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 August 7, 204)
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 2.. (a) Probability distributio fuctio for Outcome (umber of heads) = 0 = = 2 Probability 0.25 0.50 0.25 (b) Cumulative probability distributio fuctio for Outcome (umber of < 0 0 < < 2 2 heads) Probability 0 0.25 0.75.0 (c) µ = E ( ) = (0 0.25) + ( 0.50) + (2 0.25) =.00 Usig Key Cocept 2.3: 2 2 var( ) = E( ) [ E( )], ad 2 2 2 2 E ( ) = (0 0.25) + ( 0.50) + (2 0.25) =.50 so that 2 2 2 var( ) = E( ) [ E( )] =.50 (.00) = 0.50.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 2 2.3. For the two ew radom variables W = 3+ 6X ad V = 20 7, we have: (a) EV ( ) = E(20 7 ) = 20 7 E ( ) = 20 7 078. = 454,. EW ( ) = E(3+ 6 X) = 3+ 6 E( X) = 3+ 6 070. = 72.. (b) σ 2 W = var(3+ 6X ) = 6 2 σ 2 X = 36 0.2= 7.56, σ 2 V = var(20 7 ) = ( 7) 2 σ 2 = 49 0.76 = 8.4084. (c) σ WV = cov(3+ 6X, 20 7 ) = 6( 7)cov(X, ) = 42 0.084 = 3.52 corr(w, V ) = σ WV σ W σ V = 3.528 7.56 8.4084 = 0.4425.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 3 2.5. Let X deote temperature i F ad deote temperature i C. Recall that = 0 whe X = 32 ad =00 whe X = 22. This implies = (00/80) ( X 32) or = 7.78 + (5/9) X. Usig Key Cocept 2.3, µ X = 70 o F implies that µ = 7.78 + (5/9) 70 = 2. C, ad σ X = 7 o F implies σ = (5/9) 7 = 3.89 C.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 4 2.7. Usig obvious otatio, C = M + F; thus µ C = µ M + µ ad F 2 2 2 σc = σm + σf + 2cov( M, F). This implies (a) µ C = 40 + 45 = $85,000per year. (b) corr( M, F) = cov( M, F ) σ M σ F, so that cov( M, F) = σ σ corr ( M, F). Thus cov( M, F ) = 2 8 0.80 = 72.80, where the uits are squared thousads of dollars per year. M F (c) 2 2 2 2 2 2 σc = σm + σf + 2cov( M, F), so that σ C = 2 + 8 + 2 72.80 = 83.60, ad σ = 83.60 = 28.524 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 /e =.33 Dollars per Euro); each Eollar is therefore with e Euros. The mea is therefore e µ C (i uits of thousads of euros per year), ad the stadard deviatio is e σ C (i uits of thousads of euros per year). The correlatio is uit-free, ad is uchaged.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 5 2.9. Value of Probability 4 22 30 40 65 Distributio of X Value of X 0.02 0.05 0.0 0.03 0.0 0.2 5 0.7 0.5 0.05 0.02 0.0 0.40 Probability distributio of 8 0.02 0.03 0.5 0.0 0.09 0.39 0.2 0.23 0.30 0.5 0..00 (a) The probability distributio is give i the table above. E( ) = 4 0.2+ 22 0.23+ 30 0.30 + 40 0.5+ 65 0.= 30.5 E( 2 ) = 4 2 0.2+ 22 2 0.23+ 30 2 0.30 + 40 2 0.5+ 65 2 0.= 27.23 var( ) = E( 2 ) [E( )] 2 = 28.2 σ = 4.77 (b) The coditioal probability of X = 8 is give i the table below Value of 4 22 30 40 65 0.02/0.39 0.03/0.39 0.5/0.39 0.0/0.39 0.09/0.39 EX= ( 8) = 4 (0.02/0.39) + 22 (0.03/0.39) + 30 (0.5/0.39) + 40 (0.0/0.39) + 65 (0.09/0.39) = 39.2 E X= = + + 2 2 2 2 ( 8) 4 (0.02/0.39) 22 (0.03/0.39) 30 (0.5/0.39) + + = 2 2 40 (0.0/0.39) 65 (0.09/0.39) 778.7 2 var( ) 778.7 39.2 24.65 σ = = 5.54 = = X 8 (c) EX ( ) = ( 4 0.02) + ( 22:0.05) + L (8 65 0.09) = 7.7 cov( X, ) = EX ( ) EXE ( ) ( ) = 7.7 5.33 30.5 =.0 corr( X, ) = cov( X, )/( σ σ ) =.0 / (2.60 4.77) = 0.286 X
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 6 2.. (a) 0.90 (b) 0.05 (c) 0.05 (d) Whe ~ χ, the /0 ~ F0,. 2 0 (e) = Z 2, where Z ~ N(0,), thus Pr ( ) = Pr ( Z ) = 0.32.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 7 2.3. (a) E EW W 2 2 2 2 ( ) = Var( ) + µ = + 0= ; ( ) = Var( ) + µ W = 00+ 0= 00. (b) 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 E ( µ ) = ; solvig yields 3 4 σ similar calculatio yields the results for W. 4 E( ) = 3; a (d) First, coditio o X = 0, so that S = W: ESX ( = 0) = 0; ES ( X= 0) = 00, ES ( X= 0) = 0, ES ( X= 0) = 3 00 Similarly, 2 3 4 2. ESX ES X ES X ES X 2 3 4 ( = ) = 0; ( = ) =, ( = ) = 0, ( = ) = 3. From the large of iterated expectatios ES ( ) = ESX ( = 0) Pr(X= 0) + ESX ( = ) Pr( X= ) = 0 ES ES X ES X X 2 2 2 ( ) = ( = 0) Pr(X= 0) + ( = ) Pr( = ) = 00 0.0+ 0.99=.99 ES ES X ES X X 3 3 3 ( ) = ( = 0) Pr(X= 0) + ( = ) Pr( = ) = 0 ES ES X ES X X 4 4 4 ( ) = ( = 0) Pr (X = 0) + ( = ) Pr( = ) = + = 2 3 00 0.0 3 0.99 302.97 (e) µ = ES ( ) = 0, thus S ES = = from part (d). Thus skewess = 0. 3 3 ( µ S ) ES ( ) 0 2 2 2 Similarly, σ = ES ( µ ) = ES ( ) =.99, ad S 2 Thus, kurtosis = 302.97 / (.99 ) = 76.5 S ES 4 4 ( µ S ) = ES ( ) = 302.97.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 8 2.5. (a) 9.6 0 0 0.4 0 Pr (9.6 0.4) = Pr 4/ 4/ 4/ 9.6 0 0.4 0 = Pr Z 4/ 4/ where Z ~ N(0, ). Thus, (i) = 20; 9.6 0 0.4 0 Pr Z = Pr ( 0.89 Z 0.89) = 0.63 4/ 4/ (ii) = 00; 9.6 0 0.4 0 Pr Z = Pr( 2.00 Z 2.00) = 0.954 4/ 4/ (iii) = 000; 9.6 0 0.4 0 Pr Z = Pr( 6.32 Z 6.32) =.000 4/ 4/ (b) c 0 c Pr (0 c 0 + c) = Pr 4/ 4/ 4/ c c = Pr Z. 4/ 4/ As get large c 4/ gets large, ad the probability coverges to. (c) This follows from (b) ad the defiitio of covergece i probability give i Key Cocept 2.6.
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 9 2 2.7. µ = 0.4 ad σ = 0.4 0.6 = 0.24 (a) (i) P( 0.43) = 0.4 0.43 0.4 0.4 Pr = Pr 0.624 = 0.27 0.24/ 0.24/ 0.24/ (ii) P( 0.37) = 0.4 0.37 0.4 0.4 Pr = Pr.22 = 0. 0.24/ 0.24/ 0.24/ b) We kow Pr(.96 Z.96) = 0.95, thus we wat to satisfy 0.4 = >.96 ad 9220. 0.4 0.4 0.24/ <.96. Solvig these iequalities yields 0.39 0.4 0.24/
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 0 2.9. (a) (b) l Pr ( = y ) = Pr ( X = x, = y ) j i j i= l = Pr ( = y X= x)pr ( X= x) i= j i i E ( ) = ypr( = y) = y Pr( = yx = x)pr( X= x) j j j j i i j= j= i= l k j j i i= j= = y Pr ( = y X = x ) Pr ( X = x) l = EX ( = x)pr( X= x). i= k k l (c) Whe X ad are idepedet, so i i Pr ( X = x, = y ) = Pr ( X = x )Pr ( = y ), σ = E[( X µ )( µ )] X X l k i= j= i= j= i j i j = ( x µ )( y µ )Pr( X= x, = y ) l k i X j i j = ( x µ )( y µ )Pr( X= x)pr( = y ) i X j i j l k = ( xi µ X)Pr( X = xi) ( yj µ )Pr( = yj i= j= = EX ( µ ) E ( µ ) = 0 0= 0, X X 0 cor ( X, ) = σ 0 σ σ = σ σ =. X X i
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 2. 2. (a) EX E X X EX X X X X 3 2 3 2 2 2 2 3 ( µ ) = [( µ ) ( µ )] = [ 2 µ + µ µ + 2 µ µ ] = + = + 3 2 2 3 3 2 2 EX ( ) 3 EX ( ) µ 3 EX ( ) µ µ EX ( ) 3 EX ( ) EX ( ) 3 EX ( )[ EX ( )] [ EX ( = EX ( ) 3 EX ( ) EX ( ) + 2 EX ( ) 3 2 3 (b) EX E X X X X 4 3 2 2 3 ( µ ) = [( 3 µ + 3 µ µ )( µ )] = EX X + X X X + X X + 4 3 2 2 3 3 2 2 3 4 [ 3 µ 3 µ µ µ 3 µ 3 µ µ ] = EX ( ) 4 EX ( ) EX ( ) + 6 EX ( ) EX ( ) 4 EXEX ( ) ( ) + EX ( ) 4 3 2 2 3 4 = EX ( ) 4[ EX ( )][ EX ( )] + 6[ EX ( )] [ EX ( )] 3[ EX ( )] 4 3 2 2 4
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 2 2. 23. X ad Z are two idepedetly distributed stadard ormal radom variables, so µ µ σ σ σ 2 2 X = Z = 0, X = Z =, XZ = 0. (a) Because of the idepedece betwee X ad Z, Pr ( Z = zx = x) = Pr ( Z = z), ad EZX ( ) = EZ ( ) = 0. Thus EX ( ) E( X ZX ) E( X X) EZX ( ) X 0 X 2 2 2 2 = + = + = + =. (b) E X = σ + µ = ad 2 2 2 ( ) X X, µ = EX + Z = EX + µ = + =. 2 2 ( ) ( ) Z 0 (c) 3 3 EX ( ) = EX ( + ZX) = EX ( ) + EZX ( ). 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 ( ) = µµ = 0. Thus EX EX EZX 3 ( ) = ( ) + ( ) = 0. Z X (d) cov( X ) = E[( X µ )( µ )] = E[( X 0)( )] = E( X X ) = E( X ) E( X ) = 0 0= 0. σ X 0 corr ( X, ) = = = 0. σ σ σ σ X X X
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 3 2.25. (a) axi = ( ax+ ax2 + ax3+ L + ax) = a( x+ x2 + x3+ L + x) = a xi (b) i= i= i= ( x + y ) = ( x + y + x + y + L x + y ) i i 2 2 = ( x + x + L x ) + ( y + y + L y ) 2 2 = x + i i= i= y i (c) a= ( a+ a+ a+ L + a) = a i= (d) 2 2 2 2 2 2 ( a + bxi + cyi) = ( a + b xi + c yi + 2abxi + 2acyi + 2 bcxiyi) i= i= 2 2 2 2 2 i i 2 i 2 i 2 i i i= i= i= i= i= = a + b x + c y + ab x + ac y + bc x y
Stock/Watso - Itroductio to Ecoometrics - 3 rd Updated Editio - Aswers to Exercises: Chapter 2 4 2.27 (a) E(W) = E[E(W Z) ] = E[E(X! 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.