Convergence rate of eigenvector empirical spectral distribution of large Wigner matrices

Abstract

In this paper, we adopt the eigenvector empirical spectral distribution (VESD) to investigate the limiting behavior of eigenvectors of a large dimensional Wigner matrix $\mathbf {W}_n.$ In particular, we derive the optimal bound for the rate of convergence of the expected VESD of $\mathbf{W}_n$ to the semicircle law, which is of order $O(n^{-1/2})$ under the assumption of having finite 10th moment. We further show that the convergence rates in probability and almost surely of the VESD are $O(n^{-1/4})$ and $O(n^{-1/6}),$ respectively, under finite eighth moment condition. Numerical studies demonstrate that the convergence rate does not depend on the choice of unit vector involved in the VESD function, and the best possible bound for the rate of convergence of the VESD is of order $O(n^{-1/2}).$

Appendices

Appendix 1

Lemma 8

Under the conditions of Theorem 1, when $z\in {\mathcal D},$ there exists a constant C,  such that for any $l\ge 1,$

$$\begin{aligned} \text {E}\left| \xi _k\right| ^{2l}\le \left\{ \begin{array}{l@{\quad }l}\dfrac{C}{n^lv^l}\left( 1+\dfrac{\Delta }{v}\right) ^l, &{} \text{ for } \, l\le 3,\\ \dfrac{C}{n^{l+1}v^{2l-1}}\left( 1+\dfrac{\Delta }{v}\right) , &{} \text{ for }\, l\ge 4,\end{array}\right. \end{aligned}$$

where $\Delta =\Vert EF^{\mathbf{W}_n}-F\Vert .$

Proof

For Stieltjes transform $s_n(z),$ integration by parts yields

$$\begin{aligned} \left| Es_n(z)-s(z)\right| \le \pi \Delta /v. \end{aligned}$$

Combined with the fact $|s(z)|\le 1,$ implies that $|Es_n(z)|\le C(1+\Delta /v).$ Under finite eighth moment assumption, we have $E|X_{jk}|^{4l}\le Cn^{l-2}.$ Thus by Lemma 16, it follows that

$$\begin{aligned} E\left| \xi _k\right| ^{2l}\le & {} \dfrac{C}{n^{2l}}\left[ \left( Etr\mathbf{A}_k^{-1}(z)\mathbf{A}_k^{-1}(\bar{z})\right) ^l +n^{l-2}Etr\left( \mathbf{A}_k^{-1}(z)\mathbf{A}_k^{-1}(\bar{z})\right) ^l \right] \\= & {} \dfrac{C}{n^{2l}}\left[ \left( v^{-1}E\left( \mathfrak {I}tr\mathbf{A}_k^{-1}(z)\right) \right) ^l +n^{l-2}Etr\left( \mathbf{A}_k^{-1}(z)\mathbf{A}_k^{-1}(\bar{z})\right) ^l \right] \\\le & {} \dfrac{C}{n^{2l}}\left[ \dfrac{n^l}{v^l}\left| E\mathfrak {I}\left( s_n(z)\right) \right| ^l +\dfrac{n^{l-1}}{v^{2l-1}} E\left| \mathfrak {I}\left( s_n(z)\right) \right| \right] \\\le & {} \dfrac{C}{n^lv^l}\left( 1+\Delta /v\right) ^l+\dfrac{C}{n^{l+1}v^{2l-1}}(1+\Delta /v). \end{aligned}$$

$\square$

Lemma 9

Under the conditions of Theorem 1, for all $z\in {\mathcal D},$ there exists a large constant $C>0,$ such that for any $t>0$ and for all $k=1,\ldots ,n,$ we have

$$\begin{aligned} P\left( \left| z^{-1}\gamma _n\right|>C\right) =o\left( n^{-t}\right) ,\quad P\left( \left| z^{-1}\alpha _k\right| >C\right) =o\left( n^{-t}\right) . \end{aligned}$$

Proof

Recall that

$$\begin{aligned} \alpha _k=\dfrac{1}{1+z^{-1}\mathbf {w}_k^*\mathbf {A}_k^{-1}\mathbf {w}_k}, \quad \gamma _k=\dfrac{1}{1+z^{-1}n^{-1}(\text {tr}\mathbf {A}_k^{-1}+z^{-1})}. \end{aligned}$$

Define

$$\begin{aligned} b_n:= \dfrac{1}{1+z^{-1}n^{-1}\text {Etr}\mathbf {A}^{-1}(z)}, \quad \beta _n:=\dfrac{1}{1+z^{-1}n^{-1}\text {tr}\mathbf {A}^{-1}(z)}. \end{aligned}$$

First if we suppose that for any fixed $t>0,$ there exists a large constant $C>0,$ such that the following equation holds,

$$\begin{aligned} P\left( \left| z^{-1}\beta _n\right| >C\right) =o\left( n^{-t}\right) . \end{aligned}$$

(26)

Then from

$$\begin{aligned} \beta _n-\gamma _k=\beta _n\gamma _kz^{-1}n^{-1}\left( tr\mathbf{A}_k^{-1}(z)-tr\mathbf{A}^{-1}(z)+z^{-1}\right) , \end{aligned}$$

(27)

we have

$$\begin{aligned} \left| z^{-1}\gamma _k\right|= & {} \dfrac{|z^{-1}\beta _n|}{|1+z^{-1}\beta _nn^{-1}(tr(\mathbf{A}_k^{-1}(z)-\mathbf{A}^{-1}(z))+z^{-1})|}\\\le & {} \dfrac{|z^{-1}\beta _n|}{1-2|z^{-1}\beta _n|(nv)^{-1}}, \end{aligned}$$

where the last inequality comes from the facts that $|tr(\mathbf{A}_k^{-1}(z)-\mathbf{A}^{-1}(z))|\le v^{-1}$ and $|z^{-1}|\le v^{-1}.$ If (26) is true, for $v\ge O(n^{-1/2}),$ we can choose n large enough such that $2|z^{-1}\beta _n|(nv)^{-1}\le 1/2.$ Then $|z^{-1}\gamma _k|\le 2|z^{-1}\beta _n|,$ which implies that $P(|z^{-1}\gamma _k|>C)=o(n^{-t})$ holds.

Similarly, consider $z^{-1}\alpha _k,$ if $|z^{-1}\gamma _k||\xi _k|\le 1/2$ holds, then we have

$$\begin{aligned} \left| z^{-1}\alpha _k\right| =\dfrac{|z^{-1}\gamma _k|}{|1+z^{-1}\gamma _k\xi _k|} \le \dfrac{|z^{-1}\gamma _k|}{1-|z^{-1}\gamma _k||\xi _k|} \le 2\left| z^{-1}\gamma _k\right| . \end{aligned}$$

That is, for any $p\ge 1,$

$$\begin{aligned} P\left( \left| z^{-1}\alpha _k\right|>C\right)\le & {} P\left( \left| \xi _k\right| >1/(2C)\right) \le { CE}\left| \xi _k\right| ^p. \end{aligned}$$

Since $|X_{jk}|\le \varepsilon _n n^{1/4},$ choose $\eta =\varepsilon _nn^{-1/4}$ in Lemma 20, for $p\ge \log n,$

$$\begin{aligned} \text {E}\left| \xi _k\right| ^p\le & {} C\left( n\varepsilon _n^4n^{-1}\right) ^{-1}\left( v^{-1}\varepsilon _n^2n^{-1/2}\right) ^p \le C\varepsilon _n^{2p-4}\le C\varepsilon _n^p. \end{aligned}$$

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

$$\begin{aligned} \text {E}\left| \xi _k\right| ^p\le & {} C\exp \left\{ -p\log \varepsilon _n^{-1}\right\} \le C\exp \{-p(t+1)\}\\\le & {} C\exp \{-(t+1)\log n\}\\= & {} Cn^{-t-1}=o\left( n^{-t}\right) . \end{aligned}$$

This shows that Lemma 9 is true if we can prove Eq. (26).

From (8.1.13) and (8.1.18) in Bai and Silverstein (2010), we can see that

$$\begin{aligned} \left| z^{-1}b_n\right| =\dfrac{1}{|z+Es_n(z)|}=|-s(z+\delta )|, \end{aligned}$$

which implies that $|z^{-1}b_n|\le 1.$ Further from

$$\begin{aligned} b_n-\beta _n=b_n\beta _nz^{-1}n^{-1} \left( \text {tr}\mathbf {A}^{-1}(z)-\text {Etr}\mathbf {A}^{-1}(z)\right) =b_n\beta _nz^{-1}\left( s_n(z)-\text {E}s_n(z)\right) , \end{aligned}$$

we obtain

$$\begin{aligned} \left| z^{-1}\beta _n\right|= & {} \dfrac{|b_n||z^{-1}b_n|}{|1+z^{-1}b_n(s_n(z)-\text {E}s_n(z))|}\\\le & {} \dfrac{|z^{-1}b_n|}{1-|z^{-1}b_n||s_n(z)-\text {E}s_n(z)|}\\\le & {} 2\left| z^{-1}b_n\right| \le 2, \end{aligned}$$

if $|s_n(z)-Es_n(z)|\le 1/2.$ Therefore, by Lemma 19, for any $l> \max \{1,\,2t\}$ and $v\ge O(n^{-1/2}),$ we have

$$\begin{aligned} P\left( \left| z^{-1}\beta _n\right| \ge 2\right)\le & {} P\left( \left| s_n(z)-Es_n(z)\right| \ge 1/2\right) \\\le & {} { CE}\left| s_n(z)-Es_n(z)\right| ^{2l} \\\le & {} \dfrac{C}{n^{2l}v^{3l}}=o\left( n^{-t}\right) . \end{aligned}$$

This completes the proof. $\square$

Lemma 10

Under the conditions of Theorem 1, for any $l\ge 1$ and $z\in {\mathcal D},$ there exists a constant $C>0,$ such that

$$\begin{aligned} \text {E}\left| z^{-1}\left( \gamma _k+zs(z)\right) \right| ^{2l}\le \dfrac{C}{(nv)^{2l}}. \end{aligned}$$

Proof

By (27), (26), (18) and Lemmas 19 and 18, it follows that

$$\begin{aligned}&\text {E}\left| z^{-1}\left( \gamma _k+zs(z)\right) \right| ^{2l} \le { CE}\left| z^{-1}\left( \gamma _k-\beta _n\right) \right| ^{2l} +{ CE}\left| z^{-1}\left( \beta _n+zs(z)\right) \right| ^{2l}\\&\quad ={ CE}\left| z^{-2}\gamma _k\beta _n n^{-1}\left( \text {tr}\mathbf {A}_k^{-1}(z)+z^{-1}-\text {tr}\mathbf {A}^{-1}(z)\right) \right| ^{2l} +{ CE}\left| z^{-1}\beta _n-\dfrac{1}{z+s(z)}\right| ^{2l}\\&\quad \le \dfrac{C}{(nv)^{2l}}+{ CE}\left| \dfrac{z^{-1}\beta _n}{z+s(z)} \left( \dfrac{1}{n}\text {tr}\mathbf {A}^{-1}(z)-s(z)\right) \right| ^{2l}\\&\quad \le \dfrac{C}{(nv)^{2l}}+{ CE}\left| s_n(z)-\text {E}s_n(z)\right| ^{2l}+C\left| \text {E}s_n(z)-s(z)\right| ^{2l}\\&\quad \le \dfrac{C}{(nv)^{2l}}+\dfrac{C}{n^{2l}v^{3l}}+\dfrac{C}{n^l}\\&\quad \le \dfrac{C}{(nv)^{2l}}. \end{aligned}$$

$\square$

Lemma 11

Under the conditions of Theorem 1, for $z\in {\mathcal D},$ we have

$$\begin{aligned} \text {E}\left| s_n^H(z)-\text {E}s_n^H(z)\right| ^{2l}\le \dfrac{C}{n^lv^{2l}}\left( 1+\dfrac{\Delta ^H}{v}\right) ^{2l} +\dfrac{C}{n^{l/2+2}v^l}\left( 1+\dfrac{\Delta ^H}{v}\right) ^l. \end{aligned}$$

Proof

Write $\mathbf{x}_n^*\mathbf{A}^{-1}(z)\mathbf{x}_n-E\mathbf{x}_n^*\mathbf{A}^{-1}(z)\mathbf{x}_n$ as the sum of a martingale difference sequence. Denote $E_k(\cdot )$ as the conditional expectation given $\{X_{ij},\,k<i,\,j\le n\}.$

$$\begin{aligned}&\mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {x}_n-E\mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {x}_n\\&\quad =\sum _{k=1}^n\left( E_{k-1}-E_k\right) \mathbf{x}_n^*\mathbf{A}^{-1}(z)\mathbf{x}_n\\&\quad =\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\left( \mathbf {A}^{-1}(z)-\mathbf {A}_k^{-1}(z)\right) \mathbf {x}_n\\&\quad =\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}^{-1}(z)\left( \mathbf {A}_k(z)-\mathbf {A}(z)\right) \mathbf {A}_k^{-1}(z)\mathbf {x}_n\\&\quad ={-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}^{-1}(z) \left( \mathbf {w}_k\mathbf {e}_k^*+\mathbf {e}_k\mathbf {w}_k^*\right) \mathbf {A}_k^{-1}(z)\mathbf {x}_n\\&\quad :=\phi _{n1}+\phi _{n2}. \end{aligned}$$

By (14) and (12), we further decompose $\phi _{n1}$ as

$$\begin{aligned} \phi _{n1}= & {} {-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {w}_k\mathbf {e}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\\= & {} z^{-1}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) x_k\mathbf {x}_n^*\mathbf {A}_{k1}^{-1}(z)\mathbf {w}_k\alpha _k\\= & {} z^{-1}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) x_k \left( \mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\alpha _k -\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\alpha _k\right) \\= & {} z^{-1}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) x_k\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\alpha _k\\&+z^{-2}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \left| x_k\right| ^2\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\alpha _k\\= & {} z^{-1}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) x_k\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\alpha _k -z^{-1}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \left| x_k\right| ^2\alpha _k\\:= & {} \phi _{n11}+\phi _{n12}. \end{aligned}$$

Similar to $\phi _{n1},$ we obtain

$$\begin{aligned} \phi _{n2}= & {} {-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}^{-1}(z)\mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\\= & {} {-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}_{k2}^{-1}(z)\mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\= & {} {-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^* \left( \mathbf {A}_k^{-1}(z)-\mathbf {A}_k^{-1}(z)\mathbf {w}_k\mathbf {e}_k^*\mathbf {A}_k^{-1}(z)\right) \mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\= & {} {-}\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\&+\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\mathbf {e}_k^*\mathbf {A}_k^{-1}(z)\mathbf {e}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\= & {} \sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-1}\bar{x}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\&-\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-1}\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\alpha _k\\:= & {} \phi _{n21}+\phi _{n22}. \end{aligned}$$

Notice that $\phi _{nij},\,i,\,j=1,\,2$ are martingale difference sequence, for any $l\ge 1,$ we conclude from Lemmas 13(b) and 16 that

$$\begin{aligned} E\left| \phi _{n11}\right| ^{2l}\le & {} { CE} \left( \sum _{k=1}^nE_k\left| \left( E_{k-1}-E_k\right) x_k \mathbf{x}_n^*\mathbf{A}_k^{-1}(z)\mathbf{w}_kz^{-1}\alpha _k\right| ^2\right) ^l\\&+C\sum _{k=1}^nE\left| \left( E_{k-1}-E_k\right) x_k \mathbf{x}_n^*\mathbf{A}_k^{-1}(z)\mathbf{w}_kz^{-1}\alpha _k\right| ^{2l}\\\le & {} { CE}\left( \sum _{k=1}^n\left| x_k\right| ^2E_k\left| \mathbf{x}_n^*\mathbf{A}_k^{-1}(z)\mathbf{w}_k\right| ^2\right) ^l +C\sum _{k=1}^n\left| x_k\right| ^{2l}E\left| \mathbf{x}_n^*\mathbf{A}_k^{-1}(z)\mathbf{w}_k\right| ^{2l} \\\le & {} \dfrac{C}{n^lv^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l+\dfrac{C}{n^{l/2+2}v^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l. \end{aligned}$$

As in dealing with $E|\phi _{n11}|^{2l},$ similarly we obtain

$$\begin{aligned} E\left| \phi _{n21}\right| ^{2l}\le \dfrac{C}{n^lv^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l+\dfrac{C}{n^{l/2+2}v^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l. \end{aligned}$$

Since $\alpha _k=\gamma _k-z^{-1}\alpha _k\gamma _k\xi _k$ and $(E_{k-1}-E_k)\gamma _k=0,$ we have

$$\begin{aligned} \phi _{n12}=z^{-2}\sum _{k=1}^n\left( E_{k-1}-E_k\right) \left| x_k\right| ^2\alpha _k\gamma _k\xi _k. \end{aligned}$$

Again, by Lemmas 13(b) and 9, it follows that

$$\begin{aligned} E\left| \phi _{n12}\right| ^{2l}\le & {} { CE} \left( \sum _{k=1}^n\left| x_k\right| ^4E_k\left| \xi _k\right| ^2\right) ^l+C\sum _{k=1}^n\left| x_k\right| ^{4l}E\left| \xi _k\right| ^{2l}\\\le & {} \dfrac{C}{n^lv^l}\left( 1+\Delta /v\right) ^l+\dfrac{C}{n^{l+1}v^{2l-1}}(1+\Delta /v). \end{aligned}$$

Now, we turn to term $\phi _{n22}.$ Substitute $\alpha _k=\gamma _k-z^{-1}\alpha _k\gamma _k\xi _k$ into $\phi _{n22},$ and note that

$$\begin{aligned}&\left( \text {E}_{k-1}-\text {E}_k\right) \mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\gamma _k\\&\quad =\text {E}_{k-1}\gamma _k\left( \mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k -n^{-1}\mathbf {x}_n^*\mathbf {A}_k^{-1}(z) \left( \mathbf {I}-\mathbf {e}_k\mathbf {e}_k^*\right) \mathbf {A}_k^{-1}(z)\mathbf {x}_n \right) \\&\quad =\text {E}_{k-1}\gamma _k\left( \mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k -n^{-1}\left( \mathbf {x}_n^*\mathbf {A}_k^{-2}(z)\mathbf {x}_n -\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {e}_k\mathbf {e}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\right) \right) \\&\quad =\text {E}_{k-1}\gamma _k\left( \mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k -n^{-1}\left( \mathbf {x}_n^*\mathbf {A}_k^{-2}(z)\mathbf {x}_n-z^{-2}\left| x_k\right| ^2\right) \right) \\&\quad =\text {E}_{k-1}\gamma _k\eta _k. \end{aligned}$$

We further decompose $\phi _{n22}$ as

$$\begin{aligned} \phi _{n22}= & {} {-}\sum _{k=1}^nz^{-1}\text {E}_{k-1}\gamma _k\eta _k +\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-2} \left( \mathbf {w}_k^*\mathbf {A}_k^{-1}(z)\mathbf {x}_n\mathbf {x}_n^*\mathbf {A}_k^{-1}(z)\mathbf {w}_k\right) \alpha _k\gamma _k\xi _k\\&-\left( \sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-2}\dfrac{1}{n} \left( \mathbf {x}_n^*\mathbf {A}_k^{-2}(z)\mathbf {x}_n-z^{-2}\left| x_k\right| ^2\right) \alpha _k\gamma _k\xi _k\right. \\&\left. -\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-2}\dfrac{1}{n} \left( \mathbf {x}_n^*\mathbf {A}_k^{-2}(z)\mathbf {x}_n-z^{-2}\left| x_k\right| ^2\right) \alpha _k\gamma _k\xi _k\right) \\= & {} {-}\sum _{k=1}^nz^{-1}\text {E}_{k-1}\gamma _k\eta _k +\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-2}\eta _k\alpha _k\gamma _k\xi _k\\&+\sum _{k=1}^n\left( \text {E}_{k-1}-\text {E}_k\right) z^{-2}\dfrac{1}{n} \left( \mathbf {x}_n^*\mathbf {A}_k^{-2}(z)\mathbf {x}_n-z^{-2}\left| x_k\right| ^2\right) \alpha _k\gamma _k\xi _k\\:= & {} \phi _{n221}+\phi _{n222}+\phi _{n223}. \end{aligned}$$

By Lemma 16, it is easy to verify that

$$\begin{aligned} E\left| \eta _k\right| ^{2l}\le \dfrac{C}{n^{2l}v^{2l}}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^{2l} +\dfrac{C}{n^{l+2}v^{2l}}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^{2l}. \end{aligned}$$

Therefore, by Lemma 13(b), it leads to

$$\begin{aligned} E\left| \phi _{n22j}\right| ^{2l}\le \dfrac{C}{n^lv^{2l}}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^{2l}, \quad \mathrm{for}\,j=1,\,2,\,3. \end{aligned}$$

Combined the above results together, we obtain

$$\begin{aligned} E\left| s_n^H(z)-Es_n^H(z)\right| ^{2l}\le \dfrac{C}{n^{l/2+2}v^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l +\dfrac{C}{n^lv^{2l}}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^{2l}. \end{aligned}$$

Note that

$$\begin{aligned} E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^{2l} \le { CE}\left| s_n^H(z)-Es_n^H(z)\right| ^{2l}+C\left( 1+\Delta ^H/v\right) ^{2l}. \end{aligned}$$

For $z\in {\mathcal D},$ that is $v\ge C_0n^{-1/2},$ we can choose $C_0$ large enough, such that $\dfrac{C}{n^lv^{2l}}\le 2/3.$ Therefore,

$$\begin{aligned} E\left| s_n^H(z)-Es_n^H(z)\right| ^{2l} \le \dfrac{C}{n^{l/2+2}v^l}E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^l+\dfrac{C}{n^lv^{2l}}\left( 1+\Delta /v\right) ^{2l}. \end{aligned}$$

(28)

Next, we use induction to treat the term $E(\mathfrak {I}(s_n^H(z)))^l$ for $l>1.$

When $1/2<l<1,$ applying Lemma 13(a) to $E|\phi _{nij}|^{2l},$ we have

$$\begin{aligned} E\left| s_n^H(z)-Es_n^H(z)\right| ^{2l}\le & {} C\sum _{i,j=1}^2E\left| \phi _{nij}\right| ^{2l}\\\le & {} \dfrac{C}{n^lv^{2l}}\left( 1+\Delta ^H/v\right) ^l<C. \end{aligned}$$

This shows that for $1<p<2,$

$$\begin{aligned} E\left( \mathfrak {I}\left( s_n^H(z)\right) \right) ^p\le & {} 2E\left| s_n^H(z)-Es_n^H(z)\right| ^p+2\left| Es_n^H(z)\right| ^p\\\le & {} C\left( 1+\left( 1+\Delta ^H/v\right) ^p\right) . \end{aligned}$$

Thus the lemma follows for $1<l<2$ by substituting the result of $E(\mathfrak {I}(s_n^H(z)))^p$ into (28). Therefore, the lemma holds by using induction and (28). This completes the proof of lemma 11. $\square$

Lemma 12

Under the conditions in Theorem 1, for any fixed $t>0,$

$$\begin{aligned} \int _3^{\infty } \left| { EH}^{\mathbf{W}_n}(x)-F(x)\right| dx=o\left( n^{-t}\right) . \end{aligned}$$

Proof

For any fixed $t>0$ and $x>0,$ by Lemma 17, it follows that

$$\begin{aligned} P\left( \lambda _{\max }\left( \mathbf{W}_n\right) \ge 3+x\right) \le Cn^{-(t+1)}(3+x)^{-2}. \end{aligned}$$

Note that

$$\begin{aligned} 1-H^{\mathbf{W}_n}(x)\le {\mathbf {I}}\left( \lambda _{\max }\left( \mathbf{W}_n\right) \ge x\right) ,\quad \mathrm{for}\, x\ge 0. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \int _3^{\infty }\left| { EH}^{\mathbf{W}_n}(x)-F(x)\right| dx\le & {} \int _3^{\infty }P\left( \lambda _{\max }\left( \mathbf{W}_n\right) \ge x\right) dx \\\le & {} \int _0^{\infty }Cn^{-(t+1)}(3+x)^{-2}dx \\= & {} o\left( n^{-t}\right) . \end{aligned}$$

$\square$

Appendix 2

Lemma 13

(Burkholder Inequalities) Let $\{X_k\}$ be a complex martingale difference sequence with respect to the increasing $\sigma$-field $\{\mathcal {F}_k\},$ and let $\text{ E }_k$ denote the conditional expectation with respect to $\mathcal {F}_k.$ Then we have

(a) for $p>1,$

$$\begin{aligned} \text{ E } \left| \sum _{k=1}^n X_k \right| ^p \le K_p \text{ E } \left( \sum _{k=1}^n \left| X_k\right| ^2 \right) ^{p/2}; \end{aligned}$$

(b) for $p \ge 2,$

$$\begin{aligned} \text{ E }\left| \sum _{k=1}^n X_k\right| ^p \le K_p \left( \text{ E }\left( \sum _{k=1}^n \text{ E }_{k-1}\left| X_k\right| ^2 \right) ^{p/2}+\text{ E }\sum _{k=1}^n \left| X_k\right| ^p\right) , \end{aligned}$$

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

Lemma 14

(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

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

Lemma 15

((1.15) in Bai and Silverstein 2004) Let $\mathbf {Y}=(Y_1,\ldots ,Y_n),$ where $Y_i$’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:

$$\begin{aligned}&\text {E}(\mathbf {Y}^*\mathbf {AY}-\text {tr}\mathbf {A}) (\mathbf {Y}^*\mathbf {BY}-\text {tr}\mathbf {B})\\&\quad =\left( \text {E}\left| Y_1\right| ^4-\left| \text {E}Y_1^2\right| ^2-2\right) \sum _{i=1}^na_{ii}b_{ii}+ \left| \text {E}Y_1^2\right| ^2\text {tr}\mathbf {AB}^T+\text {tr}\mathbf {AB}. \end{aligned}$$

Lemma 16

(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 $\text {E}X_i=0$ and $\text {E}|X_i|^2=1,$ and for $\mathbf {C}$ an $n\times n$ complex matrix, we have, for any $p\ge 2,$

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

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

Lemma 17

((5.1.16) of Bai and Silverstein 2010) Suppose that the entries of the matrix $\mathbf {W}_n=\dfrac{1}{\sqrt{n}}(X_{ij})$ are independent (but depend on n) and satisfy

• (1) $\text{ E }X_{jk}=0,$

• (2) $\text {E}|X_{jk}|^2\le \sigma ^2,$

• (3) $\text{ E }|X_{jk}|^l\le b(\delta _n\sqrt{n})^{l-3}$ for all $l\ge 3,$

where $\delta _n\rightarrow 0$ and $b>0.$ Then, for fixed $\varepsilon >0$ and $x>0,$

$$\begin{aligned} \text{ P }\left( \lambda _{\text{ max }}\left( \mathbf {W}_n\right) \ge 2\sigma +\varepsilon +x\right) \le o\left( n^{-l}(2\sigma +\varepsilon +x)^{-2}\right) . \end{aligned}$$

Lemma 18

(Theorem 8.6 in Bai and Silverstein 2010) Under assumptions in Theorem 1, we have

$$\begin{aligned} \Delta =\left\| \text {E}F^{W_n}-F\right\| =O\left( n^{-1/2}\right) . \end{aligned}$$

Lemma 19

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

$$\begin{aligned} \text {E}\left| s_n(z)-\text {E}s_n(z)\right| ^{2l}\le Cn^{-2l}v^{-4l}(\Delta +v)^l, \end{aligned}$$

where $\Delta =\Vert \text {E}F^{W_n}-F\Vert .$

Lemma 20

(Lemma 9.1 in Bai and Silverstein 2010) Suppose that $X_i,\,i=1,\ldots , n,$ are independent, with $\text{ E }X_i=0,\,\text{ E }|X_i|^2=1,\,\sup \text{ E }|X_i|^4=\nu < \infty$ 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

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

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

Truncation part

We shall prove that the underlying random variables satisfy $X_{ii}=0$ and for $i\ne j,\,|X_{ij}|\le \varepsilon _nn^{1/4},\, \text {E}X_{ij}=0$ and $Var(X_{ij})=1,$ where $\varepsilon _n$ is a sequence decreasing to 0 and $\varepsilon _nn^{1/4}$ increasing to infinity.

1.1 Truncation

Under finite 10th moment condition, we can choose $\varepsilon _n$ decreasing to 0 and $\varepsilon _nn^{1/4}$ increasing to infinity as $n\rightarrow \infty ,$ such that

$$\begin{aligned} \varepsilon _n^{-10}\text {E}\left| X_{12}\right| ^{10}\text {I}\left( \left| X_{12}\right| >\varepsilon _nn^{1/4}\right) \rightarrow 0. \end{aligned}$$

Let $\widehat{\mathbf {X}}_n$ denote the truncated matrix whose entry on the ith row and jth column is $X_{ij}\text {I}(|X_{ij}|\le \varepsilon _nn^{1/4}),\, i=1,\ldots ,n,\,j=1,\ldots ,n.$ Define $\widehat{\mathbf {W}}_n=\dfrac{1}{\sqrt{n}}\widehat{\mathbf {X}}_n.$

Step 1(a). Truncation for Theorem1.

$$\begin{aligned} P\left( \mathbf {W}_n\ne \widehat{\mathbf {W}}_n\right)\le & {} n^2P\left( \left| X_{ij}\right|>\varepsilon _nn^{1/4}\right) \\\le & {} n^2\varepsilon _n^{-10}n^{-5/2}\text {E}\left| X_{12}\right| ^{10}\text {I}\left( \left| X_{12}\right| >\varepsilon _nn^{1/4}\right) \\= & {} o\left( n^{-1/2}\right) . \end{aligned}$$

Step 1(b). Truncation for Theorems2and3.

$$\begin{aligned} P\left( \mathbf {W}_n\ne \widehat{\mathbf {W}}_n,\,i.o.\right)= & {} \lim _{k\rightarrow \infty } P\left( \mathop \cup \limits _{n=k}^\infty \mathop \cup \limits _{i=1}^n \mathop \cup \limits _{j=1}^n \left| X_{ij}\right|>\varepsilon _nn^{1/4}\right) \\= & {} \lim _{k\rightarrow \infty }P\left( \mathop \cup \limits _{t=k}^\infty \mathop \cup \limits _{n\in [2^t,2^{t+1})} \mathop \cup \limits _{i=1}^n \mathop \cup \limits _{j=1}^n\left| X_{ij}\right|>\varepsilon _nn^{1/4}\right) \\\le & {} \lim _{k\rightarrow \infty }\sum _{t=k}^\infty P\left( \mathop \cup \limits _{i=1}^{2^{t+1}} \mathop \cup \limits _{j=1}^{2^{t+1}}\left| X_{ij}\right|>\varepsilon _{2^t}2^{t/4}\right) \\\le & {} \lim _{k\rightarrow \infty }\sum _{t=k}^\infty \left( 2^{t+1}\right) ^2 P\left( \left| X_{12}\right| >\varepsilon _{2^t} 2^{t/4}\right) \\\le & {} \lim _{k\rightarrow \infty }\sum _{t=k}^\infty \sum _{l=t}^\infty 4^t P\left( \varepsilon _{2^l}2^{l/4}<\left| X_{12}\right| \le \varepsilon _{2^{l+1}}2^{(l+1)/4}\right) \\= & {} \lim _{k\rightarrow \infty }\sum _{l=k}^\infty \sum _{t=k}^l 4^t P\left( \varepsilon _{2^l}2^{l/4}<\left| X_{12}\right| \le \varepsilon _{2^{l+1}}2^{(l+1)/4}\right) \\\le & {} \lim _{k\rightarrow \infty }\sum _{l=k}^\infty \varepsilon _{2^l}^{-8}\text {E}\left| X_{12}\right| ^8\text {I}\left( \varepsilon _{2^l}2^{l/4}<\left| X_{12}\right| \le \varepsilon _{2^{l+1}}2^{(l+1)/4}\right) \\= & {} 0. \end{aligned}$$

1.2 Removing diagonal entries

Based on step 1, we can assume that $\mathbf {W}_n=\dfrac{1}{\sqrt{n}}(X_{ij}\text {I}(|X_{ij}|\le \varepsilon _nn^{1/4}))_{i,j=1}^n.$ Let $\mathbf {W}_n^0=\dfrac{1}{\sqrt{n}}(X_{ij}\text {I}(|X_{ij}|\le \varepsilon _nn^{1/4})(1-\delta _{ij}))_{i,j=1}^n.$ Here $\delta _{ij}=1$ or 0 according to $i=j,$ or $i\ne j.$

Step 2(a). Removing diagonal entries for Theorem1.

$$\begin{aligned}&\left\| \text {E}\left( \mathbf {W}_n-\mathbf {W}_n^0\right) \right\| \\&\quad =\dfrac{1}{\sqrt{n}}\left\| \text {E}\left( \text {diag}\left( X_{11}\text {I}\left( \left| X_{11}\right| \le \varepsilon _nn^{1/4}\right) ,\ldots , X_{nn}\text {I}\left( \left| X_{nn}\right| \le \varepsilon _nn^{1/4}\right) \right) \right) \right\| \\&\quad =\dfrac{1}{\sqrt{n}}\left\| \text {E}\left( \text {diag}\left( X_{11}\text {I}\left( \left| X_{11}\right|>\varepsilon _nn^{1/4}\right) ,\ldots , X_{nn}\text {I}\left( \left| X_{nn}\right|>\varepsilon _nn^{1/4}\right) \right) \right) \right\| \\&\quad =\dfrac{1}{\sqrt{n}}\text {E}\left| X_{11}\right| \text {I}\left( \left| X_{11}\right| >\varepsilon _nn^{1/4}\right) \\&\quad =o\left( n^{-1/2}\right) . \end{aligned}$$

Step 2(b). Removing diagonal entries for Theorems2and3.

$$\begin{aligned} \left\| \mathbf {W}_n-\mathbf {W}_n^0\right\|= & {} \dfrac{1}{\sqrt{n}}\left\| \text {diag}\left( X_{11}\text {I}\left( \left| X_{11}\right| \le \varepsilon _nn^{1/4}\right) ,\ldots , X_{nn}\text {I}\left( \left| X_{nn}\right| \le \varepsilon _nn^{1/4}\right) \right) \right\| \\= & {} \dfrac{1}{\sqrt{n}}\varepsilon _nn^{1/4} =o\left( n^{-1/4}\right) . \end{aligned}$$

1.3 Centralization

Suppose that $z\in {\mathcal D}$ and denote $\widehat{\mathbf {W}}_n=\dfrac{1}{\sqrt{n}}(\widehat{X}_{ij}),$ where

$$\begin{aligned} \widehat{X}_{ij}=X_{ij}\text {I}\left( \left| X_{ij}\right| \le \varepsilon _nn^{1/4}\right) -\text {E}X_{ij}\text {I}\left( \left| X_{ij}\right| \le \varepsilon _nn^{1/4}\right) . \end{aligned}$$

Centralization for Theorem1.

$$\begin{aligned} \text {E}\left| s_{H^{W_n}}(z)-s_{H^{\widehat{W}_n}}(z)\right|= & {} \text {E}\left| \mathbf {x}_n^*\left( \mathbf {W}_n-z\mathbf {I}\right) ^{-1}\mathbf {x}_n-\mathbf {x}_n^*\left( \widehat{\mathbf {W}}_n-z\mathbf {I}\right) ^{-1}\mathbf {x}_n\right| \\\le & {} \dfrac{C}{\sqrt{n}}\left[ \left| \text {E}X_{12}\text {I}\left( \left| X_{ij}\right| \le \varepsilon _nn^{1/4}\right) \right| \left\| \mathbf {1}_n\right\| \left\| \mathbf {1}_n^{\prime }\right\| \right. \\&\left. \left( \text {E}\left\| \mathbf {x}_n^*\left( \mathbf {W}_n-z\mathbf {I}\right) ^{-1}\right\| ^2\right) ^{1/2} \left( \text {E}\left\| \left( \widehat{\mathbf {W}}_n-z\mathbf {I}\right) ^{-1}\mathbf {x}_n\right\| ^2\right) ^{1/2}\right] \\\le & {} C\sqrt{n}v^{-1}\left| \text {E}X_{12}\text {I}\left( \left| X_{ij}\right| >\varepsilon _nn^{1/4}\right) \right| \\= & {} o\left( n^{-1/2}\right) . \end{aligned}$$

Centralization for Theorems2and3.

$$\begin{aligned} \left| s_{H^{W_n}}(z)-s_{H^{\widehat{W}_n}}(z)\right|= & {} \left| \mathbf {x}_n^*\left( \mathbf {W}_n-z\mathbf {I}\right) ^{-1}\mathbf {x}_n-\mathbf {x}_n^*\left( \widehat{\mathbf {W}}_n-z\mathbf {I}\right) ^{-1}\mathbf {x}_n\right| \\\le & {} \left\| \left( \mathbf {W}_n-z\mathbf {I}\right) ^{-1}\right\| \left\| \mathbf {W}_n-\widehat{\mathbf {W}}_n\right\| \left\| \left( \widehat{\mathbf {W}}_n-z\mathbf {I}\right) ^{-1}\right\| \\\le & {} \dfrac{C}{v^2}\left\| \mathbf {W}_n-\widehat{\mathbf {W}}_n\right\| \\= & {} \dfrac{C}{v^2}\dfrac{1}{\sqrt{n}}\left| \text {E}X_{12}\text {I}\left( \left| X_{ij}\right| \le \varepsilon _nn^{1/4}\right) \right| \left\| \mathbf {1}_n\right\| \left\| \mathbf {1}_n^{\prime }\right\| \\\le & {} C\sqrt{n}v^{-2}\left| \text {E}X_{12}\text {I}\left( \left| X_{ij}\right| >\varepsilon _nn^{1/4}\right) \right| \\= & {} o\left( n^{-1/4}\right) . \end{aligned}$$

1.4 Rescaling

The rescaling procedures for the three theorems are exactly the same, and only eighth moment is required. Thus we treat them uniformly. Write $\mathbf {W}_n=\dfrac{1}{\sqrt{n}}(\widehat{X}_{ij}),$ where $\widehat{X}_{ij}=X_{ij}\text {I}(|X_{ij}|\le \varepsilon _nn^{1/4})-\text {E}X_{ij}\text {I}(|X_{ij}|\le \varepsilon _nn^{1/4}).$ And $\widetilde{\mathbf {W}}_n=\dfrac{1}{\sqrt{n}}(Y_{ij}),$ where $Y_{ij}=X_{ij}/\sigma _1$ and $\sigma _1^2=\text {E}|X_{12}\text {I}(|X_{12}|\le \varepsilon _nn^{1/4})-\text {E}X_{12}\text {I}(|X_{12}|\le \varepsilon _nn^{1/4})|^2.$ Notice that $\sigma _1\le 1,$ and $\sigma _1$ tends to 1 as n goes to $\infty .$ For $z\in {\mathcal D},$ we obtain

$$\begin{aligned} \left| s_{H^{W_n}}(z)-s_{H^{\widetilde{W}_n}}(z)\right|= & {} \left| x_n^*\left( W_n-zI\right) ^{-1}\left( W_n-\widetilde{W}_n\right) \left( \widetilde{W}_n-zI\right) ^{-1}x_n\right| \\\le & {} \dfrac{1}{v^2}\left\| \left( 1-\sigma _1^{-1}\right) W_n\right\| \, (\text {by Lemma}\,17)\\\le & {} \dfrac{C}{v^2}\left( 1-\sigma _1\right) \\\le & {} \dfrac{C}{v^2}\left( 1-\sigma _1^2\right) \\\le & {} Cv^{-2}\text {E}\left| X_{12}\right| ^2\text {I}\left( \left| X_{12}\right| >\varepsilon _nn^{1/4}\right) \\= & {} o\left( n^{-1/2}\right) . \end{aligned}$$