Functional CLT of eigenvectors for large sample covariance matrices

Abstract

In order to investigate property of the eigenvector matrix of sample covariance matrix $\mathbf {S}_n$, in this paper, we establish the central limit theorem of linear spectral statistics associated with a new form of empirical spectral distribution $H^{\mathbf {S}_n}$, based on eigenvectors and eigenvalues of sample covariance matrix $\mathbf {S}_n$. Using Bernstein polynomial approximations, we prove the central limit theorem for linear spectral statistics of $H^{\mathbf {S}_n}$, indexed by a set of functions with continuous third order derivatives over an interval including the support of Marcenko–Pastur law. This result provides further evidences to support the conjecture that the eigenmatrix of sample covariance matrix is asymptotically Haar distributed.

Appendix

Lemma 1

(Burkholder) (Lemma 2.2 of Bai and Silverstein 1998) Let $X_k$, $k=1,2,\ldots ,$ be a complex martingale difference sequence with respect to the increasing $\sigma$-fields $\mathcal {F}_k$. Then, for $p>1$,

$$\begin{aligned} \mathrm E \left| \sum X_k\right| ^p\le K_p\mathrm E \left( \sum |X_k|^2\right) ^{p/2}. \end{aligned}$$

where $K_p$ is a constant which depends upon $p$ only.

Lemma 2

(Lemma 2.6 of Silverstein and Bai 1995) Let $z\in \mathbb {C}^+$ with $v=\mathfrak {I}z$, $\mathbf {A}$ and $\mathbf {B}$ Hermitian and $\mathbf {r}\in \mathbb {C}^n$. Then

$$\begin{aligned} \left| \mathrm tr \left( (\mathbf {B}-z\mathbf {I})^{-1} -(\mathbf {B}+\mathbf {rr}^*-z\mathbf {I})^{-1}\right) \mathbf {A}\right| =\left| \frac{\mathbf {r}^*(\mathbf {B}-z\mathbf {I})^{-1}\mathbf {A} (\mathbf {B}-z\mathbf {I})^{-1}\mathbf {r}}{1+\mathbf {r}^*(\mathbf {B}-z\mathbf {I})^{-1}\mathbf {r}}\right| \le \frac{\Vert \mathbf {A}\Vert }{v}. \end{aligned}$$

Lemma 3

((1.15) in Bai and Silverstein 2004) Let $\mathbf {X}=(X_1,\ldots ,X_n)$, where $X_i$’s are i.i.d. complex random variables with mean zero and variance 1. Let $\mathbf {A}=(a_{ij})_{n\times n}$ and $\mathbf {B}=(b_{ij})_{n\times n}$ be complex matrices. Then the following identity holds:

$$\begin{aligned}&\mathrm E (\mathbf {X}^*\mathbf {AX}-\mathrm tr \mathbf {A}) (\mathbf {X}^*\mathbf {BX}-\mathrm tr \mathbf {B})\\&\qquad \qquad =(\mathrm E |X_1|^4-|\mathrm E X_1^2|^2-2)\sum \limits _{i=1}^na_{ii}b_{ii}+ |\mathrm E Y_1^2|^2\mathrm tr \mathbf {AB}^T+\mathrm tr \mathbf {AB}. \end{aligned}$$

Lemma 4

(Lemma 2.7 of Bai and Silverstein 1998) For $\mathbf {X}=(X_1,\ldots ,X_n)^{\prime }$ with i.i.d. standardized real or complex entries such that $\mathrm E X_i=0$ and $\mathrm E |X_i|^2=1$, and for $\mathbf {C}$ an $n\times n$ complex matrix, we have, for any $p\ge 2$,

$$\begin{aligned} \mathrm E |\mathbf {X}^*\mathbf {CX}-\mathrm tr \mathbf {C}|^p\le K_p\left[ \left( \mathrm E |X_i|^4\mathrm tr \mathbf {CC}^*\right) ^{p/2} +\mathrm E |X_i|^{2p}\mathrm tr (\mathbf {CC}^*)^{p/2}\right] . \end{aligned}$$

where $K_p$ is a constant which depends upon $p$ only.

Lemma 5

(Theorem 5.9 of Bai and Silverstein 2010) Suppose that the entries of the matrix $\mathbf {X}=(X_{jk},j\le n,k\le N)$ are independent (not necessarily identically distributed) and satisfy

• (1) $\hbox {E}X_{jk}=0$,

• (2) $|X_{jk}|\le \sqrt{N}\delta _N,$

• (3) $\max _{j,k}|\hbox {E}|X_{jk}|^2-\sigma ^2|\rightarrow 0$ as $N\rightarrow \infty$, and

• (4) $\hbox {E}|X_{jk}|^l\le b(\sqrt{N}\delta _N)^{l-3}$ for all $l\ge 3$,

where $\delta _N\rightarrow 0$ and $b>0$. Let $\mathbf {S}_n=\mathbf {X}\mathbf {X}^*/N$. Then, for any $x>\epsilon >0$ and integers $j,k \ge 2$, we have

$$\begin{aligned} \hbox {P}\big (\lambda _{\hbox {max}}(\mathbf {S}_n)\ge \sigma ^2(1+\sqrt{c})^2+x\big )\le KN^{-k}\big (\sigma ^2(1+\sqrt{c})^2+x-\epsilon \big )^{-k} \end{aligned}$$

for some constant $K>0$.

Lemma 6

(Lemma 8.20 in Bai and Silverstein 2010) If $|z|<A$, $v\ge CN^{-1/2}$ and $l \ge 1$, then

$$\begin{aligned} \hbox {E}\big |m_n(z)-\hbox {E}m_n(z)\big |^{2l} \le \frac{K}{N^{2l}v^{4l}c^{2l}}\left( \Delta +\frac{v}{v_c}\right) ^l, \end{aligned}$$

where A is a positive constant, $v_c=1-\sqrt{c}+\sqrt{v}$ and $\Delta \equiv \Vert \hbox {E}F^{S_n}-F_{c} \Vert$.

Lemma 7

(Lemma 9.1 in Bai and Silverstein 2010) Suppose that $X_i$, $i=1,\ldots , n$, are independent, with $\hbox {E}X_i=0$, $\hbox {E}|X_i|^2=1$, $\sup \hbox {E}|X_i|^4=\nu < \infty$ and $|X_i|\le \eta \sqrt{n}$ with $\eta >0$. Assume that $\mathbf {A}$ is a complex matrix. Then for any given $p$ such that $2\le p \le b \log (n\nu ^{-1}\eta ^4 )$ and $b>1$, we have

$$\begin{aligned} \hbox {E}\big |\mathbf {\alpha }^* \mathbf {A} \mathbf {\alpha } -\hbox {tr} (\mathbf {A})\big |^p \le \nu n^p(n\eta ^4)^{-1} \big (40b^2\Vert \mathbf {A}\Vert \eta ^2\big )^p, \end{aligned}$$

where $\mathbf {\alpha }=(X_1,\ldots , X_n)^T$.

Lemma 8

Under the conditions of Theorem 1, for all $|z|<A$ and $\mathfrak {I}(z)\ge \eta _N N^{-1/2}$, there exists a constant $K$, such that

$$\begin{aligned} |b_1(z)|\le K, \end{aligned}$$

where $A$ is a positive constant.

Proof

From (3.3.1) in Bai and Silverstein (2010),

$$\begin{aligned} m(z)=\frac{1-c-z+\sqrt{(1-c-z)^2-4zc}}{2cz}, \end{aligned}$$

where the square root of a complex number is defined as

$$\begin{aligned} \sqrt{z}=\frac{\mathfrak {I}z}{\sqrt{2(|z|-\mathfrak {R}z)}}+i\frac{|\mathfrak {I}z|}{\sqrt{2(|z|+\mathfrak {R}z)}}. \end{aligned}$$

It is easy to verify that

$$\begin{aligned} b_0(z)\triangleq \frac{1}{1+cm(z)}=1-\sqrt{c}m_\mathrm{semi }\left( \frac{z-c-1}{\sqrt{c}}\right) , \end{aligned}$$

where $m_\mathrm{semi }$denotes the Stieltjes transform of semicircular law. By the fact that $|m_\mathrm{semi }|\le 1$, we have

$$\begin{aligned} |b_0(z)|\le 1+\sqrt{c}. \end{aligned}$$

According to the relationship between $b_0(z)$ and $b_1(z)$,

$$\begin{aligned} b_1(z)-b_0(z)=-cb_1(z)b_0(z)\left( n^{-1}\mathrm Etr \mathbf {A}_1^{-1}(z)-m(z)\right) , \end{aligned}$$

we have

$$\begin{aligned} b_1(z)=\frac{b_0(z)}{1+cb_0(z)\left( n^{-1}\mathrm Etr \mathbf {A}_1^{-1}(z)-m(z)\right) }. \end{aligned}$$

For $\mathfrak {I}(z)\ge \eta _N N^{-1/2}$, from the fact that $|\mathrm Etr (\mathbf {A}_1^{-1}(z)-\mathbf {A}^{-1}(z))|\le \frac{1}{\mathfrak {I}(z)}$ and $m_n(z)=n^{-1}\mathrm tr \mathbf {A}^{-1}(z)$, we can choose $n$ large enough, such that

$$\begin{aligned} \frac{1}{n}\left| \mathrm Etr \left( \mathbf {A}_1^{-1}(z)\!-\!\mathbf {A}^{-1}(z)\right) \right| \!\le \!\frac{1}{n\mathfrak {I}(z)}\!\le \!\frac{1}{3(1+\sqrt{c})}, \quad \left| \mathrm E m_n(z)\!-\!m(z)\right| \le \frac{1}{3(1+\sqrt{c})}. \end{aligned}$$

Thus,

$$\begin{aligned} \left| \frac{1}{n}\mathrm Etr \mathbf {A}_1^{-1}(z)-m(z)\right|&\le \frac{1}{n}\left| \mathrm Etr \left( \mathbf {A}_1^{-1}(z)-\mathbf {A}^{-1}(z)\right) \right| +\left| \mathrm E m_n(z)-m(z)\right| \\&\le \frac{2}{3(1+\sqrt{c})}. \end{aligned}$$

Consequently, we obtain

$$\begin{aligned} |b_1(z)|\le 3(1+\sqrt{c})\le K. \end{aligned}$$

$\square$

Lemma 9

If $|b_1(z)|\le K$ and $v>O(N^{-1/2})$, then for any fixed $t>0$,

$$\begin{aligned} P\left( |\beta _1(z)|>2K\right) =o(N^{-t}). \end{aligned}$$

Proof

Note that if $|b_1(z)\tilde{\xi }_1(z)|\le 1/2$, then

$$\begin{aligned} |\beta _1(z)|=\frac{|b_1(z)|}{|1+b_1(z)\tilde{\xi }_1(z)|}\le \frac{|b_1(z)|}{1-|b_1(z)\tilde{\xi }_1(z)|}\le 2K. \end{aligned}$$

Thus,

$$\begin{aligned} P\left( |\beta _1(z)|>2K\right)&\le P\left( |b_1(z)\tilde{\xi }_1(z)|>\frac{1}{2}\right) \\&\le P\left( |\tilde{\xi }_1(z)|>\frac{1}{2K}\right) \\&\le (2K)^p\mathrm E |\tilde{\xi }_1(z)|^p. \end{aligned}$$

By the $C_r$-inequality, Lemmas 6 and 7, for some $\eta =\eta _NN^{-1/4}$ and $p\ge \log N$, we have

$$\begin{aligned} \hbox {E}\left| \tilde{\xi }_1(z)\right| ^p&\le \hbox {E}\left| \xi _1(z)\right| ^p +\hbox {E}\left| \frac{1}{N}\hbox {tr}\mathbf {A}_1^{-1}(z)-\frac{1}{N}\hbox {Etr}\mathbf {A}_1^{-1}(z)\right| ^p \\&\le K\big (N\eta _N^4N^{-1}\big )^{-1}\big (v^{-1}\eta _N^2N^{-1/2}\big )^p+\frac{K}{N^pv^{3p/2}v_c^{p/2}}\\&\le K\eta _N^{2p-4}\le K\eta _N^p. \end{aligned}$$

For any fixed $t>0$, when $N$ is large enough so that $\log \eta _N^{-1}>t+1$, it can be shown that

$$\begin{aligned} \hbox {E}|\xi _1(z)|^p&\le K e^{-p\log \eta _N^{-1}} \le K e^{-p(t+1)}\\&\le Ke^{-(t+1)\log N}=KN^{-t-1}\\&= o(N^{-t}). \end{aligned}$$

Similarly, we can show that $P\left( |\tilde{\beta }_1(z)|>2K\right) =o(N^{-t})$ and $\mathrm E |\xi _1(z)|^p=o(N^{-t})$, for $p\ge \log N$. Therefore, $\beta _1(z)$ and $\tilde{\beta }_1(z)$ are both bounded by a large constant almost surely by using Borel–Cantelli lemma. $\square$

Lemma 10

There exists a constant $K$, such that $\mathrm E \left| \mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {A}^{-1}(\bar{z})\mathbf {x}_n\right| \le K/v$.

Proof

In the paper, we assume that $c=\lim n/N$ is away from 1, according to (8.4.9) in Bai and Silverstein (2010) that $m(z)$ is bounded by a constant, thus we have $\mathrm E |m_n^H(z)|\le K$ for some constant $K$. Therefore,

$$\begin{aligned} \mathrm E \left| \mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {A}^{-1}(\bar{z})\mathbf {x}_n\right| =\mathrm E \left| v^{-1}\mathfrak {I}(\mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {x}_n)\right| \le K/v. \end{aligned}$$

$\square$

Lemma 11

$|\nabla _2|\le O_p(N^{-\epsilon _0})$, when $m=[N^{1/4+\epsilon _o}]$.

Proof

Using integral by parts, we get

$$\begin{aligned} |\nabla _2|&= \sqrt{N}\left| \int (f_m(x)-f(x))d(H^{S_n}(x)-F_{c_n}(x))\right| \\&= \sqrt{N}\left| \int \limits _{a_l}^{b_r}\left( f_m^{\prime }(x)-f^{\prime }(x)\right) \left( H^{S_n}(x)-F_{c_n}(x)\right) dx\right| ; \end{aligned}$$

Based on Theorem 1.6 in Xia et al. (2013), we know $\Vert H^{S_n}-F_{c_n}\Vert =O_p(N^{-1/4})$, thus

$$\begin{aligned} |\nabla _2|\le KN^{1/4}\int \limits _{a_0}^{b_0}\left| \tilde{f}_m^{\prime }(y)-\tilde{f}^{\prime }(y)\right| dy, \end{aligned}$$

where $a_0=La_l+t$, $b_0=Lb_r+t$, and $L=\frac{1-2\epsilon }{b_r-a_l}$, $t=\frac{(a_l+b_r)\epsilon -a_l}{b_r-a_l}$. For

$$\begin{aligned} \tilde{f}^{\prime }_m(y)&= \sum \limits _{k=0}^m{m \atopwithdelims ()k}y^k(1-y)^{m-k} \left( \frac{k}{y}-\frac{m-k}{1-y}\right) \tilde{f}\Big (\frac{k}{m}\Big )\\&= \sum \limits _{k=0}^m{m \atopwithdelims ()k}y^k(1-y)^{m-k} \frac{k-my}{y(1-y)} \tilde{f}\Big (\frac{k}{m}\Big ). \end{aligned}$$

Then using Taylor expansion, we obtain

$$\begin{aligned} \tilde{f}\Big (\frac{k}{m}\Big )=\tilde{f}(y)\!+\!\tilde{f}^{\prime }(y)\left( \frac{k}{m}-y\right) \!+\! \frac{1}{2}\tilde{f}^{\prime \prime }(y)\left( \frac{k}{m}-y\right) ^2\!+\! \frac{1}{6}\tilde{f}^{(3)}(\xi _{k,y})\left( \frac{k}{m}-y\right) ^3, \end{aligned}$$

where $\xi _{k,y}$ is a number between $k/m$ and $y$.

Substituting the above equality into $\tilde{f}^{\prime }_m(y)$ and again noticing $\sum \nolimits _{k=0}^m {m\atopwithdelims ()k}y^k(1-y)^{m-k}\left( \frac{k}{m}-y\right) =0$, we obtain

$$\begin{aligned}&\tilde{f}^{\prime }_m(y)-\tilde{f}^{\prime }(y)\\&\quad =\frac{1}{2}\sum \limits _{k=0}^m{m \atopwithdelims ()k}y^k(1-y)^{m-k} \frac{k-my}{y(1-y)}\\&\qquad \times \left[ \left( \frac{k}{m}-y\right) ^2\tilde{f}^{\prime \prime }(y)+\frac{1}{3}\left( \frac{k}{m}-y\right) ^3\tilde{f}^{(3)}(\xi _{k,y})\right] \\&\quad =O\left( \frac{1}{m}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} |\nabla _2|\le \frac{K}{m}N^{1/4}=O_p(N^{-\epsilon _0}). \end{aligned}$$

$\square$

Lemma 12

Under the condition of Theorem 1, we have

$$\begin{aligned} \max _{j}\left| \mathrm E _{j-1} \sum \limits _{i=1}^n\left( \left[ \mathrm E _j\mathbf {A}_j^{-1}(z_1)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z_1)\right] _{ii} \left[ \mathrm E _j\mathbf {A}_j^{-1}(z_2)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z_2)\right] _{ii}\right) \right| =o_p(1), \end{aligned}$$

uniformly in $\gamma _{mh}\cup \gamma _{mh}^{\prime }$, where the maximum is taken over all $1\le i\le n$ and $1\le j\le N$.

Proof

First, let $\mathbf {e}_i(1\le i\le n)$ be the $n$-vector whose $i$th element is one, the rest being zero and $\mathbf {e}_i^{\prime }$, the transpose of $\mathbf {e}_i$. Then by using Lemma 4 and Cauchy–Schwartz inequality, we obtain

$$\begin{aligned}&\mathrm E \left| \left[ \mathbf {A}_j^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\right] _{ii} -\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\right] _{ii}\right| \\&\quad =\mathrm E \left| \mathbf {e}_i^{\prime }\left( \mathbf {A}_j^{-1}(z)-\mathbf {A}^{-1}(z)\right) \mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\mathbf {e}_i\right| \quad \mathrm by (4) \\&\quad =\mathrm E \left| \beta _j(z)\mathbf {e}_i^{\prime }\mathbf {A}_j^{-1}(z)\mathbf {s}_j\mathbf {s}_j^*\mathbf {A}_j^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\mathbf {e}_i\right| \\&\quad \le \frac{K}{N}\left( E\left| \mathbf {X}_j^*\mathbf {A}_j^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\mathbf {e}_i\mathbf {e}_i^{\prime }\mathbf {A}_j^{-1}(z)\mathbf {X}_j -\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\mathbf {e}_i\mathbf {e}_i^{\prime }\mathbf {A}_j^{-2}(z)\mathbf {x}_n\right| ^{2}\right) ^{1/2}\\&\qquad +\frac{K}{N}\mathrm E \left| \mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\mathbf {e}_i\mathbf {e}_i^{\prime }\mathbf {A}_j^{-2}(z)\mathbf {x}_n\right| \\&\quad \le \frac{K}{Nv^2}. \end{aligned}$$

Similarly, we can show that

$$\begin{aligned} \mathrm E \left| \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z)\right] _{ii}-\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}\right| \le \frac{K}{Nv^2}. \end{aligned}$$

Second, by martingale inequality, for any $\epsilon >0$, we have

$$\begin{aligned}&P\left( \max _{i,j}\left| \mathrm E _j\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii} -\mathrm E \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}\right| >\epsilon \right) \\&\quad \le nP\left( \max _j \left| \mathrm E _j\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii} -\mathrm E \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}\right| >\epsilon \right) \\&\quad \le \frac{n}{\epsilon ^6}\mathrm E \left| \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}-\mathrm E \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}\right| ^{6}\\&\quad =\frac{n}{\epsilon ^6}\mathrm E \bigg |\sum \limits _{l=1}^N(\mathrm E _l-\mathrm E _{l-1})\bigg ( \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}-\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\\&\qquad +\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}-\left[ \mathbf {A}_l^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\bigg )\bigg |^6\\&\quad \le nK\mathrm E \bigg (\sum \limits _{l=1}^N\bigg |(\mathrm E _l-\mathrm E _{l-1})\bigg ( \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}-\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\\&\qquad +\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}-\left[ \mathbf {A}_l^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\bigg )\bigg |^2\bigg )^3\\&\quad \le nK\mathrm E \bigg (\sum \limits _{l=1}^N(\mathrm E _l-\mathrm E _{l-1}) \bigg |\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}-\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\bigg |^2\\&\qquad +\bigg |\left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}-\left[ \mathbf {A}_l^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_l^{-1}(z)\right] _{ii}\bigg |^2\bigg )^3\\&\quad \le nK\left( \sum \limits _{l=1}^N\frac{1}{N^2v^4}\right) ^3\le \frac{K}{N^2v^{12}}. \end{aligned}$$

Together with

$$\begin{aligned}&\mathrm E \left[ \mathbf {A}^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z)\right] _{ii}=\frac{1}{n}\mathrm E \mathbf {x}_n^*\mathbf {A}^{-2}(z)\mathbf {x}_n. \end{aligned}$$

Finally, we obtain

$$\begin{aligned}&\max _{j}\left| \mathrm E _{j-1} \left( \sum \limits _{i=1}^N\left[ \mathrm E _j\mathbf {A}_j^{-1}(z_1)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z_1)\right] _{ii} \left[ \mathrm E _j\mathbf {A}_j^{-1}(z_2)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_j^{-1}(z_2)\right] _{ii}\right) \right| \\&\quad =n \left| \mathrm E \left[ \mathbf {A}^{-1}(z_1)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z_1)\right] _{ii} \times \mathrm E \left[ \mathbf {A}^{-1}(z_2)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}^{-1}(z_2)\right] _{ii}\right| +o_p(1)\\&\quad =\frac{1}{n}\left| \mathrm E \mathbf {x}_n^*\mathbf {A}^{-2}(z_1)\mathbf {x}_n\right| \times \left| \mathrm E \mathbf {x}_n^*\mathbf {A}^{-2}(z_2)\mathbf {x}_n\right| +o_p(1)=o_p(1). \end{aligned}$$

The proof of Lemma 12 is complete. $\square$