Testing the Information Matrix Equality with Robust Estimators



and ASV(σ) follows from

asv( σ) =


EΦ ρPcuu)] - b2
{EΦ [pc(u)u] }2


Furthermore,

EΦ(u4IF) = f ɪ 1

Eφ [ pc ( u ) u ]


{f(1 - νc(4)) +   6


Vc (8) + νc (10)

2 c 2      6 c 4


3bc .


The computation of T , from (11), is now straightforward.

For the MAD, Eφ [pc(u)u] = 2(c) (with φ the standard normal pdf)
and
Eφ [ρ2(u)] = 2, resulting in


and


ASV ( ^) =-----1----= = 1.361

16c2 φ(c) 2

Eφ(u4IF) = r, γ ɪ 1 I3 - Vc(4)I .
φv ’   Eφ [Pc(u)uH 2   cvJ


Appendix B

Local asymptotic power of IM test

B.1 Student’s t alternative


Under Hn, Y ~ Fn = Ft(pn ) with pn = Пп/е. So


Z

U' n ,


where U ~ χ2n and Z ~ N(0, 1), with U and Z independent. By the Central
Limit Theorem, as
pn →∞,

U=1+'∕pn w+R

where W ~ N(0, 1), W and Z are independent, the remainder term R is
Op pn-1 , and E(R) = 0. Therefore σn solves


Z

ρc ɪ I

∖√1 +72/PnW + n


26


= bc,


(16)




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