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.
Rate of convergencetruncated pseudomomentsGaussian distribution60E0560F05Introduction
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 {ξn,n≥1} be a sequence of independent identically distributed random variables (iidrvs) with distribution function F(x), Eξi=0, and Dξi=σ2<∞. Let β3=∫R|x|3dF(x) be the absolute 3rd moment, Φn(x)=P{ξ1+ξ2+…+ξnσn<x}, and let Φ(x), x∈R, be the standard normal distribution function. Then the Berry–Esseen inequality states that
supx∈R|Φn(x)−Φ(x)|≤Cβ3σ3n.
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
supx∈R|Φn(x)−Φ(x)|≤Cmax(ν,ν14)n−12.
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 {ξn,n≥1} be centered iidrvs with unit variance and characteristic function f(t), μk=∫RxkdH(x) be the kth-order pseudomoment, and V(x)=V−∞xH(z) be the variation of the function H.
([9]).
LetF(x)have finite moments up to the qth order for someq≥3and satisfy the Cramer conditionlim‾|t|→∞|f(t)|<1. Thensupx∈R|Φn(x)−Φ(x)|=O(∑k=3q|μk|nk−22+1nq−221n∫0ndx∫|z|>x|z|qdV(z)).x}|z{|}^{q}dV(z)\Bigg).\]]]>
We can see that the condition μk=0, 3≤k≤r supplies the rate of convergence O(n−r+12). 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.
Generalization of Studnyev’s estimate. Main results
Let, as before, {ξn,n≥1} be a sequence of iidrvs with Eξi=0, Dξi=σ2∈(0,∞), distribution function F(x), and characteristic function f(t), and let Φn(x), x∈R, be the distribution function of the random variable
Sn=(σn)−1(ξ1+ξ2+⋯+ξn).
We assume that, for some m≥3, there exist the pseudomoments
μk=∫RxkdH(x),k=3,…,m∈N,
where H(x)=F(xσ)−Φ(x). The truncated pseudomoments are defined as
νn(1)(m)=∫|x|≤σn|x|m+1|dH(x)|
(“truncation from above”) and
νn(2)(m)=∫|x|>σn|x|m|dH(x)|\sigma \sqrt{n}}|x{|}^{m}\big|dH(x)\big|\]]]>
(“truncation from below”).
Let the following conditions hold:
The characteristic function is integrable: A=∫R|f(t)|dt<∞;
The pseudomoments up to order m equal zero, and the truncated pseudomoments are bounded:
μk=0,k=3,…,m,for somem≥3,andνn(m)=max{νn(1)(m),νn(2)(m)}<12e−32.
Then, for alln≥2,supx∈R|Φn(x)−Φ(x)|≤2Cm(1)νn(1)(m)nm−12+2Cm(2)νn(2)(m)nm−22+σAπbn−1+νn(m)4e32πe−n2n,whereCm(1)=12m+12Γ(m+12)π(m+1)!,Cm(2)=2Cm−1(1),b=exp{−π224A2σ2(2+π)2}<1.
Letξibe a random variable with bounded densityp(x)≤A1. Suppose that condition (ii) of Theorem1holds. Then, for alln≥3,supx∈R|Φn(x)−Φ(x)|≤2Cm(1)νn(1)(m)nm−12+2Cm(2)νn(2)(m)nm−22+2σA1b1n−2+νn(m)4e32πe−n2n,whereb1=exp{−196A12σ2(2+π)2}<1.
Note assumption (i) implies the existence of the density pn(x) of the random variable Sn. Also, let ϕ(x) be the density of the standard normal law.
Let the conditions of Theorem1hold. Then, for alln≥2,supx∈R|pn(x)−ϕ(x)|≤Cm(3)νn(1)(m)nm−12+Cm(4)νn(2)(m)nm−22+bn−1σn2πA+νn(m)e32πe−n2n,whereCm(3)=12m+22Γ(m2+1)4π(m+1)!,Cm(4)=2Cm−1(3).
Auxiliary results. Proofs of the main results
First we prove two auxiliary results. Denote ω(t)=|f(tσ)−e−t22|.
Letμk=0,k=3,…,m. Then, for allt∈R,ω(t)≤|t|m+1(m+1)!νn(1)(m)+2|t|mm!νn(2)(m).
Recall that f(t)=∫−∞∞eitxdF(x). Therefore, f(tσ)=∫−∞∞eitxσdF(x)=∫−∞∞eitxdF(xσ). By the condition of the lemma, the pseudomoments up to order m equal zero. Hence, it is easy to deduce that
ω(t)=|∫ReitxdF(xσ)−∫ReitxdΦ(x)|=|∫Reitxd(F(xσ)−Φ(x))|=|∫R(eitx−∑k=0m(itx)kk!)dH(x)|≤∫R|eitx−∑k=0m(itx)kk!||dH(x)|.
Using the inequality ([12], p. 372)
|eiα−1−⋯−(iα)mm!|≤21−δ|α|m+δm!(m+1)δ,m=0,1,…,δ∈[0,1],
with δ=1 and δ=0 we obtain
ω(t)≤∫|x|≤σn|eitx−∑k=0m(itx)kk!||dH(x)|+∫|x|>σn|eitx−∑k=0m(itx)kk!||dH(x)|≤∫|x|≤σn|tx|m+1(m+1)!|dH(x)|+∫|x|>σn2|tx|mm!|dH(x)|=|t|m+1(m+1)!νn(1)(m)+2|t|mm!νn(2)(m).\sigma \sqrt{n}}\Bigg|{e}^{itx}-\sum \limits_{k=0}^{m}\frac{{(itx)}^{k}}{k!}\Bigg|\big|dH(x)\big|\\{} & \displaystyle \le \int _{|x|\le \sigma \sqrt{n}}\frac{|tx{|}^{m+1}}{(m+1)!}\big|dH(x)\big|\hspace{1em}+\int _{|x|>\sigma \sqrt{n}}\frac{2|tx{|}^{m}}{m!}\big|dH(x)\big|\\{} & \displaystyle =\frac{|t{|}^{m+1}}{(m+1)!}{\nu _{n}^{(1)}}(m)+\frac{2|t{|}^{m}}{m!}{\nu _{n}^{(2)}}(m).\end{array}\]]]>
The lemma is proved. □
Now, denote T1(n,m)=−2ln(2eνn(m)). Then, in turn, we have that νn(m)=12eexp{−12T12(n,m)}. Note also that condition (ii) implies T1(n,m)≥1.
Suppose that condition (ii) of Theorem1holds.
For|t|≤T1(n,m), the characteristic function allows the following bound: |f(tσ)|≤e−t26.
For|t|>T1(n,m)T_{1}(n,m)$]]>, the characteristic function allows the following bound: |f(tσ)|≤(2e+38)νn(m)|t|m+1.
Evidently, |f(tσ)|=|f(tσ)−e−t22+e−t22|≤e−t22+ω(t). Now consider two cases.
1) Let |t|≤T1(n,m). Then we can deduce from Lemma 1 that
|f(tσ)|≤e−t24(e−t24+et24ω(t))≤e−t24(1+et24(|t|m+1(m+1)!νn(1)(m)+2|t|mm!νn(2)(m)))≤e−t24(1+eT12(n,m)4t2(T1m−1(n,m)(m+1)!νn(1)(m)+2T1m−2(n,m)m!νn(2)(m)))≤e−t24(1+t2eT12(n,m)4νn(m)(T1m−1(n,m)(m+1)!+2T1m−2(n,m)m!))=e−t24(1+12et2e−T12(n,m)4(T1m−1(n,m)(m+1)!+2T1m−2(n,m)m!)).
Consider the function f1(x)=exp{−x24}xm−1. It attains its maximal value at the point x=2(m−1), and this value equals
f1,max=exp{−m−12}(2(m−1))m−12.
Furthermore,
exp{−m−12}(2(m−1))m−12(m+1)!≤exp{−m−12}(2(m−1))m−12m(m+1)2π(m−1)(m−1)m−1e−(m−1)=(2em−1)m−1212π(m−1)1m(m+1)≤1m(m+1).
The last fraction attains its maximal value at the point m=3. Therefore,
exp{−T12(n,m)4}T1m−1(n,m)(m+1)!≤112.
Similarly,
exp{−T12(n,m)4}2T1m−2(n,m)m!≤13.
From (1) together with two last bounds it follows that
|f(tσ)|≤e−t24(1+12et2(112+13))≤e−t24(1+112t2)≤e−t24et212≤e−t26,
and the proof of the first statement follows.
2) Now, let |t|>T1(n,m)T_{1}(n,m)$]]>. Then we get from Lemma 1 that
|f(tσ)|≤e−t22+ω(t)≤e−T12(n,m)2+|t|m+1(m+1)!νn(1)(m)+2|t|mm!νn(2)(m)≤νn(m)(2e+|t|m+1(m+1)!+2|t|mm!).
Recall that T1(n,m)>11$]]>. Then |t|>T1(n,m)>1T_{1}(n,m)>1$]]>, and from (2) we get that
|f(tσ)|≤νn(m)(2e|t|m+1+|t|m+124+2|t|m+16)≤(2e+38)νn(m)|t|m+1,
whence the proof follows. □
Now we are in position to prove the main results.
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:
supx∈R|F(x)−G(x)|≤2π∫0T|f(t)−g(t)|dtt+24sup|G′|πT.
Taking F(x)=Φn(x) and G(x)=Φ(x), we have
ρn:=supx∈R|Φn(x)−Φ(x)|≤2π∫0T|fn(tσn)−e−t22|dtt+24π2πT.
Let n≥2. First, from the elementary inequality
|un−vn|≤|u−v|∑k=1n|u|k−1|v|n−k
and from Lemma 1 it follows that, for t≤T1(n,m)n,
|fn(tσn)−e−t22|≤ω(tn)∑k=1n|f(tσn)|k−1e−t22n(n−k)≤ω(tn)∑k=1ne−t26n−1n≤ω(tn)ne−t212≤n(|t|m+1(m+1)!nm+12νn(1)(m)+2|t|mm!nm2νn(2)(m))exp{−t212}=exp{−t212}(|t|m+1(m+1)!nm−12νn(1)(m)+2|t|mm!nm−22νn(2)(m)).
Second, introduce the following notation:
Cm,n(1)=12m−12Γ(m+12)2nm−12(m+1)!,Cm,n(2)=2Cm−1,n(1),T2(n,m)=12π(Cm,n(1)νn(1)(m)+Cm,n(2)νn(2)(m)).
Then
24π2πT2(n,m)=Cm(1)νn(1)(m)nm−12+Cm(2)νn(2)(m)nm−22.
Let T3(n,m)=(T1(n,m)n)∧T2(n,m). Then it follows from (3) and (5) that
ρn≤2π∫0T3(n,m)|fn(tσn)−e−t22|dtt+2π∫T3(n,m)T2(n,m)|f(tσn)|ndtt+2π∫T3(n,m)T2(n,m)e−t22dtt+24π2πT2(n,m)=I1(n,m)+I2(n,m)+I3(n,m)+Cm(1)νn(1)(m)nm−12+Cm(2)νn(2)(m)nm−22.
Since T3(n,m)≤T1(n,m)n, from (4) we get that
I1(n,m)=2π∫0T3(n,m)|fn(tσn)−e−t22|dtt≤2π∫0T3(n,m)(tm(m+1)!nm−12νn(1)(m)+2tm−1m!nm−22νn(2)(m))e−t212dt≤12m+12Γ(m+12)π(m+1)!nm−12νn(1)(m)+2·12m2Γ(m2)πm!nm−22νn(2)(m)=Cm(1)νn(1)(m)nm−12+Cm(2)νn(2)(m)nm−22.
If T3(n,m)=T2(n,m), then I2(n,m)=0 and I3(n,m)=0. Therefore, we consider the case T3(n,m)=T1(n,m)n. Then
I2(n,m)=2π∫T3(n,m)T2(n,m)|f(tσn)|ndtt=2π∫T1(n,m)/σT2(n,m)/σn|f(t)|ndtt.
Now we apply the following result of Statulevičius [8]: if a random variable with characteristic function f(t) has a density p(x)≤d<∞ and variance σ2, then, for any t∈R,
|f(t)|≤exp{−t296d2(2σ|t|+π)2}.
It follows from condition (i) that the density p(x) of any ξn can be obtained as the inverse Fourier transform p(x)=12π∫Re−itxf(t)dt and p(x)≤12π∫−∞∞|f(t)|dt=A2π. Besides, the function t2(2σt+π)2 is increasing for t>00$]]>. Therefore, for |t|≥T1(n,m)/σ (recall that T1(n,m)>11$]]>),
|f(t)|≤exp{−π224A2σ2(2+π)2}=:b,
and 0<b<1. Then
I2(n,m)=2π∫T1(n,m)/σT2(n,m)/σn|f(t)|ndtt≤2σπbn−1∫0∞|f(t)|dt=σAπbn−1.
Finally, we bound I3(n,m). Note that I3(n,m) is nonzero only if T1(n,m)n<T2(n,m). Therefore,
I3(n,m)≤2π∫T1(n,m)n∞e−t22dtt≤2πe−nT12(n,m)2nT12(n,m)≤2(2eνn(m))nπn≤4eνn(m)πn(2eνn(m))n−1≤νn(m)4e·e−n−12πn=νn(m)4e32πe−n2n.
Relations (6)–(10) supply the proof of Theorem 1. □
Let the following conditions hold: μk=0, k=3,…,m, m≥3. Then
supx∈R|Φ1(x)−Φ(x)|=supx∈R|F(xσ)−Φ(x)|≤(6π(m+1)!+2π2π)max(ν1(m),(ν1(m))1m+2).
Indeed, let n=1. Theorem is obvious when ν1(m)>11$]]>. Let ν1(m)≤1. Put T=(ν1(m))−1m+2 into (3). Then from Lemma 1 it follows that
ρ1=supx∈R|Φ1(x)−Φ(x)|=supx∈R|F(xσ)−Φ(x)|≤2π∫0T(|t|m+1(m+1)!ν1(1)(m)+2|t|mm!ν1(2)(m))dtt+24π2πT≤2π(Tm+1(m+1)·(m+1)!ν1(1)(m)+2Tmm·m!ν1(2)(m))+24π2π(ν1(m))1m+2≤(ν1(m))1m+2(2π3(m+1)!+24π2π).
Proof is similar to that of Theorem 1. We apply inequality (8) and recall again that the function t2(2σt+π)2 is increasing for t>00$]]>. Therefore, for |t|≥T1(n,m)/σ (recall that T1(n,m)>11$]]>),
|f(t)|≤exp{−196A12σ2(2+π)2}=:b1,
and 0<b1<1. It follows from [2], p. 510, that ∫−∞∞|f(t)|2dt≤2πA1. Therefore,
I2(n,m)=2π∫T1(n,m)/σT2(n,m)/σn|f(t)|ndtt≤2σπb1n−2∫0∞|f(t)|2dt=2σA1b1n−2.
Corollary 1 is proved. □
For n=2, we can get estimates similar to those in Remark 1.
As it was mentioned before, condition (i) implies the existence of a density for the random variable ξk, so the random variable Sn has the density
pn(x)=∫−∞∞e−itxfn(tσn)dt.
Since ϕ(x)=12πe−x22 is the density of the standard normal law, we have ϕ(x)=12π∫−∞∞e−itxe−t22dt and
|pn(x)−ϕ(x)|=12π|∫−∞∞e−itxfn(tσn)dt−∫−∞∞e−itxe−t22dt|≤12π∫−∞∞|fn(tσn)−e−t22|dt.
Therefore,
|pn(x)−ϕ(x)|≤12π∫|t|≤T1(n,m)n|fn(tσn)−e−t22|dt+12π∫|t|>T1(n,m)n|f(tσn)|ndt+12π∫|t|>T1(n,m)ne−t22dt=I1+I2+I3.T_{1}(n,m)\sqrt{n}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}dt\\{} & \displaystyle \hspace{1em}+\frac{1}{2\pi }\int _{|t|>T_{1}(n,m)\sqrt{n}}{e}^{-\frac{{t}^{2}}{2}}dt\\{} & \displaystyle =I_{1}+I_{2}+I_{3}.\end{array}\]]]>
From the conditions of the theorem, Lemmas 1 and 2, and from (4) (n≥2) we obtain the following: for |t|≤T1(n,m)n and νn(m)<12e−32,
I1=12π∫|t|≤T1(n,m)n|fn(tσn)−e−t22|dt≤12π∫|t|≤T1(n,m)n(|t|m+1νn(1)(m)(m+1)!nm−12+2|t|mνn(2)(m)m!nm2−1)e−t212dt≤12m+22Γ(m2+1)4π(m+1)!νn(1)(m)nm−12+2·12m+12Γ(m−12)4πm!νn(2)(m)nm−22=Cm(3)νn(1)(m)nm−12+Cm(4)νn(2)(m)nm−22.
From the conditions of the theorem, similarly to (9), we get I2=12π∫|t|>T1(n,m)n|f(tσn)|ndt=σn2π∫|z|>T1(n,m)/σ|f(z)|ndz≤bn−1σn2πA.T_{1}(n,m)\sqrt{n}}{\bigg|f\bigg(\frac{t}{\sigma \sqrt{n}}\bigg)\bigg|}^{n}dt\\{} & \displaystyle =\frac{\sigma \sqrt{n}}{2\pi }\int _{|z|>T_{1}(n,m)/\sigma }{\big|f(z)\big|}^{n}dz\le {b}^{n-1}\frac{\sigma \sqrt{n}}{2\pi }A.\end{array}\]]]>I3=12π∫|t|>T1(n,m)ne−t22dt≤12π∫T1(n,m)n∞e−t22dt≤e−nT12(n,m)22πnT1(n,m)≤(2eνn(m))n2πn≤eνn(m)πne−n−12=νn(m)e32πe−n2n.T_{1}(n,m)\sqrt{n}}{e}^{-\frac{{t}^{2}}{2}}dt\le \frac{1}{2\pi }{\int _{T_{1}(n,m)\sqrt{n}}^{\infty }}{e}^{-\frac{{t}^{2}}{2}}dt\\{} & \displaystyle \le \frac{{e}^{-\frac{n{T_{1}^{2}}(n,m)}{2}}}{2\pi \sqrt{n}T_{1}(n,m)}\le \frac{{(2e\nu _{n}(m))}^{n}}{2\pi \sqrt{n}}\\{} & \displaystyle \le \frac{e\nu _{n}(m)}{\pi \sqrt{n}}{e}^{-\frac{n-1}{2}}=\nu _{n}(m)\frac{{e}^{\frac{3}{2}}}{\pi }\frac{{e}^{-\frac{n}{2}}}{\sqrt{n}}.\end{array}\]]]> 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
F(x)=Φ(x)if|x|≥ϵ,Φ(−ϵ)if−ϵ<x<−θϵ,Φ(ϵ)ifθϵ<x<ϵ,12+Φ(ϵ)−12θϵxif|x|≤θϵ,
where 0<ϵ<1, and 0<θ<1 is the root of the equation
∫0ϵx2dΦ(x)=(θϵ)23(Φ(ϵ)−12).
This equation has a unique solution because ∫0ϵx2dΦ(x)≤ϵ23(Φ(ϵ)−12). Indeed, on one hand,
φ(x)=12π(1−x22+x4222!−x6233!+x8244!−⋯),
and therefore,
Φ(ϵ)−12=∫0ϵφ(x)dx=12π(ϵ−ϵ36+ϵ540−ϵ77·233!+⋯).
So
Φ(ϵ)−12≥12π(ϵ−ϵ36).
On the other hand,
∫0ϵx2dΦ(x)=12π∫0ϵ(x2−x42+x6222!−x8233!+⋯)dx=12π(ϵ33−ϵ510+ϵ77·222!−ϵ99·233!+⋯)≤12π(ϵ33−ϵ510+ϵ756)≤12π(ϵ33−ϵ510+ϵ556)=12πϵ3(13−23280ϵ2)=12πϵ33(1−69280ϵ2)≤12πϵ33(1−ϵ26)≤ϵ23(Φ(ϵ)−12),
and we immediately get that
∫0ϵx2dΦ(x)≤ϵ23(Φ(ϵ)−12).
It is obvious that the density function is bounded. Moreover, F(x) is symmetric. Therefore, μ1=0 and μ3=0. Furthermore, taking into account (15), consider
σ2=∫−∞∞x2dF(x)=∫|x|≥ϵx2dΦ(x)+∫−θϵθϵx2Φ(ϵ)−12θϵdx=∫|x|≥ϵx2dΦ(x)+Φ(ϵ)−12θϵ23(θϵ)3=∫−∞∞x2dΦ(x)=1.
This means that μ2=0. Consider further the pseudomoments
ν4=∫−∞∞x4|d(F(x)−Φ(x))|=∫−ϵϵx4|d(F(x)−Φ(x))|≤ϵ4∫−ϵϵ|d(F(x)−Φ(x))|≤ϵ44(Φ(ϵ)−12),νn(1)(3)=∫|x|≤σnx4|d(F(x)−Φ(x))|=∫−ϵϵx4|d(F(x)−Φ(x))|,
where ϵ can be chosen so that ν4≤12e−32 and νn(1)(3)≤12e−32. Then
νn(2)(3)=∫|x|>σn|x|3|d(F(x)−Φ(x))|=0.\sigma \sqrt{n}}|x{|}^{3}\big|d\big(F(x)-\varPhi (x)\big)\big|=0.\]]]>
Hence, condition (ii) of Theorem 1 holds. Therefore, the function F(x) satisfies the conditions of Theorem 1 with m=3.
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