In view of ecology, a good situation occurs when all species co-exist. In this section, we will consider another stochastic persistence, that is, stochastic persistence in mean. Now, we present the definition of persistence in mean.

**Definition 4** (see [7, 12])

The system (4) is said to be persistent in mean if

\underset{t\to \mathrm{\infty}}{lim}\frac{1}{t}{\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds>0,\phantom{\rule{2em}{0ex}}\underset{t\to \mathrm{\infty}}{lim}\frac{1}{t}{\int}_{0}^{t}y(s)\phantom{\rule{0.2em}{0ex}}ds>0\phantom{\rule{1em}{0ex}}\text{a.s.}

Firstly, we introduce a fundamental lemma which will be used.

**Lemma 3** *Consider the one*-*dimensional stochastic equation*

dx(t)=x(t)[a(t)-b(t)x(t)]\phantom{\rule{0.2em}{0ex}}dt+\sigma (t)x(t)\phantom{\rule{0.2em}{0ex}}dB(t),

(9)

*where* a(t), b(t), \sigma (t) *are positive*, *continuous and bounded functions*, B(t) *is a standard Brownian motion*. *Under the condition* \stackrel{\u02c6}{a}>\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}, *for any initial value* {x}_{0}>0, *the solution* x(t) *to* (9) *has the property*

\underset{t\to \mathrm{\infty}}{lim}\frac{lnx(t)}{t}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.s.}}

*Proof* We firstly show {lim\hspace{0.17em}sup}_{t\to \mathrm{\infty}}\frac{lnx(t)}{t}\le 0 a.s. To define the Lyapunov function V(t,x)={e}^{t}lnx, using the Itô formula, we obtain

d({e}^{t}lnx(t))={e}^{t}[lnx(t)+a(t)-b(t)x(t)-\frac{{\sigma}^{2}(t)}{2}]\phantom{\rule{0.2em}{0ex}}dt+{e}^{t}\sigma (t)\phantom{\rule{0.2em}{0ex}}dB(t).

Thus

{e}^{t}lnx(t)-ln{x}_{0}={\int}_{0}^{t}{e}^{s}[lnx(s)+a(s)-b(s)x(s)-\frac{{\sigma}^{2}(s)}{2}]\phantom{\rule{0.2em}{0ex}}ds+M(t),

where M(t)={\int}_{0}^{t}{e}^{s}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s), whose quadratic variation is

\u3008M(t),M(t)\u3009={\int}_{0}^{t}{e}^{2s}{\sigma}^{2}(s)\phantom{\rule{0.2em}{0ex}}ds.

By virtue of the exponential martingale inequality, for any positive constants *T*, *δ*, *β*, we have

P\{\underset{0\le t\le T}{sup}[M(t)-\frac{\delta}{2}\u3008M(t),M(t)\u3009]>\beta \}\le {e}^{-\delta \beta}.

Choose T=k\gamma, \delta =n\u03f5{e}^{-k\delta} and \beta =\frac{\theta {e}^{k\delta}lnk}{\u03f5n}, where k\in {Z}^{+}, 0<\u03f5<1, \theta >1 and \gamma >0 above. Hence

P\{\underset{0\le t\le T}{sup}[M(t)-\frac{n\u03f5{e}^{-k\delta}}{2}\u3008M(t),M(t)\u3009]>\frac{\theta {e}^{k\delta}lnk}{\u03f5n}\}\le {k}^{-\theta}.

Obviously, we know {\sum}_{k=1}^{\mathrm{\infty}}{k}^{-\theta}<\mathrm{\infty}. Applying the Borel-Cantalli lemma, we obtain that there exists some {\mathrm{\Omega}}_{i}\subset \mathrm{\Omega} with P({\mathrm{\Omega}}_{i})=1 such that for any \omega \in {\mathrm{\Omega}}_{i}, an integer {k}_{i}={k}_{i}(\omega ) such that for any k>{k}_{i}, we get

M(t)\le \frac{n\u03f5{e}^{-k\delta}}{2}\u3008M(t),M(t)\u3009+\frac{\theta {e}^{k\delta}lnk}{\u03f5n}

for all 0\le t\le k\gamma. Then

{e}^{t}lnx(t)-ln{x}_{0}\le {\int}_{0}^{t}{e}^{s}[lnx(s)+ar(s)-b(s)x(s)-\frac{{\sigma}^{2}(s)}{2}+\frac{n\u03f5{e}^{s-k\delta}}{2}{\sigma}^{2}(s)]\phantom{\rule{0.2em}{0ex}}ds+\frac{\theta {e}^{k\delta}lnk}{\u03f5}.

Note that t\in [0,k\gamma ], s\in [0,t], we have

lnx(s)+a(s)-b(s)x(s)-\frac{{\sigma}^{2}(s)}{2}+\frac{n\u03f5{e}^{s-k\delta}}{2}{\sigma}^{2}(s)\le K.

For all t\in [0,k\gamma ] with k>{k}_{0}(\omega ), we derive

{e}^{t}lnx(t)-ln{x}_{0}\le {\int}_{0}^{t}K{e}^{s}\phantom{\rule{0.2em}{0ex}}ds+\frac{\theta {e}^{k\delta}lnk}{\u03f5}=K({e}^{t}-1)+\frac{\theta {e}^{k\delta}lnk}{\u03f5}.

Thus, for (k-1)\gamma \le t\le k\gamma, we get lnx(t)\le {e}^{-t}ln{x}_{0}+K(1-{e}^{-t})+\frac{\theta {e}^{\delta}lnk}{\u03f5}. This implies

\frac{lnx(t)}{lnt}\le \frac{ln{x}_{0}}{{e}^{t}lnt}+\frac{K(1-{e}^{-t})}{lnt}+\frac{\theta {e}^{\delta}lnk}{\u03f5ln((k-1)\gamma )}.

Letting k\to \mathrm{\infty}, that is, t\to \mathrm{\infty}, we can imply {lim\hspace{0.17em}sup}_{t\to \mathrm{\infty}}\frac{lnx(t)}{lnt}\le \frac{\theta {e}^{\gamma}}{\u03f5}. By making \gamma \downarrow 0, \u03f5\uparrow 1 and \theta \downarrow 1, we get {lim\hspace{0.17em}sup}_{t\to \mathrm{\infty}}\frac{lnx(t)}{lnt}\le 1. Consequently,

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{lnx(t)}{t}=\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{lnx(t)}{lnt}\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{lnt}{t}\le \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{lnt}{t}=0.

Thus it remains to show that {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}\frac{lnx(t)}{t}\ge 0 a.s. The quadratic variation of the stochastic integral {\int}_{0}^{t}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s) is {\int}_{0}^{t}{\sigma}^{2}(s)\phantom{\rule{0.2em}{0ex}}ds\le Kt. So, the strong law of large numbers of local martingales yields that

\frac{1}{t}{\int}_{0}^{t}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s)\to 0\phantom{\rule{1em}{0ex}}\text{a.s.}t\to \mathrm{\infty}.

Hence, for any \u03f5>0, there exists some positive T<\mathrm{\infty} such that

|{\int}_{0}^{t}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s)|<\u03f5t\phantom{\rule{1em}{0ex}}\text{a.s. for any}t\ge T.

For any t>s\ge T, we have

|{\int}_{s}^{t}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s)|<\u03f5(s+t)\phantom{\rule{1em}{0ex}}\text{a.s.}

Then, for any t>T,

\begin{array}{rcl}\frac{1}{x(t)}& =& \frac{1}{x(T)}{e}^{[{\int}_{T}^{t}-(a(s)-\frac{{\sigma}^{2}(s)}{2})\phantom{\rule{0.2em}{0ex}}ds-{\int}_{T}^{t}\sigma (s)\phantom{\rule{0.2em}{0ex}}dB(s)]}+{\int}_{T}^{t}b(s){e}^{[{\int}_{s}^{t}-(a(\tau )-\frac{{\sigma}^{2}(\tau )}{2})\phantom{\rule{0.2em}{0ex}}d\tau -{\int}_{T}^{t}{\sigma}_{1}(r(\tau ))\phantom{\rule{0.2em}{0ex}}dB(\tau )]}\phantom{\rule{0.2em}{0ex}}ds\\ \le & \frac{1}{x(T)}{e}^{[{\int}_{T}^{t}-(a(s)-\frac{{\sigma}^{2}(s)}{2})\phantom{\rule{0.2em}{0ex}}ds+\u03f5(t+T)]}+{\int}_{T}^{t}b(s){e}^{[{\int}_{s}^{t}-(a(\tau )-\frac{{\sigma}^{2}(\tau )}{2})\phantom{\rule{0.2em}{0ex}}d\tau +\u03f5(t+s)]}\phantom{\rule{0.2em}{0ex}}ds.\end{array}

Therefore

\begin{array}{rcl}{e}^{-2\u03f5(t+T)}\frac{1}{x(t)}& \le & \frac{1}{x(T)}{e}^{[{\int}_{T}^{t}-(a(s)-\frac{{\sigma}^{2}(s)}{2})\phantom{\rule{0.2em}{0ex}}ds-\u03f5(t+T)]}\\ +{\int}_{T}^{t}b(s){e}^{[{\int}_{s}^{t}-(a(\tau )-\frac{{\sigma}^{2}(\tau )}{2})\phantom{\rule{0.2em}{0ex}}d\tau -\u03f5(t-s)-2\u03f5T]}\phantom{\rule{0.2em}{0ex}}ds\\ \le & K<\mathrm{\infty}.\end{array}

That is, \frac{1}{x(t)}\le K{e}^{2\u03f5(t+T)} a.s. Then \frac{ln\frac{1}{x(t)}}{t}\le \frac{1}{t}[lnK+2\u03f5(t+T)] a.s. Thus {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}\frac{lnx(t)}{t}\ge -2\u03f5 a.s. Since *ϵ* is arbitrary, we conclude that

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{lnx(t)}{t}\ge 0\phantom{\rule{1em}{0ex}}\text{a.s.}

So, the proof is complete. □

**Remark 1** Lemma 3 generalizes the works of [7] and [11].

To continue our analysis, let us impose the following hypothesis.

**Assumption 2** \stackrel{\u02c6}{a}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}>0, \stackrel{\u02c6}{b}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}>0.

Theorem 1 tells us there is a unique global solution (which is positive for any initial value {X}_{0}=({x}_{0},{y}_{0})\in {R}_{+}^{2}) to the stochastic system (4). So, we conclude the following results by the comparison theorem. We can get

dx(t)\le x(t)[\stackrel{\u02c7}{a}-{c}_{1}(t)x(t)]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{1}(t)x(t)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(t)

and

dx(t)\ge x(t)[\stackrel{\u02c6}{a}-{c}_{1}(t)x(t)]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{1}(t)x(t)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(t).

Denote that {X}_{2} is the solution to the following stochastic equation:

d{X}_{2}(t)={X}_{2}(t)[\stackrel{\u02c7}{a}-{c}_{1}(t){X}_{2}(t)]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{1}(t){X}_{2}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(t)

(10)

with {X}_{2}(0)={x}_{0}. And {X}_{1} is the solution to the equation

d{X}_{1}(t)={X}_{1}(t)(\stackrel{\u02c6}{a}-{c}_{1}(t){X}_{1}(t))\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{1}(t){X}_{1}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(t)

(11)

with {X}_{1}(0)={x}_{0}. It is obvious that {X}_{1}(t)\le x(t)\le {X}_{2}(t), t\in [0,+\mathrm{\infty}) a.s. Moreover, we can have

dy(t)\le y(t)(\stackrel{\u02c7}{b}-{c}_{2}(t)y(t))\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{2}(t)y(t)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(t)

and

dy(t)\ge y(t)(\stackrel{\u02c6}{b}-{c}_{2}(t)y(t))\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{2}(t)y(t)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(t).

We denote {Y}_{1}(t) is the solution of the stochastic differential equation

d{Y}_{1}(t)={Y}_{1}(t)[\stackrel{\u02c6}{b}-{c}_{2}(t){Y}_{1}(t)]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{2}{Y}_{1}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(t)

(12)

with {Y}_{1}(0)={y}_{0}. And the stochastic equation

d{Y}_{2}(t)={Y}_{2}(t)[(\stackrel{\u02c7}{b}-{c}_{2}(t){Y}_{2}(t)]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{2}{Y}_{2}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(t)

(13)

has the solution {Y}_{2}(t) for initial value {Y}_{2}(0)={y}_{0}. Consequently, {Y}_{1}(t)\le y(t)\le {Y}_{2}(t), t\in [0,+\mathrm{\infty}) a.s. To sum up, we have

{X}_{1}(t)\le x(t)\le {X}_{2}(t),\phantom{\rule{2em}{0ex}}{Y}_{1}(t)\le y(t)\le {Y}_{2}(t),\phantom{\rule{1em}{0ex}}t\in [0,+\mathrm{\infty})\text{a.s.}

(14)

**Lemma 4** *Under Assumption * 2, *for any initial value* {x}_{0}>0, *the solution* x(t) *to* (4) *satisfies*

\underset{t\to \mathrm{\infty}}{lim}\frac{lnx(t)}{t}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.s.}}

Lemma 3, (10), (11) and (14) can straightforward imply the assertion.

**Lemma 5** *Under Assumption * 2, *for any initial value* {y}_{0}>0, *the solution* y(t) *to* (4) *satisfies*

\underset{t\to \mathrm{\infty}}{lim}\frac{lny(t)}{t}=0\phantom{\rule{1em}{0ex}}\mathit{\text{a.s.}}

Lemma 3, (12), (13) and (14) prove the result.

**Theorem 7** *Let Assumption * 2 *hold*. *Then*, *for any initial value* ({x}_{0},{y}_{0})\in {R}_{+}^{2}, *the system* (4) *is persistent in mean*. *That is*, *the system* (4) *has the properties*

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{{\int}_{0}^{t}y(s)\phantom{\rule{0.2em}{0ex}}ds}{t}\ge \frac{g-\frac{{\sigma}_{2}^{2}}{2}}{h}>0,\phantom{\rule{2em}{0ex}}\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{{\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds}{t}\ge \frac{\stackrel{\u02c6}{a}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}}{\stackrel{\u02c7}{c}}>0\phantom{\rule{1em}{0ex}}\mathit{\text{a.s.}}

*Proof* Denote V(x)=lnx, by the Itô formula, we obtain

d(lnx(t))=[\frac{{a}_{1}(t)+{a}_{2}(t)y(t)}{1+y(t)}-{c}_{1}(t)x(t)-\frac{{\sigma}_{1}^{2}(t)}{2}]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{1}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(t).

Then

\begin{array}{rcl}lnx(t)& =& ln{x}_{0}+{\int}_{0}^{t}(\frac{{a}_{1}(s)+{a}_{2}(s)y(s)}{1+y(s)}-\frac{{\sigma}_{1}^{2}(s)}{2})\phantom{\rule{0.2em}{0ex}}ds\\ -{\int}_{0}^{t}{c}_{1}(s)x(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{t}{\sigma}_{1}(s)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(s),\end{array}

which yields

\stackrel{\u02c7}{c}{\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds\ge -lnx(t)+ln{x}_{0}+(\stackrel{\u02c6}{a}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2})t+{\int}_{0}^{t}{\sigma}_{1}(s)\phantom{\rule{0.2em}{0ex}}d{B}_{1}(s).

By virtue of the strong law of large numbers and Lemma 4, we get

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{{\int}_{0}^{t}x(s)\phantom{\rule{0.2em}{0ex}}ds}{t}\ge \frac{\stackrel{\u02c6}{a}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}}{\stackrel{\u02c7}{c}}>0\phantom{\rule{1em}{0ex}}\text{a.s.}

On the other hand, denote V(y)=lny, by the Itô formula, we obtain

d(lny(t))=[\frac{{b}_{1}(t)+{b}_{2}(t)x(t)}{1+x(t)}-{c}_{2}(t)y(t)-\frac{{\sigma}_{2}^{2}(t)}{2}]\phantom{\rule{0.2em}{0ex}}dt+{\sigma}_{2}(t)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(t).

Thus

\begin{array}{rcl}lny(t)& =& ln{y}_{0}+{\int}_{0}^{t}(\frac{{b}_{1}(s)+{b}_{2}(s)x(s)}{1+x(s)}-\frac{{\sigma}_{2}^{2}(s)}{2})\phantom{\rule{0.2em}{0ex}}ds\\ -{\int}_{0}^{t}{c}_{2}(s)y(s)\phantom{\rule{0.2em}{0ex}}ds+{\int}_{0}^{t}{\sigma}_{2}(s)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(s).\end{array}

So, we have

\stackrel{\u02c7}{c}{\int}_{0}^{t}y(s)\phantom{\rule{0.2em}{0ex}}ds\ge -lny(t)+ln{y}_{0}+(\stackrel{\u02c6}{b}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2})t+{\int}_{0}^{t}{\sigma}_{2}(s)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(s).

Dividing *t* on both sides yields

\stackrel{\u02c7}{c}\frac{{\int}_{0}^{t}y(s)\phantom{\rule{0.2em}{0ex}}ds}{t}\ge -\frac{lny(t)}{t}+\frac{ln{y}_{0}}{t}+(\stackrel{\u02c6}{b}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2})+\frac{{\int}_{0}^{t}{\sigma}_{2}(s)\phantom{\rule{0.2em}{0ex}}d{B}_{2}(s)}{t}.

Letting t\to \mathrm{\infty}, by virtue of the strong law of large numbers and Lemma 5, we have

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{{\int}_{0}^{t}y(s)\phantom{\rule{0.2em}{0ex}}ds}{t}\ge \frac{\stackrel{\u02c6}{b}-\frac{{\stackrel{\u02c7}{\sigma}}^{2}}{2}}{\stackrel{\u02c7}{c}}>0\phantom{\rule{1em}{0ex}}\text{a.s.}

The proof is complete. □