where expectations are with respect to F^. From (21),
• ∕>(⅛) d -Γ∙(
z (1 + γn ) - βn
σn
dΦ(z). (23)
Now, expanding the integrands around βn = 0, σn = 1, and γn = 0 gives
ψ Zz—βn∖ = ψ(Z) + ψ/(Z) (-βn - Z(σn - 1)) + o(βn, σn - 1) (24)
∖ σn J
and
√z(l+γn)
∖ σn
- βn
) = ψ(z) + ψ(z) (ZYn - βn - z(σn - 1))
+o(βn, σn
(25)
and so (23)
becomes
= Eφ [ψ(Z)] - βnEφ [β'(Z)] - (σn
1)Eφ [Zψ(Z)]
1, γn).
+Yn zψ ( Z ) d Φ( Z )+ O ( βn,σn
Jo
Since Eφ [ψ(Z)] = Eφ [Zψ'(Z)] = 0, it follows that βn = γnΣ2 + o(γn), with
ς = ∕o∞ zψ (z)dφ(z)
2 E [Ψ(Z)]
For σn it holds that
b Lpc YYd' ΓY
z (1 + γn ) - βn
σn
dΦ(z). (26)
Since (24) and (25) also hold with pc replacing ψ, (26) becomes
bc = eΦ [pc(z)] - βneΦ [pc(z)] - (σn
■ 1)Eφ [zp'c(Z)]
1, γn).
+Yn / Zpc(Z)dΦ(Z) + O(βn,σn-
Jo
Now Eφ [pc(Z)] = bc, Eφ [p'c(Z)] = 0 and J∞ zp'c(z)dΦ(z) = ɪEφ [Zpc(Z)],
so we get
σn = 1 + ɪ- + o ( γn ).
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