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Empirical Likelihood Method for Complete Independence Test on High Dimensional Data

\nameYongcheng Qi1 and Yingchao Zhou1 1Department of Mathematics and Statistics, University of Minnesota Duluth, 1117 University Drive, Duluth, MN 55812, USA
Abstract

Given a random sample of size nn from a pp dimensional random vector, where both nn and pp are large, we are interested in testing whether the pp components of the random vector are mutually independent. This is the so-called complete independence test. In the multivariate normal case, it is equivalent to testing whether the correlation matrix is an identity matrix. In this paper, we propose a one-sided empirical likelihood method for the complete independence test for multivariate normal data based on squared sample correlation coefficients. The limiting distribution for our one-sided empirical likelihood test statistic is proved to be Z2I(Z>0)Z^{2}I(Z>0) when both nn and pp tend to infinity, where ZZ is a standard normal random variable. In order to improve the power of the empirical likelihood test statistic, we also introduce a rescaled empirical likelihood test statistic. We carry out an extensive simulation study to compare the performance of the rescaled empirical likelihood method and two other statistics which are related to the sum of squared sample correlation coefficients.

keywords:
Empirical likelihood; complete independence test; high dimension; multivariate normal distribution
22footnotetext: This is an Accepted Manuscript version of the following article, accepted for publication in Journal of Statistical Computation and Simulation [https://doi.org/10.1080/00949655.2022.2029860]. It is deposited under the terms of the Creative Commons Attribution-NonCommercial License (http://creativecommons.org/licenses/by-nc/4.0/), which permits non-commercial re-use, distribution, and reproduction in any medium, provided the original work is properly cited.

1 Introduction

Statistical inference on high dimensional data has gained a wide range of applications in recent years. New techniques generate a vast collection of data sets with high dimensions, for example, trading data from financial market, social network data and biological data like microarray and DNA data. The dimension of these types of data is not small compared with sample size, and typically of the same order as sample size or even larger. Yet classical multivariate statistics usually deal with data from normal distributions with a large sample size nn and a fixed dimension pp, and one can easily find some classic treatments in reference books such as Anderson [1], Morrison [10] and Muirhead [11].

Under multivariate normality settings, the likelihood ratio test statistic converges in distribution to a chi-squared distribution when pp is fixed. However, when pp changes with nn and tends to infinity, this conclusion is no longer true as discovered in Bai et al. [2], Jiang et al. [6], Jiang and Yang [8], Jiang and Qi [7], Qi et al. [16], among others. The results in these papers indicate that the chi-square approximation fails when pp diverges as nn goes to infinity.

The test of complete independence of a random vector is to test whether all the components of the random vector are mutually independent. In the multivariate normal case, the test of complete independence is equivalent to the test whether covariance matrix is a diagonal matrix, or whether the correlation matrix is the identity matrix.

For more details, we assume X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector from a pp-dimensional multivariate normal distribution Np(𝝁,𝚺)N_{p}(\bm{\mu},\mathbf{\Sigma}), where 𝝁\bm{\mu} denotes the mean vector, and 𝚺\mathbf{\Sigma} is a p×pp\times p covariance matrix. Given a random sample of size nn from the normal distribution, 𝐱1,𝐱2,,𝐱n\mathbf{x}_{1},\mathbf{x}_{2},\cdots,\mathbf{x}_{n}, where 𝐱i=(xi1,xi2,,xip)\mathbf{x}_{i}=(x_{i1},x_{i2},\cdots,x_{ip}) for 1in1\leq i\leq n, Pearson’s correlation coefficient between the ii-th and jj-th components is given by

rij=k=1n(xkixi¯)(xkjx¯j)k=1n(xkix¯i)2k=1n(xkjxj¯)2,r_{ij}=\frac{\sum\limits_{k=1}^{n}(x_{ki}-\bar{x_{i}})(x_{kj}-\bar{x}_{j})}{\sqrt{\sum\limits_{k=1}^{n}{(x_{ki}-\bar{x}_{i})^{2}}\cdot{\sum\limits_{k=1}^{n}(x_{kj}-\bar{x_{j}})^{2}}}}, (1)

where x¯i=1nk=1nxki\bar{x}_{i}=\frac{1}{n}\sum\limits_{k=1}^{n}x_{ki} and x¯j=1nk=1nxkj\bar{x}_{j}=\frac{1}{n}\sum\limits_{k=1}^{n}x_{kj} for 1i,jp1\leq i,j\leq p. Now we set 𝐑n=(rij)p×p\mathbf{R}_{n}=(r_{ij})_{p\times p} as the sample correlation coefficient matrix.

The complete independence test for the normal random vector is

H0:𝚪=𝐈𝐩vsHa:𝚪𝐈p,H_{0}:\bm{\Gamma}=\mathbf{I_{p}}~{}~{}~{}vs~{}~{}~{}H_{a}:\bm{\Gamma}\neq\mathbf{I}_{p}, (2)

where 𝚪\bm{\Gamma} is the population correlation matrix and 𝐈p\mathbf{I}_{p} is p×pp\times p identity matrix. When p<np<n, the likelihood ratio test statistic for (2) is a function of |𝐑n||\mathbf{R}_{n}|, the determinant of 𝐑n\mathbf{R}_{n}, from Bartlett [3] or Morrison [10]. In traditional multivariate analysis, when pp is a fixed integer, we have under the null hypothesis in (2) that

(n12p+56)log|𝐑n|𝑑χp(p1)/22 as n,-(n-1-\frac{2p+5}{6})\log|\mathbf{R}_{n}|\overset{d}{\to}\chi^{2}_{p(p-1)/2}~{}~{}~{}~{}\mbox{ as }n\to\infty,

where χf2\chi^{2}_{f} denotes a chi-square distribution with ff degrees of freedom.

When p=pnp=p_{n} depends on nn with 2pn<n2\leq p_{n}<n and pnp_{n}\to\infty, the likelihood ratio method can still be applied to test (2). The limiting distributions of the likelihood ratio test statistics in this case have been discussed in the aforementioned papers. It is worth mentioning that Qi et al. [16] propose an adjusted likelihood ratio test statistic and show that the distribution of the adjusted likelihood test statistic can be well approximated by a chi-squared distribution whose number of degrees of freedom depends on pp regardless of whether pp is fixed or divergent.

The limitation of the likelihood ratio test is that the dimension pp of the data must be smaller than the sample size nn. Many other likelihood tests related to the sample covariance matrix or sample correlation matrix have the same problem as the sample covariance matrices are degenerate when pnp\geq n. In order to relax this constraint, a new test statistic using the sum of squared sample correction coefficients is proposed by Schott [17] as follows

tnp=1j<iprij2.t_{np}=\sum_{1\leq j<i\leq p}r_{ij}^{2}.

Assume that the null hypothesis of (2) holds. Under assumption limnpn/n=γ(0,)\lim\limits_{n\to\infty}p_{n}/n=\gamma\in(0,\infty), Schott [17] proves that tnpp(p1)2(n1)t_{np}-\frac{p(p-1)}{2(n-1)} converges in distribution to a normal distribution with mean 0 and variance γ2\gamma^{2}, that is,

tnp:=tnpp(p1)2(n1)σnp𝑑N(0,1),t_{np}^{*}:=\frac{t_{np}-\frac{p(p-1)}{2(n-1)}}{\sigma_{np}}\overset{d}{\to}N(0,1), (3)

where σnp2=p(p1)(n2)(n1)2(n+1)\sigma_{np}^{2}=\frac{p(p-1)(n-2)}{(n-1)^{2}(n+1)}.

Recently, Mao [9] proposes a different test for complete independence. His test statistic is closely related to Schott’s test and is defined by

Tnp=1j<iprij21rij2.T_{np}=\sum_{1\leq j<i\leq p}\frac{r_{ij}^{2}}{1-r_{ij}^{2}}.

It has been proved in Mao [9] that TnpT_{np} is asymptotically normal under the null hypothesis of (2) and the assumption that limnpn/n=γ(0,)\lim\limits_{n\to\infty}p_{n}/n=\gamma\in(0,\infty).

Very recently, Chang and Qi [4] investigate the limiting distributions for the two test statistics above under less restrictive conditions on nn and pp. Chang and Qi [4] show that (3) is also valid under the general condition that pnp_{n}\to\infty as nn\to\infty, regardless of the convergence rate of pnp_{n}. Thus, the normal approximation in (3) based on tnpt_{np}^{*} yields an approximate level α\alpha rejection region

t(α)={tnpp(p1)2(n1)+z1αp(p1)(n1)(n1)2(n+1)},\mathcal{R}_{t}^{*}(\alpha)=\Big{\{}t_{np}\geq\frac{p(p-1)}{2(n-1)}+z_{1-\alpha}\sqrt{\frac{p(p-1)(n-1)}{(n-1)^{2}(n+1)}}\Big{\}}, (4)

where zαz_{\alpha} is a α\alpha level critical value of the standard normal distribution.

Furthermore, Chang and Qi [4] propose adjusted test statistics whose distribution can be fitted by chi-squared distribution regardless of how pp changes with nn as long as nn is large. Chang and Qi’s [4] adjusted test statistics tnpct_{np}^{c} is defined as

tnpc=p(p1)tnp+p(p1)2.t_{np}^{c}=\sqrt{p(p-1)}t_{np}^{*}+\frac{p(p-1)}{2}. (5)

Chang and Qi show that

supx|P(tnpcx)P(χp(p1)/22x)|0\sup\limits_{x}\left|P(t_{np}^{c}\leqslant x)-P(\chi^{2}_{p(p-1)/2}\leqslant x)\right|\rightarrow 0

as long as pnp_{n}\to\infty as nn\to\infty. Let χf2(α)\chi^{2}_{f}(\alpha) denote the α\alpha level critical value of χf2\chi^{2}_{f}. Then an approximate level α\alpha rejection region based on tnpct_{np}^{c} is given by

tc(α)={tnpp(p1)2(1n2n+1)+χp(p1)22(α)n2(n1)2(n+1)}.\mathcal{R}_{t}^{c}(\alpha)=\Big{\{}t_{np}\geq\frac{p(p-1)}{2}(1-\sqrt{\frac{n-2}{n+1}})+\chi^{2}_{\frac{p(p-1)}{2}}(\alpha)\sqrt{\frac{n-2}{(n-1)^{2}(n+1)}}\Big{\}}. (6)

One can find more references on test for complete independence in Mao [9] or Chang and Qi [4].

In practice, the assumption of normality for distributions may be violated. Now we assume X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector and X1,,XpX_{1},\cdots,X_{p} are identically distributed with distribution function FF. Given a random sample of size nn, 𝐱1,𝐱2,,𝐱n\mathbf{x}_{1},\mathbf{x}_{2},\cdots,\mathbf{x}_{n}, where 𝐱i=(xi1,xi2,,xip)\mathbf{x}_{i}=(x_{i1},x_{i2},\cdots,x_{ip}) for 1in1\leq i\leq n, are drawn from the distribution of X=(X1,,Xp)X=(X_{1},\cdots,X_{p}), and define Pearson’s correlation coefficients rijr_{ij}’s as in (1). By using the Stein method, Chen and Shao [5] show that (3) holds under some moment conditions of FF if pn/np_{n}/n is bounded.

In this paper, we propose to apply empirical likelihood method to the testing problem (2). The empirical likelihood is a nonparametric statistical method proposed by Owen [12, 13], which is originally used to test the mean vector of a population based on a set of independent and identically distributed (i.i.d.) random variables. Empirical likelihood does not require to specify the family of distributions for the data and it possesses some good properties of the likelihood methods.

The rest of the paper is organized as follows. In Section 2, we first introduce a one-sided empirical likelihood method for the mean of a set of random variables with a common mean and then establish the connection between the test of complete independence and the one-sided empirical likelihood method. Our main result concerning the limiting distribution of the one-sided empirical likelihood ratio statistic is also given in Section 2. In Section 3, we carry out a simulation study to compare the performance of the empirical likelihood method and normal approximation based on Schott’s test statistic and chi-square approximation based on Chang and Qi’s adjusted test statistic. In our simulation study, we also apply these methods to some other distributions such as the exponential distributions and mixture of the exponential and normal distributions so as to compare their adaptability to non-normality. The proofs of the main results are given in Section 4.

2 Main Results

In this section, we apply the empirical likelihood method to the test of complete independence. First, we assume X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector from a pp-dimensional multivariate normal distribution. Under the null hypothesis of (2), {rij2,1i<jp}\{r_{ij}^{2},1\leq i<j\leq p\} are random variables from an identical distribution with mean 1n1\frac{1}{n-1}. As a matter of fact, it follows from Corollary 5.1.2 in Muirhead [11] that rij2r_{ij}^{2} has the same distribution as T2/(n2+T2)T^{2}/(n-2+T^{2}) under the null hypothesis of (2), where TT is a random variable having tt-distribution with n2n-2 degrees of freedom. {rij2,1i<jp}\{r_{ij}^{2},1\leq i<j\leq p\} are asymptotically independent if the sample size nn is large. We will develop a one-sided empirical likelihood test statistic and apply it to the data set {(n1)rij2,1i<jp}\{(n-1)r_{ij}^{2},1\leq i<j\leq p\}, where p=pnp=p_{n} is a sequence of positive integers such that pnp_{n}\to\infty as nn\to\infty. As an extension, we then consider the case when X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector with an identical marginal distribution function FF which is not necessarily Gaussian. When the pp components of XX are independent, we demonstrate that the empirical likelihood method we develop under normality works for general distribution FF as well if some additional conditions are satisfied.

2.1 One-sided empirical likelihood test

Consider a random sample of size NN, namely y1,,yNy_{1},\cdots,y_{N}. Assume the sample comes from a population with mean μ\mu and variance σ2\sigma^{2}. The empirical likelihood function for the mean μ\mu is defined as

L(μ)=sup{i=1Nωi|i=1Nωiyi=μ,ωi0,i=1Nωi=1}.L(\mu)=\sup\Big{\{}\prod^{N}_{i=1}\omega_{i}\Big{|}\sum^{N}_{i=1}\omega_{i}y_{i}=\mu,\omega_{i}\geq 0,\sum^{N}_{i=1}\omega_{i}=1\Big{\}}. (7)

The function L(μ)L(\mu) is well defined if μ\mu belongs to the convex hull given by

H:={i=1Nωiyi|i=1Nωi=1,ωi>0,i=1,,N};H:=\Big{\{}\sum^{N}_{i=1}\omega_{i}y_{i}\Big{|}\sum^{N}_{i=1}\omega_{i}=1,\omega_{i}>0,~{}i=1,\cdots,N\Big{\}};

otherwise, set L(μ)=0L(\mu)=0. We see that H=(min1iNyi,max1iNyi)H=(\min\limits_{1\leq i\leq N}y_{i},\max\limits_{1\leq i\leq N}y_{i}).

Assume μH\mu\in H. By the standard Lagrange multiplier technique, the supremum on the right-hand side of (7) is achieved at

ωi=1N(1+λ(yiμ)),i=1,,N,\omega_{i}=\frac{1}{N(1+\lambda(y_{i}-\mu))},~{}~{}~{}i=1,\cdots,N, (8)

where λ\lambda is the solution to equation g(λ)=0g(\lambda)=0, with g(λ)g(\lambda) defined as follows

g(λ):=i=1Nyiμ1+λ(yiμ).g(\lambda):=\sum_{i=1}^{N}\frac{y_{i}-\mu}{1+\lambda(y_{i}-\mu)}. (9)

Assume min1iNyi<max1iNyi\min\limits_{1\leq i\leq N}y_{i}<\max\limits_{1\leq i\leq N}y_{i}. When μH\mu\in H, then the function g(λ)g(\lambda) defined in (9) is strictly increasing for λ((max1iNynμ)1,(μmin1iNyi)1)\displaystyle\lambda\in(-(\max_{1\leq i\leq N}y_{n}-\mu)^{-1},(\mu-\min_{1\leq i\leq N}y_{i})^{-1}). A solution to g(λ)=0g(\lambda)=0 in this range exists and the solution must be unique.

Proposition 2.1.

Assume y1,,yNy_{1},\cdots,y_{N} are NN observations with yiyjy_{i}\neq y_{j} for some ii and jj. Then logL(μ)\log L(\mu) is strictly concave in HH, and L(y¯)=supμL(μ)=NNL(\bar{y})=\sup\limits_{\mu}L(\mu)=N^{-N}, where y¯=1Ni=1Nyi\bar{y}=\frac{1}{N}\sum\limits_{i=1}^{N}y_{i}.

Remark. The results in Proposition 2.1 are well-known among the researchers in the area of empirical likelihood methods. A short proof will be given in Section 4 for completeness.

Consider the following two-sided test problem

H0:μ=μ0vsHa:μμ0.H_{0}:\mu=\mu_{0}~{}~{}~{}vs~{}~{}~{}~{}H_{a}:\mu\neq\mu_{0}.

The empirical likelihood ratio is given by

L(μ0)supμRL(μ)=L(μ0)NN=i=1N(1+λ(yiμ0))1,\frac{L(\mu_{0})}{\sup\limits_{\mu\in R}L(\mu)}=\frac{L(\mu_{0})}{N^{-N}}=\prod^{N}_{i=1}(1+\lambda(y_{i}-\mu_{0}))^{-1},

where λ\lambda is the solution to the following equation

1Ni=1Nyiμ01+λ(yiμ0)=0.\frac{1}{N}\sum_{i=1}^{N}\frac{y_{i}-\mu_{0}}{1+\lambda(y_{i}-\mu_{0})}=0.

Therefore, the log-empirical likelihood test statistic is given by

(μ0):=2logL(μ0)supμRL(μ)=2i=1Nlog(1+λ(yiμ0)).\ell(\mu_{0}):=-2\log\frac{L(\mu_{0})}{\sup\limits_{\mu\in R}L(\mu)}=2\sum^{N}_{i=1}\log(1+\lambda(y_{i}-\mu_{0})). (10)

It is proved in Owen [14] that (μ0)\ell(\mu_{0}) converges in distribution to a chi-square distribution with one degree of freedom if y1,,yNy_{1},\cdots,y_{N} are i.i.d. random variables with mean μ0\mu_{0} and a finite second moment.

Our interest here is to consider a one-sided test

H0:μ=μ0vsHa:μ>μ0.H_{0}:\mu=\mu_{0}~{}~{}~{}vs~{}~{}~{}~{}H_{a}:\mu>\mu_{0}. (11)

According to Proposition 2.1, L(μ)L(\mu) is increasing in (,y¯)(-\infty,\bar{y}) and decreasing in (y¯,)(\bar{y},\infty), which implies supμμ0L(μ)=L(μ0)I(y¯<μ0)+NNI(y¯μ0)\sup\limits_{\mu\geq\mu_{0}}L(\mu)=L(\mu_{0})I(\bar{y}<\mu_{0})+N^{-N}I(\bar{y}\geq\mu_{0}). Therefore, the empirical likelihood ratio corresponding to test (11) is

L(μ0)supμμ0L(μ)={L(μ0)NN, if y¯μ0;1, if y¯<μ0.\frac{L(\mu_{0})}{\sup\limits_{\mu\geq\mu_{0}}L(\mu)}=\left\{\begin{array}[]{ll}\frac{L(\mu_{0})}{N^{-N}},&\hbox{ if }\bar{y}\geq\mu_{0};\\ 1,&\hbox{ if }\bar{y}<\mu_{0}.\end{array}\right.

Then the log-empirical likelihood test statistic for test (11) is

n(μ0):=2logL(μ0)supμμ0L(μ)=(μ0)I(y¯μ0),\ell_{n}(\mu_{0}):=-2\log\frac{L(\mu_{0})}{\sup\limits_{\mu\geq\mu_{0}}L(\mu)}=\ell(\mu_{0})I(\bar{y}\geq\mu_{0}), (12)

where (μ0)\ell(\mu_{0}) is defined in (10).

2.2 Empirical likelihood method for testing complete independence

Let rr denote the sample Pearson correlation coefficient based on a random sample of size nn from a bivariate normal distribution with correlation coefficient ρ\rho. From Muirhead [11], page 156,

E(r2)=1n2n1(1ρ2)2F1(1,1;12n+1;ρ2),E(r^{2})=1-\frac{n-2}{n-1}(1-\rho^{2})_{2}F_{1}(1,1;\frac{1}{2}n+1;\rho^{2}),

where

F12(a,b;c;z)=1+ab1!cz+a(a+1)b(b+1)2!c(c+1)z2+=1+k=1(a)k(b)k(c)kzkk!{}_{2}F_{1}(a,b;c;z)=1+\frac{ab}{1!c}z+\frac{a(a+1)b(b+1)}{2!c(c+1)}z^{2}+\dots=1+\sum\limits_{k=1}^{\infty}\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{z^{k}}{k!}

is the hypergeometric function, (a)k=Γ(a+k)/Γ(a)(a)_{k}=\Gamma(a+k)/\Gamma(a), and Γ(x)=0tx1et𝑑t\Gamma(x)=\int^{\infty}_{0}t^{x-1}e^{-t}dt is the gamma function. It is easy to check when ρ=0\rho=0, F12(1,1;12n+1;ρ2)=1{}_{2}F_{1}(1,1;\frac{1}{2}n+1;\rho^{2})=1, and E(r2)=1n2n1=1n1E(r^{2})=1-\frac{n-2}{n-1}=\frac{1}{n-1}; when ρ0\rho\neq 0, F12(1,1;12n+1;ρ2)<1+k=1ρ2k=11ρ2{}_{2}F_{1}(1,1;\frac{1}{2}n+1;\rho^{2})<1+\sum\limits_{k=1}^{\infty}\rho^{2k}=\frac{1}{1-\rho^{2}}, and thus, E(r2)>1n2n1=1n1E(r^{2})>1-\frac{n-2}{n-1}=\frac{1}{n-1}.

First, we assume X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector from a pp-dimensional multivariate normal distribution Np(𝝁,𝚺)N_{p}(\bm{\mu},\mathbf{\Sigma}). Review the sample correlation coefficients rijr_{ij} defined in (1). Denote the correlation matrix of 𝚺\bm{\Sigma} by 𝚪=(γij)\bm{\Gamma}=(\gamma_{ij}). From the above discussion, we have that under the null hypothesis of (2), E(rij2)=1n1E(r_{ij}^{2})=\frac{1}{n-1} for all 1i<jp1\leq i<j\leq p, E(rij2)1n1E(r_{ij}^{2})\geq\frac{1}{n-1} under the alternative of (2) and at least one of the inequalities is strict. We see that test (2) is equivalent to the following one-tailed test

H0:E(r¯ij)=1,1i<jpvsHa:E(r¯ij)>1 for some 11<jp,H_{0}:E(\bar{r}_{ij})=1,~{}1\leq i<j\leq p~{}~{}vs~{}~{}H_{a}:E(\bar{r}_{ij})>1~{}\mbox{ for some }1\leq 1<j\leq p,

where r¯ij=(n1)rij2\bar{r}_{ij}=(n-1)r_{ij}^{2}. Under the null hypothesis of (2), {(n1)rij2,1i<jp}\{(n-1)r_{ij}^{2},~{}1\leq i<j\leq p\} are identically distributed with mean 11 and variance 2(n2)(n+1)\frac{2(n-2)}{(n+1)}. We also notice from Chang and Qi [4] that {(n1)rij2,1i<jp}\{(n-1)r_{ij}^{2},~{}1\leq i<j\leq p\} behave as if they were independent and identically distributed. For these reasons, we propose a one-sided empirical likelihood ratio test as follows.

Rewrite {(n1)rij2,1i<jp}\{(n-1)r_{ij}^{2},~{}1\leq i<j\leq p\} as y1,,yNy_{1},\cdots,y_{N}, where N=p(p1)/2N=p(p-1)/2. Then y1,,yNy_{1},\cdots,y_{N} are asymptotically i.i.d with mean 11. Define the one-sided log-empirical likelihood ratio test statistics as in (12) with μ0=1\mu_{0}=1, or equivalently

n:=n(1)=2I(r¯1)1i<jplog(1+λ((n1)rij21)),\ell_{n}:=\ell_{n}(1)=2I(\bar{r}\geq 1)\sum_{1\leq i<j\leq p}\log\Big{(}1+\lambda\big{(}(n-1)r_{ij}^{2}-1\big{)}\Big{)}, (13)

where λ\lambda is the solution to the equation

1i<jp(n1)rij211+λ((n1)rij21)=0,\sum_{1\leq i<j\leq p}\frac{(n-1)r_{ij}^{2}-1}{1+\lambda\big{(}(n-1)r_{ij}^{2}-1\big{)}}=0,

and r¯=y¯=n1N1i<jprij2\bar{r}=\bar{y}=\frac{n-1}{N}\sum_{1\leq i<j\leq p}r_{ij}^{2}.

Our first result on empirical likelihood method for testing the complete independence under normality in the paper is as follows.

Theorem 1.

Assume p=pnp=p_{n}\to\infty as nn\to\infty. Then n𝑑Z2I(Z>0)\ell_{n}\overset{d}{\rightarrow}Z^{2}I(Z>0) as nn\rightarrow\infty under the null hypothesis of (2), where ZZ is a standard normal random variable.

Let Φ\Phi denote the cumulative distribution function of the standard normal distribution, i.e,

Φ(x)=12πxet2/2𝑑t,x(,).\Phi(x)=\frac{1}{\sqrt{2\pi}}\int^{x}_{-\infty}e^{-t^{2}/2}dt,~{}~{}~{}~{}x\in(-\infty,\infty).

Let GG denote the cumulative distribution function of Z2I(Z>0)Z^{2}I(Z>0). Then

G(x)={0,x<0;Φ(x),x0.G(x)=\left\{\begin{array}[]{ll}0,&\hbox{$x<0$;}\\ \Phi(\sqrt{x}),&\hbox{$x\geq 0$.}\end{array}\right.

Therefore, for any α(0,12)\alpha\in(0,\frac{1}{2}), an α\alpha-level critical value of GG is given by zα2z_{\alpha}^{2}, where zαz_{\alpha} is an α\alpha-level critical value for the standard normal distribution. Based on Theorem 1, a level α\alpha rejection region for test on (11) is

e(α)={nzα2}.\mathcal{R}_{e}(\alpha)=\Big{\{}\ell_{n}\geq z_{\alpha}^{2}\Big{\}}. (14)

Here we only consider α<12\alpha<\frac{1}{2} because Z2I(Z>0)Z^{2}I(Z>0) is nonnegative, P(Z2I(Z>0)>c|H0)<12P(Z^{2}I(Z>0)>c|H_{0})<\frac{1}{2} if c>0c>0, and P(Z2I(Z>0)>c|H0)=1P(Z^{2}I(Z>0)>c|H_{0})=1 if c0c\leq 0.

Now we consider the general case when X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a random vector with independent and identically distributed components. The one-sided empirical likelihood test statistic n\ell_{n} based on {(n1)rij2,1i<jp}\{(n-1)r_{ij}^{2},~{}1\leq i<j\leq p\} is defined as in (13). The limiting distribution for n\ell_{n} is the same as that under normality.

Theorem 2.

Assume X1,,XpX_{1},\cdots,X_{p} are independent and identically distributed and E(X124)<E(X_{1}^{24})<\infty. If p=pnp=p_{n}\to\infty as nn\to\infty and pn/np_{n}/n is bounded, then n𝑑Z2I(Z>0)\ell_{n}\overset{d}{\rightarrow}Z^{2}I(Z>0) as nn\rightarrow\infty.

Compared with Theorem 1, pnp_{n} in Theorem 2 is restricted in a smaller range and it can be of the same order as nn.

To demonstrate the performance of empirical likelihood method and two other test statistics, we have a numerical study. Our simulation study indicates that the empirical likelihood test (14) based om n\ell_{n} maintains a very stable size or type I error. In terms of size, n\ell_{n} is more accurate tnpct_{np}^{c} and tnpt_{np}^{*}. Most of the time, tnpct_{np}^{c} and tnpt_{np}^{*} have slightly larger sizes than 0.050.05 when the nominal level α\alpha is 0.050.05, and their powers are also slightly larger than that of n\ell_{n} in our simulation study. For simplicity purpose, the simulation result on n\ell_{n} is not shown in this paper.

In order to balance the size and power for the empirical likelihood method, We introduce a rescaled empirical likelihood statistic, ¯n\bar{\ell}_{n}, defined as follows

¯n=2(n1)(n+1)3(p1)(p+4)n1i<jprij4.\bar{\ell}_{n}=\frac{2(n-1)(n+1)}{3(p-1)(p+4)}\ell_{n}\sum_{1\leq i<j\leq p}r_{ij}^{4}. (15)

Under conditions of Theorems 1 or 2, ¯n\bar{\ell}_{n} and n\ell_{n} have the same limiting distribution, that is,

¯n𝑑Z2I(Z>0) as n\bar{\ell}_{n}\overset{d}{\rightarrow}Z^{2}I(Z>0)~{}~{}\mbox{ as }~{}~{}n\rightarrow\infty (16)

provided that

2(n1)(n+1)3(p1)(p+4)1i<jprij4𝑝1.\frac{2(n-1)(n+1)}{3(p-1)(p+4)}\sum_{1\leq i<j\leq p}r_{ij}^{4}\overset{p}{\to}1. (17)

This equation will be verified in Section 4. Based on (16), a level α\alpha test rejects the complete independence if ¯n\bar{\ell}_{n} falls into the rejection region

¯e(α)={¯nzα2}.\bar{\mathcal{R}}_{e}(\alpha)=\Big{\{}\bar{\ell}_{n}\geq z_{\alpha}^{2}\Big{\}}. (18)

3 Simulation

In this section, we will consider the following three test statistics for testing complete independence (2), including Schott’s test statistic tnpt_{np}^{*} given in (3), Chang and Qi’s adjusted test statistic tnpct_{np}^{c} defined in (5), and the rescaled empirical likelihood test statistic ¯n\bar{\ell}_{n} given in (15). The corresponding rejection regions are given in (4), (6), and (18), respectively. All simulations are implemented by the software R.

For sample size n=20,50,100n=20,50,100 and dimension p=10,20,50,100p=10,20,50,100, we apply the three test statistics to each of five distributions for 10000 iterations to obtain the empirical sizes and the empirical powers of the tests. We set the nominal type I error α=0.05\alpha=0.05. The five distributions include the normal, the uniform over [1,1][-1,1], the exponential, the mixture of the normal and exponential distributions, and the sum of normal and exponential distributions.

To control the dependence structure, we introduce a covariance matrix 𝚪ρ\bm{\Gamma}_{\rho} defined by

𝚪ρ=(γij)p×p, with γii=1, and γij=ρ if ij,\bm{\Gamma}_{\rho}=\big{(}\gamma_{ij}\big{)}_{p\times p},\mbox{ with }\gamma_{ii}=1,\mbox{ and }\gamma_{ij}=\rho\mbox{ if }i\neq j, (19)

which is also a correlation matrix. In our simulation study, we generate random samples from the distribution of a random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) with covariance matrix 𝚪ρ\bm{\Gamma}_{\rho} or correlation matrix 𝚪ρ\bm{\Gamma}_{\rho}. For details, see the five distributions described below. For all distributions we consider, the observations have independent components when ρ=0\rho=0 and positively dependent components when ρ>0\rho>0. We choose very small values for ρ\rho such as ρ=0.02\rho=0.02 and 0.050.05. When the value of ρ\rho is large, the resulting powers for all three methods will be too close to 11, and the comparison is meaningless. Therefore, based on 10,00010,000 replicates, the sizes for three test statistics are estimated when ρ=0\rho=0, and their powers are estimated when ρ=0.02\rho=0.02 and 0.050.05. All results are reported in Tables 1 to 5.

a.   Normal Distribution

The observations are drawn from a multivariate normal random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) with mean 𝝁=(0,,0)\bm{\mu}=(0,\cdots,0) and variance matrix 𝚪ρ\bm{\Gamma}_{\rho} specified in (19). The results on the empirical sizes and powers are given in Table 1.

b.  Uniform Distribution

We first generate p+1p+1 i.i.d. random variables Y0,Y1,,YpY_{0},Y_{1},\cdots,Y_{p} from Uniform (1,1)(-1,1) distribution, then set Xi=ρ1ρY0+YiX_{i}=\frac{\sqrt{\rho}}{\sqrt{1-\rho}}Y_{0}+Y_{i}, i=1,2,,pi=1,2,\cdots,p. It is easy to verify that random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) has mean 𝝁=(0,,0)\bm{\mu}=(0,\cdots,0) and correlation matrix 𝚪ρ\bm{\Gamma}_{\rho} as defined in (19). The results on the empirical sizes and powers are given in Table 2.

c.  Exponential Distribution

We generate p+1p+1 i.i.d. random variables Y0,Y1,,YpY_{0},Y_{1},\cdots,Y_{p} from the unit exponential distribution, then define Xi=ρ1ρY0+YiX_{i}=\frac{\sqrt{\rho}}{\sqrt{1-\rho}}Y_{0}+Y_{i}, i=1,2,,pi=1,2,\cdots,p. The random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) has a correlation matrix 𝚪ρ\bm{\Gamma}_{\rho} as defined in (19) for ρ[0,1)\rho\in[0,1). The results on the empirical sizes and powers are given in Table 3.

d.  Mixture of Normal and Exponential Distributions

The random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is sampled from a mixture of the normal and exponential distributions which is with 90% probability from the multivariate normal with mean 𝝁=(1,,1)\bm{\mu}=(1,\cdots,1) and covariance matrix 𝚪ρ\bm{\Gamma}_{\rho} given in (19) and with 10% probability from a random vector (Y1,,Yp)(Y_{1},\cdots,Y_{p}) where Y1,,YpY_{1},\cdots,Y_{p} are i.i.d. unit exponential random variables. The results on the empirical sizes and powers are given in Table 4.

e.  Sum of Normal and Exponential Distribution

The random vector X=(X1,,Xp)X=(X_{1},\cdots,X_{p}) is a weighted sum of two independent random vectors, UU and VV, X=U+0.01VX=U+0.01V, where UU is from a multivariate normal distribution with mean 𝝁=(0,,0)\bm{\mu}=(0,\cdots,0) and covariance matrix 𝚪ρ\bm{\Gamma}_{\rho} defined in (19), and V=(Y1,,Yp)V=(Y_{1},\cdots,Y_{p}) with YiY_{i}’s being i.i.d. unit exponential random variables. The results on the empirical sizes and powers are given in Table 5.

From the simulation results, the empirical sizes for all three tests are close to 0.050.05 which is the nominal type I error we set in the simulation, especially when both nn and pp are large. Test statistic tnpct_{np}^{c} has the smallest size in most cases, and it is a little bit conservative sometimes. The size of ¯n\bar{\ell}_{n} is between that of tnpct_{np}^{c} and tnpt_{np}^{*} and both ¯n\bar{\ell}_{n} and tnpt_{np}^{*} are comparable for most combinations of nn and pp.

As we expect, the powers of all three test statistics become higher as pp grows larger. The increase in nn also brings about an increase in power, but not as much as the increase in pp does, because p(p1)2\frac{p(p-1)}{2} is the number of rij2r_{ij}^{2}’s involved in the test. All test statistics achieve high power when ρ=0.05\rho=0.05. Three test statistics result in comparable powers in general, although the power of Chang and Qi’s test statistic is occasionally a little bit less than the other two test statistics. These differences may be due to the fact that Chang and Qi’s test statistic maintain a lower type I error.

In summary, in this paper, we have developed the one-sided empirical likelihood method and proposed the rescaled empirical likelihood test statistic for testing the complete independence for high dimensional random vectors. The rescaled empirical likelihood test statistic performs very well in terms of the size and power and can serve as a good alternative to the existent test statistics in the literature.

4 Proofs

Proof of Proposition 2.1. To prove the strict concavity of L(μ)L(\mu), we need to show that for μ1,μ2H\mu_{1},\mu_{2}\in H, μ1μ2,\mu_{1}\neq\mu_{2},

logL(tμ1+(1t)μ2)>tlogL(μ1)+(1t)logL(μ2),t(0,1).\log L(t\mu_{1}+(1-t)\mu_{2})>t\log L(\mu_{1})+(1-t)\log L(\mu_{2}),~{}~{}t\in(0,1). (20)

Since μjH\mu_{j}\in H for j=1,2j=1,2, we have logL(μj)=logi=1Nωji=i=1Nlogωji\log L(\mu_{j})=\log\prod\limits^{N}_{i=1}\omega_{ji}=\sum\limits^{N}_{i=1}\log\omega_{ji}, where ωji>0\omega_{ji}>0, i=1,,Ni=1,\cdots,N are determined by (8) and (9) with μ\mu being replaced by μj\mu_{j}, i=1Nωji=1\sum\limits^{N}_{i=1}\omega_{ji}=1, i=1Nωjiyi=μj\sum\limits^{N}_{i=1}\omega_{ji}y_{i}=\mu_{j} for j=1,2j=1,2.

For every t(0,1)t\in(0,1), set ωti=tω1i+(1t)ω2i\omega_{ti}=t\omega_{1i}+(1-t)\omega_{2i}, i=1,,Ni=1,\cdots,N. Then ωti>0\omega_{ti}>0, i=1Nωti=1\sum\limits^{N}_{i=1}\omega_{ti}=1, i=1Nωtiyi=tμ1+(1t)μ2H\sum\limits^{N}_{i=1}\omega_{ti}y_{i}=t\mu_{1}+(1-t)\mu_{2}\in H. Since logx\log x is strictly concave in (0,)(0,\infty), we have

log(ωti)=log(tω1i+(1t)ω2i)tlogω1i+(1t)logω2ii=1,,N,\log(\omega_{ti})=\log\Big{(}t\omega_{1i}+(1-t)\omega_{2i}\Big{)}\geq t\log\omega_{1i}+(1-t)\log\omega_{2i}~{}~{}~{}i=1,\cdots,N,

and at least one of the inequalities is strict, i.e, log(ωti)>tlogω1i+(1t)logω2i\log(\omega_{ti})>t\log\omega_{1i}+(1-t)\log\omega_{2i} for some ii, since μ1μ2\mu_{1}\neq\mu_{2} implies (ω11,ω12,,ω1N)(ω21,ω22,,ω2N)(\omega_{11},\omega_{12},\cdots,\omega_{1N})\neq(\omega_{21},\omega_{22},\cdots,\omega_{2N}). Therefore, we get

i=1Nlog(ωti)>i=1N(tlogω1i+(1t)logω2i)=tlogL(μ1)+(1t)logL(μ2),\sum^{N}_{i=1}\log(\omega_{ti})>\sum^{N}_{i=1}\Big{(}t\log\omega_{1i}+(1-t)\log\omega_{2i}\Big{)}=t\log L(\mu_{1})+(1-t)\log L(\mu_{2}),

which implies

logL(tμ1+(1t)μ2)logi=1Nωti=i=1Nlog(ωti)>tlogL(μ1)+(1t)logL(μ2),\log L(t\mu_{1}+(1-t)\mu_{2})\geq\log\prod^{N}_{i=1}\omega_{ti}=\sum^{N}_{i=1}\log(\omega_{ti})>t\log L(\mu_{1})+(1-t)\log L(\mu_{2}),

proving (20).

When μ=y¯\mu=\bar{y}, an obvious solution to (9) is λ=0\lambda=0. Since the solution to (9) is unique, we see that ωi=N1\omega_{i}=N^{-1}, and thus, L(y¯)=NNL(\bar{y})=N^{-N}. We also notice that

supμL(μ)=supμHL(μ)sup{i=1Nωi|ωi0,i=1Nωi=1}=NN.\sup_{\mu}L(\mu)=\sup_{\mu\in H}L(\mu)\leq\sup\Big{\{}\prod^{N}_{i=1}\omega_{i}\Big{|}\omega_{i}\geq 0,\sum^{N}_{i=1}\omega_{i}=1\Big{\}}=N^{-N}.

The last step is obtained by using the Lagrange multipliers. We omit the details here. Therefore, we conclude that L(y¯)=supμL(μ)=NNL(\bar{y})=\sup\limits_{\mu}L(\mu)=N^{-N}. \Box

Proof of Theorem 1. We assume the null hypothesis in (2) is true in the proof.

Define σn2=2(n2)n+1\sigma_{n}^{2}=\frac{2(n-2)}{n+1} and Sn2=1N1i<jp((n1)rij21)2S_{n}^{2}=\frac{1}{N}\sum\limits_{1\leq i<j\leq p}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}. Review that N=p(p1)/2N=p(p-1)/2. We have σnp2=Nσn2/(n1)2\sigma_{np}^{2}=N\sigma_{n}^{2}/(n-1)^{2}. Since the distribution of yjy_{j}’s depends on nn, {yj, 1jN}\{y_{j},\,1\leq j\leq N\} forms an array of random variables.

If the following three conditions are satisfied: (i). 1σnmax1jN|yj1|=op(N1/2)\frac{1}{\sigma_{n}}\max\limits_{1\leq j\leq N}|y_{j}-1|=o_{p}(N^{1/2}) as nn\to\infty; (ii). 1Nσn2j=1N(yj1)2𝑝1\frac{1}{N\sigma_{n}^{2}}\sum\limits^{N}_{j=1}(y_{j}-1)^{2}\overset{p}{\to}1 as nn\to\infty; (iii). j=1NyjNNσn2𝑑N(0,1)\displaystyle{\frac{\sum\limits^{N}_{j=1}y_{j}-N}{\sqrt{N\sigma_{n}^{2}}}}\overset{d}{\to}N(0,1) as nn\to\infty, equivalently, in term of rij2r_{ij}^{2}’s,

  1. (C1).

    1σnmax1i<jp|(n1)rij21|=op(N1/2)\displaystyle\frac{1}{\sigma_{n}}\max_{1\leq i<j\leq p}|(n-1)r_{ij}^{2}-1|=o_{p}(N^{1/2}) as nn\to\infty;

  2. (C2).

    1σn2Sn2𝑝1\displaystyle\frac{1}{\sigma_{n}^{2}}S_{n}^{2}\overset{p}{\to}1 as nn\to\infty;

  3. (C3).

    zn:=1i<jp(n1)rij2NNσn2𝑑N(0,1)\displaystyle z_{n}:=\frac{\sum\limits_{1\leq i<j\leq p}(n-1)r_{ij}^{2}-N}{\sqrt{N\sigma_{n}^{2}}}\overset{d}{\to}N(0,1) as nn\to\infty,

we can follow the same procedure as in Owen [14] or use Theorem 6.1 in Peng and Schick [15] to conclude that

(1)=(1i<jp(n1)rij2NNσn2)2(1+op(1))+op(1)=zn2(1+op(1))+op(1)\ell(1)=\Big{(}\frac{\sum\limits_{1\leq i<j\leq p}(n-1)r_{ij}^{2}-N}{\sqrt{N\sigma_{n}^{2}}}\Big{)}^{2}(1+o_{p}(1))+o_{p}(1)=z_{n}^{2}(1+o_{p}(1))+o_{p}(1)

where (1)\ell(1) is defined in (10) with μ0=1\mu_{0}=1. Again, by using condition (C3), we have

n=zn2(1+op(1))I(y¯>0)+op(1)=zn2(1+op(1))I(zn>0)+op(1)𝑑Z2I(Z>0)\ell_{n}=z_{n}^{2}(1+o_{p}(1))I(\bar{y}>0)+o_{p}(1)=z_{n}^{2}(1+o_{p}(1))I(z_{n}>0)+o_{p}(1)\overset{d}{\to}Z^{2}I(Z>0)

as nn\to\infty, where n\ell_{n} is defined in (13), proving Theorem 1.

Now we will verify conditions (C1), (C2) and (C3). (C3) has been proved by Chang and Qi [4] as we indicate below equation (3).

Assume (i,j)(i,j) is a pair of integers with for 1i<jp1\leq i<j\leq p. It is proved in Schott [17] that

E(rij2)=1n1,Var(rij2)=2(n2)(n+1)(n1)2=σn2(n1)2,E(r_{ij}^{2})=\frac{1}{n-1},~{}\text{Var}(r_{ij}^{2})=\frac{2(n-2)}{(n+1)(n-1)^{2}}=\frac{\sigma_{n}^{2}}{(n-1)^{2}}, (21)

From Chang and Qi [4], we have

E(rij4)=3(n1)(n+1),E(rij6)=15(n1)(n+1)(n+3),\displaystyle E(r_{ij}^{4})=\frac{3}{(n-1)(n+1)},~{}E(r_{ij}^{6})=\frac{15}{(n-1)(n+1)(n+3)}, (22)
E(rij8)=105(n1)(n+1)(n+3)(n+5).\displaystyle E(r_{ij}^{8})=\frac{105}{(n-1)(n+1)(n+3)(n+5)}.

By using binomial expansion, we also have

m4:=E((rij21n1)4)=E(rij8)4E(rij6)n1+6E(rij4)(n1)24E(rij2)(n1)3+1(n1)4.m_{4}:=E\big{(}(r_{ij}^{2}-\frac{1}{n-1})^{4}\big{)}=E(r_{ij}^{8})-4\frac{E(r_{ij}^{6})}{n-1}+6\frac{E(r_{ij}^{4})}{(n-1)^{2}}-4\frac{E(r_{ij}^{2})}{(n-1)^{3}}+\frac{1}{(n-1)^{4}}. (23)

Now we can verify condition (C1). By use of Chebyshev’s inequality, equations (21) and (22)

P(max1i<jp(n1)rij2σn>δN1/2)\displaystyle P(\frac{\max\limits_{1\leq i<j\leq p}(n-1)r_{ij}^{2}}{\sigma_{n}}>\delta N^{1/2}) \displaystyle\leq 1i<jpP(rij2σn>δN1/2n1)\displaystyle\sum_{1\leq i<j\leq p}P(\frac{r_{ij}^{2}}{\sigma_{n}}>\frac{\delta N^{1/2}}{n-1})
\displaystyle\leq N(n1)3δ4N3/2σn3E(r126)\displaystyle\frac{N(n-1)^{3}}{\delta^{4}N^{3/2}\sigma_{n}^{3}}E(r_{12}^{6})
=\displaystyle= O(1N1/2)0\displaystyle O(\frac{1}{N^{1/2}})\to 0

as nn\to\infty for every δ>0\delta>0. This implies 1σnmax1i<jp(n1)rij2=o(N1/2)\frac{1}{\sigma_{n}}\max\limits_{1\leq i<j\leq p}(n-1)r_{ij}^{2}=o(N^{1/2}). Hence, we have

1σnmax1i<jp|(n1)rij21|=1σnmax1i<jp(n1)rij2+O(1)=op(N1/2),\frac{1}{\sigma_{n}}\max_{1\leq i<j\leq p}|(n-1)r_{ij}^{2}-1|=\frac{1}{\sigma_{n}}\max_{1\leq i<j\leq p}(n-1)r_{ij}^{2}+O(1)=o_{p}(N^{1/2}),

proving condition (C1).

Below we will use (i,j)(i,j) and (s,t)(s,t) to denote two pair of integers with 1i<jp1\leq i<j\leq p and 1s<tp1\leq s<t\leq p. It follows from Theorem 2 in Veleval and Ignatov [18] that {rij,1i<jp}\{r_{ij},~{}1\leq i<j\leq p\} are pairwise independent, that is, If (i,j)(s,t)(i,j)\neq(s,t), then rijr_{ij} and rstr_{st} are independent, thus we have

E(((n1)rij21)2((n1)rst21)2)=E(((n1)rij21)2)E(((n1)rst21)2)=σn4.E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}=E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\Big{)}E\Big{(}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}=\sigma_{n}^{4}.

Since E(Sn2)=σn2E(S_{n}^{2})=\sigma_{n}^{2}, we have

E(Sn2σn2)2\displaystyle E\Big{(}S_{n}^{2}-\sigma^{2}_{n}\Big{)}^{2} =\displaystyle= E(Sn4)σn4\displaystyle E(S_{n}^{4})-\sigma_{n}^{4}
=\displaystyle= 1N21i<jp1s<tpE(((n1)rij21)2((n1)rst21)2)σn4.\displaystyle\frac{1}{N^{2}}\sum_{1\leq i<j\leq p}\sum_{1\leq s<t\leq p}E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}-\sigma_{n}^{4}.

We can classify the summands within the double summation above into two classes: N(N1)N(N-1) terms in class 1 when (i,j)(s,t)(i,j)\neq(s,t) and NN terms in class 2 when (i,j)=(s,t)(i,j)=(s,t). We see that

E(((n1)rij21)2((n1)rst21)2)=σn4E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}=\sigma_{n}^{4}

if (i,j)(s,t)(i,j)\neq(s,t) by using the independence, and

E(((n1)rij21)2((n1)rst21)2)=m4(n1)4E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}=m_{4}(n-1)^{4}

if (i,j)=(s,t)(i,j)=(s,t), where m4m_{4} is given by (23). Therefore, we have

E(Sn2σn2)2=1N2(N(N1)σn4+Nm4(n1)4)σn4=m4(n1)4σn4N.E\Big{(}S_{n}^{2}-\sigma^{2}_{n}\Big{)}^{2}=\frac{1}{N^{2}}(N(N-1)\sigma_{n}^{4}+Nm_{4}(n-1)^{4})-\sigma_{n}^{4}=\frac{m_{4}(n-1)^{4}-\sigma_{n}^{4}}{N}.

In view of (21), (22) and (23), some tedious calculation shows that

E(Sn2σn2)2=16(n2)(7n330n2+11n+60)p(p1)(n+1)2(n+3)(n+5)=4(7n330n2+11n+60)σn4p(p1)(n2)(n+3)(n+5),E\Big{(}S_{n}^{2}-\sigma^{2}_{n}\Big{)}^{2}=\frac{16(n-2)(7n^{3}-30n^{2}+11n+60)}{p(p-1)(n+1)^{2}(n+3)(n+5)}=\frac{4(7n^{3}-30n^{2}+11n+60)\sigma_{n}^{4}}{p(p-1)(n-2)(n+3)(n+5)},

which implies

E(Sn2σn21)2=4(7n330n2+11n+60)p(p1)(n2)(n+3)(n+5)=O(1pn2)0,E\Big{(}\frac{S_{n}^{2}}{\sigma^{2}_{n}}-1\Big{)}^{2}=\frac{4(7n^{3}-30n^{2}+11n+60)}{p(p-1)(n-2)(n+3)(n+5)}=O(\frac{1}{p_{n}^{2}})\to 0,

as nn\to\infty, and thus Condition (C2) holds. The proof of Theorem 1 is complete. \Box

Proof of Theorem 2. Theorem 2 can be proved by using similar arguments in the proof of Theorem 1. We will continue to use the notation defined in the proof of Theorem 1.

Under the conditions in Theorem 2, Chen and Shao [5] have obtained the following results:

E(rij2)=1n1,E(rij4)=3n2+O(1n3),E(rij8)=O(1n4);E(rij12rij22)=1(n1)2,E(rij14rij24)=9n4+O(1n5) if j1j2,\displaystyle\begin{aligned} &E(r_{ij}^{2})=\frac{1}{n-1},~{}~{}E(r_{ij}^{4})=\frac{3}{n^{2}}+O(\frac{1}{n^{3}}),~{}~{}E(r_{ij}^{8})=O(\frac{1}{n^{4}});\\ &E(r_{ij_{1}}^{2}r_{ij_{2}}^{2})=\frac{1}{(n-1)^{2}},~{}~{}E(r_{ij_{1}}^{4}r_{ij_{2}}^{4})=\frac{9}{n^{4}}+O(\frac{1}{n^{5}})~{}~{}\mbox{ if }j_{1}\neq j_{2},\end{aligned} (24)

where 1ijp1\leq i\neq j\leq p, 1ij1p1\leq i\neq j_{1}\leq p and 1ij2p1\leq i\neq j_{2}\leq p. It follows from the CrC_{r} inequality that

E((n1)rij21)423((n1)4rij8+1)=O(1).E\big{(}(n-1)r_{ij}^{2}-1\big{)}^{4}\leq 2^{3}\big{(}(n-1)^{4}r_{ij}^{8}+1\big{)}=O(1). (25)

We need to verify conditions (C1), (C2), and (C3) as given in the proof of theorem 1. Condition (C1) can be verified similarly by using estimates of moments in (24), and condition (C3) follows from the central limit theorem (3), which is true under the condition of the theorem in virtue of Theorem 2.2 in Chen and Shao [5].

Now we proceed to verify condition (C2). First, we have

σ¯n2:=E(Sn2)=E(((n1)r1221)2)=(n1)2E(r124)1=2+O(1n)=σn2(1+O(1n))\bar{\sigma}_{n}^{2}:=E(S_{n}^{2})=E\Big{(}\big{(}(n-1)r_{12}^{2}-1\big{)}^{2}\Big{)}=(n-1)^{2}E\big{(}r_{12}^{4}\big{)}-1=2+O(\frac{1}{n})=\sigma_{n}^{2}(1+O(\frac{1}{n})) (26)

from (24). Then

E(Sn2σ¯n2)2\displaystyle E\Big{(}S_{n}^{2}-\bar{\sigma}^{2}_{n}\Big{)}^{2} =\displaystyle= E(Sn4)σ¯n4\displaystyle E(S_{n}^{4})-\bar{\sigma}_{n}^{4}
=\displaystyle= 1N21i<jp1s<tpE(((n1)rij21)2((n1)rst21)2)σ¯n4.\displaystyle\frac{1}{N^{2}}\sum_{1\leq i<j\leq p}\sum_{1\leq s<t\leq p}E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}-\bar{\sigma}_{n}^{4}.

Considering the summands within the double summation above, we see that there are N(p2)(p3)/2N(p-2)(p-3)/2 pairs of sets {i,j}\{i,j\} and {s,t}\{s,t\} which are disjoint. For these pairs,

E(((n1)rij21)2((n1)rst21)2)=σ¯n4.E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2}\Big{)}=\bar{\sigma}_{n}^{4}.

because rijr_{ij} and rstr_{st} are independent. For all other N2N(p2)(p3)/2N^{2}-N(p-2)(p-3)/2 pairs, corresponding summands are dominated by

E(((n1)rij21)2((n1)rst21)2\displaystyle E\Big{(}\big{(}(n-1)r_{ij}^{2}-1\big{)}^{2}\big{(}(n-1)r_{st}^{2}-1\big{)}^{2} \displaystyle\leq E((n1)rij21)4E((n1)rst21)4\displaystyle\sqrt{E\big{(}(n-1)r_{ij}^{2}-1\big{)}^{4}E\big{(}(n-1)r_{st}^{2}-1\big{)}^{4}}
=\displaystyle= E((n1)rij21)4\displaystyle E\big{(}(n-1)r_{ij}^{2}-1\big{)}^{4}
=\displaystyle= O(1)\displaystyle O(1)

from the Cauchy-Schwarz inequality and equation (25). Therefore, we have

E(Sn2σ¯n2)2\displaystyle E\Big{(}S_{n}^{2}-\bar{\sigma}^{2}_{n}\Big{)}^{2} =\displaystyle= 1N2(12N(p2)(p3)σ¯n4+O(N212N(p2)(p3)))σ¯n4\displaystyle\frac{1}{N^{2}}\Big{(}\frac{1}{2}{N(p-2)(p-3)}\bar{\sigma}_{n}^{4}+O\big{(}N^{2}-\frac{1}{2}N(p-2)(p-3)\big{)}\Big{)}-\bar{\sigma}_{n}^{4}
=\displaystyle= O(1N2(N212N(p2)(p3)))\displaystyle O\Big{(}\frac{1}{N^{2}}\big{(}N^{2}-\frac{1}{2}N(p-2)(p-3)\big{)}\Big{)}
=\displaystyle= O(1p)0\displaystyle O(\frac{1}{p})\to 0

as nn\to\infty. In the estimation above, we also use the fact that σ¯n2σn22\bar{\sigma}_{n}^{2}\sim\sigma_{n}^{2}\to 2 from (26). Therefore, we have

E(Sn2σ¯n21)20E\Big{(}\frac{S_{n}^{2}}{\bar{\sigma}^{2}_{n}}-1\Big{)}^{2}\to 0

as nn\to\infty. This yields Sn2σ¯n2𝑝1\displaystyle{\frac{S_{n}^{2}}{\bar{\sigma}^{2}_{n}}\overset{p}{\to}1}, which together with (26) implies condition (C2). This completes the proof of the theorem. \Box

Proof of Equation (17). In the proofs of Theorems 2 and 2, we have obtained that Sn/σn2𝑝1S_{n}/\sigma_{n}^{2}\overset{p}{\to}1, which implies Sn𝑝2S_{n}\overset{p}{\to}2 since σn22\sigma_{n}^{2}\to 2. From condition (C3), we have

1N(1i<jp(n1)2rij2N)𝑝0.\frac{1}{N}\big{(}\sum_{1\leq i<j\leq p}(n-1)^{2}r_{ij}^{2}-N\big{)}\overset{p}{\to}0.

Therefore, we get

2(n1)(n+1)3(p1)(p+4)1i<jprij4\displaystyle\frac{2(n-1)(n+1)}{3(p-1)(p+4)}\sum_{1\leq i<j\leq p}r_{ij}^{4} =\displaystyle= 1+o(1)3N1i<jp(n1)2rij4\displaystyle\frac{1+o(1)}{3N}\sum_{1\leq i<j\leq p}(n-1)^{2}r_{ij}^{4}
=\displaystyle= 1+o(1)3(Sn2+1+2N(1i<jp(n1)2rij2N))\displaystyle\frac{1+o(1)}{3}\Big{(}S_{n}^{2}+1+\frac{2}{N}\big{(}\sum_{1\leq i<j\leq p}(n-1)^{2}r_{ij}^{2}-N\big{)}\Big{)}
𝑝\displaystyle\overset{p}{\to} 1\displaystyle 1

as nn\to\infty, proving (17). \Box

Acknowledgements

The authors would like to thank the Editor, Associate Editor, and referee for reviewing the manuscript and providing valuable comments. The research of Yongcheng Qi was supported in part by NSF Grant DMS-1916014.

Disclosure statement

No potential conflict of interest was reported by the authors.

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Table 1: Sizes and Powers of tests, Normal distribution
size (ρ=0\rho=0) power (ρ=0.02\rho=0.02) power (ρ=0.05\rho=0.05)
nn pp ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*}
20 10 0.0593 0.0486 0.0598 0.0646 0.0551 0.0657 0.0974 0.0863 0.0987
20 0.0609 0.0565 0.0618 0.0670 0.0626 0.0675 0.1357 0.1281 0.1366
50 0.0594 0.0584 0.0596 0.0836 0.0810 0.0842 0.3079 0.3030 0.3096
100 0.0524 0.0519 0.0525 0.1156 0.1142 0.1159 0.5584 0.5571 0.5588
50 10 0.0596 0.0484 0.0603 0.0750 0.064 0.0744 0.1660 0.1478 0.1656
20 0.0533 0.0481 0.0525 0.0855 0.0790 0.0859 0.3257 0.3098 0.3244
50 0.0508 0.0489 0.0508 0.1457 0.1410 0.1461 0.7372 0.7320 0.7362
100 0.0536 0.0519 0.0535 0.2684 0.2660 0.2684 0.9609 0.9604 0.9608
100 10 0.0608 0.0504 0.0607 0.0894 0.0743 0.0882 0.3305 0.3025 0.3285
20 0.0581 0.0511 0.0573 0.1265 0.1167 0.1260 0.6520 0.6345 0.6495
50 0.0523 0.0494 0.0519 0.2735 0.2675 0.2723 0.9816 0.9809 0.9816
100 0.0516 0.0506 0.0515 0.5711 0.5669 0.5703 0.9997 0.9997 0.9997
Table 2: Sizes and Powers of tests, Uniform distribution
size (ρ=0\rho=0) power (ρ=0.02\rho=0.02) power (ρ=0.05\rho=0.05)
nn pp ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*}
20 10 0.0610 0.0529 0.0615 0.0607 0.0525 0.0626 0.0913 0.0802 0.0934
20 0.0600 0.0571 0.0615 0.0668 0.0621 0.0686 0.1230 0.1147 0.1243
50 0.0595 0.0582 0.0607 0.0806 0.0785 0.0812 0.2680 0.2647 0.2693
100 0.0585 0.0582 0.0598 0.1099 0.1093 0.1109 0.5183 0.5171 0.5202
50 10 0.0572 0.0489 0.0575 0.0773 0.0655 0.0774 0.1744 0.1536 0.1749
20 0.0606 0.0548 0.0600 0.0809 0.0734 0.0816 0.3061 0.2920 0.3062
50 0.0559 0.0530 0.0562 0.1358 0.1319 0.1359 0.7449 0.7387 0.7449
100 0.0526 0.0518 0.0527 0.2417 0.2378 0.2422 0.9741 0.9736 0.9741
100 10 0.0596 0.0497 0.0583 0.0842 0.0710 0.0848 0.3225 0.2934 0.3200
20 0.0589 0.0528 0.0580 0.1269 0.1152 0.1271 0.6430 0.6265 0.6397
50 0.0516 0.0489 0.0522 0.2650 0.2587 0.2646 0.9889 0.9880 0.9887
100 0.0504 0.0494 0.0504 0.5567 0.5525 0.5566 1.0000 1.0000 1.0000
Table 3: Sizes and Powers of tests, Exponential distribution
size (ρ=0\rho=0) power (ρ=0.02\rho=0.02) power (ρ=0.05\rho=0.05)
nn pp ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*}
20 10 0.0718 0.0636 0.0745 0.0799 0.0734 0.0845 0.1419 0.1338 0.1486
20 0.0638 0.0610 0.0670 0.0867 0.0859 0.0924 0.2135 0.2116 0.2221
50 0.0581 0.0586 0.0612 0.1224 0.1266 0.1301 0.3982 0.4040 0.4080
100 0.0602 0.0622 0.0632 0.1901 0.1955 0.1975 0.5774 0.5848 0.5871
50 10 0.0734 0.0620 0.0751 0.0950 0.0840 0.0975 0.2288 0.2156 0.2335
20 0.0650 0.0606 0.0660 0.1158 0.1114 0.1205 0.3834 0.3800 0.3898
50 0.0612 0.0608 0.0643 0.1877 0.1912 0.1953 0.7102 0.7140 0.7175
100 0.0589 0.0607 0.0618 0.3230 0.3289 0.3310 0.9079 0.9107 0.9111
100 10 0.0701 0.0604 0.0698 0.1048 0.0936 0.1088 0.3636 0.3428 0.3689
20 0.0646 0.0595 0.0647 0.1555 0.1492 0.1578 0.6350 0.6281 0.6400
50 0.0636 0.0622 0.0653 0.3108 0.3125 0.3187 0.9481 0.9486 0.9495
100 0.0574 0.0583 0.0600 0.5823 0.5878 0.5899 0.9965 0.9966 0.9966
Table 4: Sizes and Powers of tests, Mixture distribution
size (ρ=0\rho=0) power (ρ=0.02\rho=0.02) power (ρ=0.05\rho=0.05)
nn pp ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*}
20 10 0.0650 0.0547 0.0664 0.0682 0.0594 0.0680 0.1042 0.0921 0.1052
20 0.0618 0.0565 0.0630 0.0704 0.0650 0.0706 0.1561 0.1477 0.1572
50 0.0605 0.0584 0.0608 0.0974 0.0955 0.0983 0.3381 0.3353 0.3394
100 0.0688 0.0677 0.0692 0.1395 0.1381 0.1407 0.5979 0.5974 0.5998
50 10 0.0624 0.0520 0.0616 0.0778 0.0657 0.0781 0.1991 0.1790 0.1988
20 0.0602 0.0548 0.0606 0.0958 0.0869 0.0952 0.3832 0.3713 0.3828
50 0.0587 0.0559 0.0578 0.1643 0.1595 0.1638 0.7827 0.7792 0.7824
100 0.0569 0.0561 0.057 0.3086 0.3063 0.3097 0.9669 0.9666 0.9672
100 10 0.0610 0.0510 0.0615 0.0990 0.0845 0.0961 0.3777 0.3484 0.3745
20 0.0599 0.0520 0.0592 0.1395 0.1299 0.1380 0.7200 0.7055 0.7183
50 0.0556 0.0522 0.0555 0.3137 0.3080 0.3137 0.9888 0.9887 0.9888
100 0.0561 0.0545 0.0559 0.6389 0.6359 0.6400 1.0000 1.0000 1.0000
Table 5: Sizes and Powers of tests, Sum
size (ρ=0\rho=0) power (ρ=0.02\rho=0.02) power (ρ=0.05\rho=0.05)
nn pp ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*} ¯n\bar{\ell}_{n} tnpct_{np}^{c} tnpt_{np}^{*}
20 10 0.0601 0.0520 0.0607 0.0625 0.0552 0.0634 0.0997 0.0884 0.1019
20 0.0556 0.0499 0.0558 0.0635 0.0584 0.0637 0.1402 0.1303 0.1401
50 0.0520 0.0499 0.0520 0.0804 0.0773 0.0807 0.3066 0.3031 0.3072
100 0.0592 0.0583 0.0593 0.1172 0.1153 0.1173 0.5509 0.5484 0.5515
50 10 0.0587 0.0493 0.0586 0.0723 0.0610 0.0723 0.1702 0.1501 0.1684
20 0.0558 0.0497 0.0557 0.0843 0.0762 0.0838 0.3315 0.3144 0.3290
50 0.0537 0.0507 0.0541 0.1429 0.1385 0.1429 0.7446 0.7377 0.7433
100 0.0557 0.0540 0.0555 0.2637 0.2614 0.2642 0.9550 0.9545 0.9550
100 10 0.0635 0.0552 0.0640 0.0897 0.0758 0.0892 0.3237 0.2955 0.3187
20 0.0593 0.0527 0.0586 0.1211 0.1097 0.1197 0.6540 0.6360 0.6513
50 0.0480 0.0459 0.0479 0.2716 0.2644 0.2706 0.9803 0.9791 0.9801
100 0.0551 0.0537 0.0549 0.5719 0.5690 0.5713 1.0000 1.0000 1.0000