Proof of Theorem 6: We denote θ$_{n}$ = ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ) and θ$_{0}$ = ( B$_{0}$ , σ 2 $_{0}$Σ$_{0}$ ). Since we assumed that the distribution of errors u Since we assumed that the distribution of errors u is elliptical with density of the form (4.1), for any function h we have E$_{H}$$_{0}$ ( x$_{j}$u$_{i}$h ( u $^{′}$Σ − 1 0 u ) ) = 0. This implies that vanishes at B = B$_{0}$ . Then θ$_{0}$ is a zero of the function Φ( θ ) = E$_{H}$$_{0}$φ ( z ; θ ). By Lemma 19 there exists a bounded subset C and a function θ ( ξ ) such that θ$_{0}$ is an interior point of θ ( C ) and since θ$_{n}$ → θ$_{0}$ a.s., θ$_{n}$ ∈ θ ( C ) for n large enough, i.e., φ$_{kj}$ ( z ; θ$_{n}$ ) and φ$_{kj}$ ( z ; θ$_{0}$ ) belong to the Euclidean class F$_{kj}$ for n sufficiently large. By (A5), the functions φ$_{kj}$ ( z ; θ$_{n}$ ) and φ$_{kj}$ ( z ; θ$_{0}$ ) are in the class [ δ ] of Lemma 20 for each δ > 0 and n sufficiently large. Hence, in probability. Then since ν$_{n}$ ( φ$_{kj}$ ( · ; θ$_{n}$ )) − ν$_{n}$ ( φ$_{kj}$ ( · ; θ$_{0}$ )) = o$_{P}$ (1 $_{/}$√$_{n}$$_{) for all}$ k = 1 , . . . , q and j = 1 , . . . , p and φ$_{kj}$ ( z ; θ ) corresponds to the element h = ( j − 1) q + k of the function φ , we conclude that Since ∂ Φ /∂ vec( B $^{′}$)′ is continuous in θ$_{0}$ , we have that where r ( θ ) → 0 when θ → θ$_{0}$ . 55 Φ( B , Σ ) = Φ( B$_{0}$ , Σ ) + ( ∂ Φ( B , Σ ) ∂ vec( B $^{′}$)$_{′}$( B$_{0}$ , Σ ) ) vec( B ′ − B ′ $_{0}$) + r ( θ )vec( B ′ − B ′ $_{0}$) (A.56) ν$_{n}$ ( φ ( · ; θ$_{n}$ )) − ν$_{n}$ ( φ ( · ; θ$_{0}$ )) = o$_{P}$ (1 $_{/}$√$_{n}$$_{)}$$_{.}$ (A.55) | √$_{n}$ { ν$_{n}$ ( φ$_{kj}$ ( · ; θ$_{n}$ )) − ν$_{n}$ ( φ$_{kj}$ ( · ; θ$_{0}$ )) }| −→ 0 (A.54) E$_{H}$$_{0}$ 1 σ 2 0 W ( d ( B , σ 2 $_{0}$Σ$_{0}$ ) ) ( y − B $^{′}$x ) x ′ (A.53)