#  Bulk universality for deformed GinUEs

Lu Zhang111School of Mathematical Sciences, University of Science and
Technology of China, Hefei 230026, P.R. China. E-mail:
<EMAIL_ADDRESS>
###### Abstract

For the deformed complex Ginibre ensemble with a mean normal matrix, under
certain assumptions on the mean matrix we prove that the same bulk statistics
holds as in the complex Ginibre matrix bulk. This is the continuation of the
previous joint papers [11] and [12], which deal with local eigenvalue
statistics at the edge.

##  1 Introduction and main results

###  1.1 Introduction

The deformed complex Ginibre ensembles are defined as follows.

######  Definition 1.1.

A random complex $N\times N$ matrix $X$, is said to belong to the deformed
complex Ginibre ensemble with mean matrix $X_{0}={\rm
diag}\left(A_{0},0_{N-r}\right)$ and time parameter $\tau>0$, denoted by
GinUE${}_{N}(A_{0})$, if the joint probability density function for matrix
entries is given by

$P_{N}(A_{0};X)=\Big{(}\frac{N}{\pi\tau}\Big{)}^{N^{2}}\
e^{-\frac{N}{\tau}{\rm Tr}(X-X_{0})(X-X_{0})^{*}}.$ (1.1)

Here $A_{0}$ is a $r\times r$ constant matrix.

The study of the deformed model (1.1) is motivated mostly from the effort
toward the understanding of the effect of perturbations on the spectrum of a
large-dimensional random matrix, see e.g. [1]. In the non-perturbed case, the
study of non-Hermitian random matrices was first initiated by Ginibre [6] for
those matrices with i.i.d. real/complex standard Gaussian entries, and then
was extended to i.i.d. entries. At a macroscopic level, the limit spectra
measure was the famous circular law, which is a uniform measure on the unit
disk, see for e.g. a recent survey [3] and references therein for more
details. At a microscopic level, local eigenvalue statistics in the bulk and
near the edge of the spectrum has already many tremendous results, see e.g.
[15], [5] and very recently [13].

Although the deformed GUEs or the general deformed Hermitian models have been
quite well understood (see e.g. [4] and references therein), not much is known
about the local eigenvalue statistics in the non-Hermitian situation. Also,
for the deformed model (1.1), the circular law still holds under a small low
rank perturbation from $X_{0}$, see [16, Corollary 1.12] as the culmination of
work by many authors and a survey [3] for further details. For general $X_{0}$
with infinite rank but certain limiting behaviour, the limit spectra measure
no longer follows the circular law, while the boundary curve of its support
can still satisfy good parametric equation, see e.g. (1.6) or [1, Proposition
1.2]. These curves can exhibit different geometries from the circle, like
degeneracy, connectedness and etc.

Our goal is to identify all the possible local eigenvalue statistics at the
edge and in the bulk of the spectrum, for the deformed model (1.1) with mean
normal matrix $X_{0}$ under certain restriction (1.2) on $A_{0}$ which has
infinite rank. The series are divided in three papers. This is the third
paper, which investigates the local eigenvalue statistics near the bulk point
of the large $N$ limit spectrum distribution of $X$. In contrast, the first
two papers [11] and [12] investigates the local eigenvalue statistics near the
edge point, which is further classified to be whether critical or not,
according to its degeneracy at the boundary curve.

It is a classical result that for complex Ginibre unitary ensembles, i.e.
model (1.1) with zero mean, local eigenvalue statistics near the bulk point
forms a class of determinantal point processes, with correlation kernels
$K(z,w)=\frac{1}{\pi}e^{-(|z|^{2}+|w|^{2})/2+z\overline{w}}$, see e.g. [14].
In [15] Tao and Vu generalise this result to more general complex random
matrix, which matches moments with complex Ginibre unitary ensembles to fourth
order and satisfies other regular conditions. Very recently in [13], Maltsev
and Osman proved the Ginibre bulk statistics both for model (1.1) with certain
restrictions on $X_{0}$ in the local law setting and a rather general
ensembles of random complex matrix with i.i.d. entries with finite moments
such that the real and imaginary parts are independent. Our present paper will
show that under certain restriction (1.2) on the mean matrix $X_{0}$ with
infinite rank, the large $N$ limit local eigenvalue statistics for model (1.1)
is still governed by the Ginibre bulk statistics.

###  1.2 Main results

Throughout the present paper, we select $A_{0}$ as a normal matrix with the
following form

$A_{0}={\rm
diag}\left(a_{1}\mathbb{I}_{r_{1}},\cdots,a_{t}\mathbb{I}_{r_{t}},z_{0}\mathbb{I}_{r_{0}},A_{t+1}\right),$
(1.2)

where $t$ is a given non-negative integer, $z_{0}$ is a spectral parameter and
$A_{t+1}$ is a normal $r_{t+1}\times r_{t+1}$ matrix with $z_{0}$ not an
eigenvalue. Put $R_{0}=N-r$ where $r=\sum_{\alpha=0}^{t+1}r_{\alpha}$, we
assume that $r_{0}$, $r_{t+1}$ and $R_{0}$ are finite, independent of $N$, and
as $N\to\infty$

$r_{\alpha}=c_{\alpha}N+R_{\alpha,N},\quad
R_{\alpha,N}=O(1),\quad\alpha=1,\cdots,t.$ (1.3)

So we have

$\sum_{\alpha=1}^{t}c_{\alpha}=1.$ (1.4)

For given complex numbers $a_{1},\cdots,a_{t}$, introduce a probabililty
measure on the complex plane

$d\nu(z)=\sum_{\alpha=1}^{t}c_{\alpha}\delta(z-a_{\alpha}),$

and a relevant expectation

$P_{00}(z_{0}):=\int\frac{1}{|z-z_{0}|^{2}}d\nu(z),$ (1.5)

the support of limit spectral measure $\mu_{\infty}$ is

$\mathrm{Supp}(\mu_{\infty}):=\Big{\\{}z_{0}\in\mathbb{C}:P_{00}(z_{0})\geq\frac{1}{\tau}\Big{\\}};$
(1.6)

see e.g. [1, Proposition 1.2]. In this paper we consider the bulk point
$z_{0}$, that is $P_{00}(z_{0})>\frac{1}{\tau}$. Then there exists a unique
positive $t_{0}$, such that

$\sum_{\alpha=1}^{t}\frac{\tau c_{\alpha}}{|a_{\alpha}-z_{0}|^{2}+t_{0}}=1.$
(1.7)

Further introduce two relevant expectations

$P_{0}:=P_{0}(z_{0})=\int\frac{z-z_{0}}{(|z-z_{0}|^{2}+t_{0})^{2}}d\nu(z),\quad
P_{1}:=P_{1}(z_{0})=\int\frac{1}{(|z-z_{0}|^{2}+t_{0})^{2}}d\nu(z)$ (1.8)

These quantities are important to characterize the rescaling of the spectral
points which are parameters in the n-point correlation functions.

Now, we are ready to state our main results.

######  Theorem 1.2.

For the ${\mathrm{GinUE}}_{N}(A_{0})$ ensemble with $R_{0}\geq n$ and the
assumptions (1.2) on $A_{0}$, if $z_{0}$ is a bulk point which satisfies
$a_{\alpha}\not=z_{0}$ for $\alpha=1,\cdots,t$ and
$\sum_{\alpha=1}^{t}\frac{\tau c_{\alpha}}{|z_{0}-a_{\alpha}|^{2}+t_{0}}=1$
with $t_{0}>0$, then as $N\to\infty$ scaled eigenvalue correlations hold
uniformly for all $\hat{z}_{1},\ldots,\hat{z}_{n}$ in a compact subset of
$\mathbb{C}$.

$\frac{1}{N^{n}\Big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\Big{)}^{n}}R_{N}^{(n)}\bigg{(}A_{0};z_{0}+\Big{(}N\Big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\Big{)}\Big{)}^{-\frac{1}{2}}\hat{z}_{1},\cdots,z_{0}+\Big{(}N\Big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\Big{)}\Big{)}^{-\frac{1}{2}}\hat{z}_{n}\bigg{)}\\\
=\det\left(\Big{[}\frac{1}{\pi}e^{-\frac{1}{2}|\hat{z}_{i}|^{2}-\frac{1}{2}|\hat{z}_{j}|^{2}+\hat{z}_{i}\overline{\hat{z}_{j}}}\Big{]}_{i,j=1}^{n}\right)+O\big{(}N^{-\frac{1}{2}}\big{)}.$
(1.9)

######  Remark 1.3.

Note in the right hand side of (1.9), the time parameter $\tau$ does not
exist, while it appears in the rescaling coefficients
$\Big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\Big{)}^{-\frac{1}{2}}$ , which is
indicated by (1.7) and (1.8). By setting $n=1$ in (1.9), we get the asymptotic
formula for the mean level density

$\frac{1}{N}R_{N}^{(1)}\big{(}A_{0};z_{0},\cdots,z_{0}\big{)}=\frac{t_{0}P_{1}}{\pi}+\frac{|P_{0}|^{2}}{\pi
P_{1}}+O\big{(}N^{-\frac{1}{2}}\big{)},$ (1.10)

this result coincides with [2, Equation (1.12)], which describes the density
of probability measure of the sum of the $\star$-free circular and normal
operators, in the terminology of free probability theory.

######  Remark 1.4.

The condition $R_{0}\geq n$ comes from [10, Proposition 1.3], which provides
an integral representation for the n-point correlations of
GinUE${}_{N}(A_{0})$ in Definition 1.1, the appearance of this condition comes
from the fact when proving [10, Proposition 1.3], we use induction on the
matrix dimension $N$, and needs to reduce the n-point correlation function to
a matrix integral consisting of expectation with respect to the same deformed
Ginibre ensemble but with dimension $N-n$; This means the mean matrix
$X_{0}={\rm diag}\left(A_{0},0_{N-r=R_{0}}\right)$(cf. Definition 1.1) should
have enough zeros to fit the truncation. The condition $a_{\alpha}\not=z_{0}$
for $\alpha=1,\cdots,t$ is essential, because by (1.2) we know the testing
point $z_{0}$ is an eigenvalue of the mean matrix $X_{0}$, if the condition
holds, then the geometric multiplicity of $z_{0}$ is finite independent of
$N$, otherwise it will tend to infinity as the dimension of the random matrix
tends to infinity. Technically, the condition $a_{\alpha}\not=z_{0}$ for
$\alpha=1,\cdots,t$ is necessary in the proof of Proposition 2.2, to be more
precise, in deriving the step (3.4).

######  Remark 1.5.

As having discussed with the authors of [13], we must admit there is overlap
between our main result Theorem 1.2 and [13, Theorem 1.1]. While we believe
this paper still has its own value, as the methodologies between our paper and
[13] are different. In the proof of [13, Theorem 1.1], the authors apply the
partial Schur decomposition to express the n-point correlation function as a
multiple integral which involves matrix and spherical integrals and an
expectation with respect to the product of characteristic polynomials, see
[13, Equation (3.19)]; Although our idea in this paper is similar, we apply a
complete(rather than partial) Schur decomposition to express the n-point
correlation function as a multiple integral which involves matrix integrals
and an expectation with respect to the product of characteristic polynomials,
this has been done in [10, Equation (1.9)](which includes me as one of the
authors), we claim that [13, Equation (3.19)] and [10, Equation (1.9)] are
different, also [10, Equation (1.9)] is an earlier result. To proceed the
analysis, the authors of [13] need to apply inequalities of concentration of
spherical integral, see [13, Lemma 6.2& 6.3]. In contrast, we are devoted to
the Laplace method for matrix integrals, see Proposition 2.1, using of two
maximum lemmas in Section 2.4 gives us the concentration reduction Proposition
2.2, which restricts the matrix variables in a small region around the maximum
point. Then we can do Taylor expansions of functions of the matrix variables.
In conclusion, proofs between our paper and [13, Theorem 1.1] are established
independently, and are different in many aspects. Our methods may be more
elementary, which could be treated as a classical example of the Laplace
method for matrix integrals. The methodology in the proof of Theorem 1.2 still
applies when considering the edge statistics, see [10, 11, 12].

##  2 Integral representation of correlation functions

###  2.1 Notation

Denote the conjugate, transpose, and conjugate transpose of a complex matrix
$A$ by $\overline{A}$, $A^{t}$ and $A^{*}$, respectively. Besides, the
following symbols may be used in this and subsequent sections. Denote the
tensor product of an $m\times n$ matrix $A=[a_{i,j}]$ and a $p\times q$ matrix
$B$ as a block matrix

$A\otimes B=\left[\begin{smallmatrix}a_{11}B&\cdots&a_{1n}B\\\
\vdots&\ddots&\vdots\\\ a_{m1}B&\cdots&a_{mn}B\end{smallmatrix}\right],$

and denote the Hilbert-Schmidt norm of a complex matrix $M$ as
$\|M\|:=\sqrt{{\rm{Tr}}(MM^{*})}$. $M=O(A_{N})$ for some matrix sequence
$A_{N}$ means that each element has the same order as that of $A_{N}$.

###  2.2 Matrix variables

To prove Theorem 1.2, we need a matrix integral representation of correlation
functions, see (2.8). To this end, recalling (1.2), we need to introduce
several matrix variables $Y$, $Q_{0}$, $Q_{t+1}$ and upper triangular
$T_{\alpha}$ with non-negative diagonal elements for $\alpha=1,\cdots,t$,
which are of sizes $n\times n$, $n\times r_{0}$, $n\times r_{t+1}$ and
$n\times n$ respectively. Introduce scaled spectral points which play the role
as parameter in correlation function

$Z=z_{0}I_{n}+N^{-\frac{1}{2}}\hat{Z},\quad
Z=\mathrm{diag}(z_{1},\ldots,z_{n}),\
\hat{Z}=\mathrm{diag}(\hat{z}_{1},\ldots,\hat{z}_{n}).$ (2.1)

Denote the following block matrices: for $\alpha=1,\cdots,t$, set

$A_{\alpha}=\begin{bmatrix}\sqrt{\gamma_{N}}(Z-a_{\alpha}\mathbb{I}_{n})&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}\big{(}Z^{*}-\overline{a}_{\alpha}\mathbb{I}_{n}\big{)}\end{bmatrix},$
(2.2) $\widehat{L}_{1}={\rm
diag}\big{(}A_{1}\otimes\mathbb{I}_{n},\cdots,A_{t}\otimes\mathbb{I}_{n},\widehat{B}_{t+1}\big{)},$
(2.3)
$\widehat{B}_{t+1}=\begin{bmatrix}\sqrt{\gamma_{N}}\big{(}Z\otimes\mathbb{I}_{\widetilde{r}_{t+1}}-\mathbb{I}_{n}\otimes\widetilde{A}_{t+1}\big{)}&-Y^{*}\otimes\mathbb{I}_{\widetilde{r}_{t+1}}\\\
Y\otimes\mathbb{I}_{\widetilde{r}_{t+1}}&\sqrt{\gamma_{N}}\big{(}Z^{*}\otimes\mathbb{I}_{\widetilde{r}_{t+1}}-\mathbb{I}_{n}\otimes\widetilde{A}_{t+1}^{*}\big{)}\end{bmatrix},$
(2.4)
$\widehat{L}_{2}=\left[\begin{smallmatrix}\left[\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes(T_{\alpha}^{*}T_{\beta})\right]_{\alpha,\beta=1}^{t}&\left[\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes(T_{\alpha}^{*}\widetilde{Q}_{t+1})\right]_{\alpha=1}^{t}\\\
\left[\begin{smallmatrix}\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}\widetilde{Q}_{t+1}^{*}T_{\beta})&\\\
&\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}^{*}\widetilde{Q}_{t+1}^{*}T_{\beta})\end{smallmatrix}\right]_{\beta=1}^{t}&\left[\begin{smallmatrix}\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}\widetilde{Q}_{t+1}^{*}\widetilde{Q}_{t+1})&\\\
&\mathbb{I}_{n}\otimes(\widetilde{A}_{t+1}^{*}\widetilde{Q}_{t+1}^{*}\widetilde{Q}_{t+1})\end{smallmatrix}\right]\end{smallmatrix}\right],$
(2.5)

where

$\widetilde{Q}_{t+1}=\left[Q_{0},Q_{t+1}\right],\quad\widetilde{A}_{t+1}={\rm
diag}\left(z_{0}\mathbb{I}_{r_{0}},A_{t+1}\right),\quad\widetilde{r}_{t+1}=r_{0}+r_{t+1},$
(2.6)

whenever $r_{0}>0$.

On the other hand, rewrite $T_{\alpha}$ as sum of a diagonal matrix
$\sqrt{T_{{\rm d},\alpha}}$ and a strictly upper triangular matrix ${T_{{\rm
u},\alpha}}$

$T_{\alpha}=\sqrt{T_{{\rm d},\alpha}}+T_{{\rm
u},\alpha}:=\begin{bmatrix}\sqrt{t_{1,1}^{(\alpha)}}&\cdots&t_{i,j}^{(\alpha)}\\\
&\ddots&\vdots\\\ &&\sqrt{t_{n,n}^{(\alpha)}}\end{bmatrix},$ (2.7)

where

$T_{{\rm d},\alpha}={\rm
diag}\big{(}t_{1,1}^{(\alpha)},\cdots,t_{n,n}^{(\alpha)}\big{)}.$

###  2.3 Sketch of the proof for Theorem 1.2

Combining integral representations for the n-point correlations given in our
previous paper [10, Proposition 1.3] and duality formulas for auto-correlation
functions of characteristic polynomials in [7, 10], we can write the n-point
correlation function $R_{N}^{(n)}(A_{0};z_{1},\cdots,z_{n})$ in an appropriate
form which is ready for asymptotic analysis by the Laplace method argument.
This has already been done in [11, Proposition 2.2]:

######  Proposition 2.1.

Let $R_{0}\geq n$, then

$\displaystyle
R_{N}^{(n)}(A_{0};z_{1},\cdots,z_{n})=\widetilde{C}_{N,n}\int\cdots\int
g(T,Y,Q)\exp\\{Nf(T,Y,Q)\\}{\rm d}Y{\rm d}Q_{0}{\rm
d}Q_{t+1}\prod_{\alpha=1}^{t}{\rm d}T_{\alpha},$ (2.8)

where

$\displaystyle
f(T,Y,Q)=\sum_{\alpha=1}^{t}c_{\alpha}\log\det(T_{\alpha}T_{\alpha}^{*})+\frac{1}{\tau}h(T,Q)-\frac{N-n}{\tau
N}{\rm
Tr}\big{(}YY^{*}\big{)}+\sum_{\alpha=1}^{t}c_{\alpha}\log\det(A_{\alpha}),$
(2.9)

with

$\displaystyle h(T,Q)=\sum_{\alpha=1}^{t}\left(-|z_{0}-a_{\alpha}|^{2}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+N^{-\frac{1}{2}}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}T_{\alpha}T_{\alpha}^{*}\big{)}\right)\right)$ (2.10)
$\displaystyle+|z_{0}|^{2}\Big{(}\sum_{\alpha=1}^{t}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}$
$\displaystyle+N^{-\frac{1}{2}}\Big{(}\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}Q_{t+1}A_{t+1}^{*}Q_{t+1}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}^{*}Q_{t+1}A_{t+1}Q_{t+1}^{*}\big{)}\Big{)}$
$\displaystyle+\sum_{1\leq i<j\leq
n}\Big{|}\Big{(}\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\Big{)}_{i,j}\Big{|}^{2}$
$\displaystyle-{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*},$

and

$\displaystyle g(T,Y,Q)=$
$\displaystyle\Big{(}\det\big{(}\mathbb{I}_{n}-\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}-Q_{0}Q_{0}^{*}-Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}^{R_{0}-n}\Bigg{(}\det\\!\begin{bmatrix}\sqrt{\gamma_{N}}Z&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}Z^{*}\end{bmatrix}\Bigg{)}^{R_{0}-n}$ (2.11)
$\displaystyle\times\Big{(}\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\big{(}t_{j,j}^{(\alpha)}\big{)}^{R_{\alpha,N}-j}\Big{)}\Big{(}\prod_{\alpha=1}^{t}\big{(}\det(A_{\alpha})\big{)}^{R_{\alpha,N}-n}\Big{)}\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)},$

Here $R_{\alpha,N}$ is defined in (1.3) and

$\gamma_{N}=\frac{N}{N-n},$ (2.12)
$\widetilde{C}_{N,n,\tau}=\Big{(}\frac{N-n}{\pi\tau}\Big{)}^{n^{2}}\frac{1}{C_{N,\tau}}e^{-\frac{N}{\tau}\sum_{k=1}^{n}|z_{k}|^{2}}\prod_{1\leq
i<j\leq
n}|z_{i}-z_{j}|^{2}\prod_{\alpha=1}^{t}\frac{\pi^{nr_{\alpha}-\frac{n(n-1)}{2}}}{\prod_{j=1}^{n}(r_{\alpha}-j)!}$
(2.13)

and

$C_{N,\tau}=\tau^{nN-\frac{n(n-1)}{2}}\pi^{n(r+1)}N^{-\frac{1}{2}n(n+1)}(N-n)^{-n(N-n)}\prod_{k=N-n-r}^{N-r-1}k!,$

the matrix variables have the restrictions

$\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n}.$
(2.14)

For proof, we refer the reader to read Section 2.2 in [11] directly, which is
our first paper in this series, and has the same setting with the present
paper.

Now we need to give a concentration reduction by restricting integration
region to a small domain. For this, we need to specify the scale of $N$ in
each quantity, from (2.2) we have

$A_{\alpha}=\begin{bmatrix}\sqrt{\gamma_{N}}(z_{0}-a_{\alpha})\mathbb{I}_{n}&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}\overline{z_{0}-a_{\alpha}}\mathbb{I}_{n}\end{bmatrix}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\begin{bmatrix}\hat{Z}&\\\
&\hat{Z}^{*}\end{bmatrix},$

and

$\log\det(A_{\alpha})=\log\det\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)+\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right),$
(2.15)

where, for $\alpha=1,\cdots,t$,

$f_{\alpha}:=|z_{0}-a_{\alpha}|^{2},$ (2.16)

and

$\displaystyle\widehat{A}_{\alpha}=\left[\begin{smallmatrix}\sqrt{\gamma_{N}}\overline{z_{0}-a_{\alpha}}\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y\right)^{-1}\hat{Z}&Y^{*}\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)^{-1}\hat{Z}^{*}\\\
-Y\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y\right)^{-1}\hat{Z}&\sqrt{\gamma_{N}}(z_{0}-a_{\alpha})\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)^{-1}\hat{Z}^{*}\end{smallmatrix}\right].$
(2.17)

Take all the leading terms of $f(T,Y,Q)$ in (2.9) we have

$\displaystyle
f_{0}(\tau;T,Y,Q)=\sum_{\alpha=1}^{t}\left(c_{\alpha}\log\det(T_{\alpha}T_{\alpha}^{*})+c_{\alpha}\log\det\big{(}f_{\alpha}\mathbb{I}_{n}+YY^{*}\big{)}\right)+\frac{1}{\tau}\Big{(}h_{0}(T,Q)-{\rm
Tr}\big{(}YY^{*})\Big{)}$ (2.18)

with

$\displaystyle h_{0}(T,Q)=-\sum_{\alpha=1}^{t}f_{\alpha}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}-{\rm
Tr}Q_{t+1}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}Q_{t+1}^{*}$
$\displaystyle+|z_{0}|^{2}\bigg{(}\sum_{\alpha=1}^{t}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}+{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}+{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}\bigg{)}$ $\displaystyle+\sum_{1\leq
i<j\leq
n}\bigg{|}\bigg{(}\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\bigg{)}_{i,j}\bigg{|}^{2}.$

Note that the two sets of matrix variables $\\{Y\\}$ and
$\\{Q_{0},Q_{t+1},T_{1},\ldots,T_{t}\\}$ in $f_{0}(\tau;T,Y)$ are separate in
$f_{0}(\tau;T,Y)$, we can do Taylor expansion near the maximum points
respectively, according to Lemma 2.3 and Lemma 2.4 in the subsequent Section
2.4 below.

Take the singular value decomposition

$Y=U_{1}\sqrt{\Lambda}U_{2},\quad\Lambda={\rm
diag}\left(\lambda_{1},\cdots,\lambda_{n}\right),\quad\lambda_{1}\geq\cdots\geq\lambda_{n}\geq
0,$ (2.19)

the Jacobian reads

${\rm d}Y=\pi^{n^{2}}\Big{(}\prod_{i=1}^{n-1}i!\Big{)}^{-2}\prod_{1\leq
i<j\leq n}(\lambda_{j}-\lambda_{i})^{2}{\rm d}\Lambda{\rm d}U_{1}{\rm
d}U_{2},$ (2.20)

where $U_{1}$ and $U_{2}$ are chosen from the unitary group $\mathcal{U}(n)$
with the Haar measure.

By restricting the integration region to

$\Omega_{N,\delta}=\Omega\cap A_{N,\delta},$ (2.21)

where

$\Omega=\bigg{\\{}\big{(}\\{T_{\alpha}\\},U_{1},\Lambda,U_{2},Q_{0},Q_{t+1}\big{)}\Big{|}\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n},\lambda_{1}\geq\cdots\geq\lambda_{n}\geq
0\bigg{\\}}$ (2.22)

and

$A_{N,\delta}=\left\\{\big{(}\\{T_{\alpha}\\},U_{1},\Lambda,U_{2},Q_{0},Q_{t+1}\big{)}\Big{|}S(T,Y,Q)\leq\delta\right\\},$
(2.23)

with positive constant $\delta$ and

$\displaystyle S(T,Y,Q)$ $\displaystyle=\sum_{\alpha=1}^{t}\Big{(}{\rm
Tr}\big{(}T_{{\rm d},\alpha}-\tau
c_{\alpha}(|z_{0}-a_{\alpha}|^{2}+t_{0})^{-1}\mathbb{I}_{n}\big{)}\big{(}T_{{\rm
d},\alpha}-\tau
c_{\alpha}(|z_{0}-a_{\alpha}|^{2}+t_{0})^{-1}\mathbb{I}_{n}\big{)}^{*}$ (2.24)
$\displaystyle+{\rm Tr}(T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*})\Big{)}+{\rm
Tr}\Big{(}Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}+\big{(}\Lambda-
t_{0}\mathbb{I}_{n}\big{)}^{2}\Big{)},$

we have a concentration reduction, the proof is given in Section 3.1.

######  Proposition 2.2.

With the same notations and conditions as in Theorem 1.2 and Proposition 2.1,
then for any positive constant $\delta$ there exists $\Delta>0$ such that

$R_{N}^{(n)}(A_{0};z_{1},\cdots,z_{n})=D_{N,n}\,\Big{(}I_{N,\delta}+O\big{(}e^{-\frac{1}{2}N\Delta}\big{)}\Big{)},$
(2.25)

where

$\displaystyle
I_{N,\delta}=\pi^{n^{2}}\Big{(}\prod_{i=1}^{n-1}i!\Big{)}^{-2}\int_{\Omega_{N,\delta}}g(T,U_{1}\sqrt{\Lambda}U_{2},Q)\prod_{1\leq
i<j\leq n}(\lambda_{j}-\lambda_{i})^{2}$ (2.26)
$\displaystyle\times\exp\Big{\\{}N\big{(}f(T,U_{1}\sqrt{\Lambda}U_{2},Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\big{)}\Big{\\}}{\rm
d}V,$

and

${\rm d}V={\rm d}\Lambda{\rm d}U_{1}{\rm d}U_{2}{\rm d}Q_{0}{\rm
d}Q_{t+1}\prod_{\alpha=1}^{t}{\rm d}T_{\alpha},\quad
D_{N,n}=\widetilde{C}_{N,n}e^{nN(|z_{0}|^{2}/\tau-1)}\prod_{\alpha=1}^{t}(\tau
c_{\alpha})^{nNc_{\alpha}}.$ (2.27)

The remaining task is to analyze the matrix integral $I_{N,\delta}$ with
sufficiently small $\delta>0$, which can be followed by a quite standard
procedure of Laplace method. To be precise, the proof will be divided into
four steps: Step 1 deals with Taylor expansion of $f(T,Y,Q)$ and the
integration region(cf (2.14)), The most relevant variables are triangular
matrices $\big{\\{}T_{\alpha}\big{\\}}$, equivalently, diagonal matrices
$\big{\\{}T_{{\rm d},\alpha}\big{\\}}$ and strictly upper triangular matrices
$\big{\\{}T_{{\rm u},\alpha}\big{\\}}$ as in (2.7). With the coming of the
change of variables

$T_{{\rm d},1},T_{{\rm d},2},\cdots,T_{{\rm
d},t}\longmapsto\widetilde{T}_{{\rm d},1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm
d},t}\longmapsto\Gamma_{1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm d},t}$

and

$T_{{\rm u},1},T_{{\rm u},2},\cdots,T_{{\rm u},t}\longmapsto G_{1},T_{{\rm
u},2},\cdots,T_{{\rm u},t};$

Step 2 deals with Taylor expansion of $g(T,Y,Q)$, which is totally based on
elementary calculations; Step 3 is a standard procedure in Laplace method to
remove the error terms resulted from the expansions in first two steps; In
Step 4, after dealing with several matrix integrals, we can complete the
proof.

###  2.4 Maximum Lemmas

Given complex numbers $z_{0}$, $a_{\alpha}$ and $c_{\alpha}>0$
($\alpha=1,\cdots,t$), assume that

$\sum_{\alpha=1}^{t}c_{\alpha}=1,\quad\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{|z_{0}-a_{\alpha}|^{2}}>\frac{1}{\tau}.$

Let $t_{0}$ be the unique positive solution of the equation

$\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{|z_{0}-a_{\alpha}|^{2}+t_{0}}=\frac{1}{\tau}.$

The following two lemmas hold and play a significant role in investigating
local eigenvalue statistics of the deformed GinUEs. See also [11, Lemma 2.4
and Lemma 2.5].

######  Lemma 2.3.

As a function of a non-negative definite matrix $H$,

$\sum_{\alpha=1}^{t}c_{\alpha}\log\det\big{(}|z_{0}-a_{\alpha}|^{2}\mathbb{I}_{n}+H\big{)}-\frac{1}{\tau}{\rm
Tr}(H)$

attains it maximum only at $H=t_{0}I_{n}$.

###### Proof.

Obviously, it’s sufficient to check the $n=1$ case. For this, take the first
and second derivatives, find the saddle point and we can easily complete the
proof. ∎

######  Lemma 2.4.

Assume that $z_{0}$ is not an eigenvalue of the $n\times n$ complex normal
matrix $D$, let

$\displaystyle J_{n}$ $\displaystyle=\sum_{\alpha=1}^{t}\Big{(}\tau
c_{\alpha}\log\det\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}-|z_{0}-a_{\alpha}|^{2}{\rm
Tr}\big{(}T_{\alpha}T_{\alpha}^{*}\big{)}\Big{)}+|z_{0}|^{2}{\rm
Tr}\Big{(}\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+AA^{*}+BB^{*}\Big{)}$
$\displaystyle-{\rm
Tr}\Big{(}B\big{(}z_{0}\mathbb{I}_{l_{2}}-D\big{)}^{*}\big{(}z_{0}\mathbb{I}_{l_{2}}-D\big{)}B^{*}\Big{)}+\sum_{1\leq
i<j\leq
n}\left|\Big{(}\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}AA^{*}+BDB^{*}\Big{)}_{i,j}\right|^{2},$

where as matrix-valued variables $T_{\alpha}$’s are $n\times n$ upper-
triangular matrices with positive diagonal elements, $A$ and $B$ are $n\times
l_{1}$ and $n\times l_{2}$ matrices respectively. Then

$J_{n}\leq n\sum_{\alpha=1}^{t}\tau c_{\alpha}\log\frac{\tau
c_{\alpha}}{|z_{0}-a_{\alpha}|^{2}+t_{0}}+n(t_{0}+|z_{0}|^{2}-\tau),$

whenever

$\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+AA^{*}+BB^{*}\leq\mathbb{I}_{n},.$

Moreover, the equality holds if and only if

$A=0,\quad B=0,\quad T_{\alpha}=\sqrt{\frac{\tau
c_{\alpha}}{|z_{0}-a_{\alpha}|^{2}+t_{0}}}\mathbb{I}_{n},\quad\alpha=1,\ldots,t.$

For proof see [11, Lemma 2.5].

##  3 Proofs of Proposition 2.2 and Theorem 1.2

###  3.1 Proof of Proposition 2.2

###### Proof of Proposition 2.2..

By Lemma 2.3 and Lemma 2.4 in Section 2.4 below, we see that
$f_{0}(\tau;T,Y,Q)$ in (2.18) attains its maximum only at

$Y=\sqrt{t_{0}}U_{1}U_{2},\ Q_{0}=0,\ Q_{t+1}=0,\quad T_{\alpha}=\sqrt{\tau
c_{\alpha}/(f_{\alpha}+t_{0}})\mathbb{I}_{n},\quad\alpha=1,\cdots,t.$ (3.1)

So there exists $\Delta>0$ such that

$f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\leq-\Delta$
(3.2)

in the domain $\Omega\cap A_{N,\delta}^{c}$. Here it should be noted that we
have taken $Y$ to be its representation of singular value decomposition
(2.19), while for simplicity we still use the symbol $Y$ in the following
argument.

On one hand, noting that

$\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}\leq\mathbb{I}_{n}$

in the domain $\Omega$ defined by (2.22), we know for $\hat{Z}$ in a compact
subset of $\mathbb{C}^{n}$ that the partial sub-leading terms of $f(T,Y,Q)$ in
(2.9) can be bounded by $O(N^{-\frac{1}{2}})$, that is,

$\displaystyle
N^{-\frac{1}{2}}\sum_{\alpha=1}^{t}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}T_{\alpha}T_{\alpha}^{*}\big{)}\right)+N^{-\frac{1}{2}}\left(\overline{z}_{0}{\rm
Tr}\big{(}\hat{Z}Q_{0}Q_{0}^{*}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}Q_{0}Q_{0}^{*}\big{)}\right)$ (3.3)
$\displaystyle+N^{-\frac{1}{2}}\left({\rm
Tr}\big{(}\hat{Z}Q_{t+1}A_{t+1}^{*}Q_{t+1}^{*}\big{)}+{\rm
Tr}\big{(}\hat{Z}^{*}Q_{t+1}A_{t+1}Q_{t+1}^{*}\big{)}\right)=O(N^{-\frac{1}{2}}).$

On the other hand, apply the singular value decomposition (2.19) for $Y$ and
note that in (2.16) all $f_{\alpha}\not=0$ because we have assumed that
$a_{\alpha}\not=z_{0}$ for $\alpha=1,\cdots,t$ in Theorem 1.2, we obtain

$\displaystyle(\sqrt{\gamma_{N}}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y)^{-1}=O(1),\quad
Y(\sqrt{\gamma_{N}}f_{\alpha}\mathbb{I}_{n}+Y^{*}Y)^{-1}=O(1),$

and

$\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right)=O(N^{-\frac{1}{2}})$
(3.4)

since $\widehat{A}_{\alpha}=O(1)$. Meanwhile,

$\log\det\left(\gamma_{N}f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)=\log\det\left(f_{\alpha}\mathbb{I}_{n}+YY^{*}\right)+O\big{(}N^{-1}\big{)}.$

Combination of (2.9), (2.15), (2.18), (3.3) and (3.4) give rise to

$\displaystyle N\Big{(}f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\Big{)}=$
(3.5) $\displaystyle
N\Big{(}f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\Big{)}+\frac{n}{\tau}{\rm
Tr}(\Lambda)+O\big{(}\sqrt{N}\big{)}.$

For $g(T,Y,Q)$ in (2.11), all factors can be controlled by a polynomial
function, so by taking a sufficiently large
$N_{0}>\sum_{\alpha=1}^{t}(n-R_{\alpha,N})/c_{\alpha}$ we know

$\displaystyle\int_{A_{N,\delta}^{c}\cap\Omega}|g(T,Y,Q)|\prod_{1\leq i<j\leq
n}(\lambda_{j}-\lambda_{i})^{2}$
$\displaystyle\times\exp\left\\{N_{0}\Big{(}f_{0}(\tau;T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\Big{)}+\frac{n}{\tau}{\rm
Tr}(\Lambda)\right\\}{\rm d}V=O(1).$

Hence,

$\displaystyle\left|\int_{A_{N,\delta}^{\complement}\cap\Omega}g(T,Y,Q)\prod_{1\leq
i<j\leq
n}(\lambda_{j}-\lambda_{i})^{2}\exp\left\\{N\Big{(}f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}\Big{)}\right\\}{\rm
d}V\right|$ $\displaystyle\leq
e^{-(N-N_{0})\Delta+O(\sqrt{N})}O(1)=O\big{(}e^{-\frac{1}{2}N\Delta}\big{)}.$

This thus completes the proof. ∎

###  3.2 Proof of Theorem 1.2

The goal in this section is to complete the proof of Theorem 1.2. With Section
2.1 in mind, we still need some extra notations: For a family of matrices
$\\{T_{{\rm u},\alpha}|\alpha=1,\cdots,t\\}$, define $\|T_{\rm
u}\|:=\sum_{\alpha=1}^{t}\|T_{{\rm u},\alpha}\|,\quad\|T_{\rm
u}\|_{2}:=\sum_{\alpha=2}^{t}\|T_{{\rm u},\alpha}\|.$ Similarly, For a family
of matrices $\\{\widetilde{T}_{{\rm d},\alpha}|\alpha=1,\cdots,t\\}$,
$\|\widetilde{T}_{{\rm d}}\|:=\sum_{\alpha=1}^{t}\|\widetilde{T}_{{\rm
d},\alpha}\|,\quad\|\widetilde{T}_{{\rm
d}}\|_{2}:=\sum_{\alpha=2}^{t}\|\widetilde{T}_{{\rm d},\alpha}\|.$ For a
complex matrix $M$, denote $M^{({\rm d})}$ the diagonal part of M, and
$M^{({\rm off},{\rm u})}$ the strictly upper triangular part of M; If the
object is $M_{\alpha}$ with index $\alpha$, then denote the corresponding
quantity as $M^{(\alpha,{\rm d})}$ and $M^{(\alpha,{\rm off},{\rm u})}$.

In view of Proposition 2.2, we will restrict all the matrix variables in the
region $\Omega_{N,\delta}$ defined in (2.21) for sufficiently small
$\delta>0$. Here we must emphasize that $\delta$ is fixed independent of $N$,
while we need $\delta>0$ to be sufficiently small so that Taylor expansions of
the relevant matrix variables are permitted. As the number of relevant matrix
variables is certainly finite, the required $\delta$ can always be realized.
In addition, in the following analysis, all error terms are written as the
form $O(A)$, which means the error terms are bounded by $CA$ for some positive
constant $C$ which is independent of $N$.

Step 1: Taylor expansion of $f(T,Y,Q)$. According to (2.26), on the exponent,
for the term
$f(T,U_{1}\sqrt{\Lambda}U_{2},Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}$,
when we have done Taylor expansions of the relevant matrix variables, we need
to do appropriate rescaling to cancel out the factor $N$, see (3.45).
Accordingly we introduce the notation of ”typical size” of each term, by means
of the rescaling. For example, in (3.45), we make the change of variables
$(Q_{0},Q_{t+1})\rightarrow N^{-\frac{1}{2}}(Q_{0},Q_{t+1})$ , so the typical
sizes of $\|Q_{0}\|$ and $\|Q_{t+1}\|$ are $N^{-\frac{1}{2}}$, and the typical
size of $\|Q_{0}\|\|Q_{t+1}\|$ is $N^{-1}$. The terms of typical size larger
than or equal to $N^{-1}$ should be reserved, otherwise the terms of typical
size smaller than or equal to $N^{-\frac{3}{2}}$ should be considered as error
terms, which can be cancelled out throughout standard method.

With the notation $f_{\alpha}$ in (2.16) noticing the integration domain and
(2.24) the decomposition of $T_{\alpha}$ as sum of a diagonal matrix
$\sqrt{T_{{\rm d},\alpha}}$ and a strictly upper triangular matrix ${T_{{\rm
u},\alpha}}$ as in (2.7), introduce new matrix variables

$T_{{\rm d},\alpha}=\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}+\widetilde{T}_{{\rm d},\alpha},$
(3.6)

It’s easy to obtain

$\displaystyle c_{\alpha}{\rm Tr}\log T_{{\rm d},\alpha}$
$\displaystyle-\frac{1}{\tau}f_{\alpha}{\rm Tr}(T_{{\rm
d},\alpha})=nc_{\alpha}\big{(}\log\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}-\frac{f_{\alpha}}{f_{\alpha}+t_{0}}\big{)}$
(3.7) $\displaystyle+\frac{1}{\tau}t_{0}{\rm Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}\big{)}-\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}+O\big{(}\|\widetilde{T}_{{\rm d},\alpha}\|^{3}\big{)},$
$\sqrt{T_{{\rm d},\alpha}}=\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}\left(\mathbb{I}_{n}+\frac{f_{\alpha}+t_{0}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right),$ $\displaystyle T_{\alpha}T_{\alpha}^{*}$
$\displaystyle=\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}+\widetilde{T}_{{\rm
d},\alpha}+\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}\left(\mathbb{I}_{n}+\frac{f_{\alpha}+t_{0}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right)T_{{\rm u},\alpha}^{*}$
$\displaystyle+\sqrt{\frac{\tau c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\left(\mathbb{I}_{n}+\frac{f_{\alpha}+t_{0}}{2\tau
c_{\alpha}}\widetilde{T}_{{\rm d},\alpha}+O\big{(}\|\widetilde{T}_{{\rm
d},\alpha}\|^{2}\big{)}\right)+T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*},$

from which we have for $i<j$,

$\left(\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}\right)_{i,j}=\sum_{\alpha=1}^{t}a_{\alpha}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}t_{i,j}^{(\alpha)}+O\left(\|T_{{\rm
u},\alpha}\|^{2}+\|T_{{\rm u},\alpha}\|\|\widetilde{T}_{{\rm
d},\alpha}\|\right),$ (3.8)

and

${\rm Tr}\big{(}\hat{Z}T_{\alpha}T_{\alpha}^{*}\big{)}=\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}{\rm Tr}\big{(}\hat{Z}\big{)}+{\rm
Tr}\big{(}\hat{Z}\widetilde{T}_{{\rm d},\alpha}\big{)}+O\left(\|T_{{\rm
u},\alpha}\|^{2}\right).$ (3.9)

At the same time,

$\left(z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\right)_{i,j}=O\left(\|Q_{0}\|^{2}+\|Q_{t+1}\|^{2}\right).$

Moreover, we see from the condition $\sum_{\alpha=1}^{t}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}=1$ that the restriction condition in (2.22)
becomes

$\sum_{\alpha=1}^{t}\widetilde{T}_{{\rm
d},\alpha}+\sum_{\alpha=1}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}+\sum_{\alpha=1}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm u},\alpha}^{*}+S\leq 0,$

where

$\displaystyle S$ $\displaystyle=\sum_{\alpha=1}^{t}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}+Q_{0}Q_{0}^{*}+Q_{t+1}Q_{t+1}^{*}$ (3.10)
$\displaystyle+\sum_{\alpha=1}^{t}T_{{\rm u},\alpha}\Big{(}\sqrt{T_{{\rm
d},\alpha}}-\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}\mathbb{I}_{n}\Big{)}+\sum_{\alpha=1}^{t}\Big{(}\sqrt{T_{{\rm
d},\alpha}}-\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}\mathbb{I}_{n}\Big{)}T_{{\rm u},\alpha}^{*},$

the first line contains all diagonal parts of $S$, while the second line only
contains the non-diagonal parts.

Let $S^{({\rm d})}$ be a diagonal matrix extracted from the diagonal part of
$S$ given in (3.10), introduce a new diagonal matrix $\Gamma_{1}$ and make
change of variables from $\widetilde{T}_{{\rm d},1},\widetilde{T}_{{\rm
d},2},\cdots,\widetilde{T}_{{\rm d},t}$ to

$\displaystyle\Gamma_{1}:=\sum_{\alpha=1}^{t}\widetilde{T}_{{\rm
d},\alpha}+S^{({\rm d})},\quad\widetilde{T}_{{\rm
d},2},\,\ldots,\widetilde{T}_{{\rm d},t}.$ (3.11)

Note that $S^{({\rm d})}$ doesn’t depend on $\widetilde{T}_{{\rm d},\alpha}$,
so this transformation is linear and

$S^{({\rm d})}=O\big{(}\|T_{{\rm
u}}\|^{2}+\|Q_{t+1}\|^{2}+\|Q_{0}\|^{2}\big{)}.$ (3.12)

We will see that the typical size for $Q_{0},Q_{t+1},\widetilde{T}_{{\rm
d},2},\,\ldots,\widetilde{T}_{{\rm d},t}$ are $N^{-1/2}$ while it is $N^{-1}$
for $\Gamma_{1}$. See (3.45) for the exact meaning.

Also let $S^{({\rm off},{\rm u})}$ be a strictly upper triangular matrix
extracted from the strictly upper triangular part of $S$ given in (3.10),
introduce a new strictly upper triangular matrix $G_{1}$ and make change of
variables from $T_{{\rm u},1},T_{{\rm u},2},\cdots,T_{{\rm u},t}$ to

$\displaystyle G_{1}:=\sum_{\alpha=1}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm u},\alpha}+S^{({\rm off},{\rm u})},\quad
T_{{\rm u},2},\,\ldots,T_{{\rm u},t}.$ (3.13)

We will see that the typical size for $T_{{\rm u},2},\cdots,T_{{\rm u},t}$ are
$N^{-1/2}$ while it is $N^{-1}$ for $G_{1}$. See (3.45) for the exact meaning.
So we will do Taylor expansions up to some proper orders accordingly.

As to the change of variables (3.13), the Jacobian determinant reads

$\det\Big{(}\frac{\partial T_{{\rm u},1}}{\partial
G_{1}}\Big{)}=\left(\frac{f_{1}+t_{0}}{\tau
c_{1}}\right)^{\frac{n(n-1)}{2}}+O\big{(}\|T_{{\rm
u},1}\|+\|\widetilde{T}_{{\rm d}}\|\big{)}.$

At the same time,

$\|\widetilde{T}_{{\rm d}}\|=O\big{(}\|\widetilde{T}_{{\rm
d}}\|_{2}+\|\Gamma_{1}\|+\|Q_{0}\|+\|T_{{\rm u}}\|+\|Q_{t+1}\|\big{)}.$ (3.14)

Here exactly, from (3.11) and (3.12) we should have

$\|\widetilde{T}_{{\rm d}}\|=O\big{(}\|\widetilde{T}_{{\rm
d}}\|_{2}+\|\Gamma_{1}\|+\|Q_{0}\|^{2}+\|T_{{\rm
u}}\|^{2}+\|Q_{t+1}\|^{2}\big{)},$

while as (2.23) and (2.26) we have

$\|T_{{\rm u}}\|+\|\widetilde{T}_{{\rm
d}}\|+\|Q_{0}\|+\|Q_{t+1}\|=O\big{(}\sqrt{\delta}\big{)}.$ (3.15)

As $\delta$ is fixed (3.14) is right. Moreover, (3.14) already is sufficient
for our analysis. From

$S^{({\rm off},{\rm u})}=O\Big{(}\|T_{{\rm u}}\|\big{(}\|T_{{\rm
u}}\|+\|\widetilde{T}_{{\rm
d}}\|\big{)}+\|Q_{t+1}\|^{2}+\|Q_{0}\|^{2}\Big{)},$

we have

$\|T_{{\rm u}}\|=O\big{(}\|G_{1}\|+\|T_{{\rm
u}}\|_{2}+\|Q_{0}\|+\|Q_{t+1}\|\big{)}+\|T_{{\rm u}}\|O\big{(}\|T_{{\rm
u}}\|+\|\widetilde{T}_{{\rm d}}\|\big{)}.$

From (3.15) and as $\delta>0$ is sufficiently small, we have

$\|T_{{\rm u}}\|=O\big{(}\|G_{1}\|+\|T_{{\rm
u}}\|_{2}+\|Q_{0}\|+\|Q_{t+1}\|\big{)}.$ (3.16)

Substitute (3.16) into (3.14), we have

$\|\widetilde{T}_{{\rm d}}\|=O\big{(}\Omega_{{\rm error}}\big{)},$

where

$\Omega_{{\rm error}}:=\|\widetilde{T}_{{\rm
d}}\|_{2}+\|\Gamma_{1}\|+\|G_{1}\|+\|T_{{\rm u}}\|_{2}+\|Q_{0}\|+\|Q_{t+1}\|,$
(3.17)

and further

$\det\Big{(}\frac{\partial T_{{\rm u},1}}{\partial
G_{1}}\Big{)}=\left(\frac{f_{1}+t_{0}}{\tau
c_{1}}\right)^{\frac{n(n-1)}{2}}+O(\Omega_{{\rm error}}).$ (3.18)

It should be noted that $\Omega_{{\rm error}}$ plays a central role in
representation of the error terms, according to (3.45), its typical size is
$N^{-\frac{1}{2}}$.

In view of (2.9) and (2.10),

$\frac{|z_{0}|^{2}}{\tau}\Big{(}\sum_{\alpha=1}^{t}{\rm
Tr}(T_{\alpha}T_{\alpha}^{*})+{\rm Tr}(Q_{0}Q_{0}^{*})+{\rm
Tr}(Q_{t+1}Q_{t+1}^{*})\Big{)}=\frac{n|z_{0}|^{2}}{\tau}+\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}(\Gamma_{1}).$ (3.19)

Also in view of (3.7),

$\frac{t_{0}}{\tau}\sum_{\alpha=1}^{t}{\rm Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}\big{)}=\frac{t_{0}}{\tau}{\rm
Tr}(\Gamma_{1})-\frac{t_{0}}{\tau}{\rm Tr}(S^{({\rm d})}).$ (3.20)

From (3.10), we have

$-\frac{t_{0}}{\tau}{\rm Tr}(S^{({\rm
d})})=-\frac{t_{0}}{\tau}\sum_{\alpha=1}^{t}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}-\frac{t_{0}}{\tau}{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}-\frac{t_{0}}{\tau}{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}.$ (3.21)

In view of (2.9), (2.10) and (3.8), from transformations (3.11) and (3.13) we
have

$\displaystyle-\sum_{\alpha=1}^{t}\frac{f_{\alpha}+t_{0}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\left|\left(\sum_{\alpha=1}^{t}a_{\alpha}T_{\alpha}T_{\alpha}^{*}+z_{0}Q_{0}Q_{0}^{*}+Q_{t+1}A_{t+1}Q_{t+1}^{*}\right)_{i,j}\right|^{2}$
(3.22) $\displaystyle=-\frac{(f_{1}+t_{0})^{2}}{\tau^{2}c_{1}}{\rm
Tr}\left(\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\right)\left(\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\right)^{*}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}+t_{0}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}$
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\left|\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\right|^{2}+O(\Omega_{{\rm
error}}^{3}+\|G_{1}\|\Omega_{{\rm error}}),$

and

$\displaystyle-\sum_{\alpha=1}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\left(\widetilde{T}_{{\rm
d},\alpha}^{2}\right)+\frac{|z_{0}|^{2}}{\tau}{\rm
Tr}(\Gamma_{1})+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=1}^{t}\left(\overline{a}_{\alpha}{\rm
Tr}\big{(}\hat{Z}\widetilde{T}_{{\rm d},\alpha}\big{)}+a_{\alpha}{\rm
Tr}\big{(}\hat{Z}^{*}\widetilde{T}_{{\rm
d},\alpha}\big{)}\right)+\frac{t_{0}}{\tau}{\rm Tr}(\Gamma_{1})$ (3.23)
$\displaystyle=\frac{|z_{0}|^{2}+t_{0}}{\tau}{\rm
Tr}(\Gamma_{1})-\frac{(f_{1}+t_{0})^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\left(\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\right)^{2}-\sum_{\alpha=2}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}$
$\displaystyle+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}+O\Big{(}\big{(}\Omega_{{\rm
error}}+N^{-\frac{1}{2}}\big{)}\big{(}\Omega_{{\rm
error}}^{2}+\|\Gamma_{1}\|\big{)}\Big{)}.$

Also, the integral region now becomes simply

$\Gamma_{1}+G_{1}+G_{1}^{*}\leq 0.$ (3.24)

In view of (2.15),

$\displaystyle\log\det\left(\mathbb{I}_{2n}+\sqrt{\gamma_{N}}N^{-\frac{1}{2}}\widehat{A}_{\alpha}\right)=N^{-\frac{1}{2}}{\rm
Tr}\big{(}\widehat{A}_{\alpha}\big{)}-\frac{1}{2}N^{-1}{\rm
Tr}\big{(}\widehat{A}_{\alpha}^{2}\big{)}+O\big{(}N^{-\frac{3}{2}}\big{)}.$
(3.25)

Now start from (2.19), we need to take a shift

$\widetilde{\Lambda}=\Lambda-t_{0}\mathbb{I}_{n}={\rm
diag}\left(\widetilde{\lambda}_{1},\cdots,\widetilde{\lambda}_{n}\right),\quad\widetilde{\lambda}_{1}\geq\cdots\geq\widetilde{\lambda}_{n}\geq
0,$ (3.26)

Combining (2.15), (2.17) and (3.25), and introducing $P_{0}$ defined in (1.5),
we have

$\displaystyle{\rm
Tr}(YY^{*})-\sum_{\alpha=1}^{t}c_{\alpha}\log\det(A_{\alpha})=\frac{nt_{0}}{\tau}-\frac{n^{2}}{N}-n\sum_{\alpha=1}^{t}c_{\alpha}\log(f_{\alpha}+t_{0})$
(3.27)
$\displaystyle+\frac{N^{-1}}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\left(\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}-2t_{0}U_{1}U_{2}\hat{Z}U_{2}^{*}U_{1}^{*}\hat{Z}^{*}\right)$
$\displaystyle-N^{-\frac{1}{2}}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{f_{\alpha}+t_{0}}\left(\overline{z_{0}-a_{\alpha}}{\rm
Tr}\big{(}\hat{Z}\big{)}+(z_{0}-a_{\alpha}){\rm
Tr}\big{(}\hat{Z}^{*}\big{)}\right)+\frac{P_{1}}{2}{\rm
Tr}\big{(}\widetilde{\Lambda}^{2}\big{)}$
$\displaystyle-N^{-\frac{1}{2}}\left(\overline{P_{0}}{\rm
Tr}\big{(}U_{2}^{*}\widetilde{\Lambda}U_{2}\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}U_{1}\widetilde{\Lambda}U_{1}^{*}\hat{Z}^{*}\big{)}\right)+O\Big{(}\big{(}N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}^{3}\Big{)}.$

From the above we know the typical size of $\widetilde{\Lambda}$ is
$N^{-\frac{1}{2}}$.

Combining (2.9), (3.7), (3.9), (3.16)-(3.23) and (3.27), we have

$f(T,Y,Q)-f_{0}\big{(}\tau;\big{\\{}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}\big{\\}},\sqrt{t_{0}}U_{1}U_{2},0\big{)}=F_{0}+F_{1}+O(F_{2}),$
(3.28)

where

$\displaystyle F_{0}$
$\displaystyle=\frac{N^{-\frac{1}{2}}}{\tau}\left(\overline{z_{0}}{\rm
Tr}\big{(}\hat{Z}\big{)}+z_{0}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}\right)+\frac{n^{2}}{N}$
$\displaystyle-\frac{N^{-1}}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\left(\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\right),$ $\displaystyle
F_{1}=\frac{|z_{0}|^{2}+t_{0}}{\tau}{\rm Tr}(\Gamma_{1})-\frac{1}{\tau}{\rm
Tr}\Big{(}Q_{t+1}\Big{(}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}+t_{0}\mathbb{I}_{r_{t+1}}\Big{)}Q_{t+1}^{*}\Big{)}$
(3.29) $\displaystyle-\frac{t_{0}}{\tau}{\rm
Tr}(Q_{0}Q_{0}^{*})-\sum_{\alpha=2}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}-\frac{(f_{1}+t_{0})^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm d},\alpha}\Big{)}^{2}$
$\displaystyle+\frac{N^{-\frac{1}{2}}}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}+t_{0}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}$
$\displaystyle-\frac{(f_{1}+t_{0})^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm u},\alpha}\Big{)}^{*}$
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\Big{|}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\Big{|}^{2}+\frac{t_{0}}{N}P_{1}{\rm
Tr}\big{(}U_{1}U_{2}\hat{Z}U_{2}^{*}U_{1}^{*}\hat{Z}^{*}\big{)}$
$\displaystyle-\frac{P_{1}}{2}{\rm
Tr}\big{(}\widetilde{\Lambda}^{2}\big{)}+N^{-\frac{1}{2}}\Big{(}\overline{P_{0}}{\rm
Tr}\big{(}U_{2}^{*}\widetilde{\Lambda}U_{2}\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}U_{1}\widetilde{\Lambda}U_{1}^{*}\hat{Z}^{*}\big{)}\Big{)},$

and

$\displaystyle
F_{2}=\big{(}N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}^{3}+\Omega_{{\rm
error}}^{3}+N^{-\frac{1}{2}}\big{(}\|\Gamma_{1}\|+\Omega_{{\rm
error}}^{2}\big{)}+\big{(}\|\Gamma_{1}\|+\|G_{1}\|\big{)}\Omega_{{\rm
error}}.$

From (3.17) and (3.45) we see that the typical size of $F_{2}$ is
$N^{-\frac{3}{2}}$, so $O(F_{2})$ is the error term.

Step 2: Taylor expansion of $g(T,Y,Q)$. For $g(T,Y,Q)$ defined in (2.11), we
write down leading terms for five relevant factors while the estimate for the
last determinant in (2.11) is non-trivial. Note $R_{\alpha,N}$ is defined in
(1.3).

  * (1)

For the first factor in (2.11), by (3.11) and (3.13),

$\Big{(}\det\big{(}\mathbb{I}_{n}-\sum_{\alpha=1}^{t}T_{\alpha}T_{\alpha}^{*}-Q_{0}Q_{0}^{*}-Q_{t+1}Q_{t+1}^{*}\big{)}\Big{)}^{R_{0}-n}=\Big{(}\det\big{(}-\Gamma_{1}-G_{1}-G_{1}^{*}\big{)}\Big{)}^{R_{0}-n},$
(3.30)

  * (2)

For the second factor in (2.11), by (2.1), (2.19) and (3.26),

$\det\begin{bmatrix}\sqrt{\gamma_{N}}Z&-Y^{*}\\\
Y&\sqrt{\gamma_{N}}Z^{*}\end{bmatrix}=(|z_{0}|^{2}+t_{0})^{n}+O\big{(}N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}.$
(3.31)

  * (3)

For the third factor in (2.11) with $\alpha=1,\cdots,t,$, it is easy to see
from (3.6) and (3.17) that

$\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\big{(}t_{j,j}^{(\alpha)}\big{)}^{R_{\alpha,N}-j}=\prod_{\alpha=1}^{t}\prod_{j=1}^{n}\Big{(}\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}+O\big{(}\Omega_{\rm
error}\big{)}\Big{)}^{R_{\alpha,N}-j}.$ (3.32)

  * (4)

For the fourth factor in (2.11), by (2.1) and (2.2)

$\big{(}\det(A_{\alpha})\big{)}^{R_{\alpha,N}-n}=\Big{(}(f_{\alpha}+t_{0})^{n}+O\big{(}N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}\Big{)}^{R_{\alpha,N}-n}.$
(3.33)

The remaining part of this subsection is devoted to the last determinant in
(2.11).

Recalling $\hat{L}_{1}$ and $\hat{L}_{2}$ defined in (2.3) and (2.5), since
all $a_{\alpha}\neq z_{0}$ and $t_{0}>0$, $A_{{\alpha}}$ is invertible when
$N$ is sufficiently large. Rewrite

$\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\left[\begin{smallmatrix}P_{1,1}&O\big{(}\|\widetilde{Q}_{t+1}\|\big{)}\\\
O\big{(}\|\widetilde{Q}_{t+1}\|\big{)}&\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\end{smallmatrix}\right],$

where

$P_{1,1}=\left[\begin{smallmatrix}A_{1}\otimes\mathbb{I}_{n}&&\\\ &\ldots&\\\
&&A_{t}\otimes\mathbb{I}_{n}\end{smallmatrix}\right]+\sqrt{\gamma_{N}}\left[\begin{smallmatrix}\mathbb{I}_{2n}\otimes
T_{1}^{*}\\\ \vdots\\\ \mathbb{I}_{2n}\otimes
T_{t}^{*}\end{smallmatrix}\right]\left[\begin{smallmatrix}\left[\begin{smallmatrix}a_{1}\mathbb{I}_{n}&\\\
&\overline{a}_{1}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{1}&,\cdots,&\left[\begin{smallmatrix}a_{t}\mathbb{I}_{n}&\\\
&\overline{a}_{t}\mathbb{I}_{n}\end{smallmatrix}\right]\otimes
T_{t}\end{smallmatrix}\right],$

change the order of the above matrix product and we obtain

$\displaystyle\det(P_{1,1})=\Big{(}\prod_{\alpha=1}^{t}\det\big{(}A_{\alpha}\big{)}\Big{)}^{n}\det\bigg{(}\mathbb{I}_{2n^{2}}+\sqrt{\gamma_{N}}\sum_{\alpha=1}^{t}\left(A_{\alpha}^{-1}\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\right)\otimes(T_{\alpha}T_{\alpha}^{*})\bigg{)},$

From (2.19) and (3.26), we have
$Y=\sqrt{t_{0}}U_{1}U_{2}+O(\|\widetilde{\Lambda}\|),$ in view of (2.2),
rewrite

$A_{\alpha}=\begin{bmatrix}Z_{\alpha}&-Y^{*}\\\
Y&Z_{\alpha}^{*}\end{bmatrix},\quad
Z_{\alpha}=\sqrt{\gamma_{N}}(Z-a_{\alpha}\mathbb{I}_{n}),$

then

$\displaystyle
A_{\alpha}^{-1}=\begin{bmatrix}\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}&Z_{\alpha}^{-1}Y^{*}\big{(}Z_{\alpha}^{*}+YZ_{\alpha}^{-1}Y^{*}\big{)}^{-1}\\\
-\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}&\big{(}Z_{\alpha}^{*}+YZ_{\alpha}^{-1}Y^{*}\big{)}^{-1}\end{bmatrix}.$

Simple calculations show that

$\Big{(}Z_{\alpha}+Y^{*}\big{(}Z_{\alpha}^{*}\big{)}^{-1}Y\Big{)}^{-1}=\overline{z_{0}-a_{\alpha}}(f_{\alpha}+t_{0})^{-1}\mathbb{I}_{n}+O\big{(}N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)},$
$T_{\alpha}T_{\alpha}^{*}=\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}\mathbb{I}_{n}+O\big{(}\Omega_{\rm
error}\big{)}.$

Rewrite
$\frac{a_{\alpha}}{z_{0}-a_{\alpha}}=\frac{a_{\alpha}\overline{z_{0}-a_{\alpha}}}{f_{\alpha}}$,
we have

$\displaystyle\mathbb{I}_{2n^{2}}$
$\displaystyle+\sqrt{\gamma_{N}}\sum_{\alpha=1}^{t}\left(A_{\alpha}^{-1}\left[\begin{smallmatrix}a_{\alpha}\mathbb{I}_{n}&\\\
&\overline{a}_{\alpha}\mathbb{I}_{n}\end{smallmatrix}\right]\right)\otimes(T_{\alpha}T_{\alpha}^{*})=$
$\displaystyle\tau\begin{bmatrix}(t_{0}P_{1}-z_{0}\overline{P}_{0})\mathbb{I}_{n^{2}}&\sqrt{t_{0}}(\overline{P}_{0}+\overline{z}_{0}P_{1})((U_{2}^{*}U_{1}^{*})\otimes\mathbb{I}_{n})\\\
-\sqrt{t_{0}}(P_{0}+z_{0}P_{1})((U_{1}U_{2})\otimes\mathbb{I}_{n})&(t_{0}P_{1}-\overline{z}_{0}P_{0})\mathbb{I}_{n^{2}}\end{bmatrix}+O\big{(}N^{-\frac{1}{2}}+\Omega_{\rm
error}+\|\widetilde{\Lambda}\|\big{)},$

where $P_{0}$ and $P_{1}$ are defined in (1.8). Also
$|t_{0}P_{1}-\overline{z}_{0}P_{0}|^{2}+t_{0}|P_{0}+z_{0}P_{1}|^{2}=(|z_{0}|^{2}+t_{0})(|P_{0}|^{2}+t_{0}P_{1}^{2}).$
Combining (3.33) we have

$\det\big{(}P_{1,1}\big{)}=\tau^{2n^{2}}(|z_{0}|^{2}+t_{0})^{n^{2}}(|P_{0}|^{2}+t_{0}P_{1}^{2})^{n^{2}}\prod_{\alpha=1}^{t}(f_{\alpha}+t_{0})^{n^{2}}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}.$ (3.34)

As $t_{0}>0$, we have $\det(P_{1,1})\not=0.$ and from
$\widetilde{Q}_{t+1}=[Q_{0},Q_{t+1}]$,
$\|\widetilde{Q}_{t+1}\|^{2}=O\big{(}\Omega_{\rm error}\big{)}.$ Therefore,

$\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\det(P_{1,1})\det\Big{(}\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\Big{)}.$

In view of (2.4) and (2.6), from elementary computations,

$\displaystyle\det\Big{(}\widehat{B}_{t+1}+O\big{(}\|\widetilde{Q}_{t+1}\|^{2}\big{)}\Big{)}=t_{0}^{nr_{0}}\Big{(}\det\big{(}$
$\displaystyle(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})^{*}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})+t_{0}\mathbb{I}_{r_{t+1}}\big{)}\Big{)}^{n}$
(3.35) $\displaystyle+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)},$

Combing (3.34)-(3.35), we have

$\displaystyle\det\\!\bigg{(}\widehat{L}_{1}+\sqrt{\gamma_{N}}\widehat{L}_{2}\bigg{)}=\tau^{2n^{2}}(|z_{0}|^{2}+t_{0})^{n^{2}}(|P_{0}|^{2}+t_{0}P_{1}^{2})^{n^{2}}t_{0}^{nr_{0}}\prod_{\alpha=1}^{t}(f_{\alpha}+t_{0})^{n^{2}}$
(3.36)
$\displaystyle\times\Big{(}\det\big{(}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})^{*}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})+t_{0}\mathbb{I}_{r_{t+1}}\big{)}\Big{)}^{n}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)},$

Now recalling (2.11), combining (3.30)-(3.33) and (3.36) we have

$\displaystyle
g(T,Y,Q)=\Big{(}\det\big{(}-\Gamma_{1}-G_{1}-G_{1}^{*}\big{)}\Big{)}^{R_{0}-n}\bigg{(}(|z_{0}|^{2}+t_{0})^{nR_{0}}\tau^{n\sum_{\alpha=1}^{t}R_{\alpha,N}-\frac{n(n+1)}{2}t+2n^{2}}$
(3.37)
$\displaystyle\times(|P_{0}|^{2}+t_{0}P_{1}^{2})^{n^{2}}t_{0}^{nr_{0}}\prod_{\alpha=1}^{t}\Big{(}c_{\alpha}^{nR_{\alpha,N}-\frac{(n+1)n}{2}}(f_{\alpha}+t_{0})^{\frac{(n+1)n}{2}}\Big{)}$
$\displaystyle\times\Big{(}\det\big{(}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})^{*}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})+t_{0}\mathbb{I}_{r_{t+1}}\big{)}\Big{)}^{n}+O\big{(}\Omega_{\rm
error}+N^{-\frac{1}{2}}+\|\widetilde{\Lambda}\|\big{)}\bigg{)}.$

Step 3: Removal of error terms: A standard procedure of Laplace method.
Finally, with those preparations in Step 1 and Step 2, we will give a compete
proof of Theorem 1.2.

Recalling (2.26), (3.18) and (3.28), we rewrite

$e^{-NF_{0}}I_{N,\delta}=J_{1,N}+J_{2,N},$ (3.38)

where

$J_{1,N}=\frac{\pi^{n^{2}}}{\big{(}\prod_{i=1}^{n-1}i!\big{)}^{2}}\Big{(}\frac{f_{1}+t_{0}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}\int_{\hat{\Omega}_{N,\delta}}g(T,Y,Q)\big{(}1+O\big{(}\Omega_{\rm
error}\big{)}\big{)}e^{NF_{1}}\prod_{1\leq i<j\leq
n}(\widetilde{\lambda}_{j}-\widetilde{\lambda}_{i})^{2}{\rm d}\widetilde{V},$
(3.39)

and

$J_{2,N}=\frac{\pi^{n^{2}}}{\big{(}\prod_{i=1}^{n-1}i!\big{)}^{2}}\Big{(}\frac{f_{1}+t_{0}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}\int_{\hat{\Omega}_{N,\delta}}g(T,Y,Q)\big{(}1+O\big{(}\Omega_{\rm
error}\big{)}\big{)}e^{NF_{1}}\prod_{1\leq i<j\leq
n}(\widetilde{\lambda}_{j}-\widetilde{\lambda}_{i})^{2}\big{(}e^{O(NF_{2})}-1\big{)}{\rm
d}\widetilde{V},$ (3.40)

with $Y=U_{1}\sqrt{t_{0}\mathbb{I}_{n}+\widetilde{\Lambda}}U_{2},$

${\rm d}\widetilde{V}={\rm d}\widetilde{\Lambda}{\rm d}U_{1}{\rm d}U_{2}{\rm
d}Q_{0}{\rm d}Q_{t+1}{\rm d}\Gamma_{1}{\rm d}G_{1}\prod_{\alpha=2}^{t}{\rm
d}\widetilde{T}_{{\rm d},\alpha}{\rm d}T_{{\rm u},\alpha}$ (3.41)

and

$\hat{\Omega}_{N,\delta}=A_{N,\delta}\cap\big{\\{}\Gamma_{1}+G_{1}+G_{1}^{*}\leq
0\big{\\}}\cap\big{\\{}\widetilde{\lambda}_{n}\leq\cdots\leq\widetilde{\lambda}_{1}\big{\\}},$
(3.42)

cf. (2.23) for definition of $A_{N,\delta}$.

Under the restriction condition of $\Gamma_{1}+G_{1}+G_{1}^{*}\leq 0$, every
principal minor of order 2 is non-positive definite, so all diagonal entries
of $\Gamma_{1}$ are zero or negative and

${\rm Tr}\big{(}G_{1}G_{1}^{*}\big{)}=\sum_{1\leq i<j\leq
n}\big{|}(G_{1})_{i,j}\big{|}^{2}\leq\sum_{1\leq i<j\leq
n}(\Gamma_{1})_{i,i}(\Gamma_{1})_{j,j}\leq\big{(}{\rm
Tr}(\Gamma_{1})\big{)}^{2}.$ (3.43)

Noticing the absence of $G_{1}$ in $F_{1}$ given in (3.29), for convergence we
need to control $G_{1}$ by $\Gamma_{1}$ in $F_{2}$, just as shown in (3.43).
Since $\|G_{1}\|=O(\|\Gamma_{1}\|)$ from (3.43), also from (2.21) and (2.23)
for sufficiently large $N$ and small $\delta$, there exists some $C>0$ such
that

$\displaystyle\frac{1}{C}NF_{2}\leq
N^{-\frac{1}{2}}+\sqrt{\delta}+\sqrt{N\delta}\big{(}\|\widetilde{\Lambda}\|+\|\widetilde{T}_{{\rm
d}}\|_{2}+\|Q_{0}\|+\|Q_{t+1}\|+\|T_{{\rm u}}\|_{2}\big{)}$
$\displaystyle+N\sqrt{\delta}\big{(}\|Q_{0}\|^{2}+\|\widetilde{\Lambda}\|^{2}+\|\widetilde{T}_{{\rm
d}}\|_{2}^{2}+\|T_{{\rm u}}\|_{2}^{2}+\|Q_{t+1}\|^{2}+\|\Gamma_{1}\|\big{)}.$

Using the inequality

$\big{|}e^{O(NF_{2})}-1\big{|}\leq O(N|F_{2}|)e^{O(N|F_{2}|)},$ (3.44)

after change of variables

$\displaystyle(G_{1},\Gamma_{1})\rightarrow
N^{-1}(G_{1},\Gamma_{1}),\quad(Q_{0},Q_{t+1},\widetilde{\Lambda})\rightarrow
N^{-\frac{1}{2}}(Q_{0},Q_{t+1},\widetilde{\Lambda}),$ (3.45)
$\displaystyle\big{(}\widetilde{T}_{{\rm d},\alpha},T_{{\rm
u},\alpha}\big{)}\rightarrow N^{-\frac{1}{2}}\big{(}\widetilde{T}_{{\rm
d},\alpha},T_{{\rm u},\alpha}\big{)}\quad\alpha=2,\cdots,t,$

the term $O(NF_{2})$ in (3.44) has an upper bound by
$N^{-\frac{1}{2}}P(\tilde{V})$ for some polynomial of variables. since $F_{1}$
can control $F_{2}$ for sufficiently small $\delta$, by the argument of
Laplace method and the dominant convergence theorem we know that $J_{2,N}$ is
typically of order $N^{-\frac{1}{2}}$ compared with $J_{1,N}$, that is,

$J_{2,N}=O\big{(}N^{-\frac{1}{2}}\big{)}J_{1,N}.$ (3.46)

For $J_{1,N}$, take a large $M_{0}>0$ such that

$\displaystyle A_{N,\delta}^{\complement}\subseteq$
$\displaystyle\bigcup_{\alpha=2}^{t}\Big{\\{}{\rm Tr}\big{(}T_{{\rm
u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\bigcup_{\alpha=2}^{t}\Big{\\{}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}\widetilde{T}_{{\rm
d},\alpha}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}\Gamma_{1}\Gamma_{1}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}$
$\displaystyle\bigcup\Big{\\{}{\rm
Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}\widetilde{\Lambda}^{2}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}\bigcup\Big{\\{}{\rm
Tr}\big{(}Q_{t+1}Q_{t+1}^{*}\big{)}>\frac{\delta}{M_{0}}\Big{\\}}.$

Here we have used (3.43) to drop out the domain $\\{{\rm
Tr}\big{(}G_{1}G_{1}^{*}\big{)}>\delta/M_{0}\\}$. For each piece of domain,
only keep the restricted matrix variable and let the others free, it’s easy to
prove that the corresponding matrix integral is exponentially small, that is,
$O\big{(}e^{-\delta_{1}N}\big{)}$ for some $\delta_{1}>0$.

So we can extend the integration region from $\hat{\Omega}_{N,\delta}$ to
$\big{\\{}\Gamma_{1}+G_{1}+G_{1}^{*}\leq
0\big{\\}}\cap\big{\\{}\widetilde{\lambda}_{n}\leq\cdots\leq\widetilde{\lambda}_{1}\big{\\}}$.
From (1.3) and (1.4), we have
$\sum_{\alpha=1}^{t}R_{\alpha,N}=-r_{0}-r_{t+1}-R_{0}$, and by the change of
variables (3.45) we have

$\displaystyle J_{1,N}$
$\displaystyle=N^{-\frac{n^{2}t}{2}-n(R_{0}+r_{0}+r_{t+1})}\frac{\pi^{n^{2}}}{\big{(}\prod_{i=1}^{n-1}i!\big{)}^{2}}\Big{(}\frac{f_{1}+t_{0}}{\tau
c_{1}}\Big{)}^{\frac{n(n-1)}{2}}\tau^{2n^{2}-n(r_{0}+r_{t+1}+R_{0})-\frac{n(n+1)}{2}t}$
(3.47)
$\displaystyle\times(|P_{0}|^{2}+t_{0}P_{1}^{2})^{n^{2}}t_{0}^{nr_{0}}\Big{(}\det\big{(}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})^{*}(z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1})+t_{0}\mathbb{I}_{r_{t+1}}\big{)}\Big{)}^{n}$
$\displaystyle\times(|z_{0}|^{2}+t_{0})^{n^{2}}\prod_{\alpha=1}^{t}c_{\alpha}^{nR_{\alpha,N}-\frac{n(n+1)}{2}}(f_{\alpha}+t_{0})^{\frac{n(n+1)}{2}}\Big{(}I_{0}+O\big{(}N^{-\frac{1}{2}}\big{)}\Big{)},$

where with $\widehat{H}=\Gamma_{1}+G_{1}+G_{1}^{*}$,

$I_{0}=\int_{\widehat{H}\leq
0}\int_{\widetilde{\lambda}_{n}\leq\cdots\leq\widetilde{\lambda}_{1}}\big{(}\det(-\widehat{H})\big{)}^{R_{0}-n}\prod_{1\leq
i<j\leq n}(\widetilde{\lambda}_{j}-\widetilde{\lambda}_{i})^{2}e^{F}{\rm
d}\widetilde{V},$ (3.48) $\displaystyle F=\frac{|z_{0}|^{2}+t_{0}}{\tau}{\rm
Tr}(\Gamma_{1})-\frac{1}{\tau}{\rm
Tr}\Big{(}Q_{t+1}\Big{(}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}+t_{0}\mathbb{I}_{r_{t+1}}\Big{)}Q_{t+1}^{*}\Big{)}$
(3.49) $\displaystyle-\frac{t_{0}}{\tau}{\rm
Tr}(Q_{0}Q_{0}^{*})-\sum_{\alpha=2}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm
d},\alpha}^{2}\big{)}-\frac{(f_{1}+t_{0})^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm d},\alpha}\Big{)}^{2}$
$\displaystyle+\frac{1}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}+t_{0}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm u},\alpha}^{*}\big{)}$
$\displaystyle-\frac{(f_{1}+t_{0})^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm u},\alpha}\Big{)}^{*}$
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\Big{|}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\Big{|}^{2}+t_{0}P_{1}{\rm
Tr}\big{(}U_{1}U_{2}\hat{Z}U_{2}^{*}U_{1}^{*}\hat{Z}^{*}\big{)}$
$\displaystyle-\frac{P_{1}}{2}{\rm
Tr}\big{(}\widetilde{\Lambda}^{2}\big{)}+\overline{P_{0}}{\rm
Tr}\big{(}U_{2}^{*}\widetilde{\Lambda}U_{2}\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}U_{1}\widetilde{\Lambda}U_{1}^{*}\hat{Z}^{*}\big{)},$

Step 4: Matrix integrals and final proof. To simplify $I_{0}$ further, we
first integrate out $Q_{0},Q_{t+1}$ and $T_{{\rm u},2},\cdots,T_{{\rm u},t}$,

$\displaystyle\int e^{-\frac{1}{\tau}{\rm
Tr}\Big{(}Q_{t+1}\Big{(}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}+t_{0}\mathbb{I}_{r_{t+1}}\Big{)}Q_{t+1}^{*}\Big{)}}{\rm
d}Q_{t+1}$ (3.50)
$\displaystyle=\Big{(}\det\Big{(}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}^{*}\big{(}z_{0}\mathbb{I}_{r_{t+1}}-A_{t+1}\big{)}+t_{0}\mathbb{I}_{r_{t+1}}\Big{)}\Big{)}^{-n}(\pi\tau)^{nr_{t+1}},$
$\int e^{-\frac{t_{0}}{\tau}{\rm Tr}\big{(}Q_{0}Q_{0}^{*}\big{)}}{\rm
d}Q_{0}=(\pi\tau)^{nr_{0}}t_{0}^{-nr_{0}}.$ (3.51)

Calculate the gaussian integrals and we obtain

$\displaystyle\int\exp\Big{\\{}-\sum_{\alpha=2}^{t}\frac{f_{\alpha}+t_{0}}{\tau}{\rm
Tr}\big{(}T_{{\rm u},\alpha}T_{{\rm
u},\alpha}^{*}\big{)}-\frac{(f_{1}+t_{0})^{2}}{\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm
u},\alpha}\Big{)}\Big{(}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}T_{{\rm u},\alpha}\Big{)}^{*}$ (3.52)
$\displaystyle+\frac{1}{\tau}\sum_{1\leq i<j\leq
n}\Big{|}\sum_{\alpha=2}^{t}\sqrt{\frac{\tau
c_{\alpha}}{f_{\alpha}+t_{0}}}(a_{\alpha}-a_{1})t_{i,j}^{(\alpha)}\Big{|}^{2}\Big{\\}}\prod_{\alpha=2}^{t}{\rm
d}T_{{\rm
u},\alpha}=\big{(}\det(A)\big{)}^{-\frac{n(n-1)}{2}}\pi^{\frac{n(n-1)}{2}(t-1)},$

where for $\alpha,\beta=2,\cdots,t$,

$A_{\alpha,\beta}=\Big{(}\frac{(f_{1}+t_{0})^{2}}{\tau
c_{1}}-(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}\Big{)}\sqrt{\frac{c_{\alpha}c_{\beta}}{(f_{\alpha}+t_{0})(f_{\beta}+t_{0})}}+\frac{1}{\tau}(f_{\alpha}+t_{0})\delta_{\alpha,\beta}.$

To compute $\det(A)$, first rewrite A as

$A={\rm
diag}\Big{(}\sqrt{\frac{c_{\alpha}}{f_{\alpha}+t_{0}}}\Big{)}_{\alpha=2}^{t}\Big{(}\frac{(f_{1}+t_{0})^{2}}{\tau
c_{1}}\vec{1}^{t}\vec{1}+\widehat{A}\Big{)}{\rm
diag}\Big{(}\sqrt{\frac{c_{\beta}}{f_{\beta}+t_{0}}}\Big{)}_{\beta=2}^{t},$

where

$\vec{1}=(1,1,\cdots,1),\quad\widehat{A}_{\alpha,\beta}=\frac{(f_{\alpha}+t_{0})^{2}}{\tau
c_{\alpha}}\delta_{\alpha,\beta}-(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}},\quad\alpha,\beta=2,\cdots,t.$

Then

$\det(A)=\det\big{(}\widehat{A}\big{)}\big{(}1+\frac{(f_{1}+t_{0})^{2}}{\tau
c_{1}}\vec{1}\widehat{A}^{-1}\vec{1}^{t}\big{)}\prod_{\alpha=2}^{t}\frac{c_{\alpha}}{f_{\alpha}+t_{0}}$
(3.53)

From basic computations, for $\alpha,\beta=2,\cdots,t$,

$\displaystyle\big{(}\widehat{A}^{-1}\big{)}_{\alpha,\beta}=\frac{\tau
c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\delta_{\alpha,\beta}+\frac{\tau^{2}c_{\alpha}c_{\beta}(a_{\alpha}-a_{1})\overline{a_{\beta}-a_{1}}}{(f_{\alpha}+t_{0})^{2}(f_{\beta}+t_{0})^{2}}\Big{(}1-\sum_{\alpha=2}^{t}\frac{\tau
c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}^{-1},$ (3.54)
$\displaystyle\det\big{(}\widehat{A}\big{)}=\Big{(}1-\sum_{\alpha=2}^{t}\frac{\tau
c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}\prod_{\alpha=2}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{\tau
c_{\alpha}}.$

Recall the definition of $P_{0}$ and $P_{1}$ in (1.8),

$\Big{|}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}(a_{\alpha}-a_{1})\Big{|}^{2}=\Big{|}\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\big{(}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\big{)}\Big{|}^{2}=|P_{0}-(a_{1}-z_{0})P_{1}|^{2}.$

Also from $\sum_{\alpha=1}^{t}\frac{\tau c_{\alpha}}{f_{\alpha}+t_{0}}=1,$

$\displaystyle\sum_{\alpha=2}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}|a_{\alpha}-a_{1}|^{2}$
$\displaystyle=\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\big{|}(a_{\alpha}-z_{0})-(a_{1}-z_{0})\big{|}^{2}$
$\displaystyle=\frac{1}{\tau}-t_{0}P_{1}-\overline{a_{1}-z_{0}}P_{0}-(a_{1}-z_{0})\overline{P}_{0}+|a_{1}-z_{0}|^{2}P_{1}.$

so from (3.54) and basic computations we obtain

$\big{(}1+\frac{(f_{1}+t_{0})^{2}}{\tau
c_{1}}\vec{1}\widehat{A}^{-1}\vec{1}^{t}\big{)}\Big{(}1-\sum_{\alpha=2}^{t}\frac{\tau
c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}|a_{\alpha}-a_{1}|^{2}\Big{)}=\tau\frac{(f_{1}+t_{0})^{2}}{c_{1}}\big{(}t_{0}P_{1}^{2}+|P_{0}|^{2}\big{)}.$

Hence, Combining (3.53) and (3.54) we obtain

$\det(A)=\tau^{-(t-2)}\frac{f_{1}+t_{0}}{c_{1}}\big{(}t_{0}P_{1}^{2}+|P_{0}|^{2}\big{)}\prod_{\alpha=1}^{t}(f_{\alpha}+t_{0}).$
(3.55)

Secondly, to integrate out $G_{1}$ and $\Gamma_{1}$, we need the following
proposition.

######  Proposition 3.1.

For a non-positive definite Hermitian matrix
$H_{n}=\left[h_{i,j}\right]_{i,j=1}^{n}$, given an integer $r_{0}\geq n$, then

$\int_{H_{n}\leq 0}\left(\det(H_{n})\right)^{r_{0}-n}\prod_{i<j}^{n}{\rm
d}h_{i,j}=\pi^{\frac{n(n-1)}{2}}((r_{0}-1)!)^{-n}\prod_{k=1}^{n}(r_{0}-k)!\prod_{j=1}^{n}(h_{j,j})^{r_{0}-1}.$
(3.56)

For proof see [11, Proposition A.2]. From this Proposition, recalling
$\widehat{H}=\Gamma_{1}+G_{1}+G_{1}^{*}$, we have

$\int_{\widehat{H}\leq 0}\big{(}\det(-\widehat{H})\big{)}^{R_{0}-n}{\rm
d}G_{1}=(-1)^{n(R_{0}-n)}\pi^{\frac{n(n-1)}{2}}\frac{\prod_{i=1}^{n}(R_{0}-i)!\big{(}\det(\Gamma_{1})\big{)}^{R_{0}-1}}{\big{(}(R_{0}-1)!\big{)}^{n}},$
(3.57)

from which

$(-1)^{n(R_{0}-n)}\int_{\Gamma_{1}\leq
0}\big{(}\det(\Gamma_{1})\big{)}^{R_{0}-1}e^{\frac{|z_{0}|^{2}+t_{0}}{\tau}{\rm
Tr}(\Gamma_{1})}{\rm
d}\Gamma_{1}=\tau^{nR_{0}}(|z_{0}|^{2}+t_{0})^{-nR_{0}}\big{(}(R_{0}-1)!\big{)}^{n}.$
(3.58)

Thirdly, we need to integrate out $\widetilde{T}_{{\rm d},\alpha}$ for
$\alpha=2,\cdots,t$. For $i=1,\cdots,n$, let

$\vec{b}_{i}=\big{(}b_{i}^{(2)},b_{i}^{(3)},\cdots,b_{i}^{(t)}\big{)},\quad
b_{i}^{(\alpha)}=\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}_{i,i},$

and introduce a square matrix

$\Sigma=[\Sigma_{\alpha,\beta}]_{\alpha,\beta=2}^{t},\quad\Sigma_{\alpha,\beta}=\frac{(f_{\alpha}+t_{0})^{2}}{2c_{\alpha}}\delta_{\alpha,\beta}+\frac{(f_{1}+t_{0})^{2}}{2c_{1}},$

Noting

$\det(\Sigma)=2P_{1}\prod_{\alpha=1}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2c_{\alpha}},\quad(\Sigma^{-1})_{\alpha,\beta}=\frac{2c_{\alpha}\delta_{\alpha,\beta}}{(f_{\alpha}+t_{0})^{2}}-\frac{2c_{\alpha}c_{\beta}}{(f_{\alpha}+t_{0})^{2}(f_{\beta}+t_{0})^{2}P_{1}},\quad\alpha,\beta=2,\cdots,t,$
(3.59)

where $P_{1}$ is given in (1.8). Calculate the gaussian integrals and we
obtain

$\displaystyle\int\exp\Big{\\{}-\frac{(f_{1}+t_{0})^{2}}{2\tau^{2}c_{1}}{\rm
Tr}\Big{(}\sum_{\alpha=2}^{t}\widetilde{T}_{{\rm
d},\alpha}\Big{)}^{2}-\sum_{\alpha=2}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2\tau^{2}c_{\alpha}}{\rm
Tr}\big{(}\widetilde{T}_{{\rm d},\alpha}^{2}\big{)}$ (3.60)
$\displaystyle+\frac{1}{\tau}\sum_{\alpha=2}^{t}{\rm
Tr}\Big{(}\big{(}\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}\big{)}\widetilde{T}_{{\rm
d},\alpha}\Big{)}\Big{\\}}\prod_{\alpha=2}^{t}{\rm d}\widetilde{T}_{{\rm
d},\alpha}$
$\displaystyle=\Big{(}2P_{1}\prod_{\alpha=1}^{t}\frac{(f_{\alpha}+t_{0})^{2}}{2c_{\alpha}}\Big{)}^{-\frac{n}{2}}\pi^{\frac{t-1}{2}n}\tau^{(t-1)n}e^{\frac{1}{4}\sum_{i=1}^{n}\vec{b}_{i}\Sigma^{-1}\vec{b}_{i}^{t}}.$

Rewrite

$\overline{a_{\alpha}-a_{1}}\hat{Z}+(a_{\alpha}-a_{1})\hat{Z}^{*}=\overline{a_{\alpha}-z_{0}}\hat{Z}+(a_{\alpha}-z_{0})\hat{Z}^{*}-\big{(}\overline{a_{1}-z_{0}}\hat{Z}+(a_{1}-z_{0})\hat{Z}^{*}\big{)},$

obviously,

$\displaystyle\sum_{i=1}^{n}\sum_{\alpha=2}^{t}\frac{c_{\alpha}\big{(}b_{i}^{(\alpha)}\big{)}^{2}}{(f_{\alpha}+t_{0})^{2}}=\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}{\rm
Tr}\big{(}\overline{a_{\alpha}-z_{0}}\hat{Z}+(a_{\alpha}-z_{0})\hat{Z}^{*}\big{)}^{2}$
$\displaystyle-2{\rm
Tr}\Big{(}\big{(}\overline{a_{1}-z_{0}}\hat{Z}+(a_{1}-z_{0})\hat{Z}^{*}\big{)}\big{(}\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*}\big{)}\Big{)}+P_{1}{\rm
Tr}\big{(}\overline{a_{1}-z_{0}}\hat{Z}+(a_{1}-z_{0})\hat{Z}^{*}\big{)}^{2},$
$\displaystyle\sum_{i=1}^{n}\Big{(}\sum_{\alpha=2}^{t}\frac{c_{\alpha}b_{i}^{(\alpha)}}{(f_{\alpha}+t_{0})^{2}}\Big{)}^{2}={\rm
Tr}\Big{(}\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*}-P_{1}\big{(}\overline{a_{1}-z_{0}}\hat{Z}+(a_{1}-z_{0})\hat{Z}^{*}\big{)}\Big{)}^{2}.$

Use (3.59), elementary calculation gives us

$\displaystyle\frac{1}{4}\sum_{i=1}^{n}\vec{b}_{i}\Sigma^{-1}\vec{b}_{i}^{t}=\frac{1}{2}\sum_{i=1}^{n}\sum_{\alpha=2}^{t}\frac{c_{\alpha}\big{(}b_{i}^{(\alpha)}\big{)}^{2}}{(f_{\alpha}+t_{0})^{2}}-\frac{1}{2P_{1}}\sum_{i=1}^{n}\Big{(}\sum_{\alpha=2}^{t}\frac{c_{\alpha}b_{i}^{(\alpha)}}{(f_{\alpha}+t_{0})^{2}}\Big{)}^{2}$
(3.61)
$\displaystyle=\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}{\rm
Tr}\big{(}\overline{a_{\alpha}-z_{0}}\hat{Z}+(a_{\alpha}-z_{0})\hat{Z}^{*}\big{)}^{2}-\frac{1}{2P_{1}}{\rm
Tr}\big{(}\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*}\big{)}^{2},$

Finally, with (2.13) and (2.27) in mind, by the Stirling’s formula we can
obtain

$\displaystyle
e^{NF_{0}}D_{N,n}=\frac{N^{\frac{n^{2}t}{2}+n(R_{0}+1+r_{0}+r_{t+1})}}{\pi^{n(n+1+r_{0}+r_{t+1})+\frac{n(n-1)}{2}t}}(2\pi)^{-\frac{nt}{2}}\frac{\prod_{\alpha=1}^{t}c_{\alpha}^{-nR_{\alpha,N}+\frac{n^{2}}{2}}}{\prod_{k=1}^{n}(R_{0}-k)!}\tau^{-\frac{n(n+1)}{2}}$
(3.62) $\displaystyle\times\prod_{1\leq i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{2}e^{-\frac{1}{\tau}{\rm
Tr}\big{(}\hat{Z}\hat{Z}^{*}\big{)}-\frac{1}{2}\sum_{\alpha=1}^{t}\frac{c_{\alpha}}{(f_{\alpha}+t_{0})^{2}}\Big{(}\overline{z_{0}-a_{\alpha}}^{2}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+(z_{0}-a_{\alpha})^{2}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\Big{)}}\big{(}1+O(N^{-1})\big{)}.$

Now combining (2.25), (3.38), (3.46), (3.47)-(3.52), (3.55), (3.57), (3.58),
(3.60), (3.61) and (3.62), we can get

$\displaystyle\frac{1}{N^{n}}R_{N}^{(n)}\big{(}\tau,A_{0};z_{0}+N^{-\frac{1}{2}}\hat{z}_{1},\cdots,z_{0}+N^{-\frac{1}{2}}\hat{z}_{n}\big{)}=\frac{\prod_{1\leq
i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{2}}{(\prod_{i=1}^{n-1}i!)^{2}(2\pi)^{\frac{n}{2}}\pi^{n}}\big{(}t_{0}P_{1}^{2}+|P_{0}|^{2}\big{)}^{\frac{n(n+1)}{2}}P_{1}^{-\frac{n}{2}}$
(3.63) $\displaystyle\times e^{-\frac{1}{2P_{1}}{\rm
Tr}\big{(}\overline{P}_{0}\hat{Z}+P_{0}\hat{Z}^{*}\big{)}^{2}-t_{0}P_{1}{\rm
Tr}\big{(}\hat{Z}\hat{Z}^{*}\big{)}}\int_{\widetilde{\lambda}_{n}\leq\cdots\leq\widetilde{\lambda}_{1}}\prod_{1\leq
i<j\leq
n}(\widetilde{\lambda}_{j}-\widetilde{\lambda}_{i})^{2}\exp\Big{\\{}t_{0}P_{1}{\rm
Tr}\big{(}U_{1}U_{2}\hat{Z}U_{2}^{*}U_{1}^{*}\hat{Z}^{*}\big{)}$
$\displaystyle-\frac{P_{1}}{2}{\rm
Tr}\big{(}\widetilde{\Lambda}^{2}\big{)}+\overline{P_{0}}{\rm
Tr}\big{(}U_{2}^{*}\widetilde{\Lambda}U_{2}\hat{Z}\big{)}+P_{0}{\rm
Tr}\big{(}U_{1}\widetilde{\Lambda}U_{1}^{*}\hat{Z}^{*}\big{)}\Big{\\}}{\rm
d}\widetilde{\Lambda}{\rm d}U_{1}{\rm
d}U_{2}+O\big{(}N^{-\frac{1}{2}}\big{)}.$

After basic computations, the exponential terms in integrand of (3.63) now
becomes

$\displaystyle\exp\Big{\\{}-\frac{P_{1}}{2}{\rm
Tr}\big{(}U_{2}^{*}\widetilde{\Lambda}U_{2}-\frac{1}{P_{1}}\big{(}\overline{P}_{0}\hat{Z}+U_{2}^{*}U_{1}^{*}\hat{Z}^{*}U_{1}U_{2}\big{)}\big{)}^{2}$
$\displaystyle+\big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\big{)}{\rm
Tr}\big{(}U_{2}^{*}U_{1}^{*}\hat{Z}^{*}U_{1}U_{2}\hat{Z}\big{)}+\frac{\overline{P}_{0}^{2}}{2P_{1}}{\rm
Tr}\big{(}\hat{Z}^{2}\big{)}+\frac{P_{0}^{2}}{2P_{1}}{\rm
Tr}\big{(}\hat{Z}^{*}\big{)}^{2}\Big{\\}}.$

Now take the matrix transformations $W=U_{1}U_{2}$, $U_{2}=U_{2}$, then we
have ${\rm d}U_{1}{\rm d}U_{2}={\rm d}W{\rm d}U_{2}$. And set
$S=U_{2}^{*}\widetilde{\Lambda}U_{2},$ with Jacobian ${\rm
d}S=\frac{\pi^{\frac{n(n-1)}{2}}}{\prod_{i=1}^{n-1}i!}\prod_{1\leq i<j\leq
n}(\widetilde{\lambda}_{j}-\widetilde{\lambda}_{i})^{2}{\rm
d}\widetilde{\Lambda}{\rm d}U_{2}.$

Now we can integrate out the Hermitian $S$ and the unitary $W$ matrix
variables

$\displaystyle\int_{S=S^{*}}e^{-\frac{P_{1}}{2}{\rm
Tr}\big{(}S-\frac{1}{P_{1}}\big{(}\overline{P}_{0}\hat{Z}+W^{*}\hat{Z}^{*}W\big{)}\big{)}^{2}}{\rm
d}S=2^{-\frac{n(n-1)}{2}}\Big{(}\frac{2\pi}{P_{1}}\Big{)}^{\frac{n^{2}}{2}},$

next, from Harish-Chandra-Itzykson-Zuber integration formula [8, 9] (see also
[14])

$\int_{\mathcal{U}(n)}\exp\\!\left\\{l{\rm
Tr}\left(AUBU^{*}\right)\right\\}{\rm
d}U=l^{-\frac{n(n-1)}{2}}\frac{\det\big{(}[e^{la_{i}b_{j}}]_{i,j=1}^{n}\big{)}}{\prod_{1\leq
i<j\leq n}(a_{j}-a_{i})(b_{j}-b_{i})}\prod_{i=1}^{n-1}i!,$ (3.64)

with $A={\rm diag}\left(a_{1},\cdots,a_{n}\right)$ and $B={\rm
diag}\left(b_{1},\cdots,b_{n}\right)$, we obtain

$\displaystyle\int_{\mathcal{U}(n)}\exp\\!\left\\{\big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\big{)}{\rm
Tr}\left(W^{*}\hat{Z}^{*}W\hat{Z}\right)\right\\}{\rm
d}W=\big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\big{)}^{-\frac{n(n-1)}{2}}$
$\displaystyle\times\prod_{i=1}^{n-1}i!\prod_{1\leq i<j\leq
n}\big{|}\hat{z}_{i}-\hat{z}_{j}\big{|}^{-2}\det\Big{(}\Big{[}e^{\big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\big{)}\hat{z}_{i}\overline{\hat{z}_{j}}}\Big{]}_{i,j=1}^{n}\Big{)}.$

Finally, take the rescaling of $\hat{Z}$ by
$\hat{z}_{i}\rightarrow\big{(}t_{0}P_{1}+\frac{|P_{0}|^{2}}{P_{1}}\big{)}^{-\frac{1}{2}}\hat{z}_{i},\quad
i=1,\cdots,n,$ we can get to the conclusion.

Acknowledgements We would like to thank Elton P. Hsu for his encouragement and
support, Dang-Zheng Liu for useful comments on this paper and Mohammed Osman
for useful discussions around the issues.

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