Nonparametric regression estimation onto a Poisson point process covariate

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Nonparametric regression estimation onto a Poisson point process covariate

Benoît Cadre, Lionel Truquet

To cite this version:

Benoît Cadre, Lionel Truquet. Nonparametric regression estimation onto a Poisson point process covariate. ESAIM: Probability and Statistics, EDP Sciences, 2015, 19, pp.251-267.

�10.1051/ps/2014023�. �hal-01715649�

Nonparametric regression estimation onto a Poisson point

process covariate

Benoˆıt CADRE^∗

IRMAR, ENS Rennes, CNRS, UEB Campus de Ker Lann

Avenue Robert Schuman, 35170 Bruz, France cadre@ens-rennes.fr

Lionel TRUQUET

IRMAR, Ensai, CNRS, UEB Campus de Ker Lann

Avenue Robert Schuman, 35170 Bruz, France truquet@ensai.fr

Abstract

Let Y be a real random variable and X be a Poisson point process. We investigate rates of convergence of a nonparametric estimate ˆr(x)of the re- gression functionr(x) =E(Y|X =x), based onnindependent copies of the pair(X,Y). The estimator ˆris constructed using a Wiener-Itˆo decomposition ofr(X). In this infinite-dimensional setting, we first obtain a finite sample bound on the expected squared differenceE(r(X)ˆ −r(X))². Then, under a condition ensuring that the model is genuinely infinite-dimensional, we ob- tain the exact rate of convergence of lnE(r(X)ˆ −r(X))².

Index Terms — Regression estimation, Poisson point process, Wiener-Itˆo decomposition, Rates of convergence.

AMS 2010 Classification— 62G05, 62G08

∗corresponding author

1 Introduction

1.1 Functional regression estimation

LetS be a measurable space, and let the data(X₁,Y₁),· · ·,(X_n,Y_n)be independent S×R-valued random variables with the same distribution as a generic pair(X,Y) such thatE|Y|<∞. In the regression estimation problem, the goal is to estimate the regression functionr(X) =E(Y|X)using the data.

In the classical setting, each covariate X_i is supposed to be a collection of numerical experiments represented by a finite-dimensional vector. Thus, to date, most of the results pertaining to regression estimation have been reported in the finite-dimensional case whereS =R^d. We refer the reader to the book by Gyo¨rfi et al [5] for a comprehensive introduction to the subject and an overview of most standard methods inR^d.

However, in an increasing number of practical applications, input data items take more complicated forms. In the functional data analysis,S is a set of curves and, in this context, regression estimation has many applications in a wide class of problems, among with speech recording, analysis of patients visits in a hospital, price of an option... Last few years have witnessed important developments in both the theory and practise of functional data analysis, and many traditional sta- tistical tools have been adapted to handle functional inputs. The book by Ramsay and Silverman [11] provides a presentation of this area.

In the infinite-dimensional setting, the regression problem is faced with new challenges which requires changing methodology. Curiously, despite a huge re- search activity in the area of infinite-dimensional data analysis, few attempts have been made to connect it with the rich theory of stochastic processes that both pro- vides a wide class of models and powerful tools. This approach, based on a use of stochastic process theory for the benefit of nonparametric estimation, has been studied in the fairly closed problem of supervised classification; in this direction, we refer the reader to the recent papers by Ba`ıllo et al [3], Biau et al [2], Cadre [4].

Among the classical models from the theory of time-dependent stochastic pro- cesses, the case of Poisson processes is of great interest. Here, S is the set of counting paths on a subset of R+. In epidemiology for example, the observed curve X_i represent the dates of the patient visits to the doctor and the variable responseY_iprovide quantitative information on the health status of the patient.

More generally, we consider in this paper the case of a Poisson point process covariate, which corresponds to the situation where each measurement report the

locations of every individual event. In this setting, the state spaceS is identified to the so-called Poisson space over some measurable spaceX, i.e.

S =

n

∑

i=1

δx_i, n∈Nandx_i∈X , whereδ_x stands for the Dirac measure onx.

1.2 Poisson point process regression estimation

In the sequel, the covariate X is a Poisson point process (see [7]) on a domain X⊂R^dwith mean measureµ, whereµ is aσ-finite measure on the Borelσ-field X of X. As seen before, this means X belongs to the space of integer-valued σ-finite measures onXand satisfies:

• for any A∈X, the number X_A of points of X lying in A has a Poisson distribution with parameterµ(A);

• for any family of disjoint setsA₁,· · ·,A_{`</sub>∈X,X<sub>A1</sub>,· · ·,X<sub>A`} are independent random variables.

In the case X=R, Itˆo’s famous chaos expansion (see [6], [13]) says that every square integrable andσ(X)-measurable random variable can be decomposed as a sum of multiple stochastic integrals, called chaos. This result has been generalized by Nualart and Vives [10], and more recently by Last and Penrose [8].

Now recall some basic facts about chaotic decomposition in the Poisson space.

Fix k≥1. Provided g∈L²(µ^⊗k), we can define the k-th chaos I_k(g)associated withg, namely

I_k(g) = Z

∆k

gd(X−µ)^⊗k, (1.1)

where ∆_k ={x∈X^k : x_i6=x_j for all i6= j}. Interestingly, if g∈L²(µ^⊗k) and h∈L²(µ<sup>⊗`</sup>)fork, `≥1, we have

EI_k(g)I_`(h) =k!

Z

X^k

¯

gh¯dµ^⊗k1_{k=`}andEI_k(g) =0, (1.2) where ¯gand ¯hare the symmetrizations ofgandh, that is, for all(x₁,· · ·,x_k)∈X^k:

g(x¯ ₁,· · ·,x_k) = 1 k!

∑

σ

g(x_σ(1),· · ·,x_σ(k)),

the sum being taken over all permutations σ = (σ(1),· · ·,σ(k)) of {1,· · ·,k}, and similarly for ¯h. In particular, note that I_k(g) is a square integrable random variable. In Nualart and Vives ([10], p. 160), it is proved that every square inte- grable σ(X)-measurable random variable can be decomposed as an infinite sum of chaos. Applied to our regression problem, this statement writes as

r(X) =EY+

∑

k≥1

1

k!I_k(f_k), (1.3)

where equality holds inL², providedEY²<∞. In the above formula, each f_k is an element of L²sym(µ^⊗k) –the subset of symmetric functions inL²(µ^⊗k)–, and the decomposition is defined in an unique way. Last and Penrose [8] proved that each f_kcan be expressed as a difference operator of orderk.

Based on independent copies of (X,Y), we shall construct a nonparametric estimate ˆr ofr with the help of decomposition (1.3). Next section is devoted to the model, and to the construction and statistical properties of ˆr. In particular, we obtain a finite sample bound on the mean squared error E(ˆr(X)−r(X))². More- over, we prove that if the model is genuinely infinite-dimensional in the sense that inf_k≥1kf_kk_L2(µ^⊗k)>0 then, under some regularity conditions:

n→∞lim

lnE r(Xˆ )−r(X)2

√lnnln2n =− rα

2d, for someα ∈]0,1[. Last two sections contains proofs.

2 Regression estimate

2.1 Model and heuristic of the estimate

Basic assumptions on the model. We assume throughout thatXis a compact set, say X⊂[−M,M]^d, and the mean measure µ has a density ϕ with respect to the Lebesgue measureλ onXthat is,ϕ :X→R+is such that for all Borel setA⊂X:

EX_A= Z

A

ϕdλ.

Heuristics. In view of a short presentation of the heuristic of the estimate of r, we assume for simplicity thatϕ is a known positive function. Suppose also that

the intensity ϕ and the f_k’s defined by (1.3) are bounded functions. LetW be a bounded density onX,h=h(n)>0 and for allx∈X:

W_h(x) = 1 h^dW x

h

. (2.1)

For a real-valued functiongdefined onX, the notationg^⊗kdenotes the real-valued function onX^ksuch that

g^⊗k(x) =

k

∏

i=1

g(x_i), x= (x₁,· · ·,x_k)∈X^k. By relations (1.2) and (1.3), we have for allx∈X^k andk≥1:

EY I_k W_h^⊗k(x− ·)

= Er(X)I_k W_h^⊗k(x− ·)

=

∑

`≥1

1

`!EI<sub>`</sub>(f_`)I_k W_h^⊗k(x− ·)

= Z

X^k

f_kW¯_h^⊗k(x− ·)ϕ^⊗kdλ^⊗k,

where ¯W_h^⊗k(x− ·)is the symmetrization of the functionW_h^⊗k(x− ·). Since f_kis a symmetric function, we can write

EY I_k W_h^⊗k(x− ·)

= Z

X^k

f_kW_h^⊗k(x− ·)ϕ^⊗kdλ^⊗k.

Thus, under smoothness assumptions on ϕ and f_k, the right-hand side converges to f_k(x)ϕ^⊗k(x), providedh→0. From the left-hand side, we thus deduce that a kernel-type estimator of f_k(x) based on independent copies(X₁,Y₁),· · ·,(X_n,Y_n) of(X,Y)is

f˜_k(x) = 1 n

n i=1

∑

Y_i Z

∆_k

W_h^⊗k(x− ·)

ϕ^⊗k(x) d(X_i−µ)^⊗k. By (1.1), thek-th chaosI_k(f_k)is thus estimated by

I˜_k= Z

∆k

f˜_k(x)d(X−µ)^⊗k, (2.2) which, by (1.3), gives the estimator ˜rofrdefined by

˜

r(X) =Y¯_n+

N

∑

k=1

1

k!I˜_k, (2.3)

whereN=N(n)tends to infinity and, as usual, Y¯_n= 1

n

n

∑

i=1

Y_i.

However, this construction requires the unrealistic assumption thatϕis known.

Thus we shall first present a nonparametric estimator ofϕ, then we shall adapt the previous idea to this context.

2.2 Construction of the estimator

Estimation of ϕ. Construction of the nonparametric estimate of ϕ is based on Mecke’s Formula (see [9]), which states in particular that

E Z

X

γdX= Z

X

γdµ = Z

X

γ ϕdλ,

provided γ : X→Ris inL¹(µ). With this respect, we define the nonparametric estimator ˆϕofϕby

ϕˆ(x) =1 n

n

∑

Z

X

K_b(x,y)X_j(dy), x∈X,

where, for allx= (x₁,· · ·,x_d)andu= (u₁,· · ·,u_d)∈X, K_b(x,u) =

d

∏

J_b(x_j−u_j) +J_b(2M+x_j+u_j) +J_b(2M−x_j−u_j)

. (2.4) In the above formula,Jis a continuous and symmetric density on[−1,1]and, for the bandwidthb=b(n)>0 such thatb→0 asn→∞:

J_b(y) =1 bJ y

b

, y∈[−b,b].

This particular construction is classical in order to avoid bias on the boundary of X(see p. 30 in the book by Silverman [12]). Then, we define for alli=1,· · ·,n the leave-one-out nonparametric estimator ˆϕ_iofϕby

ϕˆ_i(x) = ϕ(x)ˆ −1 n

Z

X

K_b(x,y)X_i(dy)

= 1 n

n j=1,

∑

Z

X

K_b(x,y)X_j(dy),

forx∈X. Leave-one-out procedure is only considered here for technical matters.

Chaos estimate. Now consider the vanishing sequences ρ =ρ(n)>0 and h= h(n)>0. Following (2.2), the leave-one-out estimator of thek-th chaosI_k(f_k)is Iˆ_k, such that

Iˆ_k= 1 n

n

∑

i=1

Y_i Z

∆k

h^Z

∆k

W_h^⊗k(x− ·)

(ϕˆ_i+ρ)^⊗k(x)(dX_i−ϕˆidλ)^⊗ki

dX−ϕˆidλ⊗k

(x), whereW_h is defined by (2.1). In the sequel, we assume for simplicity thatW has a compact support.

Regression estimate.Finally, following the idea drawn by (2.3), the estimator ˆrof ris

r(X) =ˆ Y¯_n+

N k=1

∑

1 k!Iˆ_k, whereN=N(n)tends to infinity.

2.3 Result

In the sequel, k.kis the euclidean norm on anyR^p. We assume that(X,Y)is in- dependent from the sample(X₁,Y₁),· · ·,(X_n,Y_n). Now introduce the assumptions on the model.

H1.Y is a bounded random variable.

H2. inf_Xϕ>0 and there existsL₁>0 such that for allx,y∈X:

|ϕ(x)−ϕ(y)| ≤L₁kx−yk.

H3. There exists a constantL₂>0 such that for allk≥1 andx,y∈X^k:

|f_k(x)−f_k(y)| ≤L^k₂kx−yk.

H4. There exist two constantsC₁,C₂>0 such thatnb^d+2, nh^dN+1andnb^dρ³/N are bounded below byC₁, andb²+ρ≤C₂h^dN+1.

AssumptionsH1-H3are classical in nonparametric estimation. Moreover, it is easily seen thatH4holds forb=n^−γ,ρ=n^−β andh,Ndefined by formulas (2.6)

below, under the additional constraints that 0<γ <1/(d+2), 0<3β <1−dγ andα <min(1,β,2γ). Note that last constraint is sufficient forb²+ρ≤C₂h^dN+1 to hold because, under the conditionu_N/(NlnN)→0 asn→∞:

lnh^dN+1=−αlnn+◦(lnn).

Moreover, since

lnnh^dN+1= (1−α)lnn+◦(lnn),

we see thatα <1 is a sufficient condition fornh^dN+1≥C₁to hold.

For notational simplicity, we set ln₂n=ln(lnn).

Theorem 2.1. Assume thatH1-H4hold. Then, there exists C≥1such that E r(X)ˆ −r(X)2

≤C^N h+ 1

N!

. (2.5)

In particular, for the following choices:

N=hr 2αlnn

dln₂n i

, h=e^−NlnN−uN, (2.6) whereα >0and u_N ≥0is such that u_N/(NlnN)→0as n→∞, we have:

lim sup

n→∞

lnE r(X)ˆ −r(X)2

√lnnln₂n ≤ − rα

2d.

Roughly, Theorem 2.1 entails that the rate of convergence of the mean squared error between ˆr(X)andr(X)is at least

exp

− rα

2dlnnln₂n

.

Note that this rate is obtained for that bandwidthshthat satisfy, asn→∞:

lnh∼ − rα

2dlnnln₂n.

Hence, the general finding here is that the rate of convergence of the mean squared error is much slower than the traditional finite-dimensional rate (see [5]), but faster than the rates obtained by Biau et al [1] in the infinite-dimensional setting. This,

of course, is explained by the fact that we fully exploit the particular nature of the covariate via the chaotic decomposition of square integrable σ(X)-measurable random variables whereas Biau et al [1] utilize a k-nearest neighbor estimator whose construction does not depend on the law ofX.

Our task is now to prove that the rate of Theorem 2.1 is optimal in a gen- uine infinite-dimensional setting. With this goal, we see that it is necessary to strengthen the conditions on the model. Indeed, if all the f_k’s for k≥k₀ have a null L²sym(µ^⊗k)-norm, then rcan be decomposed into a finite sum of chaos (see 1.3), and hence we are faced with a finite-dimensional estimation problem, for which the rate of theorem 2.1 may not be optimal. To avoid this situation, we assume that all the f_k’s have aL²(µ^⊗k)-norm greater than a positive constant.

Theorem 2.2. Assume thatH1-H4hold, and inf_k≥1kf_kk

L²(µ^⊗k)>0. If N and h are given by(2.6)with the additional assumption on u_N that u_N/N→∞, we have

n→∞lim

lnE r(Xˆ )−r(X)2

√lnnln₂n =− rα

2d.

To the best of our knowledge, this is the first exact rate in the regression es- timation problem with a Poisson point process covariate. However, it is an open problem to know whether this rate is optimal over the whole class of regression estimates.

3 Proof of Theorems 2.1 and 2.2

In the sequel,κ>0 is such that sup_Xϕsup_R2dK≤κand we assume for simplicity that constantsC₁,C₂of assumption H4are equal to 1. Moreover, we assume that ρ,bandhare smaller than 1 (recall that they vanish asntends to infinity). Finally, we let for allk≥1:

β_k=ρ^−kexp

−nb^dρ² 4κ

.

We start the section with the following result, whose proof is presented later in this section.

Lemma 3.1. Assume that assumptions H1-H3hold and nb^d+2≥1. Then, there exists C≥1such that for all k≥1:

E Iˆ_k−I_k(f_k)2

≤C^k

(k!)²b²+β_2k

h^dk +h+ k!

nh^dk

.

Proof of Theorem 2.1. In the sequel,C≥1 denotes a constant whose value may change from line to line. According to Jensen Inequality and Lemma 3.1:

E ^N

∑

k=1

1

k! Iˆ_k−I_k(f_k)2

≤ C^N

N

∑

k=1

b²+β_2k h^dk + h

(k!)²+ 1 k!nh^dk

≤ C^N

b²+β_2N

h^dN +h+ 1 nh^dN

. (3.1)

Moreover, according to Theorem 4.2 in Last and Penrose [8], E

k≥N+1

∑

I_k(f_k) k!

2

≤ 1 N!E

Z

X^N+2

D^N+2_x r(X)2

ϕ^⊗(N+2)(x)λ^⊗(N+2)(dx), (3.2) where for allx∈X^N+2,D^N+2_x denotes the difference operator of orderN+2, that is, ifδzis the Dirac mass onz:

D^N+2_x r(X) =

∑

S⊂{1,···,N+2}

(−1)^N+2−|S|r X+

∑

s∈S

δ_xs .

In the formula above,|S|is the number of elements ofS. But|r|is bounded since Y is bounded underH1. Hence,|D^N+2_x r(X)| ≤C2^N+2and, by (3.2):

E

k≥N+1

∑

I_k(f_k) k!

2

≤C^N N!.

Putting all pieces together, we deduce from (3.1) and above that E r(Xˆ )−r(X)2

≤C^N1

n+b²+β_2N

h^dN +h+ 1 nh^dN + 1

N!

,

becauseE(EY−Y¯_n)²≤1/n.Now, under conditionH4which in particular states thatnb^dρ³/N≥1, we have fornlarge enough:

β_2N≤ρ^−2Ne^3Nlnρ ≤ρ. (3.3)

Moreover, sincenh^dN+1≥1 andb²+ρ≤h^dN+1: E r(X)ˆ −r(X)2

≤C^N h+ 1

N!

, (3.4)

hence the first part of the theorem.

Now consider the case whereNandhare given by (2.6). We have

C^Nh=e^{NlnC−NlnN−uN} ≤e^−NlnN+NlnC. (3.5) Furthemore, according to the Stirling Formula,

C^N

N! ≤e^{NlnC+N−NlnN}. Hence by (3.4), we have for allε >0:

lnE r(X)ˆ −r(X)2

≤ ln e^−NlnN+NlnC+e^{−NlnN+N(1+lnC)}

≤ −NlnN+ln e^NlnC+e^N(1+lnC) .

Moreover, by the very definition ofN given in (2.6):

NlnN=hr 2αlnn

dln₂n i

lnhr 2αlnn

dln₂n i

= rα

2dlnnln₂n(1+◦(1)). (3.6) Putting all pieces together yield

lim sup

n→∞

lnE r(X)ˆ −r(X)2

√lnnln₂n ≤ − rα

2d, which is the desired result.

Proof of Theorem 2.2. For simplicity, we assume that inf_k≥1kf_kk

L²(µ^⊗k) ≥1.

Then, by the triangle inequality and (1.2):

q

E r(X)ˆ −r(X)2

= v u u tE

h ^N

∑

k=1

1

k! Iˆ_k−I_k(f_k)

−

∑

k≥N+1

1 k!I_k(f_k)

i2

≥ s

E

k≥N+1

∑

1 k!I_k(f_k)

2

− v u u tE

^N

∑

k=1

1

k! Iˆ_k−I_k(f_k)2

≥ 1

p(N+1)!− v u u tE

^N

∑

k=1

1

k! Iˆ_k−I_k(f_k)2

.

Applying successively (3.1), (3.3) andH4, we first deduce that E

^N

∑

k=1

1

k! Iˆ_k−I_k(f_k)2

≤C^Nh=e^{−NlnN+NlnC−uN}. Moreover, by the Stirling Formula:

1

(N+1)! ≥e−(N+3/2)ln(N+1)+N. Consequently,E(ˆr(X)−r(X)2

is greater than

e−(N/2+3/4)ln(N+1)+N/2−e^{−(N/2)lnN+(N/2)lnC−uN/2} 2

.

Then, using the conditionu_N/N→∞, we get with easy calculations that lnE r(Xˆ )−r(X)2

≥ −NlnN+cN, for some constantc>0. By (3.6), we deduce that

lim inf

n→∞

lnE r(Xˆ )−r(X)2

√lnnln2n ≥ − rα

2d, which, combined with Theorem 2.1, gives the result.

Fixk≥1 and denote for allx,y∈X^k: ˆ

g_k,i(x,y) = W_h^⊗k(x−y)

(ϕˆ_i+ρ)^⊗k(x). (3.7) We also let for alli=1,· · ·,n:

d ˆX_i=dX_i−ϕˆ_idλ, d ¯X_i=dX−ϕˆ_idλ, (3.8) d ˜X_i=dX_i−ϕdλ and d ˜X =dX−ϕdλ. (3.9) With this respect, we have:

Iˆ_k=1 n

n i=1

∑

Y_i Z

∆_k²

ˆ

g_k,i(x,y)Xˆ_i^⊗k(dy)X¯_i^⊗k(dx).

Proof of Lemma 3.1. For simplicity, we shall assume in this proof that |Y| ≤1.

Moreover,C≥1 denotes a constant whose value may change from line to line.

With the help of notations (3.7)-(3.9), we let:

Jˆ₁ = 1 n

n i=1

∑

Y_i Z

∆_k²

ˆ

g_k,i(x,y)X˜_i^⊗k(dy)X˜^⊗k(dx) Jˆ₂ = 1

n

n

∑

i=1

Y_i Z

∆_k²

g_k(x,y)X˜_i^⊗k(dy)X˜^⊗k(dx), where, forx,y∈X^k:

g_k(x,y) =W_h^⊗k(x−y) ϕ^⊗k(x) . Then, since|Y| ≤1, we get by Jensen’s Inequality:

E Iˆ_k−Jˆ₁2

≤ E ^Z

∆_k²

ˆ

g_k,1(x,y)Xˆ₁^⊗k(dy)X¯₁^⊗k(dx)−X˜₁^⊗k(dy)X˜^⊗k(dx)2

≤ 2(R_1k+R_2k),

whereR_1k andR_2kare defined in Lemma 4.4. Hence, E Iˆ_k−Jˆ₁2

≤C^k(k!)²(1+β_2k)b²

h^dk. (3.10)

Moreover, conditioning first by X₁,· · ·,X_n then byX₂,· · ·,X_n, we find with two successive applications of the isometry formula (1.2), that

E Jˆ₁−Jˆ₂2

≤ E ^Z

∆_k²

ˆ

g_k,1(x,y)−g_k(x,y)X˜₁^⊗k(dy)X˜^⊗k(dx)2

≤ (k!)² Z

X^2k

E gˆ_k,1(x,y)−g_k(x,y)2

ϕ^⊗k(x)ϕ^⊗k(y)dxdy.

Thus, using the notation of Lemma 4.3, we get:

E Jˆ₁−Jˆ₂2

≤ C^k(k!)²N_k Z

X^2k

W_h^⊗k(x−y)2

dxdy

≤ C^k(k!)²b²+ρ²+β_2k

h^dk (3.11)

Finally, formula (1.2) gives E Jˆ₂−I_k(f_k)2

= Z

X^k

E 1

n

n

∑

i=1

Z_i(x)−f_k(x)2

ϕ^⊗k(x)dx

= Z

X^k

1

nvar Z₁(x)

+ EZ₁(x)−f_k(x)2

ϕ^⊗k(x)dx,(3.12) where for allx∈X^k andi=1,· · ·,n:

Z_i(x) =Y_i Z

∆k

g_k(x,y)X˜_i^⊗k(dy).

By (1.3) and (1.2):

EZ₁(x) = Er(X) Z

∆k

g_k(x,y)X˜^⊗k(dy) = Z

X^k

f_k(y)g_k(x,y)ϕ^⊗k(y)dy

= 1

ϕ^⊗k(x) Z

X^k

f_k(y)W_h^⊗k(x−y)ϕ^⊗k(y)dy.

Then, easy calculations prove that, sinceW has a compact support:

Z

X^k

EZ₁(x)−f_k(x)2

ϕ^⊗k(x)dx≤C^kh. (3.13) Moreover, by (1.2):

var Z₁(x)

≤ E ^Z

∆k

g_k(x,y)X^⊗k(dy)2

≤ k!

Z

X^k

g²_k(x,y)ϕ^⊗k(y)dy

≤ k!

[ϕ^⊗k(x)]² Z

X^k

W_h^⊗k(x−y)2

ϕ^⊗k(y)dy.

Hence,

Z

X^k

var Z₁(x)

ϕ^⊗k(x)dx≤k!C^k

h^dk. (3.14)

We can conclude with (3.12)-(3.14) that E Jˆ₂−I_k(f_k)2

≤C^k

h+ k!

nh^dk

.

Finally, the lemma is then a consequence of (3.10), (3.11) and above, sinceband ρ vanish.

4 Auxiliary results

4.1 Intensity estimation

Recall that

β_k=ρ^−kexp

−nb^dρ² 4κ

. whereκ>0 is such that sup_Xϕsup_R2dK≤κ.

Lemma 4.1. AssumeH2holds and nb^d+2≥1. Then, there exists C≥1such that for all k≥1and x∈X:

(i) E|ϕˆ₁(x)−ϕ(x)|^k≤C^kb^k (ii) E

1

ϕˆ₁(x) +ρ − 1 ϕ(x)

k≤C^k b^k+ρ^k+β_k .

Proof. First we compute the bias of ˆϕ₁(x). For simplicity, we assumed=1 only for the computation of the biais. By the very definition of ˆϕ₁(x) (see 2.4 and below), we have

Eϕˆ₁(x) = Z _M

−M

K_b(x,y)ϕ(y)dy

=

Z _(x+M)/b

(x−M)/b J(z)ϕ(x−bz)dz+

Z _(3M+x)/b

(M+x)/b

J(z)ϕ(bz−2M−x)dz +

Z _(3M−x)/b

(M−x)/b J(z)ϕ(2M−x−bz)dz.

Distinguishing the cases x∈[−M,−M+b], x∈[−M+b,M−b] and x∈[M− b,M], it is an easy exercise to prove that under the Lipschitz conditionH2onϕ, we have

|Eϕˆ₁(x)−ϕ(x)| ≤3bL₁ Z ₁

−1

(ky|+2)J(y)dy≤Cb, (4.1) where, here and in the following,C≥1 is a constant that does not depend onx,k andn, and may change from line to line.

Our second task is to give an exponential inequality for the deviation proba- bility of ˆϕ₁(x). Fixα >0 and, for simplicity, writeF_x(y) =K_b(x,y)fory∈X. We have by independence, for alls>0:

P ϕˆ₁(x)−Eϕˆ₁(x)≥α

= P

exp s

n

∑

i=2

Z

X

F_x(dX_i−ϕdλ)

≥e^αsn

≤ e^−αsn

Eexp s

Z

X

F_x(dX₁−ϕdλ)n−1

,

using Markov Inequality. By Campbell Inequality (see [7]), we thus have P ϕˆ₁(x)−Eϕˆ₁(x)≥α

≤exp

−αsn+ (n−1) Z

X

e^sFx−sF_x−1 ϕdλ

. (4.2) But, if K₊,ϕ₊ >0 are constants such that K(x,y)≤K₊ and ϕ(x)≤ϕ₊ for all x,y∈X, we have:

Z

X

e^sFx−sF_x−1

ϕdλ ≤

∑

j≥2

s^j j!

Z

X

F_x^jϕdλ

≤

∑

j≥2

s^j j!

K₊^j−1ϕ₊ b^d(j−1)

≤ ϕ+b^d K₊

exp sK₊ b^d

−sK₊ b^d −1

.

Letting δ the function defined for all t ≥0 by δ(t) =t−(1+t)ln(1+t) and choosingsso that

s= b^d

K₊ln 1+ α ϕ₊

,

we have by (4.2) and above:

P ϕˆ₁(x)−Eϕˆ₁(x)≥α

≤ expnb^dϕ+

K₊ δ α ϕ₊

− α

ϕ₊ +b^dϕ+

K₊ ln 1+ α ϕ₊

≤ 1+ α ϕ+

expnb^dϕ₊ K₊ δ α

ϕ+

,

for n large enough, since b vanishes. Considering −F_x instead of F_x, we can conclude that

P

ϕˆ1(x)−Eϕˆ1(x) ≥α

≤2 1+ α ϕ+

exp

nb^dϕ+

K₊ δ α ϕ+

. (4.3)

We are now in a position to establish(i)and(ii). Regarding(i), we have by (4.1):

E

ϕˆ1(x)−ϕ(x)

k≤2^k C^kb^k+E

ϕˆ1(x)−Eϕˆ1(x)

k . Moreover, writing

E

ϕˆ₁(x)−Eϕˆ₁(x)

k= Z _∞

0 P

ϕˆ₁(x)−Eϕˆ₁(x)

≥u^1/k du

and observing that, sinceδ(t)is smaller than−t²/4 or−t/4, depending ont≤1 or not, we deduce from (4.3) and an obvious decomposition of the above integral that :

E

ϕˆ₁(x)−Eϕˆ₁(x)

k≤ C

√ nb^d

k

. Putting all pieces together gives(i), sincenb^d+2≥1.

Next we prove(ii). Note that, according to (4.1):

1

Eϕˆ₁(x)− 1 ϕ(x)

≤ Cb ϕ₋(ϕ₋−Cb). Sincebvanishes asntends to infinity, we thus have

1

Eϕˆ₁(x)− 1 ϕ(x)

≤Cb. (4.4)

Let now A={|ϕˆ1(x)−Eϕˆ1(x)| ≥ρ}. Since inf_Xϕ>0 by assumption, we have according to (4.1):

E

1

ϕˆ₁(x) +ρ− 1 Eϕ(x)ˆ

k ≤ E

1

ϕˆ₁(x) +ρ − 1 Eϕ(x)ˆ

k

1_A^c +E

1

ϕˆ₁(x) +ρ − 1 Eϕ(x)ˆ

k

1_A

≤ C^k ρ^k+ρ^−kP(A)

≤ C^k

ρ^k+ρ^−kexp −nb^dρ² 4κ

.

Last inequality is a consequence of (4.3),ρ→0 asn→∞and the fact thatδ(t)≤

−t²/4 provided t > 0 is small enough. We can now conclude with the above

inequality and (4.4).

4.2 Perturbated chaos

Lemma 4.2. Let g∈L²(µ^⊗k)andψ ∈L²(λ). Ifdν=ψdλ, we have E

^Z

∆k

gd

(X−ν)^⊗k−(X−µ)^⊗k2

≤

k−1

∑

i=0

i!

k i

2

kϕ−ψk^2(k−i)₂ Z

X^k

g(x)²

i

∏

ϕ(x_j)λ^⊗k(dx).

Proof. For simplicity of the proof, we assume that g is a symmetric function.

Otherwise, one only needs to consider its symmetrized version. First observe that by symmetry ofg:

Z

∆_k

gd(X−ν)^⊗k=

k i=0

∑

k i

_Z

∆_k

gd(X−µ)^⊗id(µ−ν)^⊗k−i. Consequently,

D =

Z

∆_k

gd(X−ν)^⊗k− Z

∆_k

gd(X−µ)^⊗k

=

k−1 i=0

∑

k i

_Z

∆_k

gd(µ−ν)^⊗k−id(X−µ)^⊗i. Hence, letting fori=1,· · ·,kandx∈Xⁱ,

Γ(x₁,· · ·,x_i) =

y∈∆_k−i : y<sub>`</sub>∈ {x/ <sub>1</sub>,· · ·,x<sub>i</sub>}for all`=1,· · ·,k−i , we have:

D= Z

X^k

g(ϕ−ψ)^⊗kdλ^⊗k+

k−1

∑

i=1

k i

_Z

∆i

g_id(X−µ)^⊗i, where for(x₁,· · ·,x_i)∈∆_i,

g_i(x₁,· · ·,x_i) = Z

Γ(x₁,···,xi)

g(x₁,· · ·,x_k)

k j=i+1

∏

(ϕ−ψ)(x_j)λ(dx_j).

Observe that by Cauchy-Schwarz, g_i(x₁,· · ·,x_i)²≤ kϕ−ψk^2(k−i)₂

Z

X^k−i

g(x₁,· · ·,x_k)λ(dx_i+1)· · ·λ(dx_k).

Then, since eachg_iis symmetric, we deduce from equations (1.1) and (1.2) that ED² = ^Z

X^k

g(ϕ−ψ)^⊗kdλ^⊗k2

+

k−1

∑

i=1

i!

k i

2Z

Xⁱ

g²_idµ^⊗i

≤

k−1 i=0

∑

i!

k i

2

kϕ−ψk^2(k−i)₂ Z

X^k

g(x)²

i

∏

ϕ(x_j)λ^⊗k(dx), hence the lemma.

4.3 Technical inequalities

Lemma 4.3. AssumeH2holds and nb^d+2≥1. There exists C≥1such that for all k, `,n≥1:

(i) M_k,`= sup

x∈X^k

E

kϕˆ1−ϕk^`₂ (ϕˆ1+ρ)^⊗k(x)

2

≤C^k(1+β_2k)b^2`; (ii) N_k= sup

x∈X^k

E

1

(ϕˆ₁+ρ)^⊗k(x)− 1 ϕ^⊗k(x)

2≤C^k b²+ρ²+β_2k .

Proof. We only prove(i). First observe that by Cauchy-Schwarz and since Xis bounded:

M<sub>k,`</sub><sup>2</sup> ≤ Ekϕˆ<sub>1</sub>−ϕk<sup>4`</sup>₂ sup

x∈X^k

E (ϕˆ₁+ρ)^⊗k(x)−4

≤ C^ksup

x∈X

E|ϕˆ₁(x)−ϕ(x)|^4`sup

x∈X^k

E (ϕˆ₁+ρ)^⊗k(x)−4

,

where, here and in the following,Cis a positive constant that does not depend on k, `andnand may change from line to line. Hence by Lemma 4.1:

M<sub>k,`</sub><sup>2</sup> ≤C<sup>k</sup>b<sup>4`</sup>sup

x∈X

E (ϕˆ₁+ρ)^⊗k(x)−4

. (4.5)

Thus, we only need to consider the rightmost term. Fixx∈X^k, and note that E (ϕˆ₁+ρ)^⊗k(x)−4

≤ 8

(ϕ^⊗k(x))⁴+8E

1

(ϕˆ₁+ρ)^⊗k(x)− 1 ϕ^⊗k(x)

4

. (4.6) The task is to bound the term

A=E

1

(ϕˆ1+ρ)^⊗k(x)− 1 ϕ^⊗k(x)

4

.

We shall make use of the following inequality:

k

∏

a_i−

k

∏

b_i

4≤16^k

∑

/06=I⊂{1,···,k}

∏

i∈I

|a_i−b_i|⁴

∏

i/∈I

b⁴_i,

where the a_i’s and theb_i’s are positive real numbers. Sinceϕ is bounded below by a positive constant:

A ≤ C^k

∑

/06=I⊂{1,···,k}

E

∏

i∈I

1

ϕˆ₁(x_i) +ρ − 1 ϕ(x_i)

4

≤ C^k

∑

/06=I⊂{1,···,k}

∏

i∈I

E

1 ˆ

ϕ₁(x_i) +ρ − 1 ϕ(x_i)

4|I|1/|I|

,

according to H¨older Inequality, and where|I|is the cardinality of the setI. Thus, by Lemma 4.1:

A ≤ C^k

k

∑

k j

sup

x∈X

E

1

ϕˆ₁(x) +ρ − 1 ϕ(x)

4j

≤ C^k

k

∑

k j

b^4j+ρ^4j+ρ^−4jexp −nb^dρ² 4κ

≤ C^k

1+ρ^−4kexp −nb^dρ² 4κ

.

becausebandρvanishes asn→∞. Assertion(i)is then a straightforward conse- quence of inequalities (4.5) and (4.6).

Before statement of next lemma, we recall the notations (3.7)-(3.9).

Lemma 4.4. AssumeH2andH3hold, and nb^d+2≥1. Then, there exists a con- stant C≥1such that for all k,n≥1, both quantities above:

R_1k = E ^Z

∆_k²

ˆ

g_k,1(x,y)Xˆ₁^⊗k(dy) X¯₁^⊗k(dx)−X˜^⊗k(dx)2

R_2k = E ^Z

∆_k²

ˆ

g_k,1(x,y) Xˆ₁^⊗k(dy)−X˜₁^⊗k(dy)X˜^⊗k(dx)2

are bounded by

C^k(k!)²(1+β_2k) b² h^dk.

Proof. We only prove the bound forR_1k, other proof being similar. In the sequel, C≥1 is a constant that does not depend onnandk, and may change from line to

line. Writing (see notations (3.7)-(3.9)):

R_1k=E E hn^Z

∆k

^Z

∆k

ˆ

g_k,1(x,y)Xˆ₁^⊗k(dy)

X¯₁^⊗k(dx)−X˜^⊗k(dx)o2

X₁,· · ·,X_n i

, we obtain with Lemma 4.2, using the independence ofX andX₁,· · ·,X_n and the fact thatϕ is a bounded function:

R_1k≤C^k

k−1

∑

i=0

i!

k i

2

E h

V^k−i Z

X^k

^Z

∆k

ˆ

g_k,1(x,y)Xˆ₁^⊗k(dy)2

λ^⊗k(dx)i

, (4.7) whereV =kϕˆ1−ϕk²₂. Now fixx∈X^kandi=0,· · ·,k−1. We have

EV^{k−i Z}

∆_k

ˆ

g_k,1(x,y)Xˆ₁^⊗k(dy) 2

≤2(A₁+A₂), (4.8) where

A₁ = EV^{k−i Z}

∆_k

ˆ

g_k,1(x,y)X˜₁^⊗k(dy)2

, andA₂ = EV^k−i

^Z

∆_k

ˆ

g_k,1(x,y) Xˆ₁^⊗k(dy)−X˜₁^⊗k(dy)2

.

We proceed to bound A₂. As before, we apply Lemma 4.2, but conditionally on X₂,· · ·,X_n. Hence, sinceϕ,XandW are bounded:

A₂ ≤ C^k

k−1

∑

j!

k j

2

EV^2k−i−j Z

X^k

ˆ

g²_k,1(x,y)λ^⊗k(dy)

≤ C^k h^dk

k−1

∑

j=0

j!

k j

2

E

V^2k−i−j (ϕˆ1+ρ)^⊗k(x)2. Consequently, by Lemma 4.3:

A₂≤ C^k

h^dk(1+β_2k)

k−1

∑

j=0

j!

k j

2

b^2(k−i−j). (4.9)

In a similar fashion, we get by conditioning and (1.2):

A₁ ≤ C^k

h^dkk!EV^k−i Z

X^k

ˆ

g²_k,1(x,y)λ^⊗k(dy)

≤ C^kk!(1+β_2k)b^2(k−i).

Thus, by (4.7)-(4.9) and above:

R_1k ≤ C^k

h^dk(1+β_2k)

k−1 i=0

∑

i!

k i

2

k!b^2(k−i)+

k−1 j=0

∑

j!

k j

2

b^{2(2k−i−j)}

≤ C^k(k!)²(1+β_2k)b² h^dk, hence the lemma.

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