Estimation of multivariate 3rd moment for high-dimensional data and its application for testing multivariate normality

Abstract

This paper is concerned with the multivariate 3rd moment and its estimation. Mardia (Biometrika 57:519–530, 1970) and Srivastava (Stat Probab Lett 2:263–267, 1984) proposed the multivariate skewness and its estimator, independently. However, these estimators cannot be defined for the case in which the dimension p is larger than the sample size N. In this paper, we treat the multivariate 3rd moment $\gamma$ which is defined by using Hadamard product of observation vectors, and propose an estimate of $\gamma$ which is well defined when $p>N$. Based on the estimator, we propose a new test for multivariate normality. Under the null hypothesis, the test statistic is asymptotically standard normal, which is supported by Monte Carlo simulations. We calculate some empirical powers to see the performance of the test.

Appendices

Moment of statistic

In this section, analytic forms of $\mathrm {Var}(T_1)$, $\mathrm {Var}(T_2)$ and $\mathrm {Var}(T_3)$ are proposed. We derive them by using the following lemma (Lemma 1). Proofs of Lemma 1 is tedious but simple, therefore skipped.

Lemma 1

Let $\varvec{z}$ be a random vector distributed as $N_p(\varvec{0},\varvec{I}_p)$, and $\varvec{a}$ and $\varvec{b}$ be constant vectors. Then,

$$\begin{aligned}&E[(\varvec{a}'\varvec{z})^2]=\varvec{a}'\varvec{a}, \\&E[(\varvec{a}'\varvec{z})^4]=3(\varvec{a}'\varvec{a})^2, \\&E[(\varvec{a}'\varvec{z})^2(\varvec{b}'\varvec{z})^2] =2(\varvec{a}'\varvec{b})^2+\varvec{a}'\varvec{a}\varvec{b}'\varvec{b}, \\&E[(\varvec{a}'\varvec{z})^3\varvec{b}'\varvec{z}]=3\varvec{a}'\varvec{a}\varvec{a}'\varvec{b}, \\&E[(\varvec{a}'\varvec{z})^6]=15(\varvec{a}'\varvec{a})^3, \\&E[(\varvec{a}'\varvec{z})^3(\varvec{b}'\varvec{z})^3]=6(\varvec{a}'\varvec{b})^3 +9\varvec{a}'\varvec{a}\varvec{a}'\varvec{b} \varvec{b}'\varvec{b}. \end{aligned}$$

Here, we write analytic forms of $\mathrm {Var}(T_1)$, $\mathrm {Var}(T_2)$ and $\mathrm {Var}(T_3)$ in the following lemma.

Lemma 2

For $T_1$, $T_2$ and $T_3$, as defined in Sect. 2.1,

$$\begin{aligned}&\mathrm {Var}(T_1)= \frac{6}{N} \varvec{1}_p' ( \odot ^3 \varvec{{\varSigma }} ) \varvec{1}_p, \\&\mathrm {Var}(T_2)= \frac{18}{N(N-1)} \varvec{1}_p' ( \odot ^3 \varvec{{\varSigma }} ) \varvec{1}_p, \\&\mathrm {Var}(T_3) =\frac{24}{N(N-1)(N-2)} \varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p. \end{aligned}$$

Proof

We can express that

$$\begin{aligned} T_1 = \frac{1}{\sqrt{N}} \sum _{i=1}^N \zeta _i, \end{aligned}$$

where

$$\begin{aligned} \zeta _i = \sum _{\ell =1}^p \frac{ (\varvec{a}_{\ell }'\varvec{z}_i)^3-3 \varvec{a}_{\ell }'\varvec{a}_{\ell }\varvec{a}_{\ell }'\varvec{z}_i }{ \sqrt{N}} \quad (i=1,\ldots ,N). \end{aligned}$$

Since $\zeta _1,\ldots ,\zeta _N$ are i.i.d., it holds that

$$\begin{aligned} \mathrm {Var}(T_1) = \mathrm {Var} \left( \sum _{\ell =1}^p \frac{ (\varvec{a}_{\ell }'\varvec{z})^3-3\varvec{a}_{\ell }'\varvec{a}_{\ell }\varvec{a}_{\ell }'\varvec{z} }{\sqrt{N} } \right) , \end{aligned}$$

where $\varvec{z} \sim N_p(\varvec{0},\varvec{I}_p)$. We find that

$$\begin{aligned} E \left[ \sum _{\ell =1}^p \frac{ (\varvec{a}_{\ell }'\varvec{z})^3-3\varvec{a}_{\ell }'\varvec{a}_{\ell }\varvec{a}_{\ell }'\varvec{z} }{\sqrt{N} }\right] =0, \end{aligned}$$

and so

where the third equality follows from Lemma 1.

Since $E[T_2]=0$, we have

where the fourth equality follows from Lemma 1.

Since $E[T_3]=0$, we have

where the fourth equality follows from Lemma 1. $\square$

Derivation of the unbiasedness for T and V

In this section, we show that T and V, which are defined by (5) and (6), are unbiased estimators of $E[\{\odot ^3 (\varvec{x}-\varvec{\mu })\}'\varvec{1}_p]$ and $\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p$, respectively.

Lemma 3

Assume that $\varvec{x}_1$, ..., $\varvec{x}_N$ are i.i.d., and assume that these random vectors have the multivariate linear model (3). Then T and V are unbiased estimators of $E[\{\odot ^3 (\varvec{x}-\varvec{\mu })\}'\varvec{1}_p]$ and $\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p$, respectively.

Proof

If k, $\ell$ and $\alpha$ are all different, then it is easy to see that $E[t_{k \ell \alpha ,i}^3]=(3/4)E[(x_i-\mu _i)^3]$ for $(t_{k \ell \alpha ,1},\ldots ,t_{k \ell \alpha ,p})'=\varvec{t}_{k \ell \alpha }$, and for $\begin{pmatrix} x_1 \ldots x_p \end{pmatrix}'=\varvec{x}=\varvec{\mu }+\varvec{{\varSigma }}^{1/2}\varvec{z}$ with $\varvec{z} \sim F_p(\varvec{0},\varvec{I}_p)$. Adding up over the $\mathrm {P}_{N,3}$ such triples, multiplying by 4 / 3 and dividing $\mathrm {P}_{N,3}$ gives an unbiased estimate of $E[(x_i-\mu _i)^3]$. By adding up it over i, we obtain T. It is also easy to see that $E[v_{k \ell ,i j} v_{\alpha \beta ,i j} v_{\gamma \delta ,i j} ]=\sigma _{ij}^3$ for $(v_{k \ell ,i j})=\varvec{V}_{k \ell }$, which is defined in (6), and $(\sigma _{ij})=\varvec{{\varSigma }}$ if k, $\ell$, $\alpha$, $\beta$, $\gamma$ and $\delta$ are all different. Adding up over the $\mathrm {P}_{N,6}$ such sextuples and dividing $\mathrm {P}_{N,6}$ gives an unbiased estimate of $\sigma _{ij}^3$. By adding up it over (i, j), we obtain V. $\square$

Proof of Theorem 2

In this section, we give a proof of Theorem 2. Before proving it, we prepare three lemmas. The first two lemmas (Lemma 4 and Lemma 5) treat formula about multi-sum, and the last lemma (Lemma 6) treats the uniformly boundedness on $p \in {\mathbb {N}}$ for the expectation, which is used to prove Theorem 2. Since proofs of Lemma 4 and Lemma 5 are tedious but simple, we skipped here.

Lemma 4

The following equations hold:

Lemma 5

The following equations hold:

Lemma 6

Let

$$\begin{aligned} Q_{k\ell \alpha \beta \gamma \delta } =\sum _{i=1}^p \sum _{j=1}^p \varvec{a}_i'\varvec{z}_k \varvec{z}_{\ell } \varvec{a}_j \varvec{a}_i \varvec{z}_{\alpha } \varvec{z}_{\beta } \varvec{a}_j \varvec{a}_i' \varvec{z}_{\gamma } \varvec{z}_{\delta }' \varvec{a}_j, \end{aligned}$$

where $\varvec{{\varSigma }}^{1/2}=\varvec{A}=(\varvec{a}_1,\ldots ,\varvec{a}_p)'$, and $\varvec{z}_k$, $\varvec{z}_{\ell }$, $\varvec{z}_{\alpha }$, $\varvec{z}_{\beta }$, $\varvec{z}_{\gamma }$ and $\varvec{z}_{\delta }$ are i.i.d. as $N_p(\varvec{0},\varvec{I}_p)$. Then, there exist $M>0$, which does not depend on p, such that

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} \left[ E[Q_{112233}^2] \Bigg / \max \left\{ \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2, \left| \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js} \right| \right\} \right] < M, \end{aligned}$$

(18)

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} \left[ E[Q_{112234}^2] \Bigg / \max \left\{ \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2, \left| \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js} \right| \right\} \right] <M, \end{aligned}$$

(19)

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} \left[ E[Q_{112345}^2] \bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right] <M, \end{aligned}$$

(20)

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} \left[ E[Q_{123456}^2] \bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right] <M. \end{aligned}$$

(21)

In addition, under the assumption $\mathrm {C}$,

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \left[ E[Q_{112233}^2] \bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right]< M, \quad \sup _{p \in {\mathbb {N}}} \left[ E[Q_{112234}^2] \bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right] <M. \end{aligned}$$

Proof

Since (19), (20) and (21) can be proved by using the same way to show (18), we only show (18), and omit other three here. It can be expressed that

where the second equality follows from Lemma 1, the third equality from bottom follows from Lemma 4 and the second equality from bottom follows from Lemma 5. Since $\odot ^2 \varvec{{\varSigma }}$ is positive definite, we have

$$\begin{aligned} {\mathrm{tr}}\{(\odot ^2 \varvec{{\varSigma }})\varvec{{\varSigma }}\}^2 \le \{ {\mathrm{tr}}(\odot ^2 \varvec{{\varSigma }})\varvec{{\varSigma }} \}^2, \end{aligned}$$

where the right-hand side of the inequality can be written as $\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2$. From this result, it is found that

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \left[ E[Q_{112233}^2] \Bigg / \max \left\{ \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2, \left| \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js} \right| \right\} \right] < 27. \end{aligned}$$

Note that

$$\begin{aligned} \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js}&\le \sigma _{ij} \sigma _{st} (\sqrt{|\sigma _{is}|})^2 (\sqrt{|\sigma _{jt}|})^2 \sigma _{js} \sigma _{it}\\&\le \frac{ \sigma _{ij}^2 \sigma _{st}^2 |\sigma _{is}| |\sigma _{jt}| + \sigma _{sj}^2 \sigma _{ti}^2 |\sigma _{is}| |\sigma _{jt}| }{ 2 }, \end{aligned}$$

From this inequality, we have

$$\begin{aligned} \left| \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js} \right|&\le \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p |\sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js}|\\&\le \frac{1}{2} \left[ {\mathrm{tr}}\{ (\odot ^2 \varvec{{\varSigma }}_+)\varvec{{\varSigma }}_+\}^2 + {\mathrm{tr}}\{ (\odot ^2 \varvec{{\varSigma }}_+)\varvec{{\varSigma }}_+\}^2 \right] \\&\le \left[ \varvec{1}_p' (\odot ^3 \varvec{{\varSigma }}_+) \varvec{1}_p \right] ^2. \end{aligned}$$

Hence under the assumption $\mathrm {C3}$, there exist $M>0$, which does not depend on p, such that

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \left[ \left| \sum _{i=1}^p \sum _{j=1}^p \sum _{s=1}^p \sum _{t=1}^p \sigma _{ij} \sigma _{st} \sigma _{is} \sigma _{jt} \sigma _{it} \sigma _{js} \right| \Bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right] <M, \end{aligned}$$

and so

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \left[ E[Q_{112233}^2] \Bigg / \{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2 \right] <M. \end{aligned}$$

$\square$

Proof of Theorem 2

Since V is unbiased estimator of $\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p$, it is sufficient to show that

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \mathrm {Var}(V/\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p) \rightarrow 0 \end{aligned}$$

(22)

as $N \rightarrow \infty$. Since V is invariant under location shift, without loss of generality, we may assume that $\varvec{\mu }=\varvec{0}$. Each of A, B, C and D in (16) can be written as

From (16), we can express that

$$\begin{aligned}&\left( V/\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p-1\right) ^2\\&\quad =\left[ \left\{ \frac{1}{\mathrm {P}_{N,3}} \frac{A}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} -1 \right\} \right. \\&\qquad \left. -\frac{3}{\mathrm {P}_{N,4}} \frac{B}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} +\frac{3}{\mathrm {P}_{N,5}} \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} -\frac{1}{\mathrm {P}_{N,6}} \frac{D}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right] ^2 \\&\quad \le 4 \left[ \left\{ \frac{1}{\mathrm {P}_{N,3}} \frac{A}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} -1 \right\} ^2 +\left( \frac{3}{\mathrm {P}_{N,4}}\right) ^2 \left( \frac{B}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2\right. \\&\qquad \left. +\left( \frac{3}{\mathrm {P}_{N,5}}\right) ^2 \left( \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2 +\left( \frac{1}{\mathrm {P}_{N,6}}\right) ^2 \left( \frac{D}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2 \right] , \end{aligned}$$

where the inequality follows from Jensen inequality. Thus,

$$\begin{aligned}&\mathrm {Var}\left( \frac{ V }{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right) \nonumber \\&\quad \le 4 \left[ E \left[ \left\{ \frac{1}{\mathrm {P}_{N,3}} \frac{A}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} -1 \right\} ^2 \right] +\left( \frac{3}{\mathrm {P}_{N,4}}\right) ^2 E \left[ \left( \frac{B}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2 \right] \right. \nonumber \\&\qquad \left. +\,\left( \frac{3}{\mathrm {P}_{N,5}}\right) ^2 E \left[ \left( \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2 \right] +\left( \frac{1}{\mathrm {P}_{N,6}}\right) ^2 E \left[ \left( \frac{D}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p}\right) ^2 \right] \right] . \end{aligned}$$

(23)

For the first term in the right-hand side of the inequality (23), expanding $\{.\}^2$, and excluding the term whose expectation becomes 0, we have

(24)

where

$$\begin{aligned}&E[A_1] = \frac{E[Q_{112233}^2]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[A_2] = \frac{E[Q_{112233}Q_{112244}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[A_3] = \frac{E[Q_{112233}Q_{114455}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}. \end{aligned}$$

Using the same calculation method, the second-fourth terms in the right-hand side of the inequality (23) can be written as

$$\begin{aligned}&E \left[ \left\{ \frac{1}{\mathrm {P}_{N,4}} \frac{B}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] \nonumber \\&\quad = \left( \frac{1}{ \mathrm {P}_{N,4} } \right) ^2 \left[ 2 \mathrm {P}_{N,4} E[B_1] + 2 \mathrm {P}_{N,4} E[B_2] + 4 \mathrm {P}_{N,5} E[B_3] \right. \nonumber \\&\qquad \left. +\, 4 \mathrm {P}_{N,5} E[B_4] + 2 \mathrm {P}_{N,6} E[B_5] + 2 \mathrm {P}_{N,6} E[B_6] \right] , \end{aligned}$$

(25)

$$\begin{aligned}&E \left[ \left\{ \frac{1}{\mathrm {P}_{N,5}} \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] \nonumber \\&\quad = \left( \frac{1}{ \mathrm {P}_{N,5} } \right) ^2 \left[ 4 \mathrm {P}_{N,5} E[C_1] + 8 \mathrm {P}_{N,5} E[C_2] + 8 \mathrm {P}_{N,5} E[C_3] \right. \nonumber \\&\qquad +\, 4 \mathrm {P}_{N,5} E[C_4] + 4 \mathrm {P}_{N,6} E[C_5] + 8 \mathrm {P}_{N,6} E[C_6] + 8 \mathrm {P}_{N,6} E[C_7] \nonumber \\&\qquad \left. +\, 4 \mathrm {P}_{N,6} E[C_8]\right] , \end{aligned}$$

(26)

$$\begin{aligned}&E \left[ \left\{ \frac{1}{\mathrm {P}_{N,6}} \frac{D}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] \nonumber \\&\quad = \left( \frac{1}{ \mathrm {P}_{N,6} } \right) ^2 \left[ 36 \mathrm {P}_{N,6} E[D_1] + 324 \mathrm {P}_{N,6} E[D_2] + 324 \mathrm {P}_{N,6} E[D_3] \right. \nonumber \\&\qquad \left. +\, 36 \mathrm {P}_{N,6} E[D_4] \right] , \end{aligned}$$

(27)

where

$$\begin{aligned}&E[B_1] = \frac{E[Q_{112234}^2]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[B_2] = \frac{E[Q_{112234}Q_{112243}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[B_3] = \frac{E[Q_{112234}Q_{115534}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[B_4] = \frac{E[Q_{112234}Q_{115543}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[B_5] = \frac{E[Q_{112234}Q_{556634}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[B_6] = \frac{E[Q_{112234}Q_{556643}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[C_1] = \frac{E[Q_{112345}^2]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[C_2] = \frac{E[Q_{112345}Q_{112354}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[C_3] = \frac{E[Q_{112345}Q_{112435}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[C_4] = \frac{E[Q_{112345}Q_{113254}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[C_5] = \frac{E[Q_{112345}Q_{662345}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[C_6] = \frac{E[Q_{112345}Q_{662354}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[C_7] = \frac{E[Q_{112345}Q_{662435}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[C_8] = \frac{E[Q_{112345}Q_{663254}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[D_1] = \frac{E[Q_{123456}^2]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \quad E[D_2] = \frac{E[Q_{123456}Q_{123465}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[D_3] = \frac{E[Q_{123456}Q_{124365}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}, \\&E[D_4] = \frac{E[Q_{123456}Q_{214365}]}{\{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p\}^2}. \end{aligned}$$

From Lemma 6, there exist $M>0$, which does not depend on p, such that

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} E[A_1]< M, \quad \sup _{p \in {\mathbb {N}}} E[B_1]< M, \quad \sup _{p \in {\mathbb {N}}} E[C_1]< M, \quad \sup _{p \in {\mathbb {N}}} E[D_1] < M. \end{aligned}$$

It follows from Cauchy–Schwarz inequality that

$$\begin{aligned}&\left| E[A_j]\right| \le E[A_1] \qquad (j=2,3,4),\\&\left| E[B_j]\right| \le E[B_1] \qquad (j=2,\ldots ,6),\\&\left| E[C_j]\right| \le E[C_1] \qquad (j=2,\ldots ,8),\\&\left| E[D_j]\right| \le E[D_1] \qquad (j=2,3,4). \\ \end{aligned}$$

From them, we find that there exist $M>0$, which does not depend on p, such that

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} E[|A_j|]< M \qquad (j=2,3,4),\\&\sup _{p \in {\mathbb {N}}} E[|B_j|]< M \qquad (j=2,\ldots ,6),\\&\sup _{p \in {\mathbb {N}}} E[|C_j|]< M \qquad (j=2,\ldots ,8), \\&\sup _{p \in {\mathbb {N}}} E[|D_j|] < M \qquad (j=2,3,4). \end{aligned}$$

Combining these results with (24), (25), (26) and (27), it is found that

$$\begin{aligned}&\sup _{p \in {\mathbb {N}}} E \left[ \left\{ \frac{1}{\mathrm {P}_{N,3}} \frac{A}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} -1 \right\} ^2 \right] =O(N^{-1}), \\&\sup _{p \in {\mathbb {N}}} E \left[ \left\{ \frac{1}{\mathrm {P}_{N,4}} \frac{B}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] =O(N^{-2}), \\&\sup _{p \in {\mathbb {N}}} E \left[ \left\{ \frac{1}{\mathrm {P}_{N,5}} \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] =O(N^{-4}), \\&\sup _{p \in {\mathbb {N}}} E \left[ \left\{ \frac{1}{\mathrm {P}_{N,6}} \frac{C}{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right\} ^2 \right] =O(N^{-6}), \\ \end{aligned}$$

and so

$$\begin{aligned} \sup _{p \in {\mathbb {N}}} \mathrm {Var}\left( \frac{ V }{\varvec{1}_p'(\odot ^3 \varvec{{\varSigma }})\varvec{1}_p} \right) =O(N^{-1}), \end{aligned}$$

which converges to 0 as $N \rightarrow \infty$. $\square$