# Backward Stochastic Differential Equations with Markov Chains and The
Application: Homogenization of PDEs System111This work was supported by
Natural Science Foundations of China (No. 10671112) and Shandong Province (No.
JQ200801 and 2008BS01024), the National Basic Research Program of China (973
Program, No. 2007CB814904) and the Science Fund for Distinguished Young
Scholars of Shandong University (No. 2009JQ004).

Huaibin Tang222Current address: INRIA-IRISA, Campus de Beaulieu, 35042 Rennes
Cedex, France<EMAIL_ADDRESS>School of Mathematics, Shandong
University, Jinan, Shandong 250100, P. R. China.  Zhen Wu<EMAIL_ADDRESS>
###### Abstract

Stemmed from the derivation of the optimal control to a stochastic linear-
quadratic control problem with Markov jumps, we study one kind of backward
stochastic differential equations (BSDEs) that the generator $f$ is affected
by a Markovian switching. Then, the case that the Markov chain is involved in
a large state space is considered. Following the classical approach, a
hierarchical approach is adopted to reduce the complexity and a singularly
perturbed Markov chain is involved. We will study the asymptotic property of
BSDE with the singularly perturbed Markov chain. At last, as an application of
our theoretical result, we show the homogenization of one system of partial
differential equations (PDEs) with a singularly perturbed Markov chain.

###### keywords:

BSDE , Markov chain , weak convergence , homogenization.

##  1 Introduction

The study of backward stochastic differential equations (BSDEs in short)
stemmed from stochastic control problem ([4]) in which a non-linear Ricatti
BSDE was introduced. Then, after the pioneering work of Pardoux and Peng
([17]) about the general BSDE which was driven by a Brownian motion, BSDEs
have been extensively studied in the last twenty years because of their
connections with mathematical finance, stochastic control, and partial
differential equations (PDEs in short), please refer to [10, 8, 7]. Since
then, many researchers devoted their work to more general BSDE, such as BSDE
driven by a Lévy process ([1, 15]) and BSDE with respect to both a Brownian
motion and a Poisson random measure ([3, 21, 11]). Recently, Cohen and Elliott
[5, 6] studied BSDE driven by the martingale part of a Markov chain and its
application in finance.

As studied in Zhang and Yin ([25]), we consider the stocks investment models
by virtue of hybrid geometric Brownian motion in which both the expected
return and volatility depend on a finite state Markov chain. To capture the
market trends as well as the various economic factors, a finite state Markov
chain $\alpha_{t}$, $t\geq 0$, is introduced to represent the general market
directions. If our object is to allocate assets into a number of stocks so as
to maximize an expected utility within a finite time horizon, it leads to an
optimal control problem. By virtue of maximum principle method, we need to
introduce an adjoint equation to deal with this optimization problem. The
adjoint equation and the state equation form forward backward stochastic
differential equations (FBSDEs in short) system. Here the adjoint equation
will be a kind of BSDE with a Markov chain.

Motivated by such adjoint equation, in our paper, we consider the following
BSDE with a Markov chain:

$Y_{t}=\xi+\int_{t}^{T}f(s,Y_{s},Z_{s},\alpha_{s})ds-\int_{t}^{T}Z_{s}dB_{s},$

where $\alpha=\\{\alpha_{t};0\leq t\leq T\\}$ is a continuous-time Markov
chain independent of the Brownian motion $B$. It is noted that this BSDE is
different from the one studied in Cohen and Elliott ([5]), and it can be
considered as that its generator is disturbed by random environment and takes
a set of discrete values which can be described by a Markov chain.

When study the solvability for BSDE with a Markov chain, the classic method
with the contraction mapping cannot be directly used for the lack of suitable
filtration and corresponding Itô’s representation theorem. In this paper,
inspired by the method dealing with the BSDE with doubly Brownian motion
([19]), we construct a new filtration and give the corresponding extended
Itô’s representation theorem.

When various factors are considered, the underlying Markov chain inevitably
has a large state space, and the corresponding BSDE becomes increasingly
complicated. It is rationale that the change rates of states display a two-
time-scale behavior, a fast-time scale and a slow varying one. Under this
case, a small parameter $\varepsilon>0$ can be introduced and the singularly
perturbed Markov chain is involved.

In this paper, we consider the case that the states of the underlying Markov
chain are divided into a number of weakly irreducible classes such that the
Markov chain fluctuates rapidly among different states, and jumps less
frequently among those classes. To reduce the complexity, a small parameter
($\varepsilon>0$) is introduced to reflect the different rate of changes among
different states. As shown in Zhang and Yin ([23]), it leads to a singularly
perturbed Markovian models with two-time scale, the actual time $t$ and the
stretched time $\frac{t}{\varepsilon}$. By aggregating the states in each
irreducible class into a single, a limit aggregated Markov chain with a
considerably smaller space can be obtained and its asymptotic probability
distribution is studied. Such asymptotic theory has many applications in
optimal control problem and mathematical finance. We refer interested readers
to [25, 24].

In this paper, we will focus on the asymptotic property of BSDE with a
singularly perturbed Markov chain. Following the averaged approach that
aggregating the states according to their jump rates, we will show that the
distribution of $(Y_{t},\int_{t}^{T}Z_{s}d\bar{B}_{s})$ can be seen as an
asymptotic distribution to $(Y^{\varepsilon}_{t},$
$\int_{t}^{T}Z^{\varepsilon}_{s}dB_{s})$, where
$(Y^{\varepsilon},Z^{\varepsilon})$ and $(Y,Z)$ satisfy:

$Y^{\varepsilon}_{t}=\xi+\int_{t}^{T}f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})ds-\int_{t}^{T}Z^{\varepsilon}_{s}dB_{s}$

and

$Y_{t}=\xi+\int_{t}^{T}\bar{f}(s,Y_{s},\bar{\alpha}_{s})ds-\int_{t}^{T}Z_{s}d\bar{B}_{s}.$

Here $\bar{\alpha}$ and $\bar{f}(s,Y_{s},\bar{\alpha}_{s})$ are respectively
the limit aggregated Markov chain and the averaged generator with respect to
the quasi stationary distributions of the singularly perturbed Markov chain.
Compared to the original BSDE with the singularly perturbed Markov chain, the
limit BSDE depends on a Markov chain with a much smaller state space. Thus the
complexity is reduced.

It is well known that BSDEs provide a probabilistic representation for the
solution of a large class of quasi-linear second order partial differential
equations (PDEs in short) ([3, 18, 19, 20, 13]). Thus BSDEs provide a
probabilistic tool to study the homogenization of PDEs, which is the process
of replacing rapidly varying coefficients by new ones thus the solutions are
close. In this paper, as an application of our theoretical result, after
showing the relation between our BSDE and one system of semi-linear PDE, we
will show the homogenization result of one system of semi-linear PDE with a
singularly perturbed Markov chain.

This paper is organized as following. In section 2, we study the solvability
of BSDE with a Markov chain. Section 3 is devoted to the case that Markov
chain has a large space. Under Jakubowski S-topology ([9] ) which is weaker
than Skorohod’s topology, we present the asymptotic property of BSDE with a
singularly perturbed Markov chain. In section 4, we show the application of
our theoretical results in the homogenization of one system of semi-linear PDE
with a singularly perturbed Markov chain. For the terseness of the main text
of our paper, we put part of technical proofs for some results in Appendix.

##  2 BSDEs with Markov Chains

Let $(\Omega,{\cal{F}},P)$ be a probability space and $T>0$ be fixed.
$\\{{\cal{H}}_{t},0\leq t\leq T\\}$ is a filtration on the space satisfying
the usual condition. $B=\\{B_{t};0\leq t\leq T\\}$ with $B_{0}=0$ is a
$d$-dimensional ${\cal{H}}_{t}$-Brownian motion, and
$\alpha=\\{\alpha_{t};0\leq t\leq T\\}$ is a continuous-time Markov chain
independent of $B$ with the state space ${\cal{M}}=\\{1,2,\ldots,m\\}$.
Suppose the generator of the Markov chain $Q=(q_{ij})_{m\times m}$ is given by

$\begin{split}P\\{\alpha(t+\triangle)=j|\alpha(t)=i\\}=\left\\{\begin{array}[]{l
l}q_{ij}\triangle+o(\triangle),\ \textrm{ \quad\ \ if }i\neq j\\\
1+q_{ij}\triangle+o(\triangle),\ \textrm{ if
}i=j\end{array}\right.\end{split}$

where $\triangle>0$. Here $q_{ij}\geq 0$ is the transition rate from $i$ to
$j$ if $i\neq j$, while $q_{ii}=-\sum_{j=1,i\neq j}^{m}q_{ij}$.

Throughout this paper, we introduce the following notations: $|\cdot|$ is the
norm in the corresponding space; $A^{\prime}$ is the transpose of matrix $A$;
$L^{p}({\cal{H}}_{t};R^{n})$ is the space of $R^{n}$-valued
${\cal{H}}_{t}$-adapted random variable $\xi$ satisfying
$E(|\xi|^{p})<\infty$; $M_{\mathcal{H}_{t}}^{2}(0,T;$ $R^{n})$ denotes the
space of $R^{n}$-valued $\mathcal{H}_{t}$-adapted stochastic processes
$\varphi=\\{\varphi_{t};t\in[0,T]\\}$ satisfying
$E\int_{0}^{T}|\varphi_{t}|^{2}dt<\infty$;
$S_{\mathcal{H}_{t}}^{2}(0,T;R^{n})$ is the space of $R^{n}$-valued
${\cal{H}}_{t}$-adapted continuous stochastic processes
$\varphi=\\{\varphi_{t};t\in[0,T]\\}$ satisfying $E(\sup_{0\leq t\leq
T}|\varphi_{t}|^{2})$ $<\infty$.

###  2.1 Motivation

To study the stochastic optimal control problem with a Markov chain, we
introduce an adjoint equation, then the state equation and adjoint equation
form a kind of FBSDEs with a Markov chain. Here the adjoint equation will be a
BSDE with Markov chain. We give the following linear quadratic (LQ in short)
optimal control problem as an example.

Consider the following stochastic LQ control problem with Markov jumps

min. $\displaystyle
J(v)=\frac{1}{2}E\left(\int_{0}^{T}\left((x^{v}_{t})^{\prime}R(t,\alpha_{t})x^{v}_{t}+v^{\prime}_{t}N(t,\alpha_{t})v_{t}\right)dt+(x^{v}_{T})^{\prime}Q(\alpha_{T})x^{v}_{T}\right)$
(1a) s. t. $\displaystyle\left\\{\begin{array}[]{l
l}dx^{v}_{t}=\left(A(t,\alpha_{t})x^{v}_{t}+B(t,\alpha_{t})v_{t}\right)dt+\left(C(t,\alpha_{t})x^{v}_{t}+D(t,\alpha_{t})v_{t}\right)dB_{t}\\\
x^{v}_{0}=a\in R^{n}\end{array}\right.$ (1d)

where
$A(t,\alpha_{t})=A_{i}(t),B(t,\alpha_{t})=B_{i}(t),C(t,\alpha_{t})=C_{i}(t),D(t,\alpha_{t})=D_{i}(t),$
$R(t,\alpha_{t})=R_{i}(t),N(t,\alpha_{t})=N_{i}(t)$ when $\alpha_{t}=i\
(i=1,\cdots,m)$, and they are uniformly bounded $\mathcal{F}_{t}^{B}$-adapted
processes with appropriate dimensions. $Q(\alpha_{T})=Q_{i}$ when
$\alpha_{T}=i$ $(i=1,\cdots,m)$, and it is nonnegative symmetric matrices-
valued $\mathcal{F}_{T}^{B}$-measurable random variable. Besides, $R_{i}(t)$
is nonnegative symmetric matrices-valued, $N_{i}(t)$ is positive symmetric
matrices-valued and the inverse $N_{i}(t)^{-1}$ is bounded. The set of all
$\mathcal{H}_{t}$-adapted admissible controls is $\mathcal{U}_{ad}\equiv
M^{2}_{\mathcal{H}_{t}}(0,T;R^{n_{u}\times d})$, and our aim is to find an
admissible control $u$ such that $\displaystyle
J(u)=\inf_{v\in\mathcal{U}_{ad}}J(v)$.

There are many literatures on this kind of LQ optimal control problem with
Markov jumps (1) and its application, such as [26, 12, 24] and their
references. Different to their methods that constructing the optimal control
via the solution of Riccati equation, we will use the FBSDEs approach.

######  Theorem 2.1.

If the following FBSDE admits a unique solution $(x_{t},y_{t},z_{t})$

$\begin{split}\left\\{\begin{array}[]{l l
l}~{}~{}dx_{t}=\left(A(t,\alpha_{t})x_{t}+B(t,\alpha_{t})\left(-N^{-1}(t,\alpha_{t})\left(B^{\prime}(t,\alpha_{t})y_{t}+D^{\prime}(t,\alpha_{t})z_{t}\right)\right)\right)dt\\\
~{}~{}~{}~{}+\left(C(t,\alpha_{t})x_{t}+D(t,\alpha_{t})\left(-N^{-1}(t,\alpha_{t})\left(B^{\prime}(t,\alpha_{t})y_{t}+D^{\prime}(t,\alpha_{t})z_{t}\right)\right)\right)dB_{t},\\\
-dy_{t}=\left(A^{\prime}(t,\alpha_{t})y_{t}+C^{\prime}(t,\alpha_{t})z_{t}+R(t,\alpha_{t})x_{t}\right)dt-
z_{t}dB_{t},\\\ ~{}~{}~{}x_{0}=a,\ \
y_{T}=Q(\alpha_{T})x_{T}.\end{array}\right.\end{split}$ (2)

Then

$u_{t}=-N^{-1}(t,\alpha_{t})\left(B^{\prime}(t,\alpha_{t})y_{t}+D^{\prime}(t,\alpha_{t})z_{t}\right),0\leq
t\leq T$

is the unique optimal control for the LQ problem (1)

###### Proof.

Firstly, we will prove that $\\{u=u_{t};0\leq t\leq T\\}$ is an optimal
control for the LQ problem (1).

From the forward equation of (2), we can see that $x$ is the corresponding
system state trajectory of $u$. For an arbitrary admissible control $v$,
denote $x^{v}$ as the corresponding system state trajectory, then

$\displaystyle J(v)-J(u)$ $\displaystyle=$ $\displaystyle\
\frac{1}{2}E\Big{(}\int_{0}^{T}\big{(}(x_{t}^{v}-x_{t})^{\prime}R(t,\alpha_{t})(x_{t}^{v}-x_{t})+(v_{t}-u_{t})^{\prime}N(t,\alpha_{t})(v_{t}-u_{t})$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
+2x_{t}^{\prime}R(t,\alpha_{t})(x_{t}^{v}-x_{t})+2u_{t}^{\prime}N(t,\alpha_{t})(v_{t}-u_{t})\big{)}dt$
$\displaystyle\ \ \ \ \ \ \
+(x_{T}^{v}-x_{T})^{\prime}Q(\alpha_{T})(x_{T}^{v}-x_{T})+2x^{\prime}_{T}Q(\alpha_{T})(x_{T}^{v}-x_{T})\Big{)}.$

Applying Itô’s formula to $y^{\prime}_{t}(x_{t}^{v}-x_{t})$, we have

$\displaystyle Ex^{\prime}_{T}Q(\alpha_{T})(x_{T}^{v}-x_{T})$ $\displaystyle=$
$\displaystyle
E\int_{0}^{T}\big{(}\left(y^{\prime}_{t}B(t,\alpha_{t})+z^{\prime}_{t}D(t,\alpha_{t})\right)(v_{t}+N^{-1}(t,\alpha_{t})\left(B^{\prime}(t,\alpha_{t})y_{t}+D^{\prime}(t,\alpha_{t})z_{t}\right))$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-x_{t}^{\prime}R(t,\alpha_{t})(x_{t}^{v}-x_{t})\big{)}dt$
$\displaystyle=$ $\displaystyle
E\int_{0}^{T}\left(\left(y^{\prime}_{t}B(t,\alpha_{t})+z^{\prime}_{t}D(t,\alpha_{t})\right)(v_{t}-u_{t})-x_{t}^{\prime}R(t,\alpha_{t})(x_{t}^{v}-x_{t})\right)dt.$

As $R$, $Q$ are nonnegative and $N$ is positive, we have

$\displaystyle J(v)-J(u)\geq$ $\displaystyle
E\int_{0}^{T}\left(y^{\prime}_{t}B(t,\alpha_{t})+z^{\prime}_{t}D(t,\alpha_{t})+u^{\prime}_{t}N(t,\alpha_{t})\right)(v_{t}-u_{t})dt$
$\displaystyle=$ $\displaystyle 0.$

So
$u(t)=-N^{-1}(t,\alpha_{t})\left(B^{\prime}(t,\alpha_{t})y_{t}+D^{\prime}(t,\alpha_{t})z_{t}\right)$
is an optimal control.

Uniqueness: Assume that $u^{1}$ and $u^{2}$ are both optimal controls with
$J(u^{1})=J(u^{2})=\gamma\geq 0$, and the corresponding trajectories are
$x^{1}$ and $x^{2}$. Due to the linear property of the system, the
trajectories corresponding to $\displaystyle\frac{u^{1}+u^{2}}{2}$ is
$\displaystyle\frac{x^{1}+x^{2}}{2}$.

From the classical parallelogram rule, $R$, $Q$ are nonnegative, and $N$ is
positive, there exists $\delta>0$ such that

$\displaystyle 2\gamma$ $\displaystyle=$ $\displaystyle\ J(u^{1})+J(u^{2})$
$\displaystyle=$ $\displaystyle\
2J\left(\frac{u^{1}+u^{2}}{2}\right)+E\bigg{(}\left(\frac{x_{T}^{1}-x_{T}^{2}}{2}\right)^{\prime}Q(\alpha_{T})\left(\frac{x_{T}^{1}-x_{T}^{2}}{2}\right)+\int_{0}^{T}\bigg{(}\left(\frac{x_{t}^{1}-x_{t}^{2}}{2}\right)^{\prime}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \
R(t,\alpha_{t})\left(\frac{x_{t}^{1}-x_{t}^{2}}{2}\right)+\left(\frac{u_{t}^{1}-u_{t}^{2}}{2}\right)^{\prime}N(t,\alpha_{t})\left(\frac{u_{t}^{1}-u_{t}^{2}}{2}\right)\bigg{)}dt\bigg{)}$
$\displaystyle\geq$ $\displaystyle\
2J\left(\frac{u^{1}+u^{2}}{2}\right)+E\int_{0}^{T}\left(\frac{u_{t}^{1}-u_{t}^{2}}{2}\right)^{\prime}N(t,\alpha_{t})\left(\frac{u_{t}^{1}-u_{t}^{2}}{2}\right)dt$
$\displaystyle\geq$ $\displaystyle\
2\gamma+\frac{\delta}{4}E\int_{0}^{T}|u_{t}^{1}-u_{t}^{2}|^{2}dt$

Thus $E\int_{0}^{T}|u_{t}^{1}-u_{t}^{2}|^{2}dt\leq 0$ which yields that
$u^{1}=u^{2}$. ∎

###  2.2 BSDEs with Markov chains

The derivation of the optimal control for the above LQ problem (1) can be
regarded as one motivation for us to study BSDEs with Markov jumps. In this
subsection, we will study the solvability to the following BSDE with a Markov
chain firstly:

$Y_{t}=\xi+\int_{t}^{T}f(s,Y_{s},Z_{s},\alpha_{s})ds-\int_{t}^{T}Z_{s}dB_{s}.$
(3)

Let ${\cal{N}}$ denote the class of all $P$-null sets of ${\cal{F}}$. For each
$t\in[0,T]$, we define
${\cal{F}}_{t}={\cal{F}}_{t}^{B}\vee{\cal{F}}_{t,T}^{\alpha}\vee{\cal{N}}$
where for any process $\\{\eta_{t};0\leq t\leq T\\}$,
${\cal{F}}_{t,T}^{\eta}=\sigma\\{\eta_{r};t\leq r\leq T\\}$ and
${\cal{F}}_{t}^{\eta}={\cal{F}}_{0,t}^{\eta}$. For convenience, we denote
$M^{2}(0,T;R^{n})=M_{\mathcal{F}_{t}}^{2}(0,T;R^{n})$ and $S^{2}(0,T;$
$R^{n})=$ $S_{\mathcal{F}_{t}}^{2}(0,$ $T;R^{n})$.

Thereinafter, we make the following assumption:

######  Assumption 2.1.

(i) $\xi\in L^{2}({\cal{F}}_{T};R^{k})$; (ii) $f:\Omega\times[0,T]\times
R^{k}\times R^{k\times d}\times\mathcal{M}\rightarrow R^{k}$ satisfies that
$\forall(y,z)\in R^{k}\times R^{k\times d}$, $\forall i\in\mathcal{M}$,
$f(\cdot,y,z,i)\in M^{2}_{\mathcal{F}_{t}^{B}}(0,T;R^{k})$, and
$\exists\mu>0$, such that $\forall i\in\mathcal{M}$,
$\forall(\omega,t)\in\Omega\times[0,T]$, $(y_{1},z_{1})$, $(y_{2},z_{2})\in
R^{k}\times R^{k\times d}$,

$|f(t,y_{1},z_{1},i)-f(t,y_{2},z_{2},i)|\leq\mu(|y_{1}-y_{2}|+|z_{1}-z_{2}|).$

Our main result in this section is in the following theorem.

######  Theorem 2.2.

Under Assumption 2.1, there exists a unique solution pair $(Y,Z)\in
S^{2}(0,T;R^{k})\times M^{2}(0,T;R^{k\times d})$ for BSDE $(\ref{eq:BSDE})$.

The proof of Theorem 2.2 consists of three steps.

Step 1: Extension of Itô’s representation theorem

It is noted that $\\{{\cal{F}}_{t};0\leq t\leq T\\}$ is neither increasing nor
decreasing, and it does not constitute a filtration. Inspired by the method
handling the BSDE with doubly Brown motions ([19]), we define a filtration
$({\mathcal{G}}_{t})_{0\leq t\leq T}$ by

${\cal{G}}_{t}\triangleq{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\vee\cal{N}$

For the filtration $({\mathcal{G}}_{t})_{0\leq t\leq T}$, we give the
following extension of Itô’s representation theorem. This result and its
corollary play key roles during the proof of Theorem 2.2.

######  Proposition 2.1.

For $N\in L^{2}(\mathcal{G}_{T};R^{k})$, there exist a unique random variable
$N_{0}\in L^{2}(\mathcal{F}_{T}^{\alpha};R^{k})$ and a unique stochastic
process $Z=\\{Z_{t};0\leq t\leq T\\}\in M^{2}_{\mathcal{G}_{t}}(0,T;R^{k\times
d})$ such that

$N=N_{0}+\int_{0}^{T}Z_{t}dB_{t},\quad 0\leq t\leq T.$ (4)

Actually, $N_{0}=E(N|\mathcal{F}_{T}^{\alpha})$.

During the derivation of Proposition 2.1, we need the following two lemmas.

######  Lemma 2.1.

([22]) If $X$, $Y$: $\Omega\rightarrow R^{d}$ are two given functions, $Y$ is
$\sigma(X)$-measurable if and only if there exists a Borel measurable function
$g$: $R^{d}\rightarrow R^{d}$ such that $Y=g(X)$.

######  Lemma 2.2.

(Doob’s martingale convergence theorem) Let $\\{\mathcal{F}_{t};t\geq 0\\}$ be
a filtration on the space $(\Omega,\mathcal{F},P)$, $X\in
L^{1}(\mathcal{F};R^{d})$, then

$E(X|\mathcal{F}_{t})\rightarrow E(X|\mathcal{F}_{\infty}),\quad\mbox{as
}t\rightarrow\infty,\textrm{ a.s. and in $L^{1}$ as well.}$

###### Proof.

Existence: Let $\\{t_{i}\\}_{i\geq 0}$, $\\{t^{\prime}_{j}\\}_{j\geq 0}$ be
two dense subsets of $[0,T]$ where $t_{0}=t^{\prime}_{0}=0$. For each integer
$n,m\geq 0$, let $\mathcal{G}_{n,m}$ be the $\sigma$-algebra generated by
$\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}},B_{t^{\prime}_{0}},B_{t^{\prime}_{1}},$
$\cdots,B_{t^{\prime}_{m}}$, i.e.,
$\mathcal{G}_{n,m}=\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}},B_{t^{\prime}_{0}},B_{t^{\prime}_{1}},$
$\cdots,B_{t^{\prime}_{m}}\\}$. Obviously,
$\mathcal{G}_{n,m}\subset\mathcal{G}_{n+1,m}$,
$\mathcal{G}_{n,m}\subset\mathcal{G}_{n,m+1}$,
$\mathcal{G}_{n,m}\subset\mathcal{G}_{n+1,m+1}$,
$\displaystyle\sigma(\cup_{m=1}^{\infty}\mathcal{G}_{n,m})=\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}}\\},$
and $\displaystyle\sigma(\cup_{n,m=0}^{\infty}$ $\mathcal{G}_{n,m})$
$=\mathcal{G}_{T}$.

For $N\in L^{2}(\mathcal{G}_{T};R^{k})$, $\forall n,m$, by Lemma 2.1 and Lemma
2.2, there exists a Borel measurable function $N_{n,m}:\mathcal{M}^{n+1}\times
R^{(m+1)\times d}\rightarrow R^{k}$ such that

$\begin{split}&E(N|\mathcal{G}_{n,m})=N_{n,m}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}},B_{t^{\prime}_{0}},B_{t^{\prime}_{1}},\cdots,B_{t^{\prime}_{m}})\\\
&N_{n,m}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}},B_{t^{\prime}_{0}},B_{t^{\prime}_{1}},\cdots,B_{t^{\prime}_{m}})\rightarrow
E\left(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}}\\}\right)\end{split}$

Denote $N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})\triangleq
E\left(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}}\\}\right)$,
it can be rewritten as

$N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}N_{n}(i_{0},i_{1},\cdots,i_{n})$

where $N_{n}(i_{0},i_{1},\cdots,i_{n})$ is $\mathcal{F}_{T}^{B}$-measurable.

For $(i_{0},i_{1},\cdots,i_{n})\in\mathcal{M}^{n+1}$, applying Itô’s
representation theorem,

$N_{n}(i_{0},i_{1},\cdots,i_{n})=N_{0}{(i_{0},i_{1},\cdots,i_{n})}+\int_{0}^{T}Z_{t}{(i_{0},i_{1},\cdots,i_{n})}dB_{t}$

where $N_{0}(i_{0},i_{1},$ $\cdots,i_{n})$ is a constant and
$Z{(i_{0},i_{1},\cdots,i_{n})}\in M^{2}_{\mathcal{F}_{t}^{B}}(0,T;R^{k\times
d})$.

Denote
$\displaystyle
N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}N_{0}{(i_{0},i_{1},\cdots,i_{n})}$,
and $\displaystyle
Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}Z_{t}{(i_{0},i_{1},\cdots,i_{n})}$.
Clearly, $N_{0}(\alpha_{t_{0}},$ $\alpha_{t_{1}},\cdots,\alpha_{t_{n}})$ $\in
L^{2}(\mathcal{F}_{T}^{\alpha};R^{k})$, $\\{Z_{t}(\alpha_{t_{0}},$
$\alpha_{t_{1}},\cdots,\alpha_{t_{n}});0\leq t\leq T\\}\in
M^{2}_{\mathcal{G}_{t}}(0,T;R^{k\times d})$, and we have

$\displaystyle
N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})+\int_{0}^{T}Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})dB_{t}.$
(5)

In the remained part, we will prove that as $n\rightarrow\infty$, both side of
(5) are Cauchy sequences.

For the left hand side, as $n\rightarrow\infty$, with the definition of
$N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})$ and Lemma 2.2, we
have

$N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})\rightarrow N.$

For the right hand side, since

$\displaystyle
E\left(\int_{0}^{T}Z(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})dB_{t}|\mathcal{F}_{T}^{\alpha}\right)$
(6) $\displaystyle=$
$\displaystyle\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}E\left(\int_{0}^{T}Z{(i_{0},i_{1},\cdots,i_{n})}dB_{t}|\mathcal{F}_{T}^{\alpha}\right)$
$\displaystyle=$ $\displaystyle 0.$

Thus
$N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})=E(N_{n}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})|\mathcal{F}_{T}^{\alpha})$.
As $n\rightarrow\infty$, we can conclude that

$N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})\rightarrow\
E(N|\mathcal{F}_{T}^{\alpha}).$

Now let us consider the sequence
$\\{Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}});0\leq t\leq
T\\}$. For $n,m\geq 0$,

$\begin{split}&E\int_{0}^{T}\left|Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})-Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}})\right|^{2}dt\\\
=\
&E\left(\int_{0}^{T}\left(Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})-Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}})\right)dB_{t}\right)^{2}\\\
=\
&E\big{(}E\left(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}}\\}\right)-E\left(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}}\\}\right)\\\
&-N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})+N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}})\big{)}^{2}\\\
\leq\
&2E\big{(}E(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}}\\})^{2}-E(N|\mathcal{F}_{T}^{B}\vee\sigma\\{\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}}\\})\big{)}^{2}\\\
&+2E(N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}})-N_{0}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{m}})\big{)}^{2}\\\
\rightarrow\ &0,\quad\textrm{ as }n,m\rightarrow\infty.\end{split}$

Thus $\\{Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}});0\leq
t\leq T\\}$ is a Cauchy sequence in $M^{2}_{\mathcal{G}_{t}}(0,T;$ $R^{k\times
d})$. Hence it converges to some $Z\in M^{2}_{\mathcal{G}_{t}}(0,T;R^{k\times
d})$.

Denote $N_{0}=E(N|\mathcal{F}_{T}^{\alpha})$, we can conclude that $(\ref{eq1:
formula 1 in extension of Ito})$ converges to the extended Itô’s
representation (4).

Uniqueness: By virtue of equation (6) and the fact that as
$n\rightarrow\infty$,
$\\{Z_{t}(\alpha_{t_{0}},\alpha_{t_{1}},\cdots,\alpha_{t_{n}});$ $0\leq t\leq
T\\}$ is a Cauchy sequence, we have
$E(\int_{0}^{T}Z_{t}dB_{t}|\mathcal{F}_{T}^{\alpha})$ $=0$. Then, for
$(N_{0},Z),(N_{0}^{\prime},Z^{\prime})$ satisfying the extended Itô’s
representation $(\ref{eq: extension of Ito representation})$, we get
$N_{0}=N^{\prime}_{0}$ by taking conditional expectation with respect to
$\mathcal{F}_{T}^{\alpha}$. Uniqueness of $Z$ follows easily from the fact
that

$E\int_{0}^{T}|Z_{t}-Z^{\prime}_{t}|^{2}dt=E\left(\int_{0}^{T}(Z_{t}-Z^{\prime}_{t})dB_{t}\right)^{2}=E\left(N_{0}-N_{0}^{\prime}\right)^{2}=0.$

∎

The following corollary is useful in the proof of Theorem 2.2 and its proof is
similar to Proposition 2.1.

######  Corollary 2.1.

For $t\leq T$, we consider the filtration $(\mathcal{N}_{s})_{t\leq s\leq T}$
defined by
$\mathcal{N}_{s}=\mathcal{F}_{s}^{B}\vee\mathcal{F}_{t,T}^{\alpha}$. For $N\in
L^{2}(\mathcal{N}_{T};R^{k})$, there exists a unique stochastic process
$Z=\\{Z_{s};t\leq s\leq T\\}\in M^{2}_{\mathcal{N}_{s}}(t,T;R^{k\times d})$
such that

$N=E(N|\mathcal{N}_{t})+\int_{t}^{T}Z_{s}dB_{s}.$

Step 2: The special case: the generator $f$ is independent of $y$ and $z$.

######  Proposition 2.2.

Under Assumption 2.1, the following BSDE

$Y_{t}=\xi+\int_{t}^{T}f(s,\alpha_{s})ds-\int_{t}^{T}Z_{s}dB_{s},\qquad 0\leq
t\leq T$ (7)

has a solution pair $(Y,Z)\in S^{2}(0,T;R^{k})\times M^{2}(0,T;R^{k\times
d})$.

###### Proof.

From Assumption 2.1 and Hölder inequality, we obtain

$E\left(\int_{0}^{T}f(s,\alpha_{s})ds\right)^{2}\leq
CE\int_{0}^{T}|f(s,\alpha_{s})|^{2}ds\leq
C\sum_{i=1}^{m}E\int_{0}^{T}|f(s,i)|^{2}ds<\infty$

which yields

$\xi+\int_{0}^{T}f(s,\alpha_{s})ds\in L^{2}({{\cal{G}}_{T}};R^{k})$

For the filtration $({\mathcal{G}}_{t})_{0\leq t\leq T}$ where
${\cal{G}}_{t}={\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\vee{\cal{N}}={\cal{F}}_{t}^{B}\vee{\cal{F}}_{t,T}^{\alpha}\vee{\cal{F}}_{t}^{\alpha}\vee{\cal{N}}={\cal{F}}_{t}\vee{\cal{F}}_{t}^{\alpha},$
we can define the following ${\cal{G}}_{t}$-measurable square integrable
martingale

$N_{t}=E\left(\xi+\int_{0}^{T}f(s,\alpha_{s})ds|{{\cal{G}}_{t}}\right),\qquad
0\leq t\leq T.$

By the extended Itô’s representation theorem (Proposition 2.1), there exist
$N_{0}\in L^{2}(\mathcal{F}_{T}^{\alpha};R^{k})$ and $Z=\\{Z_{t};0\leq t\leq
T\\}\in M^{2}_{{\cal{G}}_{t}}(0,T;R^{k\times d})$ such that

$N_{t}=N_{0}+\int_{0}^{t}Z_{s}dB_{s},\qquad 0\leq t\leq T.$

For $t\in[0,T]$, we define

$\begin{split}Y_{t}=N_{t}-\int_{0}^{t}f(s,\alpha_{s})ds,\textrm{ \ i.e.,
}Y_{t}=E\left(\xi+\int_{t}^{T}f(s,\alpha_{s})ds|{{\cal{G}}_{t}}\right).\end{split}$
(8)

It is easy to verify that the ${\cal{G}}_{t}$-measurable process $(Y,Z)$
satisfies BSDE $(\ref{eq:BSDE f independent of y and z})$ and $Y\in
M^{2}_{{\cal{G}}_{t}}(0,T;$ $R^{k})$. We refer interested reader to [17, 16]
for the detailed verification.

The left work is to show that the processes $Y=\\{Y_{t};0\leq t\leq T\\}$ and
$Z=\\{Z_{t};0\leq t\leq T\\}$ are ${\cal{F}}_{t}$-measurable, i.e.
${\cal{F}}_{t}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable. $\forall
t\in[0,T]$, we denote $\vartheta=\xi+\int_{t}^{T}f(s,\alpha_{s})ds$,
$\vartheta$ is ${\cal{F}}_{T}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable.

Let $\\{\bar{t}_{i}\\}_{i\geq 0}$, $\\{\bar{t}^{\prime}_{j}\\}_{j\geq 0}$ be
respectively dense subsets of $[t,T]$ and $[0,T]$, with $\bar{t}_{0}=t$ and
$\bar{t}^{\prime}_{0}=0$. For each integer $n,m\geq 0$, let
$\bar{\mathcal{G}}_{n,m}$ be the $\sigma$-algebra generated by
$\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}}$,
i.e.,
$\bar{\mathcal{G}}_{n,m}=\sigma\\{\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}},$
$B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}}\\}$.
Obviously, $\bar{\mathcal{G}}_{n,m}\subset\bar{\mathcal{G}}_{n+1,m+1}$,
$\bar{\mathcal{G}}_{n,m}\subset\bar{\mathcal{G}}_{n+1,m}$,
$\bar{\mathcal{G}}_{n,m}\subset\bar{\mathcal{G}}_{n,m+1}$, and
$\sigma(\cup_{n,m=0}^{\infty}\bar{\mathcal{G}}_{n,m})={\cal{F}}_{T}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$.

From Lemma 2.1, for each $n,m$, there exists a Borel measurable function
$\vartheta_{nm}:\mathcal{M}^{n+1}\times R^{(m+1)\times d}\rightarrow R^{k}$
such that

$\begin{split}E[\vartheta|\bar{\mathcal{G}}_{n,m}]&=\vartheta_{nm}(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}}).\end{split}$

Since
$I_{\\{(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}\in{\cal{F}}_{t,T}^{\alpha}\subset{\cal{F}}_{T}^{\alpha}$,
we have

$\begin{split}&E\left(\vartheta_{nm}(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\right)\\\
=&\
E\Bigg{(}\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}\\\
&\hskip
28.45274pt\vartheta_{nm}(i_{0},i_{1},\cdots,i_{n},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\Bigg{)}\\\
=&\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}\\\
&\hskip
28.45274ptE(\vartheta_{nm}(i_{0},i_{1},\cdots,i_{n},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}).\end{split}$

For the reason that
$\vartheta_{nm}(i_{0},i_{1},\cdots,i_{n},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})$
is $\mathcal{F}_{T}^{B}$-measurable, by Itô’s representation theorem, we know
that there exist $\nu_{0}^{n,m}(i_{0},i_{1},\cdots,i_{n})$ $\in R^{k}$ and
$Z^{n,m}(i_{0},i_{1},\cdots,i_{n})\in$ $L^{2}_{\mathcal{F}_{t}^{B}}(0,T;$
$R^{k\times d})$ such that

$\begin{split}&E\left(\vartheta_{nm}(i_{0},i_{1},\cdots,i_{n},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\right)\\\
=&E\left(\nu_{0}^{n,m}(i_{0},i_{1},\cdots,i_{n})+\int_{0}^{T}Z_{r}^{n,m}(i_{0},i_{1},\cdots,i_{n})dB_{r}|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\right)\\\
=&\nu_{0}^{n,m}(i_{0},i_{1},\cdots,i_{n})+\int_{0}^{t}Z_{r}^{n,m}(i_{0},i_{1},\cdots,i_{n})dB_{r}\end{split}$

is ${\cal{F}}_{t}^{B}$-measurable. Therefore

$\begin{split}&E\left(\vartheta_{nm}(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\right)\\\
=&\sum_{i_{0},i_{1},\cdots,i_{n}=1}^{m}I_{\\{(\alpha_{\bar{t}_{0}},\alpha_{\bar{t}_{1}},\cdots,\alpha_{\bar{t}_{n}})=(i_{0},i_{1},\cdots,i_{n})\\}}\\\
&\hskip
28.45274ptE\left(\vartheta_{nm}(i_{0},i_{1},\cdots,i_{n},B_{\bar{t}^{\prime}_{0}},B_{\bar{t}^{\prime}_{1}},\cdots,B_{\bar{t}^{\prime}_{m}})|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}\right)\end{split}$

is ${\cal{F}}_{t}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable. With Lemma 2.2,
as $n,m\rightarrow\infty$,

$E[\vartheta|\bar{\mathcal{G}}_{n,m}]\rightarrow
E[\vartheta|{\cal{F}}_{T}^{B}\vee{\cal{F}}_{t,T}^{\alpha}]=\vartheta.$

Thus $E[\vartheta|{\cal{F}}_{t}^{B}\vee{\cal{F}}_{T}^{\alpha}]$, i.e.,
$Y_{t}$, is also ${\cal{F}}_{t}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable.

Considering

$\int_{t}^{T}Z_{s}dB_{s}=-Y_{t}+\xi+\int_{t}^{T}f(s,\alpha_{s})ds,$

its right side is ${\cal{F}}_{T}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable.
With Corollary 2.1, we know $\forall t<s$, $Z_{s}$ is
${\cal{F}}_{s}^{B}\vee{\cal{F}}_{t,T}^{\alpha}$-measurable. Then, by the
continuous property of the Markov chain $\alpha$, we obtain that $Z_{s}$ is
${\cal{F}}_{s}^{B}\vee{\cal{F}}_{s,T}^{\alpha}$-measurable.

Together with the Burkholder-Davis-Gundy inequality and the form of BSDE
$(\ref{eq:BSDE f independent of y and z})$, we can conclude that
$\\{Y_{t};0\leq t\leq T\\}$ is continuous and satisfies $\displaystyle
E(\sup_{0\leq t\leq T}|Y_{t}|^{2})<\infty$. It yields that $Y\in
S^{2}(0,T;R^{k})$. ∎

Step 3: The general case: Proof of Theorem 2.2.

###### Proof.

Firstly, we define a mapping $I$ from $M^{2}(0,T;R^{k}\times R^{k\times d})$
into itself such that $(Y,Z)\in S^{2}(0,T;R^{k})\times M^{2}(0,T;R^{k\times
d})$ is the solution to BSDE $(\ref{eq:BSDE})$ iff it is a fixed point of $I$.

For a constant $\beta>0$, we introduce the following equivalent norm of
$M^{2}(0,T;R^{k}\times R^{k\times d})$

$\|v(\cdot)\|_{\beta}=\left(E\int_{0}^{T}|v_{s}|^{2}e^{\beta
s}ds\right)^{\frac{1}{2}}.$

For $(y,z)\in M^{2}(0,T;R^{k}\times R^{k\times d})$, we set

$Y_{t}=\xi+\int_{t}^{T}f(s,y_{s},z_{s},\alpha_{s})ds-\int_{t}^{T}Z_{s}dB_{s}.$

From Assumption 2.1 and H$\ddot{\textrm{o}}$lder’s inequality,

$\begin{split}&E\left(\int_{0}^{T}f(s,y_{s},z_{s},\alpha_{s})ds\right)^{2}\\\
\leq&\
2E\left(\int_{0}^{T}(f(s,y_{s},z_{s},\alpha_{s})-f(s,0,0,\alpha_{s}))ds\right)^{2}+2E\left(\int_{0}^{T}f(s,0,0,\alpha_{s})ds\right)^{2}\\\
\leq&\
C\left(E\int_{0}^{T}\left(|y_{s}|^{2}+|z_{s}|^{2}\right)ds+\sum_{i=1}^{m}E\int_{0}^{T}|f(s,0,0,i)|^{2}ds\right)<\infty\end{split}$

which yields that

$\xi+\int_{0}^{T}f(s,y_{s},z_{s},\alpha_{s})ds\in
L^{2}({{\cal{G}}_{T}};R^{k}).$

From Proposition 2.2, we can define the following contraction mapping under
the norm $\|\cdot\|_{\beta}$

$I((y,z))=(Y,Z):M^{2}(0,T;R^{k}\times R^{k\times d})\rightarrow
M^{2}(0,T;R^{k}\times R^{k\times d}).$

The proof of contraction property is similar to [17, 10, 16]. For the
compactness of the paper, the detail is omit here.

Together with the form of BSDE $(\ref{eq:BSDE})$ and Burkholder-Davis-Gundy
inequality, $Y\in S^{2}(0,T;R^{k})$. Thus, by the fixed point theorem, we know
that BSDE $(\ref{eq:BSDE})$ has a unique solution pair. ∎

##  3 BSDEs with Singularly Perturbed Markov Chains

In this section, after recalling several relevant results of singularly
perturbed Markov chains given by Zhang and Yin ([23]), we will consider the
asymptotic property of BSDE with a singularly perturbed Markov chain.
Following the averaging approach to aggregate the states according to their
jump rates and replace the actual coefficient with its average with respect to
the quasi stationary distributions of the singularly perturbed Markov chain,
we get the asymptotic probability distribution of the solution to the BSDE
with an limit averaged Markov chain which has a much smaller state space than
the original one.

###  3.1 Relevant results of singularly perturbed Markov chains

Focused on a continuous-time $\varepsilon$-dependent singularly perturbed
Markov chain $\alpha^{\varepsilon}=\\{\alpha^{\varepsilon}_{t};0\leq t\leq
T\\}$ which have the generator $\displaystyle
Q^{\varepsilon}=\frac{1}{\varepsilon}\tilde{Q}+\hat{Q},$ where $\tilde{Q}$ and
$\hat{Q}$ are time-invariant generators, with
$\tilde{Q}=\textrm{diag}(\tilde{Q}^{1},\cdots,\tilde{Q}^{l})$. The state space
can be decomposed as
${\cal{M}}=\\{1,2,\cdots,m\\}={\cal{M}}_{1}\cup\cdots\cup{\cal{M}}_{l},$
${\cal{M}}_{k}=\\{s_{k1},\cdots,s_{km_{k}}\\}$, and for
$k\in\\{1,\cdots,l\\}$, $\tilde{Q}^{k}$ is the weakly irreducible
generator333A generator $Q$ is called weakly irreducible if the system of
equations $\nu Q=0$ and $\sum_{i=1}^{m}\nu_{i}=1$ has a unique nonnegative
solution. This nonnegative solution $\nu=(\nu_{1},\cdots,\nu_{m})$ is called
the quasi-stationary distribution of $Q$. corresponding to the states in
${\cal{M}}_{k}$. The generator $\tilde{Q}$ dictates the fast motion of the
Markov chain and $\hat{Q}$ governs the slow motion, i.e., the underlying
Markov chain fluctuates rapidly in a single group $\mathcal{M}_{k}$ and jumps
less frequently among groups $\mathcal{M}_{k}$ and $\mathcal{M}_{j}$ for
$k\neq j$.

As shown in [23], when the states in $\mathcal{M}_{k}$ are lumped into a
single state, all such states are coupled by $\hat{Q}$. By defining
$\bar{\alpha}^{\varepsilon}_{t}=k$, when
$\alpha^{\varepsilon}_{t}\in{\cal{M}}_{k}$, we can obtain the aggregated
process $\bar{\alpha}^{\varepsilon}=\\{\bar{\alpha}^{\varepsilon}_{t};0\leq
t\leq T\\}$ containing $l$ states. The process $\bar{\alpha}^{\varepsilon}$ is
not necessarily Markovian, but it converges weakly to a continuous-time Markov
chain $\bar{\alpha}$.

######  Proposition 3.1.

([23]) (i) $\bar{\alpha}^{\varepsilon}$ converges weakly to $\bar{\alpha}$
generated by

$\bar{Q}=\textrm{diag}(\nu^{1},\cdots,\nu^{l})\hat{Q}\textrm{diag}(\mathbb{I}_{m_{1}},\cdots,\mathbb{I}_{m_{l}})$

as $\varepsilon\rightarrow 0$, where $\nu^{k}$ is the quasi-stationary
distribution of $\tilde{Q}^{k}$, $k=1,\cdots,l$, and
$\mathbb{I}_{k}=(1,\cdots,1)^{\prime}\in R^{k}.$

(ii) For any bounded deterministic function $\beta(\cdot)$,

$E\left(\int_{s}^{T}(I_{\\{\alpha^{\varepsilon}_{t}=s_{kj}\\}}-\nu_{j}^{k}I_{\\{\bar{\alpha}^{\varepsilon}_{t}=k\\}})\beta(t)dt\right)^{2}=O(\varepsilon),\forall\
k=1,\cdots,l,\forall\ j=1,\cdots,m_{k}.$

Here $I_{A}$ is the indicator function of a set $A$.

###  3.2 Weak convergence of BSDEs with singularly perturbed Markov chains

In this subsection, denote $D(0,T;R^{k})$ as the Skorohod space of
c$\grave{a}$dl$\grave{a}$g trajectories endowed with the Jakubowski S-topology
([9]) which is weaker than the Skorohod topology. As shown in the appendix of
[2], the tightness criteria under this S-topology is the same as the “Meyer-
Zheng tightness criteria” used in [14].

Here, we only consider the asymptotic property of the solution to the
following BSDE with a singularly perturbed Markov chain where the generator
$f$ does not depend on $Z^{\varepsilon}$,

$Y^{\varepsilon}_{t}=\xi+\int_{t}^{T}f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})ds-\int_{t}^{T}Z^{\varepsilon}_{s}dB_{s},$
(9)

For the difficulty to study the general case that the generator $f$ depends on
$Z^{\varepsilon}$, we refer interested reader to the explanation in section 6
of [16].

Firstly, we make the following assumption:

######  Assumption 3.1.

(i) $\xi\in L^{2}(\mathcal{F}_{T}^{B};R^{k}).$ (ii) For $f:[0,T]\times
R^{k}\times\mathcal{M}\rightarrow R^{k}$, there exists a constant $C>0$ such
that $\displaystyle\sup_{\begin{subarray}{c}0\leq t\leq T\\\ 1\leq i\leq
m\end{subarray}}|f(t,0,i)|\leq C$.

######  Theorem 3.3.

Under Assumption 2.1 and Assumption 3.1, the sequence of process
$(Y^{\varepsilon}_{t},$ $\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s})$ converges in
distribution to the process $(Y_{t},\int_{0}^{t}Z_{s}d\bar{B}_{s})$ as
$\varepsilon\rightarrow 0$, when probability measures on $D(0,T;R^{2k})$
equipped with the Jakubowski S-topology. Here $(Y,Z)$ is the solution pair to
the following BSDE with the limit averaged Markov chain

$Y_{t}=\xi+\int_{t}^{T}\bar{f}(s,Y_{s},\bar{\alpha}_{s})ds-\int_{t}^{T}Z_{s}d\bar{B}_{s},$
(10)

$\bar{B}=\\{\bar{B}_{t};0\leq t\leq T\\}$ with $\bar{B}_{0}=0$ is a
$d$-dimensional Brownian motion, $\bar{\alpha}$ is defined in subsection 3.1,
and $\displaystyle\bar{f}(s,y,i)=\sum_{j=1}^{m_{i}}\nu^{i}_{j}f(t,y,s_{ij})$
for $i\in\bar{{\cal{M}}}=\\{1,\cdots,l\\}$.

######  Remark 3.1.

It is obvious that the limit BSDE depends on the limit averaged Markov chain
$\bar{\alpha}$ with a state space much smaller than that of the original
singularly perturbed Markov chain $\alpha^{\varepsilon}$. Moreover, as
$\varepsilon\rightarrow 0$, the
$\mathcal{F}^{\alpha^{\varepsilon}}_{T}$-measurable random variables sequence
$(Y^{\varepsilon}_{0})$ converges in distribution to the random variable
$Y_{0}$ which is $\mathcal{F}^{\bar{\alpha}}_{T}$-measurable.

For the proof of Theorem 3.3, we follow a classical approach as in [20, 2] to
prove the weak convergence of BSDE: after showing the tightness and
convergence for $(Y_{t}^{\varepsilon},\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s})$,
we identify the limit.

Step 1: Tightness and convergence for
$(Y_{t}^{\varepsilon},\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s})$.

######  Proposition 3.2.

Under Assumption 2.1 and Assumption 3.1, BSDE $(\ref{eq:BSDE with perturbed})$
and BSDE $(\ref{eq:limit BSDE })$ have unique solutions
$(Y^{\varepsilon},Z^{\varepsilon})$ and $(Y,Z)\in S^{2}(0,T;R^{k})$ $\times
M^{2}(0,T;R^{k\times d})$. Moreover, there exists a positive constant $C$ such
that $\forall\varepsilon>0$,

$\begin{split}E\left(\sup_{0\leq t\leq
T}|Y^{\varepsilon}_{t}|^{2}+\int_{0}^{T}(Z^{\varepsilon}_{t})^{2}dt\right)&\leq
C,\\\ E\left(\sup_{0\leq t\leq
T}|Y_{t}|^{2}+\int_{0}^{T}(Z_{t})^{2}dt\right)&\leq C.\end{split}$

###### Proof.

For BSDE $(\ref{eq:BSDE with perturbed})$, by Theorem 2.2, the existence and
uniqueness of solution $(Y^{\varepsilon},Z^{\varepsilon})$ is obtained for all
$\varepsilon>0$.

Using Itô’s formula to $|Y^{\varepsilon}_{s}|^{2}$ on $[t,T]$, we get the
following from Schwartz’s inequality,

$\begin{split}&|Y^{\varepsilon}_{t}|^{2}+\int_{t}^{T}|Z^{\varepsilon}_{s}|^{2}ds\\\
=&\
|\xi|^{2}+2\int_{t}^{T}Y^{\varepsilon}_{s}f(s,Y^{\varepsilon}_{s},Z^{\varepsilon}_{s},\alpha_{s})ds-2\int_{t}^{T}Y^{\varepsilon}_{s}Z^{\varepsilon}_{s}dB_{s}\\\
\leq&\
|\xi|^{2}+2\int_{t}^{T}\left((1+\mu^{2})|Y^{\varepsilon}_{s}|^{2}+|f(s,0,0,\alpha^{\varepsilon}_{s})|^{2}\right)ds-2\int_{t}^{T}Y^{\varepsilon}_{s}Z^{\varepsilon}_{s}dB_{s}\end{split}$

here $\mu$ is the Lipschitz constant of $f$ which is independent of
$\varepsilon$. By taking expectation, we can deduce

$E\left(|Y^{\varepsilon}_{t}|^{2}+\frac{1}{2}\int_{t}^{T}|Z^{\varepsilon}_{s}|^{2}ds\right)\leq|\xi|^{2}+2\int_{t}^{T}((1+\mu^{2})|Y^{\varepsilon}_{s}|^{2}+|f(s,0,0,\alpha^{\varepsilon}_{s})|^{2})ds.$

From Gronwall’s lemma, we get

$\displaystyle
E\left(|Y^{\varepsilon}_{t}|^{2}+\int_{t}^{T}|Z^{\varepsilon}_{s}|^{2}ds\right)\leq
CE\left(|\xi|^{2}+\int_{0}^{T}|f(s,0,0,\alpha^{\varepsilon}_{s})|^{2}ds\right)\leq
C,$

and then the estimation for $(Y^{\varepsilon},Z^{\varepsilon})$ is obtained
from the Burkholder-Davis-Gundy inequality.

From the form of $\bar{f}$ presented in Theorem 3.3, we know that $\bar{f}$
also satisfies Assumption 2.1 and Assumption 3.1, thus the estimation about
$(Y,Z)$ can be obtained similarly. ∎

We set $M_{t}^{\varepsilon}=\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s}$ for the
convenience. Thus BSDE $(\ref{eq:BSDE with perturbed})$ can be rewritten as

$Y^{\varepsilon}_{t}=\xi+\int_{t}^{T}f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})ds-(M_{T}^{\varepsilon}-M_{t}^{\varepsilon}).$
(11)

######  Proposition 3.3.

The sequence of $(Y^{\varepsilon},M^{\varepsilon})$ is tight on the space
$D(0,T;$ $R^{k})\times D(0,T;R^{k}).$

###### Proof.

Let
${\cal{G}}_{t}^{\varepsilon}={\cal{F}}^{B}_{t}\vee{\cal{F}}_{T}^{\alpha^{\varepsilon}}\vee{\cal{N}},$
we define the conditional variation

$CV(Y^{\varepsilon})=\sup
E\left(\sum_{i}|E(Y^{\varepsilon}_{t_{i+1}}-Y^{\varepsilon}_{t_{i}}|{\cal{G}}^{\varepsilon}_{t_{i}})|\right)$

where the supreme is taken over all partitions of the interval $[0,T]$.

From the Proof of Proposition 2.2, we know that $M^{\varepsilon}$ is a
${\cal{G}}_{t}^{\varepsilon}$-martingale. It follows that

$CV(Y^{\varepsilon})\leq
E\int_{0}^{T}|f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})|ds.$

From $(ii)$ of Assumption 2.1, $(ii)$ of Assumption 3.1, and Proposition 3.2,
we know

$\sup_{\varepsilon}\left(CV(Y^{\varepsilon})+\sup_{0\leq t\leq
T}E|Y^{\varepsilon}_{t}|+\sup_{0\leq t\leq
T}E|M^{\varepsilon}_{t}|\right)<\infty.$

Thus the “Meyer-Zheng tightness criteria” ([2, 14]) is fully satisfied and the
result is followed. ∎

Together with the properties of $Y^{\varepsilon}$ obtained above, the
following proposition can be seen as an obvious result of Lemma 7.3 in [25].

######  Proposition 3.4.

Suppose $g(t,x)$ is a function defined on $[0,T]\times R^{m}$ satisfying that
$g(\cdot,\cdot)$ is Lipschitz continuous with $x$ and $\forall x\in R^{m}$,
either $|g(t,x)|\leq K(1+|x|)$ or $|g(t,x)|\leq K$. Denote
$\pi_{ij}^{\varepsilon}(t)=\pi_{ij}^{\varepsilon}(t,\alpha^{\varepsilon}_{t})$,
with
$\pi_{ij}^{\varepsilon}(t,\alpha)=I_{\\{\alpha=s_{ij}\\}}-\nu_{j}^{i}I_{\\{\alpha\in
M_{i}\\}}$, then for any $\ k=1,\cdots,l,j=1,\cdots,m_{k}$,

$\sup_{0<t\leq
T}E\left|\int_{0}^{t}g(t,Y_{s}^{\varepsilon})\pi_{ij}^{\varepsilon}(s,\alpha^{\varepsilon}_{s})ds\right|\rightarrow
0,\textrm{\quad as }\varepsilon\rightarrow 0.$

Step 2: Identification of the limit.
From Proposition 3.3, we know that there exists a subsequence of
$(Y^{\varepsilon},M^{\varepsilon})$, which we still denote by
$(Y^{\varepsilon},M^{\varepsilon})$, and which converges in distribution on
the space $D(0,T;R^{k})\times D(0,T;R^{k})$ toward a
c$\grave{a}$dl$\grave{a}$g process $(\bar{Y},\bar{M})$. Furthermore, there
exists a countable subset $D$ of $[0,T]$, such that
$(Y^{\varepsilon},M^{\varepsilon})$ converges in finite-distribution to
$(\bar{Y},\bar{M})$ on $D^{c}$.

######  Proposition 3.5.

For the limit process $(\bar{Y},\bar{M})$, we have

(i) For every $t\in[0,T]-D$,

$\bar{Y}_{t}=\xi+\int_{t}^{T}\bar{f}(s,\bar{Y}_{s},\bar{\alpha}_{s})ds-(\bar{M}_{T}-\bar{M}_{t}).$

(ii) For a $d$-dimensional Brownian motion $\bar{B}=\\{\bar{B}_{t};0\leq t\leq
T\\}$ with $\bar{B}_{0}=0$, $\bar{Y}$ is measurable with
${\cal{H}}_{t}={\cal{F}}^{\bar{B}}_{t}\vee{\cal{F}}^{\bar{\alpha}}_{T}$, then
$\bar{M}$ is a ${\cal{H}}_{t}$-martingale.

###### Proof.

From Proposition 3.4, as $\varepsilon\rightarrow 0$,

$\begin{split}&\sup_{0\leq t\leq
T}E\left|\int_{0}^{t}f(s,Y^{\varepsilon}_{s},s_{ij})\left(I_{\\{\alpha_{s}^{\varepsilon}=s_{ij}\\}}-\nu_{j}^{i}I_{\\{\alpha_{s}^{\varepsilon}\in{\cal{M}}_{i}\\}}\right)ds\right|\rightarrow
0.\end{split}$

Since $(Y^{\varepsilon},\bar{\alpha}^{\varepsilon})$ converge weakly to
$(\bar{Y},\bar{\alpha})$,

$\begin{split}&\int_{0}^{t}\bar{f}(s,Y^{\varepsilon}_{s},\bar{\alpha}^{\varepsilon}_{s})ds\textrm{
converges in distribution
to}\int_{0}^{t}\bar{f}(s,\bar{Y}_{s},\bar{\alpha}_{s})ds\textrm{ on
}C(0,T;R^{k}).\end{split}$

Thus

$\begin{split}&\int_{0}^{t}f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})ds\\\
=&\int_{0}^{t}\sum_{i=1}^{l}\sum_{j=1}^{m_{i}}f(s,Y^{\varepsilon}_{s},s_{ij})I_{\\{\alpha_{s}^{\varepsilon}=s_{ij}\\}}\\\
=&\int_{0}^{t}\sum_{i=1}^{l}\sum_{j=1}^{m_{i}}f(s,Y^{\varepsilon}_{s},s_{ij})\left(I_{\\{\alpha_{s}^{\varepsilon}=s_{ij}\\}}-\nu_{j}^{i}I_{\\{\alpha_{s}^{\varepsilon}\in{\cal{M}}_{i}\\}}\right)ds+\int_{0}^{t}\bar{f}(s,Y^{\varepsilon}_{s},\bar{\alpha}^{\varepsilon}_{s})ds.\end{split}$

As $\varepsilon\rightarrow 0$, passing to the limit in the backward component
of the BSDE $(\ref{eq: BSDE with perturbed in M form})$, we can derive
assertion (i).

Now, we prove assertion (ii).

For any $0\leq t_{1}\leq t_{2}\leq T$, $\Phi_{t_{1}}$ is a continuous mapping
from $C(0,t_{1};R^{d})\times D(0,t_{1};R^{k})\times D(0,T;\bar{{\cal{M}}})$.
$\forall\varepsilon>0$, since $M^{\varepsilon}$ is a martingale with respect
to
$\mathcal{G}_{t}^{\varepsilon}=\mathcal{F}_{T}^{\alpha^{\varepsilon}}\vee\mathcal{F}_{t}^{B}$,
$Y^{\varepsilon}$ and $\bar{\alpha}^{\varepsilon}$ are
$\mathcal{G}_{t}^{\varepsilon}$-adapted, we know

$E\left(\Phi_{t_{1}}(B,Y^{\varepsilon},\bar{\alpha}^{\varepsilon})\left(Y^{\varepsilon}_{t_{2}}-Y^{\varepsilon}_{t_{1}}+\int_{t_{1}}^{t_{2}}f(s,Y_{s}^{\varepsilon},\alpha_{s}^{\varepsilon})ds\right)\right)=0$

and

$E\left(\Phi_{t_{1}}(B,Y^{\varepsilon},\bar{\alpha}^{\varepsilon})\int_{0}^{\delta}(M_{t_{2}+r}^{\varepsilon}-M_{t_{1}+r}^{\varepsilon})dr\right)=0,$

here $B$ is the Brownian motion.

From the weak convergence of $(Y^{\varepsilon},\bar{\alpha}^{\varepsilon})$ to
$(\bar{Y},\bar{\alpha})$,
$\int_{0}^{t}\bar{f}(s,Y^{\varepsilon}_{s},\bar{\alpha}^{\varepsilon}_{s})ds$
converges in distribution to
$\int_{0}^{t}\bar{f}(s,\bar{Y}_{s},\bar{\alpha}_{s})ds$ on $C(0,T;R^{k})$. For
a $d$-dimensional Brownian motion $\bar{B}=\\{\bar{B}_{t};0\leq t\leq T\\}$
with $\bar{B}_{0}=0$, from the fact that $\bar{B}$ has the same probability
distribution with $B$ and $\displaystyle E(\sup_{0\leq t\leq
T}|M_{t}^{\varepsilon}|^{2})\leq C$, we obtain

$E\left(\Phi_{t_{1}}(\bar{B},\bar{Y},\bar{\alpha})\left(\bar{Y}_{t_{2}}-\bar{Y}_{t_{1}}+\int_{t_{1}}^{t_{2}}\bar{f}(s,\bar{Y}_{s},\bar{\alpha}_{s})ds\right)\right)=0$

and

$E\left(\Phi_{t_{1}}(\bar{B},\bar{Y},\bar{\alpha})\int_{0}^{\delta}(\bar{M}_{t_{2}+r}-\bar{M}_{t_{1}+r})dr\right)=0.$

Dividing the second identity by $\delta$, letting $\delta\rightarrow 0$, and
exploiting the right continuity, we obtain that

$E\left(\Phi_{t_{1}}(\bar{B},\bar{Y},\bar{\alpha})(\bar{M}_{t_{2}}-\bar{M}_{t_{1}})\right)=0.$

From the freedom choice of $t_{1}$, $t_{2}$, and $\Phi_{t_{1}}$, we deduce
that $\bar{M}$ is a ${\cal{H}}_{t}$-martingale. ∎

######  Proposition 3.6.

Let $\\{(Y_{t},Z_{t});0\leq t\leq T\\}$ be the unique solution of BSDE
$(\ref{eq:limit BSDE })$, then $\forall t\in[0,T]$,

$E|Y_{t}-\bar{Y}_{t}|^{2}+E\left([\bar{M}-\int_{0}^{\cdot}Z_{r}d\bar{B}_{r}]_{T}-[\bar{M}-\int_{0}^{\cdot}Z_{r}d\bar{B}_{r}]_{t}\right)=0.$

###### Proof.

Let $M_{t}=\int_{0}^{t}Z_{r}d\bar{B}_{r}$, by the proof of Proposition 2.2, we
know that $M_{t}$ is a
${\cal{F}}_{t}^{\bar{B}}\vee{\cal{F}}_{T}^{\bar{\alpha}}$-martingale.

From Itô’s formula and Proposition 3.5, we know that

$\begin{split}&E|Y_{t}-\bar{Y}_{t}|^{2}+E\left([M-\bar{M}]_{T}-[M-\bar{M}]_{t}\right)\\\
=\
&2E\int_{t}^{T}\left(\bar{f}(s,Y_{s},\bar{\alpha}_{s})-\bar{f}(s,\bar{Y}_{s},\bar{\alpha}_{s})\right)(Y_{s}-\bar{Y}_{s})ds\\\
\leq\ &CE\int_{t}^{T}|Y_{s}-\bar{Y}_{s}|^{2}ds.\end{split}$

From Gronwall’s lemma, we obtain $E|Y_{t}-\bar{Y}_{t}|^{2}=0$, $\forall
t\in[0,T]-D$, and the result follows. ∎

We come back to finish the Proof of Theorem 3.1:

Since $Y$ is continuous, $\bar{Y}$ is c$\grave{a}$dl$\grave{a}$g, and $D$ is
countable, we get $Y_{t}=\bar{Y}_{t}$, $P-$a.s., $\forall t\in[0,T]$.
Moreover, we can deduce that $M\equiv\bar{M}$. Hence, we get the result that
the sequence $(Y^{\varepsilon}_{t},\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s})$
converges in distribution to the process
$(Y_{t},\int_{0}^{t}Z_{s}d\bar{B}_{s})$, and the proof of Theorem 3.3 is
completed. ∎

###  3.3 Examples

######  Example 3.1.

Consider the case that $\tilde{Q}$ is weakly irreducible with the state space
$\mathcal{M}=\\{1,\cdots,m\\}$ and $\nu=(\nu_{1},\cdots,\nu_{m})$ is the quasi
stationary distribution, then $\alpha^{\varepsilon}$ can be considered as a
fast-varying noise process. As shown in the following, the noise is averaged
out with respect to the quasi stationary distribution. In this case, the
corresponding BSDE is

$Y^{\varepsilon}_{t}=\xi+\int_{t}^{T}f(s,Y^{\varepsilon}_{s},\alpha^{\varepsilon}_{s})ds-\int_{t}^{T}Z^{\varepsilon}_{s}dB_{s}.$
(12)

Under Assumption 2.1 and Assumption 3.1, from Theorem 3.3, as
$\varepsilon\rightarrow 0$, the sequence of process $(Y^{\varepsilon}_{t},$
$\int_{0}^{t}Z^{\varepsilon}_{s}dB_{s})$ converges in distribution to the
process $(Y_{t},\int_{0}^{t}Z_{s}d\bar{B}_{s})$, where $(Y,Z)$ is the unique
solution to the following BSDE

$Y_{t}=\xi+\int_{t}^{T}\sum_{i=1}^{m}\nu_{i}f(s,Y_{s},i)ds-\int_{t}^{T}Z_{s}d\bar{B}_{s}.$
(13)

It is noted that the generator of BSDE (13) depends on the quasi stationary
distribution of the Markov chain. Thus we can adopt the distribution of a
$\mathcal{F}_{t}^{\bar{B}}$-adapted process $Y$, the solution of BSDE (13), as
the asymptotic distribution for the solution of
$\mathcal{F}_{t}^{B}\vee\mathcal{F}^{\alpha^{\varepsilon}}_{t,T}$-adapted
process $Y^{\varepsilon}$.

In practical systems, the small parameter $\varepsilon$ is just a fixed
parameter and it separates different scales in the sense of order of magnitude
in the generator. It does not need to tend to 0. We give a detailed example
for interpretation.

######  Example 3.2.

Suppose the generator of the continuous-time Markov chain affected BSDE (3) is
$Q=\begin{pmatrix}{-22}&{20}&{2}\\\ {41}&{-42}&{1}\\\
{1}&{2}&{-3}\end{pmatrix}$, and the corresponding state space is
$\mathcal{M}=\\{s_{1},s_{2},s_{3}\\}$. It is obvious that the transition rate
between $s_{1}$ and $s_{2}$ is larger than the transition rate between $s_{3}$
and other states, i.e., the jumps between $s_{1}$ and $s_{2}$ are more
frequent than jumps between $s_{3}$ and other states. We can rewrite $Q$ as
following

$\displaystyle
Q=\frac{1}{0.05}\tilde{Q}+\hat{Q}=\frac{1}{0.05}\begin{pmatrix}{-1}&{1}&{0}\\\
{2}&{-2}&{0}\\\ {0}&{0}&{0}\end{pmatrix}+\begin{pmatrix}{-2}&{0}&{2}\\\
{1}&{-2}&{1}\\\ {1}&{2}&{-3}\end{pmatrix}$

It is noted that we choose suitable $\varepsilon$ to guarantee that
$\tilde{Q}$ and $\hat{Q}$ to be the generator with the same order of
magnitude.

Now, we introduce the continuous-time $\varepsilon$-dependent singularly
perturbed Markov chain $\alpha^{\varepsilon}=\\{\alpha^{\varepsilon}_{t};0\leq
t\leq T\\}$ which have the generator $\displaystyle
Q^{\varepsilon}=\frac{1}{\varepsilon}\tilde{Q}+\hat{Q}=\frac{1}{\varepsilon}\begin{pmatrix}{-1}&{1}&{0}\\\
{2}&{-2}&{0}\\\ {0}&{0}&{0}\end{pmatrix}+\begin{pmatrix}{-2}&{0}&{2}\\\
{1}&{-2}&{1}\\\ {1}&{2}&{-3}\end{pmatrix},$ and define the aggregated process

$\bar{\alpha}^{\varepsilon}=\\{\bar{\alpha}^{\varepsilon}_{t};0\leq t\leq
T\\}=\left\\{\begin{array}[]{ll}1,\
\alpha^{\varepsilon}_{t}\in\\{s_{1},s_{2}\\}\\\ 2,\
\alpha^{\varepsilon}_{t}\in\\{s_{3}\\}\end{array}\right.$

Proposition 3.1 yields that $\bar{\alpha}^{\varepsilon}$ converges in
distribution to a continuous-time Markov chain $\bar{\alpha}$ generated by
$\displaystyle\bar{Q}=\begin{pmatrix}-\frac{5}{3}&\frac{5}{3}\\\
{3}&-{3}\end{pmatrix}.$ By Theorem 3.3, we can adopt the probability
distribution of the solution to the following BSDE

$Y_{t}=\xi+\int_{t}^{T}\bar{f}(s,Y_{s},\bar{\alpha}_{s})ds-\int_{t}^{T}Z_{s}d\bar{B}_{s}$

as an asymptotic probability distribution of the solution to the original
BSDE. Here
$\displaystyle\bar{f}(t,y,1)=\frac{2}{3}f(t,y,s_{1})+\frac{1}{3}f(t,y,s_{2})$
and $\bar{f}(t,y,2)=f(t,y,s_{3}).$

Since the limit averaged Markov chain has two states and the original one has
three states, we have reduced the complexity of the model. This advantage will
be more clear when the state space of the original Markov chain is
sufficiently larger.

##  4 Homogenization of One System of PDEs

As an application of our results in previous section, we show the
homogenization of a sequence of semi-linear backward PDE with a singularly
perturbed Markov chain. In this section, after showing the relation between
BSDEs with Markov chain and one system of semi-linear PDE with Markov chain,
we derive the homogenization property of backward PDE with a singularly
perturbed Markov chain based on the weak convergence of the associated BSDE.

Here, we give some notations as follows: $C^{k}(R^{p};R^{q})$ is the space of
functions of class $C^{k}$ from $R^{p}$ to $R^{q}$, $C^{k}_{l,b}(R^{p};R^{q})$
is the space of functions of class $C^{k}$ whose partial derivatives of order
less than or equal to $k$ are bounded, and $C^{k}_{p}(R^{p};R^{q})$ is the
space of functions of class $C^{k}$ which, together with all their partial
derivatives of order less than or equal to $k$, grow at most like a polynomial
function of the variable $x$ at infinity.

###  4.1 Relation between BSDEs with Markov chains and semi-linear PDEs
systems with Markov chains

For $t\in[0,T]$, consider the following semi-linear backward PDE with a Markov
chain:

$u(t,x)=h(x)+\int_{t}^{T}\left(\mathcal{L}u(r,x)+f(r,x,u(r,x),(\nabla
u\sigma)(r,x),\alpha_{r})\right)dr,$ (14)

here $u:[0,T]\times R^{m}\rightarrow R^{k}$, and
$\displaystyle\mathcal{L}u=\left(Lu_{1},\cdots,Lu_{k}\right)^{\prime}$, with
$\allowdisplaybreaks\displaystyle
L=\frac{1}{2}\sum_{i,j=1}^{m}(\sigma\sigma^{\prime})_{ij}$
$\displaystyle(t,x)\frac{\partial^{2}}{\partial x_{i}\partial
x_{j}}+\sum_{i=1}^{m}b_{i}(t,x)\frac{\partial}{\partial x_{i}}.$

Firstly, we make the following assumption:

######  Assumption 4.1.

$b\in C^{3}_{l,b}(R^{m};R^{m})$, $\sigma\in C^{3}_{l,b}(R^{m};R^{m\times d})$,
$h\in C_{p}^{3}(R^{m};R^{k})$. For $f:[0,T]\times R^{m}\times R^{k}\times
R^{k\times d}\times\mathcal{M}\rightarrow R^{k}$, $\forall s\in[0,T]$,
$i\in\mathcal{M}$, $(x,y,z)\rightarrow f(s,x,y,z,i)$ is of class $C^{3}$.

Moreover, $f(s,\cdot,0,0,i)\in C^{3}_{p}(R^{m};R^{k})$, and the first order
partial derivatives in $y$ and $z$ are bounded on $[0,T]\times R^{m}\times
R^{k}\times R^{k\times d}\times\mathcal{M}$, as well as their derivatives of
order one and two with respect to $x$, $y$, $z$.

######  Definition 4.1.

A classical solution of PDE (14) is a $R^{k}$-valued stochastic process
$\\{u(t,x);$ $0\leq t\leq T,x\in R^{m}\\}$ which is in $C^{0,2}([0,T]\times
R^{m};R^{k})$ and satisfies that $u(t,x)$ is $\mathcal{F}_{t,T}$-measurable.

$\forall t\in[0,T]$, $x\in R^{m}$, we introduce the following FBSDE with a
Markov chain on $[t,T]$:

$X_{s}^{t,x}=x+\int_{t}^{s}b(X_{r}^{t,x})dr+\int_{t}^{s}\sigma(X_{r}^{t,x})dB_{r},$
(15)
$Y_{s}^{t,x}=h(X_{T}^{t,x})+\int_{s}^{T}f(r,X_{r}^{t,x},Y_{r}^{t,x},Z_{r}^{t,x},\alpha_{r})dr-\int_{s}^{T}Z_{r}^{t,x}dB_{r}.$
(16)

The aim of this subsection is to show that, under above assumptions, the FBSDE
(15)-(16) provides both a probabilistic representation and the unique
classical solution for PDE (14).

For SDE (15), it is well known that under Assumption 4.1, it has a unique
solution $\\{X_{s}^{t,x};t\leq s\leq T\\}$ which has a version that is a.s. of
class $C^{2}$ in $x$, the function and its derivatives are a.s. jointly
continuous in $(t,s,x)$. Moreover,

$\sup_{t\leq s\leq T}\left(|X_{s}^{t,x}|+|\nabla
X_{s}^{t,x}|+|D^{2}X_{s}^{t,x}|\right)\in\bigcap_{p\geq
1}L^{p}(R),\forall(t,x)\in[0,T]\times R^{m}$

where $\nabla X_{s}^{t,x}$, $D^{2}X_{s}^{t,x}$ denote respectively the matrix
of first order and second order derivatives of $X_{s}^{t,x}$ with respect to
$x$.

For BSDE (16), denote $\tilde{f}(s,y,z,i)=f(s,X^{t,x}_{s},y,z,i)$, $\forall
i\in\mathcal{M}$, we know that $\tilde{f}$ satisfies Assumption 2.1 since $f$
satisfies Assumption 4.1. So there exists a unique solution pair
$\\{(Y_{s}^{t,x},Z_{s}^{t,x});t\leq s\leq T\\}$ to BSDE (16).

Define $X_{s}^{t,x}=X_{s\vee t}^{t,x}$, $Y_{s}^{t,x}=Y_{s\vee t}^{t,x}$, and
$Z_{s}^{t,x}=0$, for $s\leq t$. Then
$(X,Y,Z)=(X_{s}^{t,x},Y_{s}^{t,x},Z_{s}^{t,x})$ is defined on
$(s,t)\in[0,T]^{2}$.

######  Theorem 4.4.

Under Assumption 4.1, let $\\{u(t,x);0\leq t\leq T,x\in R^{m}\\}$ be a
classical solution of PDE $(\ref{Eq:SPDEs })$. Suppose that there exists a
constant $C$ such that,

$|u(t,x)|+|\partial_{x}u(t,x)\sigma(t,x)|\leq
C(1+|x|),\quad\forall(t,x)\in[0,T]\times R^{m},$ (17)

then
$(Y_{s}^{t,x}=u(t,X_{s}^{t,x}),Z_{s}^{t,x}=\partial_{x}u(t,X_{s}^{t,x})\sigma(t,X_{s}^{t,x});t\leq
s\leq T)$ is the unique solution of BSDE $(\ref{Eq:BSDE of FBSDE})$. Here
$(X_{s}^{t,x};t\leq s\leq T)$ is the solution to SDE $(\ref{Eq:SDE of
FBSDE})$.

Proof: $\forall t\leq s\leq T$, let $s=t_{0}<t_{1}<t_{2}<\cdots<t_{n}=T$, with
Itô’s formula and PDE $(\ref{Eq:SPDEs })$, we get

$\begin{split}&Y_{s}^{t,x}-h(X_{T}^{t,x})\\\ =\ &u(s,x)-u(T,X_{T}^{t,x})\\\ =\
&\sum_{i=0}^{n-1}\left(u(t_{i},X_{t_{i}}^{t,x})-u(t_{i+1},X_{t_{i+1}}^{t,x})\right)\\\
=\
&\sum_{i=0}^{n-1}\left(u(t_{i},X_{t_{i}}^{t,x})-u(t_{i},X_{t_{i+1}}^{t,x})\right)+\sum_{i=0}^{n-1}\left(u(t_{i},X_{t_{i+1}}^{t,x})-u(t_{i+1},X_{t_{i+1}}^{t,x})\right)\end{split}$
$\begin{split}=\
&\sum_{i=0}^{n-1}\bigg{(}-\int_{t_{i}}^{t_{i+1}}\left(\mathcal{L}u(t_{i},X_{s}^{t,x})ds-(\nabla
u\sigma)(t_{i},X_{s}^{t,x})dB_{s}\right)\\\
&+\int_{t_{i}}^{t_{i+1}}\left(\mathcal{L}u(s,X_{t_{i+1}}^{t,x})+f(s,X_{t_{i+1}}^{t,x},u(r,X_{t_{i+1}}^{t,x}),(\nabla
u\sigma)(s,X_{t_{i+1}}^{t,x}),\alpha_{s})\right)ds\bigg{)}.\end{split}$

(17) yields that

$E\left(\sup_{t\leq s\leq
T}|u(t,X_{s}^{t,x})|^{2}+\int_{t}^{T}|\partial_{x}u\sigma(s,X_{s}^{t,x})|^{2}ds\right)<\infty,$

and the adaptability is obvious. The result is followed as
$\displaystyle\triangle=\sup_{0\leq i\leq n-1}|t_{i+1}-t_{i}|\rightarrow 0$. ∎

Now we deduce the converse side of Theorem 4.4.

######  Theorem 4.5.

Assume that for some $p>2$,
$E|\xi|^{p}+E\int_{0}^{T}|\tilde{f}(t,0,0,\alpha_{t})|^{p}dt<\infty,$ let
$b,\sigma,f,h,\alpha$ satisfy Assumption 4.1, then the process
$\\{u(t,x)=Y_{t}^{t,x};0\leq t\leq T,x\in R^{m}\\}$ is the unique classical
solution to PDE $(\ref{Eq:SPDEs })$.

As preliminaries for the proof, we give two propositions about the regularity
of the solution of BSDE (16) whose proofs are put in the Appendix.

######  Proposition 4.1.

Under the assumption of Theorem 4.5, $\\{Y_{s}^{t,x};(s,t)\in[0,T]^{2},x\in
R^{m}\\}$ has a version whose trajectories belong to
$C^{0,0,2}([0,T]^{2}\times R^{m})$. Hence $\forall t\in[0,T]$, $x\rightarrow
Y_{t}^{t,x}$ is of class $C^{2}$ a.s..

######  Proposition 4.2.

Under the assumption of Theorem 4.5, $\\{Z_{s}^{t,x};(s,t)\in[0,T]^{2},x\in
R^{m}\\}$ has an a.s. continuous version which is given by $Z_{s}^{t,x}=\nabla
Y_{s}^{t,x}(\nabla X_{s}^{t,x})^{-1}\sigma(X_{s}^{t,x})$. In particular,
$Z_{t}^{t,x}=\nabla Y_{t}^{t,x}\sigma(x)$. Here $\displaystyle\Big{(}\nabla
Y_{s}^{t,x}=\frac{\partial Y_{s}^{t,x}}{\partial x},$ $\displaystyle\nabla
Z_{s}^{t,x}=\frac{\partial Z_{s}^{t,x}}{\partial x}\Big{)}$ is the unique
solution of

$\begin{split}\nabla Y_{s}^{t,x}=&h^{\prime}(X_{T}^{t,x})\nabla
X_{T}^{t,x}+\int_{s}^{T}\big{(}f^{\prime}_{x}(r,X_{r}^{t,x},Y_{r}^{t,x},Z_{r}^{t,x},\alpha_{r})\nabla
X_{r}^{t,x}+f^{\prime}_{y}(r,X_{r}^{t,x},\\\
&Y_{r}^{t,x},Z_{r}^{t,x},\alpha_{r})\nabla
Y_{r}^{t,x}+f^{\prime}_{z}(r,X_{r}^{t,x},Y_{r}^{t,x},Z_{r}^{t,x},\alpha_{r})\nabla
Z_{r}^{t,x}\big{)}dr-\int_{s}^{T}Z_{r}^{t,x}dB_{r}.\end{split}$

Proof of Theorem 4.5: Let $t=t_{0}<t_{1}<\cdots<t_{n}=T$, we have

$\begin{split}&h(x)-u(t,x)\\\ =\ &u(T,x)-u(t,x)\\\
=&\sum_{i=0}^{n-1}\left(u(t_{i+1},x)-u(t_{i},x)\right)\\\
=&\sum_{i=0}^{n-1}\left(u(t_{i+1},x)-u(t_{i+1},X_{t_{i+1}}^{t_{i},x})+u(t_{i+1},X_{t_{i+1}}^{t_{i},x})-u(t_{i},x)\right).\end{split}$

Since
$u(t_{i+1},X_{t_{i+1}}^{t_{i},x})=Y_{t_{i+1}}^{t_{i+1},X_{t_{i+1}}^{t_{i},x}}=Y_{t_{i+1}}^{t_{i},x}$,
we obtain the following from BSDE (16)

$\begin{split}&u(t_{i+1},X_{t_{i+1}}^{t_{i},x})-u(t_{i},x)\\\ =\
&Y_{t_{i+1}}^{t_{i},x}-Y_{t_{i}}^{t_{i},x}\\\ =\
&-\int_{t_{i}}^{t_{i+1}}f(r,X_{r}^{t_{i},x},Y_{r}^{t_{i},x},Z_{r}^{t_{i},x},\alpha_{r})dr+\int_{t_{i}}^{t_{i+1}}Z_{r}^{t,x}dB_{r}.\end{split}$

It is known that $u(t,\cdot)\in C^{2}(R^{m})$ from Proposition 4.1. Then, with
Itô’s formula, we get

$\displaystyle\begin{split}&h(x)-u(t,x)\\\ =\
&\sum_{i=0}^{n-1}\bigg{(}\int_{t_{i}}^{t_{i+1}}\mathcal{L}u(t_{i+1},X_{r}^{t_{i},x})dr-\int_{t_{i}}^{t_{i+1}}(\nabla
u\sigma)(t_{i+1},X_{r}^{t_{i},x})dB_{r}\\\
&-\int_{t_{i}}^{t_{i+1}}f(r,X_{r}^{t_{i},x},Y_{r}^{t_{i},x},Z_{r}^{t_{i},x},\alpha_{r})dr+\int_{t_{i}}^{t_{i+1}}Z_{r}^{t,x}dB_{r}\bigg{)}\\\
=&-\sum_{i=0}^{n-1}\int_{t_{i}}^{t_{i+1}}\left(\mathcal{L}u(t_{i+1},X_{r}^{t_{i},x})+f(r,X_{r}^{t_{i},x},Y_{r}^{t_{i},x},Z_{r}^{t_{i},x},\alpha_{r})\right)dr\\\
&+\sum_{i=0}^{n-1}\int_{t_{i}}^{t_{i+1}}\left(Z_{r}^{t_{i},x}-(\nabla
u\sigma)(t_{i+1},X_{r}^{t_{i},x})\right)dB_{r}.\end{split}$

From Proposition 4.1 and Proposition 4.2, letting
$\displaystyle\triangle=\sup_{0\leq i\leq n-1}|t_{i+1}-t_{i}|\rightarrow 0$,
we have

$u(t,x)=h(x)+\int_{t}^{T}\left(\mathcal{L}u(r,x)+f(r,x,u(r,x),(\nabla
u\sigma)(r,x),\alpha_{r})\right)dr,$

here $u\in C^{0,2}([0,T]\times R^{m};R^{k})$. The uniqueness property is
followed from Theorem 4.4 and the uniqueness of the solution of BSDE (16). ∎

###  4.2 Homogenization of PDEs system with a singularly perturbed Markov
chain

Now we can give the application of our theoretical result in previous section
(Theorem 3.1): homogenization of PDEs system with a singularly perturbed
Markov chain.

Consider the following sequence of semi-linear backward PDE with a singularly
perturbed Markov chain, indexed by $\varepsilon>0$, for $t\in[0,T],x\in
R^{m},$

$u^{\varepsilon}(t,x)=h(x)+\int_{t}^{T}\left(\mathcal{L}u^{\varepsilon}(r,x)+f(r,x,u^{\varepsilon}(r,x),\alpha_{r}^{\varepsilon})\right)dr,\quad$
(18)

Here $\alpha^{\varepsilon}$ is the singularly perturbed Markov chain which is
stated in subsection 3.1. We have the following homogenization result.

######  Theorem 4.6.

Under Assumption 3.1 and Assumption 4.1, PDE $(\ref{Eq:SPDEs with parameter})$
has a classical solution $\\{u^{\varepsilon}(t,x);0\leq t\leq T,x\in
R^{m}\\}$. As $\varepsilon\rightarrow 0$, the sequence of $u^{\varepsilon}$
converges in distribution to a process $u$, where $u(t,x)$ is the classical
solution of the following PDE with the limit averaged Markov chain
$\bar{\alpha}$

$u(t,x)=h(x)+\int_{t}^{T}\left(\mathcal{L}u(r,x)+\bar{f}(r,x,u(r,x),\bar{\alpha}_{r})\right)dr,\quad
0\leq t\leq T.$ (19)

Here $\bar{f}$ is the average of $f$ defined as
$\displaystyle\bar{f}(t,x,u,i)=\sum_{j=1}^{m_{i}}\nu_{j}^{i}f(t,x,u,s_{ij})$,
for $i\in\bar{\mathcal{M}}=\\{1,\cdots,l\\}$.

###### Proof.

From Theorem 4.5, we know that
$\\{u^{\varepsilon}(t,x)=Y_{t}^{\varepsilon,t,x};0\leq t\leq T,x\in R^{m}\\}$
is the unique classical solution of PDE (18) where
$\\{Y_{s}^{\varepsilon,t,x};t\leq s\leq T\\}$ satisfies

$Y_{s}^{\varepsilon,t,x}=h(X_{T}^{t,x})+\int_{s}^{T}f(r,X_{r}^{t,x},Y_{r}^{\varepsilon,t,x},\alpha_{r}^{\varepsilon})dr-\int_{s}^{T}Z_{r}^{\varepsilon,t,x}dB_{r},$
(20)

and $\\{X_{s}^{t,x};t\leq s\leq T\\}$ satisfies SDE (15).
$\forall(t,x)\in[0,T]\times R^{m}$, from Theorem 3.3, we obtain that
$Y_{t}^{\varepsilon,t,x}$ converges in distribution to $Y_{t}^{t,x}$ as
$\varepsilon\rightarrow 0$ where $\\{Y_{s}^{t,x};t\leq s\leq T\\}$ satisfies

$Y_{s}^{t,x}=h(X_{T}^{t,x})+\int_{s}^{T}\bar{f}(r,X_{r}^{t,x},Y_{r}^{t,x},\bar{\alpha}_{r})dr-\int_{s}^{T}Z_{r}^{t,x}d\bar{B}_{r},$
(21)

Again from Theorem 4.5, we know that $u(t,x)=Y_{t}^{t,x}$ is the unique
classical solution to PDE (19), and the results are followed. ∎

##  5 Conclusion

In this paper, stemmed from the adjoint equation for deriving the optimal
control of stochastic LQ control problem with Markovian jumps, we study the
solvability of one kind of BSDE with the generator depending on a Markov
switching. Then, we consider the case that the Markov chain has a large state
space. To reduce the complexity, we adopt a hierarchical approach and study
the asymptotic property of BSDE with a singularly perturbed Markov chain.
Also, as an application, we present the homogenization property of one system
of PDE with a singularly perturbed Markov chain.

It is noted that in this paper, we only give the homogenization result of PDEs
system with Markov chains when there exists classical solution under smooth
assumptions. In the successive work, we will study the Sobolev space weak
solution for the related PDEs system and homogenization problem by virtue of
BSDEs with Markov chain. Some applications of this kind of BSDEs in optimal
control and mathematics financial problems would also be interesting to
investigate in our future research.

## Acknowledgements

It is our great pleasure to express the thankfulness to Professor Qing Zhang
in University of Georgia for many useful discussions and suggestions.

##  Appendix A Proof of Proposition 4.1 and Proposition 4.2

The proof of Proposition 4.1 and Proposition 4.2 follow a classic approach as
shown in [18, 19]. Here, we will give a sketch of the proof. Firstly, we
present a higher order moment estimation to the solution of BSDE (3).

######  Corollary A.1.

Assume that for some $p>2$, $E|\xi|^{p}+E\int_{0}^{T}|f(t,0,$
$0,\alpha_{t})|^{p}dt<\infty,$ under Assumption 2.1, we have the following
estimation for BSDE $(\ref{eq:BSDE})$,

$\displaystyle E\left(\sup_{0\leq s\leq
t}|Y_{s}|^{p}+(\int_{0}^{t}Z_{s}^{2}ds)^{\frac{p}{2}}\right)<\infty,\quad\forall
0\leq t\leq T.$

###### Proof.

Applying Itô’s formula to $|Y_{t}|^{p}$ from $t$ to $T$, we can get

$\begin{split}&|Y_{t}|^{p}+\frac{p(p-1)}{2}\int_{t}^{T}|Y_{s}|^{p-2}|Z_{s}|^{2}ds\\\
=&\
|\xi|^{p}+p\int_{t}^{T}|Y_{s}|^{p-2}Y_{s}f(s,Y_{s},Z_{s},\alpha_{s})ds-p\int_{t}^{T}|Y_{s}|^{p-2}Y_{s}Z_{s}dB_{s}.\end{split}$

By the same technique as that in Lemma 2.1 of Pardoux and Peng [18], we obtain
that

$\begin{split}&E|Y_{t}|^{p}+\frac{p(p-1)}{2}E\int_{t}^{T}|Y_{s}|^{p-2}|Z_{s}|^{2}ds\\\
\leq&\
E|\xi|^{p}+pE\int_{t}^{T}|Y_{s}|^{p-2}Y_{s}f(s,Y_{s},Z_{s},\alpha_{s})ds.\end{split}$

From Assumption 2.1, using H$\ddot{\textrm{o}}$lder and Young’s inequalities,
there exist $K>0$ and $C$ such that

$\begin{split}&E|Y_{t}|^{p}+KE\int_{t}^{T}|Y_{s}|^{p-2}|Z_{s}|^{2}ds\\\ \leq&\
E|\xi|^{p}+CE\int_{t}^{T}\left(|Y_{s}|^{p}+|f(s,0,0,\alpha_{s})|^{p}\right)ds.\end{split}$

It follows from Gronwall’s lemma that

$\sup_{0\leq t\leq
T}E|Y_{t}|^{p}+E\int_{0}^{T}|Y_{t}|^{p-2}|Z_{t}|^{2}dt<\infty.$

Since

$\begin{split}|Y_{t}|^{p}\leq\
|\xi|^{p}+p\int_{t}^{T}|Y_{s}|^{p-2}Y_{s}f(s,Y_{s},Z_{s},\alpha_{s})ds-p\int_{t}^{T}|Y_{s}|^{p-2}Y_{s}Z_{s}dB_{s},\end{split}$

Burkholder-Davis-Gundy inequality yields that $E(\sup_{0\leq t\leq
T}|Y_{t}|^{p})<\infty$.

Now we prove $E(\int_{0}^{t}Z_{s}^{2}ds)^{\frac{p}{2}}<\infty$. Since

$\int_{0}^{t}Z_{s}dB_{s}=Y_{t}-Y_{0}+\int_{0}^{t}f(s,Y_{s},Z_{s},\alpha_{s})ds,$
$\sup_{0\leq t\leq T}|\int_{0}^{t}Z_{s}dB_{s}|\leq 2\sup_{0\leq t\leq
T}|Y_{t}|+\int_{0}^{T}|f(s,Y_{s},Z_{s},\alpha_{s})|ds,$

the result is followed from Assumption 2.1 and Burkholder-Davis-Gundy
inequality. ∎

######  Lemma A.1.

(Lemma 2.7 in [18]) For any $p>2$, there exists a constant $c_{p}$ such that
for any $t,t^{\prime}\in[0,T]$, $x,x^{\prime}\in R^{m}$,
$i\in\\{1,\cdots,d\\}$, $h,h^{\prime}\in R\backslash\\{0\\}$,

$E(\sup_{0\leq s\leq T}|X_{s}^{t,x}|^{p})\leq c_{p}(1+|x|^{p}),$
$E(\sup_{0\leq s\leq T}|X_{s}^{t,x}-X_{s}^{t^{\prime},x^{\prime}}|^{p})\leq
c_{p}(1+|x|^{p})(|x-x^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}),$
$E(\sup_{0\leq s\leq T}|\triangle_{h}^{i}X_{s}^{t,x}|^{p})\leq c_{p},$
$E(\sup_{0\leq s\leq
T}|\triangle_{h}^{i}X_{s}^{t,x}-\triangle_{h^{\prime}}^{i}X_{s}^{t^{\prime},x^{\prime}}|^{p})\leq
c_{p}(|x-x^{\prime}|^{p}+|h-h^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}).$

Here $\displaystyle\triangle_{h}^{i}g(x)=\frac{g(x+he_{i})-g(x)}{h}$, $1\leq
i\leq d$, where $e_{i}$ denotes the $i$th vector of an arbitrary orthonormal
basis of $R^{m}$.

Proof of Proposition 4.1: Since

$E(\sup_{0\leq s\leq T}|X_{s}^{t,x}|^{p})\leq c_{p}(1+|x|^{p}),$

from the proof of Corollary A.1, $\forall p>2$, there exist $C_{p}$ and $q$
such that

$E\left(\sup_{0\leq s\leq
t}|Y_{s}^{t,x}|^{p}+(\int_{0}^{t}|Z_{s}^{t,x}|^{2}ds)^{\frac{p}{2}}\right)\leq
C_{p}(1+|x|^{q}).$

Note that for $t\vee t^{\prime}\leq s\leq T$

$\begin{split}&Y_{s}^{t,x}-Y_{s}^{t^{\prime},x^{\prime}}\\\ =&\
\bigg{(}\int_{0}^{1}h^{\prime}(X_{T}^{t,x}+\lambda(X_{T}^{t,x}-X_{T}^{t^{\prime},x^{\prime}}))d\lambda\bigg{)}(X_{T}^{t,x}-X_{T}^{t^{\prime},x^{\prime}})\\\
&+\int_{s}^{T}\int_{0}^{1}\bigg{(}f^{\prime}_{x}(\Xi_{r,\lambda}^{t,x,t^{\prime},x^{\prime}},\alpha_{r})(X_{r}^{t,x}-X_{r}^{t^{\prime},x^{\prime}})+f^{\prime}_{y}(\Xi_{r,\lambda}^{t,x,t^{\prime},x^{\prime}},\alpha_{r})(Y_{r}^{t,x}-Y_{r}^{t^{\prime},x^{\prime}})\\\
&+f^{\prime}_{z}(\Xi_{r,\lambda}^{t,x,t^{\prime},x^{\prime}},\alpha_{r})(Z_{r}^{t,x}-Z_{r}^{t^{\prime},x^{\prime}})\bigg{)}d\lambda
dr-\int_{s}^{T}(Z_{r}^{t,x}-Z_{r}^{t^{\prime},x^{\prime}})dB_{r},\end{split}$

where
$\displaystyle\Xi_{r,\lambda}^{t,x,t^{\prime},x^{\prime}}=(r,X_{r}^{t^{\prime},x^{\prime}}+\lambda(X_{r}^{t,x}-X_{r}^{t^{\prime},x^{\prime}}),Y_{r}^{t^{\prime},x^{\prime}}+\lambda(Y_{r}^{t,x}-Y_{r}^{t^{\prime},x^{\prime}}),Z_{r}^{t^{\prime},x^{\prime}}+\lambda(Z_{r}^{t,x}-Z_{r}^{t^{\prime},x^{\prime}})).$
Since

$E(\sup_{0\leq s\leq T}|X_{s}^{t,x}-X_{s}^{t^{\prime},x^{\prime}}|^{p})\leq
c_{p}(1+|x|^{p})(|x-x^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}),$

combing with the proof of Corollary A.1, we can deduce that $\forall p\geq 2$,
there exist $C_{p}$ and $q$ such that

$\displaystyle E\left(\sup_{0\leq s\leq
T}|Y_{s}^{t,x}-Y_{s}^{t^{\prime},x^{\prime}}|^{p}+\left(\int_{t}^{T}|Z_{s}^{t,x}-Z_{s}^{t^{\prime},x^{\prime}}|^{2}ds\right)^{\frac{p}{2}}\right)$
$\displaystyle\leq\ \ $ $\displaystyle
C_{p}(1+|x|^{q})(|x-x^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}).$

Then using Kolmogorov’s lemma, we know that
$\\{Y_{s}^{t,x};(s,t)\in[0,T]^{2},x\in R^{m}\\}$ has an a.s. continuous
version.

Next, we have

$\begin{split}\triangle_{h}^{i}Y_{s}^{t,x}=&\int_{0}^{1}h^{\prime}(X_{T}^{t,x}+\lambda
h\triangle_{h}^{i}X_{T}^{t,x})\triangle_{h}^{i}X_{T}^{t,x}d\lambda+\int_{s}^{T}\int_{0}^{1}\big{(}f^{\prime}_{x}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}X_{r}^{t,x}\\\
&+f^{\prime}_{y}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}Y_{r}^{t,x}+f^{\prime}_{z}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}Z_{r}^{t,x}\big{)}d\lambda
dr-\int_{s}^{T}\triangle_{h}^{i}Z_{r}^{t,x}dB_{r}\end{split}$

where $\Theta_{r,\lambda}^{t,x,h}=(r,X_{r}^{t,x}+\lambda
h\triangle_{h}^{i}X_{r}^{t,x},Y_{r}^{t,x}+\lambda
h\triangle_{h}^{i}Y_{r}^{t,x},Z_{r}^{t,x}+\lambda
h\triangle_{h}^{i}Z_{r}^{t,x})$.

Since for each $p\geq 2$, there exists $c_{p}$ such that

$E(\sup_{0\leq s\leq T}|\triangle_{h}^{i}X_{s}^{t,x}|^{p})\leq c_{p}.$

We can have the following estimation

$\displaystyle E\left(\sup_{t\leq s\leq
T}|\triangle_{h}^{i}Y_{s}^{t,x}|^{p}+(\int_{t}^{T}|\triangle_{h}^{i}Z_{s}^{t,x}|ds)^{\frac{p}{2}}\right)\leq
c_{p}(1+|x|^{q}+|h|^{q}).$

Then we consider

$\begin{split}&\triangle_{h}^{i}Y_{s}^{t,x}-\triangle_{h^{\prime}}^{i}Y_{s}^{t^{\prime},x^{\prime}}\\\
=&\int_{0}^{1}h^{\prime}(X_{T}^{t,x}+\lambda
h\triangle_{h}^{i}X_{T}^{t,x})\triangle_{h}^{i}X_{T}^{t,x}d\lambda-\int_{0}^{1}h^{\prime}(X_{T}^{t^{\prime},x^{\prime}}+\lambda
h\triangle_{h^{\prime}}^{i}X_{T}^{t^{\prime},x^{\prime}})\triangle_{h^{\prime}}^{i}X_{T}^{t^{\prime},x^{\prime}}d\lambda\\\
&+\int_{s}^{T}\int_{0}^{1}(f^{\prime}_{x}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}X_{r}^{t,x}-f^{\prime}_{x}(\Theta_{r,\lambda}^{t^{\prime},x^{\prime},h^{\prime}},\alpha_{r})\triangle_{h^{\prime}}^{i}X_{r}^{t^{\prime},x^{\prime}})d\lambda
dr\\\
&+\int_{s}^{T}\int_{0}^{1}(f^{\prime}_{y}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}Y_{r}^{t,x}-f^{\prime}_{y}(\Theta_{r,\lambda}^{t^{\prime},x^{\prime},h^{\prime}},\alpha_{r})\triangle_{h^{\prime}}^{i}Y_{r}^{t^{\prime},x^{\prime}})d\lambda
dr\\\
&+\int_{s}^{T}\int_{0}^{1}(f^{\prime}_{z}(\Theta_{r,\lambda}^{t,x,h},\alpha_{r})\triangle_{h}^{i}Z_{r}^{t,x}-f^{\prime}_{z}(\Theta_{r,\lambda}^{t^{\prime},x^{\prime},h^{\prime}},\alpha_{r})\triangle_{h^{\prime}}^{i}Z_{r}^{t^{\prime},x^{\prime}})d\lambda
dr\\\
&-\int_{s}^{T}(\triangle_{h}^{i}Z_{r}^{t,x}-\triangle_{h^{\prime}}^{i}Z_{r}^{t^{\prime},x^{\prime}})dB_{r}.\end{split}$

It is noted that

$E(\sup_{0\leq s\leq
T}|\triangle_{h}^{i}X_{s}^{t,x}-\triangle_{h^{\prime}}^{i}X_{s}^{t^{\prime},x^{\prime}}|^{p})\leq
c_{p}(|x-x^{\prime}|^{p}+|h-h^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}).$

$\forall i\in\mathcal{M}$, using similar arguments, we can show that

$\begin{split}&E\left(\sup_{0\leq s\leq
T}|\triangle_{h}^{i}Y_{s}^{t,x}-\triangle_{h^{\prime}}^{i}Y_{s}^{t^{\prime},x^{\prime}}|^{p}+\left(\int_{t\wedge
t^{\prime}}^{T}|\triangle_{h}^{i}Z_{s}^{t,x}-\triangle_{h^{\prime}}^{i}Z_{s}^{t^{\prime},x^{\prime}}|^{2}ds\right)^{{\frac{p}{2}}}\right)\\\
\leq&\
c_{p}(1+|x|^{q}+|x^{\prime}|^{q}+|h|^{q}+|h^{\prime}|^{q})\times(|x-x^{\prime}|^{p}+|h-h^{\prime}|^{p}+|t-t^{\prime}|^{\frac{p}{2}}).\end{split}$

The existence of a continuous derivative of $Y_{s}^{t,x}$ with respect to $x$,
and a mean-square derivative of $Z_{s}^{t,x}$ with respect to $x$ follow from
this estimation. And the existence of a continuous second derivative of
$Y_{s}^{t,x}$ with respect to $x$ can be proved in a similar scheme. Using
similar arguments as in the proof of Theorem 2.9 in [18], we can show that
$\\{Y_{s}^{t,x};(s,t)\in[0,T]^{2},x\in R^{m}\\}$ has an a.s. continuous
version. ∎

Proof of Proposition 4.2: For any random variable $F$ of the form
$F=f(\varphi,B(h_{1}),\cdots,$ $B(h_{n}))$ with $f\in C_{0}^{\infty}(R^{n})$,
$\varphi\in L_{\mathcal{F}_{T}^{\alpha}}^{2}$, $h_{1},\cdots,h_{n}\in
L^{2}_{\mathcal{F}_{t}}(0,T;R^{d})$ and $B(h_{i})=\int_{0}^{T}h_{i}(t)dB_{t}$,
where $\mathcal{F}_{t}=\mathcal{F}^{\alpha}_{t,T}\vee\mathcal{F}_{t}^{B}$, let
$\displaystyle D_{t}F=\sum_{i=1}^{n}f^{\prime}_{i}(B(h_{1}),\cdots,$
$B(h_{n}))h_{i}(t),$ $0\leq t\leq T$. For such $F$, we define its norm as

$\|F\|_{1,2}=\left(E\left(F^{2}+\int_{0}^{T}|D_{t}F|^{2}dt\right)\right)^{\frac{1}{2}}.$

Denote $S$ as the set of random variables of the above form, we can define
sobolev space: $D^{1,2}=\bar{S}^{\|\cdot\|_{1,2}}.$ Using the same argument in
Proposition 2.3 in [19], we can obtain the result.∎

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