It follows that the score test statistic for testing κ = λ =0 is
S=n
μ2 + (μ4 - 3)2 ^
6 + 24
which is the Jarque-Bera statistic (see (12)).
Let κn = k/y/n and λn = l∕y∕n and Fn = Fκn ,λn. As before, βn and σn
are the solutions of (21) and (22) where expectations are now with respect
to Fκ,λ. From (21) we have, using (24),
Eφ ψ((Z) + ψ(Z)(-βn - Z(σn - 1))}{1 + κ6n(Z3 - 3Z)
+λn ( Z4 - 6 Z 2 + 3
) + o (βn ,σn - 1,κn ,λn ) .
(28)
Since EΦ [ψ(Z)] = 0 andψ is odd, it follows that
βn = Σ3κn + o (βn, σn - 1,κn,λn) , (29)
where
ς =1 E [(Z3 - 3Z)ψ(Z)]
3 6 E [ψ(Z)]
Similarly, we have, from (22),
bc
EΦ Ppc(Z) + Pc(Z)(-βn - Z(σn - 1))}{1 + κn(Z3
- 3Z)
) + o (βn ,σn - 1,κn ,λn ) .
(30)
Now EΦ [ρc(Z)] = bc and ρc is even, so
σn =1+Σ4λn +(βn ,σn - 1,κn ,λn) , (31)
where
1 E [(Z4 - 6Z2 + 3)Pc(Z)]
24 E [ZPc(Z)]
Since
Y - βn
σn
Y (1 - (σn - 1)) - βn + o (κn, λn)
Y (1 - Σ4λn) - Σ3κn + o(κn,λn) ,
36
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