Clearly, σn → 1, since Student’s t tends to N(0,1) as pn → ∞. Rewrite (16)
as
E Wʌ = bc,
ΓV √Γ+ε Л c
where
ε = σn — 1 + σnl, Л 2/Pn W + σn, R.
Expanding ρc ( Z/ √1 + ε) around ε = 0 gives
bc = E [ρc(Z)] — 1E [(ZPc(Z)ε] + 1E [(Z2PC(Z) + 3ZρC(Z))ε2]
2 8
+o ( E ( ε2 )).
Since E [ρc(Z)] = bc, E(ε) = σ2n — 1 and E(ε2) = (σ2n — 1)2 + ‰σn + o(pn 1 ),
we obtain,
bc = bc - 2(σn - 1)e [Zρ'c(z)] + 4~~σne [z2ρ'c(z) + 3Zρc(z)]
2 4pn
+o ( σn-1 ,pn1).
So
σn = 1 + 2⅛ + o ( P-1), (17)
2 pn
where
ς e [Z2ρc(Z) + 3Z^c(Z)]
1 E [Z(Z)] .
Let ( m 1 ,m 2 ,m 3 ) ' = Wm ( Y ; θn ) with
/ u2 — 1 — d(u4 — 6u2 + 3) ʌ
W m ( Y ; θ ) = I u3 — 3 u I .
∖ u4 — 6 u2 + 3 /
By the symmetry of Student’s t distribution, EFn (ff 2) = 0. Furthermore,
EFn (Y2)
σn2
—1
EFn( γ 2) — σn + o ( P-1)
Pn
pn — 2
— 2∑-ς1 — 1 + o ( pn 1)
2Pn
4 — Σ1
2 pn
+ o ( P-1),
(18)
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