By Alexandre B. Tsybakov

Developed from lecture notes and able to be used for a path at the graduate point, this concise textual content goals to introduce the basic techniques of nonparametric estimation idea whereas preserving the exposition compatible for a primary strategy within the field.

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**Additional resources for Introduction to Nonparametric Estimation (Springer Series in Statistics)**

**Example text**

Iii) The Fourier coeﬃcients θj = 1 0 f ϕj of f satisfy ∞ |θj | < ∞. 86) holds. We will use the following property of the trigonometric basis. 7 Let {ϕj }∞ j=1 be the trigonometric basis. 92) s=1 where δjk is the Kronecker delta. √ 2 cos(2πmx), Proof. For √ brevity we consider only the case ϕj (x) = ϕk (x) = 2 sin(2πlx) where j = 2m, k = 2l + 1, j ≤ n − 1, k ≤ n − 1, n ≥ 2 and m ≥ 1, l ≥ 1 are integers. Other cases can be studied along similar lines. Put b = exp{i2πl/n}. a = exp{i2πm/n}, Then S= = 1 n n ϕj (s/n)ϕk (s/n) = s=1 n 1 2in 2 n n s=1 (as + a−s )(bs − b−s ) 4i (ab)s − (a/b)s + (b/a)s − (ab)−s .

Let us show that inf v T Bnx v ≥ λ0 v =1 for suﬃciently large n. 75) i=1 where zi = (Xi − x)/h. Observe that zi − zi−1 = (nh)−1 and z1 = x 1 1 − ≤ , nh h nh zn = 1−x ≥ 0. h If x < 1 − hΔ, then zn > Δ and the points zi form a grid with step (nh)−1 on an interval covering [0, Δ]. 76) i=1 Δ T → (v U (z))2 dz as n → ∞, 0 42 1 Nonparametric estimators since the Riemann sum converges to the integral. 74) implies that z1 < −Δ for suﬃciently large n and that the points zi form a grid with step (nh)−1 on an interval covering [−Δ, 0].

Then n ∗ ξi Wni (x) ≤ A = sup x∈[0,1] ≤ i=1 1 sup λ0 nh x∈[0,1] n ξi Si (x) i=1 n 1 λ0 nh 1≤j≤M + sup ξi Si (xj ) max i=1 n ξi (Si (x) − Si (x )) x,x : |x−x |≤1/M . i=1 Since K ∈ Σ(1, LK ) and the support of the kernel K belongs to [−1, 1], and since U (·) is a vector function with polynomial coordinates, there exists a ¯ − u |, ∀ u, u ∈ R. ¯ such that U (u)K(u) − U (u )K(u ) ≤ L|u constant L Thus A ≤ 2 2 ≤ 2 λ0 nh n 2 1 λ0 nh ¯ L ξi Si (xj ) + Mh i=1 max 1≤j≤M ηj max 1≤j≤M 2 ¯2 2L + 2 2 4 2 λ0 n h M 2 n |ξi | i=1 2 n |ξi | , i=1 where the random vectors ηj are given by 1 ηj = √ nh Therefore we have 2 E(A2 ) ≤ 2 E λ0 nh max 1≤j≤M ηj 2 n ξi Si (xj ).