Testing the Information Matrix Equality with Robust Estimators



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     ,

= — ---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) = ( EFσP3) 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∞ (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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