Hence
b = lim nEEFs [ s ( Y ; β0 ,n,σ 0 ,n, 0)]
n→∞
0
=e 0
\ 1 -∏
\ 2 π
The information matrix is
J = E [s (y ;0, 1, 0) s (y ;0, 1, 0)' ]
1 0 Eφ[u(u2 — 1)I(u > 0)] \
■ 2 Eφ[(u2 — 1)21(u> 0)]
■ ■ Eφ[(u2 — 1)21(u> 0)] J
10φ(0) \
021 ,
φ(0) 1 1
wherefrom δ = b,J 1 b = π-1 e2.
2π
C.3 Tilted normal alternative
Expanding f(y; β, σ, κ, λ) around κ =0andλ = 0 gives
f( У ;β,σ,κ,λ ) = 1 φ ( u )
σ
κλ
1 + ^( u — 3 u ) + 24( u — 6 u + 3)
+ o(κ, λ),
from which the moments given in (15) follow. The score function, evaluated
at κ = λ =0, is
s(y; β, σ, 0, 0)
σ u \
u2 -1
σ
u3 — 3 u
6
u4 — 6u 2+3
\ 24 /
The information matrix is
J = E [s ( y ; β, 1, 0, 0) s ( y ; β, 1, 0, 0) ']
/ 1 0 0 0 \
0200
= 0 0 1 0 ’
V 0 0 0 24 J
35
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