Solutions (mostly for odd-numbered exercises)

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1 Solutons (mostly for odd-numbered exercses) c 005 A. Coln Cameron and Pravn K. Trved "Mcroeconometrcs: Methods and Alcatons" 1. Chater 1: Introducton o exercses.. Chater : Causal and oncausal Models o exercses. 3. Chater 3: Mcroeconomc Data Structures o exercses. 4. Chater 4: Lnear Models 4-1 (a) For the dagonal entres = j and E[u ] =. For the rst o -dagonal = j 1 or = j + 1 so j jj = 1 and E[u u j ] =. Otherwse j jj > 1 and E[u u j ] = 0. (b) b OLS s asymtotcally normal wth mean 0 and asymtotc varance matrx V[ b OLS ] = ( 0 ) 1 0 ( 0 ) 1 ; where = : 7 5 (c) Ths examle s a smle dearture from the smlest case of = I. 1

2 Here deends on just two arameters and hence can be consstently estmated as! 1. So we use bv[ b OLS ] = ( 0 ) 1 0 b ( 0 ) 1 ; where b = 6 4 b bb 0 0 bb bb 0 0 bb b and b! f b! and d! or b!. For =E[u ] the obvous estmate s b = 1 P =1 bu, where bu = y x 0 b. For we can drectly use =E[u u 1 ] consstently estmated by d = 1 P = bu bu 1. Or use =E[u u 1 ]= E[u ]E[u 1 ] =E[u u 1 ]=E[u ] consstently estmated by b = 1 P = bu bu 1 = 1 P =1 bu and hence bb = 1 P = bu bu 1. (d) To answer (d) and (e) t s helful to use summaton notaton: " # 1 " bv[ b # " OLS ] = x x 0 b x x 0 + bb x x 0 1 x x 0 =1 =1 = =1 # 1 " # 1 " # 1 " # " = b x x 0 + bb x x 0 x x 0 1 x x 0 =1 (d) o. The usual OLS outut estmate b ( 0 ) 1 s nconsstent as t gnores the o -dagonal terms and hence the second term above. (e) o. The Whte heteroskedastcty-robust estmate s nconsstent as t also gnores the o -dagonal terms and hence the second term above. 4-3 (a) The error u s condtonally heteroskedastc, snce V[ujx] =V[x"jx] = x V["jx] = x V["] = x 1 = x whch deends on the regressor x. (b) For scalar regressor 1 0 = 1 P x. Here x are d wth mean 1 (snce E[x ] =E[(x E[x ]) ] =V[x ] = 1 usng E[x ] = 0). Alyng a LL (here Kolmogorov), 1 0 = 1 P x! E[x ] = 1, so M xx = 1: (c) V[u] =V[x"] =E[(x") ] (E[x"]) =E[x ]E[" ] (E[x]E["]) =V[x]V["] 0 0 = 1 1 = 1 where use ndeendence of x and " and fact that here E[x] = 0 and E["] = 0: =1 = =1 # 1

3 (d) For scalar regressor and dagonal, 1 0 = 1 x = 1 =1 x x = 1 usng = x from (a). Here x 4 are d wth mean 3 (snce E[x 4 ] =E[(x E[x ]) 4 ] = 3 usng E[x ] = 0 and the fact that fourth central moment of normal s 3 4 = 3 1 = 3). Alyng a LL (here Kolmogorov), 1 0 = 1 P x4! E[x 4 ] = 3, so M xx = 3. (e) Default OLS result ( b OLS ) d! 0; M 1 xx = 0; 1 (1) 1 = [0;1] : =1 =1 x 4 (f) Whte OLS result ( b OLS ) d! 0; M 1 xxm xx M 1 xx = 0; (1) 1 3 (1) 1 = [0;3]: (g) Yes. Exect that falure to control for condtonal heteroskedastcty when should control for t wll lead to nconsstent standard errors, though a ror the drecton of the nconsstency s not known. That s the case here. What s unusual comared to many alcatons s that there s a bg d erence n ths examle - the true varance s three tmes the default estmate and the true standard errrors are 3tmes larger. 4-5 (a) D erentate Set Wu by chan rule for matrx d = 0 Wu assumng W s symmetrc = 0 Wu 0 Wu = 0 ) 0 W(y )= 0 ) 0 Wy = 0 W ) = b ( 0 W) 1 0 Wy

4 where need to assume the nverse exsts. Here W s rank r > K =rank() ) 0 Z and Z 0 Z are rank K ) 0 Z(Z 0 Z) 1 Z 0 s of full rank K. (b) For W = I we have b =( 0 I) 1 0 Iy = ( 0 ) 1 0 y whch s OLS. ote that ( 0 ) 1 exsts f K matrx s of full rank K. (c) For W = 1 we have b =( 0 1 ) y whch s GLS (see (4.8)). (d) For W = Z(Z 0 Z) 1 Z 0 SLS (see (4.53)). we have b =( 0 Z(Z 0 Z) 1 Z 0 ) 1 0 Z(Z 0 Z) 1 Z 0 y whch s 4-7 Gven the nformaton, E[x] = 0 and E[z] = 0 and V[x] = E[x ] = E[(u + ") ] = u + " V[z] = E[z ] = E[(" + v) ] = " + v Cov[x; z] = E[xz] = E[(u + ")(" + v)] = " Cov[x; u] = E[xu] = E[(u + ")u] = u (a) For regresson of y on x we have b OLS = P 1 P x x y and as usual lm( b OLS ) = lm P 1 P x lm x u = E[x ] 1 E[xu] as here data are d = ( u + ") 1 u: (b) The squared correlaton coe cent s Z = [Cov[x; z] ]=[V[x]V[z]] = [ "] =[( u + ")( " + v)] (c) For sngle regressor and sngle nstrument b IV = ( P z x ) 1 P z y = ( P z x ) 1 P z (x + u ) = + ( P z x ) 1 P z u = + ( P z (u + " )) 1 P z u = + ( P z u + P z " )) 1 P z u = + (m zu + m z" ) 1 m zu b IV = m zu = (m zu + m z" )

5 where m zu = 1 P z u and m z" = 1 P z ". By a LL m zu! E[mzu ] = E[z u ] = E[(" +v )u ] = 0 snce ", u and v are ndeendent wth zero means. By a LL m z"! E[mz" ] = E[z " ] = E[(" + v )" ] = E[" ] = ": b IV! 0= 0 + ") = 0: (d) If m zu = m z" = then m zu = m z" so m zu m z" = 0 and b IV = m zu =0 whch s not de ned. (e) Frst b IV = m zu = (m zu + m z" ) = 1=( + m z" =m zu ): If m zu s large relatve to m z" = then s large relatve to m z" =m zu so + m z" =m zu s close to and 1=( + m z" =m zu ) s close to 1=: (f) Gven the de nton of Z n art (c), Z s smaller the smaller s, the smaller s ", and the larger s. So n the weak nstruments case wth small correlaton between x and z (ao Z s small), b IV s lkely to converge to 1= rather than 0, and there s nte samle bas n b IV (a) The true varance matrx of OLS s V[ b OLS ] = ( 0 ) 1 0 ( 0 ) 1 = ( 0 ) 1 0 (I +AA 0 )( 0 ) 1 = ( 0 ) 1 + ( 0 ) 1 0 AA 0 ( 0 ) 1 : (b) Ths equals or exceeds ( 0 ) 1 snce ( 0 ) 1 0 AA 0 ( 0 ) 1 s ostve semde nte. So the default OLS varance matrx, and hence standard errors, wll generally understate the true standard errors (the exceton beng f 0 AA 0 = 0). (c) For GLS V[ b GLS ] = ( 0 1 ) 1 = ( 0 [ (I + AA 0 )] = ( 0 [I + AA 0 ] 1 ) 1 1 ) 1 = ( 0 [I A(I m + A 0 A) 1 A 0 ]) 1 = ( 0 0 A(I m + A 0 A) 1 A 0 ) 1 :

6 (d) ( 0 ) 1 V[ b GLS ] snce A(I m + A 0 A) 1 A 0 n the matrx sense ) ( 0 ) 1 ( 0 0 A(I m + A 0 A) 1 A 0 ) 1 n the matrx sense. If we ran OLS and GLS and used the ncorrect default OLS standard errors we would obtan the uzzlng result that OLS was more e cent than GLS. But ths s just an artfact of usng the wrong estmated standard errors for OLS. (e) GLS requres ( 0 1 ) 1 whch from (c) requres (I m + A 0 A) 1 whch s the nverse of an m m matrx. [We also need ( 0 0 A(I m + A 0 A) 1 A 0 ) 1 but ths s a smaller k k marx gven k < m <.] 4-13 (a) Here = [1 1] and = [1 0]: From bottom of age 86 the ntercet wll be F" 1 (q) = F F 1 " (q): The sloe wll be + F" 1 (q) = F 1 The sloe should be 1 at all quantles. " (q) = 1: " (q) = The ntercet vares wth F 1 " (q). Here F 1 " (q) takes values :56, 1:68, 1:05, 0:51, 0:0, 0:51, 1:05, 1:68 and :56 for q = 0:1, 0:,..., 0:9. It follows that the ntercet takes values 1:56, 0:68, 0:05, 0:49, 1:0, 1:51, :05, :68. [For examle F 1 " (0:9) s " such that Pr[" " ] = 0:9 for " [0; 4] or equvalently " such that Pr[z " =] = 0:9 for z [0; 1]. Then " = = 1:8 so " = :56.] (b) The answers accord qute closely wth theory as the sloe and ntercets are qute recsely estmated wth sloe coe cent standard errors less than 0:01 and ntercet coe cent standard errors less than 0:04. (c) ow both the ntercet and sloe coe cents vary wth the quantle. Both ntercet and sloe coe cents ncrease wth the quantle, and for 1 = 0:5 are wthn two standard errors of the true values of 1 and 1. (d) Comared to (b) t s now the ntercet that s constant and the sloe that vares across quantles. Ths s redcted from theory smlar to that n art (a). ow = [1 1] and = [0 1]. From bottom of age 86 the ntercet wll be F" 1 (q) = F" 1 (q) = 1 and the sloe wll be + F" 1 (q) = F" 1 (q) = 1 + F" 1 (q): 4-15 (a) The OLS sloe estmate and standard error are 0:0509 and 0:0091, and the IV estmates are 0:18806 and 0:0614. The IV sloe estmate s much larger and

7 ndcates a very large return to schoolng. There s a lossn recson wth IV standard error ten tmes larger, but the coe cent s stll statststcally sgn cant. (b) OLS of wage76 on an ntercet and col4 gves sloe coe cent 0: and OLS regresson of grade76 on an ntercet and col4 gves sloe coe cent 0: From (4.46), dy=dx = (dy=dz)=(dx=dz) = 0: =0:89019 = 0: Ths s the same as the IV estmate n art (a). (c) We obtan Wald = ( ) / ( ) = Ths s the same as the IV estmate n art (a). (d) From OLS regresson of grade76 on col4, R = 0:008 and F = 60:37. Ths does not suggest a weak nstruments roblem, excet that recson of IV wll be much lower than that of OLS due to the relatvely low R. (e) Includng the addtonal regressors the OLS sloe estmate and standard error are 0:03304 and 0:00311, and the IV estmates are 0:0951 and 0:0493. The IV sloe estmate s agan much larger and ndcates a very large return to schoolng. There s a loss n recson wth IV standard error now eghteed ten tmes larger, but the coe cent s stll statststcally sgn cant usng a one-tal test at ve ercent. ow OLS of wage76 on an ntercet and col4 and other regressors gves sloe coe cent 0: and OLS regresson of grade76 on an ntercet and col4 gves sloe coe cent 0: From (4.46), dy=dx = (dy=dz)=(dx=dz) = 0: =0:89019 = 0: Ths s the same as the IV estmate n art (a) (a) The average of b OLS over 1000 smulatons was Ths s close to the theoretcal value of 1:5: lm( b OLS ) = u= u + " = (1 1)=( ) = 1= and here = 1. (b) The average of b IV over 1000 smulatons was Ths s close to the theoretcal value of 1: lm( b IV ) = 0 and here = 1. (c) The observed values of b IV over 1000 smulatons were skewed to the rght of = 1, wth lower quartle , medan and uer quartle Exercse 4-7 art (e) suggested concentraton of b IV around 1= = 1 or concetraton of b IV around + 1 = snce here = 1: (d) The R and F statstcs across smulatons from OLS regresson (wth ntercet) of z on x do ndcate a lkely weak nstruments roblem. Over 1000 smulatons, the average R was and the average F was [Asde: From Exercse 4-7 (b) Z = [ "] =[( u + ")( " + v) = [0:01] =(1 + 1)(0:01 + 1) = 0:00005:]

8 5. Chater 5: Extremum, ML, LS 5-1 Frst @x ex(1 + 0:01x)[1 + ex(1 + 0:01x)] 1 1 = 0:01 ex(1 + 0:01x)[1 + ex(1 + 0:01x)] ex(1 + 0:01x) 0:01 ex(1 + 0:01x)[1 + ex(1 + 0:01x)] ex(1 + 0:01x) = 0:01 uon sml caton [1 + ex(1 + 0:01x)] (a) The average margnal e ect over = :01 =1 ex(1 + 0:01) 1 + ex(1 + 0:01) = 0:001498: (b) The samle mean x = 1 P =1 = = 0:01 x ex(1 + 0:01 50:5) [1 + ex(1 + 0:01x 50:5)] = 0: : (c) Evaluatng at x = ex(1 + 0:01 90) = [1 + ex(1 + 0:01x 90)] = 0: : 90 (d) Usng the nte d erence method be[yjx] ex(1 + 0:01 90) = x 1 + ex(1 + 0:01x 90) 90 ex(1 + 0:01 90) 1 + ex(1 + 0:01x 90) = 0:001176: Comment: Ths examle s qute lnear, leadng to answers n (a) and (b) beng close, and smlarly for (c) and (d). A more nonlnear functon, wth greater varaton s obtaned usng be[yjx] = ex(0 + 0:04x)=[1 + ex(0 + 0:04x)] for x = 1; :::; 100. Then the answers are 0:006163, 0: , 0:000068, and 0:

9 5- (a) Here so ln f(y) = ln y ln y= wth = ex(x 0 )= and ln = x 0 ln Q ()= 1 = ln y (x 0 ln ) y=[ex(x 0 )=] = ln y x 0 + ln y ex( x 0 ) ln f(y ) = 1 fln y x 0 + ln y ex( x 0 )g: (b) ow usng x nonstochastc so need only take exectatons wrt y Q 0 () = lm Q () = lm 1 ln y lm 1 x0 + lm 1 ln lm 1 y ex( x 0 ) = lm 1 E[ln y ] lm 1 x0 + ln lm 1 E[y ] ex( x 0 ) = lm 1 E[ln y ] lm 1 x0 + ln lm 1 ex(x0 0 ) ex( x 0 ); where the last lne uses E[y ] = ex(x 0 0) n the dg and we do not need to evaluate E[ln y ] as the rst sum does not nvlove and wll therefore have dervatve of 0 wrt. (c) D erentate wrt (not 0 0 () = lm x + lm ex(x0 0 ) ex( = 0 when = 0 : x 0 )x Q 0 ()=@@ 0 = lm 1 P ex(x0 0) ex( x 0 )x x 0 s negatve de nte at 0, so local max.] Snce lm Q () attans a local maxmum at = 0, conclude that b = arg max Q () s consstent for 0. (d) Consder the last term. Snce y ex( x 0 ) s not d need to use Markov SLL. Ths requres exstence of second moments of y whch we have assumed. 5-3 (a) D erentatng Q () = 1 x + y ex( x )x = 1 fy ex( x 0 ) 1gx rearrangng = 1 y ex(x 0 ) ex(x 0 ) x multlyng by ex(x0 ) ex(x 0 )

10 (b) Then lm 0 # = lm 1 y ex(x 0 0) ex(x 0 x = 0 f E[y jx ] = ex(x 0 0 ): 0) So essental condton s correct sec caton of E[y jx ]. (c) = 1 0 y ex(x 0 0) ex(x 0 0) Aly CLT to average of the term n the sum. ow y jx has mean ex(x 0 0) and varance (ex(x 0 0)) =. So y ex(x 0 0 ) ex(x 0 0 ) x has mean 0 and varance 4 (ex(x0 0 )) = (ex(x 0 x x 0 0 )) = x x 0. Thus for Z = (V[ ]) 1= ( E[ ]) = 1 P V[ ] 1= ( 1 P ) Z = 1 1= 1 x x 0 ) 1 y ex(x 0 0) ex(x 0 0) x x : y ex(x 0 0) d ex(x 0 x! [0; I] 0) d! 0; lm 1 x x 0 (d) Here y s not d. Use Laounov CLT. Ths wll need a ( + ) th absolute moment of y. e.g. 4 th moment of y. (e) D erentatng (a) wrt 0 0 = 1 0 ex(x0 0) ex(x 0 0) x x 0! lm 1 x x 0 : (f) Combnng ( b 0 )! d [0; A( 0 ) 1 B( 0 )A( 0 ) 1 ] " d! 0; lm 1 1 x x 0 lm 1 x x 0 lm 1 " d! 0; lm 1 # 1 x x 0 : x x 0 1 #

11 (g) Test H 0 : 0j j aganst H a : 0j < j at level :05. b a ) z j = (b j j ) s j 1 ; x x 0 a [0; 1], where sj s j th dag entry n x x 0 1 : Reject H 0 at level 0:05 f z j < z :05 = 1: (a) t = b 1 =se[ b 1 ] = 5= = :5. Snce j:5j > _z :05 = 1:645 we reject H 0. (b) Rewrte as H 0 : 1 = 0 versus H 0 : 1 6= 0. Use (5.3). Test H 0 : R = r where R = [1 ] and r = 0 and 0 = [ 1 ]. Here b 5 = so R b 5 r = [1 ] = 1: Also V[ ] b = 1 C b 4 1 = usng Cov[ 1 1 b 1 ; b ] = (Cor[ b 1 ; b ]) V[ b 1 ]V[ b ] = 0:5 1 = 1. Then R 1 CR b = [1 ] = so W = (R b r) 0 R( 1 b C)R 0 1 (R b r) = Snce W = 0:5 < 1;:05 = 3:84 do not reject H 0: [Alternatvely as only one restrcton here, note that b 1 b has varance V[ b 1 ] + 4V[ b 1 ] 4Cov[ b 1 ; b ] = = 4, leadng to t = b 1 b se[ b 1 b ] = = 0:5 and do not reject as j0:5j < z :05 = 1:96. ote that t =W.] 1 0 (c) Use (5.3) Test H 0 : R = r where R = and r = 0 1 Then R b r = = 0 1 and R 1 CR b = = so W = (R b 1 r) R( 0 1 C)R b 0 (R b 4 1 r) = Snce W = 14 < ;:05 = 5:99 reject H 0: 0 0 and = 5 =

12 5-7 Results wll vary as uses generated data. Exect b 1 ' 1 and b ' 1 and standard errors smlar to those below. (a) For LS got b 1 = 1:116 and b = 1:1098 wth standard errors 0:0551 and 0:056. (b) Yes, wll need to use sandwch errors due to heteroskedastcty as V[yjx] = ex( 1 + x) =. ote that standard errors gven n (a) do not correct for heteroskedastcty. (c) For MLE got b 1 = 1:0088 and b = 1:06 wth standard errors 0:04 and 0:015. (d) Sandwch errors can be used but are not necessary snce the ML sml caton that A = B s arorate here.