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Unformatted text preview: Homework 8 8.11 a. 2 t = 1 + 2 S t + 3 D t + 4 S 2 t + 5 ( S t D t ). Note that D 2 t is not used here because it is a dummy variable. b. 2 = 3 = 4 = 5 = 0. c. Regress P t against a constant, S t , and D t and obtain u t = P t- 1- 2 S t- 3 D t . Next regress u 2 t against a constant, S t , D t , S 2 t and ( S t D t ). d. Compute LM = nR 2 , where n is the number of observations and R 2 is the unadjusted R 2 from the second regression in Step c. Under the null hypothesis of homoscedas- ticity, LM has the 2 distribution with 4 d.f. e. The critical value of 2 4 for a 5% level is 9.48773. Reject the null if LM > 9 . 48773. f. From the auxiliary regression compute 2 t = 1 + 2 S t + 3 D t + 4 S 2 t + 5 ( S t D t ). Next compute the weight as w t = 1 / q 2 t . Finally, regress w t P t against w t , w t S t and w t D t , without a constant term. 8.12 a. H : 2 = 3 = 0. H 1 : At least one of them is not zero. b. (1) Regress E t against a constant and Y t , (2) compute the residuals u t = E t- 1- 2 Y t , and (3) regress u 2 t against a constant, P t and P 2 t ....
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- Summer '07