111J. Song is partially supported by Shandong University grant 11140089963041
and National Natural Science Foundation of China grant 12071256.

, 222Y. Xiao is partially supported by NSF grant DMS-1855185. and

# On eigenvalues of the Brownian sheet matrix

Jian<EMAIL_ADDRESS>[ Yimin<EMAIL_ADDRESS>[
Wangjun<EMAIL_ADDRESS>[ Shandong University and The
University of Hong Kong  Research Center for Mathematics and Interdisciplinary
Sciences, Shandong University, Qingdao, Shandong, 266237, China; and School of
Mathematics, Shandong University, Jinan, Shandong, 250100, China
Department of Statistics and Probability, Michigan State University, A-413
Wells Hall, East Lansing, MI 48824, U.S.A.
Department of Mathematics, The University of Hong Kong

###### Abstract

We derive a system of stochastic partial differential equations satisfied by
the eigenvalues of the symmetric matrix whose entries are the Brownian sheets.
We prove that the sequence
$\left\\{L_{d}(s,t),(s,t)\in[0,S]\times[0,T]\right\\}_{d\in\mathbb{N}}$ of
empirical spectral measures of the rescaled matrices is tight on
$C([0,S]\times[0,T],\mathcal{P}(\mathbb{R}))$ and hence is convergent as $d$
goes to infinity by Wigner’s semicircle law. We also obtain PDEs which are
satisfied by the high-dimensional limiting measure.

60B20, 60B10, 60G15, 60G60,

random matrix,

Brownian sheet,

empirical spectral measure,

high-dimensional limit,

Dyson Brownian motion,

McKean-Vlasov equation,

###### keywords:

[class=AMS]

###### keywords:

###### Contents

  1. 1 Introduction
  2. 2 Stochastic calculus for the Brownian sheet
    1. 2.1 Some preliminaries on stochastic calculus on the plane
    2. 2.2 On $\psi\cdot MN$ and $J_{MN}$
    3. 2.3 Multi-dimensional Green’s formula for martingales on the plane
    4. 2.4 Quadratic covariations of $J_{MN}$ and $J_{M^{\prime}N^{\prime}}$
  3. 3 SPDEs for the eigenvalue processes
  4. 4 High-dimensional limit of the empirical spectral distributions
    1. 4.1 Tightness and high-dimensional limit
    2. 4.2 PDEs for the limit measure
  5. A Some lemmas in matrix calculus

##  1 Introduction

This paper concerns the eigenvalues of the Brownian sheet matrix
${\bm{X}}=\\{\bm{X}(s,t),0\leq s,t<\infty\\}$, which is a symmetric-matrix-
valued process with entries $X_{ij}$ for $1\leq i,j\leq d$ given by

$X_{ij}(s,t)=\begin{cases}b_{ij}(s,t),&i<j,\\\
\sqrt{2}b_{ii}(s,t),&i=j,\end{cases}$ (1.1)

where $b=\left\\{b_{ij}(s,t),0\leq s,t<\infty\right\\}_{1\leq i\leq j\leq d}$
is a family of independent Brownian sheets.

After the fundamental work [Wigner55] which established the celebrated
Wigner’s semicircle law, Brownian motion as a one-parameter stochastic process
was introduced into random matrix theory by Dyson [Dyson1962]. Since then,
there has been fruitful literature on the Dyson Brownian motion which is the
system of eigenvalues of symmetric Brownian matrix (see, e.g. [Cepa1997,
Chan1992, Rogers1993, anderson2010, EY17] and the references therein), in
which Itô’s calculus has played a key role. By studying the high-dimensional
limit of the empirical measures of the Dyson Brownian motion, one can provide
a dynamical proof for Wigner’s semicircle law (see, e.g., [anderson2010]). The
Dyson Brownian motion is also closely related to interacting particle systems,
and the equation (known as the McKean-Vlasov equation) satisfied by its
limiting empirical measure appears naturally in the study of propagation of
chaos for large systems of interacting particles (see, e.g., [BO19, JW18,
Serfaty20]).

Multiparameter stochastic processes (or random fields) are a natural extension
of one-parameter processes, they arise naturally in statistical mechanics
(e.g. Brownian sheet appears in the Ising model [KM86] and interacting
particle systems [KT88]), and systematic theories have been developed (see,
e.g., [CW75, K02] and the references therein). Motivated by the close
connection between random matrix theory and interacting particle systems, it
is natural to develop theories for random matrix with entries being random
fields. Recently, the problem on the collision of eigenvalues of symmetric
(Hermitian) matrix whose entries are independent Gaussian fields was
investigated in [Jaramillo2018, SXY20], which to our best knowledge are the
only literature on random matrix whose entries are random fields.

Another motivation for studying the Brownian sheet matrix $\bm{X}$ is from
free probability theory. As shown in [Voi91, VDN92], many theorems and
concepts in free probability have classical probability analogs, and
furthermore free probability is closely connected with random matrix theory.
In particular, free Brownian motion can be viewed as the high-dimensional
limit of rescaled Brownian motion matrix which is define by (1.1) with $b$
being a family of independent Brownian motions. Stochastic calculus for free
Brownian motion was developed in [BS98]. Free fractional Brownian motion arose
naturally in [NT14] when studying the central limit theorem for long-range
dependence time series in free probability, and the stochastic calculus was
developed in [DS19]. It was shown in [Pardo2016] that free fractional Brownian
motion is the high-dimensional limit of empirical measures of the eigenvalues
of rescaled fractional Brownian motion matrices. We remark that the free
Brownian motion and the free fractional Brownian motion in [BS98, NT14,
Pardo2016, DS19] are one-parameter stochastic processes, and we believe that
our study of the Brownian sheet matrix in this paper will provide a useful
building block for constructing free random fields.

In the present paper we shall derive a system of stochastic partial
differential equations (3) for the eigenvalue processes of the Brownian sheet
matrix $\bm{X}$ given by (1.1), obtain the tightness of the spectral empirical
measures (Theorem 4.1), and show that the limit measure satisfies a McKean-
Vlasov equation (4.15) and a Burgers’ equation (4.22). We briefly explain the
structure of the paper below.

Though the Brownian sheet is a simple multivariable extension of standard
Brownian motion, the stochastic calculus for the Brownian sheet that one needs
for deriving the stochastic partial differential equations for the eigenvalues
of the Brownian sheet matrix turns out to be highly non-trivial and cannot be
adapted directly from the classical Itô calculus. In Section 2, we follow the
approach of Cairoli and Walsh in [CW75] and develop stochastic calculus tools
for the multi-dimensional Brownian sheet on the plane for our purpose. The
main results in this section are Theorems 2.5 and 2.6 which are multi-
dimensional Green’s formulas.

In Section 3, by applying classical Itô’s formula together with Green’s
formulas (Theorems 2.5 and 2.6 ), we derive the system of stochastic partial
differential equations (3) for the eigenvalues of the Brownian sheet matrix
$\bm{X}$. Compared with the following system of SDEs for the classical Dyson
Brownian motion: for $1\leq i\leq d$,

$d\lambda_{i}(t)=\sqrt{2}dW_{i}(t)+\sum_{j\neq
i}\frac{1}{\lambda_{i}(t)-\lambda_{j}(t)}dt,$ (1.2)

where $W=(W_{1},\dots,W_{d})$ is a standard $d$-dimensional Brownian motion,
we remark that eq. (3) bears some resemblance to (1.2) but has several extra
high-order terms.

In Section 4, we study the high-dimensional limit of empirical distributions
for the eigenvalue processes of $\bm{X}$. In Section 4.1, we establish the
tightness of the set of empirical spectral measures which are viewed as
$C([0,S]\times[0,T],\mathcal{P}(\mathbb{R}))$-valued random elements (see
Theorem 4.1). This guarantees that every sequence of the empirical spectral
measures has a subsequence which converges weakly. The tightness together with
the classical Wigner’s semicircle law implies the existence and uniqueness of
the high-dimensional limit of the empirical spectral measures (see Theorem
4.2). In Section 4.2, we derive partial differential equations (4.15) and
(4.22) that are satisfied by the limiting measure, by using the property of
the semicircle distribution.

Finally, in Appendix A we provide some results in matrix analysis which are
needed in our analysis.

##  2 Stochastic calculus for the Brownian sheet

In this section, we shall apply the stochastic calculus on the plane developed
in [CW75] to derive Green’s formula for the multi-dimensional Brownian sheet,
which is a key ingredient for studying SPDEs for the eigenvalues in Section 3.

###  2.1 Some preliminaries on stochastic calculus on the plane

In this subsection, we recall from Cairoli and Walsh [CW75] some preliminaries
for stochastic calculus on the plane.

Define the partial order “$\prec$” on $\mathbb{R}^{2}$ as follows. For any
$(s_{1},t_{1}),\,(s_{2},t_{2})\in\mathbb{R}^{2}$,

$(s_{1},t_{1})\prec(s_{2},t_{2}),\text{ iff }s_{1}\leq s_{2},\,t_{1}\leq
t_{2},$

and write

$(s_{1},t_{1})\prec\prec(s_{2},t_{2}),\text{ iff }s_{1}<s_{2},\,t_{1}<t_{2}.$

Let $(\Omega,\mathcal{G},\mathbb{P})$ be a probability space and let the
filtration $\mathcal{F}=\\{\mathcal{F}_{z},z\in\mathbb{R}_{+}^{2}\\}$ be a
family of sub-$\sigma$-field of $\mathcal{G}$ satisfying

  1. 1.

$\mathcal{F}_{z}\subset\mathcal{F}_{z^{\prime}}$ if $z\prec z^{\prime}$;

  2. 2.

$\mathcal{F}_{0}$ contains all null sets of $\mathcal{G}$;

  3. 3.

for each $z,\mathcal{F}_{z}=\bigcap\limits_{z\prec\prec
z^{\prime}}\mathcal{F}_{z^{\prime}}$;

  4. 4.

for each $z,\mathcal{F}_{z}^{1}$ and $\mathcal{F}_{z}^{2}$ are conditionally
independent given $\mathcal{F}_{z}$.

Here, for $z=(s,t)\in\mathbb{R}_{+}^{2}$,

$\displaystyle\mathcal{F}_{z}^{1}=\mathcal{F}_{s\infty}:=\underset{v}{\vee}\mathcal{F}_{sv};~{}~{}~{}\mathcal{F}_{z}^{2}=\mathcal{F}_{\infty
t}:=\underset{u}{\vee}\mathcal{F}_{ut}.$

In particular, the augmented filtration generated by a finite family of
independent Brownian sheets satisfies the above conditions.

Let $Y=\\{Y_{z},z\in\mathbb{R}_{+}^{2}\\}$ be a process such that for each $z$
the random variable $Y_{z}$ is integrable. We recall the definitions of
martingale, strong martingale, weak martingale, and increasing process
relative to $\mathcal{F}$ in [CW75].

######  Definition 1.

$Y$ is a _martingale_ if

  1. 1.

$Y$ is adapted;

  2. 2.

$\mathbb{E}[Y_{z^{\prime}}|\mathcal{F}_{z}]=Y_{z},$ for each $z\prec
z^{\prime}$.

Suppose $z=(s,t)$ and $z^{\prime}=(s^{\prime},t^{\prime})$ such that
$z\prec\prec z^{\prime}$. We denote by $(z,z^{\prime}]$ the rectangle
$(s,s^{\prime}]\times(t,t^{\prime}]$. The increment of $Y$ over the rectangle
$(z,z^{\prime}]$ is

$Y((z,z^{\prime}])=Y_{s^{\prime}t^{\prime}}-Y_{st^{\prime}}-Y_{s^{\prime}t}+Y_{st}.$

######  Definition 2.

  * (a)

$Y$ is a _weak martingale_ if

    1. 1.

$Y$ is adapted;

    2. 2.

$\mathbb{E}[Y((z,z^{\prime}])|\mathcal{F}_{z}]=0$ for each $z\prec\prec
z^{\prime}.$

  * (b)

$Y$ is an _$i$ -martingale_ ($i=1,2$) if

    1. 1.

$Y$ is $\mathcal{F}_{z}^{i}$-adapted;

    2. 2.

$\mathbb{E}[Y((z,z^{\prime}])|\mathcal{F}_{z}^{i}]=0$ for each $z\prec\prec
z^{\prime}.$

  * (c)

$Y$ is a _strong martingale_ if

    1. 1.

$Y$ is adapted;

    2. 2.

$Y$ vanishes on the axes;

    3. 3.

$\mathbb{E}[Y((z,z^{\prime}])|\mathcal{F}_{z}^{1}\vee\mathcal{F}_{z}^{2}]=0$
for each $z\prec\prec z^{\prime}.$

######  Definition 3.

$Y$ is an _increasing process_ if

  1. 1.

$Y$ is right-continuous and adapted;

  2. 2.

$Y_{z}=0$ on the axes;

  3. 3.

$Y(A)\geq 0$ for each rectangle $A\subset\mathbb{R}_{+}^{2}$.

Let $M=\\{M_{z},z\in\mathbb{R}_{+}^{2}\\}$ be a martingale relative to
$\mathcal{F}$. Then $M$ is both a 1-martingale and 2-martingale, i.e.,
$\\{M_{s0},\mathcal{F}_{s0}^{1},s\in\mathbb{R}_{+}\\}$ and
$\\{M_{0t},\mathcal{F}_{0t}^{2},t\in\mathbb{R}_{+}\\}$ are martingales. The
converse is also true.

Now we assume that $M$ is a square integrable martingale. By [CW75, Theorem
1.5], there exists an increasing process $\langle M\rangle$ such that
$M^{2}-\langle M\rangle$ is a weak martingale. For each fixed $t$, let
$\\{[M]^{1}_{st},s\in\mathbb{R}_{+}\\}$ be the unique increasing process which
is predictable relative to $\\{\mathcal{F}_{st},s\in\mathbb{R}_{+}\\}$ such
that $\\{M^{2}_{st}-[M]^{1}_{st},s\in\mathbb{R}_{+}\\}$ is a martingale.
Similarly, one can define $[M]^{2}$. As pointed by [CW75, p.121], for a strong
martingale $M$, either $[M]^{1}$ or $[M]^{2}$ can serve as the process
$\langle M\rangle$. Furthermore, by [CW75, Theorem 1.9], if either
$\mathcal{F}$ is generated by the Brownian sheet or $M$ has finite fourth
moment, then $[M]^{1}=[M]^{2}$, and hence we can choose $\langle
M\rangle=[M]^{1}=[M]^{2}$. As a consequence, for any fixed $t$,
$\\{M^{2}_{st}-\langle M\rangle_{st},s\in\mathbb{R}_{+}\\}$ is a martingale,
and similarly, for any fixed $s$, $\\{M^{2}_{st}-\langle
M\rangle_{st},t\in\mathbb{R}_{+}\\}$ is a martingale. As in [CW75], we shall
use $d_{s}\langle M\rangle_{st}$ ($d_{t}\langle M\rangle_{st}$, resp.) to
denote the differential of $\langle M\rangle$ with respect to $s$ ($t$,
resp.).

For two square integrable martingales $M$ and $N$, we denote by $\langle
M,N\rangle$ any process which is the difference of two increasing processes
such that $MN-\langle M,N\rangle$ is a weak martingale. One can choose, for
instance,

$\langle M,N\rangle=\frac{1}{2}\big{(}\langle M+N\rangle-\langle
M\rangle-\langle N\rangle\big{)}.$ (2.1)

Define $[M,N]^{i}=\frac{1}{2}\left([M+N]^{i}-[M]^{i}-[N]^{i}\right)$ for
$i=1,2.$ Then either $[M,N]^{1}$ or $[M,N]^{2}$ can serve as the process
$\langle M,N\rangle.$ Two martingales $M$ and $N$ are said to be _orthogonal_
if $MN$ is a weak martingale, and we write $M\perp N$.

For $p\geq 1$, let $\mathfrak{M}^{p}$ denote the set of right-continuous
martingales $M=\\{M_{z},z\in\mathbb{R}_{+}^{2}\\}$ such that $M=0$ on the axes
and $\mathbb{E}[|M_{z}|^{p}]<\infty$ for all $z\in\mathbb{R}_{+}^{2}$. Let
$\mathfrak{M}^{p}_{c}$ (resp. $\mathfrak{M}_{s}^{p}$) be the set of continuous
(resp. strong) martingales in $\mathfrak{M}^{p}$. Similarly, let
$\mathfrak{M}^{p}(z_{0})$ (resp.
$\mathfrak{M}^{p}_{c}(z_{0}),\mathfrak{M}^{p}_{s}(z_{0})$) be the set of
right-continuous (resp. continuous, strong) martingales $M=\\{M_{z},z\prec
z_{0}\\}$ such that $M_{z}=0$ on the axes and $\mathbb{E}[|M_{z}|^{p}]<\infty$
for all $z\prec z_{0}$.

Below we recall some results which will be used in our proofs.

######  Theorem 2.1.

[CW75, Theorem 1.2] Let $\\{M_{z}:z\in\mathbb{R}_{+}^{2}\\}$ be a right-
continuous martingale. Then for $p>1$,

$\displaystyle\mathbb{E}\left[\sup_{z}|M_{z}|^{p}\right]\leq\left(\dfrac{p}{p-1}\right)^{2p}\sup_{z}\mathbb{E}\left[|M_{z}|^{p}\right].$

For any $z\in\mathbb{R}_{+}^{2}$, we denote the rectangle $(0,z]$ by $R_{z}$.
We also fix $z_{0}\in\mathbb{R}_{+}^{2}$.

######  Theorem 2.2.

[CW75, Proposition 1.6] Let $M,N\in\mathfrak{M}^{2}(z_{0})$. Then

  1. 1.

$\mathbb{E}[(MN)(D)|\mathcal{F}_{z}]=\mathbb{E}[M(D)N(D)|\mathcal{F}_{z}]$ for
each rectangle $D=(z,z^{\prime}]\subset R_{z_{0}}$;

  2. 2.

$M\perp N$ iff $\mathbb{E}[M(D)N(D)|\mathcal{F}_{z}]=0$ for each rectangle
$D=(z,z^{\prime}]\subset R_{z_{0}}$.

######  Theorem 2.3.

[CW75, Proposition 1.8] If $M\in\mathfrak{M}_{s}^{2}(z_{0})$, then $[M]^{i}$
is the unique $\mathcal{F}_{z}^{i}$-predictable increasing process such that
for $i=1,2$,

$\displaystyle\mathbb{E}\left[M(D)^{2}|\mathcal{F}_{z}^{i}\right]=\mathbb{E}\left[(M^{2})(D)|\mathcal{F}_{z}^{i}\right]=\mathbb{E}\left[[M]^{i}(D)|\mathcal{F}_{z}^{i}\right]$

for each rectangle $D=(z,z^{\prime}]\subseteq R_{z_{0}}$. Consequently, for
$M,N\in\mathfrak{M}_{s}^{2}(z_{0})$, noting that
$MN=\frac{1}{2}\left((M+N)^{2}-M^{2}-N^{2}\right)$, we have for $i=1,2$,

$\displaystyle\mathbb{E}\left[M(D)N(D)|\mathcal{F}_{z}^{i}\right]=\mathbb{E}\left[(MN)(D)|\mathcal{F}_{z}^{i}\right]=\mathbb{E}\left[[M,N]^{i}(D)|\mathcal{F}_{z}^{i}\right]$

######  Theorem 2.4.

[CW75, Theorem 1.9] Let $M\in\mathfrak{M}_{s}^{2}$. Assuming either the
filtration $\mathcal{F}$ is generated by the Brownian sheet or $M$ is
continuous with finite fourth moment, we have $[M]^{1}=[M]^{2}$.

###  2.2 On $\psi\cdot MN$ and $J_{MN}$

Let us recall from [CW75, Section 6] the notion $J_{M}$ of a continuous
martingale $M\in\mathfrak{M}_{s}^{4}$ on $\mathbb{R}_{+}^{2}$. Recall the
notation $R_{st}=(0,s]\times(0,t]$. By [CW75, Eq. (6.3)],

$\displaystyle J_{M}(s_{0},t_{0})=$
$\displaystyle\int_{0}^{s_{0}}M(s,t_{0})M(ds,t_{0})-\int_{R_{s_{0}t_{0}}}M(s,t)dM(s,t)$
$\displaystyle=$
$\displaystyle\int_{0}^{t_{0}}M(s_{0},t)M(s_{0},dt)-\int_{R_{s_{0}t_{0}}}M(s,t)dM(s,t)$
$\displaystyle=$
$\displaystyle\frac{1}{2}M^{2}(s_{0},t_{0})-\frac{1}{2}\langle
M\rangle_{s_{0},t_{0}}-\int_{R_{s_{0}t_{0}}}M(s,t)dM(s,t).$

Heuristically, one has $dJ_{M}(s,t)=M(s,dt)M(ds,t)$ (see [CW75]). Similarly,
for two $\mathcal{F}$-adapted martingales $M$ and $N$, we introduce the
following generalization $J_{MN}$ which induces the measure $M(s,dt)N(ds,t)$
on $\mathbb{R}_{+}^{2}$,

$\displaystyle
J_{MN}(s_{0},t_{0})=\int_{0}^{s_{0}}M(s,t_{0})N(ds,t_{0})-\int_{R_{s_{0}t_{0}}}M(s,t)dN(s,t),$
(2.2)

assuming that the right-hand side is well-defined. Clearly we have
$J_{M}=J_{MM}$. Analogous to $J_{M}$ in [CW75, Theorem 6.1], $J_{MN}$ will
play a key role in the multi-dimensional Green’s formula in the forthcoming
Theorems 2.5 and 2.6.

Similar to [CW75], we shall represent $J_{MN}$ by a new type of stochastic
integral denoted by $\psi\cdot MN$ which will be defined in the sequel.
Firstly, we need to introduce another order relation “$\curlywedge$” in
$\mathbb{R}_{+}^{2}$ which is complementary to “$\prec$” and plays an
essential role in the definition of $\psi\cdot MN$. For $z=(s,t)$ and
$z^{\prime}=(s^{\prime},t^{\prime})$, we say $z\curlywedge z^{\prime}$ if
$s\leq s^{\prime}$ and $t\geq t^{\prime}$, and
$z{\curlywedge\atop\curlywedge}z^{\prime}$ if $s<s^{\prime}$ and
$t>t^{\prime}$. In the $st$-plane where the $s$-axis is horizontal and the
$t$-axis is vertical, $z\curlywedge z^{\prime}$ means that $z$ is on the upper
left of $z^{\prime}$ in the plane. As a comparison, $z\prec z^{\prime}$ means
that $z$ is on the lower left of $z^{\prime}$.

######  Proposition 2.1.

Suppose $M,N\in\mathfrak{M}_{s}^{2}(z_{0})$. Let $A=(z_{A},z_{A}^{\prime}]$
and $B=(z_{B},z_{B}^{\prime}]$ be two rectangles such that $A\curlywedge B$,
i.e., $z_{1}\curlywedge z_{2}$ for all $z_{1}\in A$ and $z_{2}\in B$.

Define the process $X=\\{X_{z},~{}z\in\mathbb{R}_{+}^{2}\\}$ by

$\displaystyle X_{z}=\xi M(A\cap R_{z})N(B\cap
R_{z}),~{}z\in\mathbb{R}_{+}^{2},$

where $\xi$ is bounded and $\mathcal{F}_{z_{A}\vee z_{B}}$-measurable. Then
$X$ belongs to $\mathfrak{M}^{2}(z_{0})$, it is continuous if $M$ is, and

$\displaystyle\langle X\rangle_{z}=\xi^{2}\iint_{R_{z}\times
R_{z}}\mathbf{1}_{A}(z_{1})\mathbf{1}_{B}(z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}}.$
(2.3)

###### Proof.

We will follow the proof of [CW75, Proposition 2.4].

For $D=(z,z^{\prime}]$ with $z=(s,t)\prec\prec
z^{\prime}=(s^{\prime},t^{\prime})$, the increment of $X$ over $D$ is

$X(D)=M({\tilde{A}})N({\tilde{B}}),$ (2.4)

where $\tilde{A}=A\cap(R_{s^{\prime}t^{\prime}}\backslash R_{s^{\prime}t})$
and ${\tilde{B}}=B\cap(R_{s^{\prime}t^{\prime}}\backslash R_{st^{\prime}})$.

Suppose $z_{\tilde{A}}$ is the lower-left corner of $\tilde{A}$. Then both
$\xi$ and $N({\tilde{B}})$ are $\mathcal{F}_{z_{\tilde{A}}}^{2}$-measurable,
and hence

$\displaystyle\mathbb{E}\left[X(D)|\mathcal{F}_{z}^{2}\right]=\mathbb{E}\left[\left.\mathbb{E}\left[\xi
M({\tilde{A}})N({\tilde{B}})|\mathcal{F}_{z_{\tilde{A}}}^{2}\right]\right|\mathcal{F}_{z}^{2}\right]=\mathbb{E}\left[\left.\xi
N({\tilde{B}})\mathbb{E}\left[M({\tilde{A}})|\mathcal{F}_{z_{\tilde{A}}}^{2}\right]\right|\mathcal{F}_{z}^{2}\right]=0.$

Similarly, one can show $\mathbb{E}\left[X(D)|\mathcal{F}_{z}^{1}\right]=0$.
Hence, $X$ is a martingale.

Let $z_{\tilde{B}}$ be the lower left-hand corner of $\tilde{B}$, and denote
$z_{0}=z_{\tilde{A}}\vee z_{\tilde{B}}$. Then $z_{A}\vee z_{B}\prec z_{0}$,
and hence $\xi$ is $\mathcal{F}_{z_{0}}$-measurable. Thus, by Theorem 2.2,

$\mathbb{E}[X^{2}(D)|\mathcal{F}_{z}]=\mathbb{E}\left[\left.\xi^{2}\mathbb{E}\left[M(\tilde{A})^{2}N(\tilde{B})^{2}|\mathcal{F}_{z_{0}}\right]\right|\mathcal{F}_{z}\right].$

Now we have

$\displaystyle\mathbb{E}\left[M({\tilde{A}})^{2}N({\tilde{B}})^{2}|\mathcal{F}_{z_{0}}\right]=$
$\displaystyle\mathbb{E}\left[M({\tilde{A}})^{2}|\mathcal{F}_{z_{0}}\right]\mathbb{E}\left[N({\tilde{B}})^{2}|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[\left.\mathbb{E}\left[M({\tilde{A}})^{2}|\mathcal{F}_{z_{0}}^{2}\right]\right|\mathcal{F}_{z_{0}}\right]\mathbb{E}\left[\mathbb{E}\left.\left[N({\tilde{B}})^{2}|\mathcal{F}_{z_{0}}^{1}\right]\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[[M]^{2}({\tilde{A}})|\mathcal{F}_{z_{0}}\right]\mathbb{E}\left[[N]^{1}({\tilde{B}})|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=$
$\displaystyle\mathbb{E}\left[[M]^{2}({\tilde{A}})[N]^{1}({\tilde{B}})|\mathcal{F}_{z_{0}}\right],$

where the first and the last equalities follow from the assumption that
$\mathcal{F}_{z_{0}}^{1}$ and $\mathcal{F}_{z_{0}}^{2}$ are conditionally
independent given $\mathcal{F}_{z_{0}}$, and the third equality follows from
Theorem 2.3. Thus

$\mathbb{E}[X^{2}(D)-\xi^{2}[M]^{2}({\tilde{A}})[N]^{1}({\tilde{B}})|\mathcal{F}_{z}]=0$

and hence $X_{z}^{2}-\langle X\rangle_{z}$ is a weak martingale where $\langle
X\rangle_{z}$ is given by (2.3). The proof is concluded. ∎

With Proposition 2.1 in mind, we define a new type of stochastic integral
denoted by $\psi\cdot MN$, following the approach in [CW75].

Fix an integer $n$ and $z_{0}=(s_{0},t_{0})\in\mathbb{R}_{+}^{2}$. Divide
$R_{z_{0}}$ into rectangles ${\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}=(z_{i,j},z_{i+1,j+1}]$, where
$z_{i,j}=(is_{0}/2^{n},jt_{0}/2^{n})$ for $i,j=0,1,\dots,2^{n}-1$. We first
define $\psi\cdot MN$ for an indicator function $\psi.$ If $i,j,k,l$ are
positive integers with $1\leq i<k\leq 2^{n}$ and $1\leq l<j\leq 2^{n}$, i,e.
${\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}\curlywedge{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{k,l}$, define the so-called indicator
function

$\displaystyle\psi_{ijkl}(z_{1},z_{2})=\xi\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}}(z_{1})\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{k,l}}(z_{2}),$ (2.9)

where $\xi$ is bounded and $\mathcal{F}_{z_{k,j}}$-measurable, and define

$\displaystyle(\psi_{ijkl}\cdot MN)_{z}=\xi M({\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j}\cap
R_{z})N({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k,l}\cap R_{z}),\ \ \ \ z\in R_{z_{0}}.$

Then by Proposition 2.1, $\psi_{ijkl}\cdot MN$ is a well-defined square
integrable martingale with quadratic variation

$\displaystyle\langle\psi_{ijkl}\cdot MN\rangle_{z}=\iint_{R_{z}\times
R_{z}}\psi_{ijkl}^{2}(z_{1},z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}},$

and thus we have the following isometry

$\mathbb{E}[|\psi_{ijkl}\cdot MN|^{2}]=\mathbb{E}\left[\iint_{R_{z}\times
R_{z}}\psi_{ijkl}^{2}(z_{1},z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}}\right].$
(2.10)

We shall define $\psi\cdot MN$ for a more general class of integrands $\psi$
following the standard approximation procedure. For this purpose, one needs
the isometry (2.10) to hold for finite sum of indicator functions, and it
suffices to prove the following equality

$\displaystyle\langle\psi_{ijkl}\cdot MN,\psi_{mpqr}\cdot
MN\rangle_{z}=\iint_{R_{z}\times
R_{z}}\psi_{ijkl}(z_{1},z_{2})\psi_{mpqr}(z_{1},z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}}.$
(2.11)

Here, $\psi_{mpqr}(z_{1},z_{2})=\eta\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{m,p}}(z_{1})\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{q,r}}(z_{2})$ with $m<q\leq
2^{n},r<p\leq 2^{n}$ and $\eta$ being a bounded
$\mathcal{F}_{z_{q,p}}$-measurable random variable. To prove (2.11), we
consider the following more general situation.

Suppose $M,N,M^{\prime},N^{\prime}\in\mathcal{\mathfrak{M}}_{s}^{2}(z_{0})$,
and let $(A,B)$ and $(A^{\prime},B^{\prime})$ be two pairs of rectangles
satisfying the conditions in Proposition 2.1, i.e., $A\curlywedge B$ and
$A^{\prime}\curlywedge B^{\prime}$. Furthermore, we assume
$A,A^{\prime},B,B^{\prime}$ are from the set $\\{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j},i,j=0,1,\dots,2^{n}-1\\}$. Thus,
any two of the rectangles $A,A^{\prime},B,B^{\prime}$ are either coincide or
disjoint. Denote $z_{0}=(z_{A}\vee z_{B})\vee(z_{A^{\prime}}\vee
z_{B^{\prime}})$. We claim that the following equality holds

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})N(B)N^{\prime}(B^{\prime})\right|\mathcal{F}_{z_{0}}\right]+\mathbb{E}\left[\left.M(A^{\prime})M^{\prime}(A)N(B)N^{\prime}(B^{\prime})\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle\qquad+\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]+\mathbb{E}\left[\left.M(A^{\prime})M^{\prime}(A)N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=\mathbb{E}\left[\left.\Big{(}M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\Big{)}\Big{(}N(B)N^{\prime}(B^{\prime})+N(B^{\prime})N^{\prime}(B)\Big{)}\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=4\mathbb{E}\Big{[}\left.[M,M^{\prime}]^{2}(A\cap
A^{\prime})[N,N^{\prime}]^{1}(B\cap
B^{\prime})\right|\mathcal{F}_{z_{0}}\Big{]}.$ (2.12)

###### Proof of (2.2).

The first equality is straightforward. In the following, we shall prove the
second equality.

Recall that the four rectangles $A,A^{\prime},B,B^{\prime}$ are either
disjoint or coincide; furthermore, $A\curlywedge B$ and $A^{\prime}\curlywedge
B^{\prime}$, i.e., $A$ (resp. $A^{\prime}$) is on the upper left side of $B$
(resp. $B^{\prime}$). We prove the second inequality in (2.2) by separating
the relative locations of $A,A^{\prime},B^{\prime},B^{\prime}$ into four
cases. In the following, we denote the lower left corner of a rectangle $E$ by
$z_{E}$.

Case 1. If $A$ is on the top of $A^{\prime}$ and $A\cap A^{\prime}=\emptyset$,
noting that $A$ (resp. $A^{\prime}$) is to the upper left of $B$ (resp.
$B^{\prime}$), we have that
$M^{\prime}(A^{\prime}),N(B),N^{\prime}(B^{\prime})$ are all
$\mathcal{F}_{z_{A}}^{2}$-measurable. Since $M$ is a $2$-martingale, we have,
noting that $\mathcal{F}_{z_{0}}\subset\mathcal{F}_{z_{A}}^{2}$

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})N(B)N^{\prime}(B^{\prime})\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=\mathbb{E}\Big{[}\left.\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})N(B)N^{\prime}(B^{\prime})\right|\mathcal{F}_{z_{A}}^{2}\right]\right|\mathcal{F}_{z_{0}}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\left.M^{\prime}(A^{\prime})N(B)N^{\prime}(B^{\prime})\mathbb{E}\left[\left.M(A)\right|\mathcal{F}_{z_{A}}^{2}\right]\right|\mathcal{F}_{z_{0}}\Big{]}$
$\displaystyle=0.$

Similarly, for the other terms on the left-hand side of (2.2) we also have

$\displaystyle\mathbb{E}\left[\left.M(A^{\prime})M^{\prime}(A)N(B)N^{\prime}(B^{\prime})\right|\mathcal{F}_{z_{0}}\right]=0,$
$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]=0,$
$\displaystyle\mathbb{E}\left[\left.M(A^{\prime})M^{\prime}(A)N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]=0.$

Summing over all the above equalities, we get (2.2).

Case 2. If $A^{\prime}$ is on the top of $A$ and $A\cap A^{\prime}=\emptyset$,
the proof is the same by considering the $\sigma$-field
$\mathcal{F}_{z_{A^{\prime}}}^{2}$. If $B$ is to the right (resp. left) of
$B^{\prime}$ with $B\cap B^{\prime}=\emptyset$, then the proof is also the
same by considering the $\sigma$-field $\mathcal{F}_{z_{B}}^{1}$ (resp.
$\mathcal{F}_{z_{B^{\prime}}}^{1}$).

Case 3. Now we only have one situation left: $A$ and $A^{\prime}$ are at the
same horizontal level, which is on the top of $B$ and $B^{\prime}$, and $B$
and $B^{\prime}$ are at the same vertical level, which is to the right of $A$
and $A^{\prime}$. We denote $z_{0}:=z_{A}\vee z_{B}=z_{A^{\prime}}\vee
z_{B^{\prime}}$. Note that
$M(A),M^{\prime}(A^{\prime}),M(A^{\prime}),M^{\prime}(A)$ are
$\mathcal{F}_{z_{0}}^{1}$ measurable and
$N(B),N^{\prime}(B^{\prime}),N(B^{\prime}),N^{\prime}(B)$ are
$\mathcal{F}_{z_{0}}^{2}$ measurable. We have

$\displaystyle\quad\mathbb{E}\left[\left.\Big{(}M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\Big{)}\Big{(}N(B)N^{\prime}(B^{\prime})+N(B^{\prime})N^{\prime}(B)\Big{)}\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=\mathbb{E}\Big{[}\left.\mathbb{E}\left[M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}\right]\mathbb{E}\left[\left.N(B)N^{\prime}(B^{\prime})+N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]\Big{|}\mathcal{F}_{z_{0}}\Big{]},$
(2.13)

where the equality follows from the conditional independence of
$\mathcal{F}_{z_{0}}^{1}$ and $\mathcal{F}_{z_{0}}^{2}$ given
$\mathcal{F}_{z_{0}}$.

To compute

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}\right],$

we split it into the following three cases.

(a) If $A=A^{\prime}$, noting that
$\mathcal{F}_{z_{0}}^{2}=\mathcal{F}_{z_{A}}^{2}$, by Theorem 2.3,

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=\mathbb{E}\left[\left.(MM^{\prime})(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A)\right|\mathcal{F}_{z_{0}}^{2}\right].$
(2.14)

(b) If $A$ and $A^{\prime}$ are two adjacent disjoint rectangles on the same
horizontal level, then $A\cup A^{\prime}$ is also a rectangle. Without loss of
generality, we may assume that $A$ is to the left of $A^{\prime}$, then
$z_{A}$ is also the lower left corner of $A\cup A^{\prime}$. Thus, by Case
(a), we have

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A)\right|\mathcal{F}_{z_{0}}^{2}\right],$
$\displaystyle\mathbb{E}\left[\left.M(A^{\prime})M^{\prime}(A^{\prime})\right|\mathcal{F}_{z_{0}}^{2}\right]=\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A^{\prime})\right|\mathcal{F}_{z_{0}}^{2}\right],$
$\displaystyle\mathbb{E}\left[\left.M(A\cup A^{\prime})M^{\prime}(A\cup
A^{\prime})\right|\mathcal{F}_{z_{0}}^{2}\right]=\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A\cup
A^{\prime})\right|\mathcal{F}_{z_{0}}^{2}\right].$

Noting that $M(A\cup A^{\prime})=M(A)+M(A^{\prime})$, $M^{\prime}(A\cup
A^{\prime})=M^{\prime}(A)+M^{\prime}(A^{\prime})$ and
$[M,M^{\prime}]^{i}(A\cup
A^{\prime})=[M,M^{\prime}]^{i}(A)+[M,M^{\prime}]^{i}(A^{\prime})$, we subtract
the first two equations from the third one and obtain

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=0.$
(2.15)

(c) If $A$ and $A^{\prime}$ are two non-adjacent rectangles on the same
horizontal level, we denote by $A^{\prime\prime}$ the rectangle between $A$
and $A^{\prime}$. Note that $A^{\prime\prime}$ is the union of small
rectangles in the set $\\{{\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j},i,j=1,\dots,2^{n}\\}$. By Case (b), we have

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime\prime})+M(A^{\prime\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=0,$
$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime\prime}\cup
A^{\prime})+M(A^{\prime\prime}\cup
A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=0.$

Noting that $M^{\prime}(A^{\prime\prime}\cup
A^{\prime})=M^{\prime}(A^{\prime\prime})+M^{\prime}(A^{\prime})$, one can
subtract the first equality from the second one to obtain (2.15).

Therefore, summarizing the three cases (a-c), we can write

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}^{2}\right]=2\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A\cap
A^{\prime})\right|\mathcal{F}_{z_{0}}^{2}\right].$

Hence, by taking conditional expectation with respect to the $\sigma$-field
$\mathcal{F}_{z_{0}}$, we have

$\displaystyle\mathbb{E}\left[\left.M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right|\mathcal{F}_{z_{0}}\right]=2\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A\cap
A^{\prime})\right|\mathcal{F}_{z_{0}}\right].$ (2.16)

In the same spirit, we can also prove

$\displaystyle\mathbb{E}\left[\left.N(B)N^{\prime}(B^{\prime})+N(B^{\prime})N^{\prime}(B)\right|\mathcal{F}_{z_{0}}\right]=2\mathbb{E}\left[\left.[N,N^{\prime}]^{1}(B\cap
B^{\prime})\right|\mathcal{F}_{z_{0}}\right].$ (2.17)

Finally, substituting (2.16) and (2.17) into (2.2), we have

$\displaystyle\mathbb{E}\left[\left.\left(M(A)M^{\prime}(A^{\prime})+M(A^{\prime})M^{\prime}(A)\right)\left(N(B)N^{\prime}(B^{\prime})+N(B^{\prime})N^{\prime}(B)\right)\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=4\mathbb{E}\left[\left.\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A\cap
A^{\prime})\right|\mathcal{F}_{z_{0}}\right]\mathbb{E}\left[\left.[N,N^{\prime}]^{1}(B\cap
B^{\prime})\right|\mathcal{F}_{z_{0}}\right]\right|\mathcal{F}_{z_{0}}\right]$
$\displaystyle=4\mathbb{E}\left[\left.[M,M^{\prime}]^{2}(A\cap
A^{\prime})[N,N^{\prime}]^{1}(B\cap
B^{\prime})\right|\mathcal{F}_{z_{0}}\right],$

where the conditional independence of $\mathcal{F}_{z_{0}}^{1}$ and
$\mathcal{F}_{z_{0}}^{2}$ given $\mathcal{F}_{z_{0}}$ is used again in the
last equality. This proves (2.2). ∎

By choosing $M^{\prime}=M$ and $N^{\prime}=N$, eq. (2.2) degenerates to

$\displaystyle\mathbb{E}\left[\left.M(A)M(A^{\prime})N(B)N(B^{\prime})\right|\mathcal{F}_{z_{0}}\right]=$
$\displaystyle\mathbb{E}\Big{[}\left.[M,M]^{2}(A\cap
A^{\prime})[N,N]^{1}(B\cap B^{\prime})\right|\mathcal{F}_{z_{0}}\Big{]}.$
(2.18)

Now, as in Proposition 2.1, we can define

$X_{z}=\xi M(A\cap R_{z})N(B\cap R_{z})~{}\text{ and
}~{}X^{\prime}_{z}=\xi^{\prime}M^{\prime}(A^{\prime}\cap
R_{z})N^{\prime}(B^{\prime}\cap R_{z})$ (2.19)

for some bounded variables $\xi\in\mathcal{F}_{z_{A}\vee z_{B}}$ and
$\xi^{\prime}\in\mathcal{F}_{z_{A^{\prime}}\vee z_{B^{\prime}}}$. Denote
$z_{0}:=(z_{A}\vee z_{B})\vee(z_{A^{\prime}}\vee z_{B^{\prime}})$ and we
assume $z_{0}\prec z=(s,t)$, since otherwise at least one of $X_{z}$ and
$X_{z}^{\prime}$ is zero. Let $z^{\prime}=(s^{\prime},t^{\prime})$ be such
that $z\prec\prec z^{\prime}$ and let $D:=(z,z^{\prime}]$.

Assuming $M=M^{\prime}$ and $N=N^{\prime}$ in (2.19), following the approach
used in the proof of Proposition 2.1, we can show by (2.2),

$\mathbb{E}[(XX^{\prime})(D)|\mathcal{F}_{z}]=\xi\xi^{\prime}\mathbb{E}\Big{[}\left.[M,M]^{2}(\tilde{A}\cap\tilde{A}^{\prime})[N,N]^{1}(\tilde{B}\cap\tilde{B}^{\prime})\right|\mathcal{F}_{z}\Big{]},$

where $\tilde{A}=A\cap(R_{s^{\prime}t^{\prime}}\backslash R_{s^{\prime}t})$,
${\tilde{B}}=B\cap(R_{s^{\prime}t^{\prime}}\backslash R_{st^{\prime}})$, and
$\tilde{A}^{\prime}=A^{\prime}\cap(R_{s^{\prime}t^{\prime}}\backslash
R_{s^{\prime}t})$ and
${\tilde{B}^{\prime}}=B^{\prime}\cap(R_{s^{\prime}t^{\prime}}\backslash
R_{st^{\prime}})$. This leads to

$\displaystyle\langle
X,X^{\prime}\rangle_{z}=\xi\xi^{\prime}\iint_{R_{z}\times
R_{z}}\mathbf{1}_{A\cap A^{\prime}}(z_{1})\mathbf{1}_{B\cap
B^{\prime}}(z_{2})d[M,M]^{2}_{z_{1}}d[N,N]^{1}_{z_{2}},$ (2.20)

and hence (2.11) is verified.

Now we are ready to define $\psi\cdot MN$ for a more general integrand $\psi$.
We say $\psi$ is a simple function if it is a finite sum of $\psi_{ijkl}$
given in (2.9). Let $\mathcal{D}$ be the $\sigma$-filed on
$\mathbb{R}_{+}^{2}\times\mathbb{R}_{+}^{2}\times\Omega$ generated by all the
simple functions. We call $\mathcal{D}$ the field of predictable sets. Let
$\mathcal{L}_{MN}^{2}(z_{0})$ be the class of all predictable processes such
that

$\displaystyle\mathbb{E}\left[\iint_{R_{z_{0}}\times
R_{z_{0}}}\psi^{2}(z_{1},z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}}\right]<\infty.$
(2.21)

Then $\mathcal{L}_{MN}^{2}(z_{0})$ is a Hilbert space with the inner product

$\displaystyle(\psi,\phi)=\mathbb{E}\left[\iint_{R_{z_{0}}\times
R_{z_{0}}}\psi(z_{1},z_{2})\phi(z_{1},z_{2})d[M]^{2}_{z_{1}}d[N]^{1}_{z_{2}}\right],$
(2.22)

and the simple functions form a dense subset. By (2.11) and (2.22), the
mapping $\psi\mapsto\psi\cdot MN$ defines an isometry between the set of
simple functions and $\mathfrak{M}^{2}(z_{0})$. Then, by a standard
approximation argument, one can extend the definition of $\psi\cdot MN$ for
each process $\psi\in\mathcal{L}^{2}_{MN}(z_{0})$. Furthermore, (2.11) also
yields for $z\prec z_{0}$,

$\langle\psi\cdot MN,\phi\cdot MN\rangle_{z}=\iint_{R_{z}\times
R_{z}}\psi(z_{1},z_{2})\phi(z_{1},z_{2})d\langle M\rangle_{z_{1}}d\langle
N\rangle_{z_{2}},~{}~{}\forall\psi,\phi\in\mathcal{L}^{2}_{MN}(z_{0}).$ (2.23)

Throughout the rest of this section, we only consider continuous strong
martingales with finite fourth moments, unless otherwise stated. Then based on
Theorem 2.4, we have

$[M]^{1}=[M]^{2}=\langle M\rangle;~{}~{}[N]^{1}=[N]^{2}=\langle N\rangle.$
(2.24)

To end this subsection, we shall follow the approach used in [CW75, Section 6]
to show that $J_{MN}$ defined in (2.2) can be represented by $\psi\cdot MN$
with
$\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}$.

Recall the notations $z_{i,j}=(is_{0}/2^{n},jt_{0}/2^{n})$ and
${\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}=(z_{i,j},z_{i+1,j+1}]$. We also denote
$\epsilon_{i,j}=(z_{i0},z_{i+1,j}]$ and $\delta_{i,j}=(z_{0,j},z_{i,j+1}]$.
Denote

$\displaystyle J_{MN}^{n}(z)$
$\displaystyle:=\sum_{i,j=0}^{2^{n}-1}M(\delta_{i,j}\cap
R_{z})N(\epsilon_{i,j}\cap R_{z})$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}\left(\sum_{k=0}^{i-1}M((z_{k,j},z_{k+1,j+1}]\cap
R_{z})\right)\left(\sum_{l=0}^{j-1}N((z_{i,l},z_{i+1,l+1}]\cap R_{z})\right)$
$\displaystyle=\sum_{k<i}\sum_{l<j}\left(\psi_{kjil}\cdot MN\right)_{z},$
(2.25)

where $\psi_{kjil}$ is given in (2.9). Thus, letting $n\rightarrow\infty$, we
have

$\displaystyle J_{MN}^{n}(z)\rightarrow(\psi\cdot MN)_{z},$

where

$\displaystyle\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}=\begin{cases}1,&\text{
if }z_{1}{\curlywedge\atop\curlywedge}z_{2},\\\
0,&\mathrm{otherwise}.\end{cases}$

Define
$M^{n}=\sum\limits_{i,j=0}^{2^{n}-1}\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j}}(z)M_{z_{i,j}}$. Then $M^{n}$ is a
sequence of simple functions that approximate $M$ and

$\displaystyle\int_{R_{z_{0}}}M^{n}dN$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}M_{z_{i,j}}N({\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j})$ (2.28)
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}M_{z_{i,j}}\left(N(\epsilon_{i,j+1})-N(\epsilon_{i,j})\right)$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}\left(M_{z_{i,j+1}}N(\epsilon_{i,j+1})-M_{z_{i,j}}N(\epsilon_{i,j})\right)+\sum_{i,j=0}^{2^{n}-1}\left(M_{z_{i,j}}-M_{z_{i,j+1}}\right)N(\epsilon_{i,j+1})$
$\displaystyle=\sum_{i=0}^{2^{n}-1}M_{z_{i,2^{n}}}N(\epsilon_{i,2^{n}})-\sum_{i,j=0}^{2^{n}-1}M(\delta_{i,j})\left(N(\epsilon_{i,j})+N({\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j})\right).$ (2.31)

If we define $\widetilde{M}_{s,t_{0}}^{n}=M_{is_{0}/2^{n},t_{0}}$ for
$s\in(is_{0}/2^{n},(i+1)s_{0}/2^{n}]$, and $\delta^{n}(z)=M(\delta_{i,j})$ if
$z\in{\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j}$.
Let $H_{z_{0}}$ be the line segment with endpoints $(0,t_{0})$ and
$z_{0}=(s_{0},t_{0})$, then

$\displaystyle\int_{R_{z_{0}}}M^{n}dN=\int_{H_{z_{0}}}\widetilde{M}_{s,t_{0}}^{n}(s)N(ds,t)-J_{MN}^{n}(z_{0})-\int_{R_{z_{0}}}\delta^{n}dN.$
(2.32)

By the Cauchy-Schwarz inequality,

$\displaystyle\mathbb{E}\left[\bigg{(}\int_{R_{z_{0}}}\delta^{n}dN\bigg{)}^{2}\right]$
$\displaystyle=\mathbb{E}\left[\int_{R_{z_{0}}}\left(\delta^{n}(s,t)\right)^{2}d\langle
N\rangle_{st}\right]$
$\displaystyle\leq\mathbb{E}\left[\int_{R_{z_{0}}}\sup_{i,j}M(\delta_{i,j})^{2}d\langle
N\rangle_{st}\right]$
$\displaystyle\leq\mathbb{E}\left[\sup_{i,j}M(\delta_{i,j})^{2}\langle
N\rangle_{z_{0}}\right]$
$\displaystyle\leq\left(\mathbb{E}\left[\sup_{i,j}M(\delta_{i,j})^{4}\right]\mathbb{E}\left[\langle
N\rangle_{z_{0}}^{2}\right]\right)^{1/2}$ $\displaystyle\to 0,~{}n\to\infty,$
(2.33)

where the last step holds due to the continuity of $M$ and the dominated
convergence theorem, noting that $\mathbb{E}[\sup_{i,j,n}M(\delta_{i,j})^{4}]$
is dominated by $\mathbb{E}[\sup_{z\prec z_{0}}|M_{z}|^{4}]$, which is
dominated by $(4/3)^{8}\mathbb{E}[|M_{z_{0}}|^{4}]$ due to Theorem 2.1 and the
existence of the fourth moment of the $M$.

Furthermore, Theorem 2.1 yields

$\displaystyle\mathbb{E}\left[\sup_{n}\sup_{z\prec
z_{0}}(M_{z}^{n}-M_{z})^{4}\right]\leq 8\mathbb{E}\left[\sup_{n}\sup_{z\prec
z_{0}}|M_{z}^{n}|^{4}+\sup_{z\prec z_{0}}|M_{z}|^{4}\right]\leq
16\mathbb{E}\left[\sup_{z\prec z_{0}}|M_{z}|^{4}\right]<\infty.$

By the Cauchy-Schwarz inequality, the dominated convergence theorem and the
continuity of $M$, we have

$\displaystyle\mathbb{E}\left[\bigg{(}\int_{R_{z_{0}}}(M^{n}-M)dN\bigg{)}^{2}\right]$
$\displaystyle=\mathbb{E}\left[\int_{R_{z_{0}}}(M^{n}-M)^{2}d\langle
N\rangle\right]$
$\displaystyle\leq\mathbb{E}\left[\sup_{n,z}(M_{z}^{n}-M_{z})^{2}\langle
N\rangle_{z_{0}}\right]$
$\displaystyle\leq\left(\mathbb{E}\left[\sup_{n,z}(M_{z}^{n}-M_{z})^{4}\right]\mathbb{E}\left[\langle
N\rangle_{z_{0}}^{2}\right]\right)^{1/2}$ $\displaystyle\rightarrow
0,~{}n\rightarrow\infty.$ (2.34)

Similarly, we can show the following $L^{2}$-convergence,

$\displaystyle\int_{H_{z_{0}}}\widetilde{M}_{s,t_{0}}^{n}N(ds,t)\rightarrow\int_{H_{z_{0}}}M(s,t)N(ds,t),~{}n\rightarrow\infty.$
(2.35)

Recalling that $\lim_{n\to\infty}J_{MN}^{n}(z)=(\psi\cdot MN)_{z}$ with
$\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}$, by
(2.32), (2.33), (2.34) and (2.35), we have

$\displaystyle(\psi\cdot
MN)_{z_{0}}=\int_{H_{z_{0}}}M(s,t)N(ds,t)-\int_{R_{z_{0}}}MdN,$

and hence by the definition (2.2) of $J_{MN}$, we have

$J_{MN}(z_{0})=(\psi\cdot MN)_{z_{0}}.$ (2.36)

Therefore, we can calculate $\langle J_{MN}\rangle$ by (2.23),

$\displaystyle\langle J_{MN}\rangle_{z}=\iint_{R_{z}\times
R_{z}}\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}d\langle
M\rangle_{z_{1}}d\langle N\rangle_{z_{2}}=\int_{R_{z}}d_{t}\langle
M\rangle_{st}d_{s}\langle N\rangle_{st},$

and hence

$d\langle J_{MN}\rangle_{st}=d_{t}\langle M\rangle_{st}d_{s}\langle
N\rangle_{st}.$ (2.37)

Furthermore, the following equality holds,

$\displaystyle J_{MN}(s_{0},t_{0})=$
$\displaystyle\int_{0}^{s_{0}}M(s,t_{0})N(ds,t_{0})-\int_{R_{s_{0}t_{0}}}M(s,t)dN(s,t)$
$\displaystyle=$
$\displaystyle\int_{0}^{t_{0}}N(s_{0},t)M(s_{0},dt)-\int_{R_{s_{0}t_{0}}}N(s,t)dM(s,t).$
(2.38)

This can be deduced by rewriting (2.31) as follows

$\displaystyle\int_{R_{z_{0}}}M^{n}dN$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}M_{z_{i,j}}N({\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j})$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}M_{z_{i,j}}\left(N(\delta_{i+1,j})-N(\delta_{i,j})\right)$
$\displaystyle=\sum_{i,j=0}^{2^{n}-1}\left(M_{z_{i+1,j}}N(\delta_{i+1,j})-M_{z_{i,j}}N(\delta_{i,j})\right)+\sum_{i,j=0}^{2^{n}-1}\left(M_{z_{i,j}}-M_{z_{i+1,j}}\right)N(\delta_{i+1,j})$
$\displaystyle=\sum_{j=0}^{2^{n}-1}M_{z_{2^{n},j}}N(\delta_{2^{n},j})-\sum_{i,j=0}^{2^{n}-1}M(\epsilon_{i,j})\left(N(\delta_{i,j})+N({\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i,j})\right).$

By letting $n$ go to infinity, we get for
$\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}$,

$(\psi\cdot
NM)_{z_{0}}=\int_{0}^{t_{0}}M(s_{0},t)N(s_{0},dt)-\int_{R_{z_{0}}}MdN.$

This together with (2.36) implies (2.2).

###  2.3 Multi-dimensional Green’s formula for martingales on the plane

Now we are ready to prove Theorem 2.5, the multi-dimensional Green’s formula
on the plane. Let $\\{M^{(i)}(s,t),(s,t)\in\mathbb{R}_{+}^{2}\\}_{1\leq i\leq
d}$ be a family of independent continuous strong martingales on
$\mathbb{R}_{+}^{2}$ with finite fourth moment. We assume that the increasing
process $\langle M^{(i)}\rangle$ is deterministic for every$1\leq i\leq d$.
Let $F_{j}=F_{j}(s,t),1\leq j\leq d$ be a sequence of predictable processes of
the form,

$\displaystyle
F_{j}(s,t)=F_{j}(s,0)+\sum_{i=1}^{d}\int_{0}^{t}f_{j,i}(s,r)M^{(i)}(s,dr)+\int_{0}^{t}f_{j,0}(s,r)dr,$
(2.39)

where $f_{j,i},1\leq j\leq d,0\leq i\leq d$ are $\mathcal{F}$-predictable
processes.

######  Theorem 2.5.

Fix $s_{0},t_{0}>0$. Suppose that $\\{F_{j}(s,t)\\}_{1\leq j\leq d}$ are
predictable processes given by (2.39). Assume

$\displaystyle\mathbb{E}\left[\int_{0}^{s_{0}}\int_{0}^{t_{0}}f_{j,i}(s,t)^{2}d_{t}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}t}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}\right]<\infty,\qquad\forall
1\leq i,j\leq d,$ (2.40)

and

$\displaystyle\mathbb{E}\left[\int_{0}^{t_{0}}\int_{0}^{s_{0}}f_{j,0}(s,t)^{2}d_{s}\big{\langle}M^{(i)}\big{\rangle}_{st_{0}}dt\right]<\infty,\qquad\forall
1\leq j\leq d.$ (2.41)

Then for any rectangle $A\subseteq R_{s_{0}t_{0}}$, we have

$\displaystyle\sum_{j=1}^{d}\int_{\partial A}F_{j}(s,t)M^{(j)}(ds,t)$
$\displaystyle=\sum_{j=1}^{d}\int_{A}F_{j}(s,t)dM^{(j)}(s,t)+\sum_{i,j=1}^{d}\int_{A}f_{j,i}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)$
$\displaystyle\qquad+\sum_{j=1}^{d}\int_{A}f_{j,0}(s,t)M^{(j)}(ds,t)dt.$
(2.42)

###### Proof.

We will follow the argument in the proof of [CW75, Theorem 6.1]. Let
$A=[s_{1},s_{2}]\times[t_{1},t_{2}]\subset[0,s_{0}]\times[0,t_{0}]$. Without
loss of generality, we may assume that $F_{j}=0$ on the line segment with
endpoints $(s_{1},t_{1})$ and $(s_{2},t_{1})$. Indeed, noting
$F_{j}(s,t)=F_{j}(s,t_{1})+(F_{j}(s,t)-F_{j}(s,t_{1})),$ it follows from

$\displaystyle\int_{A}F_{j}(s,t_{1})dM^{(j)}(s,t)=\int_{s_{1}}^{s_{2}}F_{j}(s,t_{1})\left(M^{(j)}(ds,t_{2})-M^{(j)}(ds,t_{1})\right)=\int_{\partial
A}F_{j}(s,t_{1})M^{(j)}(ds,t),$

that (2.5) holds for $F_{j}(s,t)$ if and only if it holds for
$F_{j}(s,t)-F_{j}(s,t_{1})$.

Next, we consider the case that each stochastic partial derivative
$f_{j,i}(s,t)\equiv f_{j,i}\in\mathcal{F}_{s_{1},t_{1}}$ is a constant
function for $1\leq j\leq d,0\leq i\leq d$. Then by (2.39), we have

$\displaystyle
F_{j}(s,t)=\sum_{i=1}^{d}f_{j,i}\left(M^{(i)}(s,t)-M^{(i)}(s,t_{1})\right)+f_{j,0}(t-t_{1}),(s,t)\in
A,1\leq j\leq d.$ (2.43)

On one hand, noting that
$J_{MN}(A)=J_{MN}(s_{2},t_{2})-J_{MN}(s_{1},t_{2})-J_{MN}(s_{2},t_{1})+J_{MN}(s_{1},t_{1})$,
it follows from (2.2) that

$\displaystyle\int_{A}f_{j,i}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)$
$\displaystyle=\int_{\partial
A}f_{j,i}M^{(i)}(s,t)M^{(j)}(ds,t)-\int_{A}f_{j,i}M^{(i)}(s,t)dM^{(j)}(s,t)$
$\displaystyle=\int_{\partial
A}f_{j,i}\left(M^{(i)}(s,t)-M^{(i)}(s,t_{1})\right)M^{(j)}(ds,t)-\int_{A}f_{j,i}\left(M^{(i)}(s,t)-M^{(i)}(s,t_{1})\right)dM^{(j)}(s,t).$
(2.44)

Here $\int_{\partial A}$ is a line integral on $\partial A$ with clockwise as
its positive direction.

On the other hand, Itô’s formula yields

$\displaystyle\int_{A}f_{j,0}(t-t_{1})dM^{(j)}(s,t)$
$\displaystyle=f_{j,0}\int_{t_{1}}^{t_{2}}(t-t_{1})\left(M^{(j)}(s_{2},dt)-M^{(j)}(s_{1},dt)\right)$
$\displaystyle=f_{j,0}(t_{2}-t_{1})\left(M^{(j)}(s_{2},t_{2})-M^{(j)}(s_{1},t_{2})\right)-f_{j,0}\int_{t_{1}}^{t_{2}}\left(M^{(j)}(s_{2},t)-M^{(j)}(s_{1},t)\right)dt$
$\displaystyle=\int_{\partial
A}f_{j,0}(t-t_{1})M^{(j)}(ds,t)-\int_{t_{1}}^{t_{2}}\left(\int_{s_{1}}^{s_{2}}f_{j,0}M^{(j)}(ds,t)\right)dt.$
(2.45)

By (2.43), (2.3) and (2.3), we get (2.5). Thus, we have proved the theorem for
the case that all stochastic partial derivatives are constant functions. Note
that for $A=\cup_{i=1}^{k}A_{i}$ where $A_{i}$ are disjoint rectangles, one
has $\int_{\partial A}=\sum_{i=1}^{k}\int_{\partial A_{i}}$ and
$\int_{A}=\sum_{i=1}^{k}\int_{A_{i}}$. Therefore, (2.5) also holds for the
case that all stochastic partial derivatives are simple functions.

For the general case, recall that the martingales $\\{M^{(i)}\\}_{1\leq i\leq
d}$ are independent and the increasing processes $\\{\left\langle
M^{(i)}\right\rangle\\}_{1\leq i\leq d}$ are deterministic. By (2.40) and
(2.41), for $0\leq i\leq d$, $1\leq j\leq d$, we can find sequences
$\\{f_{j,i}^{(n)}\\}_{n\in\mathbb{N}}$ of bounded simple functions such that
as $n\to\infty$,

$\displaystyle\int_{0}^{s_{0}}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,i}^{(n)}(s,t)-f_{j,i}(s,t)\right)^{2}\right]d_{t}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}t}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}\rightarrow
0,~{}1\leq i,j\leq d,$ (2.46)

and

$\displaystyle\int_{0}^{s_{0}}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,0}^{(n)}(s,t)-f_{j,0}(s,t)\right)^{2}\right]dtd_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}\rightarrow
0,\qquad 1\leq j\leq d.$ (2.47)

Define

$\displaystyle
F_{j}^{(n)}(s,t)=\sum_{i=1}^{d}\int_{0}^{t}f_{j,i}^{(n)}(s,r)M^{(i)}(s,dr)+\int_{0}^{t}f_{j,0}^{(n)}(s,r)dr,~{}1\leq
j\leq d.$

Then (2.5) holds for the family $\big{\\{}F_{j}^{(n)}\big{\\}}_{1\leq j\leq
d}$, and it remains to take the limit as $n\rightarrow\infty$.

We deal with the left-hand side of (2.5) first. It follows from (2.46) that,
as $n\rightarrow\infty$, for $1\leq i,j\leq d$

$\displaystyle\mathbb{E}\left[\left(\int_{\partial
A}\int_{0}^{t}f_{j,i}(s,r)M^{(i)}(s,dr)M^{(j)}(ds,t)-\int_{\partial
A}\int_{0}^{t}f_{j,i}^{(n)}(s,r)M^{(i)}(s,dr)M^{(j)}(ds,t)\right)^{2}\right]$
$\displaystyle\leq
2\sum_{k=1,2}\mathbb{E}\left[\left(\int_{s_{1}}^{s_{2}}\int_{0}^{t_{k}}\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)M^{(i)}(s,dr)M^{(j)}(ds,t_{k})\right)^{2}\right]$
$\displaystyle=2\sum_{k=1,2}\int_{s_{1}}^{s_{2}}\int_{0}^{t_{k}}\mathbb{E}\left[\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)^{2}\right]d_{r}\big{\langle}M^{(i)}\big{\rangle}_{sr}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{k}}$
$\displaystyle\leq
2\sum_{k=1,2}\int_{s_{1}}^{s_{2}}\int_{0}^{t_{k}}\mathbb{E}\left[\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)^{2}\right]d_{r}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}r}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}$
$\displaystyle\rightarrow 0.$ (2.48)

Similarly, by (2.47), we have as $n\to\infty$, for $1\leq j\leq d$,

$\displaystyle\mathbb{E}\left[\left(\int_{\partial
A}\int_{0}^{t}f_{j,0}(s,r)drM^{(j)}(ds,t)-\int_{\partial
A}\int_{0}^{t}f_{j,0}^{(n)}(s,r)drM^{(j)}(ds,t)\right)^{2}\right]$
$\displaystyle\leq
2\sum_{k=1,2}\int_{s_{1}}^{s_{2}}\mathbb{E}\left[\left(\int_{0}^{t_{k}}\left(f_{j,0}(s,r)-f_{j,0}^{(n)}(s,r)\right)dr\right)^{2}\right]d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}$
$\displaystyle\leq
4\int_{s_{1}}^{s_{2}}t_{0}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,0}(s,r)-f_{j,0}^{(n)}(s,r)\right)^{2}\right]drd_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}$
$\displaystyle\rightarrow 0.$ (2.49)

Hence, combing (2.3) with (2.3), we get

$\displaystyle\lim_{n\rightarrow\infty}\sum_{j=1}^{d}\int_{\partial
A}F_{j}^{(n)}(s,t)M^{(j)}(ds,t)=\sum_{j=1}^{d}\int_{\partial
A}F_{j}(s,t)M^{(j)}(ds,t)$

in $L^{2}(\Omega)$.

Next, we deal with the first term on the right-hand side of (2.5). By (2.46),

$\displaystyle\mathbb{E}\left[\left(\int_{A}\int_{0}^{t}f_{j,i}(s,r)M^{(i)}(s,dr)dM^{(j)}(s,t)-\int_{A}\int_{0}^{t}f_{j,i}^{(n)}(s,r)M^{(i)}(s,dr)dM^{(j)}(s,t)\right)^{2}\right]$
$\displaystyle=\int_{A}\mathbb{E}\left[\left(\int_{0}^{t}\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)M^{(i)}(s,dr)\right)^{2}\right]d\big{\langle}M^{(j)}\big{\rangle}_{st}$
$\displaystyle=\int_{A}\int_{0}^{t}\mathbb{E}\left[\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)^{2}\right]d_{r}\big{\langle}M^{(i)}\big{\rangle}_{sr}d\big{\langle}M^{(j)}\big{\rangle}_{st}$
$\displaystyle\leq\int_{A}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)^{2}\right]d_{r}\big{\langle}M^{(i)}\big{\rangle}_{sr}d\big{\langle}M^{(j)}\big{\rangle}_{st}$
$\displaystyle\leq\int_{0}^{s_{0}}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,i}(s,r)-f_{j,i}^{(n)}(s,r)\right)^{2}\right]d_{r}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}r}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}$
$\displaystyle\rightarrow 0,\quad n\rightarrow\infty,\ \ \ \forall 1\leq
i,j\leq d.$ (2.50)

Similarly, by (2.47),

$\displaystyle\mathbb{E}\left[\left(\int_{A}\int_{0}^{t}f_{j,0}(s,r)drdM^{(j)}(s,t)-\int_{A}\int_{0}^{t}f_{j,0}^{(n)}(s,r)drdM^{(j)}(s,t)\right)^{2}\right]$
$\displaystyle\leq\int_{A}t_{0}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,0}(s,r)-f_{j,0}^{(n)}(s,r)\right)^{2}\right]drd\big{\langle}M^{(j)}\big{\rangle}_{st}$
$\displaystyle\leq
t_{0}\int_{0}^{s_{0}}\int_{0}^{t_{0}}\mathbb{E}\left[\left(f_{j,0}(s,r)-f_{j,0}^{(n)}(s,r)\right)^{2}\right]drd_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}$
$\displaystyle\rightarrow 0,\quad n\rightarrow\infty,\ \forall 1\leq j\leq d.$
(2.51)

Hence, (2.3) and (2.3) imply

$\displaystyle\lim_{n\rightarrow\infty}\sum_{j=1}^{d}\int_{A}F_{j}^{(n)}(s,t)M^{(j)}(ds,t)\rightarrow\sum_{j=1}^{d}\int_{A}F_{j}(s,t)M^{(j)}(ds,t)$

in $L^{2}(\Omega)$.

Next we deal with the limit of the second term on the right hand side of
(2.5). By (2.37) and (2.46), we have

$\displaystyle\mathbb{E}\left[\left(\int_{A}f_{j,i}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)-\int_{A}f_{j,i}^{(n)}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)\right)^{2}\right]$
$\displaystyle\leq\mathbb{E}\left[\int_{A}\left|f_{j,i}(s,t)-f_{j,i}^{(n)}(s,t)\right|^{2}d_{t}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}t}d_{s}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}\right]$
$\displaystyle\rightarrow 0,\quad n\rightarrow\infty,\ ~{}\forall 1\leq
i,j\leq d.$ (2.52)

Hence, we have

$\displaystyle\sum_{i,j=1}^{d}\int_{A}f_{j,i}^{(n)}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)\rightarrow\sum_{i,j=1}^{d}\int_{A}f_{j,i}(s,t)dJ_{M^{(i)},M^{(j)}}(s,t)$

in $L^{2}(\Omega)$.

Lastly, we deal with the limit of the third term on the right hand side of
(2.5). By the Cauchy-Schwarz inequality and (2.47),

$\displaystyle\mathbb{E}\left[\left(\int_{A}f_{j,0}(s,t)M^{(j)}(ds,t)dt-\int_{A}f_{j,0}^{(n)}(s,t)M^{(j)}(ds,t)dt\right)^{2}\right]$
$\displaystyle=\mathbb{E}\left[\left(\int_{t_{1}}^{t_{2}}\int_{s_{1}}^{s_{2}}\left(f_{j,0}(s,t)-f_{j,0}^{(n)}(s,t)\right)M^{(j)}(ds,t)dt\right)^{2}\right]$
$\displaystyle\leq\left(t_{2}-t_{1}\right)\int_{T}^{t_{2}}\mathbb{E}\left[\left(\int_{S}^{s_{2}}\left(f_{j,0}(s,t)-f_{j,0}^{(n)}(s,t)\right)M^{(j)}(ds,t)\right)^{2}\right]dt$
$\displaystyle=\left(t_{2}-t_{1}\right)\int_{T}^{t_{2}}\int_{S}^{s_{2}}\mathbb{E}\left[\left(f_{j,0}(s,t)-f_{j,0}^{(n)}(s,t)\right)^{2}\right]\big{\langle}M^{(j)}(ds,t)\big{\rangle}dt$
$\displaystyle\rightarrow 0,\quad n\rightarrow\infty,\ \forall 1\leq j\leq d.$
(2.53)

Thus, we have the following convergence in $L^{2}(\Omega)$,

$\displaystyle\sum_{j=1}^{d}\int_{A}f_{j,0}^{(n)}(s,t)M^{(j)}(ds,t)dt\rightarrow\sum_{j=1}^{d}\int_{A}f_{j,0}(s,t)M^{(j)}(ds,t)dt.$

The proof is concluded. ∎

Similarly, for predictable processes of the form

$\displaystyle
F_{j}(s,t)=F_{j}(0,t)+\sum_{i=1}^{d}\int_{0}^{s}f_{j,i}(r,t)M^{(i)}(dr,t)+\int_{0}^{s}f_{j,0}(r,t)dr,\
\ 1\leq j\leq d,$ (2.54)

where $f_{j,i},\,1\leq j\leq d,\,0\leq i\leq d$ are $\mathcal{F}$-predictable
processes, we have the following Green’s formula.

######  Theorem 2.6.

Fix $s_{0},t_{0}>0$. Suppose that $\\{F_{j}(s,t)\\}_{1\leq j\leq d}$ are
predictable processes given by (2.54). Assume

$\displaystyle\mathbb{E}\left[\int_{0}^{s_{0}}\int_{0}^{t_{0}}f_{j,i}(s,t)^{2}d_{s}\big{\langle}M^{(i)}\big{\rangle}_{st_{0}}d_{t}\big{\langle}M^{(j)}\big{\rangle}_{st_{0}}\right]<\infty,\quad\forall
1\leq i,j\leq d,$

and

$\displaystyle\mathbb{E}\left[\int_{0}^{s_{0}}\int_{0}^{t_{0}}f_{j,0}(s,t)^{2}d_{t}\big{\langle}M^{(i)}\big{\rangle}_{s_{0}t}ds\right]<\infty,\quad\forall
1\leq j\leq d.$

Then for any rectangle $A\subseteq R_{s_{0}t_{0}}$, we have

$\displaystyle\sum_{j=1}^{d}\int_{\partial A}F_{j}(s,t)M^{(j)}(s,dt)$
$\displaystyle=\sum_{j=1}^{d}\int_{A}F_{j}(s,t)dM^{(j)}(s,t)+\sum_{i,j=1}^{d}\int_{A}f_{j,i}(s,t)dJ_{M^{(j)},M^{(i)}}(s,t)$
$\displaystyle\qquad\qquad+\sum_{j=1}^{d}\int_{A}f_{j,0}(s,t)M^{(j)}(s,dt)ds.$

###### Proof.

Noting that by the second equality of (2.2), we have that for the left-hand
side of (2.3),

$\displaystyle\int_{A}f_{j,i}(s,t)dJ_{M^{(j)},M^{(i)}}(s,t)$
$\displaystyle=\int_{\partial
A}f_{j,i}M^{(i)}(s,t)M^{(j)}(s,dt)-\int_{A}f_{j,i}M^{(i)}(s,t)dM^{(j)}(s,t).$

The rest of the proof is the same as that of Theorem 2.5 and thus is omitted.
∎

###  2.4 Quadratic covariations of $J_{MN}$ and $J_{M^{\prime}N^{\prime}}$

Let $M,N,M^{\prime},N^{\prime}$ be continuous martingales belonging to
$\mathfrak{M}_{s}^{4}(z_{0}).$ In this subsection, for the completion of the
theory, we shall derive the quadratic covariation for $J_{MN}=\psi\cdot MN$
and $J_{M^{\prime}N^{\prime}}=\psi\cdot M^{\prime}N^{\prime}$ with
$\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}$
which are defined in Section 2.2. More specifically, we aim to show

$d\langle J_{MN},J_{M^{\prime},N^{\prime}}\rangle_{st}=d_{t}\langle
M,M^{\prime}\rangle_{st}d_{s}\langle N,N^{\prime}\rangle_{st}.$ (2.55)

Recall that $J_{MN}$ can be approximated by $J^{n}_{MN}$ as in (2.25), and
that one can approximate the function
$\psi(z_{1},z_{2})=\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}$ by

$\displaystyle\psi=\lim_{n\to\infty}\sum_{i,j,k,l\in\mathbf{I}_{n}}\psi_{ijkl},$

where $\psi_{ijkl}(z_{1},z_{2})=\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}}(z_{1})\mathbf{1}_{{\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{k,l}}(z_{2})$ and $\mathbf{I}_{n}$ is a
subset of $\\{(i,j,k,l),i,j,k,l\in 1,\dots,2^{n}\\}$ which consists of
$(i,j,k,l)$ satisfying $0\leq i<k\leq 2^{n}-1$ and $0\leq l<j\leq 2^{n}-1$.
Denote by $\mathbf{J}_{n}$ the subset of $\mathbf{I}_{n}\times\mathbf{I}_{n}$
such that for
$((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\in\mathbf{J}_{n}$,
the four rectangles $A={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}$, $B={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k,l}$, $A^{\prime}={\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i^{\prime},j^{\prime}}$,
$B^{\prime}={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k^{\prime},l^{\prime}}$ are of the same position as in
Case 3 in the proof of (2.2) in Section 2.2. That is, $A$ and $A^{\prime}$ are
at the same horizontal level and are at the upper left of $B$ and
$B^{\prime}$, while $B$ and $B^{\prime}$ are at the same vertical level. Now
the quadratic covariation can be computed as follows,

$\displaystyle\left\langle
J_{MN},J_{M^{\prime}N^{\prime}}\right\rangle_{z_{0}}$
$\displaystyle=\lim_{n\to\infty}\sum_{(i,j,k,l)\in\mathbf{I}_{n}}\sum_{(i^{\prime},j^{\prime},k^{\prime},l^{\prime})\in\mathbf{I}_{n}}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}$
$\displaystyle=\lim_{n\to\infty}\Bigg{(}\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\in\mathbf{J}_{n}}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}$
$\displaystyle\qquad\qquad\qquad+\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\notin\mathbf{J}_{n}}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}\Bigg{)}.$ (2.56)

For the first term on the right-hand side of (2.4), observing that the indices
$((i,j,k^{\prime},l^{\prime}),(i^{\prime},j^{\prime},k,l)),$
$((i^{\prime},j^{\prime},k,l),(i,j,k^{\prime},l^{\prime}))$, and
$((i^{\prime},j^{\prime},k^{\prime},l^{\prime}),(i,j,k,l))$ all belong to
$\mathbf{J}_{n}$ as long as
$((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))$
$\in\mathbf{J}_{n}$. Thus, we have

$\displaystyle\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\in\mathbf{J}_{n}}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}$
$\displaystyle=\dfrac{1}{4}\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\in\mathbf{J}_{n}}\Big{(}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}+\left\langle\psi_{ijk^{\prime}l^{\prime}}\cdot
MN,\psi_{i^{\prime}j^{\prime}kl}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}$
$\displaystyle\qquad\qquad\qquad\qquad+\left\langle\psi_{i^{\prime}j^{\prime}kl}\cdot
MN,\psi_{ijk^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}+\left\langle\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
MN,\psi_{ijkl}\cdot M^{\prime}N^{\prime}\right\rangle_{z_{0}}\Big{)}$
$\displaystyle=\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\in\mathbf{J}_{n}}\langle
M,M^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}\cap{\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i^{\prime},j^{\prime}})\langle
N,N^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k,l}\cap{\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k^{\prime},l^{\prime}})$ (2.65)
$\displaystyle=\sum_{(i,j,k,l)\in\mathbf{I}_{n}}\langle
M,M^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j})\langle
N,N^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k,l}),$ (2.70)

where the second equality follows from (2.2).

For the second term in (2.4), noting that when
$((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\notin\mathbf{J}_{n}$,
the four rectangles $A={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{i,j}$, $B={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k,l}$, $A^{\prime}={\ooalign{$\sqsubset\mkern
2.0mu$\cr$\mkern 2.0mu\sqsupset$\cr}}_{i^{\prime},j^{\prime}}$ and
$B^{\prime}={\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{k^{\prime},l^{\prime}}$ are of the same position as in
Case 1 or Case 2 in the proof of (2.2) in Section 2.2. Thus, we have

$\displaystyle\sum_{((i,j,k,l),(i^{\prime},j^{\prime},k^{\prime},l^{\prime}))\notin\mathbf{J}_{n}}\left\langle\psi_{ijkl}\cdot
MN,\psi_{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\cdot
M^{\prime}N^{\prime}\right\rangle_{z_{0}}=0.$ (2.71)

Therefore, substituting (2.4) and (2.71) into (2.4), one has

$\displaystyle\left\langle
J_{MN},J_{M^{\prime}N^{\prime}}\right\rangle_{z_{0}}$
$\displaystyle=\lim_{n\to\infty}\sum_{(i,j,k,l)\in\mathbf{I}_{n}}\langle
M,M^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{ij})\langle
N,N^{\prime}\rangle({\ooalign{$\sqsubset\mkern 2.0mu$\cr$\mkern
2.0mu\sqsupset$\cr}}_{kl})$
$\displaystyle=\int_{R_{z_{0}}}\int_{R_{z_{0}}}\mathbf{1}_{[z_{1}{\curlywedge\atop\curlywedge}z_{2}]}d\langle
M,M^{\prime}\rangle_{z_{1}}d\langle N,N^{\prime}\rangle_{z_{2}},$

and this implies (2.55).

######  Remark 2.1.

One can easily check that the computation is still valid if the function
$\psi(z_{1},z_{2})$ is the limit of $\psi^{(n)}(z_{1},z_{2})$ in
$\mathcal{L}_{MN}^{2}(z_{0})$ and in
$\mathcal{L}_{M^{\prime}N^{\prime}}^{2}(z_{0})$ satisfying

$\displaystyle\psi^{(n)}(z_{1},z_{2})\psi^{(n)}(z_{1}^{\prime},z_{2}^{\prime})=\psi^{(n)}(z_{1},z_{2}^{\prime})\psi^{(n)}(z_{1}^{\prime},z_{2}),$
(2.72)

for all
$z_{1}=(s_{1},t_{1}),z_{2}=(s_{2},t_{2}),z_{1}^{\prime}=(s_{1}^{\prime},t_{1}^{\prime}),z_{2}^{\prime}=(s_{2}^{\prime},t_{2}^{\prime})$
satisfying $\max\\{s_{1},s_{2}\\}\leq s_{1}^{\prime}=s_{2}^{\prime}$ and
$\max\\{t_{1}^{\prime},t_{2}^{\prime}\\}\leq t_{1}=t_{2}$. Clearly,
$\psi^{(n)}(z_{1},z_{2})=h_{1}(z_{1})h_{2}(z_{2})$ satisfies (2.72). Moreover,
by fixing $(z_{1}^{\prime},z_{2}^{\prime})$, one can check that all the
functions satisfying (2.72) must have the form
$\psi^{(n)}(z_{1},z_{2})=h_{1}(z_{1})h_{2}(z_{2})$. In this situation, we have

$\langle\psi\cdot MN,\psi\cdot
M^{\prime}N^{\prime}\rangle_{z}=\iint_{R_{z}\times
R_{z}}|\psi(z_{1},z_{2})|^{2}d\langle M,M^{\prime}\rangle_{z_{1}}d\langle
N,N^{\prime}\rangle_{z_{2}}.$ (2.73)

##  3 SPDEs for the eigenvalue processes

In this section, we will derive a system of SPDEs satisfied by the eigenvalue
processes of the Brownian sheet matrix $\bm{X}$ defined in (1.1). We assume
that the family $b(s,t)$ of independent Brownian sheets have deterministic
initial values such that the eigenvalues of the symmetric matrix $\bm{X}(0,0)$
are distinct.

Recall that the standard $1$-dimensional Brownian sheet
$\\{B(s,t),(s,t)\in\mathbb{R}_{+}^{2}\\}$ is a centered Gaussian random field
with covariance function

$\displaystyle\mathbb{E}\left[B(s_{1},t_{1})B(s_{2},t_{2})\right]=(s_{1}\wedge
s_{2})(t_{1}\wedge t_{2}).$

It follows directly from Lévy’s characterization of Brownian motion that for
any fixed $t_{1},\ t_{2}>0$,

$\displaystyle\dfrac{1}{\sqrt{t_{1}}}B(t_{1},\cdot),\
\dfrac{1}{\sqrt{t_{2}}}B(\cdot,t_{2})$

are standard $1$-dimensional Brownian motions.

Consider the Brownian sheet matrix defined in (1.1). As in Appendix A, for
$1\leq i\leq d$, let
$\lambda_{i}(s,t):=\Phi_{i}(b(s,t))=\tilde{\Phi}_{i}(\bm{X}(s,t))$ be the
$i$-th biggest eigenvalue of $\bm{X}(s,t)$, where the function
$\tilde{\Phi}_{i}:\mathbf{S}_{d}\to\mathbb{R}$ maps a $d\times d$ symmetric
matrix $A\in\mathbf{S}_{d}$ to its $i$-th biggest eigenvalue
$\tilde{\Phi}_{i}(A)$.

Let $S,T>0$ be constants. By applying Itô’s formula to $\lambda_{i}(S,\cdot)$,
we have

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)=\lambda_{i}(S,T)-\lambda_{i}(S,0)=\Phi_{i}(b(S,T))-\Phi_{i}(b(S,0))$
$\displaystyle=\sum_{k\leq h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)+\dfrac{1}{2}\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}^{2}}(b(S,t))\langle
b_{kh}(S,dt)\rangle$ $\displaystyle=\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)+\dfrac{S}{2}\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}^{2}}(b(S,t))dt.$
(3.1)

By (3) and (A.21), we have

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)=\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)+\sum_{j:j\neq
i}\int_{0}^{T}\dfrac{S}{\lambda_{i}(S,t)-\lambda_{j}(S,t)}dt.$ (3.2)

We shall express the right-hand side of (3.2) as a sum of double integrals on
$[0,S]\times[0,T]$. We first deal with the second term.

For $i\neq j$, as in Appendix A we denote for any $x\in\mathbb{R}^{d(d+1)/2}$,

$\Psi_{ij}(x)=\frac{1}{\Phi_{i}(x)-\Phi_{j}(x)}.$ (3.3)

By Itô’s formula, we have

$\displaystyle\dfrac{S}{\lambda_{i}(S,t)-\lambda_{j}(S,t)}=S\Psi_{ij}(b(S,t))$
$\displaystyle=\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}ds+\sum_{k\leq
h}\int_{0}^{S}s\dfrac{\partial\Psi_{ij}}{\partial b_{kh}}(b(s,t))b_{kh}(ds,t)$
$\displaystyle\qquad\quad+\dfrac{1}{2}\sum_{k\leq
h}\int_{0}^{S}s\dfrac{\partial^{2}\Psi_{ij}}{\partial
b_{kh}^{2}}(b(s,t))\langle b_{kh}(ds,t)\rangle$
$\displaystyle=\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}ds+\sum_{k\leq
h}\int_{0}^{S}s\dfrac{\partial\Psi_{ij}}{\partial b_{kh}}(b(s,t))b_{kh}(ds,t)$
$\displaystyle\qquad\quad+\dfrac{1}{2}\sum_{k\leq
h}\int_{0}^{S}st\dfrac{\partial^{2}\Psi_{ij}}{\partial b_{kh}^{2}}(b(s,t))ds.$
(3.4)

Substituting (A.23) into (3), we have

$\displaystyle\dfrac{S}{\lambda_{i}(S,t)-\lambda_{j}(S,t)}$
$\displaystyle=\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}ds+\sum_{k\leq
h}\int_{0}^{S}s\dfrac{\partial\Psi_{ij}}{\partial b_{kh}}(b(s,t))b_{kh}(ds,t)$
$\displaystyle\qquad+\int_{0}^{S}\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}ds$
$\displaystyle\qquad+\int_{0}^{S}\dfrac{1}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))}\sum_{l:l\neq
i,j}\dfrac{st}{\left(\lambda_{i}(s,t)-\lambda_{l}(s,t)\right)\left(\lambda_{j}(s,t)-\lambda_{l}(s,t)\right)}ds.$
(3.5)

Lastly, we substitute (3) to (3.2) to obtain

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)$ $\displaystyle=\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}dsdt$
$\displaystyle\qquad+\sum_{j:j\neq i}\sum_{k\leq
h}\int_{0}^{T}\int_{0}^{S}s\dfrac{\partial\Psi_{ij}}{\partial
b_{kh}}(b(s,t))b_{kh}(ds,t)dt$ $\displaystyle\qquad+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}dsdt$
$\displaystyle\qquad+\sum_{j\neq l:j\neq i,l\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{1}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))}\dfrac{st}{\left(\lambda_{i}(s,t)-\lambda_{l}(s,t)\right)\left(\lambda_{j}(s,t)-\lambda_{l}(s,t)\right)}dsdt.$
(3.6)

The last term on the right-hand side of (3) vanishes, noting that it sums over
all $j\neq l$ for $j,l\neq i$ and that
$\frac{1}{(\lambda_{i}-\lambda_{j})}\frac{st}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{j}-\lambda_{l}\right)}$
changes its sign by interchanging the indices $j$ and $l$. Therefore, we have

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)$ $\displaystyle=\sum_{k\leq
h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}dsdt$
$\displaystyle\qquad+\sum_{j:j\neq i}\sum_{k\leq
h}\int_{0}^{T}\int_{0}^{S}s\dfrac{\partial\Psi_{ij}}{\partial
b_{kh}}(b(s,t))b_{kh}(ds,t)dt$ $\displaystyle\qquad+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}dsdt.$
(3.7)

Now, we apply the multi-dimensional Green’s formula (Theorem 2.6) to the first
term on the right-hand side of (3). By [Jaramillo2018, Theorem 2.1] (see also
[SXY20, Theorem 1.1]), it has positive probability for the eigenvalues
$\\{\lambda_{i}(s,t),1\leq i\leq d\\}$ of the Brownian sheet matrix $\bm{X}$
to collide. To avoid the singularity at the collisions, we shall restrict
$(s,t)$ in a region where all eigenvalues keep a distance from each other.

Define the region $D_{\epsilon}$ for $\epsilon>0$ by

$\displaystyle
D_{\epsilon}=\left\\{(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}:x_{i}-x_{i+1}>\epsilon,1\leq
i\leq d-1\right\\}.$

Let $\chi_{\epsilon}\in C_{b}^{\infty}(\mathbb{R}^{d})$ such that
$\chi_{\epsilon}(x)=1$ for $x\in D_{\epsilon}$ and $\chi_{\epsilon}(x)=0$ for
$x\in\mathbb{R}^{d}\setminus D_{\frac{\epsilon}{2}}$. For simplicity, we
denote $\Phi=(\Phi_{1},\ldots,\Phi_{d})$. By Itô’s formula, we have

$\displaystyle\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\chi_{\epsilon}(\Phi)\right)(b(s,t))$
$\displaystyle=\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\chi_{\epsilon}(\Phi)\right)(b(0,t))$
$\displaystyle\qquad+\sum_{k^{\prime}\leq
h^{\prime}}\int_{0}^{s}\left(\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}}\chi_{\epsilon}(\Phi)+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}}\right)(b(r,t))b_{k^{\prime}h^{\prime}}(dr,t)$
$\displaystyle\qquad+\dfrac{t}{2}\sum_{k^{\prime}\leq
h^{\prime}}\int_{0}^{s}\left(\dfrac{\partial^{3}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}\chi_{\epsilon}(\Phi)+2\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}}\right.$
$\displaystyle\qquad\qquad\left.+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l,l^{\prime}=1}^{d}\dfrac{\partial^{2}\chi_{\epsilon}}{\partial
x_{l}\partial x_{l^{\prime}}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}}\dfrac{\partial\Phi_{l^{\prime}}}{\partial
b_{k^{\prime}h^{\prime}}}+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial^{2}\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}^{2}}\right)(b(r,t))dr.$ (3.8)

Note that the function $\chi_{\epsilon}$ and all its partial derivatives
vanish when $x\in\mathbb{R}^{d}\setminus D_{\frac{\epsilon}{2}}$, by Lemma
A.2, all the integrand functions in (3) are bounded. Hence, we can apply
Theorem 2.6 to obtain

$\displaystyle\sum_{k\leq
h}\int_{0}^{T}\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\chi_{\epsilon}(\Phi)\right)(b(S,t))b_{kh}(S,dt)$
$\displaystyle=\sum_{k\leq h}\int_{\partial
R_{ST}}\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\chi_{\epsilon}(\Phi)\right)(b(s,t))b_{kh}(s,dt)$
$\displaystyle=\sum_{k\leq
h}\iint_{R_{ST}}\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\chi_{\epsilon}(\Phi)\right)(b(s,t))db_{kh}(s,t)$
$\displaystyle\qquad+\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\left(\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}}\chi_{\epsilon}(\Phi)+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}}\right)(b(s,t))dJ_{b_{kh}b_{k^{\prime}h^{\prime}}}(s,t)$
$\displaystyle\qquad+\sum_{k\leq
h}\iint_{R_{ST}}\dfrac{t}{2}\sum_{k^{\prime}\leq
h^{\prime}}\Bigg{(}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}\chi_{\epsilon}(\Phi)+2\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial b_{k^{\prime}h^{\prime}}}$
$\displaystyle\qquad+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l,l^{\prime}=1}^{d}\dfrac{\partial^{2}\chi_{\epsilon}}{\partial
x_{l}\partial x_{l^{\prime}}}(\Phi)\dfrac{\partial\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}}\dfrac{\partial\Phi_{l^{\prime}}}{\partial
b_{k^{\prime}h^{\prime}}}+\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}\sum_{l=1}^{d}\dfrac{\partial\chi_{\epsilon}}{\partial
x_{l}}(\Phi)\dfrac{\partial^{2}\Phi_{l}}{\partial
b_{k^{\prime}h^{\prime}}^{2}}\Bigg{)}(b(s,t))b_{kh}(s,dt)ds.$ (3.9)

Denote

$\displaystyle
T_{\epsilon}=\left\\{(s,t):(\Phi_{1}(b(s,t)),\ldots,\Phi_{d}(b(s,t)))\notin
D_{\epsilon}\right\\}.$

We shall construct a sequence of adapted random time pairs
$\\{(\sigma_{\frac{1}{n}},\tau_{\frac{1}{n}})\\}_{n\geq 1}$ such that
$(\sigma_{\frac{1}{n}},\tau_{\frac{1}{n}})\prec(\sigma_{\frac{1}{n+1}},\tau_{\frac{1}{n+1}})$.
First, we choose a pair of random times $(\sigma_{1},\tau_{1})$ as follows.
For each fixed $\omega\in\Omega$, if $T_{1}(\omega)=\emptyset$, then we choose
$\sigma_{1}(\omega)=\tau_{1}(\omega)=\infty$; if
$T_{1}(\omega)\not=\emptyset$, then by Zorn’s lemma, there exists a minimal
element $(s_{1},t_{1})$ in $T_{1}(\omega)$, and we set
$(\sigma_{1}(\omega),\tau_{1}(\omega))=(s_{1},t_{1}).$ By the meaning of
minimal element, we have
$[(s,t)\prec\prec(\sigma_{1},\tau_{1})]=[(\Phi_{1}(b(s,t)),\ldots,\Phi_{d}(b(s,t)))\in
D_{1}]\in\mathcal{F}_{st}$. Next, for an arbitrary fixed $\omega\in\Omega$,
let $(\sigma_{\frac{1}{2}},\tau_{\frac{1}{2}})$ be a minimal element of the
set

$\left\\{(\sigma_{1}(\omega),\tau_{1}(\omega))\prec(s,t):(\Phi_{1}(b(s,t)),\ldots,\Phi_{d}(b(s,t)))\notin
D_{\frac{1}{2}}\right\\},$

and $(\sigma_{\frac{1}{2}},\tau_{\frac{1}{2}})=(\infty,\infty)$ if the set is
empty. Clearly
$(\sigma_{1},\tau_{1})\prec(\sigma_{\frac{1}{2}},\tau_{\frac{1}{2}})$,

$[(\sigma_{1},\tau_{1})\prec(s,t)\prec\prec(\sigma_{\frac{1}{2}},\tau_{\frac{1}{2}})]=[(\Phi_{1}(b(s,t)),\ldots,\Phi_{d}(b(s,t)))\in
D_{\frac{1}{2}}\backslash D_{1}]\in\mathcal{F}_{st},$

and hence
$[(s,t)\prec\prec(\sigma_{\frac{1}{2}},\tau_{\frac{1}{2}})]\in\mathcal{F}_{st}$.
The rest of random time pairs $(\sigma_{\frac{1}{n}},\tau_{\frac{1}{n}})$ can
be constructed in the same way. Define

$(\sigma,\tau)=\sup_{n\geq 1}(\sigma_{\frac{1}{n}},\tau_{\frac{1}{n}}).$
(3.10)

Thus, $[(s,t)\prec\prec(\sigma,\tau)]=\cup_{n\geq
1}[(s,t)\prec\prec(\sigma_{\frac{1}{n}},\tau_{\frac{1}{n}})]\in\mathcal{F}_{st}$.

For each $n\geq 1$, on the set
$[\omega\in\Omega:(S,T)\prec\prec(\sigma_{\frac{1}{n}}(\omega),\tau_{\frac{1}{n}}(\omega))]$,
we have that for $(s,t)\prec(S,T)$,
$\Phi(b(s,t))=(\Phi_{1}(b(s,t),\ldots,\Phi_{d}(b(s,t))$ belongs to
$D_{\frac{1}{n}}$ and all the partial derivatives of the function
$\chi_{\frac{1}{n}}$ vanish. Thus, by (3.9), we have for
$(S,T)\prec\prec(\sigma,\tau)$,

$\displaystyle\sum_{k\leq h}\int_{0}^{T}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(S,t))b_{kh}(S,dt)=\sum_{k\leq h}\int_{\partial
R_{ST}}\dfrac{\partial\Phi_{i}}{\partial b_{kh}}(b(s,t))b_{kh}(s,dt)$
$\displaystyle=\sum_{k\leq h}\iint_{R_{ST}}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(s,t))db_{kh}(s,t)+\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}}(b(s,t))dJ_{b_{kh}b_{k^{\prime}h^{\prime}}}(s,t)$
$\displaystyle\quad+\sum_{k\leq
h}\iint_{R_{ST}}\dfrac{t}{2}\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}(b(s,t))b_{kh}(s,dt)ds.$ (3.11)

Therefore, substitute (3) to (3), we have for $(S,T)\prec\prec(\sigma,\tau)$
and $1\leq i\leq d$,

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)$ $\displaystyle=\sum_{k\leq
h}\iint_{R_{ST}}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(s,t))db_{kh}(s,t)+\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}}(b(s,t))dJ_{b_{kh}b_{k^{\prime}h^{\prime}}}(s,t)$
$\displaystyle\qquad+\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\dfrac{t}{2}\dfrac{\partial^{3}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}(b(s,t))b_{kh}(s,dt)ds+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}dsdt$
$\displaystyle\qquad+\sum_{j:j\neq i}\sum_{k\leq
h}\int_{0}^{T}\int_{0}^{S}\dfrac{\partial\Psi_{ij}}{\partial
b_{kh}}(b(s,t))sb_{kh}(ds,t)dt+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}dsdt.$
(3.12)

Noting that by (3.3),

$\displaystyle\dfrac{\partial\Psi_{ij}}{\partial b_{kh}}(b(s,t))$
$\displaystyle=\dfrac{-1}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{2}}\left(\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}-\dfrac{\partial\Phi_{j}}{\partial b_{kh}}\right),$

we have, by (A.3),

$\displaystyle\sum_{j:j\neq i}\dfrac{\partial\Psi_{ij}}{\partial
b_{kh}}(b(s,t))=\dfrac{1}{2}\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}.$

Therefore, (3) can be written in a symmetric form: for
$(S,T)\prec\prec(\sigma,\tau)$ and $1\leq i\leq d$,

$\displaystyle\lambda_{i}(S,T)-\lambda_{i}(0,0)$ $\displaystyle=$
$\displaystyle\sum_{k\leq h}\iint_{R_{ST}}\dfrac{\partial\Phi_{i}}{\partial
b_{kh}}(b(s,t))db_{kh}(s,t)+\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}}(b(s,t))dJ_{b_{kh}b_{k^{\prime}h^{\prime}}}(s,t)$
$\displaystyle\quad+\frac{1}{2}\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\iint_{R_{ST}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}(b(s,t))\Big{(}tb_{kh}(s,dt)ds+sb_{kh}(ds,t)dt\Big{)}$
$\displaystyle\quad+\sum_{j:j\neq
i}\int_{0}^{T}\int_{0}^{S}\left(\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}+\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}\right)dsdt.$
(3.13)

Recalling that we have assumed the initial eigenvalues are distinct, by the
continuity of eigenvalue functions, we have $(0,0)\prec\prec(\sigma,\tau)$
a.s. Thus, for almost all $\omega\in\Omega$, we have the following formal
partial differential equations near the initial point $(0,0)$: for $1\leq
i\leq d$,

$\displaystyle d\lambda_{i}(s,t)=$ $\displaystyle\sum_{k\leq
h}\dfrac{\partial\Phi_{i}}{\partial b_{kh}}(b(s,t))db_{kh}(s,t)+\sum_{k\leq
h}\sum_{k^{\prime}\leq h^{\prime}}\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}\partial
b_{k^{\prime}h^{\prime}}}(b(s,t))dJ_{b_{kh}b_{k^{\prime}h^{\prime}}}(s,t)$
$\displaystyle\quad+\frac{1}{2}\sum_{k\leq h}\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}(b(s,t))\Big{(}tb_{kh}(s,dt)ds+sb_{kh}(ds,t)dt\Big{)}$
$\displaystyle\quad+\sum_{j:j\neq
i}\left(\dfrac{1}{\lambda_{i}(s,t)-\lambda_{j}(s,t)}+\dfrac{2st}{(\lambda_{i}(s,t)-\lambda_{j}(s,t))^{3}}\right)dsdt.$
(3.14)

##  4 High-dimensional limit of the empirical spectral distributions

In this section, we study the high-dimensional limit of empirical spectral
measure of the rescaled Brownian sheet matrices. In Section 4.1, we first
obtain the tightness of the empirical spectral measures (Theorem 4.1), and
then show the convergence by Wigner’s theorem (Theorem 4.2). In Section 4.2,
we derive a PDE for the Stieltjes transform of the limiting measure and also a
McKean-Vlasov equation for the limiting measure.

###  4.1 Tightness and high-dimensional limit

For every integer $d\geq 1$, let $\bm{X}^{d}(s,t)$ be a $d\times d$ matrix
given by (1.1), and $\\{\lambda_{i}^{d}(s,t):1\leq i\leq d\\}$ be the set of
eigenvalues of $\bm{X}^{d}(s,t)$. Define the empirical spectral measure of
$\bm{X}^{d}(s,t)/\sqrt{d}$

$\displaystyle
L_{d}(s,t)(dx)=\dfrac{1}{d}\sum_{i=1}^{d}\delta_{\lambda_{i}^{d}(s,t)/\sqrt{d}}(dx).$
(4.1)

For a measurable function $g:\mathbb{R}\to\mathbb{R}$, we write

$\displaystyle\langle
g,L_{d}(s,t)\rangle:=\int_{\mathbb{R}}g(x)L_{d}(s,t)(dx)=\dfrac{1}{d}\sum_{i=1}^{d}g\left(\dfrac{\lambda_{i}^{d}(s,t)}{\sqrt{d}}\right).$
(4.2)

Let $\mathcal{P}(\mathbb{R})$ be the set of probability measures on
$\mathbb{R}$ equipped with its weak topology and $C_{0}(\mathbb{R})$ be the
set of continuous functions on $\mathbb{R}$ that vanish at infinity.
Throughout this subsection, let $S$ and $T$ be two fixed positive numbers, and
recall the notation $R_{ST}=[0,S]\times[0,T].$

The following tightness criterion for probability-measure-valued stochastic
processes is a straightforward generalization of [SYY20, Proposition B.3] (see
also [Rogers1993, Section 3] where this criterion was applied implicitly).

######  Lemma 4.1.

Let $\\{\mu_{d}(s,t),(s,t)\in R_{ST}\\}_{d\in\mathbb{N}}\subset
C(R_{ST},\mathcal{P}(\mathbb{R}))$ be a sequence of probability-measure-valued
random fields. Assume the following conditions are satisfied:

  1. (A)

there exists a non-negative function $\varphi(x)$ satisfying
$\lim\limits_{|x|\to+\infty}\varphi(x)=+\infty$ and

$\displaystyle\sup_{d\in\mathbb{N}}\mathbb{E}\left[\left|\left\langle\varphi,\mu_{d}(s,t)\right\rangle\right|^{\gamma}\right]<\infty,~{}\forall(s,t)\in
R_{ST},$

for some $\gamma>0$;

  2. (B)

there exists a countable dense subset
$\\{f_{i}(x),x\in\mathbb{R}\\}_{i\in\mathbb{N}}$ of $C_{0}(\mathbb{R})$, such
that for some positive constants $a_{1}>1$ and $a_{2}>1$,

$\displaystyle\mathbb{E}\left[\left|\left\langle
f_{i},\mu_{d}(s_{2},t_{2})\right\rangle-\left\langle
f_{i},\mu_{d}(s_{1},t_{1})\right\rangle\right|^{a_{1}}\right]\leq
C_{f_{i},S,T}|(s_{2},t_{2})-(s_{1},t_{1})|^{a_{2}}$

for all $(s_{1},t_{1}),(s_{2},t_{2})\in R_{ST},d\in\mathbb{N}$ and $i\in N$,
where $C_{f_{i},S,T}$ is a constant depending only on $S$, $T$ and $f_{i}$.

Then the set $\\{\mu_{d}(s,t),(s,t)\in R_{ST}\\}_{d\in\mathbb{N}}$ of
$C(R_{ST},\mathcal{P}(\mathbb{R}))$-valued random elements is tight, i.e., it
induces a tight family of probability measures on
$C(R_{ST},\mathcal{P}(\mathbb{R}))$.

The Kolmogorov continuity theorem for random fields (see e.g. [K02, Theorem
2.5.1 in Chapter 5]) implies that, on every compact interval, the Brownian
sheet is $\beta$-Hölder continuous for $\beta\in(0,\frac{1}{2})$. The
following lemma is a direct consequence of Fernique’s theorem ([fernique]).

######  Lemma 4.2.

For any $\beta\in(0,\frac{1}{2})$, there exists a positive constant
$\delta=\delta(\beta,S,T)$ depending only on $(\beta,S,T)$ such that

$\displaystyle\mathbb{E}\left[\exp\left(\delta\left\|B\right\|_{\beta;R_{ST}}^{2}\right)\right]<\infty,$

where

$\displaystyle\left\|B\right\|_{\beta;R_{ST}}=\sup_{(s_{1},t_{1}),(s_{2},t_{2})\in
R_{ST}}\dfrac{\left|B(s_{2},t_{2})-B(s_{1},t_{1})\right|}{|(s_{2},t_{2})-(s_{1},t_{1})|^{\beta}}$
(4.3)

is the $\beta$-Hölder norm of $B$ on the rectangle $R_{ST}$.

Now we are ready to derive the following result on the tightness of the
sequence $\\{L_{d}(s,t),(s,t)\in R_{ST}\\}_{d\in\mathbb{N}}$.

######  Theorem 4.1.

Assume that there exists a nonnegative function $\varphi(x)\in
C^{1}(\mathbb{R})$ with bounded derivative, such that

$\displaystyle\lim\limits_{|x|\to+\infty}\varphi(x)=+\infty~{}\mbox{ and
}~{}\sup_{d\in\mathbb{N}}\left\langle\varphi,L_{d}(0,0)\right\rangle<\infty.$
(4.4)

Then the sequence $\\{L_{d}(s,t),(s,t)\in R_{ST}\\}_{d\in\mathbb{N}}$ is tight
on $C(R_{ST},\mathcal{P}(\mathbb{R}))$.

###### Proof.

Let $f$ be an arbitrary continuously differentiable function with bounded
derivative. By the mean value theorem and the Hoffman-Wielandt inequality (see
e.g. [anderson2010, Lemma 2.1.19]), we have for
$(s_{2},t_{2}),(s_{1},t_{1})\in R_{ST}$,

$\displaystyle\left|\left\langle
f,L_{d}(s_{2},t_{2})\right\rangle-\left\langle
f,L_{d}(s_{1},t_{1})\right\rangle\right|^{2}=\left|\dfrac{1}{d}\sum_{i=1}^{d}\left(f\left(\dfrac{\lambda_{i}^{d}(s_{2},t_{2})}{\sqrt{d}}\right)-f\left(\dfrac{\lambda_{i}^{d}(s_{1},t_{1})}{\sqrt{d}}\right)\right)\right|^{2}$
$\displaystyle\leq\dfrac{1}{d}\sum_{i=1}^{d}\left|f\left(\dfrac{\lambda_{i}^{d}(s_{2},t_{2})}{\sqrt{d}}\right)-f\left(\dfrac{\lambda_{i}^{d}(s_{1},t_{1})}{\sqrt{d}}\right)\right|^{2}\leq\dfrac{\|f^{\prime}\|_{L^{\infty}}^{2}}{d^{2}}\sum_{i=1}^{d}\left|\lambda_{i}^{d}(s_{2},t_{2})-\lambda_{i}^{d}(s_{1},t_{1})\right|^{2}$
$\displaystyle\leq\dfrac{\|f^{\prime}\|_{L^{\infty}}^{2}}{d^{2}}\sum_{i,j=1}^{d}\left|X_{ij}^{d}(s_{2},t_{2})-X_{ij}^{d}(s_{1},t_{1})\right|^{2}=\dfrac{2\|f^{\prime}\|_{L^{\infty}}^{2}}{d^{2}}\sum_{i\leq
j}\left|b_{ij}(s_{2},t_{2})-b_{ij}(s_{1},t_{1})\right|^{2}.$ (4.5)

Noting that $\\{b_{ij}(s,t)\\}_{1\leq i\leq j\leq d}$ are standard Brownian
sheets, by (4.1) and the Minkowski inequality, we have for some
$\beta\in(0,\frac{1}{2})$,

$\displaystyle\mathbb{E}\left[\left|\left\langle
f,L_{d}(s_{2},t_{2})\right\rangle-\left\langle
f,L_{d}(s_{1},t_{1})\right\rangle\right|^{4}\right]$ $\displaystyle\leq$
$\displaystyle\dfrac{4\|f^{\prime}\|_{L^{\infty}}^{4}}{d^{4}}\mathbb{E}\left[\left(\sum_{i\leq
j}\left|b_{ij}(s_{2},t_{2})-b_{ij}(s_{1},t_{1})\right|^{2}\right)^{2}\right]$
$\displaystyle\leq$
$\displaystyle\dfrac{4\|f^{\prime}\|_{L^{\infty}}^{4}}{d^{4}}\left(\sum_{i\leq
j}\left(\mathbb{E}\left[\left|b_{ij}(s_{2},t_{2})-b_{ij}(s_{1},t_{1})\right|^{4}\right]\right)^{1/2}\right)^{2}$
$\displaystyle=$
$\displaystyle\dfrac{4\|f^{\prime}\|_{L^{\infty}}^{4}}{d^{4}}\left(\dfrac{d(d+1)}{2}\left(\mathbb{E}\left[\left|b_{11}(s_{2},t_{2})-b_{11}(s_{1},t_{1})\right|^{4}\right]\right)^{1/2}\right)^{2}$
$\displaystyle=$
$\displaystyle\dfrac{(d+1)^{2}\|f^{\prime}\|_{L^{\infty}}^{4}}{d^{2}}\mathbb{E}\left[\left|b_{11}(s_{2},t_{2})-b_{11}(s_{1},t_{1})\right|^{4}\right]$
$\displaystyle\leq$ $\displaystyle
4\|f^{\prime}\|_{L^{\infty}}^{4}\mathbb{E}\left[\left\|b_{11}\right\|_{\beta;R_{ST}}^{4}|(s_{2},t_{2})-(s_{1},t_{1})|^{4\beta}\right]$
$\displaystyle=$ $\displaystyle
C(\beta,f^{\prime},S,T)|(s_{2},t_{2})-(s_{1},t_{1})|^{4\beta},$ (4.6)

where $C(\beta,f^{\prime},S,T)$ is a finite positive constant by Lemma 4.2.

As a consequence, Condition (A) in Lemma 4.1 is satisfied with $\gamma=4$.
Moreover, if we choose $\beta\in(\frac{1}{4},\frac{1}{2})$, then assumption
(4.4) and (4.1) together yield Condition (B) in Lemma 4.1 with $a_{1}=4$,
$a_{2}=4\beta$ and $\\{f_{i}\\}_{i\in\mathbb{N}}$ being a sequence of
functions in $C^{1}(\mathbb{R})$ with bounded derivative that is dense in
$C_{0}(\mathbb{R})$. Then the proof is concluded by Lemma 4.1. ∎

######  Remark 4.1.

In the above proof, the independence of the Brownian sheets $b_{ij}$ ($i\leq
j$) actually is not used.

Denote by $\mu_{sc}(dx)$ the semicircle distribution, i.e.
$\mu_{sc}(dx)=p_{sc}(x)dx$, where the density function is given by

$\displaystyle p_{sc}(x)=\dfrac{\sqrt{4-x^{2}}}{2\pi}1_{[-2,2]}(x).$

Let $\\{\tilde{\mu}(s,t),(s,t)\in R_{ST}\\}$ be an element in
$C(R_{ST},\mathcal{P}(\mathbb{R}))$ such that $\tilde{\mu}(s,t)$ is a
probability measure with density function
$\tilde{p}_{s,t}(x)=\frac{1}{\sqrt{st}}p_{sc}(x/\sqrt{st})$. That is,
$\tilde{\mu}(s,t)$ is a rescaled semicircle distribution. Here, we use the
convention that $\tilde{\mu}(s,t)(dx)=\delta_{0}(dx)$ if $st=0$.

######  Theorem 4.2.

Assume the same condition as in Theorem 4.1. Also assume that
$\\{\bm{X}^{d}(0,0),d\in\mathbb{N}\\}$ are symmetric deterministic matrices
such that

$\displaystyle
D:=\sup_{d\in\mathbb{N}}\left\|\frac{1}{\sqrt{d}}\bm{X}^{d}(0,0)\right\|<\infty,$

where $\|\cdot\|$ is the operator norm (the operator norm of a symmetric
matrix is its largest eigenvalue), and that $L_{d}(0,0)$ converges weakly to
some probability measure $\mu(0,0)$ as $d$ goes to infinity.

Then, as $d\to\infty$, $\\{L_{d}(s,t),(s,t)\in R_{ST}\\}$ converges in
probability to $\\{\mu(s,t),(s,t)\in R_{ST}\\}$ in
$C(R_{ST},\mathcal{P}(\mathbb{R}))$ which is given by

$\displaystyle\mu(s,t)=\tilde{\mu}(s,t)\boxplus\mu(0,0),$ (4.7)

where $\boxplus$ is the free additive convolution of two probability measures
([anderson2010, Definition 5.3.20]).

###### Proof.

For any fixed $(s,t)\in R_{ST}$ with $st>0$, we have

$\displaystyle\dfrac{1}{\sqrt{d}}\bm{X}^{d}(s,t)=\dfrac{1}{\sqrt{d}}\left(\bm{X}^{d}(s,t)-\bm{X}^{d}(0,0)\right)+\dfrac{1}{\sqrt{d}}\bm{X}^{d}(0,0).$

By the self-similarity property of the Brownian sheet, one can see that
$\frac{1}{\sqrt{std}}\left(\bm{X}^{d}(s,t)-\bm{X}^{d}(0,0)\right)$ is a
$d\times d$ Wigner matrix (see e.g. [anderson2010, Section 2.1] for the
definition). By Wigner’s semicircle law (see e.g. [anderson2010, Theorem
2.1.1]), the empirical spectral measure of
$\frac{1}{\sqrt{std}}\left(\bm{X}^{d}(s,t)-\bm{X}^{d}(0,0)\right)$ converges
in probability to the semicircle distribution $\mu_{sc}$ in
$\mathcal{P}(\mathbb{R})$ as $d\to\infty$. Thus, the empirical spectral
measure of $\frac{1}{\sqrt{d}}\left(\bm{X}^{d}(s,t)-\bm{X}^{d}(0,0)\right)$
converges in probability to the measure $\tilde{\mu}(s,t)$ in
$\mathcal{P}(\mathbb{R})$ as $d\to\infty$. Note that the empirical spectral
measure of the matrix $\frac{1}{\sqrt{d}}\bm{X}^{d}(0,0)$ is $L_{d}(0,0)$,
which converges to $\mu(0,0)$ in $\mathcal{P}(\mathbb{R})$. Therefore, by
[anderson2010, Theorem 5.4.5], for every $(s,t)\in R_{ST}$ with $st>0$, the
empirical spectral measure of the matrix $\frac{1}{\sqrt{d}}\bm{X}^{d}(s,t)$
converges in probability to the measure $\mu(s,t)$ given by (4.7) in
$\mathcal{P}(\mathbb{R})$ as $d$ goes to infinity. Moreover, when $st=0$,
$\frac{1}{\sqrt{d}}\bm{X}^{d}(s,t)=\frac{1}{\sqrt{d}}\bm{X}^{d}(0,0)$, and the
empirical spectral measures converge to $\mu(0,0)$ in
$\mathcal{P}(\mathbb{R})$.

By Theorem 4.1, the sequence $\\{L_{d}(s,t),(s,t)\in
R_{ST}\\}_{d\in\mathbb{N}}$ is tight. Let $\\{\nu(s,t),(s,t)\in R_{ST}\\}$ be
the weak limit of an arbitrary convergent subsequence of
$\\{L_{d}(s,t),(s,t)\in R_{ST}\\}_{d\in\mathbb{N}}$. Noting that for every
fixed $(s,t)\in R_{ST}$, $L_{d}(s,t)$ is the empirical spectral measure of the
matrix $\frac{1}{\sqrt{d}}\bm{X}^{d}(s,t)$ and it converges in probability to
the deterministic measure $\mu(s,t)$, we can conclude that $\nu(s,t)=\mu(s,t)$
for $(s,t)\in R_{ST}$.

Therefore, the limit of any convergent subsequence of $\\{L_{d}(s,t),(s,t)\in
R_{ST}\\}_{d\in\mathbb{N}}$ is the deterministic measure $\\{\mu(s,t),(s,t)\in
R_{ST}\\}$ given by (4.7). The proof is concluded. ∎

###  4.2 PDEs for the limit measure

It is known (see e.g. [anderson2010]) that the high-dimensional limit
$\hat{\mu}_{t}(dx)$ of the empirical measures of Dyson Brownian motion (1.2)
satisfies the following McKean-Vlasov equation,

$\displaystyle\dfrac{\partial}{\partial
t}\int_{\mathbb{R}}f(x)\hat{\mu}_{t}(dx)=\dfrac{1}{2}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(x)-f^{\prime}(y)}{x-y}\hat{\mu}_{t}(dx)\hat{\mu}_{t}(dy),~{}~{}\mbox{for
}f\in C_{b}^{2}(\mathbb{R}).$ (4.8)

The Stieltjes transform

$\hat{G}_{t}(z)=\int_{\mathbb{R}}\frac{1}{z-x}\hat{\mu}_{t}(dx),\mbox{ for
}z\in\mathbb{C}\backslash\mathbb{R}$

of $\hat{\mu}_{t}(dx)$ solves the following complex version of inviscid
Burgers’ equation

$\displaystyle\partial_{t}\hat{G}_{t}(z)+\hat{G}_{t}(z)\partial_{z}\hat{G}_{t}(z)=0.$

In this subsection, we will derive parallel PDEs for the limit $\mu(s,t)$ (see
Theorem 4.2) of the empirical spectral measures of the rescaled Brownian sheet
matrices. We remark that the equations are obtained by the properties the
semicircle distribution and may have other equivalent forms.

Assume $\mu(0,0)(dx)=\delta_{0}(dx)$, then the limiting measure
$\mu(s,t)(dx)=\tilde{\mu}(s,t)(dx)$, recalling that

$\tilde{\mu}(s,t)(dx)=\tilde{p}_{s,t}(x)dx=\frac{1}{\sqrt{st}}p_{sc}(x/\sqrt{st})dx$

is a rescaled semicircle distribution. Thus, for a test function $f$, we have

$\displaystyle\dfrac{\partial^{2}}{\partial s\partial t}\left\langle
f,\mu(s,t)\right\rangle=$ $\displaystyle\dfrac{\partial^{2}}{\partial
s\partial t}\int_{\mathbb{R}}\dfrac{1}{\sqrt{st}}f(x)p_{sc}(x/\sqrt{st})dx$
$\displaystyle=$ $\displaystyle\dfrac{\partial^{2}}{\partial s\partial
t}\int_{\mathbb{R}}f(\sqrt{st}x)p_{sc}(x)dx$ $\displaystyle=$
$\displaystyle\dfrac{\partial}{\partial
s}\int_{\mathbb{R}}\dfrac{\sqrt{s}x}{2\sqrt{t}}f^{\prime}(\sqrt{st}x)p_{sc}(x)dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{x^{2}}{4}f^{\prime\prime}(\sqrt{st}x)p_{sc}(x)dx+\int_{\mathbb{R}}\dfrac{x}{4\sqrt{st}}f^{\prime}(\sqrt{st}x)p_{sc}(x)dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{x^{2}}{4(st)^{3/2}}f^{\prime\prime}(x)p_{sc}(x/\sqrt{st})dx+\int_{\mathbb{R}}\dfrac{x}{4(st)^{3/2}}f^{\prime}(x)p_{sc}(x/\sqrt{st})dx$
$\displaystyle=$ $\displaystyle\dfrac{1}{4st}\left\langle
x^{2}f^{\prime\prime}(x),\mu(s,t)\right\rangle+\dfrac{1}{4st}\left\langle
xf^{\prime}(x),\mu(s,t)\right\rangle.$ (4.9)

Noting that the density of the measure $\hat{\mu}_{t}(dx)$ is
$\hat{p}_{t}(x)=\frac{1}{\sqrt{t}}p_{sc}(x/\sqrt{t})$, the left-hand side of
(4.8) can be written as

$\displaystyle\dfrac{\partial}{\partial
t}\int_{\mathbb{R}}f(x)\hat{\mu}_{t}(dx)=$
$\displaystyle\dfrac{\partial}{\partial
t}\int_{\mathbb{R}}\dfrac{1}{\sqrt{t}}f(x)p_{sc}(x/\sqrt{t})dx$
$\displaystyle=$ $\displaystyle\dfrac{\partial}{\partial
t}\int_{\mathbb{R}}f(\sqrt{t}x)p_{sc}(x)dx$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{x}{2\sqrt{t}}f^{\prime}(\sqrt{t}x)p_{sc}(x)dx.$
(4.10)

Similarly, the right-hand side of (4.8) can be written as

$\displaystyle\dfrac{1}{2}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(x)-f^{\prime}(y)}{x-y}\hat{\mu}_{t}(dx)\hat{\mu}_{t}(dy)=\dfrac{1}{2}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(\sqrt{t}x)-f^{\prime}(\sqrt{t}y)}{\sqrt{t}(x-y)}p_{sc}(x)p_{sc}(y)dxdy.$
(4.11)

Substituting (4.2) and (4.11) into (4.8), we get

$\displaystyle\int_{\mathbb{R}}xf^{\prime}(\sqrt{t}x)p_{sc}(x)dx=\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(\sqrt{t}x)-f^{\prime}(\sqrt{t}y)}{x-y}p_{sc}(x)p_{sc}(y)dxdy,~{}\forall
t>0.$ (4.12)

Taking derivative with respect to $t$ for both sides, we have

$\displaystyle\int_{\mathbb{R}}x^{2}f^{\prime\prime}(\sqrt{t}x)p_{sc}(x)dx=\iint_{\mathbb{R}^{2}}\dfrac{xf^{\prime\prime}(\sqrt{t}x)-yf^{\prime\prime}(\sqrt{t}y)}{x-y}p_{sc}(x)p_{sc}(y)dxdy,~{}\forall
t>0.$ (4.13)

Now, combining (4.2), (4.12) and (4.13), we have

$\displaystyle\dfrac{\partial^{2}}{\partial s\partial t}\left\langle
f,\mu(s,t)\right\rangle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{x^{2}}{4}f^{\prime\prime}(\sqrt{st}x)p_{sc}(x)dx+\int_{\mathbb{R}}\dfrac{x}{4\sqrt{st}}f^{\prime}(\sqrt{st}x)p_{sc}(x)dx$
$\displaystyle=$
$\displaystyle\dfrac{1}{4}\iint_{\mathbb{R}^{2}}\dfrac{xf^{\prime\prime}(\sqrt{st}x)-yf^{\prime\prime}(\sqrt{st}y)}{x-y}p_{sc}(x)p_{sc}(y)dxdy$
$\displaystyle+\dfrac{1}{4\sqrt{st}}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(\sqrt{st}x)-f^{\prime}(\sqrt{st}y)}{x-y}p_{sc}(x)p_{sc}(y)dxdy$
$\displaystyle=$
$\displaystyle\dfrac{1}{4}\iint_{\mathbb{R}^{2}}\dfrac{xf^{\prime\prime}(x)-yf^{\prime\prime}(y)}{x-y}\cdot\dfrac{1}{st}p_{sc}(x/\sqrt{st})p_{sc}(y/\sqrt{st})dxdy$
$\displaystyle+\dfrac{1}{4}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(x)-f^{\prime}(y)}{x-y}\cdot\dfrac{1}{st}p_{sc}(x/\sqrt{st})p_{sc}(y/\sqrt{st})dxdy$
$\displaystyle=$
$\displaystyle\dfrac{1}{4}\iint_{\mathbb{R}^{2}}\dfrac{xf^{\prime\prime}(x)-yf^{\prime\prime}(y)}{x-y}\mu(s,t)(dx)\mu(s,t)(dy)$
$\displaystyle+\dfrac{1}{4}\iint_{\mathbb{R}^{2}}\dfrac{f^{\prime}(x)-f^{\prime}(y)}{x-y}\mu(s,t)(dx)\mu(s,t)(dy).$
(4.14)

Therefore, we get the following McKean-Vlasov equation for $\mu(s,t)(dx)$:

$\dfrac{\partial^{2}}{\partial s\partial t}\left\langle
f,\mu(s,t)\right\rangle=\frac{1}{4}\iint_{\mathbb{R}^{2}}\frac{\left(xf^{\prime}(x)\right)^{\prime}-\left(yf^{\prime}(y)\right)^{\prime}}{x-y}(\mu(s,t))^{\otimes
2}(dx,dy).$ (4.15)

Now we consider the Stieltjes transform of $\mu(s,t)(dx)$:

$\displaystyle
G_{s,t}(z)=\left\langle\dfrac{1}{z-x},\mu(s,t)\right\rangle,\mbox{ for
}z\in\mathbb{C}\backslash\mathbb{R}.$

Note that the Stieltjes transform $G(z)$ of the semicircle distribution
$p_{sc}(x)dx$ can be written as

$\displaystyle G(z)=\left\langle(z-x)^{-1},\mu_{sc}\right\rangle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{1}{z-x}p_{sc}(x)dx$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{1}{z-x/\sqrt{st}}p_{sc}(x/\sqrt{st})\dfrac{dx}{\sqrt{st}}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\dfrac{\sqrt{st}}{\sqrt{st}z-x}\tilde{p}_{s,t}(x)dx$
$\displaystyle=$ $\displaystyle\sqrt{st}G_{s,t}(\sqrt{st}z).$ (4.16)

By [anderson2010, (2.4.6)] (see also [Tao2012, (2.103)]), $G(z)$ solves

$\displaystyle G(z)^{2}-zG(z)+1=0.$ (4.17)

Substituting (4.2) into (4.17), we have

$\displaystyle
st\left(G_{s,t}(\sqrt{st}z)\right)^{2}-z\sqrt{st}G_{s,t}(\sqrt{st}z)+1=0,$

which can be rewritten as

$\displaystyle st\left(G_{s,t}(z)\right)^{2}-zG_{s,t}(z)+1=0.$ (4.18)

Taking the derivative with respect to $z$ in (4.18), we get

$\displaystyle
2stG_{s,t}(z)\partial_{z}G_{s,t}(z)-z\partial_{z}G_{s,t}(z)-G_{s,t}(z)=0.$
(4.19)

Take the derivative with respect to $z$ in (4.19), we have

$\displaystyle
2st\left(G_{s,t}(z)\partial_{z}^{2}G_{s,t}(z)+\left(\partial_{z}G_{s,t}(z)\right)^{2}\right)-z\partial_{z}^{2}G_{s,t}(z)-2\partial_{z}G_{s,t}(z)=0.$
(4.20)

Now, by choosing $f(x)=(z-x)^{-1}$ in (4.2), we have

$\displaystyle\dfrac{\partial^{2}}{\partial s\partial t}G_{s,t}(z)=$
$\displaystyle\dfrac{1}{2st}\left\langle\dfrac{x^{2}}{(z-x)^{3}},\mu(s,t)\right\rangle+\dfrac{1}{4st}\left\langle\dfrac{x}{(z-x)^{2}},\mu(s,t)\right\rangle$
$\displaystyle=$
$\displaystyle\dfrac{1}{2st}\left\langle\dfrac{(z-x)^{2}-2z(z-x)+z^{2}}{(z-x)^{3}},\mu(s,t)\right\rangle+\dfrac{1}{4st}\left\langle\dfrac{(x-z)+z}{(z-x)^{2}},\mu(s,t)\right\rangle$
$\displaystyle=$
$\displaystyle\dfrac{1}{4st}\left\langle\dfrac{1}{z-x},\mu(s,t)\right\rangle-\dfrac{3z}{4st}\left\langle\dfrac{1}{(z-x)^{2}},\mu(s,t)\right\rangle+\dfrac{z^{2}}{2st}\left\langle\dfrac{1}{(z-x)^{3}},\mu(s,t)\right\rangle$
$\displaystyle=$
$\displaystyle\dfrac{1}{4st}G_{s,t}(z)+\dfrac{3z}{4st}\partial_{z}G_{s,t}(z)+\dfrac{z^{2}}{4st}\partial_{z}^{2}G_{s,t}(z)$
$\displaystyle=$
$\displaystyle\dfrac{1}{4st}\big{(}G_{s,t}(z)+z\partial_{z}G_{s,t}(z)\big{)}+\dfrac{z}{4st}\big{(}2\partial_{z}G_{s,t}(z)+z\partial_{z}^{2}G_{s,t}(z)\big{)}$
$\displaystyle=$
$\displaystyle\dfrac{1}{2}G_{s,t}(z)\partial_{z}G_{s,t}(z)+\dfrac{z}{2}\left(G_{s,t}(z)\partial_{z}^{2}G_{s,t}(z)+\left(\partial_{z}G_{s,t}(z)\right)^{2}\right),$
(4.21)

where the last equality follows from (4.19) and (4.20). Therefore, we have the
following generalized Burgers’ equation for $G_{s,t}(z)$

$\dfrac{\partial^{2}}{\partial s\partial
t}G_{s,t}(z)=\dfrac{1}{2}G_{s,t}(z)\partial_{z}G_{s,t}(z)+\dfrac{z}{2}\left(G_{s,t}(z)\partial_{z}^{2}G_{s,t}(z)+\left(\partial_{z}G_{s,t}(z)\right)^{2}\right).$
(4.22)

##  Appendix A Some lemmas in matrix calculus

In this Appendix, we provide some results in matrix analysis which are used in
Sections 3 and 4.

######  Lemma A.1.

Let $a=(a_{1},\dots,a_{d}),b=(b_{1},\dots,b_{d})$ be two $d$-dimensional
vectors such that $\|a\|=\|b\|=1$ and $a\cdot b=0$. Then

$\sum_{1\leq i,j\leq d}(a_{i}b_{j}+a_{j}b_{i})^{2}=\sum_{1\leq i,j\leq
d}(a_{i}a_{j}+b_{i}b_{j})^{2}=2.$

###### Proof.

This is elementary to verify:

$\displaystyle\sum_{1\leq i,j\leq
d}(a_{i}b_{j}+a_{j}b_{i})^{2}=\sum_{i,j}\left(a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}+2a_{i}b_{i}a_{j}b_{j}\right)=2\|a\|^{2}\|b\|^{2}+2(a\cdot
b)^{2}=2.$

Similarly, one can show $\sum\limits_{1\leq i,j\leq
d}(a_{i}a_{j}+b_{i}b_{j})^{2}=2$. ∎

For a $d\times d$ real symmetric matrix $X=\left(X_{ij}\right)$, we write
$X=UDU^{\textsf{T}}$, where $U$ is an orthogonal matrix and
$D=\mathrm{diag}(\lambda_{1},\ldots,\lambda_{d})$. Noting that the space of
$d\times d$ symmetric matrices can be identified with $\mathbb{R}^{d(d+1)/2}$,
we consider the $i$-th biggest eigenvalue $\lambda_{i}=\tilde{\Phi}_{i}(X)$ as
a function of $d(d+1)/2$ variables $(X_{kh},1\leq k\leq h\leq d)$ for
$i=1,\dots,d$.

######  Lemma A.2.

Suppose that $X$ is a smooth function of parameters $\theta,\xi\in\mathbb{R}$.
Then we have

$\displaystyle\partial_{\theta}\lambda_{i}=\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ii},$
(A.1)
$\displaystyle\partial_{\xi}\partial_{\theta}\lambda_{i}=\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ii}+2\sum_{j:j\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\lambda_{i}-\lambda_{j}}$
(A.2)

and

$\displaystyle\partial_{\xi}^{2}\partial_{\theta}\lambda_{i}=\left(U^{\textsf{T}}\partial_{\xi}^{2}\partial_{\theta}XU\right)_{ii}+\sum_{j:j\neq
i}\dfrac{4\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+2\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\big{(}U^{\textsf{T}}\partial_{\xi}^{2}XU\big{)}_{ij}}{\lambda_{i}-\lambda_{j}}$
$\displaystyle\qquad+2\sum_{j:j\neq i}\Bigg{[}\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}+\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}$
$\displaystyle\qquad+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{j}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}}{\left(\lambda_{j}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}$
$\displaystyle\qquad-\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{j}\right)^{2}}\left(\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ii}-\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{jj}\right)\Bigg{]}.$
(A.3)

###### Proof.

Since $D=U^{\textsf{T}}XU$, we have

$\displaystyle\partial_{\theta}D=\partial_{\theta}U^{\textsf{T}}XU+U^{\textsf{T}}\partial_{\theta}XU+U^{\textsf{T}}X\partial_{\theta}U=\partial_{\theta}U^{\textsf{T}}UD+U^{\textsf{T}}\partial_{\theta}XU+DU^{\textsf{T}}\partial_{\theta}U.$
(A.4)

Besides,

$\displaystyle
0_{d}=\partial_{\theta}I_{d}=\partial_{\theta}\left(U^{\textsf{T}}U\right)=\partial_{\theta}U^{\textsf{T}}U+U^{\textsf{T}}\partial_{\theta}U.$
(A.5)

In particular, this implies

$\left(\partial_{\theta}U^{\textsf{T}}U\right)_{ii}=\left(U^{\textsf{T}}\partial_{\theta}U\right)_{ii}=0,~{}1\leq
i\leq d.$ (A.6)

The first identity (A.1) follows from the diagonal entries of (A.4) and (A.5).

Now we deduce (A.2). By (A.4),

$\displaystyle\partial_{\xi}\partial_{\theta}D$
$\displaystyle=\partial_{\xi}\partial_{\theta}U^{\textsf{T}}UD+\partial_{\theta}U^{\textsf{T}}\partial_{\xi}UD+\partial_{\theta}U^{\textsf{T}}U\partial_{\xi}D$
$\displaystyle\quad+\partial_{\xi}U^{\textsf{T}}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\theta}X\partial_{\xi}U$
$\displaystyle\quad+\partial_{\xi}DU^{\textsf{T}}\partial_{\theta}U+D\partial_{\xi}U^{\textsf{T}}\partial_{\theta}U+DU^{\textsf{T}}\partial_{\xi}\partial_{\theta}U.$
(A.7)

By (A.5), we have

$\displaystyle\left(\partial_{\theta}U^{\textsf{T}}U\partial_{\xi}D+\partial_{\xi}DU^{\textsf{T}}\partial_{\theta}U\right)_{ii}=\partial_{\xi}\lambda_{i}\left(\partial_{\theta}U^{\textsf{T}}U+U^{\textsf{T}}\partial_{\theta}U\right)_{ii}=0.$
(A.8)

Furthermore, taking partial derivative $\partial_{\xi}$ on both sides of (A.5)
yields

$\displaystyle
0_{d}=\partial_{\xi}\left(\partial_{\theta}U^{\textsf{T}}U+U^{\textsf{T}}\partial_{\theta}U\right)=\partial_{\xi}\partial_{\theta}U^{\textsf{T}}U+\partial_{\theta}U^{\textsf{T}}\partial_{\xi}U+\partial_{\xi}U^{\textsf{T}}\partial_{\theta}U+U^{\textsf{T}}\partial_{\xi}\partial_{\theta}U,$
(A.9)

which implies

$\displaystyle\left(\partial_{\xi}\partial_{\theta}U^{\textsf{T}}UD+\partial_{\theta}U^{\textsf{T}}\partial_{\xi}UD+D\partial_{\xi}U^{\textsf{T}}\partial_{\theta}U+DU^{\textsf{T}}\partial_{\xi}\partial_{\theta}U\right)_{ii}$
$\displaystyle=\lambda_{i}\left(\partial_{\xi}\partial_{\theta}U^{\textsf{T}}U+\partial_{\theta}U^{\textsf{T}}\partial_{\xi}U+\partial_{\xi}U^{\textsf{T}}\partial_{\theta}U+U^{\textsf{T}}\partial_{\xi}\partial_{\theta}U\right)_{ii}=0.$
(A.10)

Combining (A), (A) and (A.8), we have

$\displaystyle\partial_{\xi}\partial_{\theta}\lambda_{i}=\left(\partial_{\xi}U^{\textsf{T}}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\theta}X\partial_{\xi}U\right)_{ii}.$
(A.11)

Note that the matrix identity (A.4) is also valid when $\theta$ is replaced by
$\xi$. Therefore, the non-diagonal term is

$\displaystyle 0$
$\displaystyle=\lambda_{j}\left(\partial_{\xi}U^{\textsf{T}}U\right)_{ij}+\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+\lambda_{i}\left(U^{\textsf{T}}\partial_{\xi}U\right)_{ij}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+\left(\lambda_{i}-\lambda_{j}\right)\left(U^{\textsf{T}}\partial_{\xi}U\right)_{ij},\forall
1\leq i\neq j\leq d,$ (A.12)

where the second equality follows from (A.5). Thus, by (A) and (A.5),

$\displaystyle\left(\partial_{\xi}U^{\textsf{T}}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\theta}X\partial_{\xi}U\right)_{ii}$
$\displaystyle=\left(\partial_{\xi}U^{\textsf{T}}UU^{\textsf{T}}\partial_{\theta}XU+U^{\textsf{T}}\partial_{\theta}XUU^{\textsf{T}}\partial_{\xi}U\right)_{ii}$
$\displaystyle=\sum_{j=1}^{d}\left(\left(\partial_{\xi}U^{\textsf{T}}U\right)_{ij}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ji}+\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}U\right)_{ji}\right)$
$\displaystyle=\sum_{j:j\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ji}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ji}}{\lambda_{i}-\lambda_{j}}.$
(A.13)

Substituting (A) into (A.11) and noting the symmetry of the matrices
$U^{\textsf{T}}\partial_{\theta}XU$ and $U^{\textsf{T}}\partial_{\xi}XU$, we
obtain the second identity (A.2).

Finally, we deal with (A.2). Taking $\partial_{\xi}$ for the first term on the
right-hand side of (A.2), we have by (A) and (A.5),

$\displaystyle\partial_{\xi}\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ii}$
$\displaystyle=\left(\partial_{\xi}U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ii}+\left(U^{\textsf{T}}\partial_{\xi}^{2}\partial_{\theta}XU\right)_{ii}+\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}X\partial_{\xi}U\right)_{ii}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}^{2}\partial_{\theta}XU\right)_{ii}+\sum_{j=1}^{d}\left(\partial_{\xi}U^{\textsf{T}}U\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ji}+\sum_{j=1}^{d}\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}U\right)_{ji}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}^{2}\partial_{\theta}XU\right)_{ii}+\sum_{j:j\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ji}+\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ji}}{\lambda_{i}-\lambda_{j}}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}^{2}\partial_{\theta}XU\right)_{ii}+2\sum_{j:j\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}}{\lambda_{i}-\lambda_{j}}.$
(A.14)

Similarly, it follows from (A), (A.5) and (A.6) that

$\displaystyle\partial_{\xi}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}$
$\displaystyle=\left(\partial_{\xi}U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}+\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}+\left(U^{\textsf{T}}\partial_{\theta}X\partial_{\xi}U\right)_{ij}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}+\sum_{l=1}^{d}\left(\partial_{\xi}U^{\textsf{T}}U\right)_{il}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{lj}+\sum_{l=1}^{d}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}U\right)_{lj}$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}+\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{lj}}{\lambda_{i}-\lambda_{l}}+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}}{\lambda_{j}-\lambda_{l}}$
$\displaystyle\qquad+\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left[\left(\partial_{\xi}U^{\textsf{T}}U\right)_{ii}+\left(U^{\textsf{T}}\partial_{\xi}U\right)_{jj}\right]$
$\displaystyle=\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}+\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{lj}}{\lambda_{i}-\lambda_{l}}+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}}{\lambda_{j}-\lambda_{l}}.$
(A.15)

Now we deal with the second term on the right-hand side of (A.2). By (A) and
(A.1),

$\displaystyle\partial_{\xi}\bigg{(}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\lambda_{i}-\lambda_{j}}\bigg{)}$
$\displaystyle=\dfrac{\partial_{\xi}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\partial_{\xi}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\lambda_{i}-\lambda_{j}}$
$\displaystyle\qquad-\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{j}\right)^{2}}\left(\partial_{\xi}\lambda_{i}-\partial_{\xi}\lambda_{j}\right)$
$\displaystyle=\dfrac{\left(U^{\textsf{T}}\partial_{\xi}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}+\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}^{2}XU\right)_{ij}}{\lambda_{i}-\lambda_{j}}$
$\displaystyle\qquad+\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{j}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}$
$\displaystyle\qquad+\sum_{l:l\neq
i}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}+\sum_{l:l\neq
j}\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{il}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{lj}}{\left(\lambda_{j}-\lambda_{l}\right)\left(\lambda_{i}-\lambda_{j}\right)}$
$\displaystyle\qquad-\dfrac{\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ij}}{\left(\lambda_{i}-\lambda_{j}\right)^{2}}\left(\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{ii}-\left(U^{\textsf{T}}\partial_{\xi}XU\right)_{jj}\right).$
(A.16)

Then the third equality (A.2) follows from (A.2), (A) and (A). ∎

In particular, if we choose $\theta=X_{kh}$, we have for $1\leq i,j\leq d$,

$\displaystyle\left(U^{\textsf{T}}\partial_{\theta}XU\right)_{ij}$
$\displaystyle=\left(U_{ki}U_{hj}+U_{hi}U_{kj}\right)\mathbf{1}_{[k\neq
h]}+U_{ki}U_{kj}\mathbf{1}_{[k=h]}$
$\displaystyle=\left(U_{ki}U_{hj}+U_{hi}U_{kj}\right)\left(\mathbf{1}_{[k\neq
h]}+\mathbf{1}_{[k=h]}/2\right).$ (A.17)

Applying (A) to Lemma A.2 yields

$\displaystyle\dfrac{\partial\lambda_{i}}{\partial
X_{kh}}=2U_{ki}U_{hi}\mathbf{1}_{[k\neq h]}+U_{ki}^{2}\mathbf{1}_{[k=h]},$
(A.18) $\displaystyle\dfrac{\partial^{2}\lambda_{i}}{\partial
X_{kh}^{2}}=2\sum_{j:j\neq
i}\dfrac{\left|U_{ki}U_{hj}+U_{hi}U_{kj}\right|^{2}}{\lambda_{i}-\lambda_{j}}\mathbf{1}_{[k\neq
h]}+2\sum_{j:j\neq
i}\dfrac{\left|U_{ki}U_{kj}\right|^{2}}{\lambda_{i}-\lambda_{j}}\mathbf{1}_{[k=h]},$
(A.19)

and

$\displaystyle\dfrac{\partial^{2}\lambda_{i}}{\partial X_{kh}\partial
X_{k^{\prime}h^{\prime}}}$ $\displaystyle=2\sum_{j:j\neq
i}\dfrac{(U_{ki}U_{hj}+U_{hi}U_{kj})(\mathbf{1}_{[k\neq
h]}+\mathbf{1}_{[k=h]}/2)(U_{k^{\prime}i}U_{h^{\prime}j}+U_{h^{\prime}i}U_{k^{\prime}j})(\mathbf{1}_{[k^{\prime}\neq
h^{\prime}]}+\mathbf{1}_{[k^{\prime}=h^{\prime}]}/2)}{\lambda_{i}-\lambda_{j}}.$
(A.20)

Recall that $\lambda_{i}=\tilde{\Phi}_{i}(X)=\tilde{\Phi}_{i}$ is the $i$-th
biggest eigenvalue of $X$ and that

$\tilde{\Psi}_{ij}=\tilde{\Psi}_{ij}(X)=\frac{1}{\lambda_{i}-\lambda_{j}}=\frac{1}{\tilde{\Phi}_{i}(X)-\tilde{\Phi}_{j}(X)}.$

Consider a symmetric matrix $(b_{kh})_{d\times d}$. Let
$x_{kh}=b_{kh}\mathbf{1}_{[k\neq h]}+\sqrt{2}b_{kh}\mathbf{1}_{[k=h]}$ and
define $\Phi_{i}=\Phi_{i}(b):=\tilde{\Phi}_{i}(X)$ for $i=1,\dots,d$. Thus by
the chain rule, we have for $1\leq i,k,h\leq d$,

$\frac{\partial\Phi_{i}}{\partial
b_{kh}}=\frac{\partial\tilde{\Phi}_{i}}{\partial X_{kh}}\mathbf{1}_{[k\neq
h]}+\sqrt{2}\frac{\partial\tilde{\Phi}_{i}}{\partial
X_{kh}}\mathbf{1}_{[k=h]}=\frac{\partial\lambda_{i}}{\partial
X_{kh}}\mathbf{1}_{[k\neq h]}+\sqrt{2}\frac{\partial\lambda_{i}}{\partial
X_{kh}}\mathbf{1}_{[k=h]}.$

We also define

$\Psi_{ij}=\Psi_{ij}(b)=\tilde{\Psi}_{ij}(X)=\frac{1}{\lambda_{i}-\lambda_{j}}=\frac{1}{\Phi_{i}(b)-\Phi_{j}(b)}.$

The following lemma is concerned with partial derivatives of $\Phi_{i}(b)$ and
$\Psi_{ij}(b)$.

######  Lemma A.3.

$\displaystyle\sum_{k\leq h}\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}^{2}}=2\sum_{j:j\neq i}\dfrac{1}{\lambda_{i}-\lambda_{j}},$ (A.21)
$\displaystyle\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}=$ $\displaystyle
2\left(2\times\mathbf{1}_{[k<h]}+\sqrt{2}\times\mathbf{1}_{[k=h]}\right)\sum_{j:j\neq
i}\dfrac{U_{kj}U_{hj}-U_{ki}U_{hi}}{(\lambda_{i}-\lambda_{j})^{2}}$
$\displaystyle=$ $\displaystyle 2\sum_{j:j\neq
i}\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\left(\dfrac{\partial\Phi_{j}}{\partial
b_{kh}}-\dfrac{\partial\Phi_{i}}{\partial b_{kh}}\right),$ (A.22)

$\displaystyle\sum_{k\leq h}\dfrac{\partial^{2}\Psi_{ij}}{\partial
b_{kh}^{2}}=\dfrac{4}{(\lambda_{i}-\lambda_{j})^{3}}+\dfrac{1}{(\lambda_{i}-\lambda_{j})}\sum_{l:l\neq
i,j}\dfrac{2}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{j}-\lambda_{l}\right)},\text{
for }i\neq j.$ (A.23)

###### Proof.

By (A.19) and the orthogonality of $U$, we have

$\displaystyle\sum_{k\leq h}\dfrac{\partial^{2}\Phi_{i}}{\partial b_{kh}^{2}}$
$\displaystyle=\sum_{k<h}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kh}^{2}}+2\sum_{k=1}^{d}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kk}^{2}}$ $\displaystyle=2\sum_{k<h}\sum_{j:j\neq
i}\dfrac{\left|U_{ki}U_{hj}+U_{hi}U_{kj}\right|^{2}}{\lambda_{i}-\lambda_{j}}+4\sum_{k=1}^{d}\sum_{j:j\neq
i}\dfrac{\left|U_{ki}U_{kj}\right|^{2}}{\lambda_{i}-\lambda_{j}}$
$\displaystyle=\sum_{j:j\neq
i}\dfrac{\sum_{k,h}\left|U_{ki}U_{hj}+U_{hi}U_{kj}\right|^{2}}{\lambda_{i}-\lambda_{j}}=2\sum_{j:j\neq
i}\dfrac{1}{\lambda_{i}-\lambda_{j}},$

where the last equality follows from the orthogonality of $U$ and Lemma A.1.
This proves (A.21).

Next, we show (A.3). By the chain rule, we can write

$\displaystyle\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}=(\mathbf{1}_{[k<h]}+\sqrt{2}\,\mathbf{1}_{[k=h]})\bigg{(}\sum_{k^{\prime}<h^{\prime}}\dfrac{\partial^{3}\tilde{\Phi}_{i}}{\partial
X_{kh}\partial
X_{k^{\prime}h^{\prime}}^{2}}+2\sum_{k^{\prime}=1}^{d}\dfrac{\partial^{3}\tilde{\Phi}_{i}}{\partial
X_{kh}\partial X_{k^{\prime}k^{\prime}}^{2}}\bigg{)}.$ (A.24)

We choose the parameter $\theta=X_{kh}$ and $\xi=X_{h^{\prime}k^{\prime}}$ in
(A.2). The terms with second order or third order derivative vanish and we
only need to consider the terms with only the first order derivative. Note
that for indices $1\leq p_{1},p_{2},q_{1},q_{2}\leq d$

$\displaystyle\sum_{k^{\prime}<h^{\prime}}\left(U^{\textsf{T}}\dfrac{\partial
X}{\partial
X_{k^{\prime}h^{\prime}}}U\right)_{p_{1}p_{2}}\left(U^{\textsf{T}}\dfrac{\partial
X}{\partial
X_{k^{\prime}h^{\prime}}}U\right)_{q_{1}q_{2}}+2\sum_{k^{\prime}=1}^{d}\left(U^{\textsf{T}}\dfrac{\partial
X}{\partial
X_{k^{\prime}k^{\prime}}}U\right)_{p_{1}p_{2}}\left(U^{\textsf{T}}\dfrac{\partial
X}{\partial X_{k^{\prime}k^{\prime}}}U\right)_{q_{1}q_{2}}$
$\displaystyle=\sum_{k^{\prime}<h^{\prime}}(U_{k^{\prime}p_{1}}U_{h^{\prime}p_{2}}+U_{h^{\prime}p_{1}}U_{k^{\prime}p_{2}})(U_{k^{\prime}q_{1}}U_{h^{\prime}q_{2}}+U_{h^{\prime}q_{1}}U_{k^{\prime}q_{2}})+2\sum_{k^{\prime}=1}^{d}U_{k^{\prime}p_{1}}U_{k^{\prime}p_{2}}U_{k^{\prime}q_{1}}U_{k^{\prime}q_{2}}$
$\displaystyle=\bigg{(}\sum_{k^{\prime}=1}^{d}U_{k^{\prime}p_{1}}U_{k^{\prime}q_{1}}\bigg{)}\bigg{(}\sum_{h^{\prime}=1}^{d}U_{h^{\prime}p_{2}}U_{h^{\prime}q_{2}}\bigg{)}+\bigg{(}\sum_{k^{\prime}=1}^{d}U_{k^{\prime}p_{1}}U_{k^{\prime}q_{2}}\bigg{)}\bigg{(}\sum_{h^{\prime}=1}^{d}U_{h^{\prime}p_{2}}U_{h^{\prime}q_{1}}\bigg{)}$
$\displaystyle=\mathbf{1}_{[p_{1}=q_{1}]}\mathbf{1}_{[p_{2}=q_{2}]}+\mathbf{1}_{[p_{1}=q_{2}]}\mathbf{1}_{[p_{2}=q_{1}]}.$
(A.25)

Now taking sum over $(k^{\prime},h^{\prime})$ for (A.2) (i.e. taking sum over
the non-zero terms including $U^{\textsf{T}}\partial_{\xi}XU$), and applying
(A), we have

$\displaystyle\sum_{k^{\prime}\leq
h^{\prime}}\dfrac{\partial^{3}\Phi_{i}}{\partial b_{kh}\partial
b_{k^{\prime}h^{\prime}}^{2}}=2\left(\mathbf{1}_{[k<h]}+\sqrt{2}\,\mathbf{1}_{[k=h]}\right)\sum_{j:j\neq
i}\bigg{(}\frac{(U^{\textsf{T}}\partial_{\theta}X)_{jj}}{(\lambda_{i}-\lambda_{j})^{2}}-\frac{(U^{\textsf{T}}\partial_{\theta}X)_{ii}}{(\lambda_{i}-\lambda_{j})^{2}}\bigg{)}.$

This together with (A) yields the first equality of (A.3). The second equality
of (A.3) now follows (A.18):

$\displaystyle
2\left(2\,\mathbf{1}_{[k<h]}+\sqrt{2}\,\mathbf{1}_{[k=h]}\right)\sum_{j:j\neq
i}\dfrac{U_{kj}U_{hj}-U_{ki}U_{hi}}{(\lambda_{i}-\lambda_{j})^{2}}$
$\displaystyle=2\left(\mathbf{1}_{[k<h]}+\sqrt{2}\,\mathbf{1}_{[k=h]}\right)\sum_{j:j\neq
i}\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\bigg{(}\dfrac{\partial\tilde{\Phi}_{j}}{\partial
X_{kh}}-\dfrac{\partial\tilde{\Phi}_{i}}{\partial X_{kh}}\bigg{)}$
$\displaystyle=2\sum_{j:j\neq
i}\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\left(\dfrac{\partial\Phi_{j}}{\partial
b_{kh}}-\dfrac{\partial\Phi_{i}}{\partial b_{kh}}\right).$

This proves (A.3).

Now we show (A.23). Note that for $i\neq j$,

$\displaystyle\sum_{k\leq h}\dfrac{\partial^{2}\Psi_{ij}}{\partial
b_{kh}^{2}}$ $\displaystyle=\sum_{k\leq h}\dfrac{\partial}{\partial
b_{kh}}\left(-\Psi_{ij}^{2}\dfrac{\partial(\Phi_{i}-\Phi_{j})}{\partial
b_{kh}}\right)$ $\displaystyle=\sum_{k\leq
h}2\Psi_{ij}^{3}\left(\dfrac{\partial(\Phi_{i}-\Phi_{j})}{\partial
b_{kh}}\right)^{2}-\sum_{k\leq
h}\Psi_{ij}^{2}\dfrac{\partial^{2}(\Phi_{i}-\Phi_{j})}{\partial b_{kh}^{2}}.$
(A.26)

For the first term of (A), by (A.18) and the orthogonality of the columns of
$U$, for $i\neq j$, we have

$\displaystyle\sum_{k\leq
h}2\Psi_{ij}^{3}\bigg{(}\dfrac{\partial(\Phi_{i}-\Phi_{j})}{\partial
b_{kh}}\bigg{)}^{2}$
$\displaystyle=\dfrac{2}{(\Phi_{i}-\Phi_{j})^{3}}\left(\sum_{k<h}\left(\dfrac{\partial(\Phi_{i}-\Phi_{j})}{\partial
b_{kh}}\right)^{2}+\sum_{k=1}^{d}\left(\dfrac{\partial(\Phi_{i}-\Phi_{j})}{\partial
b_{kk}}\right)^{2}\right)$
$\displaystyle=\dfrac{2}{(\Phi_{i}-\Phi_{j})^{3}}\left(\sum_{k<h}\bigg{(}\dfrac{\partial(\tilde{\Phi}_{i}-\tilde{\Phi}_{j})}{\partial
X_{kh}}\bigg{)}^{2}+2\sum_{k=1}^{d}\bigg{(}\dfrac{\partial(\tilde{\Phi}_{i}-\tilde{\Phi}_{j})}{\partial
X_{kk}}\bigg{)}^{2}\right)$
$\displaystyle=\dfrac{2}{(\Phi_{i}-\Phi_{j})^{3}}\left(4\sum_{k<h}\left(U_{ki}U_{hi}-U_{kj}U_{hj}\right)^{2}+2\sum_{k=1}^{d}\left(U_{ki}^{2}-U_{kj}^{2}\right)^{2}\right)$
$\displaystyle=\dfrac{4}{(\Phi_{i}-\Phi_{j})^{3}}\sum_{k,h=1}^{d}\left(U_{ki}U_{hi}-U_{kj}U_{hj}\right)^{2}$
$\displaystyle=\dfrac{8}{(\Phi_{i}-\Phi_{j})^{3}}=\dfrac{8}{(\lambda_{i}-\lambda_{j})^{3}},$
(A.27)

where the last step follows from Lemma A.1.

For the second term of (A), we have

$\displaystyle\sum_{k\leq
h}\Psi_{ij}^{2}\dfrac{\partial^{2}(\Phi_{i}-\Phi_{j})}{\partial b_{kh}^{2}}$
$\displaystyle=\dfrac{1}{(\Phi_{i}-\Phi_{j})^{2}}\left(\sum_{k<h}\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kh}^{2}}+\sum_{k=1}^{d}\dfrac{\partial^{2}\Phi_{i}}{\partial
b_{kk}^{2}}-\sum_{k<h}\dfrac{\partial^{2}\Phi_{j}}{\partial
b_{kh}^{2}}-\sum_{k=1}^{d}\dfrac{\partial^{2}\Phi_{j}}{\partial
b_{kk}^{2}}\right)$
$\displaystyle=\dfrac{1}{(\Phi_{i}-\Phi_{j})^{2}}\left(\sum_{k<h}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kh}^{2}}+2\sum_{k=1}^{d}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kk}^{2}}-\sum_{k<h}\dfrac{\partial^{2}\tilde{\Phi}_{j}}{\partial
X_{kh}^{2}}-2\sum_{k=1}^{d}\dfrac{\partial^{2}\tilde{\Phi}_{j}}{\partial
X_{kk}^{2}}\right).$ (A.28)

By (A.19), the orthogonality of the columns of $U$, and Lemma A.1, for $i\neq
j$, we have

$\displaystyle\sum_{k<h}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kh}^{2}}+2\sum_{k=1}^{d}\dfrac{\partial^{2}\tilde{\Phi}_{i}}{\partial
X_{kk}^{2}}$ $\displaystyle=\sum_{k<h}2\sum_{l:l\neq
i}\dfrac{\left|U_{ki}U_{hl}+U_{hi}U_{kl}\right|^{2}}{\lambda_{i}-\lambda_{l}}+2\sum_{k=1}^{d}2\sum_{l:l\neq
i}\dfrac{\left|U_{ki}U_{kl}\right|^{2}}{\lambda_{i}-\lambda_{l}}$
$\displaystyle=\sum_{l:l\neq
i}\dfrac{2\sum_{k<h}\left|U_{ki}U_{hl}+U_{hi}U_{kl}\right|^{2}+4\sum_{k}\left|U_{ki}U_{kl}\right|^{2}}{\lambda_{i}-\lambda_{l}}$
$\displaystyle=\sum_{l:l\neq
i}\dfrac{\sum_{k,h}\left|U_{ki}U_{hl}+U_{hi}U_{kl}\right|^{2}}{\lambda_{i}-\lambda_{l}}=\sum_{l:l\neq
i}\dfrac{2}{\lambda_{i}-\lambda_{l}}.$ (A.29)

Similarly, we have

$\displaystyle\sum_{k<h}\dfrac{\partial^{2}\tilde{\Phi}_{j}}{\partial
X_{kh}^{2}}+2\sum_{k=1}^{d}\dfrac{\partial^{2}\tilde{\Phi}_{j}}{\partial
X_{kk}^{2}}=\sum_{l:l\neq j}\dfrac{2}{\lambda_{j}-\lambda_{l}}.$ (A.30)

Putting (A) and (A.30) to (A) yields that the second term of (A) now is

$\displaystyle\sum_{k\leq
h}\Psi_{ij}^{2}\dfrac{\partial^{2}(\Phi_{i}-\Phi_{j})}{\partial b_{kh}^{2}}$
$\displaystyle=\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\Bigg{(}\sum_{l:l\neq
i}\dfrac{2}{\lambda_{i}-\lambda_{l}}-\sum_{l:l\neq
j}\dfrac{2}{\lambda_{j}-\lambda_{l}}\Bigg{)}.$ (A.31)

By substituting (A) and (A.31) into (A), we obtain

$\displaystyle\sum_{k\leq h}\dfrac{\partial^{2}\Psi_{ij}}{\partial
b_{kh}^{2}}$
$\displaystyle=\dfrac{8}{(\lambda_{i}-\lambda_{j})^{3}}-\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\Bigg{(}\sum_{l:l\neq
i}\dfrac{2}{\lambda_{i}-\lambda_{l}}-\sum_{l:l\neq
j}\dfrac{2}{\lambda_{j}-\lambda_{l}}\Bigg{)}$
$\displaystyle=\dfrac{4}{(\lambda_{i}-\lambda_{j})^{3}}-\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\Bigg{(}\sum_{l:l\neq
i,j}\dfrac{2}{\lambda_{i}-\lambda_{l}}-\sum_{l:l\neq
i,j}\dfrac{2}{\lambda_{j}-\lambda_{l}}\Bigg{)}$
$\displaystyle=\dfrac{4}{(\lambda_{i}-\lambda_{j})^{3}}-\dfrac{1}{(\lambda_{i}-\lambda_{j})^{2}}\sum_{l:l\neq
i,j}\dfrac{2\left(\lambda_{j}-\lambda_{i}\right)}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{j}-\lambda_{l}\right)}$
$\displaystyle=\dfrac{4}{(\lambda_{i}-\lambda_{j})^{3}}+\dfrac{1}{(\lambda_{i}-\lambda_{j})}\sum_{l:l\neq
i,j}\dfrac{2}{\left(\lambda_{i}-\lambda_{l}\right)\left(\lambda_{j}-\lambda_{l}\right)}.$

This proves (A.23). ∎

## References