On the rate of convergence in Wasserstein distance of the empirical measure Nicolas Fournier, Arnaud Guillin
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Nicolas Fournier, Arnaud Guillin. On the rate of convergence in Wasserstein distance of the empirical measure. Probability Theory and Related Fields, Springer Verlag, 2015, 162 (3-4), pp.707. hal- 00915365
HAL Id: hal-00915365 https://hal.archives-ouvertes.fr/hal-00915365 Submitted on 7 Dec 2013
HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. ON THE RATE OF CONVERGENCE IN WASSERSTEIN DISTANCE OF THE EMPIRICAL MEASURE
NICOLAS FOURNIER AND ARNAUD GUILLIN
Abstract. Let µN be the empirical measure associated to a N-sample of a given probability d distribution µ on R . We are interested in the rate of convergence of µN to µ, when measured in the Wasserstein distance of order p > 0. We provide some satisfying non-asymptotic Lp- bounds and concentration inequalities, for any values of p > 0 and d ≥ 1. We extend also the non asymptotic Lp-bounds to stationary ρ-mixing sequences, Markov chains, and to some interacting particle systems.
Mathematics Subject Classification (2010): 60F25, 60F10, 65C05, 60E15, 65D32. Keywords: Empirical measure, Sequence of i.i.d. random variables, Wasserstein distance, Concentration inequalities, Quantization, Markov chains, ρ-mixing sequences, Mc Kean-Vlasov particles system.
1. Introduction and results 1.1. Notation. Let d 1 and (Rd) stand for the set of all probability measures on Rd. For d ≥ P (R ), we consider an i.i.d. sequence (Xk)k 1 of -distributed random variables and, for N ∈ P1, the empirical measure ≥ ≥ N 1 := δ . N N Xk k =1 As is well-known, by Glivenko-Cantelli’s theorem, N tends weakly to as N (for example in probability, see Van der Vaart-Wellner [40] for details and various modes of→ convergence). ∞ The aim of the paper is to quantify this convergence, when the error is measured in some Wasserstein distance. Let us set, for p 1 and ,ν in (Rd), ≥ P p p( ,ν) = inf x y ξ(dx, dy) : ξ ( ,ν) , T Rd Rd | − | ∈ H × where ( ,ν) is the set of all probability measures on Rd Rd with marginals and ν. See VillaniH [41] for a detailed study of . The Wasserstein distance× on (Rd) is defined by Tp Wp P ( ,ν)= ( ,ν) if p (0, 1] and ( ,ν) = ( ( ,ν))1/p if p> 1. Wp Tp ∈ Wp Tp The present paper studies the rate of convergence to zero of p( N , ). This can be done in T 1 an asymptotic way, finding e.g. a sequence α(N) 0 such that limN α(N)− p( N , ) < 1 → T ∞ a.s. or limN α(N)− E( p( N , )) < . Here we will rather derive some non-asymptotic moment estimates such as T ∞ E( ( , )) α(N) for all N 1 Tp N ≤ ≥ as well as some non-asymptotic concentration estimates (also often called deviation inequalities) Pr( ( , ) x) α(N, x) for all N 1, all x> 0. Tp N ≥ ≤ ≥ 1 2 NICOLAS FOURNIER AND ARNAUD GUILLIN
They are naturally related to moment (or exponential moment) conditions on the law and we hope to derive an interesting interplay between the dimension d 1, the cost parameter p> 0 and these moment conditions. Let us introduce precisely these moment≥ conditions. For q > 0, α> 0, γ > 0 and (Rd), we define ∈P q γ x α Mq( ) := x (dx) and α,γ( ) := e | | (dx). Rd | | E Rd We now present our main estimates, the comparison with the existing results and methods will be developped after this presentation. Let us however mention at once that our paper relies on some recent ideas of Dereich-Scheutzow-Schottstedt [16].
1.2. Moment estimates. We first give some Lp bounds.
d Theorem 1. Let (R ) and let p> 0. Assume that Mq( ) < for some q>p. There exists a constant C depending∈P only on p, d, q such that, for all N 1, ∞ ≥ 1/2 (q p)/q N − + N − − if p>d/2 and q =2p, p/q 1/2 (q p)/q E ( p( N , )) CM ( ) N − log(1 + N)+ N − − if p = d/2 and q =2p, T ≤ q p/d (q p)/q N − + N − − if p (0, d/2) and q = d/(d p). ∈ − Observe that when has sufficiently many moments (namely if q > 2p when p d/2 and (q p)/q ≥ q > dp/(d p) when p (0, d/2)), the term N − − is small and can be removed. We could easily treat,− for example,∈ the case p > d/2 and q = 2p but this would lead to some logarithmic terms and the paper is technical enough. This generalizes [16], in which only the case p [1, d/2) (whence d 3) and q > dp/(d p) was treated. The argument is also slightly simplified.∈ ≥ − To show that Theorem 1 is really sharp, let us give examples where lower bounds can be derived quite precisely. d (a) If a = b R and = (δa + δb)/2, one easily checks (see e.g. [16, Remark 1]) that ∈ 1/2 E( p( N , )) cN − for all p 1. Indeed, we have N = ZN δa + (1 ZN )δb with ZN = T1 N ≥ ≥ p − N − 1 1 Xi=a , so that p( N , ) = a b ZN 1/2 , of which the expectation is of order { } T | − | | − | N 1/2. − 1/2 (b) Such a lower bound in N − can easily be extended to any (possibly very smooth) of which the support is of the form A B with d(A, B) > 0 (simply note that p( N , ) p 1 ∪N T ≥ d (A, B) ZN (A) , where ZN = N − 1 1 Xi A ). | − | { ∈ } (c) If is the uniform distribution on [ 1, 1]d, it is well-known and not difficult to prove that p/d − d for p> 0, E( p( N , )) cN − . Indeed, consider a partition of [ 1, 1] into (roughly) N cubes T 1/d ≥ − with length N − . A quick comptation shows that with probability greater than some c > 0 (uniformly in N), half of these cubes will not be charged by N . But on this event, we clearly have 1/d p( N , ) aN − for some a > 0, because each time a cube is not charged by N , a (fixed) T ≥ 1/d proportion of the mass of (in this cube) is at distance at least N − /2 of the support of N . One easily concludes. (d) When p = d/2 = 1, it has been shown by Ajtai-Koml´os-Tusn´ady [2] that for the uniform measure on [ 1, 1]d, ( , ) c(log N/N)1/2 with high probability, implying that − T1 N ≃ E( ( , )) c(log N/N)1/2. T1 N ≥ CONVERGENCEOFTHEEMPIRICALMEASURE 3
q d (e) Let (dx) = c x − − 1 x 1 dx for some q > 0. Then Mr( ) < for all r (0, q) | | {| |≥ } (q p)/q 1/q ∞ ∈ and for all p 1, E( ( , )) cN − − . Indeed, P( ( x N )=0)=( ( x < ≥ Tp N ≥ N {| | ≥ } {| | N 1/q ))N = (1 c/N)N c > 0 and ( x 2N 1/q ) c/N. One easily gets convinced that } p/q− ≥ {| | ≥ 1/q } ≥ p( N , ) N 1 ( x N 1/q )=0 ( x 2N ), from which the claim follows. T ≥ { N {| |≥ } } {| |≥ } As far as general laws are concerned, Theorem 1 is really sharp: the only possible improvements are the following. The first one, quite interesting, would be to replace log(1 + N) by something like log(1 + N) when p = d/2 (see point (d) above). It is however not clear it is feasible in full generality. The second one, which should be a mere (and not very interesting) refinement, would (q p)/q be to sharpen the bound in N − − when Mq( ) < : point (e) only shows that there is with (q ∞p)/q ε M ( ) < for which we have a lowerbound in N − − − for all ε> 0. q ∞ However, some improvements are possible when restricting the class of laws . First, when is the uniform distribution in [ 1, 1]d, the results of Talagrand [38, 39] strongly suggest that p/d− 1/2 when d 3, E( p( N , )) N − for all p > 0, and this is much better than N − when p is ≥ T ≃ 1 large. Such a result would of course immediately extend to any distribution = λ F − , for λ the uniform distribution in [ 1, 1]d and F :[ 1, 1]d Rd Lipschitz continuous. In◦ any case, a smoothness assumption for −cannot be sufficient,− see → point (b) above. p/d Second, for irregular laws, the convergence can be much faster that N − when p < d/2, see 1/2 point (a) above where, in an extreme case, we get N − for all values of p > 0. It is shown by Dereich-Scheutzow-Schottstedt [16] (see also Barthe-Bordenave [3]) that indeed, for a singular law, p/d lim N − E( ( , )) = 0. N Tp N 1.3. Concentration inequalities. We next state some concentration inequalities. Theorem 2. Let (Rd) and let p> 0. Assume one of the three following conditions: ∈P (1) α > p, γ > 0, ( ) < , ∃ ∃ Eα,γ ∞ (2) or α (0,p), γ > 0, ( ) < , ∃ ∈ ∃ Eα,γ ∞ (3) or q> 2p, M ( ) < . ∃ q ∞ Then for all N 1, all x (0, ), ≥ ∈ ∞ N P( p( , ) x) a(N, x)1 x 1 + b(N, x), T ≥ ≤ { ≤ } where exp( cNx2) if p>d/2, − a(N, x)= C exp( cN(x/ log(2 + 1/x))2) if p = d/2, − exp( cNxd/p) if p [1, d/2) − ∈ and α/p exp( cNx )1 x>1 under (1), − { } (α ε)/p α/p b(N, x)= C exp( c(Nx) − )1 x 1 + exp( c(Nx) )1 x>1 ε (0, α) under (2), − { ≤ } − { } ∀ ∈ (q ε)/p N(Nx)− − ε (0, q) under (3). ∀ ∈ The positive constants C and c depend only on p, d and either on α, γ, α,γ( ) (under (1)) or on α, γ, ( ),ε (under (2)) or on q,M ( ),ε (under (3)). E Eα,γ q We could also treat the critical case where α,γ( ) < with α = p, but the result we could obtain is slightly more intricate and not very satisfyingE for∞ small value of x (even if good for large ones). 4 NICOLAS FOURNIER AND ARNAUD GUILLIN
Remark 3. When assuming (2) with α (0,p), we actually also prove that ∈ 2 δ α/p b(N, x) C exp( cNx (log(1 + N))− )+ C exp( c(Nx) ), ≤ − − with δ =2p/α 1, see Step 5 of the proof of Lemma 13 below. This allows us to extend the inequality − α/p b(N, x) C exp( c(Nx) ) to all values of x xN , for some (rather small) xN depending on N,α,p.≤ But for very− small values of x > 0, this≥ formula is less interesting than that of Theorem 2. Despite much effort, we have not been able to get rid of the logarithmic term. We believe that these estimates are quite satisfying. To get convinced, first observe that the scales seem to be the good ones. Recall that E( ( , )) = ∞ P( ( N , ) x)dx. Tp N 0 Tp ≥ ∞ p/d 1/2 (a) One easily checks that 0 a(N, x)dx CN − if p
1.4. Comments. The control of the distance between the empirical measure of an i.i.d. sample and its true distribution is of course a long standing problem central both in probability, statistics and informatics with a wide number of applications: quantization (see Delattre-Graf-Luschgy- Pag`es [14] and Pag`es-Wilbertz [33] for recent results), optimal matching (see Ajtai-Koml´os-Tusn´ady [2], Dobri´c-Yukich [19], Talagrand [39], Barthe-Bordenave [3]), density estimation, clustering (see Biau-Devroye-Lugosi [5] and Lalo¨e[26]), MCMC methods (see [36] for bounds on ergodic averages), particle systems and approximations of partial differential equations (see Bolley-Guillin-Villani [11] and Fournier-Mischler [22]). We refer to these papers for an extensive introduction on this vast topic. If many distances can be used to consider the problem, the Wasserstein distance is quite natural, in particular in quantization or for particle approximations of P.D.E.’s. However the depth of the problem was discovered only recently by Ajtai-Koml´os-Tusn´ady [2], who considered the uniform measure on the square, investigated thoroughly by Talagrand [39]. As a review of the litterature is somewhat impossible, et us just say that the methods involved were focused on two methods inherited by the definitions of the Wasserstein distance: the construction of a coupling or by duality to control a particular empirical process. CONVERGENCEOFTHEEMPIRICALMEASURE 5
Concerning moment estimates (as in Theorem 1), some results can be found in Horowitz- Karandikar [25], Rachev-R¨uschendorf [35] and Mischler-Mouhot [32]. But theses results are far from optimal, even when assuming that is compactly supported. Very recently, strickingly clever alternatives were considered by Boissard-Le Gouic [7] and by Dereich-Scheutzow-Schottstedt [16]. Unfortunately, the construction of Boissard-Le Gouic, based on iterative trees, was a little too complicated to yield sharp rates. On the contrary, the method of [16], exposed in details in the next section, is extremely simple, robust, and leads to the almost optimal results exposed here. Some sharp moment estimates were already obtained in [16] for a limited range of parameters. Concerning concentration estimates, only few results are available. Let us mention the work of Bolley-Guillin-Villani [11] and very recently by Boissard [6], on which we considerably improve. Our assumptions are often much weaker (the reference measure was often assumed to satisfy some functional inequalities, which may be difficult to verify and usually include more ”structure” than mere integrability conditions) and Pr[ p( N , ) x] was estimated only for rather large values of x. In particular, when integrating theT concentration≥ estimates of [11], one does never find the good moment estimates, meaning that the scales are not the good ones. Moreover, the approach of [16] is robust enough so that we can also give some good moment bounds for the Wasserstein distance between the empirical measure of a Markov chain and its invariant distribution (under some conditions). This could be useful for MCMC methods because our results are non asymptotic. We can also study very easily some ρ-mixing sequences (see Doukhan [20]), for which only very few results exist, see Biau-Devroye-Lugosi [7]. Finally, we show on an example how to use Theorem 1 to study some particle systems. For all these problems, we might also obtain some concentration inequalities, but this would need further refinements which are out of the scope of the present paper, somewhat already technical enough, and left for further works.
1.5. Plan of the paper. In the next section, we state some general upper bounds of ( ,ν), Tp for any ,ν (Rd), essentially taken from [16]. Section 3 is devoted to the proof of Theorem 1. Theorem 2 is∈P proved in three steps: in Section 4 we study the case where is compactly supported and where N is replaced by a Poisson(N)-distributed random variable, which yields some pleasant independance properties. We show how to remove the randomization in Section 5, concluding the case where is compactly supported. The non compact case is studied in Section 6. The final Section 7 is devoted to dependent random variables: ρ-mixing sequences, Markov chains and a particular particle system.
2. Coupling The following notion of distance, essentially taken from [16], is the main ingredient of the paper.
d dℓ Notation 4. (a) For ℓ 0, we denote by ℓ the natural partition of ( 1, 1] into 2 translations ℓ ℓ d ≥ P d − of ( 2− , 2− ] . For two probability measures ,ν on ( 1, 1] and for p> 0, we introduce − − p 2 1 pℓ ( ,ν) := − 2− (F ) ν(F ) , Dp 2 | − | ℓ 1 F ≥ ∈Pℓ which obviously defines a distance on (( 1, 1]d), always bounded by 1. d P − n n d n 1 n 1 d (b) We introduce B0 := ( 1, 1] and, for n 1, Bn := ( 2 , 2 ] ( 2 − , 2 − ] . For Rd − ≥ − \d − ( ) and n 0, we denote by Bn the probability measure on ( 1, 1] defined as the image of∈P / (B ) by≥ the map x x/2Rn. For two probability measures ,ν− on Rd and for p> 0, we |Bn n → 6 NICOLAS FOURNIER AND ARNAUD GUILLIN introduce ( ,ν) := 2pn (B ) ν(B ) + ( (B ) ν(B )) ( , ν) . Dp | n − n | n ∧ n Dp RBn RBn n 0 ≥ A little study, using that 1 on (( 1, 1]d), shows that this defines a distance on (Rd). Dp ≤ P − P Having a look at p in the compact case, one sees that in some sense, it measures distance of the two probability measuresD simultaneously at all the scales. The optimization procedure can be made for all scales and outperforms the approach based on a fixed diameter covering of the state space (which is more or less the approach of Horowitz-Karandikar [25]). Moreover one sees that the principal control is on π(F ) (F ) which is a quite simple quantity. The next results are slightly modified versions of| estimates− found| in [16], see [16, Lemma 2] for the compact case and [16, proof of Theorem 3] for the non compact case. It contains the crucial remark that p is an upper bound (up to constant) of the Wasserstein distance. D Lemma 5. Let d 1 and p > 0. For all pairs of probability measures ,ν on Rd, ( ,ν) ≥ Tp ≤ κ ( ,ν), with κ := 2p(1+d/2)(2p + 1)/(2p 1). p,dDp p,d − Proof. We separate the proof into two steps. Step 1. We first assume that and ν are supported in ( 1, 1]d. We infer from [16, Lemma 2], in which the conditions p 1 and d 3 are clearly not used,− that, since the diameter of ( 1, 1]d ≥ ≥ − is 21+d/2, p(1+d/2) 1 pℓ (C) ν(C) ( ,ν) 2 − 2− (F ) , Tp ≤ (F ) − ν(F ) ℓ 0 F ℓ C child of F ≥ ∈P where “C child of F ” means that C and C F . Consequently, ∈Pℓ+1 ⊂ p(1+d/2) 1 pℓ ν(C) ( ,ν) 2 − 2− (F ) ν(F ) + (C) ν(C) Tp ≤ ν(F ) | − | | − | ℓ 0 F ℓ C child of F ≥ ∈P
p(1+d/2) 1 pℓ 2 − 2− (F ) ν(F ) + (C) ν(C) ≤ | − | | − | ℓ 0 F C ≥ ∈Pℓ ∈P ℓ+1 p(1+d/2) 1 p pℓ 2 − (1+2 ) 2− (F ) ν(F ) , ≤ | − | ℓ 1 F ≥ ∈Pℓ which is nothing but κp,d p( ,ν). We used that F (F ) ν(F ) = 0. D ∈P0 | − | In [16], Dereich-Scheutzow-Schottstedt use directly the formula with the children to study the rate of convergence of empirical measures. This leads to some (small) technical complications, and does not seem to improve the estimates. Step 2. We next consider the general case. We consider, for each n 1, the optimal coupling ≥ πn(dx, dy) between Bn and Bn ν for p. We define ξn(dx, dy) as the image of πn by the map (x, y) (2nx, 2ny),R which clearlyR belongsT to ( / (B ),ν /ν(B )) and satisfies x → H |Bn n |Bn n | − y pξ (dx, dy)=2np x y pπ (dx, dy)=2np ( , ν). | n | − | n Tp RBn RBn 1 Next, we introduce q := 2 n 0 ν(Bn) (Bn) and we define ≥ | − | α(dx)β(dy) ξ(dx, dy)= ( (B ) ν(B ))ξ (dx, dy)+ , n ∧ n n q n 0 ≥ CONVERGENCEOFTHEEMPIRICALMEASURE 7 where
Bn (dx) ν Bn (dy) α(dx) := ( (Bn) ν(Bn))+ | and β(dy) := (ν(Bn) (Bn))+ | . − (Bn) − ν(Bn) n 0 n 0 ≥ ≥ Using that q = (ν(B ) (B )) = ( (B ) ν(B )) =1 (ν(B ) (B )), n − n + n − n + − n ∧ n n 0 n 0 n 0 ≥ ≥ ≥ it is easily checked that ξ ( ,ν). Furthermore, we have, setting cp = 1 if p (0, 1] and p 1 ∈ H ∈ cp =2 − if p> 1, α(dx)β(dy) 1 x y p c ( x p + y p)α(dx)β(dy) | − | q ≤ q p | | | | =c x pα(dx)+ c y pβ(dy) p | | p | | c 2pn[( (B ) ν(B )) + (ν(B ) (B )) ] ≤ p n − n + n − n + n 0 ≥ =c 2pn (B ) ν(B ) . p | n − n | n 0 ≥ Recalling that x y pξ (dx, dy) 2np ( , ν), we deduce that | − | n ≤ Tp RBn RBn ( ,ν ) x y pξ(dx, dy) Tp ≤ | − | 2np (c (B ) ν(B ) + ( (B ) ν(B )) ( , ν)) . ≤ p| n − n | n ∧ n Tp RBn RBn n 0 ≥ We conclude using Step 1 and that c κ . p ≤ p,d When proving the concentration inequalities, which is very technical, it will be good to break the proof into several steps to separate the difficulties and we will first treat the compact case. On the contrary, when dealing with moment estimates, the following formula will be easier to work with. Lemma 6. Let p> 0 and d 1. There is a constant C, depending only on p, d, such that for all ,ν (Rd), ≥ ∈P pn pℓ n n ( ,ν) C 2 2− (2 F B ) ν(2 F B ) Dp ≤ | ∩ n − ∩ n | n 0 ℓ 0 F ≥ ≥ ∈Pℓ with the notation 2nF = 2nx : x F . { ∈ } n n Proof. For all n 1, we have (Bn) ν(Bn) = (2 F Bn) ν(2 F Bn) and ≥ | − | F 0 | ∩ − ∩ | ( (B ) ν(B )) ( , ν) is smaller than ∈P n ∧ n Dp RBn RBn ( (B ) ν(B )) ( , ν) n ∧ n Dp RBn RBn n n pℓ (2 F Bn) ν(2 F Bn) (Bn) 2− ∩ ∩ ≤ (Bn) − ν(Bn) ℓ 1 F ℓ ≥ ∈P pℓ n n (Bn) pℓ n 2− (2 F Bn) ν(2 F Bn) + 1 2− ν(2 F Bn). ≤ | ∩ − ∩ | − ν(Bn) ∩ ℓ 1 F ℓ ℓ 1 F ℓ ≥ ∈P ≥ ∈P p p This last term is smaller than 2− (B ) ν(B ) /(1 2− ) and this ends the proof. | n − n | − 8 NICOLAS FOURNIER AND ARNAUD GUILLIN
3. Moment estimates The aim of this section is to give the d Proof of Theorem 1. We thus assume that (R ) and that Mq( ) < for some q>p. By ∈ P ∞ q(n 1) a scaling argument, we may assume that Mq( ) = 1. This implies that (Bn) 2− − for all n 0. By Lemma 5, we have ( , ) κ ( , ), so that it suffices to study≤ E( ( , )). ≥ Tp N ≤ p,dDp N Dp N For a Borel subset A Rd, since N (A) is Binomial(N, (A))-distributed, we have ⊂ N E( (A) (A) ) min 2 (A), (A)/N . | N − | ≤ Using the Cauchy-Scharz inequality and that #( )=2 dℓ, we deduce that for all n 0, all ℓ 0, Pℓ ≥ ≥ E( (2nF B ) (2nF B ) ) min 2 (B ), 2dℓ/2( (B )/N)1/2 . | N ∩ n − ∩ n | ≤ n n F ∈Pℓ q(n 1) Using finally Lemma 6 and that (B ) 2− − , we find n ≤ pn pℓ qn dℓ/2 qn 1/2 (4) E( ( , )) C 2 2− min 2− , 2 (2− /N) . Dp N ≤ n 0 ℓ 0 ≥ ≥ Step 1. Here we show that for all ε (0, 1), all N 1, ∈ ≥ min ε, (ε/N)1/2 if p>d/2, { } pℓ dℓ/2 1/2 1/2 2− min ε, 2 (ε/N) C min ε, (ε/N) log(2 + εN) if p = d/2, ≤ { } ℓ 0 p/d ≥ min ε,ε(εN)− if p (0, d/2). { } ∈ First of all, the bound by Cε is obvious in all cases (because p > 0). Next, the case p > d/2 is dℓN,ε immediate. If p d/2, we introduce ℓN,ε := log(2 + εN)/(d log 2) , for which 2 2+ εN and get an upper≤ bound in ⌊ ⌋ ≃ 1/2 (d/2 p)ℓ pℓ (ε/N) 2 − + ε 2− . ℓ ℓ ℓ ℓ ≤ N,ε ≥ N,ε If p = d/2, we find an upper bound in 1/2 pℓ 1/2 1/2 1/2 (ε/N) ℓ + Cε2− N,ε C(ε/N) log(2 + εN)+ Cε(1 + εN)− C(ε/N) log(2 + εN) N,ε ≤ ≤ as desired. If p (0, d/2), we get an upper bound in ∈ 1/2 (d/2 p)ℓ pℓ 1/2 1/2 p/d p/d C(ε/N) 2 − N,ε + Cε2− N,ε C(ε/N) (2 + εN) − + Cε(2 + εN)− . ≤ 1/2 p/d 1/2 p/d If εN 1, then (2 + εN) − (3εN) − and the conclusion follows. If now εN (0, 1), ≥ ≤ p/d ∈ the result is obvious because min ε,ε(εN)− = ε. { } qn Step 2: p>d/2. By (4) and Step 1 (with ε =2− ), we find 1/2 pn qn qn 1/2 N − if q> 2p, E( ( , )) C 2 min 2− , (2− /N) C Dp N ≤ ≤ (q p)/q n 0 N − − if q (p, 2p). ≥ ∈ Indeed, this is obvious if q > 2p, while the case q (p, 2p) requires to separate the sum in two parts n n and n>n with n = log N/(q log 2)∈ . This ends the proof when p>d/2. ≤ N N N ⌊ ⌋ qn Step 3: p = d/2. By (4) and Step 1 (with ε =2− ), we find
pn qn qn 1/2 qn E( ( , )) C 2 min 2− , (2− /N) log(2 + 2− N) . Dp N ≤ n 0 ≥ CONVERGENCEOFTHEEMPIRICALMEASURE 9
If q> 2p, we immediately get a bound in (p q/2)n 1/2 1/2 E( ( , )) C 2 − N − log(2 + N) C log(2 + N)N − , Dp N ≤ ≤ n 0 ≥ which ends the proof (when p = d/2 and q> 2p). If q (p, 2p), we easily obtain, using that log(2 + x) 2 log x for all x 2, an upper bound in ∈ ≤ ≥ (p q)n (p q/2)n 1/2 nq E( p( N , )) C 1 N<2.2nq 2 − + C 1 N 2.2nq 2 − N − log(N2− ) D ≤ { } { ≥ } n 0 n 0 ≥ ≥ nN (q p)/q 1/2 (p q/2)n CN − − + CN − 2 − (log N nq log 2) ≤ − n=0 (q p)/q 1/2 =:CN − − + CN − KN , where n = log(N/2)/(q log 2) . A tedious exact computation shows that N ⌊ ⌋ (p q/2)(n +1) 2 − N 1 KN = log N (p q/2) − 2 − 1 − (p q/2)(n +1) (p q/2)(n +2) (p q/2) 2 − N 1 n +1 2 − N 2 − q log 2 (n + 1) − + N − . − N 2(p q/2) 1 2(p q/2) 1 − (2(p q/2) 1)2 − − − − − − Using that the contribution of the middle term of the second line is negative and the inequality log N (n + 1)q log 2 log2 (because (n + 1)q log 2 log(N/2)), we find − N ≤ N ≥ (p q/2)n p/q 1/2 K C2 − N CN − . N ≤ ≤ (q p)/q 1/2 p/q 1/2 (q p)/q We finally have checked that E( p( N , )) CN − − + CN − N − CN − − , which ends the proof when p = d/D2. ≤ ≤ Step 4: p (0, d/2). We then have, by (4) and Step 1, ∈ pn qn qn(1 p/d) p/d E( ( , )) C 2 min 2− , 2− − N − . Dp N ≤ n 0 ≥ If q > dp/(d p), which implies that q(1 p/d) > p, we immediately get an upper bound by p/d − − CN − , which ends the proof when p
d d Proposition 8. Assume that is supported in ( 1, 1] . Let ΠN be a Poisson measure on R with − d 1 intensity measure N and introduce the associated empirical measure ΨN = (ΠN (R ))− ΠN . Let p 1 and d 1. There are some positive constants C,c (depending only on d, p) such that for all N≥ 1, all x≥ (0, ), ≥ ∈ ∞ exp( Nf(cx)) if p>d/2, − P Π (Rd) (Ψ , ) Nx C exp( Nf(cx/ log(2 + 1/x))) if p = d/2, N Dp N ≥ ≤ − exp( Nf(cx)) + exp cNxd/p if p [1, d/2). − − ∈ We start with some easy and well-known concentration inequalities for the Poisson distribution. Lemma 9. For λ> 0 and X a Poisson(λ)-distributed random variable, we have (a) E(exp(θX)) = exp(λ(eθ 1)) for all θ R; (b) E(exp(θ X λ )) 2 exp(−λ(eθ 1 θ))∈ for all θ> 0; (c) P(X > λx| ) − exp(| ≤λg(x)) for all− x>− 0; (d) P( X λ >≤ λx) −2 exp( λf(x)) for all x> 0; (e) P(X| >− λx|) λ for≤ all x>−0. ≤ Proof. Point (a) is straightforward. For point (b), write E(exp(θ X λ )) eθλE(exp( θX)) + θλ θ θ | − | ≤ − e− E(exp(θX)), use (a) and that λ(e− 1+ θ) λ(e 1 θ). For point (c), write P(X > θλx − ≤ − − λx) e− E[exp(θX)], use (a) and optimize in θ. Use the same scheme to deduce (d) from (b). ≤ λ Finally, for x> 0, P(X > λx) P(X > 0)=1 e− λ. ≤ − ≤ We can now give the Proof of Proposition 8. We fix x> 0 for the whole proof. Recalling Notation 4-(a), we have
d pℓ d Π (R ) (Ψ , )= C 2− Π (F ) Π (R ) (F ) N Dp N | N − N | ℓ 1 F ≥ ∈Pℓ d pℓ C Π (R ) N + C 2− Π (F ) N (F ) ≤ | N − | | N − | ℓ 1 F ≥ ∈Pℓ ℓ0 d d pℓ pℓ C Π (R ) N + C(N + Π (R ))2− 0 + C 2− Π (F ) N (F ) ≤ | N − | N | N − | ℓ=1 F ∈Pℓ for any choice of ℓ0 N. We will choose ℓ0 later, depending on the value of x. For any nonnegative ∈ ℓ family r such that 0 r 1, we thus have ℓ 1 ℓ ≤ d ε(N, x) :=P ΠN (R ) p(ΨN , ) Nx D ≥ P Rd P Rd pℓ0 ΠN ( ) N cNx + ΠN ( ) N(cx2 1) ≤ | − |≥ ≥ − ℓ0 + P Π (F ) N (F ) cNx2pℓr . | N − |≥ ℓ ℓ=1 F ∈Pℓ By Lemma 9-(c)-(d), since Π (Rd) is Poisson(N)-distributed, P(Π (Rd) N(cx2pℓ0 1)) N N ≥ − ≤ exp( Ng(cx2pℓ0 1)) and P( Π (Rd) N cNx) 2 exp( Nf(cx)). Next, using that the − − | N − | ≥ ≤ − family (ΠN (F ))F ℓ is independent, with ΠN (F ) Poisson(N (F ))-distributed, we use Lemma 9-(a) and that #(∈P )=2ℓd to obtain, for any θ> 0, Pℓ
N (F )(eθ θ 1) 2dℓ N(eθ θ 1) E exp θ Π (F ) N (F ) 2e − − 2 e − − . | N − | ≤ ≤ F F ∈Pℓ ∈Pℓ CONVERGENCEOFTHEEMPIRICALMEASURE 11
Hence
dℓ P Π (F ) N (F ) cNx2pℓr exp cθNx2pℓr 22 exp N(eθ θ 1) . | N − |≥ ℓ ≤ − ℓ − − F ∈Pℓ pℓ Choosing θ = log(1 + cx2 rℓ), we find
dℓ P Π (F ) N (F ) cNx2pℓr 22 exp( Nf(cx2pℓr )). | N − |≥ ℓ ≤ − ℓ F ∈Pℓ We have checked that
ℓ0 dℓ ε(N, x) 2 exp( Nf(cx)) + exp( Ng(cx2pℓ0 1)) + 22 exp( Nf(cx2pℓr )). ≤ − − − − ℓ ℓ=1 At this point, the value of c> 0 is not allowed to vary anymore. We introduce some other positive constants a whose value may change from line to line. Case 1: cx > 2. Then we choose ℓ = 1 and r = 1. We have cx2pℓ0 1= 2pcx 0 1 − − 1 (2p 1)cx + 1 whence g(cx2pℓ0 1) g((2p 1)cx +1) = f((2p 1)cx). We also have dℓ d ≥ℓ0 2 − pℓ 2 − ≥ p − − ℓ=1 2 exp( Nf(cx2 rℓ)) =2 exp( Nf(2 cx)). We finally get ε(N, x) C exp( Nf(ax)), which proves the− statement (in the three− cases, when cx > 2). ≤ − Case 2: cx 2. We choose ℓ so that (1+2/(cx)) 2pℓ0 2p(1+2/(cx)), i.e. ≤ 0 ≤ ≤ ℓ := log(1 + 2/(cx))/(p log 2) +1. 0 ⌊ ⌋ This implies that cx2pℓ0 2+ cx. Hence g(cx2pℓ0 1) g(1 + cx) = f(cx). Furthermore, we pℓ pℓ0 ≥p p+2 − ≥ pℓ 2 2pℓ 2 have cx2 rℓ cx2 2 (2 + cx) 2 for all ℓ ℓ0, whence f(cx2 rℓ) ax 2 rℓ (because ≤ ≤ ≤ ≤ dℓ ≥ f(x) ax2 for all x [0, 2p+2]). We thus end up with (we use that 22 exp(2dℓ)) ≥ ∈ ≤ ℓ0 ε(N, x) 3 exp( Nf(cx)) + exp(2dℓ Nax222pℓr2). ≤ − − ℓ ℓ=1 Now the value of a> 0 is not allowed to vary anymore, and we introduce a′ > 0, whose value may change from line to line. η ηℓ Case 1.1: p > d/2. We take rℓ := (1 2− )2− for some η > 0 such that 2(p η) > d. If Nx2 1, we easily get − − ≥ ℓ0 dℓ 2 2(p η)ℓ ε(N, x) 3 exp( Nf(cx)) + exp(2 Na′x 2 − ) ≤ − − ℓ=1 2 3 exp( Nf(cx)) + C exp( a′Nx ) ≤ − − C exp( Nf(a′x)). ≤ − The last inequality uses that y2 f(y) for all y > 0. If finally Nx2 1, we obviously have ≥ ≤ ε(N, x) 1 exp(1 Nx2) C exp( Nx2) C exp( Nf(x)). ≤ ≤ − ≤ − ≤ −
We thus always have ε(N, x) C exp( Nf(a′x)) as desired. ≤ − 12 NICOLAS FOURNIER AND ARNAUD GUILLIN
Case 2.2: p = d/2. We choose r := 1/ℓ . Thus, if aN(x/ℓ )2 2, we easily find ℓ 0 0 ≥ ℓ0 ε(N, x) 3 exp( Nf(cx)) + exp(2dℓ(1 aN(x/ℓ )2) ≤ − − 0 ℓ=1 2 3 exp( Nf(cx)) + C exp( a′N(x/ℓ ) ) ≤ − − 0 3 exp( Nf(cx)) + C exp( Nf(a′x/ℓ )) ≤ − − 0 C exp( Nf(a′x/ℓ )) ≤ − 0 because ℓ 1 and f is increasing. If now aN(x/ℓ )2 < 2, we just write 0 ≥ 0 ε(N, x) 1 exp(2 aN(x/ℓ )2) C exp( aN(x/ℓ )2) C exp( Nf(ax/ℓ )). ≤ ≤ − 0 ≤ − 0 ≤ − 0 We thus always have ε(N, x) C exp( Nf(a′x/ℓ )). Using that ℓ C log(2 + 1/x), we immedi- ≤ − 0 0 ≤ ately conclude that ε(N, x) C exp( Nf(a′x/ log(2 + 1/x))) as desired. ≤ − (d/2 p)(ℓ ℓ ) p d/2 Case 2.3: p [1, d/2). We choose r := κ2 − − 0 with κ =1/(1 2 − ). For all ℓ ℓ , ∈ ℓ − ≤ 0 dl 2 2pℓ 2 2 d/p 2pℓ (d 2p)(ℓ ℓ ) 2 d/p (d 2p)ℓ d/p 2 aNx 2 r = aκ Nx 2 2 − − 0 x − 2 − /(Nax ) − ℓ − − 2 d/p 2pℓ (d 2p)ℓ (d 2p)ℓ 2 d/p aκ Nx 2 b2 − 2 − /(Naκ x ) ≤− − (d 2p )ℓ d/p 2 where the constant b > 0 is such that 2− − 0 bx − (the existence of b is easily checked). ≥ Hence if Naκ2xd/p 2/b, we find ≥ 2dl aNx222pℓr2 abκ2Nxd/p2dℓb/2 − ℓ ≤− and thus, still using that Nxd/p 2/(abκ2), ≥ ℓ0 dℓ 2 2 2pℓ 2 d/p exp(2 Nc x 2 r ) C exp( a′Nx ). − ℓ ≤ − ℓ=1 d/p 2 d/p Consequently, we have ε(N, x) 3 exp( Nf(cx)) + C exp( a′Nx ) if Naκ x 2/b. As ≤ − − ≥ usual, the case where Naκ2xd/p 2/b is trivial, since then ≤ 2 d/p d/p ε(N, x) 1 exp(2/b Naκ x ) C exp( a′Nx ). ≤ ≤ − ≤ − This ends the proof.
5. Depoissonization in the compact case We next check the following compact version of Theorem 2. Proposition 10. Assume that is supported in ( 1, 1]d. Let p> 0 and d 1 be fixed. There are some positive constants C and c (depending only on− p, d) such that for all N≥ 1, all x (0, ), ≥ ∈ ∞ exp( cNx2) if p>d/2; − 2 P [ p( N , ) x] 1 x 1 C exp cN(x/ log(2 + 1/x)) if p = d/2; D ≥ ≤ { ≤ } − exp cNxd/p if p (0, d/2). − ∈ We will need the following easy remark. Lemma 11. For all N 1, for X Poisson(N)-distributed, for all k 0,..., √N , ≥ ∈{ ⌊ ⌋} 1/2 2 P[X = N + k] κ N − where κ = e− /√2. ≥ 0 0 CONVERGENCEOFTHEEMPIRICALMEASURE 13
Proof. By Perrin [34], we have N! e√N(N/e)N . Thus ≤ N+k N+k N+k N N N 1 N 1 N k 1 P[X = N + k]=e− e− − e − . (N + k)! ≥ √N + k((N + k)/e)N+k ≥ √ N + k 2N Since log(1 + x) x on (0, 1), we have ((N + k)/N)N+k exp(k + k2/N) exp(k + 1), so that ≤ 2 ≤ ≤ P[X = N + k] e− /√2N. ≥ Proof of Proposition 10. The probability indeed vanishes if x> 1, since is smaller than 1 when Dp restricted to probability measures on ( 1, 1]d. − d Step 1. We introduce a Poisson measure ΠN on R with intensity measure N and the associated empirical measure Ψ = Π /Π (Rd). Conditionally on Π (Rd) = n ,Ψ has the N N N { N } N same law as n (the empirical measure of n i.i.d. random variables with law ). Consequently, P Π (Rd) (Ψ , ) Nx = P[Π (Rd)= n]P [n ( , ) Nx] . N Dp N ≥ N Dp n ≥ n 0 ≥ d By Lemma 11 (since ΠN (R ) is Poisson(N)-distributed),
√N ⌊ ⌋ 1 1 d P [(N + k) ( , ) Nx] κ− P Π (R ) (Ψ , ) Nx , √ Dp N+k ≥ ≤ 0 N Dp N ≥ N k=0 which of course implies that (for all N 1, all x> 0), ≥ √N ⌊ ⌋ 1 1 d P [ ( , ) x] κ− P Π (R ) (Ψ , ) Nx . √ Dp N+k ≥ ≤ 0 N Dp N ≥ N k=0 Step 2. Here we prove that there is a constant A > 0 such that for any N 1, any k 1/2 ≥ ∈ 0,..., √N , any x>AN − , { ⌊ ⌋} P [ ( , ) x] P [ ( , ) x/2] . Dp N ≥ ≤ Dp N+k ≥ Build for all values of n 1 with the same i.i.d. family of -distributed random variables n ≥ (Xk)k 1. Then a.s., ≥ k N 1 N+k k 1 δ + δ . | N+k − N |T V ≤ N(N + k) Xj N + k Xj ≤ N + k ≤ √ 1 N T V N +1 T V 1/2 This obviously implies (recall Notation 4-(a)) that p ( N , N+k) CN − a.s. (where C depends D ≤ 1/2 only on p). By the triangular inequality, ( , ) ( , )+ CN − , whence Dp N ≤ Dp N+k 1/2 P [ ( , ) x] P ( , ) x CN − P [ ( , ) x/2] Dp N ≥ ≤ Dp N+k ≥ − ≤ Dp N+k ≥ 1/2 1/2 if x CN − x/2, i.e. x 2CN − . − ≥ ≥ 1/2 Step 3. Gathering Steps 1 and 2, we deduce that for all N 1, all x>AN − , ≥ √N 1 ⌊ ⌋ P [ ( , ) x] P [ ( , ) x/2] CP Π (Rd) (Ψ , ) Nx/2 . Dp N ≥ ≤ √ Dp N+k ≥ ≤ N Dp N ≥ N k=0 We next apply Proposition 8. Observing that, for x (0, 1], (i) exp( Nf(cx/2)) exp( cNx2) (case p>d/2),∈ (ii) exp(− Nf(cx/2 log(2≤ + 2−/x))) exp( cN(x/ log(2 + 1/x)2) (case p = d/2), − ≤ − 14 NICOLAS FOURNIER AND ARNAUD GUILLIN
(iii) exp( Nf(cx/2)) + exp(cN(x/2)d/p) exp( cNxd/p) (case p (0, d/2)) − 1/2 ≤ − ∈ 1/2 concludes the proof when x>AN − . But the other case is trivial, because for x AN − , ≤
P[ ( , ) x] 1 exp(A2 Nx2) C exp( Nx2), Dp N ≥ ≤ ≤ − ≤ − which is also smaller than C exp( N(x/ log(2 + 1/x))2) and than C exp( Nxd/p) (if d> 2p). − −
6. Concentration inequalities in the non compact case Here we conclude the proof of Theorem 2. We will need some concentration estimates for the Binomial distribution.
Lemma 12. Let X be Binomial(N,p)-distributed. Recall that f was defined in Notation 7. (a) P[ X Np Npz] (1 p(1+z) 1 + 1 z 1 ) exp( Npf(z)) for all z > 0. (b) P[|X − Np|≥ Npz] ≤ Np{ for all≤ z} > 1.{ ≤ } − | − |≥ ≤ θ N θ (c) E(exp( θX)) = (1 p + pe− ) exp( Np(1 e− )) for θ> 0. − − ≤ − − Proof. Point (c) is straightforward. Point (b) follows from the fact that for z > 1, P[ X Np Npz] = P[X Np(1 + z)] P[X =0]=1 (1 p)N pN. For point (a), we use| Bennett’s− |≥ inequality [4],≥ see Devroye-Lugosi≤ [17, Exercise− 2.2− page 11],≤ together with the obvious facts that P[X Np Npz] = 0 if p(1+z) > 1 and P[X Np Npz] = 0 if z > 1. The following elementary tedious− computations≥ also works: write P[ X− Np≤− Npz] = P(X Np(1 + z)) + P(N X N(1 p+zp)) =: ∆(p,z)+∆(1 p,zp/(1 |p)),− observe| ≥ that N X ≥Binomial(N, 1 p). Use− that≥ − − − θ N − ∼ − ∆(p,z) 1 p(1+z) 1 exp( θNp(1+z))(1 p+pe ) and choose θ = log((1 p)(1+z)/(1 p pz)), ≤ { ≤ } − − − − − this gives ∆(p,z) 1 p(1+z) 1 exp( N[p(1 + z) log(1 + z)+(1 p pz) log((1 p pz)/(1 p))]). ≤ { ≤ } − − − − − − A tedious study shows that ∆(p,z) 1 p(1+z) 1 exp( Npf(z)) and that ∆(1 p,zp/(1 p)) ≤ { ≤ } − − − ≤ 1 z 1 exp( Npf(z)). { ≤ } −
We next estimate the first term when computing ( , ). Dp N Lemma 13. Let (Rd) and p > 0. Assume (1), (2) or (3). Recall Notation 4 and put p pn ∈ P ZN := n 0 2 N (Bn) (Bn) . Let x0 be fixed. For all x> 0, ≥ | − | P p 2 [ZN x] C exp( cNx )1 x x0 ≥ ≤ − { ≤ } α/p exp( cNx )1 x>x under (1), − { 0} (α ε)/p α/p + C exp( c(Nx) − )1 x x + exp( c(Nx) )1 x>x ε (0, α) under (2), − { ≤ 0} − { 0} ∀ ∈ (q ε)/p N(Nx)− − ε (0, q) under (3). ∀ ∈ Proof. Under (1) or (2), we assume that γ = 1 without loss of generality (by scaling), whence − 2(n 1)α qn α,1( ) < and thus (Bn) Ce− for all n 0. Under (3), we have (Bn) C2− for E ∞ ≤ ≥ η ηn ≤ η all n 0. For η> 0 to be chosen later (observe that n 0(1 2− )2− = 1), putting c := 1 2− ≥ ≥ − − CONVERGENCEOFTHEEMPIRICALMEASURE 15
(p+η)n and zn := cx2− / (Bn),
P p P (ZN x) 1 zn 2 [ N N (Bn) N (Bn) N (Bn)zn] 1 ≥ ≤ { ≤ } | − |≥ ∧ n 0 ≥
+ 1 z >2 P [ N N (Bn) N (Bn) N (Bn)zn] 1 { n } | − |≥ ∧ n 0 ≥ =: I (N, x) 1+ J (N, x) 1. n ∧ n ∧ n 0 n 0 ≥ ≥ From now on, the value ofc > 0 is not allowed to vary anymore. We introduce another positive constant a> 0 whose value may change from line to line.
Step 1: bound of In. Here we show that under (3) (which is of course implied by (1) or (2)), if η (0,q/2 p), there is A > 0 such that ∈ − 0 2 2 In(N, x) C exp( aNx )1 x A if Nx 1. ≤ − { ≤ 0} ≥ n 0 ≥ This will obviously imply that for all N 1, all x> 0, ≥
2 In(N, x) 1 C exp( aNx )1 x A . ∧ ≤ − { ≤ 0} n 0 ≥ qn First, n 0 In(N, x) = 0 if zn > 2 for all n 0. Recalling that (Bn) C2− , this is the case ≥ (p+η q)n ≥ ≤ if x (2C/c) supn 0 2 − = (2C/c) := A0. Next, since N N (Bn) Binomial(N, (Bn)), ≥ ≥ ∼ Lemma 12-(a) leads us to 2 In(N, x) 21 zn 2 exp( N (Bn)f(zn)) 2 exp( N (Bn)zn/4)), ≤ { ≤ } − ≤ − 2 2 2 (q 2p 2η)n because f(x) x /4 for x [0, 2]. Since finally (Bn)zn/4 ax 2 − − , we easily conclude, since q 2p ≥2η> 0 and since∈ Nx2 1, that ≥ − − ≥ 2 (q 2p 2η)n 2 In(N, x) C exp( aNx 2 − − )1 x A C exp( aNx )1 x A . ≤ − { ≤ 0} ≤ − { ≤ 0} n 0 n 0 ≥ ≥
Step 2: bound of Jn under (1) or (2) when x A. Here we fix A > 0 and prove that if η> 0 is small enough, for all x (0, A] such that Nx2 ≤ 1, ∈ ≥ exp( aNx2) under (1), Jn(N, x) C − ≤ exp( aNx2) + exp( a(Nx)(α ε)/p) ε (0, α) under (2). n 0 − ≥ − − ∀ ∈ This will imply, as usual, that for all N 1, all x> 0, ≥ exp( aNx2) under (1), − Jn(N, x) 1 C 2 (α ε)/p ∧ ≤ exp( aNx ) + exp( a(Nx) − ) ε (0, α) under (2). n 0 − − ∀ ∈ ≥ By Lemma 12-(a)-(b) (since zn > 2 implies 1 (B )(1+z ) 1 + 1 z 1 1 z 1/ (B ) ), { n n ≤ } { n≤ } ≤ { n≤ n } Jn(N, x) 1 2 − 2(n 1)α because f(y) ay log y ay log[2 y] for y> 2. Since (B ) Ce− , we get ≥ ≥ ∨ n ≤ − − − (p+η)n (p+η)n 2(n 1)α 2 (n 1)α J (N, x) C min exp( aNx2− log[2 (ax2− e )]),Ne . n ≤ { − ∨ } A straightforward computation shows that there is a constant K such that for n n1 := K(1 + − (p+η)n 2(n 1)α (n 1)α ≥ ⌊ log log(K/x)) , we have log(ax2− e ) 2 − /2. Consequently, ⌋ ≥ − (p+η)n (α p η)n 2(n 1)α J (N, x) Cn exp( aNx2− 1 )+ C min exp( aNx2 − − ),e− n ≤ 1 − − n 0 n>n1 ≥ =CJ 1(N, x)+ CJ 2(N, x). 1 aNx2 We first show that J (N, x) Ce− (here we actually could get something much better). First, ≤ (p+η)n 3/2 since n = K + K log log(K/x) and x [0, A], we clearly have e.g. x2− 1 ax . Next, 1 ⌊ ⌋ ∈ ≥ Nx2 1 implies that 1/x (Nx3/2)2. Thus ≥ ≤ J 1(N, x) C(1 + log log(C(Nx3/2)2)) exp( aNx3/2) C exp( aNx3/2) exp( aNx2). ≤ − ≤ − ≤ − We now treat J 2(N, x). Step 2.1. Under (1), we immediately get, if η (0, α p) (recall that x [0, A]), ∈ − ∈ 2 (α p η)n 2 J (N, x) exp( aNx2 − − ) C exp( aNx) C exp( aNx ), ≤ − ≤ − ≤ − n 0 ≥ where we used that x A and Nx2 1 (whence Nx 1/A). ≤ ≥ ≥ Step 2.2. Under (2), we first write − − 2 (α p η)n 2(n 1)α (α p η)n 2(n2 1)α J (N, x) min exp( aNx2 − − ),e− n exp( cNx2 − − 2 )+ Ne− . ≤ − ≤ 2 − n 0 ≥ (n2 1)α α/(p+η) 2α We choose n2 := log(Nx)/((p + η) log 2) , which yields us to 2 − (Nx) /2 and (α p η)n ⌊ α/(p+η) ⌋ ≥ (Nx)2 − − 2 (Nx) . Consequently (recall that x (0, A]), ≤ ∈ J 2(N, x) C(1 + log(Nx)+ N) exp( a(Nx)α/(p+η)) C(1 + N) exp( a(Nx)α/(p+η)). ≤ − ≤ − For any fixed ε (0, α), we choose η > 0 small enough so that α/(p + η) (α ε)/p and we conclude that (recall∈ that Nx 1/A because Nx2 1 and x A) ≥ − ≥ ≥ ≤ 2 (α ε)/p (α ε)/p J (N, x) C(1 + N) exp( a(Nx) − ) C exp( a(Nx) − ). ≤ − ≤ − The last inequality is easily checked, using that Nx2 1 implies that N (Nx)2. ≥ ≤ Step 3: bound of J under (3). Here we show that for all ε (0, q), if η > 0 is small enough, n ∈ (q ε)/p 1 − J (N, x) CN if Nx 1. n ≤ Nx ≥ n 0 ≥ As usual, this will imply that for all x> 0, all N 1, ≥ (q ε)/p 1 − J (N, x) 1 CN . n ∧ ≤ Nx n 0 ≥ Exactly as in Step 2, we get from Lemma 12-(a)-(b) that J (N, x) min exp( aN (B )z log[2 z ]) ,N (B ) . n ≤ { − n n ∨ n n } CONVERGENCEOFTHEEMPIRICALMEASURE 17 (p+η)n Hence for n3 to be chosen later, since aN (Bn)zn = aNx2− , n3 (p+η)n qn J (N, x) C exp( aNx2− )+ CN 2− n ≤ − n 0 n=0 n>n3 ≥ (p+η)n qn Cn exp( aNx2− 3 )+ CN2− 3 . ≤ 3 − qn3 q (q ε)/p We choose n3 := (q ε) log(Nx)/(pq log 2) , which implies that 2− 2 (Nx)− − and that (p+η)n ⌊ (q−ε)(p+η)/(pq) ⌋ ≤ 2− 3 (Nx)− − . Hence ≥ 1 (q ε)(p+η)/(pq) (q ε)/p J (N, x) C log(Nx) exp( a(Nx) − − )+ CN(Nx)− − . n ≤ − n 0 ≥ If η (0,pε/(q ε)), then 1 (q ε)(p + η)/(pq) > 0, and thus ∈ − − − 1 (q ε)(p+η)/(pq) (q ε)/p log(Nx) exp( a(Nx) − − ) C(Nx)− − . − ≤ This ends the step. Step 4. We next assume (1) and prove that for all x A := 2p[M ( ) + (2 log ( ))p/α], ≥ 1 p Eα,1 Pr[Zp x] C exp( aNxα/p). N ≥ ≤ − Rd pn p A simple computation shows that for any ν ( ), n 0 2 ν(Bn) 2 Mp(ν), whence ∈ P ≥ ≤ p p p 1 N p p p 1 N α p/α Z 2 M ( )+2 N − X 2 M ( )+2 [N − X ] . Thus N ≤ p 1 | i| ≤ p 1 | i| N p 1 α p α/p Pr[Z x] Pr N − X [x2− M ( )] . N ≥ ≤ | i| ≥ − p 1 Next, we note that for y 2 log ( ), ≥ Eα,1 N 1 α Pr N − X y exp( Ny + N log ( )) exp( Ny/2). | i| ≥ ≤ − Eα,1 ≤ − 1 p α/p The conclusion easily follows, since x A1 implies that y := [x2− Mp( )] 2 log α,1( ) p 1 α/p α/p≥ − ≥ E and since y [x2− − ] [M ( )] . ≥ − p Step 5. Assume (2) and put δ := 2p/α 1. Here we show that for all x> 0, N 1, − ≥ p α/p 2 δ Pr[Z x] C exp( a(Nx) )+ C exp( aNx (log(1 + N))− ). N ≥ ≤ − − Step 5.1. For R > 0 (large) to be chosen later, we introduce the probability measure R as R the law of X1 X R . We also denote by N the corresponding empirical measure (coupled with {| |≤ } R p,R N in that the Xi’s are used for N and the Xi1 Xi R ’s are chosen for N ). We set ZN := {| |≤ } pn R R p p,R p 1 N p n 0 2 N (Bn) (Bn) and first observe that ZN ZN 2 N − 1 Xi 1 Xi >R + {| | } p ≥ | p − | |p − | ≤ α |p |x α/2 2 x >R x (dx). On the one hand, x >R x (dx) exp( R /2) x e| | (dx) {| | } | | {| | } | | ≤ − | | ≤ α N p C exp( R /2) by (2) (with γ = 1). On the other hand, since α (0,p], 1 Xi 1 Xi >R − ∈ | | {| | } ≤ 18 NICOLAS FOURNIER AND ARNAUD GUILLIN N α p/α α p+1 ( 1 Xi 1 Xi >R ) . Hence if x A exp( R /2), where A := 2 C, | | {| | } ≥ − N p p,R 1 p p 1 Pr ZN ZN x Pr N − Xi 1 Xi >R x2− − | − |≥ ≤ | | {| | } ≥ 1 N α p 1 α/p Pr Xi 1 X >R (Nx2− − ) ≤ | | {| i| } ≥ 1 p 1 α/p α N exp( (Nx2− − ) /2)E exp X1 1 X >R /2 . ≤ − | | {| 1| } α α α Observing that E[exp( X1 1 X >R /2)] 1+ E[exp( X1 /2)1 X >R ] 1+ C exp( R /2) by | | {| 1| } ≤ | | {| 1| } ≤ − (2) and using that log(1 + u) u, we deduce that for all x 2p+1C exp( Rα/2), ≤ ≥ − p p,R p 1 α/p α Pr Z Z x exp (Nx2− − ) /2+ CN exp( R /2) . | N − N |≥ ≤ − − With the choice (5) R := (2 log(1 + N))1/α, we finally find p p,R p 1 α/p α/p Pr Z Z x exp (Nx2− − ) /2+ C C exp a(Nx) | N − N |≥ ≤ − ≤ − provided x A exp( Rα/2), i.e. (N + 1)x A. As usual, this immediately extends to any value of x> 0. ≥ − ≥ p,R R n 1 Step 5.2. To study ZN , we first observe that since (Bn) = 0 if 2 − R, we have pn R p α/2 αn/2 R p,R p α/2 α/≥2,R R 2 (Bn) (2R) − 2 (Bn) for all n 0. Hence ZN (2R) − ZN . But ≤ α R ≥ ≤ satisfies Rd exp( x ) (dx) < uniformly in R, so that we may use Steps 1, 2 and 4 (with | | ∞ α/2,R 2 p = α/2 < α) to deduce that for all x > 0, Pr(ZN x) C exp( aNx ). Consequently, p,R p α/2 2 ≥ ≤ − Pr(ZN x) C exp( aN(x/R − ) ). Recalling (5) and that δ := 2p/α 1, we see that that p,R ≥ ≤ − 2 δ − Pr(Z x) C exp( aNx (log(1 + N))− ). This ends the step. N ≥ ≤ − Conclusion. Recall that x0 > 0 is fixed. p α/p First assume (1). By Step 4, Pr[ZN x] C exp( aNx ) for all x A1. We deduce from ≥p ≤ − 2 ≥ Steps 1 and 2 that for x (0, A1), Pr[ZN x] C exp( aNx ). We easily conclude that for all p ∈ 2 ≥ ≤ − α/p x> 0, Pr[ZN x] C exp( aNx )1 x x0 + C exp( aNx )1 x>x0 as desired. ≥ ≤ − { ≤ } − { } p 2 δ α/p Assume next (2). By Step 5, Pr[ZN x] C exp( aNx (log(1 + N))− )+ C exp( a(Nx) ). α/p ≥ ≤ 2 − δ − p But if x x0, we clearly have (Nx) CNx (log(1+N))− because α Proof of Theorem 2. Using Lemma 5, we write ( , ) κ ( , ) Tp N ≤ p,dDp N κ 2pn (B ) (B ) + κ 2pn (B ) ( , ) ≤ p,d | N n − n | p,d n Dp RBn N RBn n 0 n 0 ≥ ≥ p p =:κp,d(ZN + VN ). Hence Pr( ( , ) x) Pr(Zp x/(2κ )) + Pr(V p x/(2κ )). Tp N ≥ ≤ N ≥ p,d N ≥ p,d p cNx2 By Lemma 13 (choosing x0 := 1/(2κp,d)), we easily find Pr(ZN x/(2κp,d)) Ce− 1 x 1 + ≥ ≤ { ≤ } b(N, x) a(N, x)1 x 1 + b(N, x), these quantities being defined in the statement of Theorem 2. We now≤ check that{ there≤ } is A> 0 such that for all x> 0, p (6) Pr[VN x/(2κp,d)] a(N, x)1 x A . ≥ ≤ { ≤ } This will end the proof, since one easily checks that a(N, x)1 x A a(N, x)1 x 1 + b(N, x) { ≤ } ≤ { ≤ } (when allowing the values of the constants to change). η Let us thus check (6). For η > 0 to be chosen later, we set c := (1 2− )/(2κp,d) and (p+η)n η ηn − zn := cx2− / (Bn). Observing that n 0(1 2− )2− = 1), we write ≥ − P (V p x/(2κ )) P [ ( , ) z ] 1 =: K (N, x) 1. N ≥ p,d ≤ Dp RBn N RBn ≥ n ∧ n ∧ n 0 n 0 ≥ ≥ From now on, the value of c > 0 is not allowed to vary anymore. We introduce another positive constant a > 0 whose value may change from line to line. We only assume (3) (which is implied by (1) or (2)). We now show that if η> 0 is small enough, (7) Kn(N, x) C exp( aNh(x))1 x A if Nh(x) 1, ≤ − { ≤ } ≥ n 0 ≥ where h(x)= x2 if p>d/2, h(x) = (x/ log(2 + 1/x))2 if p = d/2 and h(x)= xd/p if p Kn(N, x) 1 C exp( aNh(x))1 x A ∧ ≤ − { ≤ } n 0 ≥ and thus conclude the proof of (6). We thus only have to prove (7). Conditionally on N (Bn), Bn N is the empirical measure of N N (Bn) points which are Bn - distributed. Since is supportedR in ( 1, 1]d, we may apply Proposition 10 and obtainR RBn − ah(zn) Kn(N, x) CE 1 z 1 exp( aN N (Bn)h(zn)) C1 z 1 exp( N (Bn)(1 e− )) ≤ { n≤ } − ≤ { n≤ } − − by Lemma 12-(c). But the condition zn 1 implies that h(zn) is bounded (by a constant depending only on p and d), whence ≤ Kn(N, x) C1 z 1 exp( aN (Bn)h(zn)). ≤ { n≤ } − qn (q p η)n By (3), we have (B ) C2− . Hence if x > A := C/c, we have z (c/C)x2 − − > 1 for n ≤ n ≥ all n 1 (if η (0, q p)) and thus n 0 Kn(N, x) = 0 as desired. ≥ ∈ − ≥ 20 NICOLAS FOURNIER AND ARNAUD GUILLIN Next, we see that θ θh(x/θ) is decreasing, whence for all x A, → ≤ qn (q p η)n qn (q p η)n K (N, x) C exp( aN2− h(cx2 − − /C)) C exp( aN2− h(x2 − − )). n ≤ − ≤ − We now treat separately the three cases. Step 1: case p>d/2. Since h(x)= x2, we have, if η (0,q/2 p), ∈ − 2 n(q 2p 2η) 2 K (N, x) C exp( aNx 2 − − ) C exp( aNx ) n ≤ − ≤ − n 0 n 0 ≥ ≥ if Nx2 1. Step≥ 2: case p = d/2. Since h(x) = (x/ log(2 + 1/x))2, we have, if η (0,q/2 p), ∈ − 2 (q 2p 2η)n 2 (q p η)n K (N, x) C exp aNx 2 − − / log [2 + 1/(x2 − − )] n ≤ − n 0 n 0 ≥ ≥ n(q 2p 2η) C exp( aNh(x)2 − − ) ≤ − n 0 ≥ C exp( aNh(x)) ≤ − 2 n(q p η) 2 if Nh(x) 1. The third inequality only uses that log (2+1/(x2 − − )) log (2+1/x). ≥ ≤ Step 3: case p 7.1. ρ-mixing stationary sequences. A stationary sequence of random variables (Xn)n 1 with + ≥ 2 common law is said to be ρ-mixing, for some ρ : N R with ρn 0, if for all f,g L ( ) and all i, j 1 → → ∈ ≥ Cov (f(Xi),g(Xj )) ρ i j Var (f(Xi))Var (g(Xj)). ≤ | − | We refer for example to Rio [37], Doukhan [20] or Bradley [10]. Theorem 14. Consider a stationary sequence of random variables (Xn)n 1 with common law ≥ 1 N + and set := N − δ . Assume that this sequence is ρ-mixing, for some ρ : N R N 1 Xi → satisfying ρ < . Let p> 0 and assume that M (Rd) for some p > q. There exists a n 0 n ∞ ∈ q constant C depending≥ only on p,d,q,M ( ),ρ such that, for all N 1, q ≥ 1/2 (q p)/q N − + N − − if p>d/2 and q =2p, 1/2 (q p)/q E ( p( N , )) C N − log(1 + N)+ N − − if p = d/2 and q =2p, T ≤ p/d (q p)/q N − + N − − if p (0, d/2) and q = d/(d p). ∈ − CONVERGENCEOFTHEEMPIRICALMEASURE 21 This is very satisfying: we get the same estimate as in the independent case. The case n 0 ρn = can also be treated (but then the upper bounds will be less good and depend on the≥ rate of decrease∞ of ρ). Actually, the ρ-mixing condition is slightly too strong (we only need the covariance inequality when f = g is an indicator function), but it is best adapted notion of mixing we found in the litterature. Proof. We first check that for any Borel subset A Rd, ⊂ 1/2 E[ (A) (A) ] min 2 (A),C (A)N − . | N − | ≤ { } But this is immediate: E[ (A)] = (A) (whence E[ (A) (A) ] 2 (A)) and N | N − | ≤ 1 Var (A)= Cov (1 (X ), 1 (X )) N N 2 A i A i i,j N ≤ 1 ρ i j Var (1A(X1)) ≤N 2 | − | i,j N ≤ (A)(1 (A)) − ρ i j . ≤ N 2 | − | i,j N ≤ This is smaller than C (A)/N as desired, since i,j N ρ i j N k 0 ρk = CN. Once this is | − | ≤ done, it suffices to copy (without any change) the proof≤ of Theorem 1. ≥ d 7.2. Markov chains. Here we consider a R -valued Markov chain (Xn)n 1 with transition kernel ≥ d 1 N P and initial distribution ν (R ) and we set := N − δ . We assume that it admits a ∈P N 1 Xn unique invariant probability measure π and the following L2-decay property (usually related to a Poincar´einequality) 2 n (8) n 1, f L (π), P f π(f) 2 Cρ f π(f) 2 ∀ ≥ ∀ ∈ − L (π) ≤ n − L (π) for some sequence (ρn)n 1 decreasing to 0. ≥ Theorem 15. Let p 1, d 1 and r> 2 be fixed. Assume that our Markov chain (Xn)n 0 satis- ≥ ≥ ≥ fies (8) with a sequence (ρn)n 1 satisfying n 1 ρn < . Assume also that the initial distribution ≥ ≥ ∞ ν is absolutely continuous with respect to π and satisfies dν/dπ r < . Assume finally that L (π) ∞ Mq(π) < for some q > pr/(r 1). Setting qr := q(r 1)/r and dr = d(r + 1)/r, there is a constant C∞such that for all N −1, − ≥ 1/2 (q p)/q N − + N − r− r if p > d /2r and q =2p, r r 1/2 (q p)/q Eν ( p( N , π)) C N − log(1 + N)+ N − r − r if p = d /2r and q =2p, T ≤ r r p/d (qr p)/qr N − + N − − if p (0, d /2) and q = d /(d p). ∈ r r r r − Once again, we might adapt the proof to get a complete picture corresponding to other decay 2 2 than L -L and to slower mixing rates (ρn)n 1. ≥ Proof. We only have to show that for any ℓ 0, any n 0, ≥ ≥ ∆N := E ( (2nF B ) π(2nF B ) ) n,ℓ ν | N ∩ n − ∩ n | F ∈Pℓ (r 1)/r d ℓ (r 1)/r 1/2 C min (π(B )) − , [2 r (π(B )) − /N] . ≤ n n 22 NICOLAS FOURNIER AND ARNAUD GUILLIN qn Since M (π) < (whence π(B ) C2− ), we will deduce that q ∞ n ≤ N q n d ℓ/2 q n 1/2 ∆ C min 2− r , 2 r (2− r /N) . n,ℓ ≤ Then the rest of the proof is exactly the same as that of Theorem 1 , replacing everywhere q and d by qr and dr. N (r 1)/r We first check that ∆ C(π(B )) − . Using that dν/dπ r < , we write n,ℓ ≤ n L (π) ∞ N 1 dν (r 1)/r Eν ( N (Bn)) = Eπ (X0)1 X B dν/dπ Lr(π)π(Bn) − . N dπ { i∈ n} ≤ i=1 We next consider a Borel subset A of Rd and check that (r 1)/(2r) 1/2 E ( (A) π(A) ) C(π(A)) − N − . ν | N − | ≤ To do so, as is usual when working with Markov chains or covariance properties (see [7]), we introduce f =1 π(A) and write A − 1/2 N N 1 1 Eν ( N (A) π(A) )= Eν f(Xi) Eν (f(Xi)f(Xj )) . | − | N ≤ N i=1 i,j=1 For j i, it holds that ≥ j i dν j i E (f(X )f(X )) =E [f(X )P − f(X )] = E (X )f(X ).P − f(X ) . ν i j ν i i π dπ 0 i i Using the H¨older inequality (recall that dν/dπ r < with r> 2) and (8), we get L (π) ∞ E j i ν (f(Xi)f(Xj )) dν/dπ Lr(π) f L2r/(r−2)(π) P − f L2(π) Cρj i f L2r/(r−2)(π) f L2(π). ≤ ≤ − s 1/s 1/s But for s > 1, f Ls(π) Cs(π(A) + (π(A)) ) Cs(π(A)) , we find Eν (f(Xi)f(Xj)) (r 1) /r ≤ ≤ ≤ Cρj i(π(A)) − and thus − 1/2 N C (r 1)/2r (r 1)/(2r) 1/2 Eν ( N (F ) π(F ) ) ρ i j (π(F )) − C(π(F )) − N − | − | ≤ N | − | ≤ i,j=1 N as desired. We used that i,j=1 ρ i j CN. | − | ≤ We can finally conclude that N 1/2 n (r 1)/(2r) 1/2 d ℓ/2 (r 1)/(2r) ∆ CN − (π(2 F B )) − CN − 2 r (π(B )) − n,ℓ ≤ ∩ n ≤ n F ∈Pℓ by the H¨older inequality (and because # =2dℓ), where d = d(r + 1)/r as in the statement. Pℓ r 7.3. Mc Kean-Vlasov particles systems. Particle approximation of nonlinear equations has attracted a lot of attention in the past thirty years. We will focus here on the following Rd-valued nonlinear S.D.E. dX = √2dB V (X )dt W u (X )dt, X = x t t − ∇ t − ∇ ∗ t t 0 d where ut = Law(Xt) and (Bt) is and R -Brownian motion. This is a probabilistic representation of the so-called Mc Kean-Vlasov equation, which has been studied in particular by Carillo-Mac Cann-Villani [12], Malrieu [28] and Cattiaux-Guillin-Malrieu [13] to which we refer for further motivations and existence and uniqueness of solutions. We will mainly consider here the case CONVERGENCEOFTHEEMPIRICALMEASURE 23 where V and W are convex (and if V = 0 the center of mass is fixed) and W is even. To fix the ideas, let us consider only two cases: (a) HessV βId > 0, HessW 0. (b) V (x)= ≥x α for α> 2, HessW≥ 0. | | ≥ The particle system introduced to approximate the nonlinear equation is the following. Let i i,N (Bt)t 0 be N independent Brownian motion. For i =1,...,N, set X0 = x and ≥ 1 dXi,N = √2dBi V (Xi,N )dt W (Xi,N Xj,N )dt. t t − ∇ t − N ∇ t − t j Usual propagation of chaos property is usually concerned with control of (Law(X1,N ),u ) W2 t t uniformly (or not) in time. It is however very natural to consider rather a control of (ˆuN ,u ) W2 t t N 1 N whereu ˆt = δ i,N , as in Bolley-Guillin-Villani [11]. N i=1 Xt To do so, and inspired by the usual proof of propagation of chaos, let us consider nonlinear independent particles dXi = √2dBi V (Xi)dt W u (Xi)dt, Xi = x t t − ∇ t − ∇ ∗ t t 0 (driven by the same Brownian motions as the particle system) and the corresponding empirical N 1 N measure u = δ i . We then have t N i=1 Xt N N N N 2(ˆu ,ut) 2(ˆu ,u )+ 2(u ,ut). W t ≤W t t W t Then following [28] in case (a) and [13] in case (b), one easily gets (for some time-independent constant C) N 1 E( 2(ˆuN ,uN )) E Xi,N Xi 2 Cα(N) W2 t t ≤ N | t − t | ≤ i=1 1 1/(α 1) where α(n) = N − in case (a) and α(N) = N − − in case (b). It is not hard to prove here that the nonlinear particles have infinitely many moments (uniformly in time) so that combining Theorem 1 with the previous estimates gives E N sup ( 2(ˆut ,ut)) C(α(N)+ β(N)) t 0 W ≤ ≥ 1/2 1/2 1/d where β(N)= N − if d = 1, β(N)= N − log(1 + N) if d = 2 and β(N)= N − if d 3. ≥ References [1] R. Adamczak. 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Graduate Studies in Mathematics, 58. American Mathematical Society, Providence, RI, 2003. Nicolas Fournier, Laboratoire de Probabilites´ et Modeles` aleatoires,´ UMR 7599, UPMC, Case 188, 4 pl. Jussieu, F-75252 Paris Cedex 5, France, E-mail: [email protected] Arnaud Guillin, Laboratoire de Mathematiques,´ UMR 6620, Universite´ Blaise Pascal, Av. des landais, 63177 Aubiere cedex, E-mail: [email protected]