Solutions to Odd Numbered End of Chapter Exercises: Chapter 2
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1 Itroductio to Ecoometrics (3 rd Updated Editio) by James H. Stock ad Mark W. Watso Solutios to Odd Numbered Ed of Chapter Exercises: Chapter 2 (This versio July 20, 2014)
2 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) Probability distributio fuctio for Y Outcome (umber of heads) Y = 0 Y = 1 Y = 2 Probability (b) Cumulative probability distributio fuctio for Y Outcome (umber of Y < 0 0 Y < 1 1 Y < 2 Y 2 heads) Probability (c) = EY ( ) (00.25) (10.50) (20.25) 1.00 Y Usig Key Cocept 2.3: 2 2 var( Y) E( Y ) [ E( Y)], ad EY ( ) (0 0.25) (1 0.50) (2 0.25) 1.50 so that var( Y) E( Y ) [ E( Y)] 1.50 (1.00) 0.50.
3 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter For the two ew radom variables W 3 6X ad V 20 7 Y, we have: (a) EV ( ) E(207 Y) 207 EY ( ) , EW ( ) E(36 X) 36 E( X) (b) 2 W var(3 6X ) X , 2 V var(20 7Y) (7) 2 2 Y (c) WV cov (3 6X,207Y) 6(7)cov(X, Y) corr(w, V ) WV W V
4 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Let X deote temperature i F ad Y deote temperature i C. Recall that Y = 0 whe X = 32 ad Y =100 whe X = 212. This implies Y (100/180) ( X 32) or Y (5/9) X. Usig Key Cocept 2.3, X = 70 o F implies that Y (5/9) C, ad X = 7 o F implies Y (5/9) C.
5 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Usig obvious otatio, C M F; thus C M F ad C M F 2cov( M, F). This implies (a) C $85,000per year. (b) corr( M, F) cov( M, F ) M F, so that cov ( M, F) M Fcorr ( M, F). Thus cov ( M, F) , where the uits are squared thousads of dollars per year. (c) C M F 2cov( M, F), so that C , ad thousad dollars per year. C (d) First you eed to look up the curret Euro/dollar exchage rate i the Wall Street Joural, the Federal Reserve web page, or other fiacial data outlet. Suppose that this exchage rate is e (say e = 0.75 Euros per Dollar or 1/e = 1.33 Dollars per Euro); each 1 Eollar is therefore with e Euros. The mea is therefore e C (i uits of thousads of euros per year), ad the stadard deviatio is e C (i uits of thousads of euros per year). The correlatio is uit-free, ad is uchaged.
6 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Value of Y Probability Distributio of X Value of X Probability distributio of Y (a) The probability distributio is give i the table above. E(Y ) E(Y 2 ) var(y ) E(Y 2 ) [E(Y)] Y (b) The coditioal probability of Y X = 8 is give i the table below Value of Y / / / / /0.39 EYX ( 8) 14 (0.02/0.39) 22 (0.03/0.39) 30 (0.15/0.39) 40 (0.10/0.39) 65 (0.09/0.39) ( 8) 14 (0.02/0.39) 22 (0.03/0.39) 30 (0.15/0.39) EY X (0.10/0.39) 65 (0.09/0.39) var( Y ) YX 8 (c) EXY ( ) ( ) (122 : 0.05) ( ) cov( XY, ) EXY ( ) EXEY ( ) ( ) corr( XY, ) cov( XY, )/( ) 11.0 / ( ) X Y
7 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) 0.90 (b) 0.05 (c) 0.05 (d) Whe Y ~, the Y/10 ~ 10, F (e) Y Z 2, where Z ~ N(0,1), thus Pr ( Y 1) Pr ( 1 Z 1) 0.32.
8 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) EY Y EW W ( ) Var ( ) Y 10 1; ( ) Var ( ) W (b) Y ad W are symmetric aroud 0, thus skewess is equal to 0; because their mea is zero, this meas that the third momet is zero. (c) The kurtosis of the ormal is 3, so 4 EY ( ) ; solvig yields Y 3 4 Y similar calculatio yields the results for W. 4 E( Y ) 3; a (d) First, coditio o X 0, so that S W: ESX ( 0) 0; ES ( X0) 100, ES ( X0) 0, ES ( X0) Similarly, ESX ES X ES X ES X ( 1) 0; ( 1) 1, ( 1) 0, ( 1) 3. From the large of iterated expectatios ES ( ) ESX ( 0) Pr (X 0) ESX ( 1) Pr( X1) 0 ES ES X ES X X ( ) ( 0) Pr (X 0) ( 1) Pr( 1) ES ES X ES X X ( ) ( 0) Pr (X 0) ( 1) Pr( 1) 0 ES ES X ES X X ( ) ( 0) Pr (X 0) ( 1) Pr( 1) (e) ES ( ) 0, thus S ES from part (d). Thus skewess = ( S ) ES ( ) 0 Similarly, ES ( ) ES ( ) 1.99, ad S S 2 Thus, kurtosis / (1.99 ) 76.5 ES 4 4 ( S ) ES ( )
9 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) Y Pr (9.6 Y 10.4) Pr 4/ 4/ 4/ Pr Z 4/ 4/ where Z ~ N(0, 1). Thus, (i) = 20; Pr Z Pr ( 0.89 Z 0.89) / 4/ (ii) = 100; Pr Z Pr( 2.00 Z 2.00) / 4/ (iii) = 1000; Pr Z Pr( 6.32 Z 6.32) / 4/ (b) c Y 10 c Pr (10 c Y 10 c) Pr 4/ 4/ 4/ c c Pr Z. 4/ 4/ As get large c 4/ gets large, ad the probability coverges to 1. (c) This follows from (b) ad the defiitio of covergece i probability give i Key Cocept 2.6.
10 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter Y = 0.4 ad Y (a) (i) P(Y 0.43) = Y Y 0.4 Pr Pr / 0.24/ 0.24/ (ii) P(Y 0.37) = Y Y 0.4 Pr Pr / 0.24/ 0.24/ b) We kow Pr( 1.96 Z 1.96) = 0.95, thus we wat to satisfy ad / Solvig these iequalities yields / 9220.
11 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) (b) l Pr ( Y y ) Pr ( X x, Y y ) j i j i1 l Pr ( Yy Xx)Pr ( Xx) i1 j i i E ( Y) y Pr( Y y ) y Pr( Y y X x )Pr( X x ) j j j j i i j1 j1 i1 l k j j i i1 j1 y Pr ( Y y X x ) Pr ( X x) l E( Y Xx)Pr( Xx) i1 k k l (c) Whe X ad Y are idepedet, so i i Pr( X x, Y y ) Pr( X x )Pr( Y y ) E[( X )( Y )] XY X Y l k i1 j1 i1 j1 i j i j ( x )( y )Pr( Xx, Yy ) l k i X j Y i j ( x )( y )Pr( Xx)Pr( Yy ) i X j Y i j l k ( xi X) Pr ( X xi) ( yj Y) Pr ( Y yj i1 j1 EX ( ) EY ( ) 000, X Y XY 0 cor ( X, Y) 0 X Y X Y i
12 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) EX E X X EX X X X X ( ) [( ) ( )] [ 2 2 ] E( X ) 3 E( X ) 3 E( X) E( X ) 3 E( X ) E( X) 3 E( X)[ E( X)] [ E( X)] E( X ) 3 E( X ) E( X) 2 E( X) (b) EX E X X X X ( ) [( 3 3 )( )] EX X X X X X X [ ] EX ( ) 4 EX ( ) EX ( ) 6 EX ( ) EX ( ) 4 EXEX ( ) ( ) EX ( ) EX ( ) 4[ EX ( )][ EX ( )] 6[ EX ( )] [ EX ( )] 3[ EX ( )]
13 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter X ad Z are two idepedetly distributed stadard ormal radom variables, so 2 2 X Z 0, X Z 1, XZ 0. (a) Because of the idepedece betwee X ad Z, Pr ( Z zx x) Pr ( Z z), ad EZX ( ) EZ ( ) 0. Thus E( Y X) E( X Z X) E( X X) E( Z X) X 0 X (b) E X ad ( ) X X 1, EX Z EX Y 2 2 ( ) ( ) Z (c) 3 3 E( XY) E( X ZX) E( X ) E( ZX). Usig the fact that the odd momets of 3 a stadard ormal radom variable are all zero, we have EX ( ) 0. Usig the idepedece betwee X ad Z, we have EZX ( ) 0. Thus EXY EX EZX 3 ( ) ( ) ( ) 0. Z X (d) cov ( XY ) E[( X )( Y )] E[( X 0)( Y 1)] EXY ( X) EXY ( ) EX ( ) 000 XY 0 corr ( X, Y) 0 X X Y X Y Y
14 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) ax ( ax1ax2 ax3ax ) a( x1x 2 x3 x ) a x (b) i i i1 i1 i1 ( x y ) ( x y x y x y ) i i ( x x x ) ( y y y ) x i i1 i1 y i (c) a( aaa a) a i1 (d) ( a bxi cyi) ( a b xi c yi 2abxi 2acyi 2 bcxiyi) i1 i i i 2 i 2 i 2 i i i1 i1 i1 i1 i1 a b x c y ab x ac y bc x y
15 Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter (a) E(W) = E[E(W Z) ] = E[E(X ) Z] = E[ E(X Z) E(X Z) ] = 0. (b) E(WZ) = E[E(WZ Z) ] = E[ZE(W) Z] = E[ Z 0] = 0 (c) Usig the hit: V = W h(z), so that E(V 2 ) = E(W 2 ) + E[h(Z) 2 ] 2 E[W h(z)]. Usig a argumet like that i (b), E[W h(z)] = 0. Thus, E(V 2 ) = E(W 2 ) + E[h(Z) 2 ], ad the result follows by recogizig that E[h(Z) 2 ] 0 because h(z) 2 0 for ay value of z.
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