where the first line uses (1 + x)/(1 + y) ≈ 1 + x — y for x and y small,
with x = EFn (Y2) — 1 and y = σ2n — 1, and the second line uses the second
moment of Student’s t, pn/(pn — 2). We also have that
Fn 4 - 3 = EFn (Y4)(1 - 2(σ2 - 1)) - 3 + o(pn 1)
σn
3 P 2n Σ Σ1∖ , l n n 1λ
— 7------7√77------77 I 1--I — 3 + O(pn )
(pn — 2)(pn — 4) ∖ pn J
18 — 3∑1 , 1λ
= — ---1 + O ( P-1 ), (19)
pn
where the first line uses (1 + x)n2 ≈ 1 — 2x and the second line uses the
fourth moment of Student’s t, 3pП(pn — 2)n 1(pn — 4)n 1. Using (18) and (19),
EFn ( m 3) = ( EFσP—3) — 6 ( EFσY1 — ɪ)
σn σ σn
6 , z-H
= — + o ( P- )
and
EFn ( m 1)
EF^2)! — Λ — dEFn (rh3)
∖ σn /
= 2~^ (4 — ∑1 — 12d)+ o(pn 1) ∙
2 pn
To compute Σ1, note that, by partial integration and using φ'(z) = —zφ(z),
E [ZPc(Z)] = j∞ zφ(z)dpc(z)
= —У Pc ( Z )( φ ( Z ) — Z 2 φ ( Z )) dz
= E [Z2Pc(Z)] — bc∙
Along the same lines we get
E [Z2PC(Z)] = E [Z3pc(Z)] — 2E [Zp'c(Z)] ,
and
E [Z3Pc(Z)] = — 3E [Z2Pc(Z)] + E [Z4Pc(Z)] ,
28
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