Testing the Information Matrix Equality with Robust Estimators

Clearly, σ_n → 1, since Student’s t tends to N(0,1) as p_n → ∞. Rewrite (16)
as

E Wʌ = _b_c,
ΓV √Γ+ε Л c

where

ε = ^σn — 1 + σn_l, Л 2/Pn W + σn, R.

Expanding ρ_c ( Z/ √1 + ε) around ε = 0 gives

b_c = E [ρc(Z)] — ¹E [(ZPc(Z)ε] + ¹E [(Z²PC(Z) + 3ZρC(Z))ε²]
2 8

+o ( E ( ε² )).

Since E [ρc(Z)] = b_c, E(ε) = σ²n — 1 and E(ε²) = (σ²n — 1)² + ‰σn + o(p_n ¹ ),
we obtain,

^bc = ^bc ^- 2⁽^σn ^-¹)^e [Z^ρ'c⁽^z)] ⁺ 4~~^σn^e [^z²^ρ'c⁽^z) ^{+ 3}Z^ρ_c⁽^z)]

^{2 4}^pn

⁺^o⁽^σn^-¹^,pn¹⁾.

^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 ; θ_n ) 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, EF_n (ff ₂) = 0. Furthermore,

EF_n (Y²)

σ_n²

—1

EFn⁽^γ²⁾^—^σn ⁺^o⁽ P-¹)

^pn ^—²

^— 2∑^-ς1 ^—^{1 +}^o⁽ pⁿ¹)
2P_n

4 — Σ₁

²^pn

⁺^o⁽ P^-¹)^,

(18)