© Hindawi Publishing Corp. MOMENT INEQUALITIES CONNECTED WITH ACCOMPANYING POISSON LAWS IN ABELIAN GROUPS

We obtain exact inequalities which connect moments of some functions of sums of independent random variables taking values in a measurable Abelian group and those for the accompanying infinitely divisible laws. Some applications to empirical processes are studied.

1. Statement of the main results.Let X 1 ,X 2 ,... be independent random variables taking values in a measurable Abelian group (Ᏻ, Ꮽ) with respective distributions P 1 ,P 2 ,....Moreover, if these random variables are identically distributed (the i.i.d.case), then we denote by P their common distribution.We suppose that {0} ∈ Ꮽ and the group operation "+" is measurable.In other words, a sum of arbitrary random variables in (Ᏻ, Ꮽ) is a random variable too.
Denote by Pois(µ) the generalized Poisson distribution with the Lévy measure µ where µ * k is the k-fold convolution of a finite measure µ with itself; µ * 0 is the unit mass concentrated at zero.Under the measurability conditions above the convolution is well defined because we can define the convolution of probability (i.e., normed finite) measures.Put S n := i≤n X i .Generalized Poisson distribution with the Lévy measure µ := i≤n P i is called the accompanying Poisson law for S n (see, e.g., Araujo and Giné [1]).We will denote by τ µ a random variable having this distribution.
The main goal of this paper is to obtain sharp moment inequalities for some measurable functions of S n via the analogous moments of the accompanying Poisson law.Results of such kind are connected with the Kolmogorov problem of approximation of the sum distributions by infinitely divisible laws as well as with an improvement of the classical probability inequalities for the sums.
For an arbitrary measurable function f satisfying the condition E|f (τ µ )| < ∞, introduce the following notations: where S m,k := i≤k X m,i , S m,0 = S 0 = 0, and {X m,i : i ≥ 1} are independent copies of the random variable X m .We assume that all the sequences {X i }, {X 1,i }, {X 2,i },... are independent.Note that, under the moment condition above, the functions φ(k) exist and the functions φ m,z (k) are well defined at least for almost all z with respect to the distribution of S j,k for each j ≠ m and integer k ≥ 0 (for details, see Section 3).We say that a function g(k) is convex if the difference ∆g(k) := g(k + 1) − g(k) is nondecreasing.( For the initial random variables which are nondegenerate at zero, let {X 0 i } be independent random variables with respective distributions For this sequence, we introduce the notations S 0 k , S 0 m,k , φ 0 (k), and φ 0 m,z (k) as above.
Proposition 1.2.Convexity of the functions φ 0 (k) or φ 0 m,z (k) implies convexity of the functions φ(k) or φ m,z (k), respectively.The converse implication is false.Remark 1.3.If the functions in the conditions of the above two assertions are concave, then inequality (1.3) is changed to the opposite.This follows from the well-known connection between convex and concave functions.
A simple sufficient condition for the functions φ(k) and φ m,z (k) as well as φ 0 (k) and φ 0 m,z (k) to be convex is as follows: for all x ∈ Ᏻ and all z, h ∈ i≤n supp X i , the function f satisfies the inequality where supp X i denotes a measurable subset such that X i ∈ supp X i with probability 1.
For example, in the i.i.d.case, the convexity (say, of φ(k)) follows easily from (1.5): For the Banach-space-valued summands, the following result is valid.
Theorem 1.4.Let Ᏻ be a separable Banach space.Suppose that at least one of the following two conditions is fulfilled: (1) the function f is continuously differentiable in Fréchet sense (i.e., f (x) [h] is continuous in x for each fixed h), and, for each x ∈ Ᏻ and every z, h (2) EX k = 0 for all k, f is twice continuously differentiable in Fréchet sense, and f (x)[h, h] is convex in x for each fixed h ∈ i≤n supp X i .Then all the functions in the conditions of Theorem 1.1 and in Proposition 1.2 are convex.
as well as in the Hilbert space l 2 ) with nonnegative coordinates, then the function f (x) := x 2+α , where • is the corresponding Euclidean norm and α ≥ 0, satisfies inequalities (1.5) and (1.7).For the mean zero Hilbert-spacevalued summands, the function f (x) := x β , where β = 2, 4 or β ≥ 6, satisfies Theorem 1.4 (2).Therefore, in these cases, inequality (1.3) holds under the additional necessary restriction E|f (τ µ )| < ∞.Remark 1.6.In the multivariate case, conditions (1.5) and (1.7) are slightly stronger than convexity.In particular, in general, the Euclidean norm does not satisfy these conditions.Remark 1.7.There exist functions f (x) which do not satisfy the conditions of Theorem 1.4, but the corresponding functions in Theorem 1.1 and Proposition 1.2 are convex.For example, in the i.i.d.one-dimensional case, we consider the function f (x) := x 5 and the centered summands {X i }.It is clear that the conditions of Theorem 1.4 are not fulfilled.In this case we have Thus, if EX 3 1 ≥ 0, then the function φ(k) (as well as the function φ 0 (k)) is convex, otherwise it is concave.In other words, in this case we have various inequality signs in (1.3) depending on positivity or negativity of the third moment of the summands.
Theorem 1.8.In the i.i.d.case, let the function φ µ (k) be convex.Then whenever the expectation on the right-hand side of (1.9) is well defined, where ᏸ(τ µ ) = Pois(µ) and the supremum is taken over all n and P such that nP (A \ {0}) = µ(A) for all A ∈ Ꮽ.
Remark 1.9.Taking inequality (1.3) into account, we can easily reformulate Theorem 1.8 for the non-i.i.d.case.Perhaps, for the first time, the idea of employing generalized Poisson distributions for constructing upper bounds for moments of the sums was proposed by Prokhorov [25,26].In particular, relations (1.3) and (1.9) were obtained by Prokhorov [26] for the functions f (x) := x 2m (m is an arbitrary natural number) and f (x) := ch(tx), t ∈ R, and for one-dimensional symmetric {X i }.Moreover, in the case of mean zero onedimensional summands, these relations for the functions f (x) := exp(hx), h ≥ 0, can be easily deduced from Prokhorov [25] (see also Pinelis and Utev [24]).
The most general result in this direction was obtained by Utev [28] who, in fact, rediscovered and essentially employed some results of Cox and Kemperman [11] regarding lower bounds for moments of sums of independent centered random variables.Under Theorem 1.4(2), he proved extremal equality (1.9) for nonnegative functions f (x) having an exponential majorant.Moreover, he required some additional unnecessary restrictions on the sample Banach space.In our opinion, the corresponding proofs of the present paper are simpler than that of Utev and need no additional restrictions on f (x) and the sample space.
The above results deal with some type of convexity.However, we can obtain moment inequalities close to those mentioned above without any convexity conditions.
Theorem 1.10.In the i.i.d.case, for every nonnegative measurable function f , the following inequality holds: where p := Pr(X 1 = 0).
In the non-i.i.d.case, the factor (1 − p) −1 in (1.10) should be replaced by exp( i≤n p i ), where p i := Pr(X i = 0).
It is clear that inequality (1.10) provides a sufficiently good upper bound under the so-called Poissonian setting when the summand distributions have large atoms at zero (i.e., the probabilities p i are small enough).Some particular cases of inequality (1.10) are contained in Araujo and Giné [1] and in Giné et al. [14].

Applications to empirical processes.
In this section, we formulate some consequences of the above theorems as well as some new analogous results for empirical processes.For the sake of simplicity, we study the empirical processes with one-dimensional time parameter although the results below can be reformulated for empirical processes indexed by subsets of an arbitrary measurable space (moreover, for abstract empirical processes indexed by a family of measurable functions).These results are a basis for the so-called Poissonization method for empirical processes.Sometimes it is more convenient to replace an empirical process under study by the corresponding accompanying Poisson point process having a simpler structure for analysis (e.g., independent "increments").Some versions of this sufficiently popular and very effective method can be found in many papers.In particular, some probability inequalities connecting the distributions of empirical processes (in various settings) and those of the corresponding Poisson processes are contained in Borisov [3,4,5], Einmahl [13], Deheuvels and Mason [12], Giné et al. [14], and others.
Introduce the so-called tail (or local) empirical process on the interval [0,n] where F n (•) is the empirical distribution function (right-continuous version) based on a sample of size n from the (0, 1)-uniform distribution.We consider ν n as a random variable in the space LS([0,n]) which is defined as the linear span of the set of all piecewise constant right-continuous functions on [0,n] with finitely many jumps, endowed with the cylinder σ -field.It is easy to verify that the standard Poisson process π(t) on [0,n] (with right-continuous paths) has the accompanying Poisson distribution for ν n in this space.
Theorem 2.1.Let Φ(•) be a convex nonnegative functional on LS([0,n]) which is nondecreasing on the subset of all nonnegative functions with respect to the standard partial order in function spaces.Suppose that, for each function x(•) ∈ LS([0,n]), the following relation holds: (2.2) Remark 2.2.It is well known that if a convex functional defined on a topological linear space (say, on a Banach space) is bounded in a neighborhood of some point, then it is continuous (see, e.g., Kutateladze [20]).Thus, if the functional in Theorem 2.1 is defined, say, on L m ([0,n],λ), where λ is a finite measure, and satisfies the local boundedness condition, then the continuity condition connected with the step functions x (m) (t) can be omitted.
In the sequel, in the case of Banach-space-valued random variables, we consider only continuous convex functionals.For example, the functional Φ(x) with arbitrary parameters m ≥ 1 and q ≥ 1, where λ is an arbitrary finite measure on [0,n], satisfies the conditions of Theorem 2.1.
Note that the accompanying Poisson process for the centered empirical process ν 0 n (t) := ν n (t)−t, say, in L m ([0,n],λ), differs from the corresponding centered Poisson process.This process can be defined as π 0 (t) := π(t)−π(n)t/n and, by analogy with the definition of a Brownian bridge, can be called a Poissonian bridge on [0,n].For such processes, Theorem 1.4(2) can be reformulated as follows.
As an example of such a functional we can consider Φ(x) := x mq m for any m ≥ 2 and q ≥ 3 or q = 1, 2.
If we consider the processes ν n and π as random elements in LS([0,δn]), where δ < 1, then the following direct consequence of Theorems 1.10 and 2.1 above and Lemma 3.1 and Corollary 3.4 below holds.
Corollary 2.5.For every natural n and m and every t ≥ 0, the following inequality holds: where x is arbitrary for even m and x ≥ 0 for odd m.
In the following assertion which complements this inequality, the abovementioned convexity conditions need not be fulfilled.
Theorem 2.6.For every natural n and m and every t ≥ 0, the following inequality holds: where x ∈ [−t, 0) is arbitrary.
Corollary 2.7.Let f (x) be an entire function on [0, ∞), that is, an analytic function which admits Taylor expansion at all points with a converging power series on the whole positive half-line.Assume that, for a point x 0 ≥ 0, the kth derivative of this function at x 0 is nonnegative for each k ≥ 2.Then, for every t ≥ x 0 ,

Proof of the results.
First we formulate two important lemmas which play a key role in proving the above results.Lemma 3.1.In the i.i.d.case, under the above notations, the following relations hold: where the standard Poisson process π(•) is independent of {X i }, where p := Pr(X 1 = 0), ᏸ(ν(n, p)) = B n,p is the binomial distribution with parameters n and p; the pair (ν(n, p), π (np)) does not depend on the sequence {X 0 i }.
The relation (3.1) is well known.It immediately follows from the abovementioned definition of a generalized Poisson distribution: the probability law Pois(µ) may be interpreted as the distribution of i≤πµ Y i , where {Y i } are i.i.d.random variables with the common distribution µ(•)/µ(Ᏻ) and π µ is a Poissonian random variable with parameter µ(Ᏻ), which is independent of the sequence {Y i }.
The equalities in (3.2), which are more convenient in studying accuracy of Poisson approximation of the sums, are contained in various forms in many papers (see, e.g., Khintchine [18,19], Le Cam [21,22], Borovkov [9], Borisov [6,7], and others).Actually, these relations also represent versions of the total probability formula and are easily proven.
Taking into account the representations in Lemma 3.1, we can reduce the problem to the simplest one-dimensional case when we estimate the analogous moments of the binomial distribution introduced in Remark 1.3.However, in this case, we can obtain sufficiently exact inequalities for moments of arbitrary functions using the following lemma.Lemma 3.2.For each p ∈ (0, 1), Proof.For every nonnegative integer j ≤ n, we have where The following properties of H p are obvious: 3) is a part of a more general result in Borisov and Ruzankin (see [8, Lemma 1, page 1663]).It is worth noting that this upper bound is an estimate for the so-called Radon-Nikodym derivative of a binomial distribution with respect to the accompanying Poisson law.This problem was studied by a number of authors (Le Cam [21], Chen [10], Barbour et al. [2], and others).In particular, under some additional restriction on n and p, a slightly stronger estimate is contained in Le Cam [21].However, in general, estimate (3.3) cannot be essentially improved.Under some restrictions on n and p, a lower bound for the left-hand side of (3.3) has the form (1 − cp) −1 , where c is an absolute positive constant.
Corollary 3.4.Let g be an arbitrary function satisfying the condition E|g(π(λ))| < ∞ for some λ.Then, for every n and p satisfying the condition np ≤ λ, the following inequality holds: 2).In the case n = 1 in Lemma 3.2 there exists a slightly stronger upper bound for the Radon-Nikodym derivative.It is easy to see that, in this case, the right-hand side of (3.3) can be replaced by e p .In the non-i.i.d.case, evaluation of the moment Ef (S n ) can be reduced to that for a new function of n independent Bernoulli random variables ν 1 (1,p),...,ν n (1,p) (for details, see the proof of Theorem 1.1 below).In this case, the approximating moment is calculated by independent Poisson random variables π 1 (p),..., π n (p) with the same parameter p.Thus, the corresponding upper bound for the Radon-Nikodym derivative (as well as the corresponding factor on the righthand side of (1.10)) equals exp( i≤n p i ).However, in the special case when S n = i≤n ν i (1,p), there exist better upper bounds for this derivative.For example, in this case we can replace the factor exp( i≤n p i ) by (1 − p) −2 , where p = max{p i : i ≤ n} (see Barbour et al. [2], Borisov and Ruzankin [8]).
It is worth noting that, under the minimal moment condition above, we cannot replace the one-sided double limit in (3.7) by the classical double limit as well as the condition np ≤ λ in (3.6) cannot be omitted.For example, the function g(k) = (1 ∨ (k − 2))!λ −k satisfies the above-mentioned moment condition; however, it is easy to prove the relation lim sup (3.9) Proof of Theorem 1.1.In the i.i.d.case, inequality (1.3) is a simple consequence of relation (3.1) and the classical Jensen inequality (3.10) In order to prove inequality (1.3) in the non-i.i.d.case, we introduce the sequence of i.i.d.random variables {π i : i ≥ 1} having Poisson distribution with parameter 1, which is independent of all the sequences of random variables introduced in (1.2) (including the initial random variables).Then we can define the random variable τ µ in the following way: where S m,k are defined in (1.2).The further reasoning is quite analogous to the above.Put z 1 := n m=2 S m,πm .Using the above arguments, we have where the symbol E z 1 denotes the conditional expectation given z 1 .Now we put z 2 := X 1 + n m=3 S m,πm .Then, repeating the same calculation, we obtain the estimate Continuing the calculations in this way, we finally obtain inequality (1.3).Theorem 1.1 is proven.
Proof of Proposition 1.2.The first assertion is easily verified.Indeed, by Corollary 1.5 and Lemma 3.1 (see (3.2)) we have The analogous inequality holds for the functions φ m,z (k).
in x of the function f (x) [h, h] for any fixed h and using Taylor's formula, we have First we average both sides of (3.19) with respect to the distribution of X k+1 and use the fact that, for any centered (in Bochner sense) random variable X and an arbitrary linear continuous functional the equality El(X) = 0 holds.Averaging both sides of this identity with respect to the other distributions, we then obtain (with more convenient representation of the remainder in (3.19)) the equality where ζ is a random variable with the density 2(1 − t) on the unit interval, which is defined on the main probability space and independent of the sequence {X k } (we may assume here that this space is rich enough).It is worth noting that, because of integrability of the left-hand side of (3.19), the expectation on the right-hand side of (3.20) is well defined due to Fubini's theorem.In the i.i.d.case, by the classical Jensen inequality (for the conditional expectation E ζ,X k+2 ) we finally obtain the inequality we need: The proof of convexity of φ 0 (k) and φ 0 m,z (k) is the same because, for the centered initial summands, EX 0 k = 0.The theorem is proven.
Proof of Theorem 1.8.Put n > µ(Ᏻ) and consider the independent random variables X k ≡ X k (n) with the following common distribution: Then the corresponding random variables X 0 i have the common distribution P 0 (A) = µ(A)/µ(Ᏻ).Therefore, for each n, by Proposition 1.2 we have the corresponding inequality for the moments under study.It is easy to see that, in this case, the function φ 0 (k) ≡ φ µ (k) does not depend on n and we can apply and π (m) of the processes under consideration.To this end, we consider an arbitrary nonnegative function ψ(x 1 ,...,x k ) which is convex and increasing in every coordinate x i .We will study the moment Eψ(ν n (t 1 ),..., ν n (t k )), where t i ∈ [0,n) are arbitrary points and t i < t i+1 for every i < k.We will also assume that the corresponding Poisson moment exists.
The following so-called Markov property of ν n (•) is well known: given the quantity ν n (t) (the number of the sample points to the left of t/n), two new samples constituted by the points to the left and to the right of t/n, respectively, are independent and distributed as samples (of the corresponding sizes) from the uniform distributions on [0,t/n] and [t/n, 1], respectively.In other words, given ν n (t 1 ), the increment ν n (t 2 ) − ν n (t 1 ) coincides in distribution with ν * N ((t 2 − t 1 )N/n), where N := n − ν n (t 1 ), and the process ν * n (•) is an independent copy of ν n (•).Thus, taking into account Corollary 1.5 and convexity and monotonicity of the function ψ 1 (x) := Eψ(ν n (t 1 ),...,ν n (t k−1 ), ν n (t k−1 ) + x), we have where π(•) is a Poisson process independent of ν n (•).
Therefore, we reduced the problem to evaluating the moment of a function of the analogous (k−1)-dimensional projection ν n (t 1 ),...,ν n (t k−1 ).It remains to observe that the function ψ ) is convex and monotone too.In other words, to prove the assertion we may use induction on k.The theorem is proven.
Proof of Theorem 2.6.It is clear that, under the above notations, we deal with the random variable ν(n, p) having the binomial distribution B n,p .First we consider the case n = 1.Lemma 3.6.For every natural m, the function g m (t) := E(π (t) − t) m is a polynomial on [0, ∞) with nonnegative coefficients and the following inequalities due to monotonicity of the functions g m (t).The lemma is proven.
Since ν(n, p) coincides in distribution with a sum of independent copies of the random variables ν(n−1,p) and ν(1,p), the further proof of the theorem can be continued by induction on n (using (3.26) and the binomial formula).The theorem is proven.

Theorem 1 . 1 .
Let one of the following two conditions be fulfilled: (a) the random variables {X i } are identically distributed and φ(k) is a convex function; (b) for all z and m, all the functions φ m,z (k) are convex.Then, for each n, Ef S n ≤ Ef τ µ .