Now consider a even stronger version of the classical condition D2: D2'. there exists a vector 8 [0,0,0,2,01- P.) such that Wi= 0.0 +81,041,i+B2.022, i +...+p0fpitein where ed N (0,0) i=1,2,..., 2 POLS Then we have Bols ~ N(8,00), where V BjOLS ~N (Bj.0. Vajole), = 02 (xx)That is, for any j = 0.1....., B;oLs - B;8 N (0,1). (6) Or And (7) or (Bols-Bo) V (Bois-Bo) (Pols - B.) x'x (BoLs - 8 X(p+1). For any k x (p + 1) full rank matrix R, we have RBOLS NR.8., RVB..R'), (Bots-Ba)'R' (RV3_R) R (Bois - Bo) (Pols-B) 'R' (R(X'X) R') R (Pols-Bo) x2(k). of where V, is the variance of ; ols, ie, the (+1) diagonal element of V POLS is the standard deviation of BjOLS- = a (xx), where a2 = (n -p-1)-2-1 (0:- :)? Then BOLS - B2 t (n -P-1), and ,Si 1 SOL Recall that Vous (9) os$ and Sa 1 where Vos is the variance of B;Ols, i.e., the (1 +1) diagonal element of Vous is the standard deviation of Bj.0 = (XX)-!, where a2 = (n-p-1)-I-1 (0:- :)? Then BOLS-B,0 (n-p-1). Recall that y POL (9) Sons where Son SOL POLS is the standard error of Bols. For any k x (p+1) full rank matrix R, we have (Pols-A'R' RV R ') R(Bols-Bo) (10) (Pots - 8.)'R' (R(XX) R') R (Pols - B.) F(k,n-p-1). Therefore, for any mull hypothesis Ho: R8, =c, we have the following test statistics: (RBOLS - c) (RM R') (RBots-e) (11) (RBOLs -c) (R(XX) +R") Bols-c) F(k,n-p-1). For example, the null hypothesis, H: 801 Po2 = = Bop = 0, can be written as H, R8, =c, where F 0 1 0 0 0 1 0 0 0 0 R=(x, L) = 0 0 0 and c = 0,1 = 0 0 0 0 0 3 Problem 4. Consider the model: Xi = Bo +3,2,1 +812, +1+3424, +3,25,1 + #i, i=1,2,...,150. Suppose that we have the following null hypotheses. Write out Rand e in the form of Ho: R3. = c. And write down the null distribution. 1. HA"): B01 = 0.2 = 8.5=0. 2. H.2) : 30,0 + 38.3 - 680,4 = 4. 3. H: Boa +48.4 +1 = 28.1 4. H": Boo +381 + Boa +58.5 = 7. Remark 1 For the null hypothesis H : Bol = 0.2 == = 0, another way to test the wall is to consider the following statistics: (TSS-ESS)/P F= (12) ESS/(n-p-1) (1 - Ry/(n-p-1) This F statistic in equation (12) is numerically identical to the statistic in equation (11). We skip the details. Now consider a even stronger version of the classical condition D2: D2'. there exists a vector 8 [0,0,0,2,01- P.) such that Wi= 0.0 +81,041,i+B2.022, i +...+p0fpitein where ed N (0,0) i=1,2,..., 2 POLS Then we have Bols ~ N(8,00), where V BjOLS ~N (Bj.0. Vajole), = 02 (xx)That is, for any j = 0.1....., B;oLs - B;8 N (0,1). (6) Or And (7) or (Bols-Bo) V (Bois-Bo) (Pols - B.) x'x (BoLs - 8 X(p+1). For any k x (p + 1) full rank matrix R, we have RBOLS NR.8., RVB..R'), (Bots-Ba)'R' (RV3_R) R (Bois - Bo) (Pols-B) 'R' (R(X'X) R') R (Pols-Bo) x2(k). of where V, is the variance of ; ols, ie, the (+1) diagonal element of V POLS is the standard deviation of BjOLS- = a (xx), where a2 = (n -p-1)-2-1 (0:- :)? Then BOLS - B2 t (n -P-1), and ,Si 1 SOL Recall that Vous (9) os$ and Sa 1 where Vos is the variance of B;Ols, i.e., the (1 +1) diagonal element of Vous is the standard deviation of Bj.0 = (XX)-!, where a2 = (n-p-1)-I-1 (0:- :)? Then BOLS-B,0 (n-p-1). Recall that y POL (9) Sons where Son SOL POLS is the standard error of Bols. For any k x (p+1) full rank matrix R, we have (Pols-A'R' RV R ') R(Bols-Bo) (10) (Pots - 8.)'R' (R(XX) R') R (Pols - B.) F(k,n-p-1). Therefore, for any mull hypothesis Ho: R8, =c, we have the following test statistics: (RBOLS - c) (RM R') (RBots-e) (11) (RBOLs -c) (R(XX) +R") Bols-c) F(k,n-p-1). For example, the null hypothesis, H: 801 Po2 = = Bop = 0, can be written as H, R8, =c, where F 0 1 0 0 0 1 0 0 0 0 R=(x, L) = 0 0 0 and c = 0,1 = 0 0 0 0 0 3 Problem 4. Consider the model: Xi = Bo +3,2,1 +812, +1+3424, +3,25,1 + #i, i=1,2,...,150. Suppose that we have the following null hypotheses. Write out Rand e in the form of Ho: R3. = c. And write down the null distribution. 1. HA"): B01 = 0.2 = 8.5=0. 2. H.2) : 30,0 + 38.3 - 680,4 = 4. 3. H: Boa +48.4 +1 = 28.1 4. H": Boo +381 + Boa +58.5 = 7. Remark 1 For the null hypothesis H : Bol = 0.2 == = 0, another way to test the wall is to consider the following statistics: (TSS-ESS)/P F= (12) ESS/(n-p-1) (1 - Ry/(n-p-1) This F statistic in equation (12) is numerically identical to the statistic in equation (11). We skip the details
