C Durot

Y Rozenholc

AN ADAPTIVE TEST FOR ZERO MEAN

Keywords: adaptive test, minimax hypothesis testing, nonparametric alternatives, symmetrization, heteroscedasticity. 2000 Mathematics Subject Classification: 62G10, 62G08

Assume we observe a random vector y of R n and write y = f + ε, where f is the expectation of y and ε is an unobservable centered random vector. The aim of this paper is to build a new test for the null hypothesis that f = 0 under as few assumptions as possible on f and ε. The proposed test is nonparametric (no prior assumption on f is needed) and nonasymptotic. It has the prescribed level α under the only assumption that the components of ε are mutually independent, almost surely different from zero and with symmetric distribution. Its power is described in a general setting and also in the regression setting, where f i = F (x i ) for an unknown regression function F and fixed design points x i ∈ [0, 1]. The test is shown to be adaptive with respect to Hölderian smoothness in the regression setting under mild assumptions on ε. In particular, we prove adaptive properties when the ε i 's are not assumed Gaussian nor identically distributed.

Introduction

Assume we observe a random vector y ∈ R n and write y = f + ε, where f is the unknown expectation of y and ε = (ε 1 , . . . , ε n ) T is an unobservable random vector with mean zero. Assume the ε i 's are mutually independent with symmetric distribution (we mean that for every i, ε i has the same distribution as -ε i ). Our aim is to build a nonasymptotic test for the null hypothesis H 0 : f = 0 against the nonparametric composite alternative H 1 : f = 0, when no prior assumption is made on f and as few as possible assumptions are made on ε. In particular, we do not assume a priori that f belongs to a given smoothness class, and the ε i 's are not assumed identically distributed nor Gaussian. The proposed test is defined as a multi-test and is based on a symmetrization principle that exploits the symmetry assumption.

The problem of hypothesis testing in the general model y = f + ε has rarely been addressed in the literature. [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF] consider the problem of testing the null hypothesis that f belongs to a given linear subspace of R n . No prior information on f is required, but they assume that the components of ε are independent with the same Gaussian distribution. Under the same assumptions, [START_REF] Baraud | Testing convex hypotheses on the mean of a Gaussian vector. Application to testing qualitative hypotheses on a regression function[END_REF] propose a test for the null hypothesis that f belongs to a given convex subset of R n .

An interesting particular case of the general model above is the regression model, which is obtained in the case where f i = F (x i ) for all i = 1, . . . , n. Here, the x i 's are nonrandom design points and F is an unknown function. In this context, many tests have been proposed for the null hypothesis that F belongs to a given set F against a nonparametric alternative. Typically, F is a parametric set of functions or F is restricted to a single function, which amounts to F = {0}. These tests are often based on a distance between a nonparametric estimator for F and an estimator for F that is computed under the null hypothesis, see, e.g., [START_REF] Müller | Goodness-of-fit diagnostics for regression models[END_REF]. This requires the choice of a smoothing parameter such as a bandwidth. Other tests consider the smoothing parameter itself as a test statistic, see, e.g., [START_REF] Eubank | Testing goodness-of-fit in regression via order selection criteria[END_REF]. Another approach consists in building a test statistic as a function of estimators for the Fourier coefficients of F , see, e.g., [START_REF] Chen | Testing for no effect in nonparametric regression via spline smoothing techniques[END_REF]. We refer to the book of [START_REF] Hart | Nonparametric Smoothing and Lack-of-Fit Tests[END_REF] for a review of these methods.

Recently, the problem of adaptive minimax testing has been addressed. Suppose that the null hypothesis is F ≡ 0 and consider the alternative that F is bounded away from zero in the L 2 -norm, F 2 ≥ ρ(n), and possesses smoothness properties. The minimal rate of testing (that is the minimal distance ρ(n) for which testing with prescribed error probabilities is still possible) has been first derived in a white noise model with signal F . The result depends heavily on the kind of smoothness imposed, see Ingster (1982,1993) and [START_REF] Ermakov | Minimax detection of a signal in a Gaussian white noise[END_REF] for Sobolev smoothness and [START_REF] Lepski | Minimax nonparametric hypothesis testing: the case of an inhomogeneous alternative[END_REF] for Besov smoothness. The optimal rate and the structure of optimal tests depend on the smoothness parameters, whereas these parameters are usually unknown in practical applications. In the white noise model, [START_REF] Spokoiny | Adaptive hypothesis testing using wavelets[END_REF] proves that the minimal rate of testing for H 0 : F ≡ 0 is altered by a log log n factor when F belongs to a Besov functional class with unknown parameters. He builds a rate optimal adaptive test based on wavelets. [START_REF] Gayraud | Adaptive minimax testing in the discrete regression scheme[END_REF] obtain similar results in a regression model for a composite null hypothesis under Hölderian smoothness. They prove that, in the Gaussian model, the optimal rate of testing is altered by a log log n factor if the smoothness parameter is unknown. They build an adaptive test which achieves the optimal rate over a class of Hölderian functions with smoothness parameter s > 1/4 in a possibly non-Gaussian model. Other examples of adaptive tests are given by [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF], [START_REF] Härdle | Testing a regression model when we have smooth alternatives in mind[END_REF] and by [START_REF] Horowitz | An adaptive, rate-optimal test of a parametric mean-regression model against a nonparametric alternative[END_REF]. All these adaptive tests are defined as multi-tests. Roughly speaking, the authors first build a test T s that is minimax for a fixed smoothness parameter s and reject the null hypothesis if there exists s in a given grid such that T s rejects.

Most of non-adaptive tests for zero mean involve a smoothing parameter which is chosen in a somewhat arbitrary way [START_REF] Staniswalis | Diagnostics for assessing regression models[END_REF]. Adaptive tests do not show this drawback but either they require the errors to be i.i.d. Gaussian [START_REF] Härdle | Testing a regression model when we have smooth alternatives in mind[END_REF]Kneip, 1999, Baraud et al., 2003) or they are asymptotic [START_REF] Horowitz | An adaptive, rate-optimal test of a parametric mean-regression model against a nonparametric alternative[END_REF]Spokoiny, 2001, Gayraud and[START_REF] Gayraud | Adaptive minimax testing in the discrete regression scheme[END_REF]. Moreover, existing tests involve estimation of the unknown variance. This requires either homoscedasticity [START_REF] Gayraud | Adaptive minimax testing in the discrete regression scheme[END_REF] or regularity assumptions on the variance [START_REF] Horowitz | An adaptive, rate-optimal test of a parametric mean-regression model against a nonparametric alternative[END_REF]. On the contrary, our test requires very mild assumptions on the errors, is nonasymptotic and needs neither variance estimation nor arbitrary choice of a parameter. It has the prescribed level under the only assumption that the ε i 's are mutually independent with symmetric distributions (the ε i 's may have different distributions). Moreover, it achieves the optimal rate of testing over the class of Hölderian functions with smoothness parameter s > 1/4 in the case where the ε i 's satisfy a Bernstein-type condition, and in particular, in the homoscedastic Gaussian case. The test still achieves the optimal rate of testing for s ≥ 1/4 + 1/p in the case where the ε i 's possess bounded moments of order 2p for some p ≥ 2.

The paper is organized as follows. The testing procedure is described in Section 2. It is also stated in this section that the proposed test has the prescribed nonasymptotic level. In Section 3, we discuss implementation of the test. The power is studied in Section 4 under various assumptions on the ε i 's. In Section 5, we compute the rate of testing of the test in a regression model under a Hölderian assumption. A simulation study is reported in Section 6 and the proofs are given in Section 7.

The Testing Procedure

Assume we observe a random vector y of R n and write

y = f + ε,
where f is an unknown vector of R n and ε is an unobservable random vector with mean zero. Assume that the components of ε are independent and possess a symmetric distribution around zero, which means that for all i, ε i and -ε i have the same distribution. Assume furthermore that for all i, ε i is almost surely different from zero. Our aim is to build a test with nonasymptotic level α for the hypothesis H 0 : f = 0 against H 1 : f = 0. Here, α is a fixed number in (0, 1). The test is based on a symmetrization principle that exploits the symmetry assumption. Before describing the test more precisely, let us introduce some notation.

Notation:

• For every set A, let |A| denote the cardinality of A and let 1 A be the indicator function of A, which means that 1 A (x) equals 1 if x ∈ A and 0 otherwise.

• Let w be a random vector of R n , independent of y, with independent components w i distributed as random signs:

P (w i = 1) = P (w i = -1) = 1/2. • For all u, v ∈ R n , let u × v be the vector of R n with ith component (u × v) i = u i v i .
• Let • n be the Euclidean norm in R n .

• For every partition m of {1, . . . , n} into D m nonempty subsets e m,1 , . . . , e m,Dm , let t m,j (j = 1, . . . , D m ) be the vector of R n with ith component 1 em,j (i) and let Π m be the orthogonal projector onto the linear span of {t m,1 , . . . , t m,Dm }. Thus for all u ∈ R n and i = 1, . . . , n

(Π m u) i = Dm j=1 1 |e m,j | k∈em,j u k 1 em,j (i), Π m u 2 n = Dm j=1 1 |e m,j | k∈em,j u k 2 .
The test is based on the following heuristics. Under H 0 , y has the same distribution as w ×y. Hence for every partition m of {1, . . . , n}, Π m y 2 n and Π m (w ×y) 2 n have the same distribution. Under H 1 , as f = 0, consider an (unobservable) partition m such that for every e ∈ m the numbers (f i ) i∈e all have the same sign.

Then for every e ∈ m,

| i∈e f i | ≥ | i∈e w i f i | with a strict inequality if w i = -1 and f i = 0 for some i ∈ e, hence one has Π m f 2 n > Π m (w × f ) 2
n provided w i = -1 and f i = 0 for some i ∈ {1, . . . , n}. Thus under H 1 there exists m such that Π m y 2 n tends to be larger than Π m (w × y) 2 n , whereas under H 0 , Π m y 2 n and Π m (w × y) 2 n are of the same order of magnitude for every m. Therefore, we propose to reject H 0 if there exists m such that Π m y 2 n exceeds a given quantile of Π m (w × y) 2 n . Since the quantiles of Π m (w × y) 2 n cannot be computed (they depend on the unknown distribution of y) we consider conditional quantiles given y.

The precise construction of the test is as follows. Consider a collection of partitions M and positive numbers α m with m∈M α m = α. We reject H 0 if there exists m ∈ M such that Π m y 2 n exceeds the y-conditional quantile of Π m (w×y) 2 n defined by

q y m (α m ) = inf x ∈ R, P Π m (w × y) 2 n > x | y ≤ α m .
The critical region of our test is thus

(1) sup m∈M Π m y 2 n -q y m (α m ) > 0.
Note that q y m (α m ) can theoretically be computed since it only depends on the known distribution of w, but its exact calculus requires about 2 n computations which cannot be performed in practice. Hence we estimate it through Monte-Carlo simulations, see Section 3. It is stated in the following theorem that the test has a nonasymptotic level α.

Theorem 1. Assume we observe y = f + ε, where f ∈ R n and the ε i 's are independent random variables with symmetric distribution. Assume P (ε i = 0) = 0 for all i = 1, . . . , n. Let M be a collection of partitions of {1, . . . , n}, let α and (α m ) m∈M be positive numbers such that α = m∈M α m . Then

P H0 sup m∈M { Π m y 2 n -q y m (α m )} > 0 ≤ α.

Remarks.

• Like the adaptive tests mentioned in the introduction, our test is a multi-test: the null hypothesis is rejected if one of the tests with critical regions { Π m y 2 n > q y m (α m )} rejects. A given partition m allows us to detect alternatives with a given smoothness s, and adaptive properties arise from the use of many partitions.

• Our method allows us to test the more general null hypothesis H 0 : f ∈ V , where V is a linear subspace of R n , but for this problem, we need the ε i 's to be i.i.d. Gaussian. Let V be a linear subspace of R n with dimension k < n, let Π be the orthogonal projector onto V ⊥ and let ∆ be the diagonal matrix of which (nk) first diagonal components equal 1 and the others are zero. There exists an orthogonal matrix O with Π = O T ∆O. Now, let y V be the random vector that consists of the (nk) first components of ∆Oy = OΠy, let f V be the expectation of y V and let ε V = y Vf V . If the ε i 's are i.i.d. Gaussian, so are the components of ε V . Since testing f ∈ V amounts to testing f V = 0 within the model y V = f V + ε V , this may be done by applying our method. Then the theoretical results we obtain in the i.i.d. Gaussian model for H 0 : f = 0 (see Sections 4 and 5) can be generalized to H 0 : f ∈ V . We do not detail these results.

Practical Implementation

By definition, q y m (α m ) is the 1α m quantile of the discrete distribution with support Π m (u × y) 2 , u ∈ {-1, 1} n , which puts mass k2 -n at a given point x of the above set if there exist k vectors u ∈ {-1, 1} n with Π m (u × y) 2 = x. The exact calculation of q y m (α m ) thus requires about 2 n computations, which cannot be performed in practice. We thus suggest estimation of this quantile instead of computing its exact value. The practical implementation of the test then is as follows. Draw independent random vectors w 1 , . . . , w B which all are distributed like w and are independent of y. For every m ∈ M, compute the empirical quantile defined by

q B m (α m ) = inf x ∈ R, 1 B B b=1 1 Πm(w b ×y) 2 n >x ≤ α m and reject H 0 if (2) sup m∈M Π m y 2 n -q B m (α m ) > 0.
It is stated in the following theorem that the test with critical region (2) has asymptotically level α if B → ∞. Therefore, it suffices to choose B large enough so that the level of this test is close to α. But our aim is to consider quite moderate B so we also provide in Theorem 2 a control of the level in terms of α and B, under the additional assumption that the distributions of the ε i 's are continuous. This result provides a control of what we lose in terms of first kind error probability when we consider the critical region (2) instead of (1).

Theorem 2. Under the assumptions of Theorem 1,

lim B→∞ P H0 sup m∈M { Π m y 2 n -q B m (α m )} > 0 ≤ α.
If moreover the distributions of the ε i 's are continuous, then

P H0 sup m∈M { Π m y 2 n -q B m (α m )} > 0 ≤ α + m∈M 2 Dm-n + |M| √ π 2 √ 2B .
By Theorem 2, it suffices to choose B = c|M| 2 with a large enough c > 0 so that the level is close to α. The estimation of q y m (α m ) then requires about n |M| 2 computations. In practical applications, one can consider, for instance, the collection of dyadic partitions M d defined in Section 5. The cardinality of M d is about log 2 n hence, using B with order of magnitude (log 2 n) 2 , we get a level close to the nominal level α in about n(log 2 n) 2 computations.

The following theorem describes what we lose in terms of second kind error probability when we consider the critical region (2) instead of (1).

Theorem 3. For every m ∈ M, let δ m < α m . Under the assumptions of Theorem 1, we have for every

f ∈ R n lim B→∞ P f sup m∈M { Π m y 2 n -q B m (α m )} > 0 ≥ P f sup m∈M { Π m y 2 n -q y m (δ m )} > 0 .

Power

In this section, we study the power of the test with critical region [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF]. By Theorem 3, similar results hold for the test with critical region (2) provided B is large enough. Let M be a collection of partitions of {1, . . . , n}. Let α and β be fixed numbers in (0, 1). Let (α m ) m∈M be positive numbers such that α = m∈M α m . The aim of this section is to describe a subset F n (β) of R n \ {0} over which the power of the test is greater than 1β, i.e., which satisfies

(3) P f sup m∈M { Π m y 2 n -q y m (α m )} > 0 ≥ 1 -β, for all f ∈ F n (β).
For every partition m, the subsets e ∈ m which contain only one point do not contribute to the power of the test since their contributions to the norms of Π m y and Π m (w × y) are identical. Hence we restrict ourselves to collections M which do not contain what we call the trivial partition, that is the partition made up of n singletons. For every m ∈ M, we set We study the power of the test under two different assumptions on the ε i 's integrability. First, we assume that the ε i 's satisfy the following Bernstein-type condition: there exist positive real numbers γ and µ such that for all integers p ≥ 1, [START_REF] Chen | Testing for no effect in nonparametric regression via spline smoothing techniques[END_REF] max

1≤i≤n E(ε 2p i ) ≤ γp!µ p-2 .
Thus the ε i 's possess bounded moments of any order and ε 2 i possesses an exponential moment, hence the errors are much integrable. Note however that this assumption is less restrictive than the Gaussian assumption: if the ε i 's are Gaussian with bounded variance σ 2 i ≤ σ 2 , then E(ε 2p i ) ≤ p!(2σ 2 ) p , so (4) holds with µ = 2σ 2 and γ = µ 2 . It is also less restrictive than the boundedness assumption: if |ε i | ≤ c for all i, then (4) holds with µ = c 2 and γ = µ 2 . Secondly, we assume there exist p ≥ 2 and µ > 0 such that [START_REF] Ermakov | Minimax detection of a signal in a Gaussian white noise[END_REF] max

1≤i≤n E(ε 2p i ) ≤ µ p .
Note that under both assumptions, the ε i 's possess bounded variances and fourth moments: E(ε 2 i ) ≤ σ 2 and E(ε 4 i ) ≤ µ 4 for all i and some positive σ and µ 4 . It is also assumed in the sequel that |f i | ≤ L for all i = 1, . . . , n and a possibly unknown L > 0.

4.1. Power under the Bernstein-type condition. Theorem 4. Assume we observe y = f + ε, where the ε i 's are independent variables with symmetric distribution. Assume P (ε i = 0) = 0 and |f i | ≤ L for a positive number L and all i. Assume moreover (4) holds for all integers p ≥ 1 and some positive γ and µ. Let M be a collection of partitions which does not contain the trivial one. Let α, β in (0, 1) and (α m ) m∈M be positive numbers with

m∈M α m = α. For every A = (A 1 , . . . , A 5 ) ∈ (0, ∞) 5 , m ∈ M and f ∈ R n let ∆ 1 (m, f, A) = A 1 f -Π m f 2 n + A 2 γD m β (6) + A 3 ( √ γ + µ + L 2 ) 1 + 1 I m log 2D m β log 2 βα m + A 4 ( √ γ + µ + L 2 ) D m 1 + 1 I m log 2D m β log 2 βα m + A 5 L 2 |J m,1 |.
Then there exists an absolute A such that (3) holds with

F n (β) = f ∈ R n , f 2 n ≥ inf m∈M ∆ 1 (m, f, A) .
Hence our test is powerful over F n (β) provided the constants A 1 , . . . , A 5 are large enough. This set is large if there exists m ∈ M such that Π m f is close to f , D m and |J m,1 | are small, while I m is large enough. This result applies to the i.i.d. Gaussian model. Assume that the ε i 's are i.i.d. N (0, σ 2 ) and in order to make appear the signal to noise ratio, assume that there exists L > 0 such that |f i | ≤ σL for all i. Then (4) holds with µ = 2σ 2 and γ = µ 2 . Theorem 4 consequently applies with √ γ and √ γ + µ + L 2 replaced by σ 2 and σ 2 (1 + L 2 ) respectively. However (see Section 7.7), one can obtain a slightly sharper result by using Cochran's theorem: the power of the test is greater than 1β as soon as

(7) f 2 n ≥ inf m∈M A 1 f -Π m f 2 n + A 3 σ 2 (1 + L 2 ) 1 + 1 I m log 2D m β log 2 βα m + A 4 σ 2 (1 + L 2 ) D m 1 + 1 I m log 2D m β log 2 βα m + A 5 σ 2 L 2 |J m,1 |
for large enough A k 's. This condition reduces to

(8) f 2 n ≥ inf m∈M C 1 f -Π m f 2 n + C 2 σ 2 log 2 βα m + D m log 2 α m β ,
where C 1 is an absolute constant and C 2 only depends on L, provided this infimum is achieved at a partition m * ∈ M with

(9) 1 I m * log 2D m * β ≤ 1 and, e.g., |J m * ,1 | ≤ D m * .
Instead of ( 9), [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF] assume that (D m +log(1/α))/(n-D m ) remains bounded to prove that the power of their test is greater than 1β under condition [START_REF] Härdle | Testing a regression model when we have smooth alternatives in mind[END_REF]. Hence, both tests are powerful on similar sets. However, unlike the test of Baraud et al. our test requires neither Gaussian nor homoscedasticity assumptions.

4.2.

Power under the bounded moments assumption. Theorem 5. Assume we observe y = f + ε, where the ε i 's are independent variables with symmetric distribution. Assume P (ε i = 0) = 0 and |f i | ≤ L for a positive number L and all i. Assume moreover (5) holds for some p ≥ 2 and µ > 0.

Let M be a collection of partitions which does not contain the trivial one. Let α, β in (0, 1) and (α m ) m∈M be positive numbers with m∈M α m = α. For every

A = (A 1 , . . . , A 6 ) ∈ (0, ∞) 6 , m ∈ M, and f ∈ R n let ∆ 2 (m, f, A) = A 1 f -Π m f 2 n + A 2 (µ + L 2 ) D m β (10) + A 3 p(µ + L 2 ) 1 + 1 √ I m D m β 1/p log 2 α m + A 4 (µ + L 2 ) pD m 1 + 1 √ I m D m β 1/p log 2 α m + A 5 L 2 |J m,1 | + A 6 µ β .
Then there exists an absolute A such that (3) holds with

F n (β) = f ∈ R n , f 2 n ≥ inf m∈M ∆ 2 (m, f, A) .
Let us compare this with Theorem 4. There are two main differences between ∆ 1 and ∆ 2 . First, the absolute constants A 3 and A 4 in ∆ 1 are replaced in ∆ 2 by A 3 p and A 4 √ p respectively. Next, the term

1 I m log 2D m β in ∆ 1 is replaced in ∆ 2 by the greater term 1 √ I m D m β 1/p
, which depends on the error integrability. Hence, as expected, the set where our test is powerful is larger under the Bernstein-type condition than under the bounded moments condition.

Rate of Testing

In this section we assume

f i = F (x i ), i = 1, . . . , n,
for fixed numbers x i ∈ [0, 1] and an unknown F : [0, 1] → R. We aim at testing

H ′ 0 : F ≡ 0 against H ′ 1 : F ≡ 0.
For this task, we consider the so-called collection of dyadic partitions

M d = {m k , k ∈ I}, where I = {2 l , l ∈ N, 2 l ≤ n/2}
and m k consists of the nonempty sets among

i, x i ∈ j -1 k , j k , j = 1, . . . , k.
Let α and β in (0, 1) and

α m = α/|M d |.
As in Section 4, we restrict our attention to the test with critical region [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF], which still has level α in this setting. The aim of this section is to study the power of this test when the ε i 's satisfy either (4) or ( 5) and F is assumed Hölderian. In particular, we aim at proving adaptive properties of the test with respect to Hölderian smoothness. We distinguish between the cases where the Hölderian smoothness of F is at most one or greater than one. Assume the Hölderian smoothness of F to be at most one: there exist s ∈ (0, 1], R > 0, and L > 0 such that [START_REF] Yu | Minimax nonparametric detection of signals in white Gaussian noise[END_REF]

∀(u, v) ∈ [0, 1] 2 , |F (u) -F (v)| ≤ R|u -v| s , and sup u∈[0,1] |F (u)| ≤ L.
Assume moreover

(12) |J m k ,1 | ≤ a 0 D m k and I m k ≥ an k , all k ∈ I,
for absolute constants a 0 ≥ 0 and a > 0. This means that the design points are almost equidistant. In particular, the condition is satisfied with a 0 = 0 and a = 1/2 if x i = i/n for all i. The following corollary of Theorems 4 and 5 provides conditions on f under which the test is powerful, that is

(13) P f sup m∈M { Π m y 2 n -q y m (α m )} > 0 ≥ 1 -β.
For the sake of simplicity, it is assumed that α m = α/|M d | for every m but it is worth noticing that the results remain true if m∈M α m = α and α m ≥ (log 2 n) -cα for all m ∈ M and a positive c α , which only depends on α.

Corollary 1. Assume we observe y = f + ε, where the ε i 's are independent variables with symmetric distribution and f i = F (x i ) for fixed x i ∈ [0, 1] and an unknown F . Assume P (ε i = 0) = 0 for all i and F satisfies (11) for unknown s ∈ (0, 1], R > 0, and L > 0. Let M = M d be the collection of dyadic partitions. Let α and β in (0, 1). For every m ∈ M, let α m = α/|M|. Assume furthermore [START_REF] Yu | Asymptotically minimax hypothesis testing for nonparametric alternatives. I[END_REF].

1. If (4) holds for all integers p ≥ 1 and some positive numbers γ and µ, then (13) holds whenever n is large enough and one of the three following conditions is fulfilled for a large enough C (here, δ = √ γ + µ + L 2 and C only depends on a, a 0 , α, β):

(a) s > 1/4 and 1 √ n f n ≥ CR 1/(1+4s) δ √ log log n n 2s/(1+4s) ; (b) s = 1/4 and 1 √ n f n ≥ CR 2/3 n -1/4 δ 1/6 (log n × log log n) 1/12 ; (c) s < 1/4 and 1 √ n f n ≥ CRn -s .
2. If (5) holds for some p ≥ 2 and µ > 0, then (13) holds whenever n is large enough and one of the three following conditions is fulfilled for a large enough C (here, δ p = √ p(µ + L 2 ) and C only depends on a, a 0 , α, β):

(a) s ≥ 1/4 + 1/p and 1 √ n f n ≥ CR 1/(1+4s) δ p √ log log n n 2s/(1+4s) ; (b) s ∈ [1/4 -1/p, 1/4 + 1/p) and 1 √ n f n ≥ CR (2+3p)/(2+3p+8ps) δ p √ log log n n 5/4 4ps/(2+3p+8ps) ; (c) s ≤ 1/4 -1/p and 1 √ n f n ≥ CRn -s .
If x i = i/n for all i and F satisfies (11) then

1 √ n f n ≥ F 2 -Rn -s .
Hence in that case, Corollary 1 holds with f n / √ n replaced by F 2 . In particular, if s > 1/4 and (4) holds with µ = 2σ 2 and γ = µ 2 (which is indeed the case if the ε i 's are i.i.d. N (0, σ 2 )), then the power of the test is greater than a prescribed 1β whenever n is large enough and

F 2 ≥ CR 1/(1+4s) σ 2 √ log log n n 2s/(1+4s)
for a positive C, which only depends on L/σ, α, and β. This rate is precisely the minimal rate of testing obtained by [START_REF] Spokoiny | Adaptive hypothesis testing using wavelets[END_REF] in a white noise model. It is also the minimal rate of testing obtained by [START_REF] Gayraud | Adaptive minimax testing in the discrete regression scheme[END_REF] in the i.i.d. Gaussian regression model (but they do not describe the role of R and σ in this setting). This proves that our test achieves the optimal rate of testing under the Bernstein-type condition (4) if s > 1/4, and also under the less restrictive moment condition (5) if s ≥ 1/4 + 1/p. In the case where s ≤ 1/4, the optimal rate of testing is not known, even in the i.i.d. Gaussian model. Note however that the rate Rn -s (obtained when s < 1/4 and (4) holds, and also when s ≤ p -1/4 and (5) holds) was already obtained by [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF] in the i.i.d. Gaussian setting for s < 1/4. Now, assume the Hölderian smoothness of F to be greater than one: there exist d ∈ N \ {0}, κ ∈ (0, 1], R, and L such that

(14) ∀(u, v) ∈ [0, 1] 2 , |F (d) (u) -F (d) (v)| ≤ R|u -v| κ , and sup u∈[0,1] |F (u)| ≤ L.
Here, F (d) denotes the dth derivative of F (which is assumed to exist) and we denote by s = d + κ the Hölderian smoothness of F . For technical reasons, we restrict ourselves to the case where x i = i/n for all i.

Corollary 2. Assume we observe y = f + ε, where the ε i 's are independent variables with symmetric distribution and f i = F (i/n) for an unknown F . Assume P (ε i = 0) = 0 for all i and F satisfies (14) for unknown d ∈ N \ {0}, κ ∈ (0, 1], R > 0, and L > 0. Let s = d + κ > 1 and let M = M d be the collection of dyadic partitions. Let α and β in (0, 1). For every m ∈ M, let α m = α/|M|. Assume furthermore that either (4) holds for all integers p ≥ 1 and some positive numbers γ and µ, or (5) holds for some p ≥ 2 and µ > 0. Then there exists C > 0, which only depends on α, β, and s, such that (13) holds whenever n is large enough and

(15) F 2 ≥ CR 1/(1+4s) δ √ log log n n 2s/(1+4s)
.

Here, δ = √ γ + µ + L under assumption (4) and δ = √ p(µ + L 2 ) under assumption (5).

Thus our test achieves the optimal rate of testing when s > 1 under mild assumptions on the ε i 's.

Simulations

We carried out a simulation study to demonstrate the behavior of our test. We also simulated the test T P2,M dya proposed by [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF] in order to compare the performance of the two tests.

6.1. The simulation experiment. We generated a random vector ε ∈ R n from a given distribution G and computed y = f + ε, where n ∈ {64, 128, 256} and f i = F (x i ) for a given function F and fixed x i 's. We considered the collection of dyadic partitions M = M d as defined in Section 5, we set α = 5% and α m = α/|M| for every m ∈ M. Then, we generated B = 2500 independent copies of w and computed q B m (α m ) from these copies. Finally, we rejected the null hypothesis that F ≡ 0 if (2) holds. For given G, F and n, we repeated this step 1500 times in order to get 1500 independent tests, and we computed the percentage of rejections among these tests. We thus obtained an estimate for the level (if F ≡ 0) or for the power of our test. In order to assess the accuracy of the estimate, we repeated the above computations independently 100 times and computed the mean and the standard deviation of the percentages of rejection obtained over these 100 simulations. For every choice of G, F , and n, we obtained a small standard deviation (about 2.10 -3 to 5.10 -3 ), which means that the mean of the percentages of rejection is a good estimate for the actual level of the test. For each simulation, we treated likewise the T P2,M dya test of [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF].

We considered five different distributions G. In all cases, the variables ε i are mutually independent, with mean zero. The first distribution (called Gaussian) is the one under which our test and the test T P2,M dya both have prescribed level: the ε i 's are identically distributed and standard Gaussian. Then we considered two distributions, called Mixture and Heteroscedastic, under which our test has prescribed level, while T P2,M dya does not: under Mixture the ε i 's are i.i.d. mixture of Gaussian distributions as defined in Baraud et

v(i) = sin 4 4π i -1 n -1 n -i n -1 .
Finally

2 √ 5 U -2/5 - 5 3 ,
where U is uniformly distributed on [0, 1]. The distribution of the ε i 's is weakly asymmetric under Type I and strongly asymmetric under Asymmetric.

6.2. The level. We first considered the case where F ≡ 0 in order to estimate the level of the tests. For simplicity, the mean of 100 percentages of rejection (as described in Section 6.1) is called here the estimated level. The estimated levels obtained for different G and n are given in Table 1. Under Gaussian, Mixture and Heteroscedastic the estimated level of our test is, as expected, no greater than the nominal level. Under Type I, the estimated level remains less than the nominal level which means that the method is robust against a slight departure from symmetry. Under Asymmetric, the estimated level is greater than α when n is small and, despite asymmetry, it is smaller than α for n = 256. Even if we cannot explain rigorously this phenomenon, it seems interesting to us to explain it heuristically. Recall that the test consists in selecting a good partition m * and then comparing the distributions of Π m * y 2 n and Π m * (w × y) 2 n . If n is large, then either each subset in m * contains a large number of points or the number of subsets in m * is large (or even, the two properties hold simultaneously). If each subset in m * contains a large number of points, then by the central limit theorem the projections of y and w × y on these subsets have distributions, which are close to Gaussian; if the number of sets in m * is large, then the squared norm of the projections of y and w × y on m * (which are sums of projections on the subsets in m * ) have distributions close to Gaussian. It thus seems that the combination of two central limit theorems forces Π m * y n and Π m * (w × y) to have distributions, which are close to each other when f = 0 and n is large, even if the distributions of the ε i 's are not symmetric. This explains why the level of our test is less than α under Asymmetric for large n.

We now compare the performance of our test with that of T P2,M dya . Under Gaussian, Mixture, and Type I, the two tests have similar estimated levels and are conservative. Under Heteroscedastic, our test is still conservative while the estimated level of T P2,M dya is much larger than α, even for large n. Finally, under Asymmetric, both tests have estimated level larger than α when n is small. But our test performs better than T P2,M dya , all the more so that n increases. In particular, our test has the prescribed level when n = 256, whereas T P2,M dya has not. In this section, we consider cases where F ≡ 0, so the mean of 100 percentages of rejection (as described in Section 6.1) is called here estimated power. To choose regression functions F , we were inspired by [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF], where simulations of a test for linearity were performed. To adapt their alternatives to the case of testing for zero regression, we removed the linear part of their functions, so we considered the following functions:

F 1k (x) = c 1k cos(10πx), k = 1, 2, 3, 4, F 2k (x) = 5 φ(x/c 2k )/c 2k , k = 1, 2, F 3k (x) = -c 3k (x -0.1)1 x≤0.1 , k = 1, 2, 3,
where φ is the standard Gaussian density and c 11 , c 12 , . . . , c 33 are equal to 0.25, 0.5, 0.75, 1, 0.25, 1, 20, 30, 40 respectively (see Figure 1). Thus for j ∈ {1, 3}, the larger k the farther F jk is from the null function. In the case of alternatives F 1k and F 3k we set x i = (i -0.5)/n, while for F 2k the x i 's are simulated once for all as i.i.d. centered Gaussian variables with variance 25 in the range [Φ -1 (0.05), Φ -1 (0.95)], where Φ is the standard Gaussian distribution function.

Under Gaussian (see Table 2), the two tests have similar power against alternatives F 1k and F 2k : the estimated powers are good against F 2k and quite small against F 1k , especially for small values of k and n. The estimated power of T P2,M dya is greater than that of our test against F 3k , all the more so n is small and k is large (that is, the signal/noise ratio is large). This is certainly due to the small number of indices i such that F 3k (x i ) = 0. Indeed, our test can detect an alternative if Π m * y 2 n is significantly larger than Π m * (w × y) 3), the estimated powers of the two tests are smaller than under Gaussian, which is due to a greater variance of the errors: under Mixture (resp. Type I), the common variance of the ε i 's equals 4.678 (resp. 4). One can notice however that the signal/noise ratio is the same under Gaussian with F 11 (resp. F 12 , resp. F 31 ) as under Type I with F 12 (resp. F 14 , resp. F 33 ), and is close to that under Mixture with F 12 (resp. F 14 , resp. F 33 ). Comparing the estimated powers in these cases shows that, the signal/noise ratio being fixed, both tests have similar powers under these three distributions.

Finally, we studied the power of the two tests under Heteroscedastic and Asymmetric. We give in Table 4 the estimated powers only in the cases where the estimated level of the test is not much larger than the nominal level. In particular, under Heteroscedastic we only give the estimated power of our test. We can see that the estimated power of our test is smaller under Heteroscedastic than under Gaussian, but remains reasonable. Under Asymmetric, the estimated power of our test is in most cases greater than that of T P2,M dya , although the estimated level of T P2,M dya is greater than α.

In conclusion, our test is powerful against various alternatives (provided the number of indices i such that y i is significantly different from zero is large enough) and robust against departures from Gaussian distribution and homoscedasticity. It is also robust against slight departures from symmetry.

Proofs

Some of the proofs require lemmas. Lemmas are stated when needed and proved in Section 7.8.

7.1. Proof of Theorem 1. Under H 0 , we have y = ε, so our aim here is to prove that ( 16)

P sup m∈M { Π m ε 2 n -q ε m (α m )} > 0 ≤ α.
For every u ∈ R n , let |u| and sign(u) be the vectors of R n with ith component |u i | and

(17) (sign(u)) i = 1 {ui≥0} -1 {ui<0} ,
respectively. The distribution of ε i is symmetric about zero and P (ε i = 0) = 0, hence sign(ε) has the same distribution as w and is independent of |ε|. Moreover, w × sign(ε) has the same distribution as w. Since

w × ε = w × sign(ε) × |ε|,
it follows that the distribution of w × ε conditionally on ε is identical to its distribution conditionally on |ε|. In particular, q ε m (α m ) = q |ε| m (α m ), where

q |ε| m (α m ) = inf x ∈ R, P [ Π m (w × ε) 2 n > x | |ε| ] ≤ α m .
But conditionally on |ε|, w × ε has the same distribution as sign(ε) × |ε| = ε. Therefore for every m ∈ M,

P Π m ε 2 n > q ε m (α m ) | |ε| = P Π m (w × ε) 2 n > q |ε| m (α m ) | |ε| ≤ α m .
Integrating the latter inequality yields

P Π m ε 2 n > q ε m (α m ) ≤ α m .
By assumption m∈M α m = α, hence we get (16).

Proof of Theorem 2. The first kind error probability of the test with critical region (2) satisfies

P sup m∈M { Π m ε 2 n -q B m (α m )} > 0 ≤ P ∃m, q ε m (α m ) < Π m ε 2 n + P B n ,
where

P B n = P ∃m, Π m ε 2 n > q B m (α m ) and q ε m (α m ) ≥ Π m ε 2 n .
It thus follows from Theorem 1 that ( 18)

P sup m∈M { Π m ε 2 n -q B m (α m )} > 0 ≤ α + P B n .
We can have q ε m (α m ) ≥ Π m ε 2 n if and only if p ε m > α m , where

p ε m = P Π m (ω × ε) 2 n ≥ Π m ε 2 n | ε . Likewise, we can have q B m (α m ) < Π m ε 2 n if and only if 1 B B b=1 1 Πm(w b ×ε) 2 n ≥ Πmε 2 n ≤ α m .
Conditioning with respect to ε thus yields

P B n ≤ m∈M E P 1 B B b=1 1 Πm(w b ×ε) 2 n ≥ Πmε 2 n ≤ α m | ε 1 p ε m >αm .
For every m ∈ M, let S ε m be a random variable, which is distributed conditionally on ε as a binomial variable with parameter B and probability of success p ε m . The Hoeffding inequality yields

P B n ≤ m∈M E P (S ε m ≤ Bα m | ε)1 p ε m >αm (19) ≤ m∈M E exp(-2B(p ε m -α m ) 2 )1 p ε m >αm . By dominated convergence, lim B→∞ E exp(-2B(p ε m -α m ) 2 )1 p ε m >αm = 0,
so the first part of the theorem follows from [START_REF] Spokoiny | Adaptive hypothesis testing using wavelets[END_REF]. In order to prove the second part, we compute the distribution of p ε m in the case where the distributions of the ε i 's are continuous (see Section 7.8 for a proof).

Lemma 1. Under the assumptions of Theorem 1 with the additional assumption that the distributions of the ε i 's are continuous, p ε m has a discrete uniform distribution on the set

E = k2 Dm-n , k = 1, . . . , 2 n-Dm .

Combining (19) and Lemma 1 we get

P B n ≤ m∈M 2 Dm-n x∈E, x>αm exp(-2B(x -α m ) 2 ) ≤ m∈M 2 Dm-n + 1 αm exp(-2B(x -α m ) 2 ) dx .
The result now follows from ( 18) and straightforward computations.

Proof of Theorem 3.

The second kind error probability of the test with critical region (2) satisfies

P f sup m∈M { Π m y 2 n -q B m (α m )} ≤ 0 ≤ P f ∀m, q y m (δ m ) ≥ Π m y 2 n + Q B n ,
where

Q B n = P f ∃m, Π m y 2 n ≤ q B m (α m ) and q y m (δ m ) < Π m y 2 n .
Using the same arguments as in the proof of Theorem 2 (and also the same notation) we get

Q B n ≤ m∈M E f P f (S y m > Bα m | y)1 p y m ≤δm ,
where S y m is a random variable distributed conditionally on y as a binomial variable with parameter B and probability of success p y m . By assumption, δ m < α m for all m, so by dominated convergence, Q B n tends to zero as B → ∞, which proves the result.

Proof of Theorems 4 and 5.

In this section, we first describe the common line of the proof for the two theorems and then describe the specific arguments for each of them. The lemmas stated in this section are proved in Section 7.8. The following three inequalities are repeatedly used throughout the proof: for all positive real numbers a and b

(20) √ a + b ≤ √ a + √ b;
for all positive real numbers a, b and θ,

(21) 2 √ ab ≤ θa + θ -1 b;
for all positive numbers a, b and k we have

(22) (a + b) k ≤ 2 k (a k ∨ b k ).
Line of proof. Fix β ∈ (0, 1) and f ∈ F n (β). The second kind error probability of the test at f satisfies

P f sup m∈M { Π m y 2 n -q y m (α m )} ≤ 0 ≤ inf m∈M P f Π m y 2 n ≤ q y m (α m ) .
Therefore, the power of the test is at least 1β whenever there exists m ∈ M such that

(23) P f Π m y 2 n ≤ q y m (α m ) ≤ β. Since f ∈ F n (β), there exists a partition m ∈ M such that (24) f 2 n ≥ ∆(m, f, A),
where ∆ denotes either ∆ 1 or ∆ 2 . In the sequel, m denotes such a partition. We aim to prove that (23) holds for this partition, provided the A k 's are large enough. Now we show that the subsets e m,j which contain only one point cannot contribute to the power of the test. Let y ∈ R n with ith component y i = 0 if i ∈ e m,j for an index j ∈ J m,1 and y i = y i otherwise. For every j ∈ J m,1 , let i(j) be the unique element of e m,j . By definition of Π m ,

Π m y 2 n = j∈Jm,2 1 |e m,j | i∈em,j y i 2 + j∈Jm,1 y 2 i(j) = Π m y 2 n + j∈Jm,1 y 2 i(j) . Let q y m (α m ) = inf x ∈ R, P f Π m (w × y) 2 n > x | y ≤ α m .
We have

w i = ±1, so Π m (w × y) 2 n = Π m (w × y) 2 n + j∈Jm,1 y 2 i(j) ,
and we get q y m (α m ) = q y m (α m ) + j∈Jm,1

y 2 i(j) .
Hence (23) amounts to (25)

P f Π m y 2 n ≤ q y m (α m ) ≤ β,
which means that the sets with only one point can be removed from the condition. Thus we aim to prove that (25) holds provided the A k 's are large enough. We set

Z m = Π m (w × y) 2 n = j∈Jm,2 1 |e m,j | i∈em,j w i y i 2 and M m = max j∈Jm,2 1 |e m,j | i∈em,j y 2 i .
We first give an upper bound for q y m (α m ) in order to control the probability in (25). Lemma 2.

q y m (α m ) ≤ E f (Z m | y) + 8M m log(1/α m ) + 4 2E f (Z m | y)M m log(1/α m ).
The obtained upper bound depends on E f (Z m | y) and M m , which we have to control. It is proved in the following two lemmas that with high probability, these variables are not much greater than their expectation. The control we obtain depends on the assumptions on ε. Lemma 3. Under the assumptions of Theorem 1,

P f M m ≥ 18( √ γ ∨ µ ∨ L 2 ) 1 + 1 I m log 3D m β ≤ β 3
and

P f E f (Z m | y) ≥ E f (Z m ) + 8( √ γ ∨ L 2 ) D m log(3/β) + 2(µ ∨ L 2 ) log(3/β) ≤ β 3 .
Lemma 4. Under the assumptions of Theorem 5, there exists an absolute constant C ≥ 1 such that

P f M m ≥ Cp(µ ∨ L 2 ) 1 + 1 √ I m 3D m β 1/p ≤ β 3 .
Moreover,

P f E f (Z m | y) ≥ E f (Z m ) + 4(µ ∨ L 2 ) 6D m /β ≤ β 3 .
The main issue to prove Theorems 4 and 5 is to derive from the three lemmas above that (26)

P f q y m (α m ) ≤ E f (Z m ) + R m ≥ 1 -2 3 β,
where R m is a positive real number to be chosen later. Then,

P f Π m y 2 n ≤ q y m (α m ) ≤ 2 3 β + P f Π m y 2 n ≤ E f (Z m ) + R m , (27)
and it remains to prove that the right-hand side probability is less than or equal to β/3. In order to do that, we state a concentration inequality which proves that with high probability, Π m y 2 n is not much smaller than its expectation. Here again, the obtained control depends on the assumptions on ε. In the sequel, f denotes the expectation of y. Lemma 5. Under the assumptions of Theorem 5,

P f Π m y 2 n ≤ E f Π m y 2 n -6 γD m β - 1 3 Π m f 2 n -14( √ γ∨µ∨L 2 ) log 6 β ≤ β 3 .
Lemma 6. Under the assumptions of Theorem 5,

P f Π m y 2 n ≤ E f Π m y 2 n -3µ D m β - 1 3 Π m f 2 n - 9µ β ≤ β 3 .
In both cases, there is R ′ m ≥ 0 such that (28)

P f Π m y 2 n ≤ E f Π m y 2 n -R ′ m - 1 3 Π m f 2 n ≤ β 3 .
Thus the right-hand side probability in ( 27) is less than or equal to β/3 as soon as

(29) E f (Z m ) + R m ≤ E f Π m y 2 n -R ′ m - 1 3 Π m f 2 n ,
hence it suffices to prove (29). By assumption, the w i 's and the y i 's are mutually independent, and the distribution of w i is symmetric about zero. Therefore, E(w i ) = 0 and the random variables w i y i are mutually independent with zero mean and variance

E f (y 2 i ) = E(ε 2 i ) + f 2 i . Hence, ( 30
) E f (Z m ) = j∈Jm,2 1 |e m,j | i∈em,j (E(ε 2 i ) + f 2 i ).
We have |e m,j | ≥ I m for all j ∈ J m,2 , so

E f Π m y 2 n -E f (Z m ) = j∈Jm,2 1 |e m,j | i∈em,j E(ε 2 i ) + i∈em,j f i 2 - i∈em,j (E(ε 2 i ) + f 2 i ) ≥ j∈Jm,2 1 |e m,j | i∈em,j f i 2 - 1 I m j∈Jm,2 i∈em,j f 2 i .
Hence

E f Π m y 2 n -E f (Z m ) ≥ Π m f 2 n -|J m,1 | max 1≤i≤n f 2 i - 1 I m f 2 n .
In order to prove (29), it suffices to prove

(31) 2 3 Π m f 2 n ≥ 1 I m f 2 n + |J m,1 | max 1≤i≤n f 2 i + R m + R ′ m .
By definition, I m ≥ 2 and by the Pythagoras equality,

Π m f 2 n = f 2 n -f -Π m f 2 n . (32)
Thus it suffices to prove

1 6 f 2 n ≥ 2 3 f -Π m f 2 n + |J m,1 | max 1≤i≤n f 2 i + R m + R ′ m .
But f satisfies (24), so it suffices to check that for large enough A k ,

∆(m, f, A) ≥ 4 f -Π m f 2 n + 6 |J m,1 | max 1≤i≤n f 2 i + R m + R ′ m . (33)
Now to prove Theorems 4 and 5, it remains to compute R m and R ′ m and to check (33) with ∆ replaced by ∆ 1 and ∆ 2 respectively.

Proof of Theorem 4. By (30),

E f (Z m ) ≤ D m ( 2γ + L 2 ).
Using (21) we thus get

E f (Z m ) + 8( √ γ ∨ L 2 ) D m log(3/β) + 2(µ ∨ L 2 ) log(3/β) ≤ 7( √ γ ∨ µ ∨ L 2 ) D m + log(3/β) .
Combining Lemmas 2 and 3 proves that with probability greater than 1 -2β/3,

q y m (α m ) ≤ E f (Z m ) + 8( √ γ ∨ L 2 ) D m log(3/β) + 2(µ ∨ L 2 ) log(3/β) + 144( √ γ ∨ µ ∨ L 2 ) 1 + 1 I m log 3D m β log 1 α m + 24( √ γ ∨ µ ∨ L 2 ) 7 D m + log 3 β 1 + 1 I m log 3D m β log 1 α m .
Using (20) and ( 21), we get that with probability greater than 1 -2β/3,

q y m (α m ) ≤ E f (Z m ) + 8( √ γ ∨ L 2 ) D m log(3/β) + 151( √ γ ∨ µ ∨ L 2 ) log(3/β) + 151( √ γ ∨ µ ∨ L 2 ) 1 + 1 I m log 3D m β log 1 α m + 24( √ γ ∨ µ ∨ L 2 ) 7D m 1 + 1 I m log 3D m β log 1 α m .
By assumption, log(1/α m ) and log(3/β) are positive, so we obtain (26) with

R m = 151( √ γ ∨ µ ∨ L 2 ) 1 + 1 I m log 3D m β log 3 βα m + 72( √ γ ∨ µ ∨ L 2 ) D m 1 + 1 I m log 3D m β log 3 βα m .
By Lemma 5, we have (28) with

R ′ m = 6 γD m /β + 14( √ γ ∨ µ ∨ L 2 ) log(6/β), so (33)
holds with ∆ replaced by ∆ 1 provided A 1 , . . . , A 5 are large enough.

Proof of Theorem 5. By (30),

E f (Z m ) ≤ D m (µ + L 2 ).
Combining Lemmas 2 and 4 proves that with probability greater than 1 -2β/3,

q y m (α m ) ≤ E f (Z m ) + 4(µ ∨ L 2 ) 6D m β + 8Cp(µ ∨ L 2 ) 1 + 1 √ I m 3D m β 1/p log 1 α m + 18(µ ∨ L 2 ) Cp D m + D m β 1 + 1 √ I m 3D m β 1/p log 1 α m .
Using (20) and ( 21), we get (26) with

R m = 19(µ ∨ L 2 ) D m β + 17Cp(µ ∨ L 2 ) 1 + 1 √ I m 3D m β 1/p log 1 α m + 18(µ ∨ L 2 ) CpD m 1 + 1 √ I m 3D m β 1/p log 1 α m .
By Lemma 6, we have (28) with

R ′ m = 3µ D m /β + 9µ/β, so (33)
holds with ∆ replaced by ∆ 2 provided A 1 , . . . , A 6 are large enough. 7.5. Proof of Corollary 1. We define δ and δ ′ in the following way. If (4) holds for all p ≥ 1 and some γ and µ, we set δ = δ ′ = √ γ + µ + L 2 . If [START_REF] Ermakov | Minimax detection of a signal in a Gaussian white noise[END_REF] holds for some p ≥ 2 and some µ, we set δ = √ p(µ + L 2 ) and δ ′ = p(µ + L 2 ). By

Step 2 in the proof of Corollary 1 of [START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF],

f -Π m k f 2 n ≤ nR 2 k -2s
for all k ∈ I. Moreover, |M| ≤ log n/ log 2, so there exists c α > 0, which only depends on α, such that log(1/α m ) ≤ c α log log n.

We have [START_REF] Yu | Asymptotically minimax hypothesis testing for nonparametric alternatives. I[END_REF] and D m k ≤ k for all k ∈ I, so it follows from Theorem 4 that the power of the test is greater than 1β whenever (4) holds and

f 2 n ≥ A inf k∈I nR 2 k -2s + δ 1 + k n log n log log n (34) + δ k 1 + k n log n log log n ,
for a large enough A. Likewise, it follows from Theorem 5 that the power of the test is greater than 1β whenever (5) holds and

f 2 n ≥ A inf k∈I nR 2 k -2s + δ ′ 1 + k 1/p+1/2 √ n log log n (35) + δ k 1 + k 1/p+1/2
√ n log log n for a large enough A. In (34) and (35), A > 0 only depends on a, a 0 , α, and β. In the sequel, I ′ denotes the subset of I defined as follows. Under (4), I ′ is the set of those k ∈ I, which satisfy k log n ≤ n and under (5), I ′ is the set of those k ∈ I, which satisfy k 1/p+1/2 ≤ √ n, that is k ≤ n p/(p+2) .

• Assume first that (4) and 1 (a), resp. ( 5) and 2 (a), hold for a large enough C. By (34) and (35), the power of the test is greater than 1β whenever

f 2 n ≥ 2A inf k∈I ′ nR 2 k -2s + δ k log log n + δ ′ log log n . Let (36) k * = nR 2 δ √ log log n 2/(1+4s)
.

We have nR 2 k -2s ≤ δ k log log n if and only if k ≥ k * and for large enough n, there exists

k ′ ∈ I ′ such that k * ≤ k ′ ≤ 2k * . Therefore, inf k∈I ′ nR 2 k -2s + δ k log log n ≤ 2δ k ′ log log n ≤ 4δ k * log log n ≤ 4nR 2/(1+4s) δ √ log log n n 4s/(1+4s)
.

The power of the test is thus greater than 1β whenever n is large enough and

f 2 n ≥ 2A 4nR 2/(1+4s) δ √ log log n n 4s/(1+4s) + δ ′ log log n .
This indeed holds if n is large enough and either 1 (a) or 2 (a) is fulfilled for a large enough C.

• Assume [START_REF] Chen | Testing for no effect in nonparametric regression via spline smoothing techniques[END_REF]. By (34), the power is greater than 1β whenever n is large enough and (37)

f 2 n ≥ 3A inf k∈I\I ′ nR 2 k -2s + δk 1 n log n × log log n for a large enough A. Let k * = n 3/2 R 2 δ √ log n × log log n 1/(1+2s)
.

If s = 1/4 and n is large enough, there exists

k ′ ∈ I \ I ′ such that k * ≤ k ′ ≤ 2k * . Therefore, inf k∈I\I ′ nR 2 k -2s + δk 1 n log n × log log n ≤ 4δk * 1 n log n × log log n.
The power of the test is thus greater than 1β whenever n is large enough and 1 (b) holds for a large enough C. Let k 0 be a point in I with k 0 ≥ n/8. Then (37) holds whenever

f 2 n ≥ 3A nR 2 k -2s 0 + δk 0 1 n log n × log log n .
Since k 0 ≤ n/2, this indeed holds if n is large enough and 1 (c) is fulfilled for a large enough C.

• Assume (5) holds for some p ≥ 2. If n is large enough, we have for all k ≥ 1 δ ′ k 1/p+1/2 n -1/2 log log n ≤ δk 1/2p+3/4 n -1/4 log log n.

By (35), the power is thus greater than 1β whenever n is large enough and (38)

f 2 n ≥ 4A inf k∈I\I ′ nR 2 k -2s + δk 1/2p+3/4 n -1/4 log log n . Let k * = n 5/4 R 2 δ √ log log n 4p/(2+3p+8ps)
.

If s ∈ [1/4 -1/p, 1/4 + 1/p) and n is large enough, there exists k ′ ∈ I \ I ′ such that k * ≤ k ′ ≤ 2k * . Therefore, inf k∈I\I ′ nR 2 k -2s + δk 1/2p+3/4 n -1/4 log log n ≤ 4δk 1/2p+3/4 * n -1/4 log log n.
The power of the test is thus greater than 1β whenever n is large enough and 2 (b) holds for a large enough C. Let k 0 be a point in I with k 0 ≥ n/8. Then (38) holds whenever

f 2 n ≥ 4A nR 2 k -2s 0 + δk 1/2p+3/4 0 n -1/4 log log n .
This indeed holds if n is large enough and 2 (c) is fulfilled for a large enough C.

7.6. Proof of Corollary 2. Let k * be given by (36) and let k ′ ∈ I be such that k * ≤ k ′ ≤ 2k * (such a k ′ exists provided n is large enough). By Theorems 4 and 5, it suffices to prove that f 2 n exceeds ∆(m k ′ , f, A) for large enough n and a fixed A, where ∆ denotes either ∆ 1 or ∆ 2 . By (31), one can choose

A 1 = 2/3 2/3 -1/I m k ′ .
Moreover, there exists A 0 > 0, which only depends on α, β, and A such that

∆(m k ′ , f, A) ≤ A 1 f -Π m f 2 n + A 0 δ k ′ log log n.
By (32), it thus suffices to show that for large enough n,

(39) Π m k ′ f 2 n ≥ 3 2I m k ′ f 2 n + A 0 δ k ′ log log n.
We need further notation. Let F n be the function defined by F n (t) = F (x i ) for all t ∈ (x i-1 , x i ], i = 1, . . . , n, where we set x i = i/n for all i = 0, . . . , n. For every j = 0, . . . , k ′ , let

t j = 1 n i≤j |e m k ′ ,i | = 1 n nj k ′
(here, [x] denotes the integer part of x). Let Q be the orthogonal projector from L 2 [0, 1] onto the set of step functions, which are constant on each interval (t j-1 , t j ], and let Q r be the orthogonal projector from L 2 [0, 1] onto the set of step functions, which are constant on each interval ((j -1)/k ′ , j/k ′ ], j = 1, . . . , k ′ . Since t j-1 -t j ≤ 2/k ′ and |t jj/k ′ | ≤ 1/n, we have

QF 2 2 = k ′ j=1 1 t j -t j-1 tj tj-1 F (x) dx 2 ≥ 1 4 Q r F 2 2 - 2k ′ 2 F 2 ∞ n 2
(recall that (a + b) 2 ≥ a 2 /2b 2 for all real numbers a and b). Now,

Q(F -F n ) 2 ≤ F -F n 2 ≤ F -F n ∞ ≤ F ′ ∞ /n and we have Π m k ′ f n = √ n QF n 2 .
Therefore,

Π m k ′ f n ≥ √ n QF 2 -F ′ ∞ / √ n ≥ √ n 2 Q r F 2 - √ 2k ′ F ∞ √ n -F ′ ∞ / √ n.
The partition of [0, 1] associated with Q r is equispaced (all intervals have the same length 1/k ′ ), so one can prove that there exist positive numbers C 1 and C 2 , which only depend on s, such that

Q r F 2 ≥ C 1 F 2 -C 2 Rk ′ -
Π m k ′ f n ≥ C ′ 1 √ n F 2 -C ′ 2 √ nRk ′ -s . Note that f n = √ n F n 2 ≤ √ n F 2 + F ∞ / √ n
and that I m k ′ tends to infinity as n goes to infinity. Thus in order to prove (39), it suffices to prove that for large enough n and some A ′ 0 > 0,

F 2 2 ≥ A ′ 0 R 2 k ′ -2s + δ n k ′ log log n .
But this is indeed the case if F satisfies (15) for a large enough C > 0.

7.7. Power in a Gaussian homoscedastic model. In this section, we assume that the ε i 's are i.i.d. N (0, σ 2 ) and |f i | ≤ σL for all i. We will prove that the power of the test is greater than 1β whenever f satisfies (7) for large enough A k 's. Under the above assumptions, Lemmas 2 and 3 are valid with µ = 2σ 2 , γ = µ 2 , and L 2 replaced by σ 2 L 2 . In particular, we have (26) with

R m = 302σ 2 (1 + L 2 ) 1 + 1 I m log 3D m β log 6 βα m + 144σ 2 (1 + L 2 ) D m 1 + 1 I m log 3D m β log 6 βα m .
Moreover, one can improve the result given in Lemma

f 1 σ 2 Π m y 2 n ≤ 1 σ 2 E f Π m y 2 n -2 D ′ m + 2 σ 2 Π m f 2 n x ≤ exp(-x).
Using (20) and (21) one obtains

P f Π m y 2 n ≤ E f Π m y 2 n -2σ 2 D ′ m x - 1 3 Π m f 2 n -6σ 2 x ≤ exp(-x),
for all x > 0. Setting x = log(3/β) in this inequality yields

P f Π m y 2 n ≤ E f Π m y 2 n -2σ 2 D m log 3 β - 1 3 Π m f 2 n -6σ 2 log 3 β ≤ β 3 .
Therefore, we have (28) with

R ′ m = 2σ 2 D m log(3/β) + 6σ 2 log(3/β).
Let ∆(m, f, A) denote the bracketed term in [START_REF] Gayraud | Adaptive minimax testing in the discrete regression scheme[END_REF]. Then (32) holds provided the A k 's are large enough, which proves the announced result.

7.8. Proof of the lemmas. We first recall two inequalities, which will be used in this section.

Bernstein's Inequality. Let X 1 , . . . , X n be independent real-valued random variables. Assume that there exist positive numbers v and c such that for all integers k ≥ 2

n i=1 E[ |X i | k ] ≤ k! v 2 c k-2 . Let S = n i=1 (X i -E[X i ]
). Then for every x > 0, P S ≥ √ 2vx + cx ≤ exp(-x).

Rosenthal's Inequality. Let X 1 , . . . , X n be independent centered real-valued random variables with finite t-th moment, 2 ≤ t < ∞. Then there exists an absolute constant L such that

E n i=1 X i t ≤ L t t t max n i=1 E|X i | t , n i=1 E|X i | 2 t/2 .
Proof of Lemma 1. For notational convenience, we omit subscript m. For every j = 1, . . . , D and z ∈ R n , define the set A j (z) by A j (z) = {0} if D = 1 and

A j (z) = k =j 1 |e k | i∈e k u i z i 2 , u ∈ {±1} n if D > 1.
For every j = 1, . . . , D let E j = {e ⊂ e j s.t. e = ∅ and e = e j }.

Finally, let

Y = D j=1 z ∈ R n s.t. ∀u ∈ {±1} n , ∀a, a ′ ∈ A j (z), ∀e ∈ E j , 4 i∈e u i z i i∈ej \e u i z i = (a ′ -a)|e j | .
The ε i 's are independent and all have a continuous distribution, so the event {ε ∈ Y} is the intersection of a finite number of events having probability one. We thus assume in the sequel without loss of generality that ε ∈ Y. Let u and u ′ be elements of {±1} n such that

(40) Π(u × ε) 2 n = Π(u ′ × ε) 2 n .
Fix j ∈ {1, . . . , D}. We have

0 = Π(u × ε) 2 n -Π(u ′ × ε) 2 n = 1 |e j | i∈ej u i ε i 2 - i∈ej u ′ i ε i 2 + a -a ′
for some a, a ′ ∈ A j (ε). Setting e = {i ∈ e j s.t.

u i = u ′ i } we get i∈ej u i ε i 2 - i∈ej u ′ i ε i 2 = i∈ej (u i + u ′ i )ε i i∈ej (u i -u ′ i )ε i = 4 i∈e u i ε i i∈ej \e u i ε i , Here, (41)
J = j∈Jm,2
e m,j .

Denote by S m the unit sphere in R |Jm,2| . Then

Z 1/2 m = sup a∈Sm j∈Jm,2 a j X j = sup a∈Sm i∈J w i y i a j(i) |e m,j(i) | ,
where for every i, j(i) denotes the unique integer in J m,2 that satisfies i ∈ e m,j(i) .

If S ′ m denotes a finite subset of S m , then it follows from a result of Massart ( 2006) that for all x ≥ 0,

P f sup a∈S ′ m j∈Jm,2 a j X j ≥ E f sup a∈S ′ m j∈Jm,2 a j X j | y + x | y ≤ exp - x 2 8σ 2 (S ′ m )
, where

σ 2 (S ′ m ) = sup a∈S ′ m i∈J y 2 i a 2 j(i) |e m,j(i) | = sup a∈S ′ m j∈Jm,2 a 2 j |e m,j | i∈em,j y 2 i .
But S m is separable and for every subset

S ′ m of S m , σ 2 (S ′ m ) ≤ M m . Hence for all x ≥ 0, P f Z 1/2 m ≥ E f (Z 1/2 m | y) + x | y ≤ exp(-x 2 /8M m ). In particular, P f Z 1/2 m ≥ E f (Z 1/2 m | y) + 8M m log(1/α m ) | y ≤ α m .
Squaring the inequality yields

P f Z m ≥ E 2 f (Z 1/2 m | y) + 8M m log(1/α m ) + 2E f (Z 1/2 m | y) 8M m log(1/α m ) | y ≤ α m .
By definition of q y m (α m ) we thus have

q y m (α m ) ≤ E 2 f (Z 1/2 m | y) + 8M m log(1/α m ) + 2E f (Z 1/2 m | y) 8M m log(1/α m ),
and the result follows from the Jensen inequality, which implies

E 2 f (Z 1/2 m | y) ≤ E f (Z m | y).
Proof of Lemma 3. By definition of M m , we have for all real numbers x and c

P f M m ≥ c + x ≤ j∈Jm,2 P f 1 |e m,j | i∈em,j y 2 i ≥ c + x .
In particular, if c is large enough so that E f (y , where γ 0 = 2 5 (γ ∨ L 4 ) and µ 0 = 4(µ ∨ L 2 ). By Bernstein's inequality, we thus have for all x ≥ 0 P f i∈em,j

y 2 i -E f (y 2 i ) ≥ 2 γ 0 |e m,j |x + µ 0 x ≤ exp(-x).
Moreover, E f (y 2 i ) ≤ L 2 + √ 2γ, so by (42),

P f M m ≥ 2γ + L 2 + 2 γ 0 x I m + µ 0 x I m ≤ D m exp(-x).
We have

2γ + L 2 + 2 γ 0 x I m + µ 0 x I m ≤ 2γ + L 2 + (2 √ γ 0 + µ 0 ) 1 + x I m ≤ 18( √ γ ∨ µ ∨ L 2 ) 1 + x I m .
Hence for all x > 0,

P f M m ≥ 18( √ γ ∨ µ ∨ L 2 ) 1 + x I m ≤ D m exp(-x).
Setting here x = log(3D m /β) yields the first assertion in Lemma 3. The variables w i are independent with mean zero and variance 1, and these variables are independent from y. Hence conditionally on y, they are still independent with mean zero and variance 1, and we derive from the definition of Z m that (44) , so Bernstein's inequality yields for all x ≥ 0

E f (Z m | y) = j∈Jm,2
P f E f (Z m | y) ≥ E f (Z m ) + 2D m γ 0 x + 1 2 µ 0 x ≤ exp(-x).
Setting here x = log(3/β) completes the proof of the lemma.

Proof of Lemma 4. By Rosenthal's inequality, there exists an absolute constant C ′ > 0 such that for all j E f i∈em,j (y 2

i -E f (y for all x > 0. In particular,

P f M m ≥ µ + L 2 + Cp(µ ∨ L 2 ) 1 √ I m 3D m β 1/p ≤ β 3 .
We can assume without loss of generality that C ≥ 1, so the first assertion of Lemma 4 follows. We have (44), where the y i 's are independent. Therefore,

var f E f (Z m | y) ≤ max 1≤i≤n E f (y 4 i )D m ≤ 2 4 (µ 2 + L 4 )D m .
Now the second assertion of the lemma follows from the Bienaymé-Chebyshev inequality.

Proof of Lemma 5. Let J be defined by (41). We have

Π m y 2 n = Π m ε 2 n + Π m f 2 n + 2 j∈Jm,2 1 |e m,j | i∈em,j ε i i∈em,j f i ,
where ε = yf . For every i ∈ J , let j(i) denote the integer in J m,2 which satisfies i ∈ e m,j(i) and let

a i = 1 |e m,j(i) | i∈e m,j(i) f i .
Then, Using Bernstein's inequality, we obtain that for all x ≥ 0

P f j∈Jm,2 1 |e m,j | i∈em,j ε i i∈em,j f i ≤ -2 Π m f 2 n √ γx -L √ µx ≤ exp(-x).
We have L √ µ ≤ L 2 ∨ µ and by (21),

2 Π m f 2 n √ γx ≤ 1 6 Π m f 2 n + 6 √ γx.
Setting x = log(6/β) thus yields

P f j∈Jm,2 1 |e m,j | i∈em,j ε i i∈em,j f i ≤ - 1 6 Π m f 2 n -7( √ γ ∨ µ ∨ L 2 ) log 6 β ≤ β 6 .
On the other hand, the ε i 's are mutually independent and ε i and ε 3 i have mean zero, so (45) var i∈em,j

ε i 2 ≤ E i∈em,j ε i 4 ≤ 3|e m,j | 2 max 1≤i≤n E(ε 4 i ).
Therefore, var Π m ε 2 n ≤ 6γD m and it follows from the Bienaymé-Chebyshev inequality that

P f Π m ε 2 n ≤ E f Π m ε 2 n -6 γD m /β ≤ β/6.
The result follows.

Proof of Lemma 6. The distribution of ε i is symmetric, so ε i and ε 3 i have mean zero. Moreover, the ε i 's are mutually independent, so var f ( Π m y 2 n ) = By the Bienaymé-Chebyshev inequality,

P f Π m y 2 n ≤ E f Π m y 2 n - 9D m µ 2 + 12 Π m f 2 n µ β ≤ β 3 .
By ( 20) and ( 21) we have

9D m µ 2 + 12 Π m f 2 n µ β ≤ 3µ D m β + 1 3 Π m f 2 n + 9µ β
and therefore,

P f Π m y 2 n ≤ E f Π m y 2 n -3µ D m β - 1 3 Π m f 2 n - 9µ β ≤ β 3 .
This completes the proof of the lemma.

J m, 1 =

 1 {j ∈ {1, . . . , D m }, |e m,j | = 1}, J m,2 = {j ∈ {1, . . . , D m }, |e m,j | ≥ 2} (hence J m,2 = ∅) and I m = min j∈Jm,2 |e m,j |.

Figure 1 .

 1 Figure 1. Alternatives F 11 , F 12 (left), F 21 , F 22 (center) and F 31 , F 32 (right). Alternatives F j1 and F j2 are drawn in dotted and plain lines respectively

1 x p

  |e m,j | p/2 ≤ (Cp) p (µ p ∨L 2p )

j∈Jm, 2 1 1 i∈J 2 i√ 2 n√

 2122 |e m,j | i∈em,j ε i i∈em,j f i = i∈J a i ε i . Since |a i | ≤ L, we have for all integers p ≥ E|a i ε i | p ≤ L p-2 i∈J a γ p!µ p/2-1 ≤ L p-2 Π m f γp!µ p/2-1 .

  It then follows from (45) thatvar f Π m y 2 n ≤ 3D m µ 2 + 4 Π m f 2 n µ.

  al. (2003) ((b), p. 236); under Heteroscedastic, ε i is a centered Gaussian variable with variance

  , we considered two distributions, called Type I and Asymmetric under which none of the two tests have prescribed level: in both cases, the ε i 's are i.i.d., Type I is defined as in[START_REF] Baraud | Adaptive tests of linear hypotheses by model selection[END_REF] ((c), p. 236) and under Asymmetric, the ε i 's are distributed as 3

Table 1 .

 1 Estimated level of our test (Roman) and T P1,M dya (italic) for the five different distributions G when α = 5% and n ∈ {64, 128, 256}

	Distribution G n = 64 n = 128 n = 256 Gaussian 3.71 3.80 3.72
		3.53	3.54	3.46
	Mixture	3.70	3.79	3.76
		2.96	3.20	3.31
	Heteroscedastic	3.15	3.18	3.19
		16.4	19.3	21.9
	Type I	3.67	3.46	3.33
		3.94	3.89	3.70
	Asymmetric	9.10	6.52	4.75
		10.1	8.62	7.60
	6.3. The power.

Table 2 .

 2 Estimated power of our test (Roman) and T P2,M dya (italic) under Gaussian distribution. The nominal level is α = 5%

	Distribution G F	F 11 F 12 F 13 F 14	Gaussian F 21 F 22 F 31 F 32 F 33
	n = 64	4.41 8.00 19.4 41.7 99.9 100 12.2 20.2 27.5
		4.29 7.78 18.5 39.5 99.8 100 16.2 40.4 72.4
	n = 128	6.46 25.0 69.4 96.7 100 100 34.4 70.3 91.2
		6.12 24.2 68.2 96.3 100 100 43.3 89.4 99.8
	n = 256	12.3 70.1 99.5 100	100 100 84.8 99.9 100
		11.7 68.9 99.5 100 100 100 87.8 100 100

2 n for a well chosen partition m * , but Π m * y 2 n remains close to Π m * (w × y) 2 n if there are only a few indices i such that y i is significantly different from zero. Under Mixture or Type I (see Table

Table 3 .

 3 Estimated power of our test (Roman) and T P2,M dya (italic) under Mixture and Type I distributions. The nominal level is α = 5%

	Distribution G F	F 11 F 12 F 13 F 14	Mixture F 21 F 22	F 31 F 32 F 33
	n = 64	3.93 4.61 6.24 9.32 88.0 98.0 6.15 9.11 12.9
		3.12 3.63 4.63 6.80 87.1 97.5 5.06 8.43 14.4
	n =	4.31 6.63 13.0 25.6 99.8 100 9.74 19.5 34.2
		3.60 5.32 10.1 20.5 99.8 100 8.46 18.7 37.3
	n =	5.16 11.9 31.7 63.4 100 100 18.9 47.1 78.6
		4.41 10.0 27.6 59.0 100 100 17.0 45.7 79.5
	Distribution G F	F 11 F 12 F 13 F 14	Type I F 21 F 22	F 31 F 32 F 33
	n = 64	3.88 4.50 5.86 8.68 92.8 99.9 4.99 7.27 10.6
		4.15 4.69 5.93 8.38 92.4 99.9 5.67 9.08 15.6
	n =	3.99 6.21 12.4 26.4 100 100 8.04 17.5 33.8
		4.41 6.52 12.3 25.2 100 100 9.25 21.0 43.0
	n =	4.58 11.8 35.1 70.8 100 100 18.3 51.4 87.3
		4.98 11.9 33.9 69.1 100 100 19.3 53.3 88.7

Table 4 .

 4 Estimated power of our test (Roman) and T P2,M dya (italic) under Heteroscedastic and Asymmetric distributions. The nominal level is α = 5% 12 F 13 F 14 F 21 F 22 F 31 F 32 F 33

	Distribution G F	Heteroscedastic F 11 F 12 F 13 F 14 F 21 F 22 F 31 F 32 F 33
	n = 64	3.94 7.30 15.8 30.8 99.6 100 11.4 17.8 23.1
	n = 128	4.32 10.2 30.0 64.4 100 100 19.3 45.5 73.4
	n = 256	5.33 20.5 70.0 98.4 100 100 41.6 95.1 100
	Asymmetric 20.9 75.0 93.6 97.8 100 100 74.5 96.8 99.6 F 11 F n = 128 Distribution G F
		17.0 59.9 85.9 94.0 100 100 76.4 93.8 97.4
	n = 256	39.2 93.6 98.9 99.6 100 100 99.2 100 100
		30.9 85.2 96.1 98.3 100 100 93.3 98.2 99.2

  s , see Proposition 2.16 of Ingster and Suslina (2003). It then follows from the definition of k ′ and the assumption s > 1 that there exist positive numbers C ′ 1 and C ′ 2 , which only depend on s, such that for large enough n,

  5 under the Gaussian assumption, due to the Cochran theorem. Indeed, let D ′ m denote the cardinality of J m,2 . By the Cochran Theorem, Π m y 2 n /σ 2 is a non-central χ 2 variable with D ′ m degrees of freedom and non-centrality parameter Π m f 2 n /σ 2 . By Lemma 1 of Birgé (2001), we thus have for all positive x P

  2 i ) ≤ c for all i, then for all x (42)P f M m ≥ c + x ≤ j∈Jm,2 P f 1 |e m,j | i∈em,j y 2 i -E f (y 2 i ) ≥ x . Let p ≥ 1. By (22), y 2p i ≤ 2 2p (ε 2p i ∨ L 2p) and in particular,y 2p i ≤ 2 2p (ε 2p i + L 2p ). ≤ 2 2p+1 max γp!µ p-2 , L 2p ≤ γ 0 p!µ p-2

	Therefore,
	(43)	E f (y 2p i ) 0

  Moreover, by Jensen's inequality, Eε 4 i ≤ µ 2 for all i, so by (22),var f (y 2 i ) ≤ E f (y 4 i ) ≤ 2 4 (µ 2 + L 4 ). ≤ 2 5p/2 |e m,j | p/2 (µ p ∨ L 2p ). E f (y 2p i )+|E f (y 2 i )| p ≤ 2 p+1+2p |e m,j |(µ p +L 2p ), Cp) p |e m,j | p/2 (µ p ∨ L 2p ),where for instance C = 2 4 C ′ . We have E f (y 2 i ) ≤ µ + L 2 , so (42) and Markov's inequality yield P f M m ≥ µ+L 2 +x ≤ (Cp) p (µ p ∨L 2p )

			2 i ))
						i∈em,j	E f |y 2 i -E f (y 2 i )| p ,	i∈em,j	var(y 2 i )	p/2	.
	Hence,
				var f (y 2 i )	p/2
			i∈em,j
	On the other hand,
	i∈em,j	E f |y 2 i -E f (y 2 i )| p ≤ 2 p	i∈em,j
	so we get
		E f	i∈em,j	(y 2 i -E f (y 2 i ))
						j∈Jm,2

p ≤ (C ′ p) p max p ≤ (

Acknowledgments. The second author started to work on this subject while visiting the department Industrial Engineering and Management of the Technion of Haïfa, Israel. During this period, he had several valuable discussions on the subject with Felix Abramovitch from Tel Aviv University. The authors thank Felix Abramovitch for his contribution at that time.

since u i = -u ′ i for every i ∈ e j \ e. As ε ∈ Y, either e = ∅ or e = e j . Thus for all j = 1, . . . , D, either u i = u ′ i for all i ∈ e j or u i = -u ′ i for all i ∈ e j . This implies that the cardinality of the set

and that for every element a of this set, the cardinality of the set

. This proves that conditionally on ε, Π(w × ε) 2 n has a discrete uniform distribution on a set with 2 n-D distinct values. Clearly, Πε 2 n belongs to this set, so p ε = k2 D-n for some k = 1, . . . , 2 n-D . More precisely, we have p ε = k2 D-n if and only if the cardinality of the following set is equal to k:

But this set has the same cardinality as the set

where we recall that sign(ε) is the vector in R n defined by [START_REF] Staniswalis | Diagnostics for assessing regression models[END_REF]. Since sign(ε) has the same distribution as w and is independent of |ε| (see the proof of Theorem 1) we get

Moreover conditionally on |ε|, Π(w × |ε|) 2 n has a uniform discrete distribution on a set with 2 n-D distinct values. Therefore,

Integrating this inequality yields the result.

Proof of Lemma 2. First, note that Lemma 2 is trivial whenever y i = 0 for all i since in that case q y m (α m ) = 0. We thus assume in the sequel that there exists i such that y i = 0. In particular, M m > 0. Now, note that Z m can be written as the sum of squared random variables:

where for every j ∈ J m,2 , X j = i∈J w i y i |e m,j | 1 em,j (i).