Testing for Over-Identification. In testing Ho; = 0 versus H1; = 0 in section 11.4, equation (11.47): (a) Show that the second stage

Testing for Over-Identification. In testing Ho; γ = 0 versus H1; γ = 0 in section 11.4, equation

(11.47):

(a) Show that the second stage regression of 2SLS on the unrestricted model y1 = Z1δ1+W∗γ+u1 with the matrix of instruments W yields the following residual sum of squares:

URSS∗ = y



1

¯ PWy1 = y



1y1 − y



1PWy1 Hint: Use the results of problem 10 for the just-identified case.

(b) Show that the second stage regression of 2SLS on the restricted model y1 = Z1δ1 + u1 with the matrix of instruments W yields the following residual sum of squares:
RRSS∗ = y 
1 ¯ P
Z1 y1 = y 
1y1 − y 
1P
Z1 y1 where  Z1 = PWZ1 and P
Z1 = PWZ1(Z
1PWZ1)−1Z
1PW. Conclude that RRSS∗− URSS∗
yields (11.49).

(c) Consider the test statistic (RRSS∗− URSS∗)/σ11 where σ11 is given by (11.50) as the usual 2SLS residual sum of squares under Ho divided by T . Show that it can be written as Hausman’s (1983) test statistic, i.e., nR2u where R2u is the uncentered R2 of the regression of 2SLS residuals (y1−Z1δ 1,2SLS) on the matrix of all pre-determined variables W. Hint: Show that the regression sum of squares (y1 − Z1δ 1,2SLS)PW(y1 − Z1δ1,2SLS) = (RRSS∗− URSS∗)
given in (11.49).

(d) Verify that the test for Ho based on the GNR for the model given in part

(a) yields the same TR2u test statistic described in part (c).