\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{mathrsfs}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 123, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/123\hfil
 Robustness of mean-square exponential dichotomies]
{Robustness of mean-square exponential dichotomies for linear
stochastic equations}

\author[H. Zhu, Y. Jiang \hfil EJDE-2017/123\hfilneg]
{Hailong Zhu, Yongxin Jiang}

\address{Hailong Zhu \newline
Department of Mathematics,
Shanghai Normal University,
Shanghai 200234, China.\newline
School of Statistics and Applied Mathematics,
Anhui University of Finance and Economics,
Bengbu 233030, China}
\email{hai-long-zhu@163.com}

\address{Yongxin Jiang \newline
 Department of Mathematics,
College of Science, Hohai University,
Nanjing 210098, China}
\email{yxinjiang@126.com}

\dedicatory{Communicated by Zhaosheng Feng}

\thanks{Submitted July 4, 2016. Published May 5, 2017.}
\subjclass[2010]{60H10, 34D09}
\keywords{Robustness; mean-square exponential contraction;
\hfill\break\indent mean-square exponential dichotomy; 
stochastic differential equations}

\begin{abstract}
 We present the notion of mean-square exponential dichotomies for
 linear stochastic differential equations. We study the robustness of
 the mean-square exponential dichotomies, in the sense of the existence
 of a mean-square exponential dichotomy for a given linear stochastic equation
 persists under sufficiently small linear perturbations.
 As a special case, we consider mean-square exponential contractions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

The notion of exponential dichotomies \cite{per}
plays an important role in the theory of differential equations and
dynamical systems, particularly in what concerns the study of stable and
unstable invariant manifolds, and therefore has attracted much attention during
the last few decades. We refer to \cite{chi, cop, ms, ss}
for details related to exponential dichotomies. Exponential dichotomy of stochastic
cocycles was first introduced in \cite{st}. Among those results concerning
exponential dichotomies, the so-called robustness problem is very important
and has a long history. We refer to \cite{cl, chu, cop} and the references
therein for the study of robustness of exponential dichotomies.

Let $I$ be any interval on $\mathbb{R}$ and $A(t)=(A_{ij}(t))_{n \times n}$,
$G(t)=(G_{ij}(t))_{n \times n}$ be Borel-measurable, bounded functions.
In this study, we will introduce the notion of
mean-square exponential dichotomies for the
nonautonomous linear stochastic differential equations (SDEs for short)
\begin{equation} \label{a1}
dx(t)=A(t)x(t)dt+G(t)x(t)\,d\omega(t), \quad t\in I,
\end{equation}
and limit our attention to the robustness, which means that such a
mean-square exponential dichotomy persists under sufficiently small linear
perturbations. Precisely, we consider the perturbed stochastic
differential equation
\begin{equation} \label{a3}
dy(t)=(A(t)+B(t))y(t)dt+(G(t)+H(t))y(t)\,d\omega(t),
\end{equation} and we
prove that \eqref{a3} admits a mean-square exponential dichotomy for any
arbitrary small perturbations
$B,H$ if the same happens for \eqref{a1}. We also explore the continuous
dependence with the perturbation of the constants in the notion of
dichotomies. Note that in \eqref{a3} the perturbations appear in the
``drift" as well as in the ``volatility" and the proofs of the main results
will become more complicated and difficult than those for linear determined
equations.

Stochastic differential equations have been studied by many researchers
on various problems because SDEs have important applications in many scientific area.
We refer the reader to \cite{ar74, ev, ga, lad, mao, mzz} for more
information about SDEs.
Among those topics, the study of mean-square dynamical behavior of SDEs
is an important and interesting one and has attracted many researchers
\cite{fl, hig, hms, hmy, kl}.
Mean-square dynamical behavior are essentially
deterministic with the stochasticity built into or hidden in the
time-dependent state spaces. In \cite{kl}, Kloeden and Lorenz
provided a definition of mean-square random dynamical systems
and studied the existence of pullback attractors (we refer to \cite{ar98}
for details on random dynamical systems).
In \cite{fl, ls}, the concept of mean-square almost
automorphy for stochastic process is introduced and the existence, uniqueness
and asymptotic stability of mean-square almost automorphic solutions of some
linear and nonlinear stochastic differential
equations are established.  In \cite{hig}, Higham provided a stochastic
version of the theta method for mean-square asymptotic stability.

Now we introduce some notation.
Let $(\Omega, \mathscr{F}, \{\mathscr{F}_t\}_{t\geq 0},
\mathbb{P})$ be a standard filtered probability space, i.e.,
$(\Omega, \mathscr{F}, \mathbb{P})$ is a complete probability space,
$\{\mathscr{F}_t\}_{t\geq 0}$ is a filtration with
$\mathscr{F}_0$ contains all $\mathbb{P}$-null sets. For a matrix
or a vector $A$, we use $A^{T}$ to denote its transpose.  Let
$\omega(t)=(\omega_1(t),\ldots \omega_{n}(t))^{T}$ be an
$n$-dimensional Brownian motion defined on the space $(\Omega,
\mathscr{F}, \{\mathscr{F}_t\}_{t\geq 0}, \mathbb{P})$. Let
$\|\cdot\|$ be the Euclidean norm in $\mathbb{R}^{n}$ or operator norm.
In addition, let $L^2_{\mathscr{F}_{s}}(\Omega, \mathbb{R}^{n})$
denote the family of all  $\mathscr{F}_{s}$-measurable
$\mathbb{R}^{n}$-valued random variables, i.e., $\xi_{s}: \Omega \to
\mathbb{R}^{n}$ such that $\mathbb{E} \|\xi_{s}\|^2<\infty$ for all $s\geq 0$.
Let $I^2_{\geq}:=\{(t,s)\in I^2: t\geq s\}$ and
$I^2_{\leq}:=\{(t,s)\in I^2: t\leq s\}$.

The rest part of this article is organized as follows. In Section 2, we present
the robustness of mean-square exponential contractions, and the robustness of
mean-square exponential dichotomies is showed in Section 3.

\section{Robustness of mean-square exponential contractions}

In this section we consider the robustness of mean-square
exponential contractions.

\begin{definition}\label{def21} \rm
We say that \eqref{a1} admits a mean-square exponential contraction
if there exist positive constants $M$ and $\alpha$ such that, for any solution
$x(t)\in L^2_{\mathscr{F}_t}(\Omega, \mathbb{R}^{n})$ of  \eqref{a1},
 \begin{equation} \label{b1}
\mathbb{E}\|x(t)\|^2\leq M e^{-\alpha(t-s)}\mathbb{E}\|x(s)\|^2, \quad \forall
(t,s)\in I^2_{\geq}.
\end{equation}
\end{definition}

\begin{lemma} \label{lem2.2}
Let $\Phi(t)$ be a fundamental matrix solution  of
\eqref{a1}. Then \eqref{a1} admits a mean-square exponential contraction
if and only if
\[
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2 \leq M e^{-\alpha(t-s)}, 
\quad \forall (t,s)\in I^2_{\geq}.
\]
\end{lemma}

\begin{proof}
From \cite{lad, mao} it follows that $\Phi(t)$ of \eqref{a1} is invertible
with probability $1$ for all $t\in I$.
First, we have
\[
\mathbb{E}\|x(t)\|^2=\mathbb{E}[\|\Phi(t)\Phi^{-1}(s)x(s)\|^2]=\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2
\mathbb{E}\|x(s)\|^2,
\]
where $\Phi(t)\Phi^{-1}(s)$ and $x(s)$ are independent, and therefore
\[
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2=\frac{\mathbb{E}\|x(t)\|^2}{\mathbb{E}\|x(s)\|^2},
\]
where $\mathbb{E}\|x(s)\|^2\neq 0$; or else \eqref{a1} admits a ``trivial"
solution due to \eqref{b1}, i.e., $\mathbb{E}\|x(t)\|^2 = 0$ for all
$(t,s)\in I^2_{\geq}$. Thus we can obtain from \eqref{b1} that
\[
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2 \leq M e^{-\alpha(t-s)}, \quad \forall
(t,s)\in I^2_{\geq}.
\]
the proof of the converse is very similar.
\end{proof}

The following variation of parameters formula will be essential to prove
our main result of this section. The corresponding version of the nonlinear
perturbation of \eqref{a1} can be found in \cite{mao}.

\begin{lemma}[{\cite[Section 2.4.2]{lad}}] \label{lem24}
 Let $\Phi(t)$  be a fundamental matrix of
\eqref{a1}. Then the solution of \eqref{a3} is given as
\begin{equation} \label{b5}
\begin{aligned}
y(t)&= \Phi(t)\Phi^{-1}(s)y(s)+\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)\big{[}B(\tau)-
G(\tau)H(\tau)\big]y(\tau)d\tau \\
&\quad  + \int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)H(\tau)y(\tau)\,d\omega(\tau),
\end{aligned}
\end{equation}
for all $(t,s)\in I^2_{\ge}$.
\end{lemma}

\begin{theorem}\label{main21}
Assume that \eqref{a1} admits a mean-square exponential contraction with
\eqref{b1}. Furthermore, assume that $B(t), G(t)$ and $H(t)$
are all Borel-measurable and there exist nonnegative
constants $\beta, g, h$ such that
\begin{equation} \label{b2}
\|B(t)\|\le \beta,\quad \|G(t)\|\le g, \quad
\|H(t)\|\le h,\quad t\in I.
\end{equation}
Then any solution $y(t)$ of \eqref{a3} satisfies
\begin{equation} \label{m1}
\mathbb{E}\|y(t)\|^2\leq 3M e^{(-\alpha+3MK)(t-s)}\mathbb{E}\|y(s)\|^2, \quad \forall
 (t,s)\in I^2_{\geq},
\end{equation}
 where $K=2\beta^2+2g^2h^2+h^2$.
In particular, \eqref{a3} also admits a mean-square exponential
contraction if
\begin{equation} \label{bgh}
K<\frac{\alpha}{3M}.
\end{equation}
\end{theorem}

\begin{proof}
Given any initial value $y(s)$ at time
$s$,  using Lemma \ref{lem24}, the solution of \eqref{a3} can
be expressed as \eqref{b5} with $(t,s)\in I^2_{\geq}$.

Using conditions \eqref{b2}, the H\"{o}lder's inequality and the
elementary inequality
\begin{equation} \label{ele}
\|\sum_{k=1}^{m}a_{k}\|^2 \le m\sum_{k=1}^{m}\|a_{k}\|^2
\end{equation}
one can obtain that
\begin{equation} \label{b9f}
\begin{aligned}
\|y(t)\|^2&\leq 3\|\Phi(t)\Phi^{-1}(s)\|^2\|y(s)\|^2+
3\big\|\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)H(\tau)y(\tau)\,d\omega(\tau)\big\|^2  \\
&\quad  +3\int^{t}_{s}\|\Phi(t)\Phi^{-1}(\tau)\|^2\|B(\tau)-
G(\tau)H(\tau)\|^2\|y(\tau)\|^2d\tau  \\
&\leq  3\|\Phi(t)\Phi^{-1}(s)\|^2\|y(s)\|^2+
3\big\|\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)H(\tau)y(\tau)\,d\omega(\tau)\big\|^2 \\
&\quad +6(\beta^2+g^2h^2)\int^{t}_{s}\|\Phi(t)\Phi^{-1}(\tau)\|^2\|y(\tau)\|^2d\tau.
\end{aligned}
\end{equation}
 By \eqref{b9f} and
\begin{equation} \label{it}
\mathbb{E}\Big{[}\Big{(}\int^{t}_{s}x(\tau)\,d\omega(\tau)\Big{)}^2\Big{]}
=\mathbb{E}\Big{[}\int^{t}_{s}x^2(\tau)d \tau\Big{]},\quad
 x(\tau)\in L^2_{\mathscr{F}_{\tau}}(\Omega, \mathbb{R}^{n})
\text{ for } \tau\in[s,t]
\end{equation}
we have
\begin{equation} \label{b10f}
\begin{aligned}
\mathbb{E}\|y(t)\|^2
&\leq 3\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2\mathbb{E}\|y(s)\|^2+
3\int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)\|^2\mathbb{E}\|H(\tau)y(\tau)\|^2d\tau \\
&\quad  +6(\beta^2+g^2h^2)
\int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)\|^2\mathbb{E}\|y(\tau)\|^2d\tau\\
&\leq  3M e^{-\alpha(t-s)}\mathbb{E}\|y(s)\|^2+3MK
\int^{t}_{s}e^{-\alpha(t-\tau)}\mathbb{E}\|y(\tau)\|^2d\tau.
\end{aligned}
\end{equation}
Let
\begin{equation} \label{m2}
u(t)=e^{\alpha t}\mathbb{E}\|y(t)\|^2,\quad
U(t)=3Mu(s)+3MK\int^{t}_{s}u(\tau)d\tau.
\end{equation}
 We can rewrite inequality \eqref{b10f} as
\[
u(t)\le U(t),\quad \text{for all }t\ge s.
\]
On the other hand,
$\frac{d}{dt}U(t)=3MKu(t)$, and thus,
\[
\frac{d}{dt}U(t)\leq 3MK U(t).
\]
Integrating the above inequality from $s$ to $t$ and note that
$U(s)=3M u(s)$, we obtain
\[
u(t)\leq U(t)\leq 3M u(s)
e^{3MK(t-s)},\quad \text{for  all } (t,s)\in I^2_{\geq}.
\]
Now the inequality \eqref{m1} follows from \eqref{m2} and
the proof is complete.
\end{proof}

\begin{remark}\label{remark21} \rm
Assume that \eqref{a1} and \eqref{a3} have the same initial
condition, that is, $x(s)=y(s)$. By using the Theorem
\ref{main21}, for $\beta$, $g$, $h$ being sufficiently small, we have
\begin{align*}
\mathbb{E}\|y(t)-x(t)\|^2
&\leq 2K\int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)\|^2\mathbb{E}\|y(\tau)\|^2d\tau\\
&\leq  6M^2K \int^{t}_{s}e^{-\alpha(t-\tau)}e^{(-\alpha+3MK)(\tau-s)}d\tau\\
&=  6M^2Ke^{-\alpha(t-s)} \frac{e^{3MK(t-s)}-1}{3MK}\\
&\leq  2M e^{(-\alpha+3MK)(t-s)}.
\end{align*}
Thus for each $\beta$, $g$ and $h$  with \eqref{bgh}, we have
\[
\lim_{t\to +\infty}\frac{1}{t}\log(\mathbb{E}\|y(t)-x(t)\|^2)=-\alpha+3MK<0,
\]
which means that the solution of the linear perturbation equation
\eqref{a3} is forward asymptotic to the solution of \eqref{a1} in
mean-square sense if they have the same initial data.
\end{remark}

In the rest of this section, as a special case of \eqref{a3}, we
consider
\begin{equation} \label{a4}
dy(t)=(A(t)+B(t))y(t)dt+G(t)y(t)\,d\omega(t),
\end{equation}
in which the perturbed term only appears in the ``drift". Of course,
Theorem \ref{main21} can be applicable to \eqref{a4}.
In the following, we will obtain another robustness result for \eqref{a4},
in which the constants can be improved slightly. In this case, the results are
more similar to those for ordinary differential equations \cite{hale}.

\begin{lemma}[{\cite[Theorem 2.3.1]{lad}}] \label{lem27}
 Let $\Phi(t)$ be a fundamental matrix of \eqref{a1}. Then the matrix
$\Phi^{-1}(t)$ is a fundamental matrix solution of the adjoint equation
\begin{equation} \label{bff1}
dx(t)=x(t)[-A(t)+G^2(t)]dt-x(t)G(t)\,d\omega(t),\quad t\in I.
\end{equation}
\end{lemma}

As a special case of Lemma \ref{lem24}, we know that every solution of \eqref{a4}
can be written as
\[
y(t)=\Phi(t)\big{[}\Phi^{-1}(s)y(s)+\int^{t}_{s}\Phi^{-1}(\tau)B(\tau)
y(\tau)d\tau\big],\quad \tau\in I,
\]
where $\Phi(t)$ is a fundamental matrix of \eqref{a1}.

\begin{theorem}\label{main22}
Assume that the first inequality in \eqref{b2} holds. Then any solution $y(t)$
of \eqref{a4} satisfies
\begin{equation} \label{b6}
\mathbb{E}\|y(t)\|^2\leq 2M e^{(-\alpha+2M\beta^2)(t-s)}\mathbb{E}\|y(s)\|^2, \quad
\forall  (t,s)\in I^2_{\geq}.
\end{equation}
In particular, if $\beta<\sqrt{\alpha/(2M)}$, then \eqref{a4} also admits
a mean-square exponential contraction.
\end{theorem}


\begin{proof}
Given any initial value $y(s)$ at time $s$, the solution of \eqref{a4}
can be expressed as
\[
 y(t)= \Phi(t)\Phi^{-1}(s)y(s)+\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)
B(\tau)y(\tau)d\tau,
\]
for all $(t,s)\in I^2_{\geq}$, where $\Phi(t)$ is the fundamental
matrix of \eqref{a1}.

Using the elementary inequality $\|a+b\|^2\le 2(\|a\|^2+\|b\|^2)$,
the H\"{o}lder's inequality, we obtain that
\begin{equation} \label{b7}
\begin{aligned}
\|y(t)\|^2&\leq 2\|\Phi(t)\Phi^{-1}(s)\|^2\|y(s)\|^2+2 \beta^2
\int^{t}_{s}\|\Phi(t)\Phi^{-1}(\tau)\|^2\|y(\tau)\|^2d\tau.
\end{aligned}
\end{equation}
 It follows from \eqref{b7} and \eqref{it} that
\begin{equation} \label{b8}
\begin{aligned}
\mathbb{E}\|y(t)\|^2
&\leq  2\mathbb{E}\|\Phi(t)\Phi^{-1}(s)\|^2\mathbb{E}\|y(s)\|^2+
2 \beta^2 \int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)\|^2\mathbb{E}\|y(\tau)\|^2d\tau \\
&\leq  2M  e^{-\alpha(t-s)}\mathbb{E}\|y(s)\|^2+2 M\beta^2
\int^{t}_{s}e^{-\alpha(t-\tau)}\mathbb{E}\|y(\tau)\|^2d\tau.
\end{aligned}
\end{equation}
Let
\[
u(t)=e^{\alpha t}\mathbb{E}\|y(t)\|^2,\quad
U(t)=2Mu(s)+2M\beta^2\int^{t}_{s}u(\tau)d\tau.
\]
Then \eqref{b8} can be rewritten as
 \[
u(t)\le U(t),\quad \text{for all } t\ge s.
\]
On the other hand,
$\frac{d}{dt}U(t)=2M\beta^2 u(t)$, and thus
\[
\frac{d}{dt}U(t)\leq 2M\beta^2 U(t).
\]
Integrating the above inequality from $s$ to $t$ and using the
relation $U(s)=2M u(s)$, we obtain that
\[
u(t)\leq U(t)\leq 2M u(s)
e^{2M\beta^2(t-s)},\quad \text{for all } (t,s)\in I^2_{\geq},
\]
which implies  \eqref{b6} and completes the proof.


Note that in Theorem \ref{main22} we do not impose any condition on $G(t)$.

\begin{example}[Geometric Brownian motion \cite{ar74, mao}]\label{ex1} \rm
 Consider the equation
\begin{equation} \label{b16}
dx(t)=-ax(t)dt+\sigma x(t)\,d\omega(t),
\end{equation}
with  initial data $x(0)$, where $a$, $\sigma$ are constants satisfying $a>0$ and
$\sigma^2<2a$. Then the solution of \eqref{b16} is given as
\[
x(t)=x(0)\exp[\big(-a-\frac{\sigma^2}{2})t+\sigma \omega(t)\big].
\]
Further, we can obtain
 \[
\mathbb{E}\|x(t)\|^2\le e^{(-2a+\sigma^2)(t-s)}\mathbb{E}\|x(s)\|^2
\]
with $t\ge s$. Using Theorem \ref{main21} and Theorem \ref{main22}, we know that
\begin{gather*}
dx(t)=(-a+b)x(t)dt+(\sigma+\eta) x(t)\,d\omega(t),\\
dx(t)=(-a+b)x(t)dt+\sigma x(t)\,d\omega(t)
\end{gather*}
admits a mean-square exponential contraction if $|b|$ and $|\eta|$ are all
sufficiently small.
\end{example}

\section{Robustness of mean-square exponential dichotomies}

In this section we consider the robustness of mean-square exponential
dichotomies. We assume that the phase space $\mathbb{R}^{n}$ can be
split as
\[
\mathbb{R}^{n}=X_1\oplus X_{2},
\]
where $X_1$ is a linear subspace of
$\mathbb{R}^{n}$ and $X_{2}$ is the complementary subspace of $X_1$.

\begin{definition}\label{def41}\rm
 We say that \eqref{a1} admit a mean-square exponential dichotomy if
 there exist positive constants $M$ and $\alpha$ such that, for any
solution $x(t)$ with initial data in $X_1$,
 \begin{equation} \label{d1}
\mathbb{E}\|x(t)\|^2\leq M e^{-\alpha(t-s)}\mathbb{E}\|x(s)\|^2, \quad \forall
(t,s)\in I^2_{\geq},
\end{equation}
 and for any solution $x(t)$ with initial data in $X_2$,
\begin{equation} \label{d2}
\mathbb{E}\|x(t)\|^2\leq M e^{-\alpha(s-t)}\mathbb{E}\|x(s)\|^2, \quad \forall
(t,s)\in I^2_{\leq}.
\end{equation}
\end{definition}

The subspaces $X_1$ and $X_{2}$ are called the stable and instable
spaces, respectively \cite{mzz,st}. Let $P(t)$ be the projections for
each $t\in I$ such that
\[
\Phi(t)\Phi^{-1}(s)P(s)=P(t)\Phi(t)\Phi^{-1}(s),  \quad
\forall  (t,s)\in I^2_{\geq}
\]
and
\[
x(t)=\Phi(t)\Phi^{-1}(s)P(s)x(s)
\]
for any solution $x(t)$ with initial data in $X_1$. Thus,
\[
\mathbb{E}\|x(t)\|^2=\mathbb{E}[\|\Phi(t)\Phi^{-1}(s)P(s)x(s)\|^2]
=\mathbb{E}\|\Phi(t)\Phi^{-1}(s)P(s)\|^2\mathbb{E}\|x(s)\|^2,
\]
which is just
\[
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)P(s)\|^2=\frac{\mathbb{E}\|x(t)\|^2}{\mathbb{E}\|x(s)\|^2}.
\]
Hence  from \eqref{d1} we obtain
\begin{equation} \label{d3}
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)P(s)\|^2 \leq M e^{-\alpha(t-s)},
\quad \forall  (t,s)\in I^2_{\geq}.
\end{equation}
Similarly, we can obtain
\begin{equation} \label{d4}
\mathbb{E}\|\Phi(t)\Phi^{-1}(s)Q(s)\|^2 \leq M e^{-\alpha(s-t)}, \quad
\forall  (t,s)\in I^2_{\leq},
\end{equation}
 where $Q(t)={\rm Id}-P(t)$ is the
complementary projection of $P(t)$. We will use the estimates
\eqref{d1}-\eqref{d2} as well as the equivalent formulation
\eqref{d3}-\eqref{d4}.

\begin{theorem}\label{main3}
Assume that \eqref{a1} admits a mean-square exponential dichotomy in $I$ and
\begin{equation} \label{con-2}
K<\frac{\alpha}{10M}.
\end{equation}
Then \eqref{a3} also admits a mean-square exponential
dichotomy and for any solution $y(t)$ with initial data in $X_1$,
\[
\mathbb{E}\|y(t)\|^2\leq M_1 e^{-\sqrt{\alpha(\alpha-10
MK)}(t-s)}\mathbb{E}\|y(s)\|^2,
\quad \forall  (t,s)\in I^2_{\geq},
\]
and for any solution $y(t)$ with initial data in $X_2$,
\[
 \mathbb{E}\|y(t)\|^2\leq M_1 e^{-\sqrt{\alpha(\alpha-10 MK)}(s-t)}\mathbb{E}\|y(s)\|^2,
\quad \forall  (t,s)\in I^2_{\leq},
\]
where the positive constant $M_1$ is given as
\[
M_1=\max\Big\{\frac{5M\big(\alpha+\sqrt{\alpha(\alpha-10 MK)}\big)}
{\alpha+\sqrt{\alpha(\alpha-10 MK)}-5MK},1\Big\}.
\]
\end{theorem}

\begin{proof}
Firstly, we introduce the spaces
\[
\mathscr{L}_{c}:=\{\hat{\Phi}: I^2_{\geq} \to
L^2_{\mathscr{F}_t}(\Omega, \mathbb{R}^{n}): \|\hat{\Phi}\|_{c}<\infty\},
\]
with the norm
\[
\|\hat{\Phi}\|_{c}=\sup\left\{(\mathbb{E}\|\hat{\Phi}(t)\hat{\Phi}^{-1}(s)
\hat{P}(s)\|^2)^{\frac{1}{2}}: (t,s)\in I^2_{\geq}\right\},
\]
and
\[
\mathscr{L}_{d}:=\{\hat{\Phi}: I^2_{\leq} \to L^2_{\mathscr{F}_t}(\Omega, \mathbb{R}^{n}):
\|\hat{\Phi}\|_{d}<\infty\},
\]
with the norm
\[
\|\hat{\Phi}\|_{d}=\sup\left\{(\mathbb{E}\|\hat{\Phi}(t)\hat{\Phi}^{-1}(s)
\hat{Q}(s)\|^2)^{\frac{1}{2}}:
(t,s)\in I^2_{\leq}\right\},
\]
where $\hat{\Phi}(t)$ is the fundamental
matrix solution of \eqref{a3}, $\hat{P}(t)$ are projections for each $t\in I$ and
$\hat{Q}(t)={\rm Id}-\hat{P}(t)$ is the complementary projection.
One can verify that both $(\mathscr{L}_{c},\|\cdot\|_{c})$ and
$(\mathscr{L}_{d},\|\cdot\|_{d})$  are Banach spaces.
\end{proof}

Next we show several auxiliary results.

\begin{lemma}\label{lem43}
For each $(t,s)\in I^2_{\geq}$, it holds
\begin{equation} \label{d5}
\begin{aligned}
y(t)&= \Phi(t)\Phi^{-1}(s)P(s)y(s)+\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)P(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau   \\
&\quad   +\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)P(\tau)
H(\tau)y(\tau)\,d\omega(\tau)  \\
&\quad -\int^{\infty}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
 H(\tau)y(\tau)\,d\omega(\tau)   \\
&\quad  -\int^{\infty}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau\in\mathscr{L}_{c},
\end{aligned}
\end{equation}
and for each $(t,s)\in I^2_{\leq}$,
\begin{equation} \label{d6}
\begin{aligned}
y(t)&=\Phi(t)\Phi^{-1}(s)Q(s)y(s)-\int^{s}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau \\
&\quad   -\int^{s}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
H(\tau)y(\tau)\,d\omega(\tau) \\
&\quad + \int^{t}_{-\infty}\Phi(t)\Phi^{-1}(\tau)P(\tau)
H(\tau)y(\tau)\,d\omega(\tau)   \\
&\quad  +\int^{t}_{-\infty}\Phi(t)\Phi^{-1}(\tau)P(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau\in\mathscr{L}_{d}.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Set \begin{align*}
  \hat{\xi}(t)
&=  \Phi^{-1}(s)P(s)y(s)+\int^{t}_{s}\Phi^{-1}(\tau)P(\tau)
  [B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau    \\
&\quad   +\int^{t}_{s}\Phi^{-1}(\tau)P(\tau)  H(\tau)y(\tau)\,d\omega(\tau)
  -  \int^{\infty}_t\Phi^{-1}(\tau)Q(\tau)  H(\tau)y(\tau)\,d\omega(\tau)   \\
&\quad  -   \int^{\infty}_t\Phi^{-1}(\tau)Q(\tau)
  [B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau.
\end{align*}
We can obtain
\begin{equation} \label{d7}
d\hat{\xi}(t)= \Phi^{-1}(t)\big{[}B(t)-
G(t)H(t)\big]y(t)dt+\Phi^{-1}(t)H(t)y(t)\,d\omega(t).
\end{equation}
 Let $y(t)=\Phi(t)\hat{\xi}(t)$. Using \eqref{d7}, It\^{o}
product rule and the following fact
\[
d\Phi(t)=A(t)\Phi(t)dt+G(t)\Phi(t)\,d\omega(t),
\]
we obtain that
\begin{align*}
 dy(t)&= d\Phi(t)\hat{\xi}(t)+\Phi(t)d\hat{\xi}(t)
 +G(t)\Phi(t)\Phi^{-1}(t)H(t)y(t)dt\\
&=  A(t)y(t)dt+G(t)y(t)\,d\omega(t)+\big{[}B(t)- G(t)H(t)\big]y(t)dt\\
&\quad +H(t)y(t)\,d\omega(t)+G(t)H(t)y(t)dt\\
&= (A(t)+B(t))y(t)dt+(G(t)+H(t))y(t)\,d\omega(t),
\end{align*}
which means that $y(t)$ given by \eqref{d5} is a solution of \eqref{a3}. Now we
show that $y(t)$ is unique in the space
$(\mathscr{L}_{c}, \|\cdot\|_{c})$. Set
\begin{align*}
\hat{\mathcal {T}}y(t)
&=\Phi(t)\Phi^{-1}(s)P(s)y(s)+\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)P(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau  \\
&\quad   +\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)P(\tau)
H(\tau)y(\tau)\,d\omega(\tau) \\
&\quad -\int^{\infty}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)H(\tau)y(\tau)\,d\omega(\tau)   \\
&\quad  -\int^{\infty}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
[B(\tau)-G(\tau)H(\tau)]y(\tau)d\tau.
\end{align*}

Using  \eqref{b2}, the H\"{o}lder's
inequality and the inequality \eqref{ele}, we have
\begin{align*}
&\|\hat{\mathcal{T}}y(t)\|^2 \\
&\leq 5\|\Phi(t)\Phi^{-1}(s)P(s)\|^2\|y(s)\|^2 +10(\beta^2+g^2h^2)
\int^{t}_{s}\|\Phi(t)\Phi^{-1}(\tau)P(\tau)\|^2\|y(\tau)\|^2d\tau   \\
&\quad  + 5\big\|\int^{t}_{s}\Phi(t)\Phi^{-1}(\tau)P(\tau)H(\tau)y(\tau)
 \,d\omega(\tau)\big\|^2 \\
&\quad + 5\big\|\int^{\infty}_t\Phi(t)\Phi^{-1}(\tau)Q(\tau)
H(\tau)y(\tau)\,d\omega(\tau)\big\|^2   \\
&\quad  +10(\beta^2+g^2h^2)
\int^{\infty}_t\|\Phi(t)\Phi^{-1}(\tau)Q(\tau)\|^2\|y(\tau)\|^2d\tau.
\end{align*}
Using \eqref{it} we can show that
\begin{align*}
\mathbb{E}\|\hat{\mathcal {T}}y(t)\|^2
&\leq 5\mathbb{E}\|\Phi(t)\Phi^{-1}(s)P(s)\|^2\mathbb{E}\|y(s)\|^2 \\
&\quad +10(\beta^2+g^2h^2) \int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)P(\tau)\|^2
 \mathbb{E}\|y(\tau)\|^2d\tau \\
&\quad +5\int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)P(\tau)\|^2
 \mathbb{E}\|H(\tau)y(\tau)\|^2d\tau \\
&\quad +5\int^{\infty}_t\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)Q(\tau)\|^2
 \mathbb{E}\|H(\tau)y(\tau)\|^2d\tau   \\
&\quad  +10(\beta^2+g^2h^2)
\int^{\infty}_t\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)Q(\tau)\|^2\mathbb{E}\|y(\tau)\|^2d\tau \\
&\leq  5M  e^{-\alpha(t-s)}\mathbb{E}\|y(s)\|^2+5MK
\int^{t}_{s}e^{-\alpha(t-\tau)}\mathbb{E}\|y(\tau)\|^2d\tau   \\
&\quad  +5MK \int^{\infty}_te^{-\alpha(\tau-t)}\mathbb{E}\|y(\tau)\|^2d\tau.
\end{align*}
Note that $y(t)=\hat{\Phi}(t)\hat{\Phi}^{-1}(s)\hat{P}(s)y(s)$. Hence,
\begin{align*}
&\mathbb{E}\|\hat{\mathcal {T}}\hat{\Phi}(t)\hat{\Phi}^{-1}(s)\hat{P}(s)\|^2\mathbb{E}\|y(s)\|^2 \\
&\leq  5MK \int^{t}_{s}e^{-\alpha(t-\tau)}\mathbb{E}\|\hat{\Phi}(\tau)
 \hat{\Phi}^{-1}(s)\hat{P}(s)\|^2\mathbb{E}\|y(s)\|^2d\tau  \\
&\quad  +5MK \int^{\infty}_te^{-\alpha(\tau-t)}
 \mathbb{E}\|\hat{\Phi}(\tau)\hat{\Phi}^{-1}(s)\hat{P}(s)\|^2\mathbb{E}\|y(s)\|^2d\tau
  + 5M  e^{-\alpha(t-s)}\mathbb{E}\|y(s)\|^2.
\end{align*}
 Thus,
\[
\|\hat{\mathcal {T}}\hat{\Phi}\|^2_{c}
\le 5 M +\frac{10 MK}{\alpha}\|\hat{\Phi}\|^2_{c}<\infty,
\]
and $\hat{\mathcal {T}}:\mathscr{L}_{c}\to
\mathscr{L}_{c}$ is well-defined. Proceeding in the
same procedure above, for any $\hat{\Phi}_1, \hat{\Phi}_{2}\in
\mathscr{L}_{c}$, we have
\[
\|\hat{\mathcal {T}}\hat{\Phi}_1-\mathcal
{T}\hat{\Phi}_{2}\|_{c}\le \sqrt{\frac{10
MK}{\alpha}}\|\tilde{\Phi}_1-\tilde{\Phi}_{2}\|_{c}.
\]

When condition \eqref{con-2} holds, $\hat{\mathcal {T}}$ is a contraction
operator. Hence, there exist a
unique $\hat{\Phi} \in \mathscr{L}_{c}$ such that
$\hat{\mathcal {T}}\hat{\Phi}=\hat{\Phi}$.
In a similar manner, we can also prove \eqref{d6}.
\end{proof}

\begin{lemma}\label{lemma44}
Let $x(t)$ be  a  bounded,  continuous  real-valued  function such
that
\begin{equation} \label{d10}
x(t)\le D  e^{-\alpha(t-s)}\zeta+\delta D
  \int^{t}_{s}e^{-\alpha(t-\tau)}x(\tau)d\tau +\delta D
  \int^{\infty}_te^{-\alpha(\tau-t)}x(\tau)d\tau,
\end{equation}
where $D,\alpha, \delta$ are all positive constants. If
$\alpha>2\delta D$, then
\[
x(t)\leq \tilde{K} \zeta e^{-\tilde{\alpha}(t-s)}, \quad
(t,s)\in I^2_{\ge},
\]
where
\[
\tilde{\alpha}= \sqrt{\alpha(\alpha-2\delta D )},\quad
\tilde{K}=\max\big\{\frac{D(\alpha+\tilde{\alpha})}{\alpha+\tilde{\alpha}-\delta
D}, 1\big\}.
\]
\end{lemma}

\begin{proof}
Let $\tilde{x}(t)$ be any bounded continuous function satisfying the integral
equation
\begin{equation} \label{d11}
\tilde{x}(t)=D  e^{-\alpha(t-s)}\zeta+\delta D
\int^{t}_{s}e^{-\alpha(t-\tau)}\tilde{x}(\tau)d\tau  +\delta D
\int^{\infty}_te^{-\alpha(\tau-t)}\tilde{x}(\tau)d\tau,
\end{equation}
with the initial condition $x(s)=\tilde{x}(s)$. It is easy to verify that
\[
\tilde{x}'(t)=-\alpha D e^{-\alpha(t-s)}\zeta-\alpha\delta D
\int^{t}_{s}e^{-\alpha(t-\tau)}\tilde{x}(\tau)d\tau  +\alpha \delta
D \int^{\infty}_te^{-\alpha(\tau-t)}\tilde{x}(\tau)d\tau,
\]
and
\begin{align*}
\tilde{x}''(t)
&=  \alpha^2 De^{-\alpha(t-s)}\zeta+\alpha^2
\delta D\int^{t}_{s}e^{-\alpha(t-\tau)}\tilde{x}(\tau)d\tau \\
&\quad +\alpha^2 \delta D
\int^{\infty}_te^{-\alpha(\tau-t)}\tilde{x}(\tau)d\tau-2\alpha\delta
D \tilde{x}(t).
\end{align*}
Then it is easy to verify that $\tilde{x}(t)$
is a solution of differential equation
\[
\tilde{x}''=\alpha(\alpha-2\delta D)\tilde{x}.
\]
Note that $\alpha>0$, $\alpha-2\delta D>0$ and $\tilde{x}(t)$ is a
bounded continuous function, then
\[
\tilde{x}(t)=\tilde{x}(s)e^{-\tilde{\alpha}(t-s)}.
\]
In addition, setting $t=s$ in \eqref{d11} gives
\[
\tilde{x}(s)=D  \zeta+\delta D \tilde{x}(s)
\int^{\infty}_{s}e^{-(\alpha+\tilde{\alpha})(\tau-s)}d\tau.
\]
Note that $\alpha+\tilde{\alpha}>0$, we obtain that
\[
\tilde{x}(s)\leq \frac{D(\alpha+\tilde{\alpha})}{\alpha+\tilde{\alpha}
-\delta D}\zeta.
\]
Thus for any $(t,s)\in I^2_{\ge}$, it has
\[
\tilde{x}(t)\leq \tilde{K} \zeta e^{-\tilde{\alpha}(t-s)}.
\]
Set $\Upsilon(t)=x(t)-\tilde{x}(t)$ for $(t,s)\in I^2_{\ge}$. It
follows from \eqref{d10} and \eqref{d11} that
\begin{equation} \label{x1}
\Upsilon(t)\le \delta D\int^{t}_{s}e^{-\alpha(t-\tau)}\Upsilon(\tau)d\tau
+\delta D \int^{\infty}_te^{-\alpha(\tau-t)}\Upsilon(\tau)d\tau.
\end{equation}
 Let
$\Upsilon=\sup\{\Upsilon(t): (t,s)\in I^2_{\ge}\}$. Then
$\Upsilon$ is finite. It follows from \eqref{x1} that
\begin{align*}
\Upsilon
&\leq \delta D \Upsilon \sup_{t\ge s}\int^{t}_{s}
e^{-\alpha(t-\tau)}d\tau +\delta D\Upsilon
\sup_{t\ge s} \int^{\infty}_te^{-\alpha(\tau-t)}d\tau \\
&\leq  \frac{2\delta D}{\alpha}\Upsilon.
\end{align*}
 Since $\alpha>2\delta D$, then $\Upsilon\le 0$ and thus
$x(t)\le \tilde{x}(t)$ for $(t,s)\in I^2_{\ge}$,
which means that
\[
x(t)\leq \tilde{K} \zeta  e^{-\tilde{\alpha}(t-s)},\quad (t,s)\in I^2_{\ge},
\]
and the proof  is complete.
\end{proof}

The proof of the following lemma is similar to that
of Lemma \ref{lemma44}, so we omit it.

\begin{lemma}\label{lemma45}
Let $y(t)$ be  a  bounded,  continuous  real-valued  function such that
\[
y(t)\le D  e^{-\alpha(s-t)}\zeta+\delta D\int^{s}_t
e^{-\alpha(\tau-t)}y(\tau)d\tau
+\delta D \int^{t}_{-\infty}e^{-\alpha(t-\tau)}y(\tau)d\tau,
\]
where $D,\alpha, \delta$ are all positive constants. If
$\alpha>2\delta D$, then
\[
y(t)\leq \tilde{K} \zeta e^{-\tilde{\alpha}(s-t)}, \quad (t,s)\in I^2_{\le}.
\]
\end{lemma}

As in the proof for Theorem \ref{main3}, we consider
$\hat{\Phi}\in \mathscr{L}_{c}$. Then it follows from Lemma
\ref{lem43} that the unique solution of \eqref{a3} in the
space $(\mathscr{L}_{c}, \|\cdot\|_{c})$ is given as \eqref{d5}.
Then we have
\begin{equation}  \label{d15}
\begin{aligned}
\mathbb{E}\|y(t)\|^2
&\leq  5\mathbb{E}\|\Phi(t)\Phi^{-1}(s)P(s)\|^2\mathbb{E}\|y(s)\|^2 \\
&\quad +10(\beta^2+g^2h^2) \int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}
 (\tau)P(\tau)\|^2\mathbb{E}\|y(\tau)\|^2d\tau \\
&\quad +5\int^{t}_{s}\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)P(\tau)\|^2
 \mathbb{E}\|H(\tau)y(\tau)\|^2d\tau \\
&\quad +5\int^{\infty}_t\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)Q(\tau)\|^2
 \mathbb{E}\|H(\tau)y(\tau)\|^2d\tau   \\
&\quad  +10(\beta^2+g^2h^2) \int^{\infty}_t\mathbb{E}\|\Phi(t)\Phi^{-1}(\tau)Q(\tau)\|^2
 \mathbb{E}\|y(\tau)\|^2d\tau  \\
&\leq  5M  e^{-\alpha(t-s)}\mathbb{E}\|y(s)\|^2+5MK
\int^{t}_{s}e^{-\alpha(t-\tau)}\mathbb{E}\|y(\tau)\|^2d\tau   \\
&\quad  +5MK \int^{\infty}_te^{-\alpha(\tau-t)}\mathbb{E}\|y(\tau)\|^2d\tau.
\end{aligned}
\end{equation}
Applying Lemma \ref{lemma44} to \eqref{d15} and note the condition
\eqref{con-2}, we have
\[
\mathbb{E}\|y(t)\|^2\leq M_1 e^{-\sqrt{\alpha(\alpha-10MK)}(t-s)}\mathbb{E}\|y(s)\|^2,
\quad \forall  (t,s)\in I^2_{\geq}.
\]
Similarly, consider $\hat{\Phi}\in \mathscr{L}_{d}$, then  from Lemma
\ref{lem43} it follows that  the unique solution of \eqref{a3} in the space
$(\mathscr{L}_{d},\|\cdot\|_{d})$ is given as \eqref{d6},
and we have
\[
\mathbb{E}\|y(t)\|^2\leq  M_1 e^{-\sqrt{\alpha(\alpha-10MK)}(s-t)}\mathbb{E}\|y(s)\|^2,
\quad \forall  (t,s)\in I^2_{\leq}.
\]
Now the proof is complete.
\end{proof}

 The following theorem is equivalent to Theorem \ref{main3}.

\begin{theorem} \label{main-p}
Assume that \eqref{a1} admits a mean-square exponential dichotomy and
condition \eqref{con-2} holds. Then there exist projections
 $\hat{P}(t)$ and $\hat{Q}(t)={\rm Id} -\hat{P}(t)$ such that
\begin{gather}\label{imp}
\|\hat{\Phi}(t)\hat{\Phi}^{-1}(s)\hat{P}(s)\|
\leq M_1 e^{-\sqrt{\alpha(\alpha-10 MK)}(t-s)},\quad \forall  (t,s)\in I^2_{\geq}, \\
\label{imq}
\|\hat{\Phi}(t)\hat{\Phi}^{-1}(s)\hat{Q}(s)\|
\leq M_1e^{-\sqrt{\alpha(\alpha-10 MK)}(s-t)},\quad \forall  (t,s)\in I^2_{\leq},
\end{gather}
where $\hat{\Phi}(t)$ is the fundamental matrix solution of \eqref{a3}.
\end{theorem}

The following result is a direct consequence of Theorem \ref{main3}.

\begin{theorem}\label{main4}
Assume that  \eqref{a1} admits a mean-square exponential dichotomy
in $I$ and $\beta^2<\frac{\alpha}{20M}$. Then \eqref{a4} also
admits a mean-square exponential dichotomy with
\[
\mathbb{E}\|y(t)\|^2\leq M_{2} e^{-\sqrt{\alpha(\alpha-20M\beta^2)}(t-s)}\mathbb{E}\|y(s)\|^2, \quad
\forall  (t,s)\in I^2_{\geq},
\]
for any solution $y(t)$ with initial data from $X_1$, and
\[
\mathbb{E}\|y(t)\|^2\leq M_{2}
e^{-\sqrt{\alpha(\alpha-20M\beta^2)}(s-t)}\mathbb{E}\|y(s)\|^2, \quad
\forall  (t,s)\in I^2_{\leq},
\]
for any solution $y(t)$ with initial data from $X_2$, where
\[
M_{2}=\max\Big\{\frac{5M(\alpha+\sqrt{\alpha(\alpha-20M\beta^2)})}
{\alpha+\sqrt{\alpha(\alpha-20M\beta^2)}-5M\beta^2}, 1\Big\}.
\]
\end{theorem}

Next, we present an example to illustrate the
robustness of mean-square exponential dichotomies.

\begin{example}\label{ex2}\rm
Consider the  equation
\begin{equation} \label{d16}
\begin{gathered}
dx(t) =-ax(t)dt+\sigma x(t)\,d\omega(t), \\
dy(t) =ay(t)dt+\sigma y(t)\,d\omega(t),
\end{gathered}
\end{equation}
with  initial data $(x(0),y(0))$, where $a$, $\sigma$ are constants
satisfying $a>0$ and $\sigma^2<2a$. Then the solution of \eqref{d16} is
given as
\begin{gather*}
x(t) =x(0)\exp\big{[}\big(-a-\frac{\sigma^2}{2}\big)t+\sigma \omega(t)\big], \\
y(t) =y(0)\exp\big{[}\big(a-\frac{\sigma^2}{2}\big)t+\sigma
\omega(t)\big].
\end{gather*}
It is easy to verify that
\begin{gather*}
 \mathbb{E}\|x(t)\|^2 \le
e^{(-2a+\sigma^2)(t-s)}\mathbb{E}\|x(s)\|^2,\quad t\geq s, \\
\mathbb{E}\|y(t)\|^2 \le e^{-(2a+\sigma^2)(s-t)}\mathbb{E}\|y(s)\|^2,\quad s\geq t,
\end{gather*}
and therefore \eqref{d16} admits a mean-square exponential dichotomy.
Using Theorem \ref{main3} and Theorem \ref{main4}, we know that
\begin{gather*}
dx(t) =(-a+b)x(t)dt+(\sigma +\eta) x(t)\,d\omega(t), \\
dy(t) =(a+b)y(t)dt+(\sigma +\eta) y(t)\,d\omega(t),
\end{gather*}
and
\begin{gather*}
dx(t) =(-a+b)x(t)dt+\sigma x(t)\,d\omega(t), \\
dy(t) =(a+b)y(t)dt+\sigma y(t)\,d\omega(t),
\end{gather*}
also admit a mean-square exponential dichotomy if
$|b|$ and $|\eta|$ are small enough.
\end{example}

\subsection*{Acknowledgments}
Hailong Zhu was supported by the National Natural Science
Foundation of China (No. 11301001), the China Postdoctoral Science Foundation
(No. 2016M591697), the Excellent Youth Scholars
Foundation and the Natural Science Foundation of Anhui Province of
China (No. 1708085MA17, No. 1208085QG131, No. KJ2014A003).
 Yongxin Jiang was supported by the National Natural Science
Foundation of China (No. 11401165, No. 11671118).

The author would like to thank Professor Jifeng Chu for his careful
reading of the manuscript and valuable suggestions.

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\end{document}