The rate of convergence to the normal law in terms of pseudomoments

We establish the rate of convergence of distributions of sums of independent identically distributed random variables to the Gaussian distribution in terms of truncated pseudomoments by implementing the idea of Yu. Studnyev for getting estimates of the rate of convergence of the order higher than $n^{-1/2}$.


Introduction
Applications of the central limit theorem and other weak limit theorems are closely connected to the rate of convergence to the limit distribution. The rate of convergence was studied by many authors; see [6] and the references therein. The simplest result in this direction is the Berry-Esseen inequality. Let {ξ n , n ≥ 1} be a sequence of independent identically distributed random variables (iidrvs) with distribution function F (x), Eξ i = 0, and Dξ i = σ 2 < ∞. Let β 3 = R |x| 3 dF (x) be the absolute 3rd moment, Φ n (x) = P { ξ1+ξ2+...+ξn σ √ n < x}, and let Φ(x), x ∈ R, be the standard normal distribution function. Then the Berry-Esseen inequality states that This estimate gives the rate of convergence O(n −1/2 ) and the asymptotic expansions of the distribution function of the sum of iidrvs in terms of semiinvariants, presented in the book [6]. The same rate of convergence was obtained by Paulauskas [5] in terms of pseudomoments. Let σ = 1. Then the "pseudomoment" function is defined as H(x) = F (x) − Φ(x), the (absolute) third pseudomoment is defined as ν = R |x| 3 |dH(x)|, and we have However, this rate of convergence is slow, for instance, from the point of view of financial applications. The conditions that allow one to improve the rate of convergence were formulated by several authors. After the introduction of pseudomoments in [10], they are widely used in limit theorems. Zolotarev [11] obtained very general estimates in the central limit theorem using a different type of pseudomoments. Studnyev [9] obtained the following estimate of the rate of convergence in terms of pseudomoments. Let {ξ n , n ≥ 1} be centered iidrvs with unit variance and characteristic function f (t), µ k = R x k dH(x) be the kth-order pseudomoment, and be the variation of the function H. Proposition 1 ([9]). Let F (x) have finite moments up to the qth order for some q ≥ 3 and satisfy the Cramer condition lim |t|→∞ |f (t)| < 1. Then We can see that the condition µ k = 0, 3 ≤ k ≤ r supplies the rate of convergence O(n −r+1 2 ). The rate of convergence was also studied in [1,3,7]. In our work, we use a different type of pseudomoments and get the same rate of convergence avoiding the Cramer condition. Instead, we impose the boundedness of the truncated pseudomoments and integrability of the characteristic function.
We assume that, for some m ≥ 3, there exist the pseudomoments The truncated pseudomoments are defined as ("truncation from above") and ("truncation from below").

Theorem 1. Let the following conditions hold:
(i) The characteristic function is integrable: (ii) The pseudomoments up to order m equal zero, and the truncated pseudomoments are bounded: Then, for all n ≥ 2,

Corollary 1.
Let ξ i be a random variable with bounded density p(x) ≤ A 1 . Suppose that condition (ii) of Theorem 1 holds. Then, for all n ≥ 3, Note assumption (i) implies the existence of the density p n (x) of the random variable S n . Also, let φ(x) be the density of the standard normal law.

Auxiliary results. Proofs of the main results
First we prove two auxiliary results. Denote . By the condition of the lemma, the pseudomoments up to order m equal zero. Hence, it is easy to deduce that Using the inequality ( [12], p. 372) with δ = 1 and δ = 0 we obtain The lemma is proved.
1) For |t| ≤ T 1 (n, m), the characteristic function allows the following bound: 2) For |t| > T 1 (n, m), the characteristic function allows the following bound: . Now consider two cases.
1) Let |t| ≤ T 1 (n, m). Then we can deduce from Lemma 1 that Consider the function f 1 (x) = exp{− x 2 4 }x m−1 . It attains its maximal value at the point x = 2(m − 1), and this value equals The last fraction attains its maximal value at the point m = 3. Therefore, Similarly, From (1) together with two last bounds it follows that 12 ≤ e − t 2 6 , and the proof of the first statement follows.
Now we are in position to prove the main results.
Proof of Theorem 1. Let F and G be two distribution functions with characteristic functions f and g, respectively, and suppose that G has a density function, which we denote G ′ . We shall use the following inequality from [4], p. 297: Taking F (x) = Φ n (x) and G(x) = Φ(x), we have Let n ≥ 2. First, from the elementary inequality u n − v n ≤ |u − v| n k=1 |u| k−1 |v| n−k and from Lemma 1 it follows that, for t ≤ T 1 (n, m) √ n, (m + 1)!n Second, introduce the following notation: .

Remark 2.
For n = 2, we can get estimates similar to those in Remark 1.
Proof of Theorem 2. As it was mentioned before, condition (i) implies the existence of a density for the random variable ξ k , so the random variable S n has the density Therefore, From the conditions of the theorem, Lemmas 1 and 2, and from (4) (n ≥ 2) we obtain the following: for |t| ≤ T 1 (n, m) √ n and ν n (m) < 1 2 e − 3 2 , From the conditions of the theorem, similarly to (9), we get A. (13) Relations (11)-(14) supply the proof of Theorem 2.

Example
We give an example of application of Theorem 1. It is similar to the example of [12], p. 375, where the discrete distribution was considered. Define the distribution function F (x) as if θǫ < x < ǫ, where 0 < ǫ < 1, and 0 < θ < 1 is the root of the equation This equation has a unique solution because . Indeed, on one hand, and therefore, On the other hand, and we immediately get that ǫ 0 It is obvious that the density function is bounded. Moreover, F (x) is symmetric. Therefore, µ 1 = 0 and µ 3 = 0. Furthermore, taking into account (15), consider This means that µ 2 = 0. Consider further the pseudomoments where ǫ can be chosen so that ν 4 ≤ 1 2 e − 3 2 and ν Hence, condition (ii) of Theorem 1 holds. Therefore, the function F (x) satisfies the conditions of Theorem 1 with m = 3.