)1.000 4/4/ ZZ nn = 10 Stock/Watson - Introduction to Econometrics - Second Edition (b) 10 Pr(1010)Pr 4/4/4/ Pr. 4/4/ cYc cYc nnn cc Z nn += = As n get large 4/ c n gets large, and the probability converges to 1. (c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6. 16. There are several ways to do this. Here is one way. Generate n draws of Y, Y1, Y2, Yn. Let Xi = 1 if Yi and 0.39 0.4 0.24/ 1.96. n = , , = . . .= .= Solutions to Exercises in Chapter 2 11 19. (a) 1 1 Pr()Pr(,) Pr(|)Pr() l jij i l jii i YyXx Yy Y y X xX x = = = = (b) 111 1 1 1 ( )Pr()Pr(|)Pr() Pr(|) Pr() ( |)Pr() i ii kkl jjjjii jji k l jji j i l i E YyYyyYy XxXx yYy XxX x E Y X xX x = = = = = = = =. (c) When X and Y are independent, Pr(,)Pr()Pr() ijij Xx YyXxYy=, so 11 11 11 ()() ()()Pr(,) ()()Pr()Pr() ()Pr()()Pr( () ()000, XYXY lk iXjYij ij lk iXjYij ij lk iXijYj ij XY E XY xyX x Y y xyX xY y xXxyYy E XE Y = = = = = = = = 0 ( ,)0 XY XYXY cor X Y = . 20. (a) 11 Pr()Pr(|,)Pr(,) lm iijhjh jh YyYy XxZzXxZz = = (b) 1 111 111 11 ( )Pr()Pr() Pr(|,)Pr(,) Pr(|,) Pr(,) ( |,)Pr(,) k iii i klm iijhjh ijh lmk iijhjh jhi lm jhjh jh E YyYyYy yYy XxZzXxZz yYy XxZzXxZz E Y XxZzXxZz = = = = = = = = 12 Stock/Watson - Introduction to Econometrics - Second Edition 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. 21. (a) 32322223 32233223 323 ()() ()22 ()3 ()3 ()()3 () ()3 () () () ()3 () ()2 () E XE XXE XXXXX E XE XE XE XE XE XE XE XE X E XE XE XE X =+ =+=+ =+ (b) 43223 4322332234 432234 43224 ()(33)() 3333 ()4 () ()6 () ()4 () ()() ()4 () ()6 () ()3 () E XE XXXX E XXXXXXX E XE XE XE XE XE X E XE X E XE XE XE XE XE X