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] = 2cφ(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 + Rσn
26
= bc,
(16)