PART ONE. Solutions to Exercises

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1 PART ONE Soutions to Exercises

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3 Chapter Review of Probabiity Soutions to Exercises. (a) Probabiity distribution function for Outcome = 0 = = (number of heads) probabiity Cumuative probabiity distribution function for Outcome < 0 0 < < (number of heads) Probabiity (c) µ = E ( ) = (0 0.5) + ( 0.50) + ( 0.5) =.00 Using Key Concept.3: var( ) E( ) [ E( )], = and E ( ) = (0 0.5) + ( 0.50) + ( 0.5) =.50 so that var( ) = E( ) [ E( )] =.50 (.00) = We know from Tabe. that Pr( = 0) = 0., Pr( = ) = 0. 78, Pr( = 0) = 0. 30, Pr( = ) = So (a) µ = E ( ) = 0 Pr( = 0) + Pr( = ) = = 0. 78, µ = E ( ) = 0 Pr( = 0) + Pr( = ) = = = E[( µ ) ] = = + = (0 0.70) Pr ( 0) ( 0.70) Pr ( ) = ( 0. 70) = 0., = E[( µ ) ] = = + = (0 0.78) Pr ( 0) ( 0.78) Pr ( ) = ( 0. 78) =

4 4 Stock/Watson - Introduction to Econometrics - Second Edition (c) Tabe. shows Pr ( = 0, = 0) = 0. 5, Pr ( = 0, = ) = 0. 5, Pr ( =, = 0) = 0. 07, Pr ( =, = ) = So = cov(, ) = E[( µ )( µ )] = (0-0.70)(0-0.78)Pr( = 0, = 0) + ( )( 0. 78)Pr ( = 0, = ) + ( 0. 70)( )Pr ( =, = 0) + ( 0. 70)( 0. 78) Pr ( =, = ) =. ( 070). ( 078) ( 070) ( 078) = , cor (, ) = = = For the two new random variabes W = 3+ 6 and V = 0 7, we have: (a) (c) EV ( ) = E(0 7 ) = 0 7 E ( ) = = 4. 54, EW ( ) = E(3+ 6 ) = 3+ 6 E ( ) = = 7.. = var (3 + 6 ) = 6 = = 7. 56, = var (0 7 ) = ( 7) = = W V WV = cov (3 + 6, 0 7 ) = 6( 7) cov (, ) = = WV. 358 cor ( W, V ) = = = (a) E ( ) = 0 ( p) + p= p E ( k ) = 0 k ( p) + k p= p (c) E ( ) = 0.3 var( ) = E( ) [ E( )] = = 0. Thus, = 0. = W V To compute the skewness, use the formua from exercise.: E ( µ ) = E ( ) 3[ E ( )][ E ( )] + [ E ( )] 3 = = Aternativey, E ( µ ) = [( 0.3) 0.3] + [(0 0.3) 0.7] = Thus, skewness = E ( µ ) / =.084/0.46 = 0.87.

5 Soutions to Exercises in Chapter 5 To compute the kurtosis, use the formua from exercise.: E ( µ ) = E ( ) 4[ E ( )][ E ( )] + 6[ E ( )] [ E ( )] 3[ E ( )] 3 4 = = Aternativey, E ( µ ) = [( 0.3) 0.3] + [(0 0.3) 0.7] = Thus, kurtosis is E ( µ ) / =.0777/0.46 = Let denote temperature in F and denote temperature in C. Reca that = 0 when = 3 and = 00 when = ; this impies = (00/80) ( 3) or = (5/9). Using Key Concept.3, µ = 70 F impies that µ = (5/9) 70 =. C, and = 7F impies = (5/9) 7 = 3.89 C. 6. The tabe shows that Pr ( = 0, = 0) = , Pr ( = 0, = ) = , Pr ( =, = 0) = , Pr ( =, = ) = 0. 4, Pr ( = 0) = , Pr( = ) = 0. 46, Pr( = 0) = , Pr( = ) = (a) E ( ) = µ = 0 Pr( = 0) + Pr( = ) = = #(unempoyed) Unempoyment Rate = #(abor force) = Pr( = 0) = = = E( ). (c) Cacuate the conditiona probabiities first: Pr ( = 0, = 0) Pr ( = 0 = 0) = = = , Pr ( = 0) Pr ( = 0, = ) Pr ( = = 0) = = = , Pr ( = 0) Pr ( =, = 0) Pr ( = 0 = ) = = = , Pr ( = ) Pr ( =, = ) 0. 4 Pr ( = = ) = = = Pr ( = ) The conditiona expectations are E ( = ) = 0 Pr ( = 0 = ) + Pr ( = = ) = = , E ( = 0) = 0 Pr( = 0 = 0) + Pr( = = 0) = =

6 6 Stock/Watson - Introduction to Econometrics - Second Edition (d) Use the soution to part, Unempoyment rate for coege grads = E ( = ) = = Unempoyment rate for non-coege grads = E ( = 0) = = (e) The probabiity that a randomy seected worker who is reported being unempoyed is a coege graduate is Pr ( =, = 0) Pr ( = = 0) = = = 0.. Pr ( = 0) The probabiity that this worker is a non-coege graduate is Pr ( = 0 = 0) = Pr ( = = 0) = 0. = (f) Educationa achievement and empoyment status are not independent because they do not satisfy that, for a vaues of x and y, For exampe, Pr ( = y = x) = Pr ( = y). Pr ( = 0 = 0) = Pr ( = 0) = Using obvious notation, C = M + F ; thus µ C = µ M + µ F and C = M + F + cov( MF, ). This impies (a) µ = = $85,000 per year. (c) C Cov( M, F ) cor ( M, F) =, M so that Cov( M, F) = (, ). F MFcor M F Thus Cov( M, F ) = = 7.80, where the units are squared thousands of doars per year. = + + cov(, ), so that = = 83.60, and C M F MF C C = = 8.54 thousand doars per year. (d) First you need to ook up the current Euro/doar exchange rate in the Wa Street Journa, the Federa Reserve web page, or other financia data outet. Suppose that this exchange rate is e (say e = 0.80 euros per doar); each $ is therefore with e. The mean is therefore eµ 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 correation is unit-free, and is unchanged. 8. µ = E ( ) =, = var( ) = 4. With Z = ( ), µ Z = E ( ) ( ) ( ) 0, = µ = = Z = var ( ) 4 = = =. 4 4

7 Soutions to Exercises in Chapter 7 9. Vaue of Probabiity distribution of Vaue of Probabiity Distribution of (a) The probabiity distribution is given in the tabe above. E ( ) = = 30.5 E Var() = E ( ) [ E ( )] = 8. = 4.77 ( ) = = 7.3 Conditiona Probabiity of = 8 is given in the tabe beow Vaue of / / / / /0.39 E= ( 8) = 4 (0.0/0.39) + (0.03/0.39) + 30 (0.5/0.39) + 40 (0.0/0.39) + 65 (0.09/0.39) = 39. E ( = 8) = 4 (0.0/0.39) + (0.03/0.39) + 30 (0.5/0.39) + 40 (0.0/0.39) + 65 (0.09/0.39) = Var( ) = = 4.65 = = (c) E ( ) = ( 4 0.0) + ( : 0.05) + + ( ) = 7.7 Cov(, ) = E ( ) EE ( ) ( ) = =.0 Corr(, ) = Cov(, )/( ) =.0 /( ) = Using the fact that if N µ, then µ ~ N (0,) (a) and Appendix Tabe, we have 3 Pr( 3) = Pr =Φ () = Pr( > 0) = Pr( 0) = Pr = Φ( ) =Φ () =

8 8 Stock/Watson - Introduction to Econometrics - Second Edition (c) (d) Pr (40 5) = Pr =Φ (0. 4) Φ( ) =Φ (0. 4) [ Φ()] = = Pr (6 8) = Pr =Φ (. 3) Φ (0. 707) = = (a) (c) 0.05 (d) When ~ χ 0, then /0 ~ F0,. (e) = Z, where Z ~ N(0,), thus Pr( ) = Pr( Z ) = (a) (c) (d) The t df distribution and N(0, ) are approximatey the same when df is arge. (e) 0.0 (f) (a) E ( ) = Var ( ) + µ = + 0 = ; EW ( ) = Var ( W) + µ W = = 00. and W are symmetric around 0, thus skewness is equa to 0; because their mean is zero, this means that the third moment is zero. 4 E ( µ ) 4 (c) The kurtosis of the norma is 3, so 3 = $ ; soving yieds E( ) = 3; a simiar cacuation yieds the resuts for W. (d) First, condition on = 0, so that S= W: ES ( = 0) = 0; ES ( = 0) = 00, ES ( = 0) = 0, ES ( = 0) = 3 00 Simiary, 3 4. ES ES ES ES 3 4 ( = ) = 0; ( = ) =, ( = ) = 0, ( = ) = 3. From the arge of iterated expectations ES ( ) = ES ( = 0) Pr( = 0) + ES ( = ) Pr( = ) = 0 ES ES ES ( ) = ( = 0) Pr ( = 0) + ( = ) Pr( = ) = =.99 ES ES ES ( ) = ( = 0) Pr( = 0) + ( = ) Pr( = ) = 0 ES ES ES ( ) = ( = 0) Pr ( = 0) + ( = ) Pr( = ) = = 30.97

9 Soutions to Exercises in Chapter 9 (e) µ = ES ( ) = 0, thus S 3 3 ES ( µ S ) ES ( ) 0 S ES ( S) ES ( ).99, = = from part d. Thus skewness = 0. Simiary, = µ = = and ES Thus, kurtosis = /(.99 ) = ( µ S ) = ES ( ) = The centra imit theorem suggests that when the sampe size (n) is arge, the distribution of the sampe average ( ) is approximatey N µ, with =. Given µ = 00, = 43. 0, n (a) n = 00, 43 n = = = 043,. and Pr( 0) = Pr Φ (. 55) = n = 65, 43 n = = = , and Pr ( > 98) = Pr ( 98) = Pr Φ( ) =Φ (3. 978) =. 000 (rounded to four decima paces). (c) n = 64, 43 = = = 0679,. and Pr (0 03) = Pr Φ ( ) Φ (. 00) = = (a) Pr( ) = Pr 4/ n 4/ n 4/ n = Pr Z 4/ n 4/ n where Z ~ N(0, ). Thus, (i) n = 0; Pr Z = Pr ( 0.89 Z 0.89) = / n 4/ n (ii) n = 00; Pr Z = Pr(.00 Z.00) = / n 4/ n (iii) n = 000; Pr Z = Pr( 6.3 Z 6.3) =.000 4/ n 4/ n

10 0 Stock/Watson - Introduction to Econometrics - Second Edition As n get arge c 4/ n c 0 c Pr (0 c 0 + c) = Pr 4/ n 4/ n 4/ n c c = Pr Z. 4/ n 4/ n gets arge, and the probabiity converges to. (c) This foows from and the definition of convergence in probabiity given in Key Concept There are severa ways to do this. Here is one way. Generate n draws of,,, n. Let i = if i < 3.6, otherwise set i = 0. Notice that i is a Bernoui random variabes with µ = Pr( = ) = Pr( < 3.6). Compute. Because converges in probabiity to µ = Pr( = ) = Pr( < 3.6), wi be an accurate approximation if n is arge. 7. µ = 0.4 and = = (a) (i) P( 0.43) = Pr = Pr 0.64= / n 0.4/ n 0.4/ n (ii) P( 0.37) = Pr = Pr.= / n 0.4/ n 0.4/ n We know Pr(.96 Z.96) = 0.95, thus we want n to satisfy 0.4 = >.96 and <.96. Soving these inequaities yieds n / n 8. Pr( = $ 0) = 0. 95, Pr( = $ 0000) = (a) The mean of is µ = 0 Pr ( = $ 0) + 0, 000 Pr ( = $ 0000) = $ 000. The variance of is ( ) = E µ ( ) = (0 000) = + = = ( 000) =. 9 0, 0.4/ n Pr 0 ( ) Pr ( 0000) 7 $ 7 so the standard deviation of is = (. 9 0 ) = (i) E ( ) = µ = $ 000, = = = n 00 (ii) Using the centra imit theorem, Pr ( > 000) = Pr ( 000) 000, 000, 000 = Pr Φ (. 94) = =

11 Soutions to Exercises in Chapter 9. (a) Pr ( = y ) = Pr ( = x, = y ) j i j i= = Pr ( = y = x)pr ( = x) i= j i i E ( ) = ypr ( = y) = y Pr ( = y = x)pr ( = x) j j j j i i j= j= i= k j j i i= j= = y Pr ( = y = x ) Pr ( = x) = E ( = x)pr( = x). i= k k (c) When and are independent, so i i Pr( = x, = y ) = Pr( = x )Pr( = y ), = E[( µ )( µ )] k i= j= i= j= i j i j = ( x µ )( y µ )Pr( = x, = y ) k i j i j = ( x µ )( y µ )Pr( = x)pr( = y ) i j i j k = ( xi µ )Pr ( = xi) ( yj µ )Pr ( = yj i= j= = E ( µ ) E ( µ ) = 0 0= 0, 0 cor (, ) = 0 = =. i 0. (a) m Pr ( = y) = Pr ( = y = x, Z = z )Pr ( = x, Z = z ) i i j h j h j= h= k E ( ) = ypr( = y)pr( = y) i= = y Pr ( = y = x, Z = z )Pr ( = x, Z = z ) i i j h j h i= j= h= m k = yipr ( = yi = xj, Z = zh) Pr ( = xj, Z = zh) j= h= i= = E( = x, Z = z )Pr( = x, Z = z ) j= h= i i i k m m j h j h

12 Stock/Watson - Introduction to Econometrics - Second Edition. (a) where the first ine in the definition of the mean, the second uses (a), the third is a rearrangement, and the fina ine uses the definition of the conditiona expectation. E E E = E ( ) 3 E ( ) µ + 3 E ( ) µ µ = E ( ) 3 E ( ) E ( ) + 3 E ( )[ E ( )] [ E ( )] 3 3 = E ( ) 3 E ( ) E ( ) + E ( ) ( µ ) = [( µ ) ( µ )] = [ µ + µ µ + µ µ ] E ( µ ) = E[( 3 µ + 3 µ µ )( µ )] = E [ 3µ + 3µ µ µ + 3µ 3 µ + µ ] = E ( ) 4 E ( ) E ( ) + 6 E ( ) E ( ) 4 EE ( ) ( ) + E ( ) = E ( ) 4[ E ( )][ E ( )] + 6[ E ( )] [ E ( )] 3[ E ( )]. The mean and variance of R are given by µ = w ( w) 0.05 = w ( w) w ( w) [ ] where = Cov( Rs, Rb) foows from the definition of the correation between R s and R b. (a) µ = 0.065; = µ = 0.075; = (c) w = maximizes µ ; = 0.07 for this vaue of w. (d) The derivative of with respect to w is = w w + w d dw.07 ( ) 0.04 ( 4 ) [ ] = 0.00w soving for w yieds w = 8 /0 = 0.8. (Notice that the second derivative is positive, so that this is the goba minimum.) With w = 0.8, R = and Z are two independenty distributed standard norma random variabes, so µ µ = Z = 0, = Z =, Z = 0. (a) Because of the independence between and Z, Pr ( Z = z = x) = Pr ( Z = z), and EZ ( ) = EZ ( ) = 0. Thus E( ) = E( + Z ) = E( ) + E( Z ) = + 0=. E ( ) = + µ =, and µ = E ( + Z) = E ( ) + µ = + 0=. (c) 3 3 ( ) ( ) ( ) ( ). E = E + Z = E + E Z Using the fact that the odd moments of a standard norma 3 random variabe are a zero, we have E ( ) = 0. Using the independence between and Z, we 3 have EZ ( ) = µ µ = 0. Thus E ( ) = E ( ) + EZ ( ) = 0. Z Z

13 Soutions to Exercises in Chapter 3 (d) 4. (a) i Cov( ) = E[( µ )( µ )] = E[( 0)( )] = E ( ) = E ( ) E ( ) = 0 0= 0. 0 cor (, ) = = = 0. E ( ) = + µ = and the resut foows directy. n ( i /) is distributed i.i.d. N(0,), W = ( / ), i i and the resut foows from the definition of a = χ n random variabe. n n i i (c) E(W) = EW ( ) = E =. E = n i= i= (d) Write V = = / n i= i n i= ( / ) n n which foows from dividing the numerator and denominator by. / ~ N(0,), χn, and / and the t distribution. n ( / ) i i ~ = n i i are independent. The resut then foows from the definition of = ( / )