# Testing Parameter Restrictions

8.5.1. The Pseudo t-Test and the Wald Test

In view of Theorem 8.2 and Assumption 8.3, the matrix Й can be estimated consistently by the matrix Й in (8.53):

If we denote the ith column of the unit matrix Im by ei it follows now from (8.53), Theorem 8.4, and the results in Chapter 6 that

Theorem 8.5: (Pseudo t-test) under Assumptions 8.1-8.3, ti = л/не^в/ JejЙ-1ei ^dN(0, 1) ifeTeo = 0.

Thus, the null hypothesis Й0 : eTв0 = 0, which amounts to the hypothesis that the ith component of в0 is zero, can now be tested by the pseudo t-value ti in the same way as for M-estimators.

Next, consider the partition

в0 = (вз °) , в1,0 Є Rm-r, в2,0 Є Rr (8.54)

and suppose that we want to test the null hypothesis в2,0 = 0. This hypothesis corresponds to the linear restriction Rв0 = 0, where R = (O, Ir). It follows from Theorem 8.4 that under this null hypothesis

4nR^ ^d Nr(0, R^1 RT).

is the restricted ML estimator. Note that к is always between 0 and 1. The intuition behind the LR test is that, if 02,0 = 0, then к will approach 1 (in probability) as n because then both the unrestricted ML estimator 0 and the restricted ML estimator 0 are consistent. On the other hand, if the null hypothesis is false, then к will converge in probability to a value less than 1.

Theorem 8.7: (LR test) Under Assumptions 8.1-8.3, —2 In (к) ^d хГ if

02,0 = 0.

Proof: As in (8.38) we have

where H1,1 is the upper-left (m — r) x (m — r) block of H ‘Hi, i HiO

H2,1 H2,2

and consequently

Гв л 4 (H— 0 (дЩі„(0o))A/^V,14 ,0,04

■ло – ад = – 0 о) ^—щ—) + °>(1)- (8 58)

Subtracting (8.58) from (8.34) and using condition (8.33) yield

гл Л4 (r-r-і (H – 0\( d ln(Ln(0o))/Vn

– ад = – H 0 o))[—————— Щ———– )

+ Op ( 1 ) ^dNm (0, A),

where

The last equality in (8.60) follows straightforwardly from the partition (8.56).

Next, it follows from the second-order Taylor expansion around the unrestricted ML estimator в that, for some fj є [0, 1],

Thus, we have

-2 ln(X) = (A-1/Vn(0 – 9)) (A1/2HA1/2) (A-1/Vn'(0 – 9))

+ Op (1). (8.63)

Because, by (8.59), A-1/2 «Jn(§ – §) ^d Nm(0, Im), and by (8.60) the matrix A1/2 H A1/2 is idempotent with rank (Д1/2 Й A1/2) = trace(A1/2 Й A1/2) = r, the theorem follows from the results in Chapters 5 and 6. Q. E.D.

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