It is shown that the absolute constant in the Berry–Esseen inequality for i.i.d. Bernoulli random variables is strictly less than the Esseen constant, if 1≤n≤500000, where n is a number of summands. This result is got both with the help of a supercomputer and an interpolation theorem, which is proved in the paper as well. In addition, applying the method developed by S. Nagaev and V. Chebotarev in 2009–2011, an upper bound is obtained for the absolute constant in the Berry–Esseen inequality in the case under consideration, which differs from the Esseen constant by no more than 0.06%. As an auxiliary result, we prove a bound in the local Moivre–Laplace theorem which has a simple and explicit form.
Despite the best possible result, obtained by J. Schulz in 2016, we propose our approach to the problem of finding the absolute constant in the Berry–Esseen inequality for two-point distributions since this approach, combining analytical methods and the use of computers, could be useful in solving other mathematical problems.
Optimal value of absolute constant in Berry–Esseen inequalitybinomial distributionnumerical methods60F0565-04Introduction
Let us consider the class V of all probability distributions on the real line R, which have zero mean, unit variance and finite third absolute moment. Let X,X1,X2,…,Xn be i.i.d. random variables, where the distribution of X belongs to V. Denote
Φ(x)=12π∫−∞xe−t2/2dt,β3=E|X|3.
According to the Berry–Esseen inequality [2, 5], there exists such an absolute constant C0 that for all n=1,2,…,
supx∈R|P(1n∑j=1nXj<x)−Φ(x)|≤C0β3n.
The first upper bounds for the constant C0 were obtained by C.-G. Esseen [5] (1942), H. Bergström [1] (1949) and K. Takano [30] (1951).
In 1956 C.-G. Esseen [6] showed that
limn→∞nβ3supx∈R|P(1n∑j=1nXj<x)−Φ(x)|≤CE,
where CE=3+1062π=0.409732…. He has also found a two-point distribution, for which the equality holds in (2). He has proved the uniqueness of such a distribution (up to a reflection).
Consequently, C0≥CE. The result of Esseen served as an argument for the conjecture
C0=CE,
that V.M. Zolotarev advanced in 1966 [38]. The question whether the conjecture is correct remains open up to now.
Since then, a number of upper bounds for C0 have been obtained. A historical review can be found, for example, in [11, 17, 28]. We only note that recent results in this field were obtained by I.S. Tyurin (see, for example, [31–35]), V.Yu. Korolev and I.G. Shevtsova (see, for example, [11, 13]), and I.G. Shevtsova (see, for example, [25–29]). The best upper estimate, known to date, belongs to Shevtsova: C0≤0.469 [28]. Note that in obtaining upper bounds, beginning from the estimates in [38, 39], calculations play an essential role. In addition, because of the large amount of computations, it was necessary to use computers.
The present paper is devoted to estimation of C0 in the particular case of i.i.d. Bernoulli random variables. In this case we will use the notation C02 instead of C0. Let us recall the chronology of the results along these lines.
In 2007 C. Hipp and L. Mattner published an analytical proof of the inequality C02≤12π in the symmetric case [8].
In 2009 the second and third authors of the present paper have suggested the compound method in which a refinement of C.L.T. for i.i.d. Bernoulli random variables was used along with direct calculations [17]. In unsymmetric case this method allows to obtain majorants for C02, arbitrarily close to CE, provided that the computer used is of sufficient power. The main content of the preprint [17] was published in 2011, 2012 in the form of the papers [18, 19]. In these papers, the following bound was proved, C02<0.4215.
In 2015 we obtained the bound
C02≤0.4099539,
by applying the same approach as in [17–19], with the only difference that this time a supercomputer was used instead of an ordinary PC. We announced bound (4) in [20], but for a number of reasons, delayed publishing the proof, and do it just now. While the present work being in preparation, we have detected a small inaccuracy in the calculations, namely, bound (4) must be increased by 10−7. Thus the following statement is true.
The boundC02≤0.409954holds.
Meanwhile, in 2016 J. Schulz [23] obtained the unimprovable result: if the symmetry condition is violated, C02=CE. As it should be expected, J. Schulz’s proof turned out to be very long and complicated. It should be said that methods based on the use of computers, and analytical methods complement each other. The former ones cannot lead to a final result, but they do not require so much effort. On the other hand, they allow us to predict the exact result, and thus facilitate theoretical research.
Shortly about the proof of Theorem <xref rid="j_vmsta113_stat_001">1</xref>Some notations. On the choice of the left boundary of the interval for <italic>p</italic>
Let X,X1,X2,…,Xn be a sequence of independent random variables with the same distribution:
P(X=1)=p,P(X=0)=q=1−p.
In what follows we use the following notations,
Fn,p(x)=P(∑i=1nXi<x),Gn,p(x)=Φ(x−npnpq),Δn(p)=supx∈R|Fn,p(x)−Gn,p(x)|,ϱ(p)=E|X−p|3(E(X−p)2)3/2=p2+q2pq,Tn(p)=Δn(p)nϱ(p),E(p)=2−p32π[p2+(1−p)2].
Obviously,
C02=supn≥1supp∈(0,0.5]Tn(p).
In this paper we solve, in particular, the problem of computing the sequence T(n)=supp∈(0,0.5)Tn(p) for all n such that 1≤n≤N0. Here and in what follows,
N0=5·105.
Note that for fixed n and p, the quantity supx∈R|Fn,p(x)−Gn,p(x)| is achieved at some discontinuity point of the function Fn,p(x) (see Lemma 2). We consider distribution functions that are continuous from the left. Consequently,
Δn(p)=max0≤i≤nΔn,i(p),
where i are integers, Δn,i(p)={|Fn,p(i)−Gn,p(i)|,|Fn,p(i+1)−Gn,p(i)|}.
Note also that we can vary the parameter p in a narrower interval than [0,0.5], namely, in
I:=[0.1689,0.5].
This conclusion follows from the next statement.
If0<p≤0.1689, then for alln≥1,Tn(p)<0.4096.
Lemma 1 is proved in Section 4 with the help of some modification of the Berry – Esseen inequality (with numerical constants) obtained in [10, 12].
By the same method that is used to prove inequality (10), the estimate Tn(p)≤0.369 is found in [19] in the case 0<p<0.02 (n≥1) (see the proof of (1.37) in [19]), where an earlier estimate of V. Korolev and I. Shevtsova [11] is used, instead of [10, 12]. Note that the use of modified inequalities of the Berry – Esseen type, obtained in [10, 12, 11], is not necessary for obtaining estimates of Tn(p) in the case when p are close to 0.
An alternative approach, using Poisson approximation, is proposed in the preprint [17]. Let us explain the essence of this method.
An alternative bound is found in the domain {(p,n):0.258≤λ≤6,n≥200}, where λ=np. Under these conditions, we have p≤0.03, i.e. p are small enough. Consequently, the error arising under replacement of the binomial distribution by Poisson distribution Πλ with the parameter λ is small.
Next, the distance d(Πλ,Gλ) between Πλ and normal distribution Gλ with the mean λ and the variance λ is estimated, where d(U,V)=supx∈R|U(x)−V(x)| for any distribution functions U(x) and V(x). Then the estimate of the distance between Gλ and the normal distribution Gn,p with the mean λ and variance npq is deduced. Summing the obtained estimates, we arrive at an estimate for the distance between the original binomial distribution and Gn,p. As a result, in [17, Lemma 7.8, Theorem 7.2] we derive the estimate Tn(p)<0.3607, which is valid for all points (p,n) in the indicated domain.
On calculations
Define
C02(N)=max1≤n≤Nsupp∈(0,0.5]Tn(p),C‾02(N)=supn≥Nsupp∈(0,0.5]Tn(p).
Obviously, C02=max{C02(N),C‾02(N+1)} for every N≥1.
It was proved in [19] that C‾02(200)<0.4215. By that time it was shown with the help of a computer (see the preprint [9]) that C02(200)<0.4096, i.e.
C02(200)<CE,
and thus, C02<0.4215 for all n≥1.
Some words about bound (11). By (8), to get C02(N) it is enough to calculate T(n)=supp∈(0,0.5]Tn(p) for every 1≤n≤N, and then find max1≤n≤NT(n). The calculation of T(n) is reduced to two problems. The first problem is to calculate maxpj∈STn(pj), where S is a grid on (0,0.5], and the second one is to estimate Tn(p) in intermediate points p. Both problems were solved in [9] for 1≤n≤200.
It should be noted here that, according to the method, the quantity C02(N) is calculated (with some accuracy), and C‾02(N) is estimated from above. In both cases, a computer is required. The power of an ordinary PC is sufficient for calculating majorants for C‾02(N) whereas to calculate C02(N) a supercomputer is needed if N is sufficiently large. Moreover, an additional investigation of the interpolation type is required for the convincing conclusion from computer calculations of C02(N). In our paper, Theorem 2 plays this role.
Denote by symbol S the uniform grid on I with the step h=10−12. The values of Tn(pj) for all pj∈S and 1≤n≤N0 were calculated on a supercomputer.
The result of the calculations. For all1≤n≤N0,
maxpj∈STn(pj)=TN1(p‾)=0.40973212897643…<0.40973213.
The counting algorithm is a triple loop: a loop with respect to the parameter i (see (9)) is nested in a loop with respect to the parameter p, which in turn is nested in the loop with respect to the parameter n.
With the growth of n, the computation time increased rapidly. For example, for 2000≤n≤2100 calculations took more than 3 hours on a computer with processor Core2Due E6400. For 2101≤n≤N0 calculations were carried out on the supercomputer Blue Gene/P.
It follows from [20, Corollary 7] that for n>200200$]]> in the loop with respect to i, one can take not all values of i from 0 to n, but only those, which satisfy the inequality
np−(ν+1)npq≤i≤np+νnpq,
where ν=3+6. This led to a significant reduction of computation time. We give information about the computer time (without waiting for the queue) in Table 1.
Dependence of computer time on n (supercomputer Blue Gene/P)
n∈[N1,N2]:
[10000,11024]
[30000,50000]
[300000,320000]
[490000,N0]
computer time:
3 min
2 hrs + 5 min
4 hrs + 50 min
7 hrs
Calculations were carried out on the supercomputer Blue Gene/P of the Computational Mathematics and Cybernetics Faculty of Lomonosov Moscow State University. After some changes in the algorithm, the calculations for n such that 490000≤n≤N0, were also performed on the CC FEB RAS Computing Cluster [41]. The corresponding computer time was 6 hours and 40 minutes.
The program is written in C+MPI and registered [40].
Interpolation type results
Let p∗∈(0,0.5). Consider a uniform grid on [p∗,0.5] with a step h. The following statement allows to estimate the value of the function 1ϱ(p)Δn,k(p) at an arbitrary point from the interval [p∗,0.5] via the value of this function at the nearest grid node and h.
Denote
c1=0.516,c2=0.121,c3=0.271.
Let0<p∗<p≤0.5,p′be a node of a grid with a step h on the interval[p∗,0.5], closest to p. Then for alln≥1and0≤k≤n,|1ϱ(p)Δn,k(p)−1ϱ(p′)Δn,k(p′)|≤h2L(p∗),whereL(p)=1(1−2pq)pq(c1p+c2+c3(1−2p)(1+2pq)1−2pq).
The next statement follows from Theorem 2. Note that without it the proof of Theorem 1 would be incomplete.
Ifp∈I, andp′is a node of the grid S, closest to p, then for all1≤n≤N0,|Tn(p)−Tn(p′)|≤4.6·10−9.
It follows from Theorem 2 that for 0≤k≤n≤N0,
|nϱ(p)Δn,k(p)−nϱ(p′)Δn,k(p′)|≤N01210−12L(0.1689).
Since L(0.1689)<12.98, the right-hand side of inequality (15) is majorized by the number 4.6·10−9. This implies the statement of Corollary 1. □
On the proof of Theorem <xref rid="j_vmsta113_stat_001">1</xref>
It follows from (12), Corollary 1 and Lemma 1 that for all 1≤n≤N0 and p∈(0,0.5], the following inequality holds, Tn(p)<0.4097321346<CE (for details, see (64)). It is easy to verify that this inequality is true for p∈(0.5,1) as well. Hence, inequality (5) implies Theorem 1.
About structure of the paper
The structure of the paper is as follows. The proof of Theorem 2, the main analytical result of the paper, is given in Section 3. The proof consists of 12 lemmas.
In Section 4, Theorem 1 is proved. The section consists of three subsections. In the first one, the formulation of Theorem 1.1 [19] is given. Several corollaries from the latter are also deduced here. The second subsection discusses the connection between the result of K. Neammanee [21], who refined and generalized Uspensky’s estimate [36], and the problem of estimating C02. It is shown that one can obtain from the result of K. Neammanee the same estimate for C02 as ours, but for a much larger N. This means that calculating C02(N) requires much more computing time if to use Neammanee’s estimate.
In the third subsection, we give, in particular, the proof of Lemma 1.
Proof of Theorem <xref rid="j_vmsta113_stat_004">2</xref>
We need the following statement, which we give without proof.
LetG(x)be a distribution function with a finite number of discontinuity points, andG0(x)a continuous distribution function. Denoteδ(x)=G(x)−G0(x). There exists a discontinuity pointx0ofG(x)such that the magnitudesupx|δ(x)|is attained in the following sense: if G is continuous from the left, thensupx|δ(x)|=max{δ(x0+),−δ(x0)}, and if G is continuous from the right, thensupx|δ(x)|=max{δ(x0),−δ(x0−)}.
Define f(t)=Eeit(X−p)≡qe−itp+peitq.
For allt∈R,|f(t)|≤exp{−2pqsin2t2}.
Taking into account the difference in the notations, we obtain the statement of Lemma 3 from [19, Lemma 8]. □
Further, we will use the following notations:
σ=npq,β3(p)=E|X−p|3,Y is a standard normal random variable. Note that ϱ(p)=β3(p)(pq)3/2.
The following bound is true for alln≥2,∫|t|≤π|fn(t)−e−npqt2/2|dt<1σ2(f(p,n)+πσ2e−σ2+4πe−π2σ2/8),wheref(p,n)=(p2+q2)π496(nn−1)2+3π5πpq210n(nn−1)5/2.
Using the equalities e−pqt2/2=EeitpqY, E(X−p)j=E(Ypq)j, j=0,1,2, and the Taylor formula, we get
|f(t)−e−pqt2/2|=|E[∑j=12(it(X−p))jj!+(it(X−p))32∫01(1−θ)2eitθ(X−p)dθ]−E[∑j=131j!(itpqY)j+(itpqY)43!∫01(1−θ)3eitθpqYdθ]|=|E[(it(X−p))32∫01(1−θ)2eitθ(X−p)dθ−(itpqY)43!∫01(1−θ)3eitθpqYdθ]|≤|t|36β3(p)+t48(pq)2.
Since for |x|≤π4 the inequality |sinx|≥22|x|π is fulfilled, then with the help of Lemma 3 we arrive at the following bound for |t|≤π/2,
|f(t)|≤exp{−2pqsin2(t/2)}≤exp{−4t2pqπ2}.
Then, taking into account the elementary equality an−bn=(a−b)∑j=0n−1ajbn−1−j and the estimate (16), we obtain for |t|≤π/2 that
|fn(t)−e−npqt2/2|≤|f(t)−e−pqt2/2|∑j=0n−1|f(t)|je−(n−1−j)t2pq/2≤≤(|t|36β3(p)+t48(pq)2)∑j=0n−1exp{[j(1−8/π2)−(n−1)]t2pq/2}≤≤(|t|36β3(p)+t48(pq)2)nexp{−4(n−1)t2pqπ2}.
Using the well-known formulas E|Y|3=42π and EY4=3, we deduce from the previous inequality that for n≥2,
∫|t|≤π/2|fn(t)−e−npqt2/2|dt≤n2π(β3(p)6m2E|Y|3+(pq)28m5/2EY4)|m=8(n−1)pqπ2=n(π4ϱ(p)96pq(n−1)2+3π5π210pq(n−1)5/2)=f(p,n)σ2.
Applying Lemma 3 again, we get
∫π/2≤|t|≤π|fn(t)|dt≤2∫π/2πe−2σ2sin2(t/2)dt<πe−σ2.
Moreover, by virtue of the known inequality
∫c∞e−t2/2dt≤1ce−c2/2,
which holds for every c>00$]]>, we have
∫|t|≥π/2e−σ2t2/2dt≤4πσ2e−σ2π2/8.
Collecting the estimates (17)–(20), we obtain the statement of Lemma 4. □
For everyn≥1and0≤k≤nthe following bound holds,|δn(k,p)|<min{1σ2e,c1σ2},wherec1is defined in (13).
It was proved in [7] that Pn(k)≤12enpq. Moreover, 1npqφ(k−npnpq)≤12πnpq. Hence,
|δn(k,p)|≤12enpq=1σ2e.
Let us find another bound for δn(k,p). Let σ>11$]]>. Then n>1pq≥4\frac{1}{pq}\ge 4$]]>, i.e. n≥5.
By the inversion formula for integer random variables,
Pn(k)=12π∫−ππ(q+eitp)ne−itkdt=12π∫−ππfn(t)e−it(k−np)dt.
Moreover, by the inversion formula for densities,
1σφ(x−μσ)=12π∫−∞∞e−t2σ2/2−it(x−μ)dt.
Consequently,
δn(k,p)=12π(J1−J2),
where
J1=∫−ππ[fn(t)−e−σ2t2/2]e−it(k−np)dt,J2=∫|t|≥πe−σ2t2/2e−it(k−np)dt.
Note that the function f(p,n) from Lemma 4 decreases in n. Hence, f(p,n)≤f(p,5). It is not hard to verify that maxp∈[0,1]f(p,5)<1.707. Thus, for σ>11$]]>,
|J1|≤1σ2(1.707+πe+4πe−π2/8)<3.234σ2.
Using inequality (19), we get the estimate
|J2|≤2πσ2e−π2σ2/2<0.005σ2.
Thus, we get from (23) that for σ>11$]]>,
|δn(k,p)|≤3.242πσ2<0.516σ2.
Since 1σ2e≤c1σ2 for 0<σ≤c12e=1.203…>11$]]>, the statement of Lemma 5 follows from (22) and (24). □
The following equality holds,∂∂pGn,p(x)=−x(1−2p)+np2pqnpqφ(x−npnpq).
We have
ddpp−1/2(1−p)−1/2=−q−p2pqpq,ddpp1/2(1−p)−1/2=12p−1/2(1−p)−1/2+12p1/2(1−p)−3/2=12qpq.
Hence,
∂∂px−npnpq=−x(q−p)2pqnpq−n2qpq=−x(q−p)+np2pqnpq,
and we arrive at (25). □
For alln≥1and0≤k≤nthe following bound holds,|∂∂pFn,p(k+1)−∂∂pGn,p(k)|≤L1(p)≡1pq(c1q+c2).
It is shown in [22] that
∂∂pFn,p(k+1)=−nCn−1kpkqn−1−k=−n−kqPn(k).
By Lemma 5,
n−kq|Pn(k)−1σφ(k−npσ)|≤nc1qσ2=c1pq2.
In turn, it follows from Lemma 6 that
n−kqσφ(k−npσ)+∂∂pGn,p(k)=(n−kqσ−k(1−2p)+np2pqσ)φ(k−npσ)=−k−np2pqσφ(k−npσ).
Since
∂∂pFn,p(k+1)−∂∂pGn,p(k)=−n−kq[Pn(k)−1σφ(k−npσ)]−[n−kqσφ(k−npσ)+∂∂pGn,p(k)]
and maxx|x|φ(x)=12πe<0.242, the statement of the lemma follows from (26) and (27). □
For alln≥1and0≤k≤nthe following bound holds,|∂∂pFn,p(k)−∂∂pGn,p(k)|≤L2(p)≡1pq(c1p+c2),wherec1,c2are from (13).
Similarly to the proof of Lemma 7 we obtain
∂∂pFn,p(k)=−nCn−1k−1pk−1qn−k=−kpPn(k),kp|Pn(k)−1σφ(k−npσ)|≤kc1pσ2≤c1p2q.
Hence,
∂∂pFn,p(k)−∂∂pGn,p(k)=−kp[Pn(k)−1σφ(k−npσ)]−k−np2pqσφ(k−npσ).
Since the last summand on the right-hand side of the equality is less than 0.121pq, then by using (28) we get the statement of the lemma. □
For every0<p<0.5,ddp1ϱ(p)=12A(p):=12(1−2p)(1+2pq)pq(1−2pq)2.
The lemma follows from the equalities:
ddp1ϱ(p)=ddxx1−2x2|x=pq×ddpp(1−p),ddpp(1−p)=1−2p2pq,ddxx1−2x2=11−2x2+4x2(1−x2)2=1+2x2(1−2x2)2.
□
The functionA(p)decreases on the interval(0,0.5).
Denote x=x(p)=p(1−p), A1(t)=1−4t(1+2t)t(1−2t)2. Taking into account the equality 1−2p=1−4pq, we obtain A(p)=A1(x).
Since x(p) increases for 0<p<0.5, it remains to prove the decrease of the function A1(x) for 0<x<0.25. We have
ddxlnA1(x)=−21−4x+21+2x−12x+41−2x=−32x3+36x2−12x+12x(1−4x)(1−4x2).
On the interval [0,0.25] the polynomial A2(x)≡32x3+36x2−12x+1 has the single minimum point x1=−3+178=0.140…. Since A2(x1)=0.11…>00$]]>, we have ddxlnA1(x)<0 for 0≤x<0.25, i.e. the function A1(x) decreases on (0,0.25). The lemma is proved. □
The functionL(p), defined in (14), decreases on[0,0.5].
Taking into account the equality p2+q2=1−2pq, it is not difficult to see that
L(p)=1ϱ(p)L2(p)+c3A(p).
According to Lemma 10, the function A(p) decreases. Consequently, it remains to prove that the function L3(p):=1ϱ(p)L2(p)=c1+c2pppq(1−2pq) decreases on [0,0.5]. We have
ddplnL3(p)=c2c1+c2p−32p+12(1−p)+2(1−2p)1−2p+2p2=A3(p)2pq(c1+c2p)(1−2pq),
where A3(p)=−3c1+(14c1−c2)p−(26c1−8c2)p2+(16c1−18c2)p3+12c2p4. Let us prove that
A3(p)<0,0<p<0.5.
We have
A3′(p)=14c1−c2−4(13c1−4c2)p+6(8c1−9c2)p2+48c2p3,A3″(p)=−4(13c1−4c2)+12(8c1−9c2)p+144c2p2.
As a result of calculations, we find that the equation A3′(p)=0 has the single root p0=0.478287… on [0,0.5]. The roots of the equation A3″(p)=0 have the form
p1,2=124c2(−8c1+9c2±(8c1−9c2)2+16c2(13c1−4c2)),
and are equal to p1=−2.6…, p2=0.54… respectively. Hence, A3″(p)<0 for p∈[0,0.5]. Thus, the function A3(p), considered on [0,0.5], takes a maximum value at the point p0. Since A3(p0)=−0.257…, inequality (31) is proved. This implies that L3(p) decreases on (0,0.5). □
Let f(x) be an arbitrary function. Denote by D+f(x) and D−f(x) its right-side and left-side derivatives respectively (if they exist).
Letg(x)=max{f1(x),f2(x)}, wheref1(x)andf2(x)are functions, differentiable on a finite interval(a,b). Then at every pointx∈(a,b)there exist both one-side derivativesD+g(x)andD−g(x), each of which coincides with eitherf1′(x)orf2′(x).
Let x be a point such that f1(x)≠f2(x). Then the function g is differentiable at x, and in this case the statement of the lemma is trivial.
Now let for a point x∈(a,b),
f1(x)=f2(x).
First, consider the case f1′(x)≠f2′(x). Let, for instance, f1′(x)>f2′(x){f^{\prime }_{2}}(x)$]]>. Then there exists h0>00$]]> such that f1(x+h)>f2(x+h),0<h≤h0,{f_{2}}(x+h),\hspace{1em}0f2(x+h)>f1(x+h),−h0≤h<0.{f_{1}}(x+h),\hspace{1em}-{h_{0}}\le h<0.\end{aligned}\]]]> From differentiability of the functions f1 and f2 it follows that for h→0,
fi(x+h)=fi(x)+fi′(x)h+o(h),i=1,2.
Then using (33) we obtain the equality
g(x+h)=f1(x+h)=f1(x)+f1′(x)h+o(h),h>0,0,\]]]>
and using (34),
g(x+h)=f2(x+h)=f2(x)+f2′(x)h+o(h),h<0.
Thus, existence of D+g(x) and D−g(x) follows.
Now let
f1′(x)=f2′(x).
It follows from (32), (35) and (36) that for h→0,
g(x+h)=fi(x)+fi′(x)h+o(h),i=1,2.
Hence, g′(x)=f1′(x)=f2′(x). The lemma is proved. □
Denote
ϱ=ϱ(p),qi=1−pi,ϱi=ϱ(pi)≡ω(pi)piqi.
Let0<p1<p<p2≤0.5. Then for alln≥1and0≤k≤n,|1ϱΔn,k(p)−1ϱ1Δn,k(p1)|≤L(p1)(p−p1),and|1ϱΔn,k(p)−1ϱ2Δn,k(p2)|<L(p1)(p2−p).
Note that Δn,k(p)<0.541 (see [3]). Consequently,
|1ϱΔn,k(p)−1ϱ1Δn,k(p1)|≤1ϱ1|Δn,k(p)−Δn,k(p1)|+0.541(1ϱ−1ϱ1).
It is obvious that Fn,p(k) and Gn,p(k), considered as functions of the argument p, are differentiable. Then, according to Lemma 12, the one-side derivatives of the functions Δn,k(p) exist at each point p∈[0,0.5] and coincide with ∂∂p(Fn,p(k+1)−Gn,p(k)) or ∂∂p(Gn,p(k)−Fn,p(k)).
Taking into account that L1(p)≤L2(p) for 0<p≤0.5, we obtain from Lemmas 7 and 8|Δn,k(p)−Δn,k(p1)|≤(p−p1)maxp1≤s≤p|D+Δn,k(s)|≤(p−p1)maxp1≤s≤pL2(s).
The function L2(s) decreases on (0,0.5]. Hence,
maxp1≤s≤pL2(s)=L2(p1).
The inequality
1ϱ1|Δn,k(p)−Δn,k(p1)|≤p−p1ϱ1L2(p1)
follows from (40) and (41). Taking into account Lemmas 9 and 10, we have
1ϱ−1ϱ1≤(p−p1)maxp1<s<pdds1ϱ(s)<2−1A(p1)(p−p1).
Collecting the estimates (39), (42), (43), we obtain with the help of (30) that for 0≤p1<p≤0.5,
|1ϱΔn,k(p)−1ϱ1Δn,k(p1)|≤(p−p1)(1ϱ1L2(p1)+0.271A(p1))=(p−p1)L(p1).
Hence, for 0<p<p2≤0.5,
|1ϱΔn,k(p)−1ϱ2Δn,k(p2)|<(p2−p)L(p).
Inequality (37) coincides with (44), and inequality (38) follows from (45) and Lemma 11. Lemma 13 is proved. □
It follows from the definition of p′ that either 0<p−p′<h/2 or 0<p′−p<h/2. In the first case the statement of the theorem follows from (37) and Lemma 11, and in the second one from (38) and Lemma 11 again. □
Proof of Theorem <xref rid="j_vmsta113_stat_001">1</xref>Theorem 1.1 [<xref ref-type="bibr" rid="j_vmsta113_ref_019">19</xref>] and some its consequences
First we formulate Theorem 1.1 from [19]. To do this, we need to enter a rather lot of notations from [19]:
ω3(p)=q−p,ω4(p)=|q3+p3−3pq|,ω5(p)=q4−p4,ω6(p)=q5+p5+15(pq)2,K1(p,n)=ω3(p)4σ2π(n−1)(1+14(n−1))+ω4(p)12σ2π(nn−1)2+ω5(p)40σ32π(nn−1)5/2+ω6(p)90σ4π(nn−1)3;ω(p)=p2+q2,ζ(p)=(ω(p)6)2/3,e(n,p)=exp{124σ2/3ζ2(p)},e5=0.0277905,ω˜5(p)=p4+q4+5!e5(pq)3/2,V6(p)=ω32(p),V7(p)=ω3(p)ω4(p),V8(p)=2ω˜5(p)ω3(p)5!3!+(ω4(p)4!)2,V9(p)=ω˜5(p)ω4(p),V10(p)=ω˜52(p),Ak(n)=(nn−2)k/2n−1n,γ6=19,γ7=52π96,γ8=24,γ9=72π4!16,γ10=26·3(5!)2,γ˜6=23,γ˜7=78,γ˜8=109,γ˜9=118,γ˜10=53,K2(p,n)=1πσ∑j=15γj+5Aj+5(n)Vj+5(p)σj[1+γ˜j+5e(n,p)nσ2(n−2)];A1=5.405,A2=7.521,A3=5.233,μ=3π2−16π4,χ(p,n)=2ζ(p)σ2/3ifp∈(0,0.085),andχ(p,n)=0ifp∈[0.085,0.5],K3(p,n)=1π{112σ2+(136+μ8)1σ4+(136eA1/6+μ8)1σ6+5μ24eA2/61σ8+13exp{−σA1+A16}+(π−2)μexp{−σA2+A26}+exp{−σA3+A36}14ln(π4σ24A3)+exp{−σ2/32ζ(p)}[2ζ(p)σ2/3+eA3/61+χ(p,n)24ζ(p)σ4/3]};R(p,n)=K1(p,n)+K2(p,n)+K3(p,n).
Let4n≤p≤0.5,n≥200.ThenΔn(p)≤ϱ(p)nE(p)+R(p,n),and the sequenceR0(p,n):=nϱ(p)R(p,n)tends to zero for every0<p≤0.5, decreasing in n.
Denote
E(p,n)=E(p)+R0(p,n).
Figure 1 shows the mutual location of the following functions: E(p,n) for n=200 and 800, E(p) and Tn(p)|n=50. Note that, as a consequence of the definition of the binomial distribution, the behavior of these functions is symmetric with respect to p=0.5.
Graphs of the functions (from top to down): E(p,200),E(p,800),E(p),T50(p)
Since E(p,n) decreases in n, we obtain the statement of Corollary A by finding the maximal value of E(p,N0) directly using a computer. □
In order to verify the plausibility of the previous numerical result, we estimate the function E(p,N0), making preliminary estimates of some of the terms that enter into it. This leads to the following somewhat more coarse inequality.
Forp∈[0.1689,0.5], andn≥N0,E(p,n)<0.409954153.
Separate the proof of (49) into four steps. First we rewrite R0(p,n) in the following form,
R0(p,n)=K1(p,n)σω(p)+K2(p,n)σω(p)+K3(p,n)σω(p).
In each function Ki(p,n)σω(p), i=1,2,3, we will select the principal term, and estimate the remaining ones.
Step 1. Note that for n≥N0 and 0<a≤3,
(nn−1)a≤(nn−1)3<e1:=1.00000601,1+14(n−1)<e2:=1.000000501.
Then
K1(p,n)σω(p)=ω4(p)12πω(p)σ(nn−1)2+r1(p,n),
where
r1(p,n)<r˜1(p,n):=e1ω(p)(e2ω3(p)42π(n−1)+ω5(p)402πσ2+ω6(p)90πσ3).
Using a computer, we get the estimate r˜1(p,n)≤r˜1(0.1689,N0)<2.78·10−7.
Step 2. We have
K2(p,n)σω(p)=γ6A6(n)V6(p)πω(p)σ+r2(p,n),
where
r2(p,n)=∑j=25γj+5Aj+5(n)Vj+5(p)πω(p)σj[1+γ˜j+5e(n,p)nσ2(n−2)]+γ6γ˜6A6(n)e(n,p)nπω(p)σ3(n−2).
Taking into account that for n≥N0, 1≤j≤5 and p∈[0.1689,0.5], we have
Aj+5(n)<A10(N0)<e3:=1.00001801,e(n,p)≤e(N0,0.5)<1.02316,1+γ˜j+5e(n,p)nσ2(n−2)<1+(5/3)·1.02316pq(N0−2)|p=0.1689<e4:=1.0000243.
Then, taking into account as well that A6(N0)<1.0000101, we get
r2(p,n)<r˜2(p,n):=e3·e4πω(p)∑j=25γj+5Vj+5(p)σj+(1/9)(2/3)1.0000101·1.02316πω(p)(pq)3/2n(n−2).
We find with the help of a computer: r˜2(p,n)≤r˜2(0.1689,N0)<8.852·10−8.
Step 3. Let us write up
K3(p,n)σω(p)=112πω(p)σ+r3(p,n),
where
r3(p,n)=σπω(p){(136+μ8)1σ4+(136eA1/6+μ8)1σ6+5μ24eA2/61σ8+13exp{−σA1+A16}+(π−2)μexp{−σA2+A26}+exp{−σA3+A36}14ln(π4σ24A3)+exp{−σ2/32ζ(p)}[2ζ(p)σ2/3+eA3/61+χ(p,n)24ζ(p)σ4/3]}.
Using a computer, we get r3(p,n)≤r3(0.1689,N0)<1.08·10−9.
Thus, for p∈[0.1689,0.5], n≥N0, we have
r1(p,n)+r2(p,n)+r3(p,n)<2.78·10−7+8.852·10−8+1.08·10−9<3.676·10−7.
Step 4. Now consider the function
B(p,n)=E(p)+112πω(p)σ(ω4(p)(nn−1)2+12γ6A6(n)V6(p)+1).
We find with the help of a computer that for p∈[0.1689,0.5], n≥N0,
maxp∈[0.1689,0.5]B(p,n)=maxp∈[0.1689,0.5]B(p,N0)=B(0.418886928…,N0)=0.40995378459….
Consequently,
E(p,n)=B(p,n)+∑j=13rj(p,n)<0.4099537846+3.676·10−7<0.409954153.
□
Let us introduce the following notations:
E1(p)=(p2+q2)E(p)=2−p32π,D2(p,n) is the coefficient at 1σ2 in the expansion of R(p,n) in powers of 1σ, D‾2(p,n)=σ2R(p,n), where the remainder R(p,n) is defined by equality (46). One can rewrite bound (48) in the following form,
Δn(p)≤E1(p)σ+D‾2(p,n)σ2.
Define D2I(n)=maxp∈ID2(p,n), D‾2I(n)=maxp∈ID‾2(p,n), where I is an interval.
The quantitiesmaxn≥ND2I(n)andmaxn≥ND‾2I(n)take the following values depending onN=200,N0and intervalsI=[0.02,0.5],[0.1689,0.5]:
Some values of maxn≥ND2I(n) and maxn≥ND‾2I(n)
∫I=[0.02,0.5]
I=[0.1689,0.5]
N=200
N=200
N=N0
maxn≥ND2I(n)=
0.083592…
0.046656…
0.0462198…
maxn≥ND‾2I(n)=
0.1940…
0.05986…
0.05531…
Since
maxn≥ND‾2I(n)=maxn≥Nmaxp∈Iσ2R(p,n)=maxp∈Iσ2R(p,N),
then by using a computer, we get the tabulated values of maxn≥ND‾2I(n).
Proceed to the derivation of the values of maxn≥ND2I(n). It follows from the definitions of K1(p,n), K2(p,n), and K3(p,n) that the coefficient at 1σ2 in R(p,n) is
D2(p,n)=ω4(p)12π(nn−1)2+1πγ6A6(n)V6(p)+112π
or, in more detail,
D2(p,n)=136π(3|q3+p3−3pq|(nn−1)2+4A6(n)(q−p)2+3)=:G2(p,n)36π.
First we consider G2(p):=limn→∞G2(p,n). We have
G2(p)=3|q3+p3−3pq|+4(q−p)2+3≡3|6p2−6p+1|+4(1−2p)2+3.
Taking into account that
|6p2−6p+1|=6p2−6p+1ifp≤p1:=3−36=0.211324…,−6p2+6p−1ifp>p1,{p_{1}},\end{array}\right.\]]]>
we obtain
G2(p)=2(17p2−17p+5)ifp≤p1,−2(p2−p−2)ifp>p1.{p_{1}}.\end{array}\right.\]]]>
Since G2(p) decreases for p<p1, and increases for p>p1{p_{1}}$]]>, then the maximum value of this function is achieved either at the left bound or at the right bound of the interval. We have
G2(0.02)=9.3336,G2(0.1689)=5.2273251…,G2(0.5)=4.5.
Thus,
136πmax0.02≤p≤0.5G2(p)=G2(0.02)36π=0.0825271…,136πmax0.1689≤p≤0.5G2(p)=G2(0.1689)36π=0.04621970….
Similarly, with more efforts only, we get
max0.02≤p≤0.5G2(p,200)=G2(0.02,200)=9.4541…,max0.1689≤p≤0.5G2(p,200)=G2(0.1689,200)=5.2767…,G2(0.5,200)=4.515…,max0.02≤p≤0.5G2(p,N0)=G2(0.02,N0)=9.33364…,max0.1689≤p≤0.5G2(p,N0)=G2(0.1689,N0)=5.227344…,G2(0.5,N0)=4.00006….
1. One can observe from the previous proof that G2(p,N0)≈G2(p), therefore, D2(p,N0)≈G2(p)36π.
2. With increasing N, the sequence aI(N):=maxn≥ND2I(n) approaches to aI:=136πmaxp∈IG2(p). For instance, by Table 2, we have for the interval I=[0.1689,0.5] that aI(200)=0.046656…, aI(N0)=0.0462198… while aI=0.0462197…. The sequence a‾I(N):=maxn≥ND‾2I(n) tends to 0.0462197… as well, but slowly, since the main term of the difference D‾2(p,n)−G2(p)36π has the order 1n.
The following bound for Δn(p), simpler than Theorem A, follows from (50) and Table 2.
For allp∈I=[0.1689,0.5]andn≥N0,Δn(p)≤E1(p)σ+0.05532σ2.
Corollary C allows to obtain the same estimate for C02 as (4), but for larger n. Really, it is easy to verify with the help of a computer that
supp∈[0.1689,0.5](E(p)+0.05532npq(p2+q2)|n=971000)<0.409954,
but
supp∈[0.1689,0.5](E(p)+0.05532npq(p2+q2)|n=970000)>0.409954.0.409954.\]]]>
On the connection between Uspensky’s result and its refinements with the problem of estimating <inline-formula id="j_vmsta113_ineq_325"><alternatives>
<mml:math><mml:msub><mml:mrow><mml:mi mathvariant="italic">C</mml:mi></mml:mrow><mml:mrow><mml:mn>02</mml:mn></mml:mrow></mml:msub></mml:math>
<tex-math><![CDATA[${C_{02}}$]]></tex-math></alternatives></inline-formula>
First we recall Uspensky’s estimate, published by him in 1937 in [36]. To this end we introduce the following notations: Sn is the number of occurrences of an event in a series of n Bernoulli trials with a probability of success p, μ=np,
G(x)=Φ(x)+q−p62πσ(1−x2)e−x2/2.
For every x∈R, define
xn±=x−μ±12σ,
where σ=npq, as before.
Uspensky’s result can be formulated in the following form.
([<xref ref-type="bibr" rid="j_vmsta113_ref_036">36</xref>, p. 129]).
Letσ2≥25. Then for arbitrary integersa<b,|P(a≤Sn≤b)−(G(bn−)−G(an+))|≤0.13+0.18|p−q|σ2+e−3σ/2.
A lot of works are devoted to generalizations and refinements of (55), for example, [4, 14–16, 21, 24, 37].
In 2005 K. Neammanee [21] refined and generalized (55) to the case of non-identically distributed Bernoulli random variables. Let us formulate his result as applied to the case of Bernoulli trials: ifσ2≥100, then|P(a≤Sn≤b)−(G(bn−)−G(an+))|<0.1618σ2,wherean+,bn−are defined by the formula (54).
It follows from (56) that under condition σ2≥100,
|P(Sn≤b)−G(bn−)|≤0.1618σ2.
We may consider p∈(0,0.5]. Denote for brevity, d=0.1618. It follows from (57) and the definition of G(·) that
|P(Sn≤b)−Φ(bn−)|<|(1−(bn−)2)(q−p)|e−(bn−)2/262πσ+dσ2.
Taking into account that maxt|t2−1|e−t2/2=1, we get
|P(Sn≤b)−Φ(bn−)|≤|q−p|62πσ+dσ2.
Denote xn=x−μσ. It is easily seen that
|Φ(bn)−Φ(bn−)|<bn−bn−2π=122πσ.
It follows from (58), (59) that
|P(Sn≤b)−Φ(bn)|<(|q−p|6+12)1σ2π+dσ2=E1(p)σ+dσ2,
provided that 0<p≤0.5. Thus,
Δn(p)≤E1(p)σ+0.1618σ2.
Note that our bound (51) is more accurate than (60). To get the bound 0.409954 for C02 from (60), we should take n almost five times larger than in (52). Really, with the help of a computer we have
supp∈[0.1689,0.5](E(p)+0.1618npq(p2+q2)|n=4.6·106)<0.410031,
and
supp∈[0.1689,0.5](E(p)+0.1618npq(p2+q2)|n=4.2·106)>0.4100440.410044\]]]>
(cf. (52), (53)).
In 2014 V. Senatov obtained non-uniform estimates of the approximation accuracy in the central limit theorem, and, in particular, generalized Uspensky’s result (55) to lattice distributions [24].
Proof of Theorem <xref rid="j_vmsta113_stat_001">1</xref>
Before proving Theorem 1, we first prove Lemma 1.
By [10, Theorem 1],
Δn(p)≤0.33477n(ϱ(p)+0.429).
Therefore, Tn(p)≡nΔn(p)ϱ(p)≤0.33477(1+0.429ϱ(p)). Since ϱ(p) decreases on (0,0.5], then maxp∈(0,0.1689]1ϱ(p)=1ϱ(0.1689)=0.52090548…. Consequently,
maxp∈(0,0.1689]Tn(p)≤0.33477(1+0.429·0.52090549)<0.409581.
□
If instead of [10, Theorem 1] we will use other modifications of the Berry–Esseen inequality by I. Shevtsova [25], the interval (0,0.1689] for which Lemma 1 is true can be extended, i.e. one can find b>0.16890.1689$]]> such that the inequality maxp∈(0,b]Tn(p)<CE will be fulfilled. This will narrow the interval I (see (12)), which in turn will reduce the computation time on the supercomputer.
Let us indicate such b. The estimates found in [25] as applied to the particular case of Bernoulli trials can be written in the following form, Δn(p)≤0.33554n(ϱ(p)+0.415),Δn(p)≤0.3328n(ϱ(p)+0.429). It is easy to verify that inequality (62) implies b=0.174, and (63) implies that b=0.177.
It follows from Corollary 1 and (12) that for all p∈I the following bound holds,
Tn(p)<0.40973213+4.6·10−9<0.4097321346,1≤n≤N0.
Then by Lemma 1, this inequality is fulfilled for all p∈(0,0.5] as well. It is not hard to see that the bound (64) is also true for all p∈(0.5,1). Hence, bound (5) implies Theorem 1. □
Acknowledgments
We thank the following colleagues from Lomonosov Moscow State University for providing the opportunity to use supercomputer Blue Gene/P: V. Yu. Korolev, Head of the Department of Mathematical Statistics of the Faculty of Computational Mathematics and Cybernetics, Professor, I. G. Shevtsova, Assistant Professor of the same Department, A. V. Gulyaev, Deputy Dean of the same Faculty, and S. V. Korobkov, the Data Center administrator.
We also thank our colleagues from Computing Center FEB RAS for the opportunity to use the Center for the Collective Use “Data Center FEB RAS”.
We also would like to thank reviewers for useful comments.
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