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

3.4. Lower bound 49With these preliminaries, we can now prove inequality (3.7). For any f ∈ Σ(β, L) andfor any estimator ̂f n , using the monotonicity of w and the Markov inequality, we obtainthat(E f[w ψn −1 ‖ ̂f)][n − f‖ ∞ ≥w(C 0 (1 − ε))P f ψn −1 ‖ ̂f]n − f‖ ∞ ≥ C 0 (1 − ε) .Since Σ ′ ⊂ Σ(β, L), it is enough to prove that lim n→∞ Λ n = 1, where[Λ n = inf sup P f ψn −1 ‖ ̂f]n − f‖ ∞ ≥ C 0 (1 − ε) .̂f n f∈Σ ′We have max j=1,...,M | ̂f n (a j ) − f(a j )| ≤ ‖ ̂f n − f‖ ∞ . Setting ˆθ j = ̂f n (a j )C 0 ψ n and usingthe fact that f(a j , θ) = C 0 ψ n θ j for θ ∈ [−1, 1] M , we see thatwhere C n =Λ n ≥ inf sup P θ (C n ),ˆθ∈R M θ∈[−1,1] M{}max j=1,...,M |ˆθ j − θ j | ≥ 1 − ε and ˆθ = (ˆθ 1 , . . . , ˆθ M ) ∈ R Mwith respect to y = {Y t , t ∈ [0, 1] d }. We have∫Λ n ≥ infP θ (C n )π(dθ),ˆθ∈R M {−(1−ε),1−ε} Mis measurablewhere π is the prior distribution on θ, π(dθ) = ∏ Mj=1 π j(dθ j ), where π j is the Bernoullidistribution on {−(1 − ε), 1 − ε} that assigns probability 1/2 to −(1 − ε) and to (1 − ε).Since P θ is absolutely continuous with respect to P 0 (see Proposition 3.3), we have∫Λ n ≥ infˆθ∈R M∫= infˆθ∈R M( )dP θE 0 I Cn π(dθ)dP 0( M)∏ ϕ vn (y j − θ j )E 0 I Cnπ(dθ).ϕ vn (y j )By the Fubini and Fatou theorems, we can write∫( ∫ M)1 ∏Λ n ≥1 − sup ∏ Mˆθ∈R M j=1 ϕ I {|θj −ˆθv n(y j )j |