wherefrom, using (4),
ξ ( Y ; θ )= 1
σ2
u2 - 1
u3 - 3u
u4 - 5u2 +2
2 / IF( Y ; ^; Fθ )
⅛ ( θ
σ ∖ IF( Y ; ^; Fθ )
Note that ξ ( Y ; θ ) does not depend on IF( Y ; /; Fθ ). Take ^ to be equivariant,
i.e. σ ( aY1 + b, ■ ■ ■ , aYn+b ) = ∣α∣σ ( Y1, ∙ ∙ ∙ , Yn ), so IF( Y ; ^; Fθ ) = σ IF( u ; σ; Φ)
and
1 / u2 - 1 - 2IF(u; ^; Φ)
ξ ( Y ; θ ) = — I u3 — 3 u
(6)
σ \—4 — 5u2 + 2 — 2IF(u; ^; Φ)
A straightforward calculation shows that V = σ-4B , where B is a 3 × 3
matrix with elements Bij given by
B u = — 2 + 4ASV( ^),
B22 =6,
B33 = 46 + 4ASV( ^) — 4 E φ ( u 4IF),
B13 = 10 + 4ASV( ^) — 2 E φ ( u 4IF) = B 31,
B12 = B21 = B23 = B32 =0,
with IF = IF(u; ^; Φ) and ASV(σ) = Eφ(IF2), the asymptotic variance of
σ when σ = 1. Note that V does not depend on the estimator β that is
chosen. For a given estimator θ = (β, σ)1, let
ʌ .
u i = ( Yi — β)/σ,
n
-1 j
μj = n X"i,
i=1
. ) ʌ
and write M as σ 2 N with
/ μ2 — 1
I .ʌ n .ʌ
I μ3 — 3 μ1
ξ μ4 — 5μ2 + 2
Taking V+ equal to V+ with σ replaced by ^ yields the test statistic
T = nM '1V+MM = nN^ B+N^.
(7)
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