Cauchy Combination Test for Sparse Signals

Mingya Long1,2, Zhengbang Li3, Wei Zhang1,2, Qizhai Li1,2,∗

1 Academy of Mathematics and Systems Science, Chinese Academy of Sciences,
Beijing 100190, China.

2 School of Mathematical Science, University of Chinese Academy of Sciences,
Beijing 100049, China.

3 School of Mathematics and Statistics, Central China Normal University, Wuhan
430079, China.

*Corresponding author. E-mail<EMAIL_ADDRESS>
Abstract

Aggregating multiple effects is often encountered in large-scale data analysis
where the fraction of significant effects is generally small. Many existing
methods cannot handle it effectively because of lack of computational accuracy
for small $p$-values. The Cauchy combination test (abbreviated as CCT) (J Am
Statist Assoc, 2020, 115(529):393-402) is a powerful and computational
effective test to aggregate individual $p$-values under arbitrary correlation
structures. This work revisits CCT and shows three key contributions including
that (i) the tail probability of CCT can be well approximated by a standard
Cauchy distribution under much more relaxed conditions placed on individual
$p$-values instead of the original test statistics; (ii) the relaxation
conditions are shown to be satisfied for many popular copulas formulating
bivariate distributions; (iii) the power of CCT is no less than that of the
minimum-type test as the number of tests goes to infinity with some regular
conditions. These results further broaden the theories and applications of
CCT. The simulation results verify the theoretic results and the performance
of CCT is further evaluated with data from a prostate cancer study.

Keywords:  Cauchy combination test; $p$-value combinations; Sparse signals.

##  1 Introduction

Combining individual p-values to perform an overall test is a long-standing
problem in statistics, which has a wide range of applications in genetics,
genomics and functional magnetic resonance imaging. Suppose that there are $m$
individual $p$-values, $p_{1},\cdots,p_{m}$. A few well-known conventional
combination approaches include $-2\sum_{i=1}^{m}\ln p_{i}$ [Fisher(1932)] ,
$-\sum_{i=1}^{m}\ln(1-p_{i})$ [Pearson(1933)],
$\sum_{i=1}^{m}\Phi^{-1}(1-p_{i})$ [Liptak(1958)], and $\sum_{i=1}^{m}p_{i}$
[Edgington(1972)]. These methods are appropriate to handle dense signals that
many $p_{i}$s are significant under a given nominal significance level. In the
recent decades, with the development of science and technology, high-
dimensional data are frequently collected in many fields. A typical feature of
such data is that the significant effects account for a small proportion of
signals. In this case, the conventional tests may lose substantial powers.
Some approaches were subsequently developed to address this issue, for
example, the Tippett’s minimum $p$-value test (abbreviated as MINP)
[Tippett(1931)], the Berk-Jones test [Berk & Jones(1979)], the higher
criticism test [Donoho & Jin(2004)], the group-combined test [Hu et
al.(2016)], and the generalized higher criticism test [Barnett et al.(2016)],
among others. Most of these tests have no analytic methods for $p$-value
calculation when the individual $p$-values are correlated. Throughout the
paper, $p$-value calculation means computing the $p$-value of a combination
test instead of the individual $p$-values which are assumed to be readily
used. Although permutation or bootstrap approaches can be used to accommodate
correlation structure, they are computationally intensive or even intractable
in the analysis of large-scale data, especially when the $p$-value of a
combination test is extremely small. To overcome this drawback, Liu and Xie
[Liu & Xie(2020)] recently proposed a Cauchy combination test as
$\text{CCT}=\sum_{i=1}^{m}\omega_{i}\tan\big{(}(0.5-p_{i})\pi\big{)}$, where
the weights $\omega_{i}$s are nonnegative and $\sum_{i=1}^{m}\omega_{i}=1$.
They showed that the tail of the null distribution of CCT can be well
approximated by a standard Cauchy distribution under the null hypothesis that
$p_{1},\cdots,p_{m}$ follow the uniform distribution on $(0,1)$ and two key
assumptions mentioned next paragraph. So far, CCT has been widely used in many
fields. For example, Gorfine, Schlesinger and Hsu [Gorfine et al.(2020)]
applied CCT to present novel and powerful tests for right censoring data in
survival analysis; McCaw et al. [McCaw et al.(2020)] constructed a powerful
test based on CCT for quantitative trait genome-wide association studies; Xu
et al. [Xu et al.(2020)] applied CCT to analyze whole-genome sequencing data;
Khalid et al. [Khalid et al.(2020)] emplyed CCT to do inter-module
communications for runtime hardware trojan detection.

Let $Z_{1},\cdots,Z_{m}$ be the potential test statistic corresponding to
$p_{1},\cdots,p_{m}$, respectively. Two key assumptions for
$Z_{1},\cdots,Z_{m}$ are required for the standard Cauchy distribution
approximation. They are (C1) the joint distribution of the test statistic
$(Z_{i},Z_{j})^{\top}$ follows a bivariate normal distribution, $1\leq i\neq
j\leq m$, and (C2) the maximal eigenvalue of the correlation matrix
$\boldsymbol{R}$ of the test statistics $Z_{1},\cdots,Z_{m}$ is bounded, where
⊤ denotes the transpose of a matrix or a vector. These assumptions are not
general and may be violated in practice. On one hand, the joint distribution
of two potential test statistics can be an arbitrary bivariate distribution,
rather than a bivariate normal distribution. On the other hand, the commonly
used spiked covariance model [Liptak(1958), Paul(2007), Lee et al.(2014),
Zhang et al.(2020)] does not satisfy the assumption (C2). As for the spiked
model, the first $d$ largest eigenvalues of the covariance matrix is an
increasing function of $m$, which converges to infinity as
$m\rightarrow\infty$, and the other eigenvalues are finite. For example, we
considered a spiked model where the first $d$ largest eigenvalues are
$\\{\lambda_{i}=m/3^{i},~{}i=1,\cdots,d\\}$ and the others are
$\\{\lambda_{i}=1,~{}i=d+1,\cdots,m\\}$. The other special case is the
compound symmetry covariance structure
$\left(\rho^{\text{Sgn}(|i-j|)}\right)_{m\times m}$ whose maximal eigenvalues
is $m\rho+(1-\rho)$ and the other eigenvalues are all equal to $1-\rho$, where
$\text{Sgn}(|i-j|)$ is a sign function equal to 0 if $i=j$ and $1$ otherwise,
$i,j=1,\cdots,m$, and $|\rho|\in(0,1)$.

A natural question is that whether the tail distribution of CCT under the null
hypothesis can be still approximated by a standard Cauchy distribution when
the two assumptions are violated. Moreover, as is well known, the MINP or
equivalently the maximum-type test (abbreviated as MAX) have good power under
the sparse alternatives. Although extensive numerical studies have been
conducted in the literature to investigate the performances of CCT and MINP, a
theoretic justification is still lacking.

To fill this gap, the present article further studies the theories of CCT and
makes the following useful contributions. We show that the approximation of
the standard Cauchy distribution for the tail of the null distribution of CCT
is still valid under the spiked correlation models and a broad range of
bivariate distributions, and the power of CCT is no less than that of MINP
when the number of tests goes to infinity and the number of zero signals
belongs to $(0,1)$. The extension is nontrivial. These theoretical results can
broaden the applications of CCT.

The rest of the paper is organized as follows. The main results are presented
in Section 2. Simulation studies are conducted in Section 3 to examine the
approximation accuracy. In Section 4, the data from a real study are analyzed
to further investigate the performance of MINP and CCT. Proofs of the theorems
are provided in Section 5.

##  2  Main Results

Let $Z_{i}$ be the potential test statistic corresponding to individual
$p$-value $p_{i}$ with zero mean and unit standard deviation, respectively,
and $\rho_{ij}$ be the correlation coefficient between $Z_{i}$ and $Z_{j}$,
$i,j=1,\cdots,m$. Write $\boldsymbol{R}=(\rho_{ij})_{m\times m}$ and assume
that

  1. (C1)

$(Z_{i},Z_{j})^{\top}$ follows a bivariate normal distribution for $1\leq
i<j\leq m$;

  2. (C2)

$\lambda_{max}(\boldsymbol{R})<\infty$, where $\lambda_{max}(\boldsymbol{R})$
is the largest eigenvalue of $\boldsymbol{R}$;

  3. (C3)

$\underset{{1\leq i<j\leq m}}{\max}|\rho_{ij}|\leq\rho_{max}<1$;

Under the assumption (C1) with fixed $m$ or the assumptions (C1)-(C3) with
$m=t^{\delta}$, $\delta\in\big{(}0,1/2\big{)}$, Liu and Xie [Liu & Xie(2020)]
showed that

$\lim\limits_{t\rightarrow\infty}\frac{P\left(\text{CCT}>t\right)}{0.5-\arctan(t)/\pi}=1,$
(1)

where the denominator is the tail probability of the standard Cauchy
distribution, that is, $P(X>t)=0.5-\arctan(t)/\pi$ and $X$ follows the
standard Cauchy distribution.

###  2.1  Tail null distribution of CCT for fixed $m$

Instead of making assumptions (C1) and (C2) on potential test statistics, we
transfer to consider the following condition on the individual $p$-values
$p_{1},\cdots,p_{m}$.

  1. (D1)

For any $1\leq i<j\leq m$, as $t\rightarrow\infty$, there exists a positive
number $\delta_{t}$ with $\delta_{t}\rightarrow 0$ and
$\delta_{t}t\rightarrow\infty$, such that

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/t)$

and

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/t).$

Under the assumption (D1), we will show the standard Cauchy approximation
still holds, which is shown in Theorem 1 and the proof is given in Section 5.

###### Theorem 1

Assume $p_{i}\sim U(0,1)$, $i=1,\cdots,m$, and the assumption (D1) is
satisfied. Then (1) holds.

The assumption (D1) imposes no restriction on the type of joint distribution
of $Z_{i}$ and $Z_{j}$, $i,j=1,\cdots,m$. Such an assumption is satisfied not
only for bivariate normal distribution of $(T_{i},T_{j})^{\top}$, $i\neq
j=1,\cdots,m$, but also holds for other bivariate distributions. With loss of
generality, we consider six types of bivariate copula functions $\mathbb{C}$
supported on $[0,1)^{2}$ and the univariate marginal distribution on $[0,1]$,
that is, the joint distribution function of $(p_{i},p_{j})^{\top}$ is
$\mathbb{C}(u_{i},v_{j}),1\leq i\neq j\leq m$.

1) Product Copula:

$\mathbb{C}(u_{i},v_{j})=u_{i}v_{j},\;\text{where}\;1\leq i\neq j\leq m.$

2) Farlie-Gumbel-Morgenstern (FGM) Copula:

$\mathbb{C}(u_{i},v_{j})=u_{i}v_{j}[1+\theta(1-u_{i})(1-v_{j})],\;\text{where}\;\theta\in[-1,1],\;1\leq
i\neq j\leq m.$

3) Cuadras-Augé Copula:

$\mathbb{C}(u_{i},v_{j})=\big{(}\min{(u_{i},v_{j})}\big{)}^{\theta}(u_{i}v_{j})^{1-\theta},\;\text{where}\;\theta\in[0,1],\;1\leq
i\neq j\leq m.$

4) Normal Copula:

$\displaystyle\mathbb{C}(u_{i},v_{j})=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int_{-\infty}^{\Phi^{-1}(u_{i})}\int_{-\infty}^{\Phi^{-1}(v_{j})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy,$
$\displaystyle\text{where}\;1\leq i\neq j\leq m\;,\,\underset{{1\leq i<j\leq
m}}{\max}|\rho_{ij}|\leq\rho_{max}<1.$

5) Ali-Mikhail-Haq (AMH) Copula:

$\mathbb{C}(u_{i},v_{j})=\frac{u_{i}v_{j}}{1-\theta(1-u_{i})(1-v_{j})},\;\text{where}\;\theta\in[-1,1],\;1\leq
i\neq j\leq m.$

6) Survival Copula:

$\mathbb{C}(u_{i},v_{j})={u_{i}v_{j}}\exp{(-\theta\ln u_{i}\ln
v_{j})},\;\text{where}\;\theta\in[0,1],\;1\leq i\neq j\leq m.$

The product copula is for two independent uniform distributions [Sklar(1959)].
The FGM copula is always used in risk models, for example, Zhang and Yang
[Zhang & Yang(2011)] studied the Gerber-Shiu functions for a perturbed
compound Poisson risk model with dependence between the claim size and the
interclaim times based on the FGM copula. Cuadras-Augé copula models the
dependence structure of some bivariate distributions [Cuadras(2006)]. The
Normal copula is the most prominent choice for copulas and it was used in the
valuation of some products’ declining during the financial crisis of 2007 and
2008 [Meyer(2013)]. As for the AMH copula, it could be extensively used in
data analysis due to good statistical properties [Kumar(2010)]. In reality,
AMH copula was applied to obtain the most suitable bivariate distribution
between two rain gauge stations [Yee et al.(2014)]. Finally, the survival
copula is used in censored data frequently. For instance, Geerdens, Acar and
Janssen proposed a model to infer the impact of a covariate on the dependence
structure of right-censored clustered event time data based on survival copula
[Geerdens et al.(2018)]. In Section 5, we will show that (D1) is satisfied for
the above six types of copulas, thus, the conclusion (1) holds.

We point out that the original results shown in Liu and Xie [Liu & Xie(2020)]
had no requirement on the covariance matrix. Typically the spiked model is one
of the special cases where the maximal eigenvalues of $\boldsymbol{R}$ change
with $m$. As for the condition (D1), this condition has no requirement for the
covariance matrix, which apparently means that spiked model is applicable.

###  2.2 Tail null distribution of CCT for divergent $m$

Next we establish the theory of the tail of the null distribution of CCT for
for divergent $m$, in other word, for high-dimensional data with large $m$.
The following condition is needed.

  1. (D2)

For any $1\leq i<j\leq m$, as $t\rightarrow\infty$, there exists a positive
number $\delta_{t}$ with $\delta_{t}\rightarrow 0$ and
$\delta_{t}t\rightarrow\infty$, such that

$\sup_{1\leq i<j\leq
m}P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}})$

and

$\sup_{1\leq i<j\leq
m}P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/{t^{1+\gamma}}),$

where $0<\gamma\leq 1$.

###### Theorem 2

Assume $p_{i}\sim U(0,1)$, $i=1,\cdots,m$, $m=t^{\gamma/2}$, and the
assumption (D2) is satisfied. Then (1) holds.

The proof of Theorem 2 is given in Section 5. As commented in Liu and Xie [Liu
& Xie(2020)], CCT has some advantages in high-dimensional data, such as fast
$p$-value computation and relaxing the assumption on the positive
definitiveness of the correlation matrix $\boldsymbol{R}$, which is often
violated in high-dimensional data with small sample size and a large number of
feature variables. As mentioned in the introduction section, it is unknown
whether the conclusion (1) holds for the spiked covariance models since the
maximal eigenvalue $\lambda_{max}(\boldsymbol{R})$ goes to infinity as $m$
goes to infinity, and other bivariate distributions. On one hand, similar to
the case of fixed $m$, there are no restrictions on the type of joint
distribution of $Z_{i}$ and $Z_{j}$, $i,j=1,\cdots,m$. Only the condition (D2)
is assumed. So the conclusion (1) might hold for other type of bivariate
distributions. We point out that the copula functions given in Section 2.1 can
result in (D2), thus, the conclusion (1) holds. The proof of these results are
given in Section 5. On the other hand, based on the assumption (C2) of Liu and
Xie [Liu & Xie(2020)], the conclusion (1) does not hold for the spiked model
with the maximal eigenvalue increasing with $m$. Theorem 2 relaxes the
assumption (C2) by using (D2) which imposes no restriction on the covariance
matrix. So, the conclusion (1) still holds for the spiked model, which can
also be seen in the process of Theorem 2.

###  2.3 Power comparison between CCT and MINP

As a conventional $p$-value combination test, MINP [Tippett(1931)] has been
widely used in genomic analysis and genetic association studies. Since the
powers of MINP and MAX are equivalent, we then compare the performance of CCT
to MAX in this part. By following the theoretical settings of Donoho and Jin
[Donoho & Jin(2004)] and Liu and Xie [Liu & Xie(2020)], we assume that
$(Z_{1},\cdots,Z_{m})^{\top}\sim N_{m}(\boldsymbol{\mu},\boldsymbol{R})$,
where $\boldsymbol{\mu}=(\mu_{1},\cdots,\mu_{m})^{\top}$. Our interest is
testing for the hypothesis
$H_{0}:\boldsymbol{\mu}={\boldsymbol{0}}_{m}\leftrightarrow
H_{1}:\boldsymbol{\mu}\neq{\boldsymbol{0}}_{m}$, where ${\boldsymbol{0}}_{m}$
is $m$-dimensional column vector with all zero units. Under $H_{1}$, denote
the index set of nonzero signals by $\Omega=\\{1\leq i\leq m:\mu_{i}\neq 0\\}$
and let the total number of nonzero signals be $\|\Omega\|=m^{\nu}$, where
$\|\cdot\|$ is the cardinality of a set and the parameter $0<\nu<1$ measures
the sparsity magnitude of signals. Similarly, denote the number of zero
signals by $\|\Omega^{c}\|=m^{\kappa}$, where $1>\kappa>0$ and
$\Omega^{c}=\\{1,\cdots,m\\}\backslash\Omega$. To derive the asymptotic
distribution of the MAX test statistic, where
$\text{MAX}=\max\left\\{Z_{1}^{2},\cdots,Z_{m}^{2}\right\\}$, we first give
the following assumption.

  1. (D3)

$\varrho_{1}<1~{}\hbox{and}~{}\varrho_{k}(\log
k)^{2+s}{\rightarrow}0~{}\hbox{for}~{}s>0~{}\hbox{as}~{}{k\rightarrow\infty}~{}\hbox{,
where the }~{}\varrho_{k}=\underset{|i-j|\geq
k}{\sup}|\rho_{ij}|,k=1,\cdots,m-1.$

###### Corollary 1

Suppose that the assumption (D3) holds, for a large enough $t>0$, we have

$\lim\limits_{m\rightarrow\infty}P\left(\frac{\text{MAX}-\tilde{a}_{m}-O\left(\frac{1}{\log
m}\right)}{\tilde{b}_{m}}<t\right)=\exp\big{(}\exp(-t)\big{)},$

where $\tilde{a}_{m}=2\log{m}-\\{\log{(\log{m})}+\log(4\pi)-\log
4\\}+\\{\log{(\log{m})}+\log(4\pi)-\log 4\\}/(2\log{m})$ and
$\tilde{b}_{m}=2-1/(\log{m})$.

Deo [Deo(1972)] and Pakshirajan et al. [Pakshirajan & Hebbar(1977)] considered
$\\\ \max\left\\{|Z_{1}|,\cdots,|Z_{m}|\right\\},$ and showed that for a large
enough $t>0$,

$\lim\limits_{m\rightarrow\infty}P\left(\frac{\max\left\\{|Z_{1}|,\cdots,|Z_{m}|\right\\}-a_{m}-O\left(\frac{1}{\log
m}\right)}{b_{m}}<t\right)=\exp\big{(}\exp(-t)\big{)},$

where $a_{m}=(2\log m)^{\frac{1}{2}}-\frac{\log\log m+4\pi-4}{(8\log
m)^{1/2}}$ and $b_{m}=(2\log m)^{-\frac{1}{2}}$ by assuming (D3). Of note,
corollary 1 is similar to Deo [Deo(1972)] and Pakshirajan et al. [Pakshirajan
& Hebbar(1977)]’s. However, our current proof is totally different and more
challenging since the technical complications are under a square term test to
support the theorem 3.

Let $\xi_{1-\alpha}$ and $\eta_{1-\alpha}$ be the $1-\alpha$ quantile of the
standard Cauchy distribution and the standard Gumbell distribution,
respectively, i.e., $\xi_{1-\alpha}=\cot(\pi\alpha)$ and
$\eta_{1-\alpha}=-\log(\log\alpha)^{-1}$. The asymptotic powers of CCT and MAX
are thus given by

$\beta_{\text{CCT}}=P_{H_{1}}(\text{CCT}>\xi_{1-\alpha})~{}~{}\text{and}~{}~{}\beta_{\text{MAX}}=P_{H_{1}}\bigg{(}\frac{\sqrt{\text{MAX}}-a_{m}-O\left(\frac{1}{\log
m}\right)}{b_{m}}>\eta_{1-\alpha}\bigg{)}.$

###### Theorem 3

Assume that $\varrho_{1}<1$, $\varrho_{k}(\log k)^{2+d}{\rightarrow}0$ for
$d>0$ as ${k\rightarrow\infty}$ and
$\min\limits_{i=1,\cdots,m}\omega_{i}=O(1/m)$. Then as $m\rightarrow\infty$,

$\beta_{\text{CCT}}\geq\beta_{\text{MAX}}+o(1).$

Theorem 3 shows that when $m$ is large enough, the asymptotical power of CCT
is no less than that of MAX or MINP. This provides another good theoretic
justification for CCT.

##  3  Simulation Studies

###  3.1  Accuracy of p-value calculation for spiked models and copulas

To evaluate the calculation accuracy of tail probability based on the standard
Cauchy distribution, we first conduct simulation studies under two spiked
models with equal and unequal covariance matrices. We consider $m=10$, 50 and
500. The potential test statistics $(Z_{1},\cdots,Z_{m})^{\top}$ are generated
from the $m$-dimensional normal distribution $N({\bf 0}_{m},{\boldsymbol{R}})$
with mean vector ${\bf 0}_{m}=(0,\cdots,0)^{\top}$ and covariance matrix
${\boldsymbol{R}}=\big{(}\rho_{ij}\big{)}_{m\times m}$. Let the eigenvalues of
${\boldsymbol{R}}$ be $\lambda_{i}$, $i=1,\cdots,m$ and the correlation matrix
${\boldsymbol{R}}$ is specified using the following two models.

  * •

Model 1 (Equal correlation spiked model): $\rho_{ij}=\rho$ for $1\leq i\neq
j\leq m$ and $\rho_{ii}=1$ for $1\leq i\leq m$, where $\rho=0.2,0.5$ and
$0.8$.

  * •

Model 2 (Unequal correlation spiked model): $\lambda_{i}=m/3^{i}$ for
$i=1,\cdots,d$ and $\lambda_{i}=1$ for $i=d+1,\cdots,m$. Set $d=4,5$, and $6$.

The individual $p$-value is obtained as $p_{i}=2(1-\Phi(|Z_{i}|))$, where
$\Phi(\cdot)$ is the cumulative distribution function of the standard normal
distribution. Figures 1 and 2 display the tail probabilities of $P(X>t)$ and
$P(\text{CCT}>t)$, where $X$ is a standard Cauchy distributed random variable
and $P(\text{CCT}>t)$ is calculated based on 100,000 simulated samples. Since
the exact distribution of CCT is unknown, we use this resampling strategy as a
”golden” method throughout. In addition, we set the endpoints of the
horizontal axis in the figures to be the 95% and 99.97% quantiles,
respectively and the corresponding values are $6.314$ and 1000. The figures
show that the standard Cauchy distribution can well approximate the tail
probability of CCT, especially for large values of CCT corresponding to small
$p$-values. For example, in Figure 2 with $m=10$, and $d=5$, when $t=306.214$,
the ”golden” tail probability of CCT is $0.00098$, while it is $0.00103$ using
the standard Cauchy distribution approximation.

Secondly, we conduct simulation studies under two copula models including FGM
copula mixed with product copula and AHM copula mixed with product copula. We
consider $m=10$, 50 and 500. The individual $p$-values
$(p_{1},\cdots,p_{m})^{\top}$ are generated from the following two models.

  * •

Model 3 (Mixed FGM copula model):
$(p_{i},p_{i+1})^{\top}\sim\mathbb{C}(u_{i},v_{i+1})=u_{i}v_{i+1}[1+\theta(1-u_{i})(1-v_{i+1})]$,
for $i=1,3,\cdots,2\lfloor m/2\rfloor-1$, where $\lfloor m\rfloor$ means the
maximal integer less than $m$ and
$(p_{i},p_{i+1})^{\top}\sim\mathbb{C}(u_{i},v_{i+1})=u_{i}v_{j}$ for other
$i,j$, where $\theta=0.2,0.5$ and $0.8$.

  * •

Model 4 (Mixed AHM copula model):
$(p_{i},p_{i+1})^{\top}\sim\mathbb{C}(u_{i},v_{i+1})=\frac{u_{i}v_{i+1}}{1-\theta(1-u_{i})(1-v_{i+1})}$,
for $i=1,3,\cdots,2\lfloor m/2\rfloor-1$ where
$(p_{i},p_{i+1})^{\top}\sim\mathbb{C}(u_{i},v_{i+1})=u_{i}v_{j}$ for other
$i,j$, where $\theta=0.2,0.5$ and $0.8$.

Figures 3 and 4 show the tail probabilities of $P(X>t)$ and $P(\text{CCT}>t)$,
where $P(\text{CCT}>t)$ is also calculated based on 100,000 simulated samples.
We can draw the same conclusion from Figures 3 and 4 that the standard Cauchy
distribution can well approximate the tail probability of CCT under two mixed
copula models. For example, in Figure 3 with $\theta=0.8$ and $m=50$, when
$t=406.214$, the tail probability of CCT is $0.00074$, while it is $0.00078$
using the standard Cauchy distribution approximation.

###  3.2  Power comparison

Next we conduct simulation studies to verify Theorem 3 under the assumption
that $\varrho_{1}<1$ and $\varrho_{k}(\log k)^{2+d}{\rightarrow}0$ for $d>0$
as ${k\rightarrow\infty}$, by comparing the powers of CCT and MAX for various
$m$. Since the above two spiked models do not satisfy the assumption, we use
another two common models to specify the covariance matrix. The potential test
statistics $(Z_{1},\cdots,Z_{m})^{\top}$ are generated from the
$m$-dimensional normal distribution
$N({\boldsymbol{\mu}}_{m},{\boldsymbol{R}})$ with mean vector
${\boldsymbol{\mu}}_{m}$ and covariance matrix
${\boldsymbol{R}}=\big{(}\rho_{ij}\big{)}_{m\times m}$. Consider a sparse mean
vector ${\boldsymbol{\mu}}_{m}$ by letting the number of nonzero signals
$\|\Omega\|/m=0.1,0.2$ and $0.3$. The two correlation models for
${\boldsymbol{R}}$ are specified as

  * •

Model 1 (AR(1) correlation): $\rho_{ij}=\rho^{|i-j|}$ for $1\leq i\neq j\leq
m$ and $\rho_{ii}=1$ for $1\leq i\leq m$, where $\rho=0.2,0.5$ and $0.8$.

  * •

Model 2 (Polynomial decay): $\rho_{ij}=1/{(1+|i-j|^{a})}$ for $1\leq i\neq
j\leq m$, where $a=0.5,1.5$ and $2.5$.

We choose $m$ from $\\{1000,1100,1200,1300,1400,1500\\}$. The significance
level is $0.05$. Figures 5 and 6 display the powers of CCT and MAX for Models
1 and 2, respectively. From the figures, we can see that the powers of MAX are
uniformly less than those of CCT. For example, in Figure 6, when the number of
nonzero signals $\|\Omega\|/m$ is $0.3$, $a=2.5$ and $m=1200$, the power of
MAX is 0.3130, which is by far smaller than that of CCT with $0.5165$. For the
considered scenarios, the largest power difference between MAX and CCT is
$0.2735$.

##  4  Real Data Analysis

In this section, we apply the MINP and CCT to analyze the data from a prostate
cancer study, with the aim being to investigate whether gene expression level
differences contribute to common clinical and pathological features of
prostate cancer. The data consists of expression profiles of approximately
12,600 genes from 52 tumor samples and 50 non-tumor prostate specimens
[Sklar(1959)]. The individual p-values were obtained from the Wilcoxon test
for the comparison of gene expression profiles between the tumor and normal
prostate samples. For illustration, we confine our analysis to the Axon
guidance pathway (map04360), which consists of 163 genes.

The organism name produced through the Axon guidance pathway is Axon guidance.
Axon guidance occupies a pivotal place in the formation of neuronal network.
Multitudinous guidance factors guide the Axons, such as netrins, ephrins,
Slits, and semaphorins [Negishi et al.(2005)]. These guidance cues are seized
by growth cone receptors, and downstream signal transduction pathways of these
receptors converge onto the Rho GTPases to attract changes in cytoskeletal
organization that conclude which way the growth cone will transform [Govek et
al.(2005)]. Therefore, a variety of diseases are reported to be associated
with the Axon guidance pathway, such as Nemaline myopathy, Chromosme 5p
deletion syndrome, L1 syndrome, Congenital mirror movements, Congenital
hydrocephalus, Capillary malformation-arteriovenous malformation and
schizophrenia [Dickson(2003), Kim et al.(2011)].

With the Wicoxon test, the individual p-values for $m=163$ genes are
calculated. For an overall comparison between the tumor and normal groups, we
use the MINP and CCT to combine these individual p-values. The p-values are
0.0015 for MINP and 0.0005 for CCT, indicating that CCT is more powerful than
MINP, where wWe calculate the p-value of MINP and CCT based on the 2000 Monte
Carlo replicates and the approximation formula (1), respectively.

##  5 Technical Details

To develop the theories, we list some useful Lemmas.

###### Lemma 1

(i) As $t\rightarrow\infty$,
$0<0.5-\frac{\arctan(t)}{\pi}=\frac{\arctan(1/t)}{\pi}<\frac{1}{\pi t}.$

(ii) Suppose random variables $X$ and $U$ follow a standard Cauchy
distribution and Uniform distribution in $[-{\pi}/{2},{\pi}/{2}]$,
respectively. Then,

$P\left(X>t\right)=P\left(\tan(U)>t\right)=0.5-\frac{\arctan(t)}{\pi}=\dfrac{1}{\pi
t}+o(1/t).$

######  Proof 1 ( of Lemma 1)

(i) By the equation $0.5=\frac{\arctan(t)}{\pi}+\frac{\arctan(1/t)}{\pi}$, we
can obtain

$0.5-\frac{\arctan(t)}{\pi}=\frac{\arctan(1/t)}{\pi}.$

Since
$\frac{\mathrm{d}\big{(}(1/t)-\arctan(1/t)\big{)}}{\mathrm{d}(1/t)}=1-\frac{1}{1+(1/t)^{2}}>0$
and $(1/t)-\arctan(1/t)\big{|}_{t\rightarrow\infty}\rightarrow 0,$
$(1/t)>\arctan(1/t)$. Hence, $\frac{\arctan(1/t)}{\pi}<\frac{1}{\pi t}$.

(ii) Using Taylor’s series expansion,

$\frac{\arctan(1/t)}{\pi}=\frac{1}{\pi
t}+\frac{\frac{\mathrm{d}^{2}\big{(}(1/t)-\arctan(1/t)\big{)}}{\mathrm{d}(1/t)^{2}}|_{1/t_{\epsilon}}}{2}(1/t)^{2},$

where $1/t_{\epsilon}\in(0,1/t)$ and
$\frac{\frac{\mathrm{d}^{2}\big{(}(1/t)-\arctan(1/t)\big{)}}{\mathrm{d}(1/t)^{2}}|_{1/t_{\epsilon}}}{2}(1/t)^{2}$
is an error term and equals to $o(1/t)$.

###### Lemma 2

(i) For any $x>0$,

$\Phi^{-1}\left(1-\dfrac{\phi(x)}{x}\right)\leq
x\leq\Phi^{-1}\left(1-\dfrac{\phi(x)}{x}\cdot\dfrac{x^{2}}{1+x^{2}}\right).$

(ii) For $y\rightarrow 0+$,

$\sqrt{\log(1/y^{2})+1}-1\leq\Phi^{-1}(1-y)\leq\sqrt{\log(1/y^{2})}.$

(iii) [Cramer(1951)] For positive integer $m\rightarrow\infty$ and positive
fixed number $a>0$,

$\Phi^{-1}\left(1-\dfrac{a}{m}\right)=\sqrt{2\log(m)}-\frac{\log[\log(m)]+\log(4\pi)}{\sqrt{8\log(m)}}-\frac{\log(a)}{\sqrt{2\log(m)}}+O\left(\frac{1}{\log(m)}\right).$

(iv) For $y\rightarrow 0+$,

$\Phi^{-1}(0.5+y)={\sqrt{2\pi}}y+o(y)\leq\pi y.$

######  Proof 2 ( of Lemma 2)

(i) According to that $\Phi(\cdot)$ is monotone increasing and the Mill’s
ratio inequality [Birnbaum(1942)], it can be verified that

$\Phi^{-1}\left(1-\dfrac{\phi(x)}{x}\right)\leq
x\leq\Phi^{-1}\left(1-\dfrac{\phi(x)}{x}\times\dfrac{x^{2}}{1+x^{2}}\right).$

(ii) When $x\rightarrow\infty$ and
$y=\dfrac{\phi(x)}{x}\times\dfrac{x^{2}}{1+x^{2}}$, $x\leq\Phi^{-1}(1-y)$ and

$\log(y)=-\log(\sqrt{2\pi})-x^{2}/2-\log(x)+\log(\dfrac{x^{2}}{1+x^{2}})\geq-x^{2}/2-x.$

Based on some algebras, we deduce that $x>\sqrt{\log(1/y^{2})+1}-1,$ which
means

$\sqrt{\log(1/y^{2})+1}-1<\Phi^{-1}(1-y).$

Similarly, when $x\rightarrow\infty$ and $y=\dfrac{\phi(x)}{x}$, we have
$\Phi^{-1}(1-y)\leq x$ and

$\log(y)=-\log(\sqrt{2\pi})-x^{2}/2-\log(x)\leq-x^{2}/2.$

We can conclude $x<\sqrt{\log(1/y^{2})}$ and
$\Phi^{-1}(1-y)<\sqrt{\log(1/y^{2})}$ in the same way.

(iv) According to Taylor expansion, we have

$\Phi^{-1}(0.5+y)=\Phi^{-1}(0.5)+\frac{1}{\sqrt{2\pi}}y+o(y).$

As $y\rightarrow 0+$, we can get $\Phi^{-1}(0.5+y)\leq\pi y$ directly.

###### Proposition 1

[Berman(1962)] If $X$ and $Y$ have a bivariate normal distribution with
expectation $0$, variance $1$ and correlation coefficient $\rho$, then

$\lim\limits_{c\rightarrow\infty}\frac{P(X>c,Y>c)}{[2\pi(1-\rho)^{1/2}c^{2}]^{-1}\exp(-\frac{c^{2}}{1+\rho})(1+\rho)^{3/2}}=1,$

uniformly for all $\rho$ such that $\lvert\rho\rvert\leq\delta$, for any
$\delta$, $0<\delta<1$.

######  Proof 3 ( of Theorem 1)

Without loss of generality, we assume $\omega_{i}>0$ for any $1\leq i\leq m$.
Let

$A_{i,t}=\\{\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/\omega_{i},\text{CCT}>t\\}$

and

$B_{i,t}=\\{\tan[(0.5-p_{i})\pi]\leq(1+\delta_{t})t/\omega_{i},\text{CCT}>t\\},$

where $\delta_{t}$ satisfies that $\delta_{t}>0,\delta_{t}\rightarrow 0$ and
$\delta_{t}t\rightarrow+\infty$ as $t\rightarrow+\infty$. Further, denote

$A_{t}=\bigcup_{i=1}^{m}A_{i,t},B_{t}=\bigcap_{i=1}^{m}B_{i,t},$

then we can decompose the set $\\{\text{CCT}>t\\}$ into the two disjoint sets
$A_{t}$ and $B_{t}$,

$P(\text{CCT}>t)=P(A_{t})+P(B_{t}).$

Firstly, we will show that $P(B_{t})=o(1/t)$. The event $\\{\text{CCT}>t\\}$
implies that there exists at least one $i$ such that
$\tan[(0.5-p_{i})\pi]>t/(\omega_{i}m)$. Then we have

$\displaystyle P(B_{t})$
$\displaystyle\leq\sum\limits_{i=1}^{m}P\big{(}B_{i,t}\cap\\{\tan[(0.5-p_{i})\pi]>t/(\omega_{i}m)\\}\big{)}$
$\displaystyle=\sum\limits_{i=1}^{m}P\left(\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1+\delta_{t})t}{\omega_{i}},\text{CCT}>t\right)$
$\displaystyle\leq\sum\limits_{i=1}^{m}P\left(\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1-\delta_{t})t}{\omega_{i}},\text{CCT}>t\right)$
$\displaystyle+\sum\limits_{i=1}^{m}P\left(\frac{(1-\delta_{t})t}{\omega_{i}}<\tan[(0.5-p_{i})\pi]\leq\frac{(1+\delta_{t})t}{\omega_{i}}\right)$
$\displaystyle\widehat{=}I_{1}+I_{2}.$

It is easy to see the $I_{2}=\frac{\omega_{i}}{(1-\delta_{t})\pi
t}-\frac{\omega_{i}}{(1+\delta_{t})\pi t}+o(1/t)=o(1/t)$ by noting that
$\delta_{t}\rightarrow 0$ and using (ii) of Lemma 1. As to $I_{1}$, we can
obtain that

$\displaystyle I_{1}$
$\displaystyle\leq\sum_{i=1}^{m}P\Bigg{(}\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1-\delta_{t})t}{\omega_{i}},\sum_{j\neq
i}^{m}\omega_{j}\tan[(0.5-p_{j})\pi]>\delta_{t}t\Bigg{)}$
$\displaystyle\leq\sum_{i=1}^{m}\sum_{j\neq{i}}^{m}P\Bigg{\\{}\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1-\delta_{t})t}{\omega_{i}},\tan[(0.5-p_{j})\pi]>\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{\\}}.$

It remains to show for any $1\leq{i}\neq{j}\leq{m}$, $I_{1,ij}=o(1/t).$

$\displaystyle I_{1,ij}$
$\displaystyle=P\Bigg{(}\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1-\delta_{t})t}{\omega_{i}},\tan[(0.5-p_{j})\pi]>\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}\arctan\left(\frac{\omega_{i}}{(1-\delta_{t})t}\right)\leq
p_{i}<\frac{1}{\pi}\arctan\left(\frac{\omega_{i}m}{t}\right),0<p_{j}<\frac{1}{\pi}\arctan\left(\frac{(m-1)\omega_{j}}{\delta_{t}t}\right)\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m-1}{\delta_{t}t}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Bigg{)}$
$\displaystyle=o(1/t).$

Next, we will verify that $P(A_{t})=1/(t\pi)+o(1/t)$. By Bonferroni inequality
[Dohmen(2003)],

$\sum_{i=1}^{m}P(A_{i,t})-\sum_{1\leq{i}<j\leq{m}}P(A_{i,t}\cap A_{j,t})\leq
P(A_{t})\leq\sum_{i=1}^{m}P(A_{i,t}).$

Similarly, $P(A_{i,t}\cap A_{j,t})=o(1/t)$ for any $1\leq{i}<j\leq{m}$.
Furthermore,

$\displaystyle P(A_{i,t})$
$\displaystyle=P\big{(}\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/\omega_{i}\big{)}-P\big{(}\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/\omega_{i},\text{CCT}\leq{t}\big{)}$
$\displaystyle\widehat{=}J_{1,i}-J_{2,i}.$

$J_{1,i}=\frac{\omega_{i}}{\pi(1+\delta_{t})t}+o(\frac{1}{(1+\delta_{t})t})=\frac{\omega_{i}}{\pi
t}+o(1/t),$ and
$\begin{array}[]{cl}J_{2,i}&\leq
P\bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{\omega_{i}},\sum_{j\neq
i}^{m}\omega_{j}\tan[(0.5-p_{j})\pi]\leq-\delta_{t}t\bigg{)}\\\
&\leq\sum_{j\neq{i}}^{m}P\Bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{m\omega_{i}},\tan[(0.5-p_{j})\pi]\leq-\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}\\\
&=\sum_{j\neq{i}}^{m}J_{2,ij}.\end{array}$
So it remains to prove that for any $1\leq{i}\neq{j}\leq{m}$,
$J_{2,ij}=o(1/t).$

$\displaystyle J_{2,ij}$
$\displaystyle=P\Bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{m\omega_{i}},\tan[(0.5-p_{j})\pi]\leq-\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{1}{\pi}\arctan\left(\frac{m\omega_{i}}{(1+\delta_{t})t}\right),1-\frac{1}{\pi}\arctan\left(\frac{(m-1)\omega_{j}}{t}\right)<p_{j}<1\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{m}{\pi}\frac{\omega_{i}}{(1+\delta_{t})t},1-\frac{1}{\pi}\frac{(m-1)\omega_{j}}{\delta_{t}t}<p_{j}<1\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{m}{\pi}\frac{\omega_{i}}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m-1}{\delta_{t}t}<p_{j}<1\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Bigg{)}$
$\displaystyle=o(1/t).$

######  Proof 4 ( of Theorem 2)

Similar to the proof of the Theorem 1 and the notations of the Theorem 1, we
will also show that $P(B_{t})=o(1/t)$ at first.

$\displaystyle P(B_{t})$ $\displaystyle\leq I_{1}+I_{2}.$

It is easy to see $I_{2}=\frac{\omega_{i}}{(1-\delta_{t})\pi
t}-\frac{\omega_{i}}{(1+\delta_{t})\pi t}+o(1/t)=o(1/t)$ by noting that
$\delta_{t}\rightarrow 0$ and using (ii) of Lemma 2. As for $I_{1}$, we can
obtain that

$\displaystyle I_{1}$
$\displaystyle\leq\sum_{i=1}^{m}\sum_{j\neq{i}}^{m}I_{1,ij}.$

It remains to show for any $1\leq{i}\neq{j}\leq{m}$,
$I_{1,ij}=o(1/{t^{1+\gamma}}).$

$\displaystyle I_{1,ij}$
$\displaystyle=P\Bigg{(}\frac{t}{\omega_{i}m}<\tan[(0.5-p_{i})\pi]\leq\frac{(1-\delta_{t})t}{\omega_{i}},\tan[(0.5-p_{j})\pi]>\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}\arctan\left(\frac{\omega_{i}}{(1-\delta_{t})t}\right)\leq
p_{i}<\frac{1}{\pi}\arctan\left(\frac{\omega_{i}m}{t}\right),0<p_{j}<\frac{1}{\pi}\arctan\left(\frac{(m-1)\omega_{j}}{\delta_{t}t}\right)\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m-1}{\delta_{t}t}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Bigg{)}$
$\displaystyle=o(1/t^{1+\gamma}).$

Next, we will verify that $P(A_{t})=1/(t\pi)+o(1/t)$. By Bonferroni inequality
[Dohmen(2003)],

$\sum_{i=1}^{m}P(A_{i,t})-\sum_{1\leq{i}<j\leq{m}}P(A_{i,t}\cap A_{j,t})\leq
P(A_{t})\leq\sum_{i=1}^{m}P(A_{i,t}).$

In this situation, we have to prove that
$\sum\limits_{1\leq{i}<j\leq{m}}P(A_{i,t}\cap A_{j,t})=o(1/t)$.

$\displaystyle P(A_{i,t}\cap A_{j,t})$
$\displaystyle<P\Big{(}\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/(m\omega_{i}),\tan[(0.5-p_{j})\pi]>(1+\delta_{t})t/\omega_{j}\Big{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{1}{\pi}\arctan\left(\frac{m\omega_{i}}{(1+\delta_{t})t}\right),0<p_{j}<\frac{1}{\pi}\arctan\left(\frac{m\omega_{j}}{(1+\delta_{t})t}\right)\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Bigg{)}$
$\displaystyle=o(1/t^{1+\gamma}).$

On the other hand,

$\displaystyle P(A_{i,t})$
$\displaystyle=P\big{(}\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/\omega_{i}\big{)}-P\big{(}\tan[(0.5-p_{i})\pi]>(1+\delta_{t})t/\omega_{i},\text{CCT}\leq{t}\big{)}$
$\displaystyle\widehat{=}J_{1,i}+J_{2,i}.$

$J_{1,i}=\frac{\omega_{i}}{\pi(1+\delta_{t})t}+o(\frac{\omega_{i}}{(1+\delta_{t})t})=\frac{\omega_{i}}{\pi
t}+o(\omega_{i}/t).$
$\begin{array}[]{cl}J_{2,i}&\leq
P\bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{\omega_{i}m},\sum_{j\neq
i}^{m}\omega_{j}\tan[(0.5-p_{j})\pi]\leq-\delta_{t}t\bigg{)}\\\
&\leq\sum_{j\neq{i}}^{m}P\Bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{\omega_{i}m},\tan[(0.5-p_{j})\pi]\leq-\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}\\\
&=\sum_{j\neq{i}}^{m}J_{2,ij}.\end{array}$
So it remains to prove that for any $1\leq{i}\neq{j}\leq{m}$,
$J_{2,ij}=o(1/t^{1+\gamma}).$

$\displaystyle J_{2,ij}$
$\displaystyle=P\Bigg{(}\tan[(0.5-p_{i})\pi]>\frac{(1+\delta_{t})t}{\omega_{i}m},\tan[(0.5-p_{j})\pi]\leq-\frac{\delta_{t}t}{(m-1)\omega_{j}}\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{1}{\pi}\arctan\left(\frac{\omega_{i}m}{(1+\delta_{t})t}\right),1-\frac{1}{\pi}\arctan\left(\frac{(m-1)\omega_{j}}{\delta_{t}t}\right)<p_{j}<1\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{(m-1)}{\delta_{t}t}<p_{j}<1\Bigg{)}$
$\displaystyle\leq
P\Bigg{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Bigg{)}$
$\displaystyle=o(1/t^{1+\gamma}).$

######  Proof 5 ( of (D1) and (D2) for Six Copulas)

In this part, we will prove that when the joint distribution of
$(p_{i},p_{j})^{\top}$ follows six commonly used copula functions, the
assumptions (D1) and (D2) are satisfied.
1) Product Copula: $\mathbb{C}(u_{i},v_{j})=u_{i}v_{j},1\leq i\neq j\leq m.$
We can show that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{2}\frac{1}{\delta_{t}},$
(2)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}.$
(3)

When $m$ is fixed, from (2) we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (3) we can see

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, then from (2) we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

and from (3) we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}).$

2) Farlie-Gumbel-Morgenstern (FGM) Copula:
$\mathbb{C}(u_{i},v_{j})=u_{i}v_{j}[1+\theta(1-u_{i})(1-v_{j})],\theta\in[-1,1],1\leq
i\neq j\leq m.$
We can show that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{2}\frac{1}{\delta_{t}}[1+\theta(1-\frac{\omega_{i}}{\pi}\frac{m}{t})(1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t})],$
(4)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}\left[1+\theta(1-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t})\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\right]$
$\displaystyle\leq\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}.$
(5)

When $m$ is fixed, from (4) we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (5), we can see

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, then from (4) we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

and from (5), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/{t^{1+\gamma}}).$

3) Cuadras-Augé Copula:
$\mathbb{C}(u_{i},v_{j})=\big{(}\min{(u_{i},v_{j})}\big{)}^{\theta}(u_{i}v_{j})^{1-\theta},\theta\in[0,1],,1\leq
i\neq j\leq m.$
We can show that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{1-\theta}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t},$
(6)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}^{1-\theta}$
$\displaystyle\leq\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}.$
(7)

When $m$ is fixed, from (6) we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (7), we can see

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, then from (6) we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

and from (7), then we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}=o(1/{t^{1+\gamma}}).$

4) Normal Copula:

$\mathbb{C}(u_{i},v_{j})=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int_{-\infty}^{\Phi^{-1}(u_{i})}\int_{-\infty}^{\Phi^{-1}(v_{j})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy,1\leq
i\neq j\leq m.$

We can show that,

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int_{-\infty}^{\Phi^{-1}(\frac{\omega_{i}}{\pi}\frac{m}{t})}\int_{-\infty}^{\Phi^{-1}(\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy$
$\displaystyle=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int^{\infty}_{\Phi^{-1}(1-\frac{\omega_{i}}{\pi}\frac{m}{t})}\int^{\infty}_{\Phi^{-1}(1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy.$

Denote $d\gg 0$ satisfying the following equation

$\frac{\exp\\{-d^{2}/2\\}}{d\sqrt{2\pi}}\frac{d^{2}}{1+d^{2}}=\sup_{i\in
1,\cdots,m}\frac{\omega_{i}m}{\pi\delta_{t}t}.$

By some algebras, we can obtain that $d\rightarrow\infty$ as
$t\rightarrow\infty$. Then

$d^{2}=O\Big{(}\log(\pi(\delta_{t}t)^{2}/(\omega m)^{2})\Big{)},\quad
t\rightarrow\infty,$

where $\omega=\arg\sup\limits_{i\in
1,\cdots,m}\frac{\omega_{i}m}{\pi\delta_{t}t}.$ According to (i) in Lemma 2
and Lemma 1, we can obtain that $\Phi^{-1}(1-\frac{\omega_{i}m}{\pi t})>d$,
$\Phi^{-1}(1-\frac{\omega_{j}m}{\pi\delta_{t}t})>d$ and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle\leq\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int^{\infty}_{d}\int^{\infty}_{d}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy$
$\displaystyle=O\left(\frac{1}{(\delta_{t}t)^{{}^{\frac{2}{1+\rho_{ij}}}}\log(2\pi{(\delta_{t}t)}^{2}/m^{2})}\right),$
(8)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int_{-\infty}^{\Phi^{-1}(\frac{\omega_{i}}{\pi}\frac{m}{t})}\int^{\infty}_{\Phi^{-1}(1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy$
$\displaystyle=\frac{1}{2\pi\sqrt{1-\rho_{ij}}}\int^{\infty}_{\Phi^{-1}(1-\frac{\omega_{i}}{\pi}\frac{m}{t})}\int^{\infty}_{\Phi^{-1}(1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t})}\exp\left\\{-\frac{x^{2}-2\rho_{ij}xy+y^{2}}{2(1-\rho^{2}_{ij})}\right\\}dxdy.$
(9)

When $m$ is fixed and $0\leq\rho_{ij}<\rho_{\max}<1$, from (8), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (9), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

When $m$ is divergent, $0\leq\rho_{ij}<\rho_{\max}<1$ and $m=o(t^{\gamma/2})$,
we take $\delta_{t}t=t^{\beta}$ where $(\frac{1+\rho_{max}}{2}\leq\beta<1)$,
namely, $\delta_{t}=\frac{1}{t^{1-\beta}}$,
$\gamma=\frac{2}{1+\rho_{\max}}\beta-1(0<\gamma\leq 1)$, from (8) then we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

and from (9), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}).$

5) Ali-Mikhail-Haq (AMH) Copula: (i) When $\theta\in[0,1]$, we have
$\mathbb{C}(u_{i},v_{j})=\frac{u_{i}v_{j}}{1-\theta(1-u_{i})(1-v_{j})},\\\
1\leq i\neq j\leq m.$
We can show that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle\leq\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{2}\frac{1}{\delta_{t}(1-\theta)},$
(10)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle\leq\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}.$
(11)

When $m$ is fixed, from (10), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (11), we can have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, from (10), we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

and from (11), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}).$

(ii) When $\theta\in[-1,0]$, we have

$\mathbb{C}(u_{i},v_{j})=\frac{u_{i}v_{j}}{1-\theta(1-u_{i})(1-v_{j})},1\leq
i\neq j\leq m.$

We can know that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}\leq\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{2}\frac{1}{\delta_{t}},$
(12)

and

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle\leq\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{2\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}-\frac{\omega_{i}}{2\pi}\frac{m}{(1+\delta_{t})t}.$
(13)

When $m$ is fixed, from (12), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}),$

and from (13), we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, from (12), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

from (13), we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}).$

6) Survival Copula: $\mathbb{C}(u_{i},v_{j})={u_{i}v_{j}}\exp{(-\theta\ln
u_{i}\ln v_{j})},\theta\in[0,1],1\leq i\neq j\leq m.$
We can show that

$\displaystyle
P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t},\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}\leq\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{t}\Big{)}^{2}\frac{1}{\delta_{t}},$
(14)

and

$\displaystyle P\Big{(}$ $\displaystyle
0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}<p_{j}<1\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\mathbb{C}\Big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t},1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}\exp{\Big{(}-\theta\ln\big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\big{)}\ln\big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\big{)}\Big{)}}$
$\displaystyle=\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}-\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\Big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}\Big{(}1+\theta\ln\big{(}\frac{\omega_{i}}{\pi}\frac{m}{(1+\delta_{t})t}\big{)}\ln\big{(}1-\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\big{)}\Big{)}$
$\displaystyle=\frac{\omega_{i}\omega_{j}m^{2}}{\pi^{2}(1+\delta_{t})\delta_{t}t^{2}}\Big{(}1+\theta\ln\big{(}\frac{\omega_{i}m}{\pi(1+\delta_{t})t}\big{)}\ln\big{(}1-\frac{\omega_{j}m}{\pi\delta_{t}t}\big{)}\Big{)}-\frac{\omega_{i}m\theta}{\pi(1+\delta_{t})t}\ln\big{(}\frac{\omega_{i}m}{\pi(1+\delta_{t})t}\big{)}\ln\big{(}1-\frac{\omega_{j}m}{\pi\delta_{t}t}\big{)}$
(15)

When $m$ is fixed, from (14), we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

and from (15), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t}).$

When $m$ is divergent and $m=o(t^{\gamma/2})$, let $\delta_{t}=t^{\gamma-1}$,
where $\gamma\in(0,1]$, from (14), we have

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}),$

from (15), we can get

$P\Big{(}0<p_{i}<\frac{\omega_{i}}{\pi}\frac{m}{t},0<p_{j}<\frac{\omega_{j}}{\pi}\frac{m}{\delta_{t}t}\Big{)}=o(1/{t^{1+\gamma}}).$

###### Proof 6

Corollary 1 Let $x$ be an arbitrary positive real number and

$c_{m}(x)=[2-(\log{m})^{-1}]x+\big{(}2\log{m}-[\log{(\log{m})}+\log(4\pi)-\log
4]+\frac{\log{(\log{m})}+\log(4\pi)-\log 4}{2\log{m}}\big{)}.$

Let $Y_{1},\cdots,Y_{p}$ iid $\sim N(0,1)$. Under (D3), we have to prove that

$\left|P\left([\max\limits_{s=1,\cdots,m}(|Z_{s}|)]^{2}<c_{m}(x)\right)-P\left([\max\limits_{s=1,\cdots,m}(|Y_{s}|)]^{2}<c_{m}(x)\right)\right|\rightarrow
0,~{}~{}as~{}~{}m\rightarrow\infty.$

We can get

$\displaystyle\left|P\left([\max\limits_{s=1,\cdots,m}(|Z_{s}|)]^{2}<c_{m}(x)\right)-P\left([\max\limits_{s=1,\cdots,m}(|Y_{s}|)]^{2}<c_{m}(x)\right)\right|$
$\displaystyle=\left|P\left(\max\limits_{s=1,\cdots,m}(|Z_{s}|)<\sqrt{c_{m}(x)}\right)-P\left(\max\limits_{s=1,\cdots,m}(|Y_{s}|)<\sqrt{c_{m}(x)}\right)\right|$
$\displaystyle=\left|P\Big{(}-\sqrt{c_{m}(x)}<|Z_{s}|<\sqrt{c_{m}(x)}~{};1\leq
s\leq m\Big{)}-P\Big{(}-\sqrt{c_{m}(x)}<Y_{s}<\sqrt{c_{m}(x)}~{};1\leq s\leq
m\Big{)}\right|.$

Let
$\phi(x,y,r)=\frac{1}{2\pi\sqrt{1-r}}\exp\left\\{-\frac{x^{2}-2rxy+y^{2}}{2(1-r^{2})}\right\\}$.
According to the inequality (4.5) in Berman [Berman(1971)], we have

$\displaystyle\left|P\left([\max\limits_{s=1,\cdots,m}(|Z_{s}|)]^{2}<c_{m}(x)\right)-P\left([\max\limits_{s=1,\cdots,m}(|Y_{s}|)]^{2}<c_{m}(x)\right)\right|$
$\displaystyle\geq 2\sum_{1\leq q<s\leq
m}|\rho_{j,k}|\phi(\sqrt{c_{m}(x)}~{},\sqrt{c_{m}(x)}~{};|\rho_{q,s}|)+2\sum_{1\leq
q<s\leq
m}|\rho_{j,k}|\phi(\sqrt{c_{m}(x)}~{},-\sqrt{c_{m}(x)}~{};|\rho_{q,s}|).$

Let $c_{A}$ be a generic finite positive constant. Note that $c_{m}(x)=2\log
m-\log(\log m)+o_{p}(1)$, we have

$\displaystyle 2\sum_{1\leq q<s\leq
m}|\rho_{q,s}|\phi(\sqrt{c_{m}(x)}~{},\sqrt{c_{m}(x)}~{};|\rho_{q,s}|)$
$\displaystyle\leq c_{A}\sum_{1\leq q<s\leq
m}|\rho_{q,s}|e^{\frac{-c_{m}(x)}{1+|\rho_{q,s}|}}$ $\displaystyle\leq
c_{A}~{}\log m\sum_{1\leq q<s\leq
m}|\rho_{q,s}|m^{-\frac{2}{1+|\rho_{q,s}|}}.$

Using the technique in Berman [Berman(1964)], the left expression of the above
inequation is

$\displaystyle 2\sum_{1\leq q<s\leq m}|\rho_{q,s}|$
$\displaystyle\phi(\sqrt{c_{m}(x)}~{},-\sqrt{c_{m}(x)}~{};|\rho_{q,s}|)\leq
c_{A}\sum_{1\leq q<s\leq m}|\rho_{q,s}|e^{\frac{-c_{m}(x)}{1-|\rho_{q,s}|}}$
$\displaystyle\leq c_{A}\sum_{1\leq q<s\leq m}|\rho_{q,s}|e^{-c_{m}(x)}\leq
c_{A}(\log m)m^{-2}\sum_{i=1}^{m-1}(m-i)\delta_{i},$

and the right one goes to 0 as $m\rightarrow\infty$. Under (D3),
$\delta_{m}(\log m)^{2+\beta}\rightarrow 0$, as $m\rightarrow\infty$, we have

$c_{A}(\log m)m^{-2}\sum_{i=1}^{m-1}[(m-i)\delta_{i}]\leq c_{A}(\log
m)m^{-1}\sum_{i=1}^{m-1}\delta_{i}\rightarrow 0.$

###### Proof 7

Theorem 3 Assume that $\min\limits_{i=1,\cdots,m}\omega_{i}=\frac{c_{0}}{m}$,
where the constant $c_{0}>0$ and according to the definition of CCT, we can
figure out the conclusion below

CCT
$\displaystyle=\sum\limits_{j\in\Omega}\omega_{j}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(|Z_{j}|)\big{)}\Big{)}\pi\Big{]}+\sum\limits_{j\in\Omega^{c}}\omega_{j}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(|Z_{j}|)\big{)}\Big{)}\pi\Big{]}$
$\displaystyle\geq\frac{c_{0}}{m}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\max\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}+\frac{m^{\eta}-c_{0}}{m}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\min\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}$
$\displaystyle+\sum\limits_{j\in\Omega^{c}}\omega_{j}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(|Z_{j}|)\big{)}\Big{)}\pi\Big{]}.$

According to Theorem 1, we can deduce that

$\sum\limits_{j\in\Omega^{c}}\omega_{j}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(|Z_{j}|)\big{)}\Big{)}\pi\Big{]}$

approximates a standard Cauchy distribution as $m$ goes to infinity, which
indicates that

$\sum\limits_{j\in\Omega^{c}}\omega_{j}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(|Z_{j}|)\big{)}\Big{)}\pi\Big{]}=O_{p}(1).$

Based on Theorem 3 of Liu and Xie (2020), we can get

$\frac{m^{\eta}-c_{0}}{m}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\min\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}\geq
o_{p}(1).$

As $m$ goes to infinity, we have

$\text{CCT}\geq\frac{c_{0}}{m}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\max\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}+O_{p}(1).$

The power of CCT is

$\displaystyle\beta_{\text{CCT}}$
$\displaystyle=P\left(\text{CCT}>C_{\alpha}\right)$ $\displaystyle\geq
P\left(\frac{c_{0}}{m}\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\max\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}+O_{p}(1)>C_{\alpha}\right)$
$\displaystyle=P\left(\tan\Big{[}\Big{(}0.5-2\big{(}1-\Phi(\max\limits_{j\in\Omega}|Z_{j}|)\big{)}\Big{)}\pi\Big{]}>\big{(}C_{\alpha}-O_{p}(1)\big{)}m/c_{0}\right)$
$\displaystyle=P\left(\max\limits_{s\in\Omega}|Z_{j}|>\Phi^{-1}\Big{[}1-\frac{1}{2\pi}\arctan\frac{c_{0}}{\big{(}C_{\alpha}-O_{p}(1)\big{)}m}\Big{]}\right)$
$\displaystyle=P\left(\max\limits_{s\in\Omega}|Z_{j}|>(2\log
m)^{1/2}-\frac{\log\log m+\log(4\pi)}{(8\log
m)^{1/2}}+\frac{\log[2\pi\big{(}C_{\alpha}-O_{p}(1)\big{)}]}{(2\log
m)^{1/2}}+O(\frac{1}{\log m})\right),$

where the last equation is based on (iii) of Lemma 2 . The power of MAX is

$\displaystyle\beta_{\text{MAX}}$
$\displaystyle=P\left(\frac{\sqrt{\text{MAX}}-a_{m}-O\left(\frac{1}{\log
m}\right)}{b_{m}}>G_{\alpha}\right)$
$\displaystyle=P\left(\sqrt{\text{MAX}}>G_{\alpha}b_{m}+a_{m}\right)$
$\displaystyle=P\left(\max\limits_{j=1,\cdots,m}|Z_{j}|>G_{\alpha}b_{m}+a_{m}+O\Big{(}\frac{1}{\log
m}\Big{)}\right)$
$\displaystyle=P\left(\max\\{\max\limits_{j\in\Omega}|Z_{j}|,\max\limits_{j\in\Omega^{c}}|Z_{j}|\\}>G_{\alpha}b_{m}+a_{m}+O\Big{(}\frac{1}{\log
m}\Big{)}\right)$ $\displaystyle\leq
P\left(\max\limits_{j\in\Omega}|Z_{j}|>G_{\alpha}b_{m}+a_{m}+O\Big{(}\frac{1}{\log
m}\Big{)}\right)+P\left(\max\limits_{j\in\Omega^{c}}|Z_{j}|>G_{\alpha}b_{m}+a_{m}+O\Big{(}\frac{1}{\log
m}\Big{)}\right).$

Denote $m^{\star}=m^{\kappa}$, $b_{m^{\star}}=(2\log m^{\star})^{-1/2}$ and
$a_{m^{\star}}=(2\log m^{\star})^{1/2}-\frac{\log\log m^{\star}+4\pi-4}{(8\log
m^{\star})^{1/2}}$, then

$\displaystyle
P\left(\max\limits_{j\in\Omega^{c}}|Z_{j}|>G_{\alpha}b_{m}+a_{m}+O\left(\frac{1}{\log
m}\right)\right)$
$\displaystyle=P\left(\frac{\max\limits_{j\in\Omega^{c}}|Z_{j}|-a_{m^{\star}}}{b_{m^{\star}}}>\frac{G_{\alpha}b_{m}+a_{m}+O\left(\frac{1}{\log
m}\right)-a_{m^{\star}}}{b_{m^{\star}}}\right).$

According to Lemma 1, we have

$\displaystyle\frac{a_{m}-a_{m^{\star}}}{b_{m^{\star}}}$ $\displaystyle\geq
2[\sqrt{\log(m)\log(m^{\star})}-\log(m^{\star})].$

Due to $1>\kappa>0$, as both $m^{\star}$ and $m$ go to $\infty$, we have

$\displaystyle\frac{a_{m}-a_{m^{\star}}}{b_{m^{\star}}}$ $\displaystyle\geq
2\left[\sqrt{{\frac{1}{\kappa}}\log\left({m^{\star}}\right)\log(m^{\star})}-\log(m^{\star})\right]=2\left[\big{(}\sqrt{{\frac{1}{\kappa}}-1}\big{)}\log(m^{\star})\right]\rightarrow\infty$

and

$\displaystyle\frac{G_{\alpha}b_{m}}{b_{m^{\star}}}={{(\log
m^{\star})^{1/2}}\over{(\log m)^{1/2}}}\rightarrow 0,$

which means that

$\displaystyle
P\left(\max\limits_{j\in\Omega^{c}}|Z_{j}|>G_{\alpha}b_{m}+a_{m}+O\Big{(}\frac{1}{\log
m}\Big{)}\right)$ $\displaystyle\rightarrow
1-\exp\big{\\{}\exp\\{-\infty\\}\big{\\}}\text{, as }m^{\star}\text{ and
}m\rightarrow\infty$ $\displaystyle=o(1).$

Therefore, we have

$\beta_{\text{MAX}}\leq P\left(\max\limits_{j\in\Omega}|Z_{j}|>(2\log
m)^{1/2}-\frac{\log\log m+\log(4\pi)-\log(4)}{(8\log
m)^{1/2}}+\frac{G_{\alpha}}{(2\log m)^{1/2}}+O\Big{(}\frac{1}{\log
m}\Big{)}\right)+o(1).$

According to above inequalities, as $m$ goes infinity, we can draw the
conclusion

$\beta_{\text{MAX}}\leq\beta_{\text{CCT}}+o(1).$

## Acknowledgement

This work was partially supported by Beijing Natural Science Foundation
(Z180006).

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Figure 1: The tail distributions of CCTs (red lines) and standard Cauchy
distribution (black lines), where the test statistics $Z_{i},i=1,\cdots,m$ are
generated from $N\big{(}{\bf 0}_{m},(\rho^{\text{Sgn}(|i-j|)})_{m\times
m}\big{)}$, $i$ and $j=1,\cdots,m$, $\rho=0.2$, $0.5$, $0.8$.
$\text{Sgn}(\cdot)$ is a sign function. Figure 2: The tail distributions of
CCTs (red lines) and standard Cauchy distribution (black lines), where the
test statistics $Z_{i},i=1,\cdots,m$ are generated from the $N\big{(}{\bf
0}_{m},{\boldsymbol{R}}\big{)}$, where ${\boldsymbol{R}}$ belongs to spiked
model 2 whose parameter space are $m=10,50$, and $500$, and $d=4,5,$ and $6$.
Figure 3: The tail distributions of CCTs (red lines) and standard Cauchy
distribution (black lines), where the individual $p$-values
$p_{i},i=1,\cdots,m$ are generated from the mixed FGM copula model, whose
parameter setting is $\theta=0.2$, $0.5$ and $0.8$. Figure 4: The tail
distributions of CCTs (red lines) and standard Cauchy distribution (black
lines), where the individual $p$-values $p_{i},i=1,\cdots,m$ are generated
from the mixed AHM copula model, whose parameter setting is $\theta=0.2$,
$0.5$ and $0.8$. Figure 5: The powers of CCTs and MAXs, where the test
statistics $Z_{i},i=1,\cdots,m$ are generated from $N\big{(}{\bf
0}_{m},(\rho^{|i-j|})_{m\times m}\big{)}$, $i$ and $j=1,\cdots,m$. Let the
number of nonzero signals be $\|\Omega\|=0.1m,0.2m,0.3m$ from first row to
bottom row. The circular red spots represent the power of CCTs and the
triangle blue spots represent the power of MAX.  Figure 6: The powers of CCTs
and MAXs, where the test statistics $Z_{i},i=1,\cdots,m$ are generated from
$N\big{(}{\bf 0}_{m},(\rho_{ij})_{m\times m}\big{)}$, where
$\rho_{ij}=1/{(1+|i-j|^{a})}$ for $1\leq i\neq j\leq m$. Let the number of
nonzero signals be $\|\Omega\|=0.1m,0.2m,0.3m$ from first row to bottom row.
The circular red spots represent the power of CCTs and the triangle blue spots
represent the power of MAX.


# Which numbers are not the sum plus the product of three positive integers?

Brian Conrey & Neil Shah  American Institute of Mathematics, San Jose, CA
95112, USA<EMAIL_ADDRESS>Morgan Hill, CA 95037, USA<EMAIL_ADDRESS>
###### Abstract.

We investigate the number $R_{3}(n)$ of representations of $n$ as the sum plus
the product of three positive integers. On average, $R_{3}(n)$ is
$\frac{1}{2}\log^{2}n$. We give an upper bound for $R_{3}(n)$ and an upper
bound for the number of $n\leq N$ such that $R_{3}(n)=0$. We conjecture that
$R_{3}(n)=0$ infinitely often.

This work is partially supported by an NSF grant.

##  1\. Introduction

It is well-known that the number of representations of an integer as a product
of $k$ positive integers, denoted $d_{k}(n)$, satisfies
$d_{k}(n)\ll_{\epsilon}n^{\epsilon}$ for any $\epsilon>0$ as $n$ goes to
infinity (see Section 1.1 for notation). The proof of this standard result
relies crucially on the fact that $d_{k}$ is a multiplicative function. For
$k=3$ this amounts to counting integral solutions of

$n=xyz.$

The slightest perturbation of this problem turns it from an exercise into a
seriously hard research question. One such example is the problem of trying to
estimate the number of representations of a number $n$ as the sum plus the
product of three positive integers, i.e.

(1) $n=xyz+x+y+z.$

Let $R_{3}(n)$ denote the number of positive integral solutions to (1). Using
this definition, our motivating problem was the following conjecture:

######  Conjecture 1.

For any $\epsilon>0$ there is a $C(\epsilon)>0$ such that

$R_{3}(n)\leq C(\epsilon)n^{\epsilon}$

for all positive integers $n$.

While we were unable to prove this conjecture, we were able to find and prove
a number of other interesting results as well as formulate some new
conjectures.

###  1.1. Notation

We use some notation that is standard in analytic number theory but perhaps
not standard to everybody. In this section, we define this notation for the
reader.

Suppose $f(n)$ and $g(n)$ are functions defined on the positive integers.
Then, $f(n)=O(g(n))$ means that there exists some $M>0$ and $N_{0}$ (which
could depend on $M$) such that $|f(n)|\leq Mg(n)$ for all $n\geq N_{0}$. Also,
$f(n)=\Omega(g(n))$ means that there exists some $M>0$ and a sequence
$n_{1}<n_{2}<\ldots$ such that $f(n_{j})\geq Mg(n_{j})$ for all $j$. Lastly,
$f(n)\ll g(n)$ is equivalent to $f(n)=O(g(n))$ and $f(n)\gg g(n)$ is
equivalent to $g(n)\ll f(n)$. Note that $\ll_{a}$, $\gg_{a}$, and $\Omega_{a}$
indicate that the implicit constant $M$ is dependent on $a$.

##  2\. Statement of Results

We can prove Conjecture 1 on average.

######  Theorem 1.

As $N$ goes to infinity,

$\frac{1}{N}\sum_{n\leq N}R_{3}(n)\sim\frac{1}{2}\log^{2}N.$

For an upper bound, we can show the following.

######  Theorem 2.

For any $\epsilon>0$,

$R_{3}(n)\ll_{\epsilon}n^{1/3+\epsilon}.$

If the following conjecture111For the purposes of this conjecture, we
interpret $\tau_{k}(0)$ to be $0$. is true, we can replace the right side with
$n^{\frac{1}{3}}\log n$.

######  Conjecture 2.

Let $f(x,y)$ be a fixed polynomial in $x$ and $y$ with integer coefficients.
Then for any $\epsilon>0$,

$\sum_{N-M<n\leq N}\tau_{k}(f(N,n))\ll M\log N^{k-1}$

as $N\to\infty$ provided that $N^{\epsilon}<M<N^{1-\epsilon}$.

######  Remark 1.

This conjecture is a combination of the results of [E] and [S]; it may be
well-known to experts but we could not find a reference.

In the course of proving an upper bound for $R_{3}(n)$, we discovered some
interesting properties of $R_{3}(n)$ which we state here.

######  Proposition 1.

If $R_{3}(n)=0$, then $n$ is prime.

###### Proof.

Let

$f_{3}(x,y,z)=xyz+x+y+z.$

Observe that $f_{3}(x,y,1)=(x+1)(y+1)$. Thus, if $n$ is composite, there
exists a triple $(x,y,1)$ such that $f_{3}(x,y,1)=n$. ∎

This result led us to wonder how many primes $p\leq x$ there are for which
$R_{3}(p)=0$. Define

$U_{3}(N)=\\#\\{n\leq N:R_{3}(n)=0\\}.$

Using the large sieve we can show the following.

######  Theorem 3.

We have

$U_{3}(N)\ll\frac{N}{e^{\sqrt{\log N}}}.$

We expect that this estimate is far from optimal. We did calculations for $N$
up to the 250 millionth prime ($5336500537$) and found 2014 prime numbers
which are not the sum plus the product of three positive integers.

Based on numerical evidence, we believe that $U_{3}(N)\to\infty$ as
$N\to\infty$ but very slowly, maybe like a power of $\log N$.

######  Conjecture 3.

There exist positive constants $A<B$ such that

$\log^{A}N\ll U_{3}(N)\ll\log^{B}N.$

It is interesting to compare this problem with the analogous problem in four
variables. Let $R_{4}(n)$ denote the number of positive integral solutions to

$n=xyzw+x+y+z+w$

and let

$U_{4}(N)=\\#\\{n\leq N:R_{4}(n)=0\\}.$

######  Conjecture 4.

For any $\epsilon>0$,

$R_{4}(n)\ll_{\epsilon}n^{\epsilon}.$

Again, we can prove this conjecture on average.

######  Theorem 4.

As $N$ goes to infinity,

$\frac{1}{N}\sum_{n\leq x}R_{4}(n)\sim\frac{1}{6}\log^{3}N.$

We can also obtain an upper bound of $R_{4}(n)$.

######  Theorem 5.

For any $\epsilon>0$,

$R_{4}(n)\ll_{\epsilon}n^{\frac{1}{2}+\epsilon}.$

If an analog of Conjecture 2 for a polynomial of $3$ variables is true, we can
replace the right side with $n^{\frac{1}{2}}\log^{2}n$.

By the large sieve inequality we can show the following.

######  Theorem 6.

As $N$ goes to infinity,

$U_{4}(N)\ll N^{1/2}.$

Unlike the situation for $R_{3}(n)$, we believe that only a finite number of
$n$ have $R_{4}(n)=0$.

######  Conjecture 5.

If $n>45752$, then

$R_{4}(n)>0.$

###  2.1. Omega results

If $n\geq 5$, by consideration of solutions where at least one of $x,y,z$ is
equal to $1$, it is easy to see that

$R_{3}(n)\geq 6d(n)-6$

where $d(n)=d_{2}(n)$ is the number of divisors of $n$. Using what we know
about how large $d(n)$ can get, it follows that for every $\epsilon>0$,

(2) $R_{3}(n)=\Omega_{\epsilon}\left(\exp(\frac{\log(2-\epsilon)\log
n}{\log\log n})\right).$

One could also consider a solution where at least one of $x,y,z$ is equal to
$2$, in which case we can see

$R_{3}(n)\geq 6d(n)+6d(2n-3)+O(1).$

It is not clear to us how big $d(n)+d(2n-3)$ can be. One could do similarly
for $3,4,\ldots$ and obtain apparently stronger omega results.

We wonder whether

$R_{3}(n)=\Omega_{\epsilon}\left(\exp(\frac{\log(3-\epsilon)\log n}{\log\log
n})\right),$

which would be the case if $R_{3}(n)$ behaves like $d_{3}(n)$.

###  2.2. Algorithm for calculating $R_{3}(n)$

In order to find all the $n\leq 5\times 10^{9}$ with $R_{3}(n)=0$, we need a
reasonable algorithm for determining when $R_{3}(n)>0$. Suppose that
$xyz+x+y+z=n$ and that $x\leq y\leq z$. Multiplying through by $x$ and adding
$1$ to both sides yields

(3) $(xy+1)(xz+1)=nx-x^{2}+1.$

Therefore, if we can find the proper divisors $d$ of $nx-x^{2}+1$ such that
$d\equiv 1\pmod{x}$, then from this set of divisors we can easily determine
the complete set of solutions of $f_{3}(x,y,z)=n$. If there are no proper
divisors $d\equiv 1\bmod x$, then $R_{3}(n)=0$. The time complexity for prime
factorization of a number $m$ is $O(\exp((\log m)^{\frac{1}{3}+\epsilon}))$.
We apply this to $m=nx-x^{2}+1$, for all $x\leq n^{\frac{1}{3}}$ and so for
each $n$, there are $n^{\frac{1}{3}+\epsilon}$ steps to determine the value of
$R_{3}(n)$. Since the number of divisors of $nx-x^{2}+1$ is
$O((nx-x^{2}+1)^{\epsilon})=O(n^{\epsilon})$, this argument also proves
Theorem 2.

For the primes $p$ up to $5336500537$, we checked using this method whether or
not $R_{3}(p)=0$. We computed all $2014$ primes up to $5336500537$ such that
$R_{3}(p)=0$:

$\displaystyle 2,3,5,7,11,13,17,23,31,37,41,43,53,67,71,83,97,101,107,113,$
$\displaystyle\qquad\dots,5178563387,5220047297,5284333573,5322410117$

See this link for the full list.

###  2.3. Algorithm for calculating $R_{4}(n)$

Let

$f_{4}(x,y,z,w)=xyzw+x+y+z+w.$

Suppose $f_{4}(x,y,z,w)=n$ and $x\leq y\leq z\leq w$. Multiplying through by
$xy$ and rearranging yields

$(xyz+1)(xyw+1)=nxy+1-x^{2}y-xy^{2}.$

So, to find $R_{4}(n)$, it is a matter of finding proper divisors $d$ of
$nxy+1-x^{2}y-xy^{2}$ such that $d\equiv 1\pmod{xy}$. This involves
$n^{\frac{1}{2}}$ factorizations of numbers that are less than
$n^{\frac{3}{2}}$. So, the total time to calculate $R_{4}(n)$ is
$O(n^{\frac{1}{2}+\epsilon})$.

Since the number of divisors of $nxy+1-x^{2}y-xy^{2}$ is
$O((nxy+1-x^{2}y-xy^{2})^{\epsilon})=O(n^{\epsilon})$, this argument also
proves Theorem 5.

Next, note that

$f_{4}(x,y,z,1)=f_{3}(x,y,z)+1.$

Therefore, if $R_{3}(n)>0$, then $R_{4}(n+1)>0$. Thus, to verify that
$R_{4}(n)>0$ for $45752<n\leq 5336500538$, it suffices to check that
$R_{4}(p+1)>0$ for the $2014$ values of $p\leq 5336500537$ for which
$R_{3}(p)=0$. For each of these, $R_{4}(p+1)>0$.

We believe that the complete list of numbers $n$ for which $R_{4}(n)=0$ is:

$\displaystyle 1,2,3,4,5,6,8,11,12,14,18,23,32,38,39,44,54,$ $\displaystyle
68,102,108,119,182,192,194,224,252,299,374,422,$ $\displaystyle
432,908,1043,1092,1202,1278,2468,2768,$ $\displaystyle
3182,4508,7208,10763,16104,21998,26348,45752$

###  2.4. Residue Classes

The expression $f_{3}(x,y,z)=xyz+x+y+z$ is also equal to

$z(xy+1)+x+y.$

As a result, every positive integer $n\equiv x+y\pmod{xy+1}$ with $n>xy+1$ has
$R_{3}(n)>0$. For example, if $x=2$ and $y=2$, we see that every $n>5$ such
that $n\equiv 4\pmod{5}$ has $R_{3}(n)>0$. Similarly, if $n\equiv 5\pmod{7}$,
if $n\equiv 7\pmod{11}$ or if $n\equiv 7\text{ or }8\pmod{13}$, then
$R_{3}(n)>0$. In general, there are

(4) $f(p)=\frac{1}{2}(d(p-1)-2)$

residue classes modulo any odd prime $p$ such that if $n$ is congruent to one
of these residue classes mod $p$, then $R_{3}(n)>0$.

Consequently, to give a bound on the number of $n\leq N$ for which
$R_{3}(n)=0$, we can count how many $n\leq N$ are not in any of these residue
classes. This idea suggests using the large sieve to estimate $U_{3}(n)$,
which we will do in the proof of Theorem 3.

##  3\. Proofs

###  3.1. Proofs of Theorems 1 and 4

###### Proof.

Observe that we can split the sum into two parts:

$\sum_{n\leq N}R_{3}(n)=\sum_{xyz+x+y+z\leq
N}1=\sum_{xyz+x+y+z\leq\frac{N}{\log N}}1+\sum_{\frac{N}{\log
N}<xyz+x+y+z<N}1$

For the second part,

$\sum_{\frac{N}{\log N}<xyz+x+y+z<N}1=\sum_{xy<N}\sum_{\frac{\frac{N}{\log
N}-x-y}{xy+1}<z<\frac{N-x-y}{xy+1}}1$

If $x+y+z\gg\frac{N}{\log N}$ and $xyz<N$, then $\max(x,y,z)\gg\frac{N}{\log
N}$. Without loss of generality, let $z$ be the maximum of $x,y,z$. Then,

$N\gg xyz\gg xy~{}\frac{N}{\log N}.$

Therefore,

$x+y\ll xy\ll\log N.$

Thus,

$\sum_{xy<N}\sum_{\frac{\frac{N}{\log
N}-x-y}{xy+1}<z<\frac{N-x-y}{xy+1}}1\ll\sum_{\begin{subarray}{c}xyz\leq 2N\\\
xy\ll\log N\end{subarray}}1\ll\sum_{xy\ll\log N}\frac{N}{xy}\ll N(\log\log
N)^{2}$

Therefore, the terms with $x+y+z\gg\frac{N}{\log N}$ are negligible, and we
can assume that $x+y+z\ll\frac{N}{\log N}$. Then, we have

$\sum_{xyz+x+y+z<N}1=\sum_{xyz<N+O(\frac{N}{\log
N})}1=\sum_{n<N+O(\frac{N}{\log N})}d_{3}(n)\sim\frac{N\log^{2}N}{2}$

(see 12.1.4 of [T]), which proves Theorem 1. ∎

The proof of Theorem 4 is similar.

###  3.2. Proof of Theorem 3

###### Proof.

Define

$a_{n}=\left\\{\begin{array}[]{cc}0&\mbox{if $R_{3}(n)>0$}\\\ 1&\mbox{if
$R_{3}(n)=0$}\end{array}\right.$

and define $Z=\sum\limits_{n=1}^{N}a_{n}$. Note that $Z=U_{3}(N)$. By the
discussion in section 1.4 and the large sieve inequality (see Theorem 7.11 of
[O]), we have that

(5) $Z\leq\frac{(N^{\frac{1}{2}}+X)^{2}}{Q}$

where $X$ is a free parameter and

$Q=\sum_{q\leq X}\mu^{2}(q)\prod_{p|q}\frac{f(p)}{p-f(p)},$

where $\mu$ denotes the Möbius function and $f(p)$ is defined as in (4). We
now prove the following lemma:

######  Lemma 1.

Let

$f(p)=\frac{\tau(p-1)-2}{2}$

for a prime $p>3$. Then, as $X\to\infty$, we have

$\displaystyle Q:=\sum_{q\leq X\atop(q,6)=1}\mu^{2}(q)\prod_{p\mid
q}\frac{f(p)}{p-f(p)}\gg\exp\bigg{(}\sqrt{\log X}\bigg{)}.$

###### Proof.

First of all

$Q\geq\sum_{q\leq X\atop(q,6)=1}\mu^{2}(q)\prod_{p\mid q}\frac{f(p)}{p}.$

Let $g(p)=\tau(p-1)$. Then for $p>3$ we have $f(p)\geq\frac{g(p)}{6}$.
Therefore, for any $k$, by restricting only to squarefree integers $q$ with
$k$ distinct prime factors, we have

$Q\geq 6^{-k}\prod_{p_{1}\cdot p_{2}\dots\cdot p_{k}\leq X\atop
3<p_{1}<p_{2}<\dots<p_{k}}\frac{g(p_{1})\dots g(p_{k})}{p_{1}\dots p_{k}}.$

Now we use an easy consequence of the Titchmarsh divisor problem (as found in
Theorem 3.9 of [HR]), namely that there exists a constant $c>0$ such that

$\sum_{3<p\leq X}\frac{\tau(p-1)}{p}\geq c\log X$

for any $X\geq 10$. This implies that

$Q\geq\frac{\log^{k}X}{(c_{1})^{k}k!}$

for $X\geq 10$ and all $k$ where $c_{1}>0$ is an absolute constant. Keeping in
mind that, by Stirling’s formula,

$k!\ll(k/e)^{k+1/2}$

we obtain

$Q\gg\frac{\left(\frac{e\log X}{c_{1}k}\right)^{k}}{\sqrt{k}}.$

Now we choose

$k=\sqrt{\log X}$

and have

$Q\gg k^{-1/2}\exp\left(\sqrt{\log X}\log\left(\frac{e\sqrt{\log
X}}{c_{1}}\right)\right)\gg\exp(\sqrt{\log X})$

as claimed. ∎

Using the result of Lemma 1 in (5) and choosing $X=N^{\frac{1}{2}}$, we obtain
the bound

$U_{3}(N)\ll\frac{N}{e^{\sqrt{\log N}}}.$

∎

##  4\. Conclusion and Open Questions

Here are some related problems we think are worth exploring further:

  1. (1)

Prove Conjecture 1, that $R_{3}(n)$ grows slower than $n^{\epsilon}$ for any
$\epsilon>0$.

  2. (2)

Let $g_{3}(x,y,z)=xy+yz+zx+1$ and $S_{3}(n)$ denote the number of positive
integer solutions to $S_{3}(n)=g_{3}(x,y,z)$. Prove that there are only
finitely many positive integers $n$ for which
$S_{3}(n)=0$222$f_{3}(x,y,z)+g_{3}(x,y,z)=(x+1)(y+1)(z+1)$. Also, if
$S_{3}(n)=0$, then $n$ is prime. (We think this might be possible using the
techniques used in [FI].)

  3. (3)

Prove Conjecture 5.

  4. (4)

Prove that there are infinitely many positive integers $n$ for which
$R_{3}(n)=0$.

  5. (5)

Give any improvement on the omega result in (2).

  6. (6)

Give omega results for $d(n)+d(n+1)$ and $d(n^{2}+1)$.

We don’t think this question has been researched before, and we hope that this
work stimulates the reader to pursue some of these questions further.

## References

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  * [HR] Halberstam, H.; Richert, H.-E. Sieve methods. London Mathematical Society Monographs, No. 4. Academic Press [Harcourt Brace Jovanovich, Publishers], London-New York, 1974. xiv+364 pp.
  * [O] Odlyzko, Andrew Michael SIEVE METHODS. Thesis (B.S.)– California Institute of Technology. 1971. 146 pages.
  * [S] Shiu, P. A Brun-Titchmarsh theorem for multiplicative functions. J. Reine Angew. Math. 313 (1980), 161–170.
  * [T] Titchmarsh, E. C. The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. x+412 pp.