Using a suitable change of variables, for all Σ ∈ S q we have that Since this holds for all k = 1 , . . . , q and j = 1 , . . . , p so that for all Σ ∈ S $_{q}$. By (2.10), the pair θ$_{n}$ = ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ) is a zero of the function (1 /n ) ∑ n i $_{=1}$φ ( z$_{i}$ , θ ). Using this, after doing some simple operations of sum and subtraction and using (A.55), we have Since E$_{H}$$_{0}$φ ( z , θ$_{n}$ ) is equal to Φ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ), we can solve for E$_{H}$$_{0}$φ ( z , θ$_{n}$ ) in the 56 0 = (1 /n ) n ∑ i =1 φ ( z$_{i}$ , θ$_{n}$ ) = E$_{H}$$_{0}$φ ( z , θ$_{n}$ ) + [ 1 n n ∑ i =1 φ ( z$_{i}$ , θ$_{0}$ ) − E$_{H}$$_{0}$φ ( z , θ$_{0}$ ) ] + { 1 n n i =1 [ φ ( z$_{i}$ , θ$_{n}$ ) − φ ( z$_{i}$ , θ$_{0}$ )] − E$_{H}$$_{0}$ [ φ ( z , θ$_{n}$ ) − φ ( z , θ$_{0}$ )] } = E$_{H}$$_{0}$φ ( z , θ$_{n}$ ) + [ 1 n n ∑ i =1 φ ( z$_{i}$ , θ$_{0}$ ) − E$_{H}$$_{0}$φ ( z , θ$_{0}$ ) ] + [ ν$_{n}$ ( φ ( · ; θ$_{n}$ )) − ν$_{n}$ ( φ ( · ; θ$_{0}$ ))] = E$_{H}$$_{0}$φ ( z , θ$_{n}$ ) + [ 1 n n ∑ i =1 φ ( z$_{i}$ , θ$_{0}$ ) − E$_{H}$$_{0}$φ ( z , θ$_{0}$ ) ] + o$_{P}$ (1 $_{/}$√$_{n}$$_{)}$$_{.}$ Φ( B$_{0}$ , Σ ) = 0 (A.57) E$_{H}$$_{0}$W ( d ( B$_{0}$ , Σ )) u$_{k}$x$_{j}$ = ( Ex$_{j}$ ) E$_{F}$$_{0}$W ( u $^{′}$Σ− $^{1}$u ) u$_{k}$ = ( Ex$_{j}$ ) (∫ { u : u$_{k}$> 0 } u$_{k}$W ( u $^{′}$Σ− $^{1}$u ) f ∗ $_{0}$( u $^{′}$Σ− 1 0 u ) d u + ∫ { u : u$_{k}$< 0 } W ( u $^{′}$Σ− $^{1}$u ) u$_{k}$f ∗ $_{0}$( u $^{′}$Σ− 1 0 u ) d u ) = ( Ex$_{j}$ ) (∫ { u : u$_{k}$> 0 } u$_{k}$W ( u $^{′}$Σ− $^{1}$u ) f ∗ $_{0}$( u $^{′}$Σ− 1 0 u ) d u − ∫ { u : u$_{k}$> 0 } u$_{k}$W ( u $^{′}$Σ− $^{1}$u ) f ∗ $_{0}$( u $^{′}$Σ− 1 0 u ) d u ) = 0 .