UNIVERSITY OF CAMBRIDGE FACULTY OF ECONOMICS M.Phil. in Economics M.Phil. in Economic Research Subject M300 Econometric Methods Suggested solutions to Exercise Sheet 2 1. Consider the textbook example of IV regression ln Qcigarettes = i 0 + 1 ln Picigarettes + ui with instrument Z1i = SalesT axi : Here ln Qcigarettes is the logarithm of i the number of packs of cigarettes sold per capita in state i; ln Picigarettes is the logarithm of the average real price per pack of cigarettes in state i including all taxes, SalesT axi is the portion of the tax on cigarettes arising from the general sales tax in state i: (a) The results of the …rst and the second stage regressions of the 2SLS method of estimating 1 are: d ln Picigarettes d ln Qcigarettes i = = 4:63 + 0:031 SalesTaxi (0:03) 9:72 (0:005) d 1:08ln Picigarettes ; where the standard errors of the estimates in the …rst stage regression 2SLS are given below the coe¢ cient estimates. What is the value of ^ 1 ? 2SLS Suggested solution to (a) ^ 1 = 1:08 (b) Check whether we have a weak instrument problem in the 2SLS regressions from a). Suggested solution to (b) F-statistic from the …rst stage regression equals the square of t-statistic because there is only one instrument. 2 Therefore, F = 0:031 = 38:44 > 10: It does not seem that there is 0:005 a weak instrument problem. (c) Given the 2SLS results from a), what is the value of the J statistic for testing the instrument’s exogeneity? Suggested solution to (c) The value of J statistic is zero because we have a just-identi…ed case here. 1 (d) Discuss why SalesTaxi is a good or bad instrument for ln Picigarettes from the point of view of the two requirements for the instrument validity. Suggested answer to (d) Answers here can vary. Students should discuss both instrument relevance and exogeneity. The relevance is more or less obvious. Sales taxes vary from state to state and passed on to consumers, so that they represent a part of the price. The exogeneity is somewhat more tricky. In this particular example, income is an omitted variable, and it may well be correlated with sales taxes violating exogeneity. 2. This problem is the empirical exercise E12.2 from the Stock and Watson’s textbook. We reproduce it here for convenience. How does fertility a¤ect labour supply? That is, how much does a woman’s labour supply fall when she has an additional child? In this exercise you will estimate this e¤ect using data for married women from the 1980 U.S. Census. The data are available on the textbook Web site www.pearsonhighered.com/stock_watson in the …le Fertility and described in the …le Fertility_Description (from the page with the above address, you will need to go to the Companion Website for the 3-rd edition of the textbook. From there, you’ll go to Student Resources/Data for Empirical Exercises). The data set contains information on married women aged 21-35 with two or more children. (a) Regress weeksworked on the indicator variable morekids using OLS. On average, do women with more than two children work less than women with two children? How much less? Suggested solutions to (a) As can be seen from Table 1, the coe¢ cient is 5:387; which indicates that women with more than 2 children work 5.387 fewer weeks per year than women with 2 or fewer children. (b) Explain why the OLS regression estimated in (a) is inappropriate for estimating the causal e¤ect of fertility (morekids) on labour supply (weeksworked). Suggested solution to (b) Both fertility and weeks worked are choice variables. A woman with a positive labor supply regression error (a Estimation method OLS IV 5:387 6:313 Regressor Morekids (0:087) Additional regressors IV 5:821 (1:275) const const First stage F 1238.2 (1:246) const; agem1; black; hispan; othrace 1280.9 Table 1: Results using full dataset 2 woman who works more than average) may also be a woman who is less likely to have an additional child. This would imply that Morekids is negatively correlated with the regression error, so that the OLS estimator of M orekids is negatively biased. (c) The data set contains the variable samesex, which is equal to 1 if the …rst two children are of the same sex (boy-boy or girl-girl) and equal to 0 otherwise. Are couples whose …rst two children are of the same sex more likely to have a third child? Is the e¤ect large? Is it statistically signi…cant? Suggested solution to (c) The linear regression of morekids on samesex (a linear probability model) yields d M orekids = 0:346 + 0:066 (0:001) (0:002) samesex so that couples with samesex=1 are 6.6% more likely to have an additional child that couples with samesex=0. The e¤ect is highly signi…cant (t-statistic=35.2). (d) Explain why samesex is a valid instrument for the instrumental variable regression of weeksworked on morekids. Suggested solution to (d) Samesex is random and is unrelated to any of the other variables in the model including the error term in the labor supply equation. Thus, the instrument is exogenous. From (c), the …rst stage F-statistic is large (F=1238) so the instrument is relevant. Together, these imply that samesex is a valid instrument (e) Is samesex a weak instrument? Suggested solution to (e) No, see the answer to (d). (f ) Estimate the regression of weeksworked on morekids using samesex as an instrument. How large is the fertility e¤ect on labour supply? Suggested solution to (f ) See column (2) of Table 1. The estimated value of M orekids is 6:313. This is somewhat surprising because we expected a negative bias of OLS, and IV is supposed to correct for this. (g) Do the results change when you include the variables agem1, black, hispan, and othrace in the labour supply regression (treating these variables as exogenous)? Explain why or why not. Suggested solution to (g) See column (3) of Table 1. The results do not change in an important way. The reason is that samesex is unrelated to agem1, black, hispan, othrace, so that there is no omitted variable bias in IV regression in column (2). 3. Consider a regression yi = 0 3 + 1 xi + ui with one endogenous regressor xi . Let zi be a valid instrument for xi : Assume that (yi ; xi ; zi ) form an i.i.d. sequence. Denote the …tted value ^ 0 + ^ 1 zi from the …rst stage regression xi = 0 + 1 zi + ei as x ^i : (a) Denote the vector (1; x ^i )0 as Vi , and let n ^ 2SLS = 1X Vi Vi0 n i=1 + =( ! 1 0; 1) 0 : Show that n 1X Vi ui n i=1 ! : Suggested solution to (a) Since ^ 2SLS is an OLS estimator from the second stage regression, we have n ^ 2SLS 1X Vi Vi0 n i=1 = ! 1 n 1X Vi yi n i=1 ! : On the other hand, yi = 0 + 1 xi + ui = Vi0 + ui + Using this in the above formula, we get n ^ 2SLS = + 1X Vi Vi0 n i=1 ! 1 1 (xi x ^i ) : ! n 1X Vi (ui + n i=1 1 (xi x ^i )) : Pn It remains to prove that n1 i=1 Vi (xi x ^i ) = 0: But this is true because xi x ^i is the residual from the …rst stage regression, and Vi is the vector of regressors from the …rst stagePregression. Residuals n ^i ) = 0: are always orthogonal to regressors, hence, n1 i=1 Vi (xi x (b) Let V~i = (1; 0 Vi Vi0 = V~i V~i0 + + 1 zi ) ^0 0 : Show that 0 0 + (^ 1 1 ) zi ^0 0 + (^ 1 2 (^ 0 + ^ 1 zi ) ( 1 ) zi 0 + 1 zi ) 2 Using the Slutsky theorem, prove that n 1X p Vi Vi0 ! E V~i V~i0 : n i=1 Suggested solution to (b) Note that Vi = V~i + ^0 4 0 + (^ 0 1 1 ) zi : (1) : Therefore, Vi Vi0 = V~i V~i0 + V~i (0; ^ 0 + + 0 + (^ 1 1 ) zi ) ^0 0 + (^ 0 1 1 ) zi ^0 0 + (^ 0 1 1 ) zi V~i0 (0; ^ 0 0 + (^ 1 1 ) zi ) : Opening up the brackets, we get Vi Vi0 = V~i V~i0 + 0 + (^ 0 1 ^0 ^0 0 + (^ 1 2 (^ 0 + ^ 1 zi ) ( 1 ) zi 1 ) zi 0+ : 2 1 zi ) (2) Now, n n 1X (^ 0 n i=1 p since ^ 0 ! theorem 0 0; + (^ 1 p ^1 ! 1; 1 ) zi ) and 1 n n 1X (^ 0 n i=1 Further, 1 Xh 2 (^ 0 + ^ 1 zi ) n i=1 = (^ 0 Pn 0 i=1 zi 2 0+ 1 zi ) i + (^ 1 1X zi n i=1 1) p ! Ezi ; we have, by Slutsky’s + (^ 1 n ( 0) 1 ) zi ) p ! 0: (3) n ^ 20 + 2^ 0 ^ 1 = n 1X 2 1X zi + ^ 21 z n i=1 n i=1 i n 2 0 2 ^ 20 = 0 1 2 0 n 1X zi n i=1 + (2^ 0 ^ 1 2 1 1X 2 z n i=1 i n 2 0 1) n + ^ 21 2 1 p 1X 2 z : n i=1 i p By Slutsky’s theorem, ^ 20 20 ! 0; 2^ 0 ^ 1 2 0 1 ! 0; and ^ 21 0: Therefore, using Slutsky once again, we conclude that 1 Xh 2 (^ 0 + ^ 1 zi ) n i=1 n Finally, by LLN, ( 0 + 2 1 zi ) i p ! 0: 2 p 1 ! (4) n 1 X ~ ~0 p Vi Vi ! E V~i V~i0 : n i=1 5 (5) 1X zi n i=1 Combining (2, 3, 4, and 5), we obtain n 1X p Vi Vi0 ! E V~i V~i0 : n i=1 (c) Show that Vi ui = V~i ui + 0 ) u + (^ 1 0 i (^ 0 1 ) zi ui : Using the Slutsky theorem, prove that n 1 X d p Vi ui ! N 0; E u2i V~i V~i0 n i=1 : Suggested solution to (c) Equality Vi ui = V~i ui + (^ 0 0 ) ui 0 + (^ 1 1 ) zi ui follows from (1). Next, n 1 X p (^ 0 n i=1 = (^ 0 0) 0 ) ui + (^ 1 1 ) zi ui n n 1 X p ui + (^ 1 n i=1 1) p 1 X p zi ui : n i=1 p On the other hand, ^ 0 0 ! 0; ^ 1 1 ! 0; whereas, by CLT, Pn Pn d d 1 p1 p u ! N (0; V ar (u )) and i i=1 i i=1 zi ui ! N (0; V ar (zi ui )) : n n By the Slutsky theorem, n (^ 0 0) 1 X p ui + (^ 1 n i=1 n 1) 1 X p p zi ui ! 0; n i=1 and Vi ui converges in distribution to the same limit as V~i ui . Specifically, n 1 X d p Vi ui ! N 0; E u2i V~i V~i0 n i=1 (d) Using your results in (a), (b), and (c), show that p n ^ 2SLS d !N h i 0; E V~i V~i0 6 1 E u2i V~i V~i0 h i E V~i V~i0 1 Suggested solution to (d) From (a), p n n ^ 2SLS = 1X Vi Vi0 n i=1 ! 1 n 1 X p V i ui n i=1 ! : From (b), n 1X p Vi Vi0 ! E V~i V~i0 : n i=1 From (c), n 1 X d p Vi ui ! N 0; E u2i V~i V~i0 n i=1 : Combining this (using the Slutsky theorem and the continuous mapping theorem), we get p d n ^ 2SLS !N h i 0; E V~i V~i0 1 E u2i V~i V~i0 h i E V~i V~i0 1 : 4. Consider a general IV regression model Yi = 0 + 1 X1i + ::: + k Xki + k+1 W1i + ::: + k+r Wri + ui with instruments Z1i ; :::; Zmi . ^ 1i ; :::; X ^ ki be …tted values from the …rst stage regressions, let u (a) Let X ~i be the residuals from the second stage regression, and let 2SLS ^ 2SLS X1i ::: ^ 2SLS Xki ^ 2SLS W1i ::: ^ 2SLS Wri : u ^i = Yi ^ 0 1 k k+1 k+r Show that u ^i = u ~i ^ 2SLS X1i 1 ^ 1i X ::: ^ 2SLS Xki k ^ ki X Suggested solution to (a) By de…nition, 2SLS ^ 2SLS X ^ 2SLS Wri : ^ 1i ::: ^ 2SLS ^ ki ^ 2SLS u ~i = Yi ^ 0 X 1 k k+1 W1i ::: k+r It follows that u ^i = u ~i ^ 2SLS X1i 1 ^ 1i X ::: ^ 2SLS Xki k ^ ki : X (b) Show that, for any j = 1; :::; k; the OLS estimates of all the coe¢ cients ^ 1i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri in the regression of X1i X equal zero. 7 ^ 1i are residuals from the …rst stage Suggested solution to (b) X1i X regression. Z1i ; :::; Zmi ; W1i ; :::; Wri are explanatory variables in the …rst stage regression. OLS residuals are always orthogonal to the corresponding regressors. (c) Show that the OLS estimates of all the coe¢ cients in the regression ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri equal zero. of u ~i on constant and X Suggested solution to (c) u ~i is the OLS residual from a regression of ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri : Since OLS residuals are Yi on constant and X orthogonal to regressors, all the coe¢ cients in the regression of u ~i on ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri equal zero. constant and X 0 (d) Let Z be an n m matrix with i-th row (Z1i ; :::; Zmi ) and let X 0 be an n k matrix with i-th row (X1i ; :::; Xki ) : Assume that the number of instruments m equals the number of the endogenous regressors k; and that both Z 0 Z and Z 0 X are invertible matrices. Argue that the …tted value from the regression of u ~i on constant and ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri must be the same as the …tted value from the X regression of u ~i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri : Using this and the result from (c), show that the OLS estimates of all the coe¢ cients in the regression of u ~i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri equal zero. Suggested solution to (d) Let W be the n r matrix with i-th row 0 0 ^ the n k matrix with i-th row X ^ 1i ; :::; X ^ ki ; (W1i ; :::; Wri ) ; X u ~ = (~ u1 ; :::; u ~n ) ; and 1 = (1; :::; 1) (an n-dimensional vector of ones). Further, let ^ ; ^ ; and ^ be the vectors of the OLS coe¢ cient estimates in the regression of u ~i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri ; and let ~ ; ~ ; and ~ be the OLS coe¢ cient estimates in the regression of ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri That is, u ~i on constant and X n o 0 ^ ; ^ ; ^ = arg min (~ u 1 Z W ) (~ u 1 Z W ) ; ; ; (6) and n o ~ ; ~ ; ~ = arg min ; ; u ~ 1 ^ X W 0 u ~ 1 ^ X W : In particular, the …tted values from the regression of u ~i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri equal ^ 1 + Z ^ + W ^; ^ 1i ; :::; X ^ ki ; W1i ; :::; Wri and that from the regression of u ~i on constant and X equal ^ ~ + W ~: ~1 + X 8 ^ = Z (Z 0 Z) Now, by de…nition, X be invertible, we can write 1 Z 0 X: Since Z 0 X is assumed to ^ (Z 0 X) Z=X 1 ^ Z 0 Z = XA; 1 where A = (Z 0 X) Z 0 Z is an invertible k-dimensional matrix. Re^ we get placing Z in (6) by XA; n o ^ ; ^ ; ^ = arg min u ~ ; ; 1 ^ XA W ^ X W 0 u ~ 1 ^ XA u ~ 1 ^ X W : This implies that n o ^ ; A ^ ; ^ = arg min ; ; u ~ 1 0 W ; and hence ~ = ^ ; ~ = A ^ ; and ~ = ^ : ^ imply that This, and the equality Z = XA ^ ~ + W ~: ^1 + Z ^ + W ^ = ~1 + X Furthermore, from (c), ^ = 0; ^ = 0; and ^ = 0: Therefore, ~ = 0; ~ = 0; and ~ = 0 too. (e) Combining the results from (a), (b), and (d), prove that the value of J statistic in the just-identi…ed case (m = k) must be zero. Suggested solution to (e) From (a), (b) and (d), the OLS estimates of the coe¢ cients in the regression of u ^i on constant and Z1i ; :::; Zmi ; W1i ; :::; Wri must be zero. Therefore, the F statistic for testing the null that the coe¢ cients on Z are zero must be equal to zero. This implies that J = 0: 5. This problem is the empirical exercise E12.3 from the Stock and Watson textbook. We reproduce it here for convenience. On the textbook Web site www.pearsonhighered.com/stock_watson you will …nd the data set WeakInstrument that contains 200 observations on (Yi ; Xi ; Zi ) for the instrumental variable regression Yi = 0 + 1 Xi + ui : 2SLS (a) Construct ^ 1 ; its standard error, and the 95% con…dence interval for 1 : Suggested solution to (a) The 2SLS estimate of beta1 is 1.1577. Assuming homoskedasticity, the standard error of the 2SLS estimate of beta1 equals 0.4269, and the 95 percent con…dence interval is [0.3210, 1.9945]. (b) Compute the F -statistic for the regression of Xi on Zi : Is there evidence of a "weak instrument" problem? 9 12 10 8 6 4 2 0 -5 0 5 Figure 1: Values of Anderson-Rubin statistic. The horizontal line at the critical level 3.84. Circles on the horizontal axis represent the non-rejected points, which span the con…dence set. Suggested solution to (b) Assuming homoskedasticity, F = 4:5661 < 10: This suggests that there may be a weak instrument problem. (c) Compute a 95% con…dence interval for 1 using the Anderson-Rubin procedure. (To implement the procedure, assume that 5 5:) 1 Suggested solution to (c) Assuming homoskedasticity, the con…dence interval is [ 5; 1:6667] : Figure 1 shows a plot of the values of the AR-statistic for various 5 5: The non-rejected points are 1 denoted as circles on the horizontal axis. The non-rejection happens when the value of the F-statistic is above the horizontal line with ordinate 3.84. (d) Comment on the di¤erences in the con…dence intervals in (a) and (c). Which is more reliable? Suggested solution to (d) The con…dence interval in (a) is not reliable because of the weak instrument problem. The con…dence interval in (c) is reliable even when instruments are weak. Students were welcome to do the problem using a program of their choice, be it Excel, stata, E-views, or anything else. I have done the problem in Matlab. Here is the corresponding code. Matlab code for problem 5 %…rst stage regression %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% n=200;%number of observations c=ones(200,1); %create constant Z=[c,z]; %create matrix of explanatory variables pihat=inv(Z’*Z)*Z’*x; %OLS estimates of the coe¢ cients xhat=Z*pihat; %get …tted values 10 ehat=x-xhat;% residuals from the …rst stage sigehat=ehat’*ehat/(n-2);%estimate of the variance of the %error in the …rst stage regression varpi=sigehat*inv(Z’*Z); %covariance matrix of the %…rst stage estimates tstat=pihat(2,1)/sqrt(varpi(2,2));%t-statistic from the …rst stage Fstat=tstat^2;%F-statistic from the …rst stage disp(’F-statistic from the …rst stage regression equals’) disp(Fstat) %second stage regression %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Xhat=[c,xhat]; %create matrix of explanatory variables beta2SLS=inv(Xhat’*Xhat)*Xhat’*y; %obtain 2SLS estimates beta12SLS=beta2SLS(2,1); %2SLS estimate of beta1 disp(’the 2SLS estimate of beta1 is’) disp(beta12SLS); %asymptotic covariance matrix %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% X=[c,x]; %create matrix of explanatory variables uhat=y-X*beta2SLS; %compute 2SLS residuals %assuming homoscedasticity sigmahat=uhat’*uhat/(n-2); %estimate of the variance of error Omegahat=sigmahat*inv(Xhat’*Xhat); %estimate of the asymptotic % variance %standard error %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% stanerr=sqrt(Omegahat(2,2));%standard error estimate disp(’the standard error of the 2SLS estimate of beta1 equals’) disp(stanerr) %con…dence interval %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp(’a 95 percent con…dence interval is’) disp([beta12SLS-1.96*stanerr,beta12SLS+1.96*stanerr]) %Inverting Anderson Rubin test %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% betagrid=linspace(-5,5,100);%creat a grid of equally spaced %point in between -5 and 5. keep=zeros(100,1);% create an indicator variable, whose values % will be changed to one for those points on the % grid not rejected by AR test for i=1:100 %start a loop 11 ynew=y-x*betagrid(1,i);%transform left-hand variable betaAR=inv(Z’*Z)*Z’*ynew; %run Anderson-Rubin regression resid=ynew-Z*betaAR;%get residuals from A-R regression sigAR=resid’*resid/(n-2);%estimate of the variance of the error term OmegaAR=sigAR*inv(Z’*Z);%covariance matrix of estimates tAR=betaAR(2,1)/sqrt(OmegaAR(2,2));%t-statistic Fstatistic(i,1)=tAR^2;% F-statistic if Fstatistic(i,1)>3.84 %do nothing else %do not reject keep(i,1)=1;%remember that you’ve not rejected this value end end plot(betagrid,Fstatistic)%plot the values of AR statistic hold on%use this to keep previous plot line([-5 5],[3.84,3.84])%%critical value for AR statistic plot(betagrid(keep==1),zeros(sum(keep),1),’o’)% plot ’o’for non-rejected 6. Consider a regression yi = x0i + ui ; i = 1; :::; n; 0 where xi = (x1i ; :::; xki ) and E (ui jxi ) 6= 0: Suppose you have m instru0 ments zi = (z1i ; :::; zmi ) ; and m > k: Assume that (yi ; xi ; zi ) is an i.i.d. sequence. Denote the n m matrix with i-th row zi0 as Z; the n k matrix with i-th row x0i as X; and the n 1 vector with i-th element yi as y. (a) State the assumptions that z1i ; :::; zmi must satisfy to be valid instru^ = Z (Z 0 Z) 1 Z 0 X: ments. Suppose these assumptions hold. Let W Prove that p 1 ^ ^0 ^ 0y ! W 2SLS = W X Suggested solution to (a) The weakest IV assumptions implying consistency are 1 0 p Z Z ! C > 0; n 1 0 p Z X ! D; which is full column rank (relevance) n 1 0 p Z u ! 0 (exogeneity). n 12 The following shows consistency ^ X 0 Z (Z 0 Z) = 2SLS p ! D0 C 1 1 Z 0X 1 Z 0Z n X 0Z n = 1 1 D 1 X 0 Z (Z 0 Z) Z 0 u ! 1 X 0Z Z 0Z Z 0X n n n D0 C 1 1 Z 0u n 0=0 (b) Suppose that heteroskedasticity is present, with E u2i jzi = 2i : Prove p that the variance of the asymptotic distribution of n ^ 2SLS is given by D0 C 1 1 D D0 C 1 1 BC D D0 C 1 1 D ; where C = E (zi zi0 ) ; D = E (zi x0i ) ; and B = E u2i zi zi0 : Suggested solution to (b) p n ^ 2SLS = Z 0Z n X 0Z n 1 Z 0X n ! 1 X 0Z n 1 Z 0Z n Z 0u p n On the other hand, X 0Z n Z 0Z n 1 Z 0X n ! 1 X 0Z n 1 Z 0Z n p ! D0 C 1 D 1 D0 C 1 and Z 0u d p ! N (0; B) n Therefore, by Slutsky’s theorem p 1 d n ^ 2SLS ! N 0; D0 C 1 D D0 C (c) Take the residuals u ^=y 1 BC X ^ 2SLS and form the m 1 D D0 C 1 D 1 n matrix ^ = [z1 u Q ^1 ; :::; zn u ^n ] : ^=Q ^Q ^ 0 : Consider Then form R n ^ = arg min (y X )0 Z R ^ 1 Z 0 (y o X ) : Find the …rst order conditions. Using these conditions, or otherwise, show that 1 ^ = X 0Z R ^ 1Z 0X ^ 1 Z 0 y: X 0Z R Prove that ^ is consistent. 13 Suggested solution to (c) The …rst order conditions are ^ 2X 0 Z R 1 ^ = X 0Z R ^ 1 X^ Z0 y =0 Therefore, 1 Z 0X ^ X 0Z R 1 Z 0 y: Since y = X + u; we have ^ = ^ X 0Z R = X 0Z ^ R n 1 p ! D0 B 1 Z 0X 1Z ^ X 0Z R 1 0 X n 1 1 D 1 Z 0u X 0Z ^ R n 1Z D0 B 1 u n 0=0 (d) Show that the variance of the asymptotic distribution of equals DB 1 0 p n ^ 1 D Suggested solution to (d) p n ^ = X 0Z ^ R n 1Z 1 0 X n X 0Z ^ R n 0 u p : n 1Z On the other hand, X 0Z ^ R n 1Z 0 X n 1 X 0Z ^ R n 1 p ! D0 B 1 1 D D0 B 1 ^ to B one needs to apply Slutsky the(to show the convergence of R orem similarly to how this is done in problem 3) and Z 0u d p ! N (0; B) : n By the Slutsky theorem, we have p n ^ d ! N 0; D0 B 1 D 1 D0 B 1 BB 1 D D0 B 1 D 1 Performing the cancellations, we obtain p n ^ d ! N 0; D0 B 1 D 1 (e) Bonus question Prove that, under heteroskedasticity, ^ is asymptotically more e¢ cient than ^ 2SLS . 14 Suggested solution to (e) Note that D0 C 1 1 D D0 C 1 1 BC D D0 C 1 1 D D0 B 1 1 D can be represented in the form 1 D0 C where H = B 1=2 D0 C 1 D 1 1 H (H 0 H) B 1=2 I H 0 B 1=2 C 1 D D0 C 1 D 1 ; D: On the other hand, H (H 0 H) I 1 H0 0 because this is an idempotent matrix (its square is equal to itself). Indeed I 1 H (H 0 H) H0 H (H 0 H) I 1 H0 = +H (H 0 H) = Any matrix of the form M10 M2 M1 ; where M2 de…nite matrix. In particular, D0 C 1 D 1 D0 C 1 BC 1 D D0 C 1 D 1 1 H (H 0 H) I 1 1 D 1 D0 C 1 BC 1 15 D D0 C 1 D H0 H0 1 H 0: 0 is a positive semi- 1 D0 B 1 D D0 B 1 1 0; which implies that D0 C H0 H 0 H (H 0 H) 2H (H 0 H) I +H (H 0 H) = 1 2H (H 0 H) I 1 D 1 : 1 H0

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