THÈSE DE DOCTORAT DE L'UNIVERSITÉ PARIS 6 Spécialité ...

2.3. Proofs 29[for ε introduced in Section 2.2. Let M =γ n2h(2 1/β +1)]and define the points a 1 , . . . , a M ina neighbourhood of x 0 in the following way. For n large enough and if x 0 ∈ (0, 1), we puta 1 = x 0 − γ n /2 + ( 2 1/β + 1 ) h, a j+1 − a j = 2 ( 2 1/β + 1 ) h.If x 0 = 0 (respectively x 0 = 1), we define the points a j in the same way except that a 1 is(2 1/β + 1 ) h (respectively 1 − γ n + ( 2 1/β + 1 ) h). We define the set Σ ′ aswhere for θ = (θ 1 , . . . , θ M ) ∈ [−1, 1] M and x ∈ [0, 1]Σ ′ = { f(·, θ), θ ∈ [−1, 1] M} , (2.12)f(x, θ) = Lh βM∑j=1θ j(1 −x − a j∣ hFor all θ ∈ [−1, 1] M , f(·, θ) ∈ Σ(β, L)(cf. the appendix of Chapter 2) and ‖f‖ ∞ ≤ Q forn large enough. Therefore for n large enough Σ ′ ⊂ Σ(β, L, Q).Remark: For β > 1 the subspace Σ ′ should be defined in a similar way:{∑ M ( )}x −Σ ′ = f(x, θ) = Lh βajθ j f β , θ ∈ [−1, 1] Mhj=1and the values (a j ) j=1,...,M should satisfy a j+1 − a j = 2A β h(2 1/β + 1) where [−A β , A β ] isthe support of f β .∣β ) +Then we need to introduce an event N n that satisfies the following lemma.Lemma 2.2. The event⎧⎨N n =⎩ (X 1, . . . , X n ) :supj=1,...,M∣(β + 1)(2β + 1)4µ 0 β 2 nh(n∑1 −X k − a j∣ hk=1∣β ) 2+⎫ ⎬− 1∣ < ε ⎭ ,satisfies lim n→∞ P X (N n ) = 1.The proof is in Subsection 2.3.3.Finally, we study a set of statistics. Let θ ∈ [−1, 1] M . We suppose that f(·) = f(·, θ).The model (2.1) is then written in the formY k = f(X k , θ) + ξ k , k = 1, . . . , n,and the vector (X 1 , Y 1 , . . . , X n , Y n ) follows the law P f(·,θ) that we will denote for brevityP θ . For X ∈ N n , consider the statisticsy j =∑ nk=1 Y kf j (X k )∑ nk=1 f j 2(X , j = 1, . . . , M (2.13)k)