Introduction to Econometrics (4th Edition) Solutions to Odd-Numbered End-of-Chapter Exercises: Chapter 2*
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1 Itroductio to Ecoometrics (4th Editio by James H. Stock ad Mark W. Watso Solutios to Odd-Numbered Ed-of-Chapter Exercises: Chapter 2* (This versio September 14, 2018
2 (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 ³ 2 heads Probability (c µ Y = E(Y = ( ( ( = 1.00 Usig Key Cocept 2.3: var(y = E(Y 2 [E(Y ] 2, ad E(Y 2 = ( ( ( = 1.50 so that var(y = E(Y 2 [E(Y ] 2 = 1.50 ( = 0.50.
3 For the two ew radom variables W = 3+ 6X ad V = 20 7Y, we have: (a E(V = E(20 7Y = 20 7E(Y = = 14.54, E(W = E(3+ 6X = 3+ 6E( X = = 7.2. (b σ 2 W = var(3+ 6X = 6 2 σ 2 X = = 7.56, σ 2 V = var(20 7Y = ( 7 2 σ 2 Y = = (c σ WV = cov(3+ 6X, 20 7Y = 6( 7cov(X, Y = = 3.52 corr(w, V = σ WV σ W σ V = =
4 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, µx = 70 o F implies that µ Y = (5/9 70 = C, ad sx = 7 o F implies σ Y = (5/9 7 = 3.89 C.
5 Usig obvious otatio, C = M + F; thus µ C = µ M + µ F ad σ C 2 = σ M 2 + σ F 2 + 2cov( M, F. This implies (a µ C = = $85,000 per year. (b corr( M, F = cov( M, F so that Thus σ M σ F, cov( M, F = σ M σ F corr( M, F. cov( M, F = = , where the uits are squared thousads of dollars per year. (c σ 2 C = σ 2 M + σ 2 F + 2cov( M, F, so that σ 2 C = = , ad σ C = = thousad dollars per year. (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 = 0.85 Euros per Dollar or 1/e = 1.18 Dollars per Euro; each 1 Eollar is therefore with e Euros. The mea is therefore e µc (i uits of thousads of euros per year, ad the stadard deviatio is e sc (i uits of thousads of euros per year. The correlatio is uit-free, ad is uchaged.
6 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 ] 2 = σ Y = (b The coditioal probability of Y X = 8 is give i the table below Value of Y / / / / /0.39 E(Y X = 8 = 14 (0.02/ (0.03/ (0.15/ (0.10/ (0.09/0.39 = E(Y 2 X = 8 = 14 2 (0.02/ (0.03/ (0.15/ (0.10/ (0.09/0.39 = var(y = = σ Y X =8 = (c E( XY = ( (1 22 :0.05 +!( = cov( X,Y = E( XY E( X E(Y = = 11.0 corr( X,Y = cov( X,Y/(σ X σ Y = 11.0 / ( = 0.286
7 (a 0.90 (b 0.05 (c 0.05 (d Whe Y ~ χ 2, 10 the Y /10 ~ F. 10, (e Y = Z 2, where Z ~ N(0,1, thus Pr(Y 1 = Pr( 1 Z 1 = 0.32.
8 (a E(Y 2 = Var(Y + µ Y 2 = 1+ 0 = 1; E(W 2 = Var(W + µ W 2 = = 100. (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 3 = E(Y µ Y 4 ; solvig yields a σ E(Y 4 = 3; 4 Y similar calculatio yields the results for W. (d First, coditio o X = 0, so that S = W: E(S X = 0 = 0; E(S 2 X = 0 = 100, E(S 3 X = 0 = 0, E(S 4 X = 0 = Similarly, E(S X = 1 = 0; E(S 2 X = 1 = 1, E(S 3 X = 1 = 0, E(S 4 X = 1 = 3. From the large of iterated expectatios E(S = E(S X = 0 Pr(X = 0 + E(S X = 1 Pr( X = 1 = 0 E(S 2 = E(S 2 X = 0 Pr(X = 0 + E(S 2 X = 1 Pr( X = 1 = = 1.99 E(S 3 = E(S 3 X = 0 Pr(X = 0 + E(S 3 X = 1 Pr( X = 1 = 0 E(S 4 = E(S 4 X = 0 Pr(X = 0 + E(S 4 X = 1 Pr( X = 1 = = (e µ S = E(S = 0, thus E(S µ S 3 = E(S 3 = 0 from part (d. Thus skewess = 0. Similarly, σ 2 S = E(S µ S 2 = E(S 2 = 1.99, ad E(S µ S 4 = E(S 4 = Thus, kurtosis = / ( = 76.5
9 (a where Z ~ N(0, 1. Thus, Pr(9.6 Y 10.4 = Pr Y = Pr Z (i = 20; Pr Z = Pr( 0.89 Z 0.89 = 0.63 (ii = 100; Pr Z = Pr( 2.00 Z 2.00 = (iii = 1000; Pr Z = Pr( 6.32 Z 6.32 = (b Pr(10 c Y 10 + c = Pr = Pr c Y 10 c c Z c. c As get large gets large, ad the probability coverges to 1. 4 / (c This follows from (b ad the defiitio of covergece i probability give i Key Cocept 2.6.
10 µy = 0.4 ad σ 2 Y = = 0.24 (a (i P( Y ³ 0.43 = Pr Y = Pr Y = 0.27 (ii P( Y 0.37 = Pr Y = Pr Y = 0.11 b We kow Pr( 1.96 Z 1.96 = 0.95, thus we wat to satisfy 0.41= > ad < Solvig these iequalities yields ³ 0.2
11 (a (b (c Whe X ad Y are idepedet, so k l Pr(Y = y j = Pr( X = x i, Y = y j l = Pr(Y=y X=x Pr( X=x j i i k E (Y = y j Pr(Y = y j = y j Pr(Y = y j X = x i Pr( X = x i j=1 j=1 = l k j=1 l y j Pr(Y = y j X = x i = E(Y X=x ipr( X=x i. l Pr( X =x i Pr(X = x i, Y = y j = Pr(X = x i Pr(Y = y j, σ XY = E[( X µ X (Y µ Y ] = = l l k (x i µ X ( y j µ Y Pr( X=x i, Y=y j j=1 k (x i µ X ( y j µ Y Pr( X=x i Pr(Y=y j j=1 l = (x i µ X Pr( X = x i k j=1 = E( X µ X E(Y µ Y = 0 0 = 0, cor (X, Y = σ XY 0 = = 0. σ X σ Y σ X σ Y ( y j µ Y Pr(Y = y j
12 (a E( X µ 3 = E[( X µ 2 ( X µ] = E[X 3 2X 2 µ + X µ 2 X 2 µ + 2X µ 2 µ 3 ] = E( X 3 3E( X 2 µ + 3E( X µ 2 µ 3 = E( X 3 3E( X 2 E( X + 3E( X [E( X ] 2 [E( X ] 3 = E( X 3 3E( X 2 E( X + 2E( X 3 (b E( X µ 4 = E[( X 3 3X 2 µ + 3X µ 2 µ 3 ( X µ] = E[X 4 3X 3 µ + 3X 2 µ 2 X µ 3 X 3 µ + 3X 2 µ 2 3X µ 3 + µ 4 ] = E( X 4 4E( X 3 E( X + 6E( X 2 E( X 2 4E( X E( X 3 + E( X 4 = E( X 4 4[E( X ][E( X 3 ]+ 6[E( X ] 2 [E( X 2 ] 3[E( X ] 4
13 X ad Z are two idepedetly distributed stadard ormal radom variables, so µ X = µ Z = 0,σ 2 X = σ 2 Z = 1,σ XZ = 0. (a Because of the idepedece betwee X ad Z, Pr(Z = z X = x = Pr(Z = z, ad E(Z X = E(Z = 0. Thus E(Y X = E( X 2 + Z X = E( X 2 X + E(Z X = X = X 2. (b E( X 2 = σ 2 X + µ 2 X = 1, ad µ Y = E( X 2 + Z = E( X 2 + µ Z = 1+ 0 = 1. (c E( XY = E( X 3 + ZX = E( X 3 + E(ZX. Usig the fact that the odd momets of a stadard ormal radom variable are all zero, we have E( X 3 = 0. Usig the idepedece betwee X ad Z, we have E(ZX = µ µ = 0. Thus Z X E( XY = E( X 3 + E(ZX = 0. (d cov(xy = E[( X µ X (Y µ Y ] = E[( X 0(Y 1] = E( XY X = E( XY E( X = 0 0 = 0. corr(x, Y = σ XY 0 = = 0. σ X σ Y σ X σ Y
14 (a ax i = (ax 1 + ax 2 + ax 3 +!+ ax = a(x 1 + x 2 + x 3 +!+ x = a x i (b (x i + y i = (x 1 + y 1 + x 2 + y 2 +!x + y = (x 1 + x 2 +!x + ( y 1 + y 2 +! y = x i + y i (c a = (a + a + a +!+ a = a (d (a + bx i + cy i 2 = (a 2 + b 2 x 2 i + c 2 y 2 i + 2abx i + 2acy i + 2bcx i y i = a 2 + b 2 2 x i + c 2 2 y i + 2ab x i + 2ac y i + 2bc x i y i
15 (a E(u = E[E(u X ] = E[E(Y Ŷ 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 Usig the hit: v = (Y - Ŷ h(x = u - h(x, so that E(v 2 = E[u 2 ] + E[h(X 2 ] 2 E[u h(x]. Usig a argumet like that i (b, E[u h(x] = 0. Thus, E(v 2 = E(u 2 + E[h(X 2 ], ad the result follows by recogizig that E[h(X 2 ] 0 because h(x 2 0 for ay value of x.
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