Exact asymptotic behavior of VarΠ

The proof of Theorem 2.1 is organized as follows. First Subsection 3.1, we determine the exact asymptotic behavior of the variance of Π n ǫ n . Then Subsection 3.2, we prove a central limit theorem for Π n ǫ n . By means of a de-Poissonization result Subsection 3.3, we then infer 3.3. In a final step Subsection 3.4 we prove 3.4, which completes the proof of Theorem 2.1. This Poissonizationde-Poissonization methodology goes back to at least Beirlant, Györfi, and Lugosi 1994.

3.1 Exact asymptotic behavior of VarΠ

n ǫ n Let ∆ n x = 1{π n x 0} − 1{ f x 0} . In the sequel, the letter C will denote a positive constant, the value of which may vary from line to line. Let ǫ n be the sequence of positive real numbers defined in 3.1. In this subsection, we intend to prove that, under the conditions of Theorem 2.1, lim n →∞ r n r d n Var Π n ǫ n = σ 2 f , 3.5 where σ 2 f is as in 2.3. Towards this goal, observe first that Π n ǫ n = Z ˜ E n 1{π n x 0} − 1{ f x 0} d x , where we set ˜ E n = E n ∩ S r n f , with S r n f = ¦ x ∈ R d : distx, S f ≤ r n © . Clearly, VarΠ n ǫ n = Z ˜ E n Z ˜ E n C ∆ n x, ∆ n y dxd y, where here and elsewhere C denotes ‘covariance’. Since ∆ n x and ∆ n y are independent when- ever kx − yk 2r n , we may write Var Π n ǫ n = Z ˜ E n Z ˜ E n 1 kx − yk ≤ 2r n C ∆ n x, ∆ n y dxd y. Using the change of variable y = x + 2r n u, we obtain VarΠ n ǫ n 2624 = 2 d r d n Z R d Z R d 1 ˜ E n x1 ˜ E n x + 2r n u 1 B0,1 uC ∆ n x, ∆ n x + 2r n u dxdu. By construction, whenever n is large enough, ˜ E n is included in the tubular neighborhood V ∂ S f , ρ of ∂ S f of radius ρ 0. In this case, each x ∈ ˜ E n may be written as x = p + ve p as described in 2.1. Hence, for all large enough n, we obtain VarΠ n ǫ n = 2 d r d n Z ∂ S f Z ρ −r n Z B0,1 1 ˜ E n p + ve p 1 ˜ E n p + ve p + 2r n u × Θp, vC € ∆ n p + ve p , ∆ n p + ve p + 2r n u Š dudv v σ dp. For all p ∈ ∂ S f , let κ p ǫ n be the distance between p and the point x of the set {x ∈ R d : f x = ǫ n } such that the vector x − p is orthogonal to ∂ S f . Using the change of variable v = t p nr d n , we may write Var Π n ǫ n = 2 d r d n p nr d n Z ∂ S f Z p nr d n κ p ǫ n − p nr d+2 n Z B0,1 1 ˜ E n   p + t p nr d n e p + 2r n u   Θ   p, t p nr d n   × C   ∆ n   p + t p nr d n e p   , ∆ n   p + t p nr d n e p + 2r n u     dudt v σ dp. For a justification of this change of variable, refer to equation 2.2 and equation 4.2 in the Ap- pendix. By conditions r.i and r.iii, nr d+2 n → 0. Consequently, r n r d n Var Π n ǫ n = o1 + 2 d Z ∂ S f Z p nr d n κ p ǫ n Z B0,1 1 ˜ E n   p + t p nr d n se p + 2r n u   Θ   p, t p nr d n   × C   ∆ n   p + t p nr d n e p   , ∆ n   p + t p nr d n e p + 2r n u     dudt v σ dp. 3.6 To get the limit as n → ∞ of the above integral, we will need the following lemma, whose proof is deferred to the end of the subsection. Lemma 3.1. Let p ∈ ∂ S f , t 0 and u ∈ B0, 1 be fixed. Suppose that the conditions of Theorem 2.1 hold. Then lim n →∞ C   ∆ n   p + t p nr d n e p   , ∆ n   p + t p nr d n e p + 2r n u     = Φp, t, u, where Φp, t, u is defined in Theorem 2.1. 2625 Returning to the proof of 3.5, we notice that by 4.4 in the Appendix and e.ii we have p nr d n κ p ǫ n → ∞ as n → ∞ and Θp, 0 = 1. Therefore, using Lemma 3.1 and the fact that for all t 0 and u ∈ B0, 1 1 ˜ E n   p + t p nr d n + 2r n u   → 1 as n → ∞, we conclude that the function inside the integral in 3.6 converges pointwise to Φp, t, u as n → ∞. We now proceed to sufficiently bound the function inside the integral in 3.6 to be able to apply the Lebesgue dominated convergence theorem. Towards this goal, fix p ∈ ∂ S f , u ∈ B0, 1 and t ≤ p nr d n κ p ǫ n . Since ∆ n x ≤ 1 for all x ∈ R d , using the inequality |CY 1 , Y 2 | ≤ 2E|Y 1 | whenever |Y 2 | ≤ 1, we have C   ∆ n   p + t p nr d n e p   , ∆ n   p + t p nr d n e p + 2r n u     ≤ 2 E ∆ n   p + t p nr d n e p   . 3.7 By the bound in 4.3 in the Appendix, we see that sup p ∈∂ S f κ p ǫ n → 0 as n → ∞. 3.8 Then, since e p is a normal vector to ∂ S f at p which is directed towards the interior of S f , there exists an integer N independent of p, t and u such that, for all n ≥ N , the point p + t p nr d n e p belongs to the interior of S f . Therefore, f p + t p nr d n e p 0 and, letting ϕ n x = P X ∈ Bx, r n , we obtain E ∆ n   p + t p nr d n e p   = P   π n   p + t p nr d n e p   = 0   = E  P   ∀i ≤ N n : X i ∈ B   p + t p nr d n e p , r n   N n     = E   1 − ϕ n   p + t p nr d n e p     N n = exp   −nϕ n   p + t p nr d n e p     , 3.9 where we used the fact that N n is a mean n Poisson distributed random variable independent of the sample. By a slight adaptation of the proof of Lemma A.1 in Biau, Cadre, and Pelletier 2008 and 2626 under Assumption 1-b, one deduces that for all x ∈ V ∂ S f , ρ ∩ ◦ S f , there exists a quantity K n x such that ϕ n x = r d n ω d f x + r d+2 n K n x and sup n sup x ∈V ∂ S f , ρ∩ ◦ S f |K n x| ∞. 3.10 For all x in V ∂ S f , ρ written as x = p + ue p with p ∈ ∂ S f and 0 ≤ u ≤ ρ, a Taylor expansion of f at p gives the expression f x = 1 2 D 2 e p f p + ξe p u 2 , for some 0 ≤ ξ ≤ u since, by Assumption 1-b, D e p f p = 0. Thus, in our context, expanding f at p, we may write n ϕ n   p + t p nr d n e p   = ω d D 2 e p f p + ξe p t 2 2 + nr d+2 n R n p, t, for some 0 ≤ ξ ≤ κ p ǫ n , and where R n p, t satisfies sup n sup § R n p, t : p ∈ ∂ S f and 0 ≤ t ≤ Æ nr d n κ p ǫ n ª ∞. Furthermore, by 3.8, each point p + ξe p falls in the tubular neighborhood V ∂ S f , ρ for all large enough n. Consequently, by Assumption 2-c there exists α 0 independent of n and N 1 ≥ N independent of p, t and u such that, for all n ≥ N 1 , inf p ∈∂ S f D 2 e p f p + ξe p 2α. This, together with identity 3.9 and r.i, r.iii, which imply nr d+2 n → 0, leads to E ∆ n   p + t p nr d n e p   ≤ C exp−ω d αt 2 3.11 for n ≥ N 1 and all ≤ t ≤ Æ nr d n sup p ∈∂ S f κ p ǫ n . Thus, using inequality 3.11, we deduce that the function on the left hand side of 3.7 is dominated by an integrable function of p, t, u, which is independent of n provided n ≥ N 1 . Finally, we are in a position to apply the Lebesgue dominated convergence theorem, to conclude that lim n →∞ r n r d n Var Π n ǫ n = 2 d Z ∂ S f Z ∞ Z B0,1 Φp, t, ududt v σ dp = σ 2 f . To be complete, it remains to prove Lemma 3.1. Proof of Lemma 3.1 Let x n = p + t p nr d n e p . Since nr d n → ∞ and nr d+2 n → 0, both x n and x n + 2r n u lie in the interior of S f for all large enough n. As a consequence, f x n 0 and f x n + 2r n u 0 for all large enough n. Thus, C ∆ n x n , ∆ n x n + 2r n u 2627 = C 1{π n x n = 0}, 1{π n x n + 2r n u = 0 } = P π n x n = 0, π n x n + 2r n u = 0 − P π n x n = 0 P π n x n + 2r n u = 0 = P ∀i ≤ N n : X i ∈ Bx n , r n ∪ Bx n + 2r n u, r n − P ∀i ≤ N n : X i ∈ Bx n , r n P ∀i ≤ N n : X i ∈ Bx n + 2r n u, r n = exp −nµ Bx n , r n ∪ Bx n + 2r n u, r n − exp −nµBx n , r n − nµBx n + 2r n u, r n , where µ denotes the distribution of X . Let B n = Bx n , r n ∩ Bx n + 2r n u, r n . Using the equality µ Bx n , r n ∪ Bx n + 2r n u, r n = ϕ n x n + ϕ n x n + 2r n u − µB n , we obtain C ∆ n x n , ∆ n x n + 2r n u 3.12 = exp −n ϕ n x n + ϕ n x n + 2r n u exp nµB n − 1 . Now, µB n may be expressed as µB n = f x n λB n + Z B n f v − f x n dv. Since f is of class C 1 on R d , by developing f at x n in the above integral, we obtain Z B n f v − f x n dv = r d+1 n R n , where R n satisfies R n ≤ C sup K kgrad f k, and K is some compact subset of R d containing ∂ S f and of nonempty interior. Next, note that λB n = r d n βu, where βu = λ B0, 1 ∩ B2u, 1 . Therefore, expanding f at p in the direction e p , we obtain µB n = βu t 2 2n D 2 e p f p + ξ t p nr d n e p + r d+1 n R n , where ξ ∈ 0, 1. Hence by r.iii, lim n →∞ n µB n = βuD 2 e p f p t 2 2 . The above limit, together with identity 3.12 and 3.10, leads to the desired result. ƒ 2628

3.2 Central limit theorem for Π