Differentiating ρ$_{1}$ ( d$_{i}$ ( B , Γ ) / ˆ σ$_{n}$ ) with respect to Σ , we get From (A.14) and (A.15) we have Then, by (A.22) and (A.23), the equation (A.20) results equivalent to Rearranging and using that W ( u ) = ψ ( u ) /u and that ̂ Σ$_{n}$ = ˆ σ 2 $_{n}$Γ ( C$_{n}$ ) we have that and solving for ̂ Σ$_{n}$ we get (2.11). ■ Before showing Theorem 3 we will prove the following lemma: Lemma 10. Let Z = { z$_{1}$ , . . . , z$_{n}$ } , with z$_{i}$ = ( y ′ $_{i}$, x ′ $_{i}$) that satisfy (1.1) and con- sider a ρ -funtion ρ$_{0}$ . Then the explosion breakdown point of the M-estimate of scale of the Mahalanobis norms (2.3) , ˆ σ$_{n}$ := s ( d ( ˜ B$_{n}$ , ˜ Σ$_{n}$ )) , is bounded below by min( ε ∗ $_{n}$( Z , ˜ B$_{n}$ , ˜ Σ$_{n}$ ) , 0 . 5) . 35 ̂ Σ$_{n}$ ( n ∑ i =1 ψ$_{1}$ ( d$_{i}$ ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ) d$_{i}$ ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ) ) = q n ∑ i =1 W ( d$_{i}$ ( ̂ B$_{n}$ , ̂ Σ$_{n}$ ) ̂ u$_{i}$ ( ̂ B$_{n}$ ) ̂ u$_{i}$ ( ̂ B$_{n}$ ) ′ n ∑ i =1 ψ$_{1}$ ( d$_{i}$ ( ̂ B$_{n}$ , Γ ( C$_{n}$ )) ˆ σ$_{n}$ ) ( d$_{i}$ ( ̂ B$_{n}$ , Γ ( C$_{n}$ )) ˆ σ$_{n}$ − q ̂ u$_{i}$ ( ̂ B$_{n}$ ) ̂ u$_{i}$ ( ̂ B$_{n}$ ) $^{′}$Γ ( C$_{n}$ ) − 1 d$_{i}$ ( B$_{n}$ , Γ ( C$_{n}$ ))ˆ σ$_{n}$ ) = 0 . ∂ ( d$_{i}$ ( B , Σ ) | Σ | 1 / (2 q $^{)}$) ∂ Σ = ∂ | Σ | 1 / (2 q ) ∂ Σ d$_{i}$ ( B , Σ ) + ∂d$_{i}$ ( B , Σ ) ∂ Σ | Σ | 1 / (2 q ) = | Σ | 1 / (2 q $^{)}$Σ − 1 2 q d$_{i}$ ( B , Σ ) − Σ − $^{1}$̂ u$_{i}$ ( B ) ̂ u$_{i}$ ( B ) $^{′}$Σ− 1 2 d$_{i}$ ( B , Σ ) | Σ | 1 / (2 q ) = Σ − 1 2 q ( d$_{i}$ ( B , Γ ( Σ )) − q ̂ u$_{i}$ ( B ) ̂ u$_{i}$ ( B ) $^{′}$Γ ( Σ ) − 1 d$_{i}$ ( B , Γ ( Σ )) ) . (A.23) ∂ρ$_{1}$ ( d$_{i}$ ( B , Γ ( Σ )) / ˆ σ$_{n}$ ) ∂ Σ = 1 ˆ σ$_{n}$ ψ$_{1}$ ( d$_{i}$ ( B , Γ ( Σ )) ˆ σ$_{n}$ ) ∂ ( d$_{i}$ ( B , Σ ) | Σ | 1 / (2 q $^{)}$) ∂ Σ . (A.22)