text:
\section{Introduction}

The leader-follower game involves two players with \emph{asymmetric} roles,
one called the leader and the other called the follower. In the game,
the leader first announces her action, and the follower, according to the
leader's action, chooses an optimal response to minimize his cost functional.
Next, the leader has to take the follower's optimal response into account and
chooses an optimal action to minimize her cost functional. Yong \cite{Yong2002}
first considered a linear quadratic (LQ) leader-follower stochastic differential
game. Then, within the LQ framework, the result was extended by, e.g.,
\cite{SWX2016,Moon2021,LXY2021} in different settings.

Mean-field stochastic differential equations (SDEs) were initially suggested to
describe physical systems involving a \emph{large} number of \emph{interacting}
particles. In the dynamics of a mean-field SDE, one replaces the interactions
of all the particles by their \emph{average} or \emph{mean} to reduce the complexity
of the problem. In the last decade, since Buchdahn et al. \cite{BDLP2009,BLP2009}
and Carmona and Delarue \cite{CD2013-1,CD2013-2,CD2015} introduced the mean-field
backward SDEs (BSDEs) and mean-field forward-backward SDEs (FBSDEs), optimal control
problems, especially stochastic maximum principles (SMPs), for mean-field systems
have become a popular topic; see, for example,
\cite{Li2012,Yong2013,WZZ2014,CDL2016,CZ2016,CD2018,ZSX2018,LSX2019,ABC2019,WangWu2022}.

Another feature of this paper is the use of a regime switching model, in which
the \emph{continuous} state of the LQ problem and the \emph{discrete} state of
the Markov chain coexist; see
\cite{Zhang2001,ZhouYin2003,SSZ2011,Zhu2011,ZLX2021,LWZ2022,LX2022}
for more information and applications of regime switching models.
Recently, Nguyen, Yin, and Hoang \cite{NYH2020} established the law of large numbers
for systems with regime switching and mean-field interactions, where the mean-field
limit was characterized as the \emph{conditional expectation} of the solution to
a conditional mean-field SDE with regime switching (see also Remark \ref{mean field limit}).
This work paves the way for treating mean-field optimal control problems with
regime switching; see \cite{NNY2020,NYN2021,BDTY2020,JLSY}.

In this paper, we consider an LQ leader-follower stochastic differential game
for mean-field switching diffusions. Based on the SMP in Nguyen, Nguyen, and
Yin \cite{NNY2020}, an open-loop optimal control for the follower is obtained.
Then, by applying the \emph{four-step scheme} developed by Ma, Protter, and Yong
\cite{MPY1994}, we derive its (anticipating) state feedback representation
in terms of two Riccati equations and an auxiliary BSDE. Knowing the follower's
optimal control, the leader faces a state equation which is a conditional mean-field
FBSDE with regime switching. We also utilize the SMP to obtain an open-loop optimal
control for the leader. Then, by the \emph{dimensional augmentation approach}
in Yong \cite{Yong2002}, a \emph{non-anticipating} state feedback representation
is derived in terms of two Riccati equations. As a consequence, the follower's
optimal control can be also represented in a non-anticipating way.

The rest of this paper is organized as follows. In the next, we present an example
which motivates us to study the leader-follower problem in this paper.
In Section \ref{Problem formulation}, we formulate the problem and provide some
preliminary results. In Sections \ref{follower problem} and \ref{leader problem},
we solve the LQ problems for the follower and the leader, respectively. Finally,
Section \ref{conclusion} concludes the paper.

\emph{Motivation: a pension fund optimization problem}.
Typically, in a defined benefit (DB) scheme pension fund there are two participants
who make contributions: one is the leader (such as the company) with contribution
rate $u_{2}(\cdot)$, the other one is the follower (such as the individual) with
contribution rate $u_{1}(\cdot)$. The dynamics of the pension fund is described as
\begin{equation*}
\begin{aligned}
dF(t)=F(t)d\Delta(t)+\{u_{1}(t)+u_{2}(t)-\xi_{0}\}dt,
\end{aligned}
\end{equation*}
where $d\Delta(t)$ is the return rate of the fund and $\xi_{0}$ is the
pension scheme benefit outgo. The pension fund is invested in a bond
$S_{0}(t)$ and a stock $S(t)$, which are given by
\begin{equation*}
\left\{
\begin{aligned}
dS_{0}(t)=&r(\alpha(t))S_{0}(t)dt,\\
dS(t)=&b(\alpha(t))S(t)dt+\sigma(\alpha(t))S(t)dW(t),
\end{aligned}
\right.
\end{equation*}
where $r(i)$ is the interest rate, $b(i)$ is the appreciation rate,
and $\sigma(i)$ is the volatility corresponding to the market regime
$\alpha(t)=i$. Assume the proportions $\pi(\cdot)$ and $1-\pi(\cdot)$
of the fund are to be allocated in the stock and the bond, respectively.
Then we have
\begin{equation*}
\begin{aligned}
d\Delta(t)=\{r(\alpha(t))+[b(\alpha(t))-r(\alpha(t))]\pi(t)\}dt
+\sigma(\alpha(t))\pi(t)dW(t).
\end{aligned}
\end{equation*}
Therefore, the dynamics of the pension fund can be written as
\begin{equation*}
\begin{aligned}
dF(t)=\{r(\alpha(t))&F(t)+[b(\alpha(t))-r(\alpha(t))]\pi(t)F(t)\\
+&u_{1}(t)+u_{2}(t)-\xi_{0}\}dt+\sigma(\alpha(t))\pi(t)F(t)dW(t).
\end{aligned}
\end{equation*}
The cost functionals for the follower and the leader to minimize
are defined as
\begin{equation*}
\begin{aligned}
J_{k}(u_{1}(\cdot),u_{2}(\cdot))
=\frac{1}{2}E\bigg[\int_{0}^{T}\Big(u_{k}(t)-\xi_{k}\Big)^{2}dt
+\Big(E[F(T)|\mathcal{F}_{T}^{\alpha}]-\xi_{T}\Big)^{2}\bigg],\quad k=1,2,
\end{aligned}
\end{equation*}
respectively, where $\xi_{k}$, $k=1,2$, are the running benchmark, and $\xi_{T}$
is the terminal wealth target; both are introduced to measure the stability and
performance of the pension scheme.

The above pension fund optimization problem formulates naturally a special case of
the LQ leader-follower game considered in this paper. For more pension fund
optimization problems under various contexts, see \cite{JR2001,HWX2009,ZhengShi2020};
for a conditional mean-variance portfolio selection problem (as an application of
conditional mean-field control theory), see \cite{NYN2021}.

\section{Problem formulation and preliminaries}\label{Problem formulation}

Let $R^{n}$ be the $n$-dimensional Euclidean space with Euclidean norm $|\cdot|$
and Euclidean inner product $\langle\cdot,\cdot\rangle$. Let $R^{n\times m}$ be
the space of all $(n\times m)$ matrices. $A^{\top}$ denotes the transpose of
a vector or matrix $A$.
$I_{n}$ denotes the $(n\times n)$ identity matrix.

Let $[0,T]$ be a finite time horizon and $(\Omega,\mathcal{F},P)$ be a fixed
probability space on which a one-dimensional standard Brownian motion $W(t)$,
$t\in [0,T]$, and a Markov chain $\alpha(t)$, $t\in [0,T]$, are defined.
The Markov chain $\alpha(\cdot)$ takes values in a finite state space $\mathcal{M}$.
Let $Q=(\lambda_{ij})_{i,j\in\mathcal{M}}$ be the generator (i.e., the matrix of
transition rates) of $\alpha(\cdot)$ with $\lambda_{ij}\geq 0$ for $i\neq j$ and
$\sum_{j\in \mathcal{M}}\lambda_{ij}=0$ for each $i\in \mathcal{M}$. Assume that
$W(\cdot)$ and $\alpha(\cdot)$ are independent. For $t\geq0$, denote
$\mathcal{F}^{\alpha}_{t}=\sigma\{\alpha(s):0\leq s\leq t\}$
and $\mathcal{F}_{t}=\sigma\{W(s),\alpha(s):0\leq s\leq t\}$.
Let $\mathcal{L}_{\mathcal{F}}^{2}(R^{n})$ be the set of all $R^{n}$-valued
$\mathcal{F}_{t}$-adapted processes $x(\cdot)$ on $[0,T]$ such that
$E\int_{0}^{T}|x(t)|^{2}dt<\infty$.

The state of the system is described by the following linear conditional
mean-field SDE with regime switching on $[0,T]$:
\begin{equation}\label{system}
\left\{
\begin{aligned}
dx(t)=&\Big[A(\alpha(t))x(t)+\widehat{A}(\alpha(t))E[x(t)|\mathcal{F}_{t}^{\alpha}]
+B_{1}(\alpha(t))u_{1}(t)+B_{2}(\alpha(t))u_{2}(t)\Big]dt\\
+&\Big[C(\alpha(t))x(t)+\widehat{C}(\alpha(t))E[x(t)|\mathcal{F}_{t}^{\alpha}]
+D_{1}(\alpha(t))u_{1}(t)+D_{2}(\alpha(t))u_{2}(t)\Big]dW(t),\\
x(0)=&x_{0},
\end{aligned}
\right.
\end{equation}
where $x(\cdot)$ is the state process with values in $R^{n}$, $u_{1}(\cdot)$ and
$u_{2}(\cdot)$ are control processes taken by the follower and the leader, with
values in $R^{m_{1}}$ and $R^{m_{2}}$, respectively, and $A(i)$, $\widehat{A}(i)$,
$B_{1}(i)$, $B_{2}(i)$, $C(i)$, $\widehat{C}(i)$, $D_{1}(i)$, $D_{2}(i)$,
$i\in\mathcal{M}$, are constant matrices of suitable dimensions. It follows from
Nguyen et al. \cite{NNY2020,NYN2021} that,
for any $u_{1}(\cdot)\in \mathcal{L}_{\mathcal{F}}^{2}(R^{m_{1}})$
and $u_{2}(\cdot)\in \mathcal{L}_{\mathcal{F}}^{2}(R^{m_{2}})$,
SDE (\ref{system}) admits a unique solution
$x(\cdot)\in \mathcal{L}_{\mathcal{F}}^{2}(R^{n})$.
Then, $\mathcal{U}_{1}\doteq \mathcal{L}_{\mathcal{F}}^{2}(R^{m_{1}})$
and $\mathcal{U}_{2}\doteq \mathcal{L}_{\mathcal{F}}^{2}(R^{m_{2}})$
are called the admissible control sets for the follower and the leader, respectively.

The cost functionals for the follower and the leader to minimize are defined as
\begin{equation}\label{cost functionals}
\begin{aligned}
J_{k}(u_{1}(\cdot),u_{2}(\cdot))
=&\frac{1}{2}E\bigg[\int_{0}^{T}\bigg(\Big\langle Q_{k}(\alpha(t))x(t),x(t)\Big\rangle
+\Big\langle \widehat{Q}_{k}(\alpha(t))E[x(t)|\mathcal{F}_{t}^{\alpha}],E[x(t)|\mathcal{F}_{t}^{\alpha}]\Big\rangle\\
&+\Big\langle N_{k}(\alpha(t))u_{k}(t),u_{k}(t)\Big\rangle\bigg)dt+\Big\langle G_{k}(\alpha(T))x(T),x(T)\Big\rangle\\
&+\Big\langle \widehat{G}_{k}(\alpha(T))E[x(T)|\mathcal{F}_{T}^{\alpha}],E[x(T)|\mathcal{F}_{T}^{\alpha}]\Big\rangle\bigg],
\quad k=1,2,
\end{aligned}
\end{equation}
respectively, where $Q_{k}(i)$, $\widehat{Q}_{k}(i)$, $N_{k}(i)$, $G_{k}(i)$, $\widehat{G}_{k}(i)$,
$k=1,2$, $i\in\mathcal{M}$, are constant symmetric matrices of suitable dimensions.
\begin{remark}\label{mean field limit}
In fact, SDE (\ref{system}) is obtained as the mean-square limit as $N\rightarrow\infty$
of a system of interacting particles in the following form:
\begin{equation*}
\left\{
\begin{aligned}
dx^{l,N}(t)=&\bigg[A(\alpha(t))x^{l,N}(t)+\widehat{A}(\alpha(t))\frac{1}{N}\sum_{l=1}^{N}x^{l,N}(t)
+B_{1}(\alpha(t))u_{1}(t)+B_{2}(\alpha(t))u_{2}(t)\bigg]dt\\
+\bigg[C(&\alpha(t))x^{l,N}(t)+\widehat{C}(\alpha(t))\frac{1}{N}\sum_{l=1}^{N}x^{l,N}(t)
+D_{1}(\alpha(t))u_{1}(t)+D_{2}(\alpha(t))u_{2}(t)\bigg]dW^{l}(t),\\
x^{l,N}(0)=&x_{0},\quad 1\leq l\leq N,
\end{aligned}
\right.
\end{equation*}
where $\{W^{l}(\cdot)\}_{l=1}^{N}$ is a collection of independent standard Brownian motions
and the Markov chain $\alpha(\cdot)$ serves as a common noise for all particles, which leads
to the conditional expectations rather than expectations in (\ref{system}).

Intuitively, since all the particles depend on the history of $\alpha(\cdot)$, their average
and thereby its limit as $N\rightarrow\infty$ should also depend on this process. This intuition
has been justified by the law of large numbers established by Nguyen et al. \cite[Theorem 2.1]{NYH2020},
which shows that the joint process $(\frac{1}{N}\sum_{l=1}^{N}x^{l,N}(\cdot),\alpha(\cdot))$
converges weakly to a process $(\mu_{\alpha}(\cdot),\alpha(\cdot))$, where
$(\mu_{\alpha}(t),\alpha(t))=(E[x(t)|\mathcal{F}_{t}^{\alpha}],\alpha(t))$, $0\leq t\leq T$,
and $x(\cdot)$ is exactly the solution of (\ref{system}).
\end{remark}
\begin{remark}
Note that the cost functionals $J_{k}$, $k=1,2$, defined by (\ref{cost functionals})
are standard in the LQ mean-field control literature (see \cite{Yong2013,NNY2020,NYN2021})
and, if we assume the Assumptions (A1) and (A2) given in Sections \ref{follower problem}
and \ref{leader problem} hold, then $J_{k}$ is convex with respect to $u_{k}$, $k=1,2$,
respectively. However, for LQ mean-field games of large-population systems, the tracking-type
cost functionals where one wants to keep the system states stay as much close as possible
to a function of the mean-field term are more frequently adopted (see \cite{HCM2007,LZ2008,CZ2016}).
\end{remark}
Now we explain the leader-follower feature of the game; see also Yong \cite{Yong2002}.
In the game, for any $u_{2}(\cdot)\in \mathcal{U}_{2}$ of the leader, the follower
would like to choose an optimal control $u_{1}^{*}(\cdot)\in \mathcal{U}_{1}$ so that
$J_{1}(u_{1}^{*}(\cdot),u_{2}(\cdot))$ achieves the minimum of $J_{1}(u_{1}(\cdot),u_{2}(\cdot))$
over $u_{1}(\cdot)\in \mathcal{U}_{1}$. Knowing the follower's optimal control $u_{1}^{*}(\cdot)$
(depending on $u_{2}(\cdot)$), the leader would like to choose an optimal control
$u_{2}^{*}(\cdot)\in \mathcal{U}_{2}$ to minimize $J_{2}(u_{1}^{*}(\cdot),u_{2}(\cdot))$
over $u_{2}(\cdot)\in \mathcal{U}_{2}$.

In a more rigorous way, the follower wants to find an optimal map
$\Pi_{1}^{*}:\mathcal{U}_{2}\mapsto\mathcal{U}_{1}$ and the leader
wants to find an optimal control $u_{2}^{*}(\cdot)\in \mathcal{U}_{2}$ such that
\begin{equation*}
\left\{
\begin{aligned}
J_{1}(\Pi_{1}^{*}[u_{2}(\cdot)](\cdot),u_{2}(\cdot))
=&\inf_{u_{1}(\cdot)\in \mathcal{U}_{1}}J_{1}(u_{1}(\cdot),u_{2}(\cdot)),
\quad\forall u_{2}(\cdot)\in\mathcal{U}_{2},\\
J_{2}(\Pi_{1}^{*}[u_{2}^{*}(\cdot)](\cdot),u_{2}^{*}(\cdot))
=&\inf_{u_{2}(\cdot)\in \mathcal{U}_{2}}J_{2}(\Pi_{1}^{*}[u_{2}(\cdot)](\cdot),u_{2}(\cdot)).
\end{aligned}
\right.
\end{equation*}
If the above optimal pair $(\Pi_{1}^{*}[\cdot],u_{2}^{*}(\cdot))$ exists, it is called
an \emph{open-loop} Stackelberg equilibrium of the leader-follower stochastic differential game.

Then we present some preliminary results on the martingales associated with a Markov chain,
which are needed to establish the conditional mean-field BSDEs with regime switching.
For each pair $(i,j)\in \mathcal{M}\times \mathcal{M}$ with $i\neq j$, define
$[M_{ij}](t)=\sum_{0\leq s\leq t}1_{\{\alpha(s-)=i\}}1_{\{\alpha(s)=j\}}$
and $\langle M_{ij}\rangle(t)=\int_{0}^{t}\lambda_{ij}1_{\{\alpha(s-)=i\}}ds$,
where $1_{A}$ denotes the indicator function of a set $A$.
It follows from \cite{NNY2020,NYN2021} that the process
$M_{ij}(t)\doteq[M_{ij}](t)-\langle M_{ij}\rangle(t)$
is a purely discontinuous and square-integrable martingale with respect to
$\mathcal{F}_{t}^{\alpha}$, which is null at the origin. In this sense,
$[M_{ij}](t)$ and $\langle M_{ij}\rangle(t)$ are the optional and predictable
quadratic variations of $M_{ij}(t)$, respectively. In addition, we denote
$M_{ii}(t)=[M_{ii}](t)=\langle M_{ii}\rangle(t)\equiv0$ for each $i\in\mathcal{M}$.

Let $\mathcal{S}_{\mathcal{F}}^{2}(R^{n})$ be the set of all $R^{n}$-valued
$\mathcal{F}_{t}$-adapted c\`{a}dl\`{a}g processes $y(\cdot)$ on $[0,T]$ such that
$E\int_{0}^{T}|y(t)|^{2}dt<\infty$. Let $\mathcal{K}_{\mathcal{F}}^{2}(R^{n})$
be the set of all collections of $R^{n}$-valued $\mathcal{F}_{t}$-adapted processes
$\{k_{ij}(\cdot)\}_{i,j\in \mathcal{M}}$ on $[0,T]$ such that
$\sum_{i,j\in \mathcal{M}}E\int_{0}^{T}|k_{ij}(t)|^{2}d[M_{ij}](t)<\infty$
with $k_{ii}(t)\equiv0$ for each $i\in \mathcal{M}$. For convenience, we also denote
$k(\cdot)=\{k_{ij}(\cdot)\}_{i,j\in \mathcal{M}}$ and
\begin{equation*}
\begin{aligned}
\int_{0}^{t}k(s)\bullet dM(s)=\sum_{i,j\in \mathcal{M}}\int_{0}^{t}k_{ij}(s)dM_{ij}(s),\quad
k(s)\bullet dM(s)=\sum_{i,j\in \mathcal{M}}k_{ij}(s)dM_{ij}(s).
\end{aligned}
\end{equation*}
The following two lemmas play an important role in the subsequent analysis.
The proof of the first lemma is elementary and the proof of the second one
is similar to that of Xiong \cite[Lemma 5.4]{Xiong2008}. For completeness
and readers' convenience, their proofs are provided here.
\begin{lemma}\label{Lemma conditional E}
For any $\mathcal{F}_{t}$-adapted and square-integrable processes
$x(\cdot)$ and $y(\cdot)$, we have
\begin{equation*}
\begin{aligned}
E\Big[x(t)E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big]
=E\Big[E[x(t)|\mathcal{F}_{t}^{\alpha}]y(t)\Big]
=E\Big[E[x(t)|\mathcal{F}_{t}^{\alpha}]E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big].
\end{aligned}
\end{equation*}
\end{lemma}
\begin{proof}
Note that
\begin{equation*}
\begin{aligned}
E\Big[x(t)E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big]
=E\Big[E\Big(x(t)E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big|\mathcal{F}_{t}^{\alpha}\Big)\Big]
=E\Big[E[x(t)|\mathcal{F}_{t}^{\alpha}]E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big].
\end{aligned}
\end{equation*}
Similarly,
\begin{equation*}
\begin{aligned}
E\Big[E[x(t)|\mathcal{F}_{t}^{\alpha}]y(t)\Big]
=E\Big[E[x(t)|\mathcal{F}_{t}^{\alpha}]E[y(t)|\mathcal{F}_{t}^{\alpha}]\Big].
\end{aligned}
\end{equation*}
Consequently, the desired conclusion follows.
\end{proof}
\begin{lemma}\label{Lemma filtering}
For any $\mathcal{F}_{t}$-adapted and square-integrable process $x(\cdot)$,
we have
\begin{equation*}
\begin{aligned}
E\bigg[\int_{0}^{t}x(s)ds\bigg|\mathcal{F}_{t}^{\alpha}\bigg]
=\int_{0}^{t}E[x(s)|\mathcal{F}_{s}^{\alpha}]ds,
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
E\bigg[\int_{0}^{t}x(s)dW(s)\bigg|\mathcal{F}_{t}^{\alpha}\bigg]=0.
\end{aligned}
\end{equation*}
\end{lemma}
\begin{proof}
For the first equation, from the Markov property of $\alpha(\cdot)$
and the independence of $W(\cdot)$ and $\alpha(\cdot)$,
it follows that
\begin{equation*}
\begin{aligned}
E\bigg[\int_{0}^{t}x(s)ds\bigg|\mathcal{F}_{t}^{\alpha}\bigg]
=\int_{0}^{t}E[x(s)|\mathcal{F}_{t}^{\alpha}]ds
=\int_{0}^{t}E[x(s)|\mathcal{F}_{s}^{\alpha}]ds.
\end{aligned}
\end{equation*}
Now we proceed to prove the second equation. We first suppose $x(\cdot)$ is simple, namely
\begin{equation*}
\begin{aligned}
x(s)=\sum_{m\geq1}x_{m}1_{[t_{m},t_{m+1})}(s),
\end{aligned}
\end{equation*}
where, for each $m\geq1$, $x_{m}$ is an $\mathcal{F}_{t_{m}}$-measurable random variable.
As $W(t_{m+1})-W(t_{m})$ is independent of
$\mathcal{F}_{t}^{\alpha}\vee\sigma(x_{m})\doteq\sigma(\mathcal{F}_{t}^{\alpha}\cup\sigma(x_{m}))$,
we have
\begin{equation*}
\begin{aligned}
E\bigg[\int_{0}^{t}x(s)dW(s)\bigg|\mathcal{F}_{t}^{\alpha}\bigg]
=&\sum_{m\geq1}E\bigg[x_{m}[W(t_{m+1})-W(t_{m})]\bigg|\mathcal{F}_{t}^{\alpha}\bigg]\\
=&\sum_{m\geq1}E\bigg[x_{m}E\bigg(W(t_{m+1})-W(t_{m})
\bigg|\mathcal{F}_{t}^{\alpha}\vee\sigma(x_{m})\bigg)\bigg|\mathcal{F}_{t}^{\alpha}\bigg]=0.
\end{aligned}
\end{equation*}
For general $x(\cdot)$, we can approximate $x(\cdot)$ by a sequence of simple processes
$\{x_{n}(\cdot):n\geq1\}$ such that $|x_{n}(s)|\leq|x(s)|$,
a.s., for each $n\geq1$ and all $s\leq t$. Note that
\begin{equation*}
\begin{aligned}
E\bigg[\bigg|\int_{0}^{t}x_{n}(s)dW(s)\bigg|^{2}\bigg]
=E\bigg[\int_{0}^{t}|x_{n}(s)|^{2}ds\bigg]
\leq E\bigg[\int_{0}^{t}|x(s)|^{2}ds\bigg]<\infty,
\end{aligned}
\end{equation*}
which implies that $\{\int_{0}^{t}x_{n}(s)dW(s):n\geq1\}$ is uniformly integrable. Therefore,
\begin{equation*}
\begin{aligned}
E\bigg[\int_{0}^{t}x(s)dW(s)\bigg|\mathcal{F}_{t}^{\alpha}\bigg]
=\lim_{n\rightarrow\infty}
E\bigg[\int_{0}^{t}x_{n}(s)dW(s)\bigg|\mathcal{F}_{t}^{\alpha}\bigg]
=0.
\end{aligned}
\end{equation*}
This completes the proof.
\end{proof}

\section{The problem for the follower}\label{follower problem}

In this section, we deal with the problem for the follower.
For convenience, we denote
\begin{equation*}
\begin{aligned}
\widehat{\phi}(t)=E[\phi(t)|\mathcal{F}_{t}^{\alpha}],
\end{aligned}
\end{equation*}
for a process $\phi(\cdot)$. We will apply the SMP obtained by Nguyen et al.
\cite[Theorem 3.7]{NNY2020} to solve the follower's problem. Besides the
open-loop optimal control, we would like further to find its state feedback
representation. We make the following assumption:

(A1) $Q_{1}(i)\geq0$, $\widehat{Q}_{1}(i)\geq0$, $N_{1}(i)>0$,
$G_{1}(i)\geq0$, $\widehat{G}_{1}(i)\geq0$, $i\in\mathcal{M}$.
\begin{lemma}
Let Assumption (A1) hold. For any given $u_{2}(\cdot)\in\mathcal{U}_{2}$
for the leader, let $u_{1}^{*}(\cdot)$ be an optimal control for the follower,
then $u_{1}^{*}(\cdot)$ should have the following form:
\begin{equation}\label{follower optimal control}
\begin{aligned}
u_{1}^{*}(t)=-\widetilde{N}_{1}^{-1}(t,\alpha(t))
\Big[S_{1}(t,\alpha(t))x(t)+\widehat{S}_{1}(t,\alpha(t))\widehat{x}(t)+\Phi(t)\Big],
\end{aligned}
\end{equation}
where, for notational simplicity, we denote
\begin{equation*}
\begin{aligned}
&\widetilde{N}_{1}(t,i)=N_{1}(i)+D_{1}^{\top}(i)P_{1}(t,i)D_{1}(i),\\
&S_{1}(t,i)=B_{1}^{\top}(i)P_{1}(t,i)+D_{1}^{\top}(i)P_{1}(t,i)C(i),\quad
\widehat{S}_{1}(t,i)=B_{1}^{\top}(i)\widehat{P}_{1}(t,i)+D_{1}^{\top}(i)P_{1}(t,i)\widehat{C}(i),\\
&\Phi(t)=B_{1}^{\top}(\alpha(t))\varphi(t)+D_{1}^{\top}(\alpha(t))\theta(t)
+D_{1}^{\top}(\alpha(t))P_{1}(t,\alpha(t))D_{2}(\alpha(t))u_{2}(t),\quad i\in\mathcal{M},
\end{aligned}
\end{equation*}
and $P_{1}(\cdot,i)$ and $\widehat{P}_{1}(\cdot,i)$, $i\in\mathcal{M}$, are the solutions
of Riccati equations (\ref{follower 1}) and (\ref{follower 2}), respectively, and
$(\varphi(\cdot),\theta(\cdot),\eta(\cdot))\in \mathcal{S}_{\mathcal{F}}^{2}(R^{n})
\times \mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times \mathcal{K}_{\mathcal{F}}^{2}(R^{n})$
is the solution of BSDE (\ref{follower 3}).
\end{lemma}
\begin{proof}
From \cite[Theorem 3.7]{NNY2020}, the adjoint equation for the follower is given by
\begin{equation}\label{follower adjoint equation}
\left\{
\begin{aligned}
dp(t)=&-\Big[A^{\top}(\alpha(t))p(t)+\widehat{A}^{\top}(\alpha(t))\widehat{p}(t)
+C^{\top}(\alpha(t))q(t)+\widehat{C}^{\top}(\alpha(t))\widehat{q}(t)\\
&+Q_{1}(\alpha(t))x(t)+\widehat{Q}_{1}(\alpha(t))\widehat{x}(t)\Big]dt
+q(t)dW(t)+r(t)\bullet dM(t),\\
p(T)=&G_{1}(\alpha(T))x(T)+\widehat{G}_{1}(\alpha(T))\widehat{x}(T),
\end{aligned}
\right.
\end{equation}
which, from \cite[Theorem 3.4]{NNY2020}, admits a unique solution
$(p(\cdot),q(\cdot),r(\cdot))\in\mathcal{S}_{\mathcal{F}}^{2}(R^{n})
\times \mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times \mathcal{K}_{\mathcal{F}}^{2}(R^{n})$,
and an optimal control $u_{1}^{*}(\cdot)$ for the follower should satisfy
\begin{equation}\label{follower open loop}
\begin{aligned}
N_{1}(\alpha(t))u_{1}^{*}(t)+B_{1}^{\top}(\alpha(t))p(t)
+D_{1}^{\top}(\alpha(t))q(t)=0.
\end{aligned}
\end{equation}
Inspired by the terminal condition of the adjoint equation
(\ref{follower adjoint equation}), it is natural to guess
\begin{equation}\label{follower FSS}
\begin{aligned}
p(t)=P_{1}(t,\alpha(t))x(t)
+\widehat{P}_{1}(t,\alpha(t))\widehat{x}(t)+\varphi(t),
\end{aligned}
\end{equation}
for some $R^{n\times n}$-valued deterministic, differentiable, and symmetric
functions $P_{1}(t,i)$ and  $\widehat{P}_{1}(t,i)$,\  $i\in \mathcal{M}$,
and an $R^{n}$-valued $\mathcal{F}_{t}$-adapted process $\varphi(t)$ with
\begin{equation*}
\begin{aligned}
d\varphi(t)=\gamma(t)dt+\theta(t)dW(t)+\eta(t)\bullet dM(t).
\end{aligned}
\end{equation*}
Then,
\begin{equation}\label{follower p hat}
\begin{aligned}
\widehat{p}(t)=\Big(P_{1}(t,\alpha(t))
+\widehat{P}_{1}(t,\alpha(t))\Big)\widehat{x}(t)
+\widehat{\varphi}(t).
\end{aligned}
\end{equation}
From Lemma \ref{Lemma filtering}, we have
\begin{equation*}
\begin{aligned}
d\widehat{x}(t)=&\Big[\Big(A(\alpha(t))+\widehat{A}(\alpha(t))\Big)\widehat{x}(t)
+B_{1}(\alpha(t))\widehat{u}_{1}(t)+B_{2}(\alpha(t))\widehat{u}_{2}(t)\Big]dt.
\end{aligned}
\end{equation*}
In the rest of this paper, the arguments $t$ and $\alpha(t)$ will be dropped to save space,
if needed and when no confusion arises.	Applying It\^{o}'s formula for Markov-modulated
processes (see Zhou and Yin \cite[Lemma 3.1]{ZhouYin2003}) to (\ref{follower FSS}), we obtain
\begin{equation}\label{follower dp}
\begin{aligned}
dp=&\bigg(\dot{P}_{1}+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[P_{1}(t,j)-P_{1}(t,\alpha(t))]\bigg)xdt
+\sum_{i,j\in\mathcal{M}}[P_{1}(t,j)-P_{1}(t,i)]xdM_{ij}\\
&+P_{1}[Ax+\widehat{A}\widehat{x}+B_{1}u_{1}+B_{2}u_{2}]dt+P_{1}[Cx+\widehat{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}]dW\\
&+\bigg(\dot{\widehat{P}}_{1}
+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[\widehat{P}_{1}(t,j)-\widehat{P}_{1}(t,\alpha(t))]\bigg)\widehat{x}dt
+\sum_{i,j\in\mathcal{M}}[\widehat{P}_{1}(t,j)-\widehat{P}_{1}(t,i)]\widehat{x}dM_{ij}\\
&+\widehat{P}_{1}\Big[(A+\widehat{A})\widehat{x}+B_{1}\widehat{u}_{1}+B_{2}\widehat{u}_{2}\Big]dt
+\gamma dt+\theta dW+\eta\bullet dM.
\end{aligned}
\end{equation}
Comparing the coefficients of $dW$ parts in (\ref{follower adjoint equation})
and (\ref{follower dp}), it follows that
\begin{equation}\label{follower q}
\begin{aligned}
q=P_{1}\Big[Cx+\widehat{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}\Big]+\theta,
\end{aligned}
\end{equation}
and then,
\begin{equation}\label{follower q hat}
\begin{aligned}
\widehat{q}=P_{1}\Big[(C+\widehat{C})\widehat{x}+D_{1}\widehat{u}_{1}+D_{2}\widehat{u}_{2}\Big]+\widehat{\theta}.
\end{aligned}
\end{equation}
Inserting (\ref{follower FSS}) and (\ref{follower q}) into (\ref{follower open loop}) yields
\begin{equation*}
\begin{aligned}
0=&\Big(N+D_{1}^{\top}P_{1}D_{1}\Big)u_{1}^{*}
+\Big(B_{1}^{\top}P_{1}+D_{1}^{\top}P_{1}C\Big)x
+\Big(B_{1}^{\top}\widehat{P}_{1}+D_{1}^{\top}P_{1}\widehat{C}\Big)\widehat{x}\\
&+B_{1}^{\top}\varphi+D_{1}^{\top}\theta+D_{1}^{\top}P_{1}D_{2}u_{2},
\end{aligned}
\end{equation*}
i.e., $u_{1}^{*}=-\widetilde{N}_{1}^{-1}[S_{1}x+\widehat{S}_{1}\widehat{x}+\Phi]$,
provided $\widetilde{N}_{1}$ is invertible. So we have (\ref{follower optimal control}). Also,
\begin{equation}\label{follower optimal control hat}
\begin{aligned}
\widehat{u}_{1}^{*}=-\widetilde{N}_{1}^{-1}\Big[(S_{1}+\widehat{S}_{1})\widehat{x}+\widehat{\Phi}\Big].
\end{aligned}
\end{equation}
On the one hand, substituting (\ref{follower FSS}), (\ref{follower p hat}),
(\ref{follower q}), (\ref{follower q hat}), and (\ref{follower optimal control}),
(\ref{follower optimal control hat}) into (\ref{follower adjoint equation}),
we have
\begin{equation}\label{follower comparison 1}
\begin{aligned}
dp=&-\Big[\Big(A^{\top}P_{1}+C^{\top}P_{1}C-C^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}S_{1}+Q_{1}\Big)x\\
&+\Big(\widehat{A}^{\top}P_{1}+(A+\widehat{A})^{\top}\widehat{P}_{1}
+C^{\top}P_{1}\widehat{C}+\widehat{C}^{\top}P_{1}C+\widehat{C}^{\top}P_{1}\widehat{C}\\
&-C^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}
-\widehat{C}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}(S_{1}+\widehat{S}_{1})+\widehat{Q}_{1}\Big)\widehat{x}\\
&+\Big(A-B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}C\Big)^{\top}\varphi
+\Big(\widehat{A}-B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}\widehat{C}\Big)^{\top}\widehat{\varphi}\\
&+\Big(C-D_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}C\Big)^{\top}\theta
+\Big(\widehat{C}-D_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}\widehat{C}\Big)^{\top}\widehat{\theta}\\
&+\Big(C^{\top}P_{1}D_{2}-C^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}D_{2}\Big)u_{2}\\
&+\Big(\widehat{C}^{\top}P_{1}D_{2}
-\widehat{C}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}D_{2}\Big)\widehat{u}_{2}\Big]dt\\
&+qdW+r\bullet dM.
\end{aligned}
\end{equation}
On the other hand, substituting (\ref{follower optimal control})
and \eqref{follower optimal control hat} into (\ref{follower dp}),
we have
\begin{equation}\label{follower comparison 2}
\begin{aligned}
dp=&\Big[\Big(\dot{P}_1+P_{1}A-P_{1}B_{1}\widetilde{N}_{1}^{-1}S_{1}
+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[P_{1}(t,j)-P_{1}(t,\alpha(t))]\Big)x\\
&+\Big(\dot{\widehat{P}}_1+P_{1}\widehat{A}+\widehat{P}_{1}(A+\widehat{A})
-P_{1}B_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}
-\widehat{P}_{1}B_{1}\widetilde{N}_{1}^{-1}(S_{1}+\widehat{S}_{1})\\
&+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[\widehat{P}_{1}(t,j)-\widehat{P}_{1}(t,\alpha(t))]\Big)\widehat{x}\\
&+\gamma-P_{1}B_{1}\widetilde{N}_{1}^{-1}B_{1}^{\top}\varphi
-\widehat{P}_{1}B_{1}\widetilde{N}_{1}^{-1}B_{1}^{\top}\widehat{\varphi}
-P_{1}B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}\theta
-\widehat{P}_{1}B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}\widehat{\theta}\\
&+\Big(P_{1}B_{2}-P_{1}B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}D_{2}\Big)u_{2}
+\Big(\widehat{P}_{1}B_{2}-\widehat{P}_{1}B_{1}\widetilde{N}_{1}^{-1}D_{1}^{\top}P_{1}D_{2}\Big)\widehat{u}_{2}\Big]dt\\
&+\Big\{\cdots\Big\}dW+\Big\{\cdots\Big\}\bullet dM.
\end{aligned}
\end{equation}
By equalizing the coefficients of $x$ and $\widehat{x}$
as well as the non-homogeneous terms in the $dt$ parts of
(\ref{follower comparison 1}) and (\ref{follower comparison 2}),
we obtain two Riccati equations:
\begin{equation}\label{follower 1}
\left\{
\begin{aligned}
\dot{P}_{1}(t,i)
=&-\Big[P_{1}(t,i)A(i)+A^{\top}(i)P_{1}(t,i)
+C^{\top}(i)P_{1}(t,i)C(i)+Q_{1}(i)\\
&-S_{1}^{\top}(t,i)\widetilde{N}_{1}^{-1}(t,i)S_{1}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[P_{1}(t,j)-P_{1}(t,i)]\Big],\\
P_{1}(T,i)=&G_{1}(i),\quad i\in\mathcal{M},
\end{aligned}
\right.
\end{equation}
and
\begin{equation}\label{follower 2}
\left\{
\begin{aligned}
\dot{\widehat{P}}_{1}(t,i)
=&-\Big[\widehat{P}_{1}(t,i)(A(i)+\widehat{A}(i))
+(A(i)+\widehat{A}(i))^{\top}\widehat{P}_{1}(t,i)\\
&+P_{1}(t,i)\widehat{A}(i)+\widehat{A}^{\top}(i)P_{1}(t,i)
+C^{\top}(i)P_{1}(t,i)\widehat{C}(i)\\
&+\widehat{C}^{\top}(i)P_{1}(t,i)C(i)
+\widehat{C}^{\top}(i)P_{1}(t,i)\widehat{C}(i)+\widehat{Q}_{1}(i)\\
&-S_{1}^{\top}(t,i)\widetilde{N}_{1}^{-1}(t,i)\widehat{S}_{1}(t,i)
-\widehat{S}_{1}^{\top}(t,i)\widetilde{N}_{1}^{-1}(t,i)S_{1}(t,i)\\
&-\widehat{S}_{1}^{\top}(t,i)\widetilde{N}_{1}^{-1}(t,i)\widehat{S}_{1}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[\widehat{P}_{1}(t,j)-\widehat{P}_{1}(t,i)]\Big],\\
\widehat{P}_{1}(T,i)=&\widehat{G}_{1}(i),\quad i\in\mathcal{M},
\end{aligned}
\right.
\end{equation}
and an auxiliary BSDE:
\begin{equation}\label{follower 3}
\left\{
\begin{aligned}
d\varphi(t)
=&-\Big[\mathbb{A}^{\top}(t,\alpha(t))\varphi(t)+\widehat{\mathbb{A}}^{\top}(t,\alpha(t))\widehat{\varphi}(t)
+\mathbb{C}^{\top}(t,\alpha(t))\theta(t)+\widehat{\mathbb{C}}^{\top}(t,\alpha(t))\widehat{\theta}(t)\\
&+\mathbb{F}_{2}^{\top}(t,\alpha(t))u_{2}(t)+\widehat{\mathbb{F}}_{2}^{\top}(t,\alpha(t))\widehat{u}_{2}(t)\Big]dt
+\theta(t)dW(t)+\eta(t)\bullet dM(t),\\
\varphi(T)=&0,
\end{aligned}
\right.
\end{equation}
where, for simplicity of presentation, we denote
\begin{equation*}
\begin{aligned}
\mathbb{A}(t,i)=&A(i)-B_{1}(i)\widetilde{N}_{1}^{-1}(t,i)S_{1}(t,i),\quad
\widehat{\mathbb{A}}(t,i)=\widehat{A}(i)-B_{1}(i)\widetilde{N}_{1}^{-1}(t,i)\widehat{S}_{1}(t,i),\\
\mathbb{C}(t,i)=&C(i)-D_{1}(i)\widetilde{N}^{-1}(t,i)S_{1}(t,i),\quad
\widehat{\mathbb{C}}(t,i)=\widehat{C}(i)-D_{1}(i)\widetilde{N}^{-1}(t,i)\widehat{S}_{1}(t,i),\\
S_{2}(t,i)=&B_{2}^{\top}(i)P_{1}(t,i)+D_{2}^{\top}(i)P_{1}(t,i)C(i),\quad
\widehat{S}_{2}(t,i)=B_{2}^{\top}(i)\widehat{P}_{1}(t,i)+D_{2}^{\top}(i)P_{1}(t,i)\widehat{C}(i),\\
\mathbb{F}_{2}(t,i)=&S_{2}(t,i)-D_{2}^{\top}(i)P_{1}(t,i)D_{1}(i)\widetilde{N}_{1}^{-1}(t,i)S_{1}(t,i),\\
\widehat{\mathbb{F}}_{2}(t,i)=&\widehat{S}_{2}(t,i)
-D_{2}^{\top}(i)P_{1}(t,i)D_{1}(i)\widetilde{N}_{1}^{-1}(t,i)\widehat{S}_{1}(t,i),\quad i\in\mathcal{M}.
\end{aligned}
\end{equation*}
Further, let $\widetilde{P}_{1}(t,i)=P_{1}(t,i)+\widehat{P}_{1}(t,i)$,
$i\in\mathcal{M}$, then we have
\begin{equation}\label{follower 5}
\left\{
\begin{aligned}
\dot{\widetilde{P}}_{1}(t,i)
=&-\Big[\widetilde{P}_{1}(t,i)\widetilde{A}(i)+\widetilde{A}^{\top}(i)\widetilde{P}_{1}(t,i)
+\widetilde{C}^{\top}(i)P_{1}(t,i)\widetilde{C}(i)+\widetilde{Q}_{1}(i)\\
&-\widetilde{S}_{1}^{\top}(t,i)\widetilde{N}_{1}^{-1}(t,i)\widetilde{S}_{1}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[\widetilde{P}_{1}(t,j)-\widetilde{P}_{1}(t,i)]\Big],\\
\widetilde{P}_{1}(T,i)=&\widetilde{G}_{1}(i),\quad i\in\mathcal{M},
\end{aligned}
\right.
\end{equation}
where $\widetilde{\Lambda}\doteq\Lambda+\widehat{\Lambda}$ for $\Lambda=A,C,Q_{1},S_{1},G_{1}$;
so we can use (\ref{follower 5}) instead of (\ref{follower 2}).
Similar to \cite[Theorem 4.1]{Yong2013}, under Assumption (A1), (\ref{follower 1}) and (\ref{follower 5})
have unique solutions $P_{1}(\cdot,i)$ and $\widetilde{P}_{1}(\cdot,i)$, $i\in\mathcal{M}$, respectively,
which are positive definite. From \cite[Theorem 3.4]{NNY2020}, (\ref{follower 3}) also admits
a unique solution $(\varphi(\cdot),\theta(\cdot),\eta(\cdot))\in \mathcal{S}_{\mathcal{F}}^{2}(R^{n})
\times \mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times \mathcal{K}_{\mathcal{F}}^{2}(R^{n})$.
\end{proof}
\begin{remark}
Note that $P_{1}$ and $\widetilde{P}_{1}$ do not depend on $u_{2}$, whereas
$(\varphi,\theta,\eta)$ does depend on $u_{2}$. Moreover, since (\ref{follower 3})
is a BSDE, the value $(\varphi(t),\theta(t),\eta(t))$ of $(\varphi,\theta,\eta)$
at time $t$ depends on $\{u_{2}(s):s\in [0,T]\}$. Then, $\Phi$ and hence $u_{1}^{*}$
defined by (\ref{follower optimal control}) depend on $\{u_{2}(s):s\in [0,T]\}$ as well,
which means $u_{1}^{*}$ is anticipating in nature. Thus, it is important to find a ``real"
state feedback representation for $u_{1}^{*}$ only in terms of $x$ and $\widehat{x}$.
\end{remark}
In the following theorem, based on the so-called \emph{completion of the squares method},
we verify the optimality of (\ref{follower optimal control}) and compute the minimal cost
for the follower.
\begin{lemma}\label{Theorem follower}
Let Assumption (A1) hold. For any given $u_{2}(\cdot)\in\mathcal{U}_{2}$ for the leader,
$u_{1}^{*}(\cdot)$ defined by (\ref{follower optimal control}) is indeed an optimal control
for the follower, and
\begin{equation*}
\begin{aligned}
J_{1}(u_{1}^{*}(\cdot),&u_{2}(\cdot))
=\frac{1}{2}\langle \widetilde{P}_{1}(0,i)x_{0},x_{0}\rangle+\langle\varphi(0),x_{0}\rangle\\
&+\frac{1}{2}E\bigg[\int_{0}^{T}\Big(-|\widetilde{N}_{1}^{-\frac{1}{2}}\Phi|^{2}
+\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi+D_{2}^{\top}\theta,u_{2}\rangle\Big)dt\bigg].
\end{aligned}
\end{equation*}
\end{lemma}
\begin{proof}
Note that $x(0)=\widehat{x}(0)=x_{0}$, then for any $u_{1}\in\mathcal{U}_{1}$, we have
\begin{equation}\label{follower proof 1}
\begin{aligned}
&J_{1}(u_{1}(\cdot),u_{2}(\cdot))\\
=&J_{1}(u_{1}(\cdot),u_{2}(\cdot))
-\frac{1}{2}\langle P_{1}(0,i)(x(0)-\widehat{x}(0)),x(0)-\widehat{x}(0)\rangle\\
&-\frac{1}{2}\langle \widetilde{P}_{1}(0,i)\widehat{x}(0),\widehat{x}(0)\rangle
+\frac{1}{2}\langle \widetilde{P}_{1}(0,i)\widehat{x}(0),\widehat{x}(0)\rangle
-\langle\varphi(0),x(0)\rangle+\langle\varphi(0),x(0)\rangle\\
=&J_{1}(u_{1}(\cdot),u_{2}(\cdot))
+\frac{1}{2}\langle \widetilde{P}_{1}(0,i)x_{0},x_{0}\rangle
+\langle\varphi(0),x_{0}\rangle\\
&-\frac{1}{2}E\bigg[\langle P_{1}(T,\alpha(T))(x(T)-\widehat{x}(T)),x(T)-\widehat{x}(T)\rangle
-\int_{0}^{T}d\langle P_{1}(x-\widehat{x}),x-\widehat{x}\rangle\bigg]\\
&-\frac{1}{2}E\bigg[\langle \widetilde{P}_{1}(T,\alpha(T))\widehat{x}(T),\widehat{x}(T)\rangle
-\int_{0}^{T}d\langle \widetilde{P}_{1}\widehat{x},\widehat{x}\rangle\bigg]
-E\bigg[\langle\varphi(T),x(T)\rangle-\int_{0}^{T}d\langle\varphi,x\rangle\bigg]\\
=&\frac{1}{2}\langle \widetilde{P}_{1}(0,i)x_{0},x_{0}\rangle+\langle\varphi(0),x_{0}\rangle\\
&+\frac{1}{2}E\bigg[\int_{0}^{T}\Big(\langle Q_{1}(x-\widehat{x}),x-\widehat{x}\rangle
+\langle \widetilde{Q}_{1}\widehat{x},\widehat{x}\rangle
+\langle N_{1}u_{1},u_{1}\rangle\Big)dt\bigg]\\
&+\frac{1}{2}E\bigg[\int_{0}^{T}\Big(d\langle P_{1}(x-\widehat{x}),x-\widehat{x}\rangle
+d\langle \widetilde{P}_{1}\widehat{x},\widehat{x}\rangle
+2d\langle\varphi,x\rangle\Big)\bigg].
\end{aligned}
\end{equation}
On the one hand, applying It\^{o}'s formula for Markov modulated processes to $P_{1}(x-\widehat{x})$,
\begin{equation}\label{follower proof +1}
\begin{aligned}
d[P_{1}(x-\widehat{x})]
=&-[A^{\top}P_{1}+C^{\top}P_{1}C+Q_{1}-S_{1}^{\top}\widetilde{N}_{1}^{-1}S_{1}](x-\widehat{x})dt\\
&+P_{1}[B_{1}u_{1}-B_{1}\widehat{u}_{1}+B_{2}u_{2}-B_{2}\widehat{u}_{2}]dt\\
&+P_{1}[C(x-\widehat{x})+\widetilde{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}]dW\\
&+\sum_{i,j\in\mathcal{M}}[P_{1}(t,j)-P_{1}(t,i)](x-\widehat{x})dM_{ij}.
\end{aligned}
\end{equation}
Applying It\^{o}'s formula for semi-martingales
(see Karatzas and Shreve \cite[Theorem 3.3]{KS}) to
$\langle P_{1}(x-\widehat{x}),x-\widehat{x}\rangle$
(only the $dt$ part is preserved),
\begin{equation}\label{follower proof 2}
\begin{aligned}
&d\langle P_{1}(x-\widehat{x}),x-\widehat{x}\rangle\\
=&\langle d[P_{1}(x-\widehat{x})],x-\widehat{x}\rangle
+\langle P_{1}(x-\widehat{x}),d(x-\widehat{x})\rangle
+\langle d[P_{1}(x-\widehat{x})],d(x-\widehat{x})\rangle\\
=&\langle [-C^{\top}P_{1}C-Q_{1}+S_{1}^{\top}\widetilde{N}_{1}^{-1}S_{1}](x-\widehat{x})
+P_{1}[B_{1}u_{1}-B_{1}\widehat{u}_{1}+B_{2}u_{2}-B_{2}\widehat{u}_{2}],x-\widehat{x}\rangle dt\\
&+\langle P_{1}(x-\widehat{x}),B_{1}u_{1}-B_{1}\widehat{u}_{1}+B_{2}u_{2}-B_{2}\widehat{u}_{2}\rangle dt\\
&+\langle P_{1}[C(x-\widehat{x})+\widetilde{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}],
C(x-\widehat{x})+\widetilde{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}\rangle dt.
\end{aligned}
\end{equation}
On the other hand,
applying It\^{o}'s formula for Markov modulated processes to $\widetilde{P}_{1}\widehat{x}$,
\begin{equation}\label{follower proof +2}
\begin{aligned}
d[\widetilde{P}_{1}\widehat{x}]
=&-[\widetilde{A}^{\top}\widetilde{P}_{1}+\widetilde{C}^{\top}P_{1}\widetilde{C}+\widetilde{Q}_{1}
-\widetilde{S}_{1}^{\top}\widetilde{N}_{1}^{-1}\widetilde{S}_{1}]\widehat{x}dt
+\widetilde{P}_{1}[B_{1}\widehat{u}_{1}+B_{2}\widehat{u}_{2}]dt\\
&+\sum_{i,j\in\mathcal{M}}[\widetilde{P}_{1}(t,j)-\widetilde{P}_{1}(t,i)]\widehat{x}dM_{ij}.
\end{aligned}
\end{equation}
Applying It\^{o}'s formula for semi-martingales to
$\langle \widetilde{P}_{1}\widehat{x},\widehat{x}\rangle$,
\begin{equation}\label{follower proof 3}
\begin{aligned}
d\langle \widetilde{P}_{1}\widehat{x},\widehat{x}\rangle
=&\langle d(\widetilde{P}_{1}\widehat{x}),\widehat{x}\rangle
+\langle \widetilde{P}_{1}\widehat{x},d\widehat{x}\rangle
+\langle d(\widetilde{P}_{1}\widehat{x}),d\widehat{x}\rangle\\
=&\langle [-\widetilde{C}^{\top}P_{1}\widetilde{C}-\widetilde{Q}_{1}
+\widetilde{S}_{1}^{\top}\widetilde{N}_{1}^{-1}\widetilde{S}_{1}]\widehat{x}
+\widetilde{P}_{1}[B_{1}\widehat{u}_{1}+B_{2}\widehat{u}_{2}],\widehat{x}\rangle dt\\
&+\langle\widetilde{P}_{1}\widehat{x},B_{1}\widehat{u}_{1}+B_{2}\widehat{u}_{2}\rangle dt.
\end{aligned}
\end{equation}
Finally, applying It\^{o}'s formula for semi-martingales to $2\langle\varphi,x\rangle$,
\begin{equation}\label{follower proof 4}
\begin{aligned}
2d\langle\varphi,x\rangle=&2(\langle d\varphi,x\rangle
+\langle\varphi,dx\rangle+\langle d\varphi,dx\rangle)\\
=&2\langle -[\mathbb{A}^{\top}\varphi+\widehat{\mathbb{A}}^{\top}\widehat{\varphi}
+\mathbb{C}^{\top}\theta+\widehat{\mathbb{C}}^{\top}\widehat{\theta}
+\mathbb{F}_{2}^{\top}u_{2}+\widehat{\mathbb{F}}_{2}^{\top}\widehat{u}_{2}],x\rangle dt\\
&+2\langle\varphi,Ax+\widehat{A}\widehat{x}+B_{1}u_{1}+B_{2}u_{2}\rangle dt\\
&+2\langle\theta,Cx+\widehat{C}\widehat{x}+D_{1}u_{1}+D_{2}u_{2}\rangle dt.
\end{aligned}
\end{equation}
We first look at the terms involving $u_{1}$ and $\widehat{u}_{1}$
in (\ref{follower proof 1})--(\ref{follower proof 4}):
\begin{equation*}
\begin{aligned}
&u_{1}^{\top}(N_{1}+D_{1}^{\top}P_{1}D_{1})u_{1}\\
&+2u_{1}^{\top}[B_{1}^{\top}P_{1}(x-\widehat{x})
+D_{1}^{\top}P_{1}(C(x-\widehat{x})+\widetilde{C}\widehat{x}+D_{2}u_{2})
+B_{1}\widetilde{P}_{1}\widehat{x}+B_{1}^{\top}\varphi+D_{1}^{\top}\theta]\\
=&|\widetilde{N}_{1}^{\frac{1}{2}}u_{1}+\widetilde{N}_{1}^{-\frac{1}{2}}[S_{1}(x-\widehat{x})+\widetilde{S}_{1}\widehat{x}+\Phi]|^{2}
-|\widetilde{N}_{1}^{-\frac{1}{2}}[S_{1}(x-\widehat{x})+\widetilde{S}_{1}\widehat{x}+\Phi]|^{2},
\end{aligned}
\end{equation*}
in which we have used Lemma \ref{Lemma conditional E} to get
\begin{equation*}
\begin{aligned}
&E\langle P_{1}B_{1}\widehat{u}_{1},x-\widehat{x}\rangle
=E\langle P_{1}B_{1}u_{1},\widehat{x}-\widehat{x}\rangle=0,\\
&E\langle \widetilde{P}_{1}B_{1}\widehat{u}_{1},\widehat{x}\rangle
=E\langle \widetilde{P}_{1}B_{1}u_{1},\widehat{x}\rangle.
\end{aligned}
\end{equation*}
For the terms involving no $u_{1}$ or $\widehat{u}_{1}$
in (\ref{follower proof 1})--(\ref{follower proof 4}):
\begin{equation*}
\begin{aligned}
&\langle S_{1}^{\top}\widetilde{N}_{1}^{-1}S_{1}(x-\widehat{x}),x-\widehat{x}\rangle
+\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle\\
&+2\langle B_{2}^{\top}P_{1}(x-\widehat{x}),u_{2}\rangle
+2\langle D_{2}^{\top}P_{1}[C(x-\widehat{x})+\widetilde{C}\widehat{x}],u_{2}\rangle\\
&+\langle \widetilde{S}_{1}^{\top}\widetilde{N}_{1}^{-1}\widetilde{S}_{1}\widehat{x},\widehat{x}\rangle
+2\langle B_{2}^{\top}\widetilde{P}_{1}\widehat{x},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi,u_{2}\rangle+2\langle D_{2}^{\top}\theta,u_{2}\rangle\\
&+2\langle B_{1}\widetilde{N}_{1}^{-1}S_{1}x,\varphi\rangle
+2\langle B_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\varphi\rangle
+2\langle D_{1}\widetilde{N}_{1}^{-1}S_{1}x,\theta\rangle
+2\langle D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\theta\rangle\\
&-2\langle[S_{2}-D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}S_{1}]x,u_{2}\rangle
-2\langle[\widehat{S}_{2}-D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}]\widehat{x},u_{2}\rangle\\
=&\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi,u_{2}\rangle+2\langle D_{2}^{\top}\theta,u_{2}\rangle\\
&+\langle S_{1}^{\top}\widetilde{N}_{1}^{-1}S_{1}(x-\widehat{x}),x-\widehat{x}\rangle
+\langle \widetilde{S}_{1}^{\top}\widetilde{N}_{1}^{-1}\widetilde{S}_{1}\widehat{x},\widehat{x}\rangle\\
&+2\langle D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}S_{1}x,u_{2}\rangle
+2\langle D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},u_{2}\rangle\\
&+2\langle B_{1}\widetilde{N}_{1}^{-1}S_{1}x,\varphi\rangle
+2\langle B_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\varphi\rangle
+2\langle D_{1}\widetilde{N}_{1}^{-1}S_{1}x,\theta\rangle
+2\langle D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\theta\rangle\\
=&\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi,u_{2}\rangle+2\langle D_{2}^{\top}\theta,u_{2}\rangle\\
&+|\widetilde{N}_{1}^{-\frac{1}{2}}[S_{1}(x-\widehat{x})+\widetilde{S}_{1}\widehat{x}+\Phi]|^{2}
-|\widetilde{N}_{1}^{-\frac{1}{2}}\Phi|^{2},
\end{aligned}
\end{equation*}
in which we have also used Lemma \ref{Lemma conditional E} to get
\begin{equation*}
\begin{aligned}
&E\langle P_{1}B_{2}\widehat{u}_{2},x-\widehat{x}\rangle
=E\langle P_{1}B_{2}u_{2},\widehat{x}-\widehat{x}\rangle=0,\\
&E\langle P_{1}C(x-\widehat{x}),\widetilde{C}\widehat{x}\rangle
=E\langle P_{1}C(\widehat{x}-\widehat{x}),\widetilde{C}x\rangle=0,\\
&E\langle \widetilde{P}_{1}B_{2}\widehat{u}_{2},\widehat{x}\rangle
=E\langle \widetilde{P}_{1}B_{2}u_{2},\widehat{x}\rangle,\\
&E\langle B_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}x,\widehat{\varphi}\rangle
=E\langle B_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\varphi\rangle,\\
&E\langle D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}x,\widehat{\theta}\rangle
=E\langle D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}\widehat{x},\theta\rangle,\\
&E\langle[\widehat{S}_{2}-D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}]x,\widehat{u}_{2}\rangle
=E\langle[\widehat{S}_{2}-D_{2}^{\top}P_{1}D_{1}\widetilde{N}_{1}^{-1}\widehat{S}_{1}]\widehat{x},u_{2}\rangle.
\end{aligned}
\end{equation*}
Then, (\ref{follower proof 1}) reduces to
\begin{equation*}
\begin{aligned}
J_{1}(u_{1}(\cdot),&u_{2}(\cdot))
=\frac{1}{2}\langle \widetilde{P}_{1}(0,i)x_{0},x_{0}\rangle
+\langle\varphi(0),x_{0}\rangle\\
&+\frac{1}{2}E\bigg[\int_{0}^{T}\Big(|\widetilde{N}_{1}^{\frac{1}{2}}
(u_{1}+\widetilde{N}_{1}^{-1}[S_{1}(x-\widehat{x})+\widetilde{S}_{1}\widehat{x}+\Phi])|^{2}
-|\widetilde{N}_{1}^{-\frac{1}{2}}\Phi|^{2}\\
&+\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi+D_{2}^{\top}\theta,u_{2}\rangle\Big)dt\bigg].
\end{aligned}
\end{equation*}
It follows that $u_{1}^{*}$ defined by (\ref{follower optimal control})
is indeed an optimal control for the follower, and
\begin{equation*}
\begin{aligned}
J_{1}(u_{1}^{*}(\cdot),&u_{2}(\cdot))
=\frac{1}{2}\langle \widetilde{P}_{1}(0,i)x_{0},x_{0}\rangle
+\langle\varphi(0),x_{0}\rangle\\
&+\frac{1}{2}E\bigg[\int_{0}^{T}\Big(-|\widetilde{N}_{1}^{-\frac{1}{2}}\Phi|^{2}
+\langle D_{2}^{\top}P_{1}D_{2}u_{2},u_{2}\rangle
+2\langle B_{2}^{\top}\varphi+D_{2}^{\top}\theta,u_{2}\rangle\Big)dt\bigg].
\end{aligned}
\end{equation*}
The proof is completed.
\end{proof}

\section{The problem for the leader}\label{leader problem}

After the follower's problem being solved and the follower taking his optimal control
(\ref{follower optimal control}), the leader faces a state equation, which is a
conditional mean-field FBSDE with regime switching, consisting of the state equation
(\ref{system}) of the LQ problem and the auxiliary BSDE (\ref{follower 3}) of the follower:
\begin{equation}\label{leader system}
\left\{
\begin{aligned}
dx=&\Big[\mathbb{A}x+\widehat{\mathbb{A}}\widehat{x}
+\mathbb{F}_{1}\varphi+\mathbb{B}_{1}\theta+\mathbb{B}_{2}u_{2}\Big]dt
+\Big[\mathbb{C}x+\widehat{\mathbb{C}}\widehat{x}
+\mathbb{B}_{1}^{\top}\varphi+\mathbb{D}_{1}\theta+\mathbb{D}_{2}u_{2}\Big]dW,\\
d\varphi=&-\Big[\mathbb{A}^{\top}\varphi+\widehat{\mathbb{A}}^{\top}\widehat{\varphi}
+\mathbb{C}^{\top}\theta+\widehat{\mathbb{C}}^{\top}\widehat{\theta}
+\mathbb{F}_{2}^{\top}u_{2}+\widehat{\mathbb{F}}_{2}^{\top}\widehat{u}_{2}\Big]dt
+\theta dW+\eta\bullet dM,\\
x(0)=&x_{0},\quad \varphi(T)=0,
\end{aligned}
\right.
\end{equation}
where, for convenience, we denote
\begin{equation*}
\begin{aligned}
\mathbb{B}_{1}(t,i)=&-B_{1}(i)\widetilde{N}_{1}^{-1}(t,i)D_{1}^{\top}(i),\quad
\mathbb{B}_{2}(t,i)=B_{2}(i)-B_{1}(i)\widetilde{N}_{1}^{-1}(t,i)D_{1}^{\top}(i)P_{1}(t,i)D_{2}(i),\\
\mathbb{D}_{1}(t,i)=&-D_{1}(i)\widetilde{N}_{1}^{-1}(t,i)D_{1}^{\top}(i),\quad
\mathbb{D}_{2}(t,i)=D_{2}(i)-D_{1}(i)\widetilde{N}_{1}^{-1}(t,i)D_{1}^{\top}(i)P_{1}(t,i)D_{2}(i),\\
\mathbb{F}_{1}(t,i)=&-B_{1}(i)\widetilde{N}_{1}^{-1}(t,i)B_{1}^{\top}(i),\quad i\in\mathcal{M}.
\end{aligned}
\end{equation*}
Note that the FBSDE (\ref{leader system}) is \emph{decoupled} in the sense that
one can first solve the backward equation for $(\varphi,\theta,\eta)$
and then solve the forward equation for $x$,
so the unique solvability of (\ref{leader system}) is guaranteed.
The leader's problem is to find an optimal control $u_{2}^{*}(\cdot)\in\mathcal{U}_{2}$
to minimize her cost functional (\ref{cost functionals}) for $k=2$.
We will also utilize the SMP approach to solve the leader's problem.
In addition to Assumption (A1), we further make the following assumption:

(A2) $Q_{2}(i)\geq0$, $\widehat{Q}_{2}(i)\geq0$, $N_{2}(i)>0$,
$G_{2}(i)\geq0$, $\widehat{G}_{2}(i)\geq0$, $i\in\mathcal{M}$.

The adjoint equation for the leader is given by
\begin{equation}\label{leader adjoint equation}
\left\{
\begin{aligned}
dy=&-\Big[\mathbb{A}^{\top}y+\widehat{\mathbb{A}}^{\top}\widehat{y}
+\mathbb{C}^{\top}z+\widehat{\mathbb{C}}^{\top}\widehat{z}
+Q_{2}x^{*}+\widehat{Q}_{2}\widehat{x}^{*}\Big]dt+zdW+k\bullet dM,\\
d\psi=&\Big[\mathbb{A}\psi+\widehat{\mathbb{A}}\widehat{\psi}
+\mathbb{F}_{1}y+\mathbb{B}_{1}z\Big]dt
+\Big[\mathbb{C}\psi+\widehat{\mathbb{C}}\widehat{\psi}
+\mathbb{B}_{1}^{\top}y+\mathbb{D}_{1}z\Big]dW,\\
y(T)=&G_{2}(\alpha(T))x^{*}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),\quad \psi(0)=0,
\end{aligned}
\right.
\end{equation}
where $(x^{*},\varphi^{*},\theta^{*},\eta^{*})$ is the corresponding solution of
(\ref{leader system}) under an optimal control $u_{2}^{*}$ for the leader.
Note that (\ref{leader adjoint equation}) is also a decoupled conditional mean-field
FBSDE with regime switching, and thereby its unique solvability is guaranteed.
Based on Yong \cite[Theorem 3.2]{Yong2002} and Nguyen et al. \cite[Theorem 3.7]{NNY2020},
one can establish the following SMP for the leader's problem.
\begin{lemma}\label{leader theorem 1}
Let Assumptions (A1) and (A2) hold. Then $u_{2}^{*}\in\mathcal{U}_{2}$ is an optimal control
for the leader if and only if the adjoint equation (\ref{leader adjoint equation}) admits
a unique solution $(y,z,k,\psi)\in\mathcal{S}_{\mathcal{F}}^{2}(R^{n})
\times\mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times\mathcal{K}_{\mathcal{F}}^{2}(R^{n})
\times\mathcal{L}_{\mathcal{F}}^{2}(R^{n})$ such that
\begin{equation}\label{leader original optimal condition}
\begin{aligned}
N_{2}u_{2}^{*}+\mathbb{B}_{2}^{\top}y+\mathbb{D}_{2}^{\top}z+\mathbb{F}_{2}\psi
+\widehat{\mathbb{F}}_{2}\widehat{\psi}=0.
\end{aligned}
\end{equation}
\end{lemma}
\begin{proof}
Let $(x^{*},\varphi^{*},\theta^{*},\eta^{*})\in
\mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times\mathcal{S}_{\mathcal{F}}^{2}(R^{n})
\times\mathcal{L}_{\mathcal{F}}^{2}(R^{n})\times\mathcal{K}_{\mathcal{F}}^{2}(R^{n})$
be the corresponding solution of (\ref{leader system}) under $u_{2}^{*}$.
For any $u_{2}^{0}\in\mathcal{U}_{2}$, we introduce the following state equation:
\begin{equation}\label{leader proof 1-1}
\left\{
\begin{aligned}
dx^{0}=&\Big[\mathbb{A}x^{0}+\widehat{\mathbb{A}}\widehat{x}^{0}
+\mathbb{F}_{1}\varphi^{0}+\mathbb{B}_{1}\theta^{0}+\mathbb{B}_{2}u_{2}^{0}\Big]dt\\
&+\Big[\mathbb{C}x^{0}+\widehat{\mathbb{C}}\widehat{x}^{0}
+\mathbb{B}_{1}^{\top}\varphi^{0}+\mathbb{D}_{1}\theta^{0}+\mathbb{D}_{2}u_{2}^{0}\Big]dW,\\
d\varphi^{0}=&-\Big[\mathbb{A}^{\top}\varphi^{0}+\widehat{\mathbb{A}}^{\top}\widehat{\varphi}^{0}
+\mathbb{C}^{\top}\theta^{0}+\widehat{\mathbb{C}}^{\top}\widehat{\theta}^{0}
+\mathbb{F}_{2}^{\top}u_{2}^{0}+\widehat{\mathbb{F}}_{2}^{\top}\widehat{u}_{2}^{0}\Big]dt\\
&+\theta^{0}dW+\eta^{0}\bullet dM,\\
x^{0}(0)=&0,\quad \varphi^{0}(T)=0,
\end{aligned}
\right.
\end{equation}
and the adjoint equation:
\begin{equation}\label{leader proof 1-2}
\left\{
\begin{aligned}
dy^{0}=&-\Big[\mathbb{A}^{\top}y^{0}+\widehat{\mathbb{A}}^{\top}\widehat{y}^{0}
+\mathbb{C}^{\top}z^{0}+\widehat{\mathbb{C}}^{\top}\widehat{z}^{0}
+Q_{2}x^{0}+\widehat{Q}_{2}\widehat{x}^{0}\Big]dt\\
&+z^{0}dW+k^{0}\bullet dM,\\
d\psi^{0}=&\Big[\mathbb{A}\psi^{0}+\widehat{\mathbb{A}}\widehat{\psi}^{0}
+\mathbb{F}_{1}y^{0}+\mathbb{B}_{1}z^{0}\Big]dt\\
&+\Big[\mathbb{C}\psi^{0}+\widehat{\mathbb{C}}\widehat{\psi}^{0}
+\mathbb{B}_{1}^{\top}y^{0}+\mathbb{D}_{1}z^{0}\Big]dW,\\
y^{0}(T)=&G_{2}(\alpha(T))x^{0}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),\quad \psi^{0}(0)=0.
\end{aligned}
\right.
\end{equation}
Note that the initial condition $x^{0}(0)=0$ in (\ref{leader proof 1-1}),
which is the only difference compared with (\ref{leader system}).
Also, the FBSDEs (\ref{leader proof 1-1}) and (\ref{leader proof 1-2})
have a unique solution
$(x^{0},\varphi^{0},\theta^{0},\eta^{0},y^{0},z^{0},k^{0},\psi^{0})$
in the usual space.

For any $\lambda\in R$, consider $u_{2}\doteq u_{2}^{*}+\lambda u_{2}^{0}\in\mathcal{U}_{2}$
and denote $(x,\varphi,\theta,\eta)$ the corresponding solution of (\ref{leader system}).
From the linearity of the above FBSDEs, we have $x=x^{*}+\lambda x^{0}$. Then,
\begin{equation}\label{leader proof 2}
\begin{aligned}
&J_{2}(u_{1}^{*},u_{2})-J_{2}(u_{1}^{*},u_{2}^{*})\\
=&\frac{\lambda^{2}}{2}E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{0},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{0},\widehat{x}^{0}\rangle
+\langle N_{2}u_{2}^{0},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{0}(T),x^{0}(T)\rangle
+\langle \widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),\widehat{x}^{0}(T)\rangle\bigg]\\
&+\lambda E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{*},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{*},\widehat{x}^{0}\rangle
+\langle N_{2}u_{2}^{*},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{*}(T),x^{0}(T)\rangle
+\langle \widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),\widehat{x}^{0}(T)\rangle\bigg]\\
=&\frac{\lambda^{2}}{2}E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{0},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{0},\widehat{x}^{0}\rangle
+\langle N_{2}u_{2}^{0},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{0}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),x^{0}(T)\rangle\bigg]\\
&+\lambda E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{*},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{*},\widehat{x}^{0}\rangle
+\langle N_{2}u_{2}^{*},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{*}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),x^{0}(T)\rangle\bigg].
\end{aligned}
\end{equation}
On the one hand,
\begin{equation}\label{leader proof 3}
\begin{aligned}
&E[\langle G_{2}(\alpha(T))x^{0}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),x^{0}(T)\rangle]\\
=&E[\langle y^{0}(T),x^{0}(T)\rangle]\\
=&E[\langle y^{0}(T),x^{0}(T)\rangle-\langle y^{0}(0),x^{0}(0)\rangle
-\langle \psi^{0}(T),\varphi^{0}(T)\rangle+\langle \psi^{0}(0),\varphi^{0}(0)\rangle]\\
=&E\bigg[\int_{0}^{T}\Big(-\langle Q_{2}x^{0},x^{0}\rangle
-\langle \widehat{Q}_{2}\widehat{x}^{0},x^{0}\rangle
+\langle u_{2}^{0},\mathbb{B}_{2}^{\top}y^{0}+\mathbb{D}_{2}^{\top}z^{0}
+\mathbb{F}_{2}\psi^{0}+\widehat{\mathbb{F}}_{2}\widehat{\psi}^{0}\rangle\Big)dt\bigg].
\end{aligned}
\end{equation}
Therefore,
\begin{equation}\label{leader proof 4}
\begin{aligned}
&E\bigg[\int_{0}^{T}\langle u_{2}^{0},N_{2}u_{2}^{0}+\mathbb{B}_{2}^{\top}y^{0}+\mathbb{D}_{2}^{\top}z^{0}
+\mathbb{F}_{2}\psi^{0}+\widehat{\mathbb{F}}_{2}\widehat{\psi}^{0}\rangle dt\bigg]\\
=&E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{0},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{0},x^{0}\rangle
+\langle N_{2}u_{2}^{0},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{0}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),x^{0}(T)\rangle\bigg]\\
=&E\bigg[\int_{0}^{T}\Big(\langle Q_{2}x^{0},x^{0}\rangle
+\langle \widehat{Q}_{2}\widehat{x}^{0},\widehat{x}^{0}\rangle
+\langle N_{2}u_{2}^{0},u_{2}^{0}\rangle\Big)dt\\
&+\langle G_{2}(\alpha(T))x^{0}(T),x^{0}(T)\rangle
+\langle \widehat{G}_{2}(\alpha(T))\widehat{x}^{0}(T),\widehat{x}^{0}(T)\rangle\bigg]\geq0,
\end{aligned}
\end{equation}
where we have used Assumption (A2)
and the following facts (noting Lemma \ref{Lemma conditional E}):
\begin{equation*}
\begin{aligned}
&E\langle\widehat{Q}_{2}\widehat{x}^{0},x^{0}\rangle
=E\langle\widehat{Q}_{2}\widehat{x}^{0},\widehat{x}^{0}\rangle\geq0,\\
&E\langle\widehat{G}_{2}\widehat{x}^{0}(T),x^{0}(T)\rangle
=E\langle\widehat{G}_{2}\widehat{x}^{0}(T),\widehat{x}^{0}(T)\rangle\geq0.
\end{aligned}
\end{equation*}
On the other hand,
\begin{equation}\label{leader proof 5}
\begin{aligned}
&E[\langle G_{2}(\alpha(T))x^{*}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),x^{0}(T)\rangle]\\
=&E[\langle y(T),x^{0}(T)\rangle]\\
=&E[\langle y(T),x^{0}(T)\rangle-\langle y(0),x^{0}(0)\rangle
-\langle \psi(T),\varphi^{0}(T)\rangle+\langle \psi(0),\varphi^{0}(0)\rangle]\\
=&E\bigg[\int_{0}^{T}\Big(-\langle Q_{2}x^{*},x^{0}\rangle
-\langle \widehat{Q}_{2}\widehat{x}^{*},x^{0}\rangle
+\langle u_{2}^{0},\mathbb{B}_{2}^{\top}y+\mathbb{D}_{2}^{\top}z
+\mathbb{F}_{2}\psi+\widehat{\mathbb{F}}_{2}\widehat{\psi}\rangle\Big)dt\bigg].
\end{aligned}
\end{equation}
Thus, combining (\ref{leader proof 2}), (\ref{leader proof 3}), and (\ref{leader proof 5})
leads to
\begin{equation*}
\begin{aligned}
&J_{2}(u_{1}^{*},u_{2})-J_{2}(u_{1}^{*},u_{2}^{*})\\
=&\frac{\lambda^{2}}{2}
E\bigg[\int_{0}^{T}\langle u_{2}^{0},N_{2}u_{2}^{0}+\mathbb{B}_{2}^{\top}y^{0}+\mathbb{D}_{2}^{\top}z^{0}
+\mathbb{F}_{2}\psi^{0}+\widehat{\mathbb{F}}_{2}\widehat{\psi}^{0}\rangle dt\bigg]\\
&+\lambda E\bigg[\int_{0}^{T}\langle u_{2}^{0},N_{2}u_{2}^{*}
+\mathbb{B}_{2}^{\top}y+\mathbb{D}_{2}^{\top}z
+\mathbb{F}_{2}\psi+\widehat{\mathbb{F}}_{2}\widehat{\psi}\rangle dt\bigg].
\end{aligned}
\end{equation*}
From (\ref{leader proof 4}), we deduce that $u_{2}^{*}$ is optimal
if and only if
\begin{equation*}
\begin{aligned}
N_{2}u_{2}^{*}+\mathbb{B}_{2}^{\top}y+\mathbb{D}_{2}^{\top}z
+\mathbb{F}_{2}\psi+\widehat{\mathbb{F}}_{2}\widehat{\psi}=0.
\end{aligned}
\end{equation*}
The proof is completed.
\end{proof}
Similar to the follower's problem, we also expect to derive a state feedback representation
for $u_{2}^{*}$ defined by (\ref{leader original optimal condition}), which, as shown later,
is non-anticipating. To apply the dimensional augmentation approach by Yong \cite{Yong2002},
we denote
\begin{equation*}
\begin{aligned}
			X=\left[
			\begin{array}{c}
				x^{*} \\
				\psi \\
			\end{array}
			\right],\quad
			Y=\left[
			\begin{array}{c}
				y \\
				\varphi^{*} \\
			\end{array}
			\right],\quad
			Z=\left[
			\begin{array}{c}
				z \\
				\theta^{*} \\
			\end{array}
			\right],\quad
			K=\left[
			\begin{array}{c}
				k \\
				\eta^{*} \\
			\end{array}
			\right],\quad
			X_{0}=\left[
			\begin{array}{c}
				x_{0} \\
				0 \\
			\end{array}
			\right],
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
			\mathbf{A}=\left[
			\begin{array}{cc}
				\mathbb{A} & 0 \\
				0 & \mathbb{A} \\
			\end{array}
			\right],\quad
			\widehat{\mathbf{A}}=\left[
			\begin{array}{cc}
				\widehat{\mathbb{A}} & 0 \\
				0 & \widehat{\mathbb{A}} \\
			\end{array}
			\right],\quad
			\mathbf{C}=\left[
			\begin{array}{cc}
				\mathbb{C} & 0 \\
				0 & \mathbb{C} \\
			\end{array}
			\right],\quad
			\widehat{\mathbf{C}}=\left[
			\begin{array}{cc}
				\widehat{\mathbb{C}} & 0 \\
				0 & \widehat{\mathbb{C}} \\
			\end{array}
			\right],
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
			\mathbf{B}_{1}=\left[
			\begin{array}{cc}
				0 & \mathbb{B}_{1} \\
				\mathbb{B}_{1} & 0 \\
			\end{array}
			\right],\quad
			\mathbf{B}_{2}=\left[
			\begin{array}{c}
				\mathbb{B}_{2} \\
				0 \\
			\end{array}
			\right],\quad
			\mathbf{D}_{1}=\left[
			\begin{array}{cc}
				0 & \mathbb{D}_{1} \\
				\mathbb{D}_{1} & 0 \\
			\end{array}
			\right],\quad
			\mathbf{D}_{2}=\left[
			\begin{array}{c}
				\mathbb{D}_{2} \\
				0 \\
			\end{array}
			\right],
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
			\mathbf{F}_{1}=\left[
			\begin{array}{cc}
				0 & \mathbb{F}_{1} \\
				\mathbb{F}_{1} & 0 \\
			\end{array}
			\right],\quad
			\mathbf{F}_{2}=\left[
			\begin{array}{cc}
				0 & \mathbb{F}_{2} \\
			\end{array}
			\right],\quad
			\widehat{\mathbf{F}}_{2}=\left[
			\begin{array}{cc}
				0 & \widehat{\mathbb{F}}_{2} \\
			\end{array}
			\right],
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
			\mathbf{Q}_{2}=\left[
			\begin{array}{cc}
				Q_{2} & 0 \\
				0 & 0 \\
			\end{array}
			\right],\quad
			\widehat{\mathbf{Q}}_{2}=\left[
			\begin{array}{cc}
				\widehat{Q}_{2} & 0 \\
				0 & 0 \\
			\end{array}
			\right],\quad
			\mathbf{G}_{2}=\left[
			\begin{array}{cc}
				G_{2} & 0 \\
				0 & 0 \\
			\end{array}
			\right],\quad
			\widehat{\mathbf{G}}_{2}=\left[
			\begin{array}{cc}
				\widehat{G}_{2} & 0 \\
				0 & 0 \\
			\end{array}
			\right].
\end{aligned}
\end{equation*}
Then, (\ref{leader system}) and (\ref{leader adjoint equation})
can be rewritten as
\begin{equation}\label{leader Hamiltonian system}
\left\{
\begin{aligned}
dX=&\Big[\mathbf{A}X+\widehat{\mathbf{A}}\widehat{X}
+\mathbf{F}_{1}Y+\mathbf{B}_{1}Z+\mathbf{B}_{2}u_{2}^{*}\Big]dt\\
&+\Big[\mathbf{C}X+\widehat{\mathbf{C}}\widehat{X}
+\mathbf{B}_{1}^{\top}Y+\mathbf{D}_{1}Z+\mathbf{D}_{2}u_{2}^{*}\Big]dW,\\
dY=&-\Big[\mathbf{A}^{\top}Y+\widehat{\mathbf{A}}^{\top}\widehat{Y}
+\mathbf{C}^{\top}Z+\widehat{\mathbf{C}}^{\top}\widehat{Z}
+\mathbf{Q}_{2}X+\widehat{\mathbf{Q}}_{2}\widehat{X}\\
&+\mathbf{F}_{2}^{\top}u_{2}^{*}+\widehat{\mathbf{F}}_{2}^{\top}\widehat{u}_{2}^{*}\Big]dt
+ZdW+K\bullet dM,\\
X(0)=&X_{0},\quad Y(T)=\mathbf{G}_{2}(\alpha(T))X(T)
+\widehat{\mathbf{G}}_{2}(\alpha(T))\widehat{X}(T),
\end{aligned}
\right.
\end{equation}
and (\ref{leader original optimal condition}) becomes
\begin{equation}\label{leader optimal condition}
\begin{aligned}
0=N_{2}u_{2}^{*}+\mathbf{B}_{2}^{\top}Y
+\mathbf{D}_{2}^{\top}Z+\mathbf{F}_{2}X
+\widehat{\mathbf{F}}_{2}\widehat{X}.
\end{aligned}
\end{equation}
\begin{theorem}
Let Assumptions (A1) and (A2) hold. An optimal control $u_{2}^{*}$ for the leader
is given by
\begin{equation}\label{leader optimal control}
\begin{aligned}
u_{2}^{*}(t)=-\widetilde{N}_{2}^{-1}(t,\alpha(t))\Big[\mathbf{S}_{2}(t,\alpha(t))X(t)
+\widehat{\mathbf{S}}_{2}(t,\alpha(t))\widehat{X}(t)\Big],
\end{aligned}
\end{equation}
where, for the sake of simplicity, we denote
\begin{equation*}
\begin{aligned}
\widetilde{N}_{2}(t,i)=&N_{2}(i)+\mathbf{D}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\mathbf{D}_{2}(t,i),\\
\mathbf{J}_{2}(t,i)=&\mathbf{B}_{1}^{\top}(t,i)P_{2}(t,i)+\mathbf{C}(t,i),\quad
\widehat{\mathbf{J}}_{2}(t,i)=\mathbf{B}_{1}^{\top}(t,i)\widehat{P}_{2}(t,i)+\widehat{\mathbf{C}}(t,i),\\
\mathbf{S}_{2}(t,i)=&\mathbf{D}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\mathbf{J}_{2}(t,i)
+\mathbf{B}_{2}^{\top}(t,i)P_{2}(t,i)+\mathbf{F}_{2}(t,i),\\
\widehat{\mathbf{S}}_{2}(t,i)=&\mathbf{D}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\widehat{\mathbf{J}}_{2}(t,i)
+\mathbf{B}_{2}^{\top}(t,i)\widehat{P}_{2}(t,i)+\widehat{\mathbf{F}}_{2}(t,i),\quad i\in\mathcal{M},
\end{aligned}
\end{equation*}
provided $\widetilde{N}_{2}$ and $(I-P_{2}\mathbf{D}_{1})$ are invertible
and $P_{2}(\cdot,i)$ and $\widehat{P}_{2}(\cdot,i)$, $i\in\mathcal{M}$,
are solutions of Riccati equations (\ref{leader 1}) and (\ref{leader 2}), respectively.
\end{theorem}
\begin{proof}
In the light of the terminal condition of (\ref{leader Hamiltonian system}),
it is natural to set
\begin{equation}\label{leader FSS}
\begin{aligned}
Y(t)=P_{2}(t,\alpha(t))X(t)+\widehat{P}_{2}(t,\alpha(t))\widehat{X}(t),
\end{aligned}
\end{equation}
for some $R^{2n\times 2n}$-valued deterministic, differentiable, and symmetric
functions $P_{2}(t,i)$ and $\widehat{P}_{2}(t,i)$, $i\in \mathcal{M}$.
Applying It\^{o}'s formula for Markov-modulated processes to (\ref{leader FSS}), we have
\begin{equation}\label{leader dY}
\begin{aligned}
&dY=\Big(\dot{P}_{2}+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[P_{2}(t,j)-P_{2}(t,\alpha(t))]\Big)Xdt
+\sum_{i,j\in\mathcal{M}}[P_{2}(t,j)-P_{2}(t,i)]XdM_{ij}\\
&+P_{2}\Big[\mathbf{A}X+\widehat{\mathbf{A}}\widehat{X}
+\mathbf{F}_{1}Y+\mathbf{B}_{1}Z+\mathbf{B}_{2}u_{2}^{*}\Big]dt
+P_{2}\Big[\mathbf{C}X+\widehat{\mathbf{C}}\widehat{X}
+\mathbf{B}_{1}^{\top}Y+\mathbf{D}_{1}Z+\mathbf{D}_{2}u_{2}^{*}\Big]dW\\
&+\Big(\dot{\widehat{P}}_{2}
+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[\widehat{P}_{2}(t,j)-\widehat{P}_{2}(t,\alpha(t))]\Big)\widehat{X}dt
+\sum_{i,j\in\mathcal{M}}[\widehat{P}_{2}(t,j)-\widehat{P}_{2}(t,i)]\widehat{X}dM_{ij}\\
&+\widehat{P}_{2}\Big[(\mathbf{A}+\widehat{\mathbf{A}})\widehat{X}
+\mathbf{F}_{1}\widehat{Y}+\mathbf{B}_{1}\widehat{Z}+\mathbf{B}_{2}\widehat{u}_{2}^{*}\Big]dt.
\end{aligned}
\end{equation}
Comparing the coefficients of $dW$ parts in (\ref{leader Hamiltonian system})
and (\ref{leader dY}), we obtain
\begin{equation}\label{leader Z}
\begin{aligned}
Z=(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}
\Big[\mathbf{J}_{2}X+\widehat{\mathbf{J}}_{2}\widehat{X}+\mathbf{D}_{2}u_{2}^{*}\Big].
\end{aligned}
\end{equation}
Substituting (\ref{leader FSS}) and (\ref{leader Z}) into (\ref{leader optimal condition})
and observing that $(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}$ is symmetric, we get
\begin{equation*}
\begin{aligned}
u_{2}^{*}=&-\widetilde{N}_{2}^{-1}\Big[\mathbf{S}_{2}X
+\widehat{\mathbf{S}}_{2}\widehat{X}\Big].
\end{aligned}
\end{equation*}
Inserting (\ref{leader FSS}), (\ref{leader Z}), and (\ref{leader optimal control})
into (\ref{leader Hamiltonian system}) and (\ref{leader dY}), respectively, we have
\begin{equation}\label{leader compare 1}
\begin{aligned}
dY=&-\Big[\Big(\mathbf{A}^{\top}P_{2}+\mathbf{Q}_{2}
+\mathbf{C}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{J}_{2}\\
&-\mathbf{C}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}
-\mathbf{F}_{2}^{\top}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}\Big)X\\
&+\Big(\mathbf{A}^{\top}\widehat{P}_{2}+\widehat{\mathbf{A}}^{\top}(P_{2}+\widehat{P}_{2})+\widehat{\mathbf{Q}}_{2}\\
&+\mathbf{C}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\widehat{\mathbf{J}}_{2}
+\widehat{\mathbf{C}}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}(\mathbf{J}_{2}+\widehat{\mathbf{J}}_{2})\\
&-\mathbf{C}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}\\
&-\widehat{\mathbf{C}}^{\top}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}
(\mathbf{S}_{2}+\widehat{\mathbf{S}}_{2})\\
&-\mathbf{F}_{2}^{\top}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}
-\widehat{\mathbf{F}}_{2}^{\top}\widetilde{N}_{2}^{-1}(\mathbf{S}_{2}
+\widehat{\mathbf{S}}_{2})\Big)\widehat{X}\Big]dt\\
&+\Big\{\cdots\Big\}dW+\Big\{\cdots\Big\}\bullet dM,
\end{aligned}
\end{equation}
and
\begin{equation}\label{leader compare 2}
\begin{aligned}
dY=&\Big[\Big(\dot{P}_{2}+P_{2}\mathbf{A}+P_{2}\mathbf{F}_{1}P_{2}
+P_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{J}_{2}\\
&-P_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}
-P_{2}\mathbf{B}_{2}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}\\
&+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[P_{2}(t,j)-P_{2}(t,\alpha(t))]\Big)X\\
&+\Big(\dot{\widehat{P}}_{2}+P_{2}\widehat{\mathbf{A}}+\widehat{P}_{2}(\mathbf{A}+\widehat{\mathbf{A}})
+P_{2}\mathbf{F}_{1}\widehat{P}_{2}+\widehat{P}_{2}\mathbf{F}_{1}(P_{2}+\widehat{P}_{2})\\
&+P_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\widehat{\mathbf{J}}_{2}
+\widehat{P}_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}(\mathbf{J}_{2}+\widehat{\mathbf{J}}_{2})\\
&-P_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}
-P_{2}\mathbf{B}_{2}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}\\
&-\widehat{P}_{2}\mathbf{B}_{1}(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}(\mathbf{S}_{2}+\widehat{\mathbf{S}}_{2})
-\widehat{P}_{2}\mathbf{B}_{2}\widetilde{N}_{2}^{-1}(\mathbf{S}_{2}+\widehat{\mathbf{S}}_{2})\\
&+\sum_{j\in\mathcal{M}}\lambda_{\alpha(t),j}[\widehat{P}_{2}(t,j)-\widehat{P}_{2}(t,\alpha(t))]\Big)\widehat{X}\Big]dt\\
&+\Big\{\cdots\Big\}dW+\Big\{\cdots\Big\}\bullet dM.
\end{aligned}
\end{equation}
By equalizing the coefficients of $X$ and $\widehat{X}$ in
(\ref{leader compare 1}) and (\ref{leader compare 2}),
we obtain the following two Riccati equations
\begin{equation}\label{leader 1}
\left\{
\begin{aligned}
\dot{P}_{2}(t,i)
=&-\Big[P_{2}(t,i)\mathbf{A}(t,i)+\mathbf{A}^{\top}(t,i)P_{2}(t,i)
+P_{2}(t,i)\mathbf{F}_{1}(t,i)P_{2}(t,i)+\mathbf{Q}_{2}(i)\\
&+\mathbf{J}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\mathbf{J}_{2}(t,i)\\
&-\mathbf{S}_{2}^{\top}(t,i)\widetilde{N}_{2}^{-1}(t,i)\mathbf{S}_{2}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[P_{2}(t,j)-P_{2}(t,i)]\Big],\\
P_{2}(T,i)=&\mathbf{G}_{2}(i),\quad i\in\mathcal{M},
\end{aligned}
\right.
\end{equation}
and
\begin{equation}\label{leader 2}
\left\{
\begin{aligned}
\dot{\widehat{P}}_{2}(t,i)
=&-\Big[\widehat{P}_{2}(t,i)(\mathbf{A}(t,i)+\widehat{\mathbf{A}}(t,i))
+(\mathbf{A}(t,i)+\widehat{\mathbf{A}}(t,i))^{\top}\widehat{P}_{2}(t,i)\\
&+P_{2}(t,i)\widehat{\mathbf{A}}(t,i)+\widehat{\mathbf{A}}^{\top}(t,i)P_{2}(t,i)
+P_{2}(t,i)\mathbf{F}_{1}(t,i)\widehat{P}_{2}(t,i)\\
&+\widehat{P}_{2}(t,i)\mathbf{F}_{1}(t,i)P_{2}(t,i)
+\widehat{P}_{2}(t,i)\mathbf{F}_{1}(t,i)\widehat{P}_{2}(t,i)+\widehat{\mathbf{Q}}_{2}(i)\\
&+\mathbf{J}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\widehat{\mathbf{J}}_{2}(t,i)\\
&+\widehat{\mathbf{J}}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\mathbf{J}_{2}(t,i)\\
&+\widehat{\mathbf{J}}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\widehat{\mathbf{J}}_{2}(t,i)\\
&-\mathbf{S}_{2}^{\top}(t,i)\widetilde{N}_{2}^{-1}(t,i)\widehat{\mathbf{S}}_{2}(t,i)
-\widehat{\mathbf{S}}_{2}^{\top}(t,i)\widetilde{N}_{2}^{-1}(t,i)\mathbf{S}_{2}(t,i)\\
&-\widehat{\mathbf{S}}_{2}^{\top}(t,i)\widetilde{N}_{2}^{-1}(t,i)\widehat{\mathbf{S}}_{2}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[\widehat{P}_{2}(t,j)-\widehat{P}_{2}(t,i)]\Big],\\
\widehat{P}_{2}(T,i)=&\widehat{\mathbf{G}}_{2}(i),\quad i\in\mathcal{M}.
\end{aligned}
\right.
\end{equation}
As the follower's problem, we can also let $\widetilde{P}_{2}(t,i)=P_{2}(t,i)+\widehat{P}_{2}(t,i)$,
$i\in\mathcal{M}$, to get an equation that is structurally similar to (\ref{leader 1}) and can be used
instead of (\ref{leader 2}), i.e.,
\begin{equation}\label{leader 4}
\left\{
\begin{aligned}
\dot{\widetilde{P}}_{2}(t,i)
=&-\Big[\widetilde{P}_{2}(t,i)\widetilde{\mathbf{A}}(t,i)+\widetilde{\mathbf{A}}^{\top}(t,i)\widetilde{P}_{2}(t,i)
+\widetilde{P}_{2}(t,i)\mathbf{F}_{1}(t,i)\widetilde{P}_{2}(t,i)+\widetilde{\mathbf{Q}}_{2}(i)\\
&+\widetilde{\mathbf{J}}_{2}^{\top}(t,i)(I-P_{2}(t,i)\mathbf{D}_{1}(t,i))^{-1}P_{2}(t,i)\widetilde{\mathbf{J}}_{2}(t,i)\\
&-\widetilde{\mathbf{S}}_{2}^{\top}(t,i)\widetilde{N}_{2}^{-1}(t,i)\widetilde{\mathbf{S}}_{2}(t,i)
+\sum_{j\in\mathcal{M}}\lambda_{ij}[\widetilde{P}_{2}(t,j)-\widetilde{P}_{2}(t,i)]\Big],\\
\widetilde{P}_{2}(T,i)=&\widetilde{\mathbf{G}}_{2}(i),\quad i\in\mathcal{M},
\end{aligned}
\right.
\end{equation}
where $\widetilde{\mathbf{H}}\doteq\mathbf{H}+\widehat{\mathbf{H}}$
for $\mathbf{H}=\mathbf{A},\mathbf{Q}_{2},\mathbf{J}_{2},\mathbf{S}_{2},\mathbf{G}_{2}$.
\end{proof}
Then, we compute the minimal cost for the leader under $u_{2}^{*}$
defined by (\ref{leader optimal control}), and derive the \emph{non-anticipating}
state feedback representation of the follower's optimal control (\ref{follower optimal control}).
\begin{theorem}\label{leader theorem 2}
Let Assumptions (A1) and (A2) hold. Suppose that the Riccati equations (\ref{leader 1})
and (\ref{leader 4}) have solutions $P_{2}(\cdot,i)$ and $\widetilde{P}_{2}(\cdot,i)$,
$i\in\mathcal{M}$, respectively, such that $\widetilde{N}_{2}$ and $(I-P_{2}\mathbf{D}_{1})$
are invertible. Then,
\begin{equation}\label{leader optimal cost}
\begin{aligned}
J_{2}(u_{1}^{*}(\cdot),u_{2}^{*}(\cdot))
=\langle \widetilde{P}_{2}^{(11)}(0,i)x_{0},x_{0}\rangle,
\end{aligned}
\end{equation}
where $\widetilde{P}_{2}^{(11)}(0,i)$ is taken from
$$
\widetilde{P}_{2}(0,i)=\left(
\begin{array}{cc}
\widetilde{P}_{2}^{(11)}(0,i) & \widetilde{P}_{2}^{(12)}(0,i) \\
(\widetilde{P}_{2}^{(12)})^{\top}(0,i) & \widetilde{P}_{2}^{(22)}(0,i) \\
\end{array}
\right).
$$
Moreover, the non-anticipating state feedback representation of the follower's
optimal control (\ref{follower optimal control}) is given by (\ref{follower non anticipating}).
\end{theorem}
\begin{proof}
Note that
\begin{equation*}
\begin{aligned}
&E[\langle y(T),x^{*}(T)\rangle-\langle y(0),x^{*}(0)\rangle
-\langle \psi(T),\varphi^{*}(T)\rangle+\langle \psi(0),\varphi^{*}(0)\rangle]\\
=&E[\langle G_{2}(\alpha(T))x^{*}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),x^{*}(T)\rangle
-\langle y(0),x^{*}(0)\rangle].
\end{aligned}
\end{equation*}
By applying It\^{o}'s formula for semi-martingales to
$\langle x^{*},y \rangle-\langle \psi,\varphi^{*}\rangle$, we have
\begin{equation*}
\begin{aligned}
&E[\langle G_{2}(\alpha(T))x^{*}(T)+\widehat{G}_{2}(\alpha(T))\widehat{x}^{*}(T),x^{*}(T)\rangle
-\langle y(0),x^{*}(0)\rangle]\\
=&E\bigg[\int_{0}^{T}\Big(-\langle Q_{2}x^{*},x^{*}\rangle
-\langle \widehat{Q}_{2}\widehat{x}^{*},x^{*}\rangle
+\langle u_{2}^{*},\mathbb{B}_{2}^{\top}y+\mathbb{D}_{2}^{\top}z
+\mathbb{F}_{2}\psi+\widehat{\mathbb{F}}_{2}\widehat{\psi}\rangle\Big)dt\bigg]\\
=&E\bigg[\int_{0}^{T}\Big(-\langle Q_{2}x^{*},x^{*}\rangle
-\langle \widehat{Q}_{2}\widehat{x}^{*},x^{*}\rangle
+\langle u_{2}^{*},\mathbf{B}_{2}^{\top}Y
+\mathbf{D}_{2}^{\top}Z+\mathbf{F}_{2}X
+\widehat{\mathbf{F}}_{2}\widehat{X}\rangle\Big)dt\bigg],
\end{aligned}
\end{equation*}
which implies that (noting (\ref{leader optimal condition}))
\begin{equation*}
\begin{aligned}
J_{2}(u_{1}^{*}(\cdot),u_{2}^{*}(\cdot))
=\langle Y(0),X(0)\rangle=\langle \widetilde{P}_{2}(0,i)X(0),X(0)\rangle
=\langle \widetilde{P}_{2}^{(11)}(0,i)x_{0},x_{0}\rangle.
\end{aligned}
\end{equation*}
On the other hand, note that $u_{2}^{*}$ defined by (\ref{leader optimal control})
for the leader is non-anticipating, thereby $u_{1}^{*}$ defined by
(\ref{follower optimal control}) for the follower can be also represented
in a non-anticipating way, i.e.,
\begin{equation}\label{follower non anticipating}
\begin{aligned}
u_{1}^{*}=&-\widetilde{N}_{1}^{-1}\Big[S_{1}x+\widehat{S}_{1}\widehat{x}+\Phi\Big]\\
=&-\widetilde{N}_{1}^{-1}\Big[(
\begin{array}{cc}
S_{1} & 0 \\
\end{array}
)X+(
\begin{array}{cc}
\widehat{S}_{1} & 0 \\
\end{array}
)\widehat{X}
+(
\begin{array}{cc}
0 & B_{1}^{\top} \\
\end{array}
)Y+(
\begin{array}{cc}
0 & D_{1}^{\top} \\
\end{array}
)Z+D_{1}^{\top}P_{1}D_{2}u_{2}^{*}\Big]\\
=&-\widetilde{N}_{1}^{-1}\Big[(
\begin{array}{cc}
S_{1} & 0 \\
\end{array}
)+(
\begin{array}{cc}
0 & B_{1}^{\top} \\
\end{array}
)P_{2}+(
\begin{array}{cc}
0 & D_{1}^{\top} \\
\end{array}
)(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{J}_{2}\\
&-(
\begin{array}{cc}
0 & D_{1}^{\top} \\
\end{array}
)(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}
-D_{1}^{\top}P_{1}D_{2}\widetilde{N}_{2}^{-1}\mathbf{S}_{2}\Big]X\\
&-\widetilde{N}_{1}^{-1}\Big[(
\begin{array}{cc}
\widehat{S}_{1} & 0 \\
\end{array}
)+(
\begin{array}{cc}
0 & B_{1}^{\top} \\
\end{array}
)\widehat{P}_{2}+(
\begin{array}{cc}
0 & D_{1}^{\top} \\
\end{array}
)(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\widehat{\mathbf{J}}_{2}\\
&-(
\begin{array}{cc}
0 & D_{1}^{\top} \\
\end{array}
)(I-P_{2}\mathbf{D}_{1})^{-1}P_{2}\mathbf{D}_{2}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}
-D_{1}^{\top}P_{1}D_{2}\widetilde{N}_{2}^{-1}\widehat{\mathbf{S}}_{2}\Big]\widehat{X}.
\end{aligned}
\end{equation}
The proof is completed.
\end{proof}
\begin{remark}
Up to now, we have completely solved our LQ leader-follower stochastic differential game for
mean-field switching diffusion. It turns out that the game admits an open-loop Stackelberg
equilibrium $(u_{1}^{*},u_{2}^{*})$ with a non-anticipating state feedback representation
(\ref{follower non anticipating}) and (\ref{leader optimal control}), respectively.
\end{remark}
Finally, we provide a numerical example to illustrate the effectiveness of our
theoretical results. Note that the optimal controls (\ref{follower non anticipating})
for the follower and (\ref{leader optimal control}) for the leader as well as
the value of the game (\ref{leader optimal cost}) depend only on the solutions
$P_{1}$, $\widetilde{P}_{1}$, $P_{2}$, $\widetilde{P}_{2}$ to Riccati equations
(\ref{follower 1}), (\ref{follower 5}), (\ref{leader 1}), (\ref{leader 4}), respectively.
So, in order to implement our control policies in practice, the \emph{whole} task for us
is to compute $P_{1}$, $\widetilde{P}_{1}$, $P_{2}$, $\widetilde{P}_{2}$.
\begin{example}
Let $n=m_{1}=m_{2}=1$ and $T=1$. Consider the following state equation:
\begin{equation*}
\left\{
\begin{aligned}
dX(t)=&[B_{1}(\alpha(t))u_{1}(t)+B_{2}u_{2}(t)]dt+CX(t)dW(t),\\
X(0)=&x_{0},
\end{aligned}
\right.
\end{equation*}
where $\alpha(\cdot)$ is a two-state Markov chain taking values in $\mathcal{M}=\{1,2\}$
with generator
\begin{equation*}
\begin{aligned}
\left[
  \begin{array}{cc}
    -1 & 1 \\
    1 & -1 \\
  \end{array}
\right],
\end{aligned}
\end{equation*}
and $B_{1}(1)=2$, $B_{1}(2)=1$, $B_{2}=1$, $C=0.5$.

The cost functionals for the follower and the leader are given by
\begin{equation*}
\begin{aligned}
J_{k}(u_{1}(\cdot),u_{2}(\cdot))=\frac{1}{2}
E\bigg[\int_{0}^{1}N_{k}u_{k}^{2}(t)dt+G_{k}X^{2}(1)
+\widehat{G}_{k}(E[X(1)|\mathcal{F}_{1}^{\alpha}])^{2}\bigg],
\end{aligned}
\end{equation*}
where $N_{k}=1$, $G_{k}=1$, $\widehat{G}_{k}=0.5$, $k=1,2$, respectively.
Note that in this example, to exhibit the effect of regime switching more clearly,
we only let $B_{1}$ vary depending on the Markov chain and keep all the other parameters
fixed as constants.

Then, $P_{1}(t,i)$, $\widetilde{P}_{1}(t,i)$, $P_{2}^{(11)}(t,i)$, $\widetilde{P}_{2}^{(11)}(t,i)$,
$i\in\{1,2\}$, on $[0,1]$ are computed and plotted in Figures \ref{FR} and \ref{LR}, respectively.
It is mentioned that the other elements of the matrix-valued functions $P_{2}(t,i)$ and
$\widetilde{P}_{2}(t,i)$, $i\in\{1,2\}$, are not plotted for simplicity.
\begin{figure}
\centering
\includegraphics[width=5.5in]{FollowerRiccati.eps}
\caption{Riccati equations for the follower}\label{FR}
\includegraphics[width=5.5in]{LeaderRiccati.eps}
\caption{Riccati equations for the leader}\label{LR}
\end{figure}
\end{example}

\section{Concluding remarks}\label{conclusion}

In this paper, we studied an LQ leader-follower stochastic differential game
with regime switching and mean-field interactions. Conditional mean-field terms
are included due to the presence of a Markov chain (just like a \emph{common noise}).
Some new-type Riccati equations are introduced for the first time in the literature.
The open-loop Stackelberg equilibrium and its non-anticipating state feedback
representation are obtained. There are several interesting problems that deserve
further investigation, in particular, the existence and uniqueness results of
the Riccati equations (\ref{leader 1}) and (\ref{leader 4}).

redpajama_set_name: RedPajamaArXiv


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I plan to start adding this to Dexter's dinner meal. As a test, we put some on his lunch time meal and he never noticed. It is liver flavored, and most dogs I know love liver! I also learned that a quality probiotic like this can help firm up a dog's stool. Who doesn't love firm dog poop?!
There are 10x more bacteria in your body than cells! Most of those bacteria reside in your gut, too.
Probonix is made up of 100% natural ingredients.
All probiotics do not need to be refrigerated.
Humarian also makes a human version, called Probonix (*Women). I came home with a bottle of it and will be using the cherry flavored version daily for myself!
What strains are in the pet version? I had to know, and if you want the intel on the strains, click here.
Keep in mind that the Probonix for Dogs is for use in dogs only. If your dog's condition worsens or does not improve, stop product administration and consult your veterinarian.
I loved hearing learning about this product, so I decided to become an affiliate. You can order from Humarian for probiotics for your dog and all members of the family.
Let me know if you start using this line for yourself, your family, and/or your dog. I am looking forward to seeing how they all help keep our guts thriving to their maximum efficacy!
Have you ever tried a probiotic for your dog(s) or yourself? Share your thoughts in the comments below.
We brought some home from BlogPaws, too, and are looking forward to trying it with Ruby. She has allergies and some immune system-related issues so I'm hopeful that it will help.
And LOL, the reason I haven't started her on it yet is because I couldn't read the label! The print is tiny and I thought I might need a magnifying glass to read it. Thanks for giving the dosage for small dogs.
I am going to make a note of the fact the label is so small and let Humarian know. I am curious to see how it will help Dex and me, too, since they gave me the women's sample to try.
Good gut health is important for dogs.
Good gut health is so important for everybody!
I have never tried a probiotic for my rescues and had no idea that based on All the info you shared in your post that they could both really benefit from them! Thank you so much for sharing your findings! I feel so good learning about ways I can make my fur babies happier and healthier!
You are very welcome. I learned so much talking to Humarian and cannot wait to see how it helps Dexter…and I started the human version, too!
A good gut is healthy for everyone humans and fur babies. I must admit I would never have thought to give a dog probotics so thank you for this great information.
We need probiotics so it totally makes sense that our 4 legged friends do too! I know how good my family and I feel when we take them regularly so super happy you were able to do this with your dog!
We take probiotics but I never thought about dogs being able to take them too! I think it's great that they can, and they seem to help with so many different problems!
At our vet's suggestion, we've had Tavish on a probiotic for the past few months, and in concert with some slight diet modifications, it does seem to be making a difference in keeping things more even for him. Good to know about this particular brand, thanks!
I like the convenience of this one and I am taking the human version with my dog. Glad to hear Tavish is doing well!
I have started my cats and dogs on Humarian probiotics!
Not only is it a wonderful brand but the people of Humarian are so kind.
That makes a big difference to me.
My family have been firm believers in probiotics for humans for over a decade and I was so thrilled to learn someone made them for pets!
We are excited to see how it helps us. I am so glad you all are on Humarian, too, Joely!
My Husky was prescribed a probiotic when one of her tummy upsets occurred – Auntie Janice fed her a donut during a family gathering – BAD Auntie Janice!! It helped tremendously right away. I've wanted to start giving my dogs a probiotic but I felt I didn't know enough about it. I've also heard you should provide a prebiotic along with it? Thanks for this great info!
Any update on how you Humarian is working for Dexter? Curious because we're about to reorder our probiotic supply and am considering giving this brand a try. We like our current brand which is in powder form (and it does seem to be helping), but Tavish is very picky and doesn't always like his food sprinkled with it.
Yes, as a matter of fact, I have a blog post this week with a video on how it is working. In short, famously. We love it and he doesn't even know I put it in his food. I hear you on pickiness. Stay tuned. Thanks for asking!


The Ombudsman has called on government to buy urban land to resettle those who do not want to farm, while at the same time cautioning it to stop buying farms without water and infrastructure.
This, the Ombudsman John Walters shared as recommendations in a lengthy report on the Land Reform and Resettlement Programme in Namibia.
Collected from public hearings and written submissions, Walters said the lack of available land especially urban land, adding to the inflated high prices are key issues for consideration. At the same time resettled farmers, particularly small scale farmers call for the government to support them since many of them are dumped in farms not fit for farming.
Other submissions enlisted in the report include ideas that farms should be subdivided into more units so that more people can be resettled. There are also calls for the resettlement of persons above the age of 80 to stop.
"The current Resettlement Program forces us to share land with people who have never lost land. The return of ancestral land must be part of the policy and such a policy should not prioritize the number of livestock but rather restitution," reads another submission.
The Ombudsman also criticized the Land Reform Act, saying the primary objectives of the national resettlement policy have thus far been achieved but only in part.
According to the report, recent research conducted has revealed that since the inception of the resettlement programme, the set objective of attaining self-sufficiency by means of 'creating employment through full-time farming and bringing smallholder farmers into the mainstream of the Namibian economy by producing for the market' has not been accomplished.
The process of applying for resettlement also often bypasses the San, one reason being that most San people in Namibia live in remote areas and are not made aware of resettlement projects planned in their areas until it is too late to apply.
Another reason is that many are illiterate and unable to submit a written application, the report further highlights.
The Ombudsman also encourages the state party to implement its policies on land reform in such a way to ensure the equal exercise by the different ethnic communities of the rights enshrined in the Convention within the framework of a democratic system.
Without prejudice to the overall objective of the land reforms and the "willing seller – willing buyer" approach, the Ombudsman is concerned that the land reform programme of the State party has not addressed poverty or that security of tenure remains an enduring challenge, as a large number of individual and communal land owners are without title.
The Ombudsman is also concerned that many resettled farmers have not been able to restore their livelihoods and earn an adequate standard of living, even when support has been provided. Moreover, the Committee is concerned that the Communal Land Reform Act of 2002 has had little effect on women's access to land.
In this regard, the Ombudsman expressed its concern at women's limited access to land and the lack of information on rural women's access to microfinance and microcredit schemes.
He further stressed the need for clarity about resettlement while pointing out the inconsistencies in application of selection criteria for resettlement as well as inadequate selection criteria.
The non-productive use of resettlement farms by beneficiaries, affordability of farm land for resettlement purposes and inadequate support for resettled farmers and groups also comes out as a matter of concern.
"While we intended to submit our report to the Land Reform minister before the start of the Second National Land Conference, it turned out not to be possible for various reasons, among others, the extend of the research, the inordinate delay by the ministry to provide the Ombudsman with the full list of beneficiaries, the proofreading and printing of the report," said Walters.


Wellllcoooommmeee Ladies and Gentleman, let's get ready to ruummmmmbbbblllleee!
Its Round 2 of SO Obsessed Sundays!
1. Contact your local architect now.
May I recommend Kustom Home Design (shameless plug for my mama y'all, sorry). But seriously, I'm saving these 36 things that I didn't know I ever wanted in my dream home, but now I absolutely have to have.
Gah I can't wait for my multiple waterfalls (including the bath), two story closet and I just know Cotton and Mallie are going to love that dog pool!
2. Looking at Married Life through GOLD Colored Glasses.
It's okay to have a second wedding within 3 years of being married right? Because I can't imagine a more perfect wedding day – gold and white accessories, a pool, monograms and extremely tall flower arrangements. Oh, did I mention ribbon chiavari chairs?!
4. Sucker for a Sale!
Man, I'm a sucker for a sale! And this is the perfect time of year to get your shop on, a lot of my favorite places are giving you an EXTRA percentage off their sale items. And while we're on the royal subject, I have to believe Mama Middleton would give this blue polka dot number from J.Crew a thumbs up, especially at an extra 40% off!
Also check out Kate Spade, Tory Burch, World Market and Piperlime!
Quck TIp: You'll need to sign up for their emails to get the EXTRA, EXTRA discount!
The amount of excitement I get from receiving this little gem of a publication each month is unreal. I feel like this is the most approachable and useful glossy that I get when it comes to home design and decorating. If you don't get this already, go, run, frolic, whatever(!), subscribe now!
Hope you're as obsessed as I am! Happy Sunday!


Located in the heart of Lebanon in picturesque Wilson County, Colonial Village presents classically styled homes.. Just 25 miles east of Nashville, you can enjoy the peaceful solitude of a country locale, with easy access to the urban activities, with commuter rail service a short drive from Colonial Village.
Goodall Homes is offering six different floor plans for the remaining homes in the final phase of Colonial Village. The homes range from 1,638 to 2,647 square feet of indoor living space, with one- and two-story designs. We'll help you find the plan that best suits your style. Every home includes a two-car garage, hardwood flooring, energy efficient design, and the option of adding a bonus room. Our standard features already reflect upgraded quality!
Spend your free time enjoying Colonial Villages' community amenities! Catch some rays down by the swimming pool, build up endurance at the fitness center and have lunch in the picnic pavilion, and so much more! Save time and energy, and ask us about our plans that have landscaping and lawn maintenance included with your home.
With only a few homesites left, don't miss this opportunity to become part of this unique, affordable Goodall Homes neighborhood. Visit Colonial Village and discover why so many people are happy to come home here.
Awesome all brick home in prime location! Convenient to everything but lots of privacy, sits way back off the road and backs up to corps property! Huge 24x40 detached shop with electric and 100 amp subpanel. Beautiful setting!


The SMPTC will assist in obtaining rooms and entrance fees to attractions for writers with commissioned work. It is up to the individual property to decide if services will be complimentary or at a reduced rate.
Costs incurred by family members/friends traveling with a travel writer will be paid for by the writer.
Writers must give the SMPTC at least two weeks notice before their arrival.


text:
Jirečkova linie je imaginární linie přes starověký Balkán, která rozdělovala vliv latiny (na severu) a řečtiny (na jihu) až do 4. století. Prvně toto označení použil český historik Konstantin Jireček v roce 1911 v díle Geschichte der Serben (Historie Srbů).

Linie začínala poblíž města Laç v současné Albánii a vedla přes Serdiku (nyní Sofie v Bulharsku) a pak podél pohoří Stara planina do Odesu (Varna) k Černému moři. Umístění linie je založeno na archeologických nálezech: většina nápisů nalezených na sever od linie byla psána v latině, zatímco většina nápisů nalezených na jihu byla v řečtině.

Tato linie je důležitá k určení oblasti, kde se tvořil rumunský a arumunský lid, má se za to, že latinsky mluvící lid se mohl zformovat poblíž této oblasti, oproti tomu Valaši se nacházeli na sever a na jih od ní.

Několik vědců později vyslovilo poněkud odlišný názor: finský historik Jorma Kaimio (1979) umístil Dalmácii a Horní Moesii do latinské oblasti a Dolní Moesii do oblasti řeckého vlivu. M. D. MacLeod (1982) naznačil, že oficiální politický jazyk nemusel být používán pro každý aspekt běžného života, ale že jednotliví římští úředníci se rozhodovali podle své úvahy pro každý určitý případ. Také poukázal na to, že dokonce i v řeckých oblastech byla latina dominantním jazykem na nápisech zaznamenávajících veřejné práce, na milnících a v armádě.

Literatura
 Jireček, Konstantin, Geschichte der Serben ' The history of the Serbians ', Gotha, 1911.
 Kaimio, Jorma, The Romans and the Greek Language, Commentationes Humanarum Litterarum 64. Helsinki: Societas Scientiarum Fennica, 1979.
 MacLeod, M.D., review of Kaimio, 1979 in The Classical Review, New Ser., 32:2:216-218, 1982.

Dějiny Balkánu
Česká pozitivistická historiografie
Latina
Řečtina
Izoglosy

redpajama_set_name: RedPajamaWikipedia


Most investors tend to think that hedge funds and other asset managers are worthless, as they cannot beat even simple index fund portfolios. In fact, most people expect hedge funds to compete with and outperform the bull market that we have witnessed over the past few years. However, hedge funds are generally partially hedged and aim at delivering attractive risk-adjusted returns rather than following the ups and downs of equity markets, hoping that they will outperform the broader market. Our research shows that hedge funds do have great stock picking skills, so let's take a glance at the smart money sentiment towards Wells Fargo & Co (NYSE:WFC).
Wells Fargo & Co (NYSE:WFC) was in 85 hedge funds' portfolios at the end of September. Wells Fargo & Co (NYSE:WFC) shareholders have witnessed a decrease in support from the world's most elite money managers lately. There were 91 hedge funds in our database with Wells Fargo & Co (NYSE:WFC) positions at the end of the previous quarter. At the end of this article, we will also compare Wells Fargo & Co (NYSE:WFC) to other stocks, including Johnson & Johnson (NYSE:JNJ), General Electric Company (NYSE:GE), and Facebook Inc (NASDAQ:FB) to get a better sense of its popularity.
In the eyes of most traders, hedge funds are viewed as worthless, old investment vehicles of yesteryear. While there are over 8000 funds with their doors open at present, our researchers look at the moguls of this group, about 700 funds. These money managers command the lion's share of all hedge funds' total capital, and by monitoring their best investments, Insider Monkey has determined many investment strategies that have historically defeated the S&P 500 index. Insider Monkey's small-cap hedge fund strategy surpassed the S&P 500 index by 12 percentage points a year for a decade in their back tests.
Now, let's review the latest action regarding Wells Fargo & Co (NYSE:WFC).
How have hedgies been trading Wells Fargo & Co (NYSE:WFC)?
At the end of September, a total of 85 of the hedge funds tracked by Insider Monkey were long this stock, a decline of 7% from the second quarter. With the smart money's positions undergoing their usual ebb and flow, there exists a select group of key hedge fund managers who were upping their stakes considerably (or already accumulated large positions).
When looking at the institutional investors followed by Insider Monkey, Berkshire Hathaway has the most valuable position in Wells Fargo & Co (NYSE:WFC), worth close to $24.15 billion, comprising 19% of its total 13F portfolio. The second largest stake is held by Ken Fisher of Fisher Asset Management, with a $965.8 million position; 2% of its 13F portfolio is allocated to the stock. Remaining peers that hold long positions encompass Tom Russo's Gardner Russo & Gardner, Patrick Degorce's Theleme Partners, and Cliff Asness' AQR Capital Management.


Join us for our fifth New Black Showcase and discover the exciting playwriting talent of the future. Once again the Belgrade is providing a platform for the voices of new writers from Black, Asian and Minority Ethnic backgrounds. Previous New Black Showcase writers include Liz Mytton, whose play Red Snapper went on to be professionally produced by the Belgrade in 2016.
Mother and Bibi speak Punjabi. Miah and Roma speak Emoji. As phone calls and texts flood the home, can the family figure out what happened to Miah?
Maroon rebel Accompong attempts to woo the Queen of Blue Mountain whilst Chief Cudjoe makes a deal with an infamous Jewish trader. Join us for Shakespeare's The Merchant of Venice with a Caribbean twist.
Midlands-based Futuretheatre presents extracts from Invisible by Daniel Anderson and Ninety Days by Ashok Patel as a tantalising taster before their showcase at mac Birmingham.


The Xavier University Newswire students can sell Television Production Handbook (ISBN# 0495898848) written by Herbert (Herbert Zettl) Zettl, Herbert Zettl and receive a check, along with a free pre-paid shipping label. Once you have sent in Television Production Handbook (ISBN# 0495898848), your Xavier University Newswire textbook will be processed and your check will be sent out to you within a matter days. You can also sell other The Xavier University Newswire textbooks, published by CENGAGE Learning and written by Herbert (Herbert Zettl) Zettl, Herbert Zettl and receive checks.