Testing high-dimensional mean vector with applications

Abstract

A centered $L^2$-norm based test statistic is used for testing if a high-dimensional mean vector equals zero where the data dimension may be much larger than the sample size. Inspired by the fact that under some regularity conditions the asymptotic null distributions of the proposed test are the same as the limiting distributions of a chi-square-mixture, a three-cumulant matched chi-square-approximation is suggested to approximate this null distribution. The asymptotic power of the proposed test under a local alternative is established and the effect of data non-normality is discussed. A simulation study under various settings demonstrates that in terms of size control, the proposed test performs significantly better than some existing competitors. Several real data examples are presented to illustrate the wide applicability of the proposed test to a variety of high-dimensional data analysis problems, including the one-sample problem, paired two-sample problem, and MANOVA for correlated samples or independent samples.

Appendix: Technical proofs

We first prove the following useful lemma.

Lemma 1

Let $\varvec{{W}}\sim W_{p}(v,\varvec{{\varSigma }}/v)$ denote a Wishart distribution with v degrees of freedom and a covariance matrix $\varvec{{\varSigma }}/v$. Assume that $p/v^2\longrightarrow 0$ as $v, p\rightarrow \infty$. Then the unbiased and ratio-consistent estimator of ${\text {tr}}(\varvec{{\varSigma }}^{3})$ is given by

$$\begin{aligned} \widehat{{\text {tr}}(\varvec{{\varSigma }}^{3})}=\frac{v^{4}}{(v-1)(v+4)(v^{2}-4)}\left[ {\text {tr}}(\varvec{{W}}^{3})-\frac{3}{v}{\text {tr}}(\varvec{{W}}){\text {tr}}(\varvec{{W}}^{2})+\frac{2}{v^{2}}{\text {tr}}^{3}(\varvec{{W}})\right] . \end{aligned}$$

(A.33)

Proof of Lemma 1

Let $\varvec{{V}}=v\varvec{{W}}$. Then $\varvec{{V}}\sim W_{p}(v,\varvec{{\varSigma }})$. To show (A.33) is equivalent to show

$$\begin{aligned} \widehat{{\text {tr}}(\varvec{{\varSigma }}^{3})}=\frac{v}{(v-1)(v+4)(v^{2}-4)}\left[ {\text {tr}}(\varvec{{V}}^{3})-\frac{3}{v}{\text {tr}}(\varvec{{V}}){\text {tr}}(\varvec{{V}}^{2})+\frac{2}{v^{2}}{\text {tr}}^{3}(\varvec{{V}})\right] \end{aligned}$$

(A.34)

is unbiased and ratio consistent for ${\text {tr}}(\varvec{{\varSigma }}^{3})$. By the proof of Theorem 2.1 of Srivastava and Yanagihara (2010), we can express the above expression as $\widehat{{\text {tr}}(\varvec{{\varSigma }}^{3})}=I_{1}+I_{2}+I_{3}+I_{4},$ where

$$\begin{aligned} \begin{array}{rcl} I_{1} &{} = &{} \frac{1}{v(v+2)(v+4)}\sum _{i=1}^{p}\lambda _{i}^{3}w_{ii}^{3},\\ I_{2} &{} = &{} \frac{3}{(v-1)(v+2)(v+4)}\sum _{i\ne j}\lambda _{i}^{2}\lambda _{j}\left( w_{ii}w_{ij}^{2}-v^{-1}w_{ii}^{2}w_{jj}\right) ,\\ I_{3} &{} = &{} \frac{v}{(v-1)(v+4)(v^{2}-4)}\sum _{i\ne j\ne k}\lambda _{i}\lambda _{j}\lambda _{k}\left( w_{ij}w_{jk}w_{ki}-v^{-2}w_{ii}w_{jj}w_{kk}\right) ,\\ I_{4} &{} = &{} \frac{-3}{(v-1)(v+4)(v^{2}-4)}\sum _{i\ne j\ne k}\lambda _{i}\lambda _{j}\lambda _{k}\left( w_{ij}w_{jk}^{2}-v^{-1}w_{ii}w_{jj}w_{kk}\right) , \end{array} \end{aligned}$$

where $w_{ij}=\varvec{{u}}_{i}^{\top }\varvec{{u}}_{j},\ i,j=1,\ldots ,p$ and $\varvec{{u}}_{1},\ldots ,\varvec{{u}}_{p}{\mathop {\sim }\limits ^{\text {i.i.d.}}}\mathcal {N}_{v}(\varvec{{0}},\varvec{{I}}_{v})$. By Lemma 6.2 (a) and (e) of Srivastava and Yanagihara (2010), we have ${\text {E}}(w_{ii}^{3})=v(v+2)(v+4)$ and ${\text {Var}}(w_{ii}^{3})=6v(v+2)(v+4)(3v^{2}+30v+80)$. Therefore,

$$\begin{aligned} \begin{array}{rcl} {\text {E}}(I_{1}) &{} = &{} \frac{1}{v(v+2)(v+4)}\sum _{i=1}^{p}\lambda _{i}^{3}{\text {E}}(w_{ii}^{3})={\text {tr}}(\varvec{{\varSigma }}^{3}),\\ {\text {Var}}(I_{1}) &{} = &{} [\frac{1}{v(v+2)(v+4)}]^{2}\sum _{i=1}^{p}\lambda _{i}^{6}{\text {Var}}(w_{ii}^{3})=v^{-1}{\text {tr}}(\varvec{{\varSigma }}^{6})[1+o(1)]. \end{array} \end{aligned}$$

Noting that

$$\begin{aligned} {\text {tr}}(\varvec{{A}}^{r})/{\text {tr}}^{r}(\varvec{{A}})\le 1,\ r=1,2,\ldots \end{aligned}$$

(A.35)

hold for any nonnegative matrix $\varvec{{A}}$, we have ${\text {Var}}[I_{1}/{\text {tr}}(\varvec{{\varSigma }}^{3})]<v^{-1}[1+o(1)]$. It follows that $I_{1}/{\text {tr}}(\varvec{{\varSigma }}^{3})\longrightarrow 1$ as $v\rightarrow \infty$.

Following the proof of Theorem 2.1 of Srivastava and Yanagihara (2010) closely, let $r_{ij}=w_{ii}w_{ij}^{2}-v^{-1}w_{ii}^{2}w_{jj}$. Then we have ${\text {E}}(r_{ij})=0$ and ${\text {Cov}}(r_{ij},r_{kl})=0$ where $(i\ne j),\ (k\ne l)$ and $(i,j)\ne (k,l)$. In addition, by Lemma 6.2 (f), (g) and (h) of Srivastava and Yanagihara (2010), we have

$$\begin{aligned} \begin{array}{rcl} {\text {Var}}(r_{ij}) &{} = &{} {\text {E}}\left( w_{ii}w_{ij}^{2}-v^{-1}w_{ii}^{2}w_{jj}\right) ^{2}={\text {E}}\left( w_{ii}^{2}w_{ij}^{4}-2v^{-1}w_{ii}^{3}w_{ij}^{2}w_{jj}+v^{-2}w_{ii}^{4}w_{jj}^{2}\right) \\ &{} = &{} O(v^{4})-O(v^{-1}v^{5})+O(v^{-2}v^{6})=O(v^{4}). \end{array} \end{aligned}$$

It follows that ${\text {E}}(I_{2})=0$ and

$$\begin{aligned} \begin{array}{rcl} {\text {Var}}(I_{2})&{}=&{}9\left[ (v-1)(v+2)(v+4)\right] ^{-2}\sum _{i\ne j}\lambda _{i}^{4}\lambda _{j}^{2}{\text {Var}}(r_{ij})\\ &{}\le &{} 9v^{-2}{\text {tr}}(\varvec{{\varSigma }}^{4}){\text {tr}}(\varvec{{\varSigma }}^{2})[1+o(1)]. \end{array} \end{aligned}$$

It follows from (A.35) that ${\text {Var}}\left[ I_{2}/{\text {tr}}(\varvec{{\varSigma }}^{3})\right] \le [{\text {tr}}(\varvec{{\varSigma }}^{4})(9\kappa )]/[{\text {tr}}^{2}(\varvec{{\varSigma }}^{2})v^{2}][1+o(1)]\le {9\kappa }/{v^{2}}[1+o(1)],$ where

$$\begin{aligned} \kappa =\frac{{\text {tr}}^{3}(\varvec{{\varSigma }}^{2})}{{\text {tr}}^{2}(\varvec{{\varSigma }}^{3})}. \end{aligned}$$

(A.36)

By Theorem 5 of Zhang et al. (2020), we have $\kappa /p\le 1$. Since $p/v^2\longrightarrow 0$ as $v,p\rightarrow \infty$, we have $\kappa /v^{2}\longrightarrow 0$ as $v,p\rightarrow \infty$. Therefore, we have that $I_{2}=o_{p}[{\text {tr}}(\varvec{{\varSigma }}^{3})]$.

Similarly, we have ${\text {E}}(I_{3})=0,\ {\text {E}}(I_{4})=0$ and

$$\begin{aligned} {\text {Var}}[I_{3}/{\text {tr}}(\varvec{{\varSigma }}^{3})]\le \kappa v^{-3}[1+o(1)]\longrightarrow 0,\;\;{\text {Var}}[I_{4}/{\text {tr}}(\varvec{{\varSigma }}^{3})]\le \kappa v^{-4}[1+o(1)]\longrightarrow 0, \end{aligned}$$

showing that $I_{3}=o_{p}[{\text {tr}}(\varvec{{\varSigma }}^{3})]$ and $I_{4}=o_{p}[{\text {tr}}(\varvec{{\varSigma }}^{3})]$. It follows that $\widehat{{\text {tr}}(\varvec{{\varSigma }}^{3})}$ is an unbiased and ratio-consistent estimator of ${\text {tr}}(\varvec{{\varSigma }}^{3})$. The lemma is then proved. $\square$

We now prove the main results.

Proof of Theorem 1

We first prove (a). We shall use the characteristic function ($\psi _{X}(t)={\text {E}}(e^{itX})$ for a random variable X) method. Set $\varvec{{x}}_i=\varvec{{y}}_i-\varvec{{\mu }},\ i=1,\ldots ,n$ and hence $\bar{\varvec{{x}}}=\bar{\varvec{{y}}}-\varvec{{\mu }}$. Set $\varvec{{w}}_{n,p}=\sqrt{n}\bar{\varvec{{x}}}$. By (10), we have

$$\begin{aligned} \tilde{T}_{n,p,0}=\left[ \Vert \varvec{{w}}_{n,p}\Vert ^2-{\text {tr}}(\varvec{{\varSigma }})\right] /\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}[1+o(1)], \end{aligned}$$

(A.37)

since ${\text {tr}}(\hat{\varvec{{\varSigma }}})/{\text {tr}}(\varvec{{\varSigma }})\longrightarrow 1$ as $n,p\rightarrow \infty$ (See Proof of Theorem 9 in Zhang et al. 2020). Further, we have ${\text {E}}(\varvec{{w}}_{n,p})=\varvec{{0}}$ and ${\text {Cov}}(\varvec{{w}}_{n,p})=\varvec{{\varSigma }}$. Write $\varvec{{w}}_{n,p}=\sum _{r=1}^{p}\xi _{n,p,r}\varvec{{u}}_{p,r}$, where $\xi _{n,p,r}=\varvec{{w}}_{n,p}^{\top }\varvec{{u}}_{p,r}$, and $\varvec{{u}}_{p,1},\ldots ,\varvec{{u}}_{p,p}$ denote the eigenvectors associated with the eigenvalues $\lambda _{p,1},\ldots ,\lambda _{p,p}$ of $\varvec{{\varSigma }}$ in the descending order. We have ${\text {E}}(\xi _{n,p,r})=0$, and ${\text {Var}}(\xi _{n,p,r})=\lambda _{p,r},\ r=1,2,\ldots$, and $\xi _{n,p,r},\ r=1,\ldots ,p$, are uncorrelated.

By Lemma S.4 of Zhang et al. (2020), we have

$$\begin{aligned} {\text {Var}}(\xi _{n,p,r}^{2})=2\lambda _{p,r}^{2}+[{\text {E}}(\varvec{{x}}_{1}^{\top }\varvec{{u}}_{p,r})^{4}-3\lambda _{p,r}^{2}]/n. \end{aligned}$$

Under Conditions C1 and C2, by some simple algebra, we have ${\text {E}}(\varvec{{x}}_{1}^{\top }\varvec{{u}}_{p,r})^{4}\le (3+\varDelta )\lambda _{p,r}^{2}$. Thus, we have

$$\begin{aligned} {\text {Var}}(\xi _{n,p,r}^{2})\le (2+\varDelta /n)\lambda _{p,r}^{2},\ r=1,2,\ldots . \end{aligned}$$

(A.38)

It follows from (A.37) that

$$\begin{aligned} \tilde{T}_{n,p,0}=\sum _{r=1}^p (\xi _{n,p,r}^2-\lambda _{p,r})/\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}[1+o(1)]. \end{aligned}$$

Set $\tilde{T}_{n,p,0}^{(q)}=\sum _{r=1}^q (\xi _{n,p,r}^2-\lambda _{p,r})/\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}$ where $q<p$. Then

$$\begin{aligned}&{\text {E}}\big (\tilde{T}_{n,p,0}-\tilde{T}_{n,p,0}^{(q)}\big )^{2}={\text {E}}\bigg [\sum _{r=q+1}^{p}(\xi _{n,p,r}^{2}-\lambda _{p,r})/\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}\bigg ]^{2}[1+o(1)]\\&\qquad \qquad \qquad \qquad \qquad \qquad ={\text {Var}}\bigg (\sum _{r=q+1}^{p}\xi _{n,p,r}^{2}\bigg )/\left[ 2{\text {tr}}(\varvec{{\varSigma }}^2)\right] [1+o(1)]\\&\qquad \qquad \qquad \qquad \qquad \qquad \le \bigg [\sum _{r=q+1}^{p}\sqrt{{\text {Var}}(\xi _{n,p,r}^{2})}\bigg ]^{2}/\left[ 2{\text {tr}}(\varvec{{\varSigma }}^2)\right] [1+o(1)]. \end{aligned}$$

By (A.38), we have

$$\begin{aligned} \bigg [\sum _{r=q+1}^p \sqrt{{\text {Var}}(\xi _{n,p,r}^{2})} \bigg ]^2/\left[ 2{\text {tr}}(\varvec{{\varSigma }}^2)\right] \le \left( 1+\varDelta /n\right) \bigg (\sum _{r=q+1}^p \rho _{p,r}\bigg )^2. \end{aligned}$$

It follows that

$$\begin{aligned} \begin{array}{rcl} |\psi _{\tilde{T}_{n,p,0}}(t)-\psi _{\tilde{T}_{n,p,0}^{(q)}}(t)| &{}\le &{} |t|\big [{\text {E}}(\tilde{T}_{n,p,0}-\tilde{T}_{n,p,0}^{(q)})^{2}\big ]^{1/2}\\ &{}\le &{} |t|\left( 1+\varDelta /n\right) ^{1/2} \bigg (\sum _{r=q+1}^{p}\rho _{p,r}\bigg )[1+o(1)]. \end{array} \end{aligned}$$

(A.39)

Set $\zeta ^{(q)}=\sum _{r=1}^{q}\rho _{r}(A_{r}-1)/\sqrt{2}$, we have

$$\begin{aligned} \big |\psi _{\tilde{T}_{n,p,0}}(t)-\psi _{\zeta }(t)\big |\le & {} \big |\psi _{\tilde{T}_{n,p,0}}(t)-\psi _{\tilde{T}_{n,p,0}^{(q)}}(t)\big | + \big |\psi _{\tilde{T}_{n,p,0}^{(q)}}(t)-\psi _{\tilde{T}_{p,0}^{(q)}}(t)\big | \\&+\big |\psi _{\tilde{T}_{p,0}^{(q)}}(t)-\psi _{\zeta ^{(q)}}(t)\big |+\big |\psi _{\zeta ^{(q)}}(t)-\psi _{\zeta }(t)\big |. \end{aligned}$$

By similar arguments in Proof of Theorem 2 of Zhang et al. (2020), we can show under Conditions C1–C3 all the four terms on the right hand side of the previous inequality converge to zero as $n, p\rightarrow \infty$, so the first expression of (11) follows. In particular, the convergence of the first term can be derived from (A.39) and Condition C3, the convergence of the second term is ensured by the standard central limit theorem, and the convergence of the last two terms is due to Condition C3.

Notice that when the data (2) are normally distributed, Conditions C1–C2 are automatically satisfied so that under Condition C3, the second expression of (11) follows immediately since under the normality assumption, we have $T_{n,p,0}{\mathop {=}\limits ^{d}}T_{n,p,0}^*$.

We now prove (b). By Lemma 8.1 of Pauly et al. (2015), Condition C5 is equivalent to the condition “${\text {tr}}(\varvec{{\varSigma }}^4)/{\text {tr}}^2(\varvec{{\varSigma }}^2)=o(1)$” imposed by Chen and Qin (2010). Therefore, under Conditions C1, C2 and C5, the proofs of the asymptotic normality of $\tilde{T}_{n,p,0}$ and $\tilde{T}_{n,p,0}^*$ are along the same lines as the one given by Chen and Qin (2010) for the asymptotic normality of their test statistic.

Finally, we prove (13). We have $\sup _{x} \big |\Pr (T_{n,p,0}\le x)-\Pr (T_{n,p,0}^*\le x)\big |\le \sup _{x} \big |\Pr (\tilde{T}_{n,p,0}\le \tilde{x})-\Pr (\zeta \le \tilde{x})\big |+ \sup _{x} \big |\Pr (\tilde{T}_{n,p,0}^*\le \tilde{x})-\Pr (\zeta \le \tilde{x})\big |$, where $\tilde{x}=[x-{\text {tr}}(\hat{\varvec{{\varSigma }}})]/\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}$. Under the conditions of (a), the two terms on the right hand side of previous inequality both converge to zero because both $\tilde{T}_{n,p,0}$ and $\tilde{T}_{n,p,0}^*$ converge to $\zeta$ in distribution. Under the conditions of (b), the proof of (13) are along the same lines as the above and hence are omitted for space saving. $\square$

Proof of Theorem 2

Under the local alternative (21), $T_{n,p}=\left( T_{n,p,0}+n\Vert \varvec{{\mu }}\Vert ^2\right) [1+o_p(1)]$. In addition, under the given conditions, we have $\hat{\beta }_0/\beta _0{\mathop {\longrightarrow }\limits ^{P}}1,\ \hat{\beta }_1/\beta _1{\mathop {\longrightarrow }\limits ^{P}}1$ and $\hat{d}/d{\mathop {\longrightarrow }\limits ^{P}}1$ as $n, p\rightarrow \infty$. We first prove (a). Under Conditions C1, C2 and C3, Theorem 1(a) indicates that as $n,p\rightarrow \infty$, we have $\tilde{T}_{n,p,0}=T_{n,p,0}/\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}{\mathop {\longrightarrow }\limits ^{\mathcal {L}}}\zeta$. It follows that as $n, p\rightarrow \infty$, we have

$$\begin{aligned} \begin{array}{rcl} \Pr \left[ T_{n,p}\ge \hat{\beta }_0+\hat{\beta }_1\chi _{\hat{d}}^2(\alpha )\right] &{}=&{} \Pr \left[ \tilde{T}_{n,p,0} \ge \frac{\beta _0+ \beta _1 \chi ^2_{d}(\alpha )}{\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}}-\frac{n\Vert \varvec{{\mu }}\Vert ^2}{\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}} \right] [1+o(1)] \\ &{}= &{} \Pr \left[ \zeta \ge \frac{\chi _{d}^2(\alpha )-d}{\sqrt{2d}}-\frac{n\Vert \varvec{{\mu }}\Vert ^2}{\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}} \right] [1+o(1)]. \end{array} \end{aligned}$$

(A.40)

We now prove (b). Under the given conditions, Theorem 1(b) indicates that as $n\rightarrow \infty$, we have $\tilde{T}_{n,p,0} {\mathop {\longrightarrow }\limits ^{\mathcal {L}}}\mathcal {N}(0,1)$ and by Theorem 5, as $n,p\rightarrow \infty$, we have $d\longrightarrow \infty$ and $[\chi _d^2(\alpha )-d]/\sqrt{2d}\longrightarrow z_{\alpha }$ where $z_{\alpha }$ denotes the upper $100\alpha$-percentile of $\mathcal {N}(0,1)$. Then by (A.40), as $n,p\rightarrow \infty$, we have

$$\begin{aligned} \begin{array}{rcl} \Pr \left[ T_{n,p}\ge \hat{\beta }_0+\hat{\beta }_1\chi _{\hat{d}}^2(\alpha )\right] &{}=&{} \Pr \left[ \tilde{T}_{n,p,0} \ge \frac{\beta _0+ \beta _1 \chi ^2_{d}(\alpha )}{\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}}-\frac{n\Vert \varvec{{\mu }}\Vert ^2}{\sqrt{\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^2)}} \right] [1+o(1)] \\ &{}= &{} \varPhi \left[ -z_{\alpha }+\frac{n\Vert \varvec{{\mu }}\Vert ^2}{\sqrt{2{\text {tr}}(\varvec{{\varSigma }}^2)}} \right] [1+o(1)], \end{array} \end{aligned}$$

where $\varPhi (\cdot )$ denotes the cumulative distribution function of $\mathcal {N}(0,1)$. The proof is complete. $\square$

Proof of Theorem 3

Let $\varvec{{x}}_{i}=\varvec{{y}}_{i}-\varvec{{\mu }},\ i=1,\ldots ,n$. Then $\varvec{{x}}_{i},\ i=1,\ldots ,n$ are i.i.d. with ${\text {E}}(\varvec{{x}}_{i})=\varvec{{0}}$ and ${\text {Cov}}(\varvec{{x}}_{i})=\varvec{{\varSigma }}$. It is easy to verify that $T_{n,p,0}=2(n-1)^{-1}\sum _{i<j}\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j}$. It follows that ${\text {E}}(T_{n,p,0})=0$ and

$$\begin{aligned} {\text {Var}}(T_{n,p,0})={\text {E}}(T_{n,p,0}^{2})=4(n-1)^{-2}\sum _{i<j}{\text {E}}(\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j})^{2}=\frac{2n}{(n-1)}{\text {tr}}(\varvec{{\varSigma }}^{2}). \end{aligned}$$

Furthermore, we have

$$\begin{aligned} {\text {E}}(T_{n,p,0}^3)= & {} 8(n-1)^{-3}{\text {E}}\left[ \sum _{i<j}(\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j})\right] ^{3}\\= & {} 8(n-1)^{-3}{\text {E}}\left[ \sum _{i<j}(\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j})^{3}+3\sum _{*}(\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j})^{2}(\varvec{{x}}_{u}^{\top }\varvec{{x}}_{v})\right. \\&\left. +6\sum _{**}(\varvec{{x}}_{i}^{\top }\varvec{{x}}_{j})(\varvec{{x}}_{u}^{\top }\varvec{{x}}_{v})(\varvec{{x}}_{\alpha }^{\top }\varvec{{x}}_{\beta })\right] \\= & {} 8(n-1)^{-3}\left\{ \frac{n(n-1)}{2}{\text {E}}(\varvec{{x}}_{1}^{\top }\varvec{{x}}_{2})^{3}+n(n-1)(n-2){\text {E}}[(\varvec{{x}}_{1}^{\top }\varvec{{x}}_{2})(\varvec{{x}}_{2}^{\top }\varvec{{x}}_{3})(\varvec{{x}}_{3}^{\top }\varvec{{x}}_{1})]\right\} \\= & {} \frac{8n(n-2)}{(n-1)^{2}}{\text {tr}}(\varvec{{\varSigma }}^{3})+\frac{4n\varUpsilon }{(n-1)^{2}}, \end{aligned}$$

where $\varUpsilon ={\text {E}}(\varvec{{x}}_{1}^{\top }\varvec{{x}}_{2})^{3}={\text {E}}[(\varvec{{y}}_{1}-\varvec{{\mu }})^{\top }(\varvec{{y}}_{2}-\varvec{{\mu }})]^{3}$, $*$ means “$i<j,\ u<v$” and “$(i,j)\ne (u,v)$” while $**$ means “$i<j,\ u<v,\ \alpha <\beta$” and “$(i,j),\ (u,v),\ (\alpha ,\beta )$ are not mutually equal to each other.” The proof is complete. $\square$

Proof of Theorem 4

We first show (a). Under Condition C1, we have $\varvec{{y}}_i=\varvec{{\mu }}+\varvec{{\varGamma }}\varvec{{z}}_i,\ i=1,\ldots , n$ where $\varvec{{z}}_i,\ i=1,\ldots , n$ are i.i.d. with ${\text {E}}(\varvec{{z}}_i)=\varvec{{0}}$ and ${\text {Cov}}(\varvec{{z}}_i)=\varvec{{I}}_p$ and $\varvec{{\varSigma }}=\varvec{{\varGamma }}\varvec{{\varGamma }}^{\top }$. It follows that $\varUpsilon ={\text {E}}[(\varvec{{y}}_{1}-\varvec{{\mu }})^{\top }(\varvec{{y}}_{2}-\varvec{{\mu }})]^{3}={\text {E}}(\varvec{{z}}_1^{\top }\varvec{{\varOmega }}\varvec{{z}}_2)^3$ where $\varvec{{\varOmega }}=\varvec{{\varGamma }}^{\top }\varvec{{\varGamma }}$. By Jensen’s inequality, we have

$$\begin{aligned} \varUpsilon ={\text {E}}(\varvec{{z}}_1^{\top }\varvec{{\varOmega }}\varvec{{z}}_2)^3\le \left[ {\text {E}}(\varvec{{z}}_1^{\top }\varvec{{\varOmega }}\varvec{{z}}_2)^4\right] ^{3/4}. \end{aligned}$$

(A.41)

Denote the (i, j)-th entry of $\varvec{{\varOmega }}$ as $w_{ij},\ i,j,=1,\ldots ,p$. Under Conditions C1 and C2, from the proof of Lemma 6.2 of Srivastava and Kubokawa (2013) (p. 215), we have

$$\begin{aligned} {\text {E}}\left( \varvec{{z}}_1^{\top }\varvec{{\varOmega }}\varvec{{z}}_2 \right) ^4=\varDelta ^2\sum _{i=1}^p\sum _{j=1}^p w_{ij}^4+6\varDelta \sum _{i=1}^p\left( \sum _{j=1}^p w_{ij}^2 \right) ^2 +6{\text {tr}}(\varvec{{\varOmega }}^4)+3{\text {tr}}^2(\varvec{{\varOmega }}^2),\qquad \end{aligned}$$

(A.42)

where $\varDelta$ is given in Condition C2. Notice that we have

$$\begin{aligned} \begin{array}{rcl} \sum _{i=1}^p\sum _{j=1}^p w_{ij}^4&{}\le &{} \left( \sum _{i=1}^p\sum _{j=1}^p w_{ij}^2 \right) ^2={\text {tr}}^2(\varvec{{\varOmega }}^2),\\ \sum _{i=1}^p \left( \sum _{j=1}^p w_{ij}^2 \right) ^2 &{}\le &{} \left( \sum _{i=1}^p \sum _{j=1}^p w_{ij}^2 \right) ^2={\text {tr}}^2(\varvec{{\varOmega }}^2),\\ {\text {tr}}(\varvec{{\varOmega }}^4) &{}\le &{}{\text {tr}}^2(\varvec{{\varOmega }}^2). \end{array} \end{aligned}$$

These, together with (A.42), imply that ${\text {E}}\left( \varvec{{z}}_1^{\top }\varvec{{\varOmega }}\varvec{{z}}_2\right) ^4\le (\varDelta ^2+6\varDelta +9){\text {tr}}^2(\varvec{{\varOmega }}^2)$. Then by (A.41), we have

$$\begin{aligned} \varUpsilon \le (\varDelta ^2+6\varDelta +9)^{3/4}{\text {tr}}^{3/2}(\varvec{{\varOmega }}^2)=(\varDelta ^2+6\varDelta +9)^{3/4}{\text {tr}}^{3/2}(\varvec{{\varSigma }}^2), \end{aligned}$$

(A.43)

where we use the fact that ${\text {tr}}(\varvec{{\varOmega }}^2)={\text {tr}}(\varvec{{\varGamma }}^{\top }\varvec{{\varGamma }}\varvec{{\varGamma }}^{\top }\varvec{{\varGamma }})={\text {tr}}(\varvec{{\varGamma }}\varvec{{\varGamma }}^{\top }\varvec{{\varGamma }}\varvec{{\varGamma }}^{\top })={\text {tr}}(\varvec{{\varSigma }}^2).$

We now show (b). First of all, under Conditions C1 and C2, by (22) and (A.43), we have

$$\begin{aligned} \delta \le 1+\frac{(\varDelta ^2+6\varDelta +9)^{3/4}{\text {tr}}^{3/2}(\varvec{{\varSigma }}^2)}{2(n-2){\text {tr}}(\varvec{{\varSigma }}^3)} = 1+(\varDelta ^2+6\varDelta +9)^{3/4} \frac{\sqrt{\kappa }}{2(n-2)},\nonumber \\ \end{aligned}$$

(A.44)

where $\kappa$ is given in (A.36). By Theorem 5 of Zhang et al. (2020), we have

$$\begin{aligned} 1\le \kappa \le \frac{{\text {tr}}^2(\varvec{{\varSigma }})}{{\text {tr}}(\varvec{{\varSigma }}^2)}\le p. \end{aligned}$$

(A.45)

Under Condition C3, as $p\rightarrow \infty$, we have ${{\text {tr}}^2(\varvec{{\varSigma }})}/{{\text {tr}}(\varvec{{\varSigma }}^2)}=(\sum _{r=1}^p \rho _{p,r})^2\longrightarrow (\sum _{r=1}^{\infty } \rho _{r})^2<\infty .$ It follows that under Conditions C1, C2 and C3, $\kappa$ is bounded as $p\rightarrow \infty$. This, together with (A.44), implies that under Conditions C1, C2 and C3, as $n, p\rightarrow \infty$, we have $\delta =1+o(1)$ as $n, p\rightarrow \infty$.

Under Conditions C1, C2 and C4, by (A.45), as $n, p\rightarrow \infty$, we have $\sqrt{\kappa }/[2(n-2)]\le \sqrt{p/[2(n-2)]^2}=o(1).$ This, together with (A.44), implies that under Conditions C1, C2 and C4, as $n, p\rightarrow \infty$, we always have $\delta =1+o(1)$. The proof is complete. $\square$

Proof of Theorem 5

By (23), the skewness of $T_{n,p,0}$ is $\sqrt{8/f}$ where f is defined in (22) and under Conditions C1, C2 and C4, by Theorem 4, we have $f=d[1+o(1)]$. On the one hand, when $\tilde{T}_{n,p,0}{\mathop {\longrightarrow }\limits ^{\mathcal {L}}}\mathcal {N}(0,1)$, the skewness of $T_{n,p,0}$ must tend to 0 as $n,p\rightarrow \infty$, it follows that we must have $d\longrightarrow \infty$ as $n,p\rightarrow \infty$. On the other hand, by (15), as $n\rightarrow \infty$, we have $d=\kappa [1+o(1)]$ where $\kappa$ is defined in (A.36). We have $\kappa ={{\text {tr}}^{3}(\varvec{{\varSigma }}^{2})}/{{\text {tr}}^{2}(\varvec{{\varSigma }}^{3})}\ge {{\text {tr}}(\varvec{{\varSigma }}^{2})}/{\lambda _{p,\max }^{2}}$ where $\lambda _{p,\max }$ is the largest eigenvalue of $\varvec{{\varSigma }}$. Therefore, as $d\rightarrow \infty$, we have $\kappa \longrightarrow \infty$ and ${\lambda _{p,\max }^{2}}/{{\text {tr}}(\varvec{{\varSigma }}^{2})}\ge \kappa ^{-1}\longrightarrow 0$ as $p\rightarrow \infty$. That is, Condition C5 holds. This, together with Conditions C1, C2 and C4, implies that by Theorem 1(b), we have $\tilde{T}_{n,p,0}{\mathop {\longrightarrow }\limits ^{\mathcal {L}}}\mathcal {N}(0,1)$. The proof is then complete. $\square$