### Appendix

In this appendix, we will give some proofs of the main results.

### 1.1 Proof of Theorem 2.1

### Proof

Let \(J(E_{0}),J(E_{1})\) denote the Jacobian matrix of system (1.2) at the equilibria \(E_{0}\) and \(E_{1}\), respectively, then we have

$$\begin{aligned} \mathbf{J}(X)= \begin{pmatrix} -(\beta _{1}-\frac{\beta _{2}I}{\eta +I}){I}-\mu & -(\beta _{1}- \frac{\beta _{2}I}{\eta +I}){S}+ \frac{\beta _{2}\eta {SI}}{(\eta +I)^{2}}-\mu (1-\alpha )q \\ (\beta _{1}-\frac{\beta _{2}I}{\eta +I}){I} & -(p\mu +\gamma )+ ( \beta _{1}-\frac{\beta _{2}I}{\eta +I}){S}- \frac{\beta _{2}\eta {IS}}{(\eta +I)^{2}} \end{pmatrix}, \end{aligned}$$

thus

$$\begin{aligned} \mathbf{J}(E_{0})={}& \begin{pmatrix} -\mu & -\beta _{1}{S_{0}}-\mu (1-\alpha )q \\ 0 & -(p\mu +\gamma )+\beta _{1}{S_{0}} \end{pmatrix} \\ ={}& \begin{pmatrix} -\mu & -\beta _{1}(1-\alpha )-\mu (1-\alpha )q \\ 0 & (p\mu +\gamma )(R_{0}-1) \end{pmatrix} \end{aligned}$$

(A.1)

and

$$\begin{aligned} \mathbf{J}(E_{1})={}& \begin{pmatrix} -(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){I^{*}}-\mu & -( \beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){S^{*}}+ \frac{\beta _{2}\eta {S^{*}I^{*}}}{(\eta +I^{*})^{2}}-\mu (1-\alpha )q \\ (\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){I^{*}} & -(p\mu + \gamma )+ (\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){S^{*}}- \frac{\beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}} \end{pmatrix} \\ ={}& \begin{pmatrix} -(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){I^{*}}-\mu & -(p \mu +\gamma )+\frac{\beta _{2}\eta {S^{*}I^{*}}}{(\eta +I^{*})^{2}}- \mu (1-\alpha )q \\ (\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}}){I^{*}} & - \frac{\beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}} \end{pmatrix}. \end{aligned}$$

(A.2)

From (A.1), we can know that all characteristic values have negative real parts if and only if \(R_{0}<1\), according to Routh–Hurwitz criteria, \(E_{0}\) is locally stable if \(R_{0}<1\).

For (A.2), we have

$$\begin{aligned} \operatorname{tr} \bigl(J(E_{1}) \bigr)={}&{-} \biggl(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}}- \mu -(p\mu +\gamma )+ \biggl(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){S^{*}}- \frac{\beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}} \\ ={}&{- }\biggl(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}}-\mu - \frac{\beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}}< 0, \\ \det \bigl(J(E_{1}) \bigr)={}& \biggl(\beta _{1}- \frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}} \frac{\beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}}+ \frac{\mu \beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}}+(p\mu + \gamma ) \biggl(\beta _{1}- \frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}} \\ &{}- \biggl(\beta _{1}-\frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}} \frac{\beta _{2}\eta {S^{*}I^{*}}}{(\eta +I^{*})^{2}}+\mu (1-\alpha )q \biggl( \beta _{1}- \frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}} \\ ={}&\frac{\mu \beta _{2}\eta {I^{*}S^{*}}}{(\eta +I^{*})^{2}}+ \bigl[p\mu + \gamma +\mu (1-\alpha )q \bigr] \biggl(\beta _{1}- \frac{\beta _{2}I^{*}}{\eta +I^{*}} \biggr){I^{*}}>0. \end{aligned}$$

Thus, \(E_{1}\) is locally asymptotically stable if \(R_{0}>1\).

Choose a Dulac function \(B=\frac{1}{SI}\). Denote

$$\begin{aligned} &F=- \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){SI}-\mu {S}-(1-\alpha ) \mu {q}I+(1-\alpha ){\mu }, \\ &G= \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){SI}-(p\mu +\gamma )I. \end{aligned}$$

Note that

$$\begin{aligned} &\frac{\partial {(BF)}}{\partial {S}}=-\frac{(1-\alpha ){\mu }}{S^{2}I}+ \frac{(1-\alpha )\mu {q}}{S^{2}}=- \frac{\mu (1-\alpha )(1-qI)}{S^{2}I}, \\ &\frac{\partial {(BG)}}{\partial {I}}=- \frac{\mu \beta _{2}}{(\eta +I)^{2}}, \\ &\frac{\partial {(BF)}}{\partial {S}}+\frac{\partial {(BG)}}{\partial {I}}=- \frac{\mu (1-\alpha )(1-qI)}{S^{2}I}- \frac{\mu \beta _{2}}{(\eta +I)^{2}}< 0, \end{aligned}$$

in the interior of the positive invariant set Γ. The Dulac criterion holds and there are no close orbits in Γ. Incorporating local stability of \(E_{0}\) and \(E_{1}\), this proves that \(E_{0}\) is globally asymptotically stable if \(R_{0} < 1\) and \(E_{1}\) is globally asymptotically stable if \(R_{0} > 1\). The proof is complete. □

### 1.2 Proof of Theorem 3.1

### Proof

Owing to system (1.3), we get

$$\begin{aligned} d \bigl(S (s )+I (s ) \bigr)={}& \bigl\{ -\mu S (s )- \bigl[ \mu (1- \alpha q)+\gamma \bigr]I (s )+\mu (1-\alpha ) \bigr\} \,ds \\ < {}& \bigl\{ -\mu S (s )-\mu (1-\alpha q)I (s )+\mu (1- \alpha ) \bigr\} \,ds \\ < {}& \bigl\{ -\mu (1-\alpha q) \bigl(S (s )+I (s ) \bigr)+\mu (1- \alpha ) \bigr\} \,ds, \end{aligned}$$

so, by integration we check

$$ S (s )+I (s )< \frac{1-\alpha }{1-\alpha q}+ \biggl[S(0)+I(0)- \frac{1-\alpha }{1-\alpha q} \biggr]e^{-\mu (1-\alpha q)s} \quad \text{for all }s \in [0,t]\text{ a.s.} $$

Then \(S (s )+I (s )<\frac{1-\alpha }{1-\alpha q}<1\). In addition

$$\begin{aligned} d \bigl(S (s )+I (s ) \bigr)={}& \bigl\{ -\mu S (s )- \bigl[ \mu (1- \alpha q)+\gamma \bigr]I (s )+\mu (1-\alpha ) \bigr\} \,ds \\ >{}& \bigl\{ -\mu S (s )-[\mu +\gamma ]I (s )+\mu (1- \alpha ) \bigr\} \,ds \\ >{}& \bigl\{ -[\mu +\gamma ] \bigl(S (s )+I (s ) \bigr)+\mu (1- \alpha ) \bigr\} \,ds, \end{aligned}$$

so, by integration we check

$$ S (s )+I (s )> \frac{\mu (1-\alpha )}{\mu +\gamma }+ \biggl[S(0)+I(0)- \frac{\mu (1-\alpha )}{\mu +\gamma } \biggr]e^{-(\mu +\gamma )s} \quad \text{for all }s\in [0,t]\text{ a.s.} $$

Hence \(S (s )+I (s )> \frac{\mu (1-\alpha )}{\mu +\gamma }\). So

$$ S (s ),I (s )\in \biggl( \frac{\mu (1-\alpha )}{\mu +\gamma },1 \biggr) \quad \text{for all }s \in [0,t]\quad \text{a.s.} $$

(A.3)

We can easily see that the coefficients of system (1.3) are locally Lipschitz continuous for any given initial value \((S(0),I(0))\in \mathbb{R}_{+}^{2}\). Hence, there is a unique local solution \((S(t),I(t))\) on \(t\in [0,\tau _{e})\), where \(\tau _{e}\) is the explosion time (see [41]). To show that this solution is global, we only need to prove that \(\tau _{e}=\infty \) a.s. Let \(k_{0}\geq 0\) be sufficiently large so that \((S(0),I(0))\) all lie within the interval \([\frac{1}{k_{0}},k_{0}]\). For each integer \(k\geq {k_{0}}\), define the following stopping time:

$$ \tau _{k}=\inf \biggl\{ t\in [0,\tau _{e}): \min \bigl\{ \bigl(S (t ),I (t ) \bigr) \bigr\} \leq \frac{1}{k} \text{ or } \max \bigl\{ \bigl(S (t ),I (t ) \bigr) \bigr\} \geq {k} \biggr\} , $$

where throughout this paper, we set \(\inf {\O }{=}\infty \) (and as usual Ø denotes the empty set). According to the definition, \(\tau _{k}\) is increasing as \(k\rightarrow \infty \). Set \(\tau _{\infty }{=}\lim_{k\rightarrow \infty }\tau _{k}\), whence \(\tau _{\infty }\leq \tau _{e}\) a.s. Namely, we need to show that \(\tau _{\infty }{=}\infty \) a.s. We assumed that there exist a pair of constants \(T>0\) and \(\epsilon \in (0,1 )\) such that

$$ P\{\tau _{\infty }\leq {T}\}>\epsilon. $$

As a result, there is an integer \(k_{1}\geq {k_{0}}\) such that

$$ P\{\tau _{k}\leq {T}\}>\epsilon \quad \text{for all }k\geq {k_{1}}. $$

(A.4)

Now define a \(C^{2}\)-function \(V:\mathbb{R}_{+}^{2}\rightarrow \overline{\mathbb{R}}_{+}\), where \(\overline{\mathbb{R}}_{+}=\{x\in \mathbb{R}:x\geq 0\}\), by

$$ V(t)=S-1-\log S+ I-1-\log I. $$

The nonnegativity of this function can be seen from \(u-1-\log {u}\geq {0}\), \(\forall {u}\geq 0\). Let \(k\geq {k_{0}}\) and \(T>{0}\) be arbitrary. Applying to It\(\hat{o}^{,}s\) formula, we obtain

$$ dV(t)=LV(t)\,dt- \bigl[\sigma (S-1)I-\sigma (I-1)S \bigr]\,dB(t), $$

where

$$\begin{aligned} LV(t)={}& \biggl(1-\frac{1}{S} \biggr) \biggl(- \biggl(\beta _{1}- \frac{\beta _{2}I}{\eta +I} \biggr){SI}-\mu {S}-(1-\alpha )\mu {q}I+(1- \alpha ){ \mu } \biggr) \\ &{}+ \biggl(1-\frac{1}{I} \biggr) \biggl( \biggl(\beta _{1}- \frac{\beta _{2}I}{\eta +I} \biggr){SI}-(p\mu +\gamma )I \biggr)+ \frac{\sigma ^{2}S^{2}I^{2}}{2S^{2}}+ \frac{\sigma ^{2}S^{2}I^{2}}{2I^{2}} \\ ={}&(1-\alpha )\mu + \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){I}+\mu +p \mu +\gamma +\frac{(1-\alpha )\mu {q}I}{S}+ \frac{\sigma ^{2}(S^{2}+I^{2})}{2} \\ &{}-\mu S- \bigl[(1-\alpha )\mu {q}+p\mu +\gamma \bigr]I- \biggl(\beta _{1}- \frac{\beta _{2}I}{\eta +I} \biggr){S} \\ \leq {}&(1-\alpha )\mu +\beta _{1} I+\mu +p\mu +\gamma + \frac{(1-\alpha )\mu {q}I}{S}+\frac{\sigma ^{2}(S^{2}+I^{2})}{2}+ \beta _{2}{S}, \end{aligned}$$

owing to (A.3), we have thus

$$ LV(t)\leq (1-\alpha )\mu +\beta _{1}+\mu +p\mu +\gamma +( \mu + \gamma )q+\sigma ^{2}+\beta _{2}\doteq B. $$

Then

$$ dV(t)=B\,dt- \bigl[\sigma (S-1)I-\sigma (I-1)S \bigr]\,dB(t) .$$

(A.5)

Integrating both sides (A.5) from 0 to \(T\wedge {\tau _{k}}\) and taking expectations, then we can obtain

$$ \mathbb{E}V \bigl(S(T\wedge {\tau _{k}}),I(T\wedge {\tau _{k}}) \bigr) \leq {V \bigl(S (0 ),I (0 ) \bigr)}+BT< \infty. $$

(A.6)

Set \(\Omega _{k}=\{\tau _{k}\leq {t}\}\) for \(k\geq {k_{1}}\) by (A.4), \(P (\Omega _{k} )\geq \epsilon \). Notice that, for every \(\omega \in \Omega _{k}\), there is at least one of \(S (\tau _{k},\omega )\), \(I (\tau _{k},\omega )\) that equals either *k* or \(\frac{1}{k}\). Hence \(V(S (\tau _{k},\omega )\), \(I (\tau _{k},\omega )\) is no less than

$$ k-1-\log {k} \quad\text{or}\quad \frac{1}{k}-1-\log {\frac{1}{k}}= \frac{1}{k}-1+\log {k}. $$

Consequently,

$$ V \bigl(S(\tau _{k},\omega ),I(\tau _{k},\omega ) \bigr)\geq {(k-1- \log {k})}\wedge \biggl(\frac{1}{k}-1+\log {k} \biggr), $$

(A.7)

where \(a\wedge {b}\) denotes the minimum of *a* and *b*. In view of (A.6) and (A.7), we have

$$\begin{aligned} V \bigl(S (0 ),I (0 ) \bigr)+BT&\geq { \mathbb{E}} \bigl[I_{\Omega _{k}}V \bigl(S(\tau _{k},\omega ),I(\tau _{k}, \omega ) \bigr) \bigr] \\ &\geq \epsilon \biggl[(k-1-\log {k})\wedge \biggl(\frac{1}{k}-1+\log {k} \biggr) \biggr], \end{aligned}$$

where \(I_{\Omega _{k}}\) is the indicator function of \(\Omega _{k}\). Let \(k\rightarrow \infty \) leads to the contradiction

$$ \infty >V \bigl(S (0 ),I (0 ) \bigr)+BT= \infty. $$

Therefore, we must have \(\tau _{\infty }=\infty \) a.s. □

### 1.3 Proof of Theorem 3.2

### Proof

Making use of It\(\hat{o}^{,}s\) formula for ln*I*, we have

$$\begin{aligned} d\ln I&= \biggl[ \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){S}-(p\mu +\gamma )- \frac{\sigma ^{2}S^{2}}{2}) \biggr]\,dt+\sigma S\,dB(t) \\ &\leq \biggl[\beta _{1} S-(p\mu +\gamma )-\frac{\sigma ^{2}S^{2}}{2} \biggr] \,dt+ \sigma S\,dB(t) \\ &= \biggl\{ - \biggl[\frac{\sigma S}{\sqrt{2}}- \frac{\sqrt{2}\beta _{1}}{2\sigma } \biggr]^{2}+ \frac{\beta ^{2}_{1}}{2\sigma ^{2}}-(p\mu +\gamma ) \biggr\} \,dt+ \sigma S \,dB(t) \\ &\leq \biggl\{ \frac{\beta _{1}^{2}}{2\sigma ^{2}}-(p\mu +\gamma ) \biggr\} \,dt+\sigma S\,dB(t). \end{aligned}$$

(A.8)

Integrating Eq. (A.8) from 0 to *t* and dividing by *t* on both sides, we have

$$ \frac{\ln I(t)-\ln I(0)}{t}\leq \frac{\beta _{1}^{2}}{2\sigma ^{2}}-(p \mu +\gamma )+ \frac{M(t)}{t}, $$

(A.9)

where \(M(t)=\int _{0}^{t}\sigma S \,dB(t)\) is a real-value continuous local martingale, since we have the quadratic variations, we can have

$$ \limsup_{t\rightarrow \infty }\langle {M,M\rangle }_{t} \leq \sigma ^{2}< \infty. $$

By the large number theorem for the martingale (see [41]), we can get

$$ \lim_{t\rightarrow \infty }\frac{M(t)}{t}=0,\quad \text{a.s.} $$

(A.10)

According to (A.9) and (A.10), we have

$$ \limsup_{t\rightarrow \infty }\frac{\ln I(t)}{t}\leq \frac{\beta _{1}^{2}}{2\sigma ^{2}}-(p\mu +\gamma ), \quad \text{a.s.} $$

(A.11)

That is to say, if \(\sigma ^{2}>\frac{\beta _{1}^{2}}{2(p\mu +\gamma )}\), we obtain

$$ \lim_{t\rightarrow \infty }I(t)=0, \quad \text{a.s.} $$

On the other hand, let \(x= S,x\in (0,1],f(x)=\beta _{1} x -\frac{\sigma ^{2}x^{2}}{2}=-( \frac{\sigma x}{\sqrt{2}}-\frac{\beta _{1}\sqrt{2}}{2\sigma })^{2}\), if \(\frac{\sigma }{\sqrt{2}}\leq \frac{\beta _{1}\sqrt{2}}{2\sigma }\), that is, \(\sigma ^{2}\leq \beta _{1}\), \(f(x)\) has the max value \(f(1)=\beta _{1}-\frac{\sigma ^{2}}{2}\), namely, \(S=1\) by Eq. (A.8), we have

$$\begin{aligned} d\ln I&= \biggl[ \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){S}-(p\mu +\gamma )- \frac{\sigma ^{2}S^{2}}{2}) \biggr]\,dt+\sigma S\,dB(t) \\ &\leq \biggl[\beta _{1} S-(p\mu +\gamma )-\frac{\sigma ^{2}S^{2}}{2} \biggr] \,dt+ \sigma S\,dB(t) \\ &= \biggl\{ - \biggl[\frac{\sigma S}{\sqrt{2}}- \frac{\sqrt{2}\beta _{1}}{2\sigma } \biggr]^{2}+ \frac{\beta ^{2}_{1}}{2\sigma ^{2}}-(p\mu +\gamma ) \biggr\} \,dt+ \sigma S \,dB(t) \\ &\leq \biggl\{ \beta _{1}-\frac{\sigma ^{2}}{2}-(p\mu +\gamma ) \biggr\} \,dt+\sigma S\,dB(t). \end{aligned}$$

(A.12)

Integrating Eq. (A.12) from 0 to *t* and dividing by *t* on both sides, we have

$$ \frac{\ln I(t)-\ln I(0)}{t}\leq \beta _{1}-\frac{\sigma ^{2}}{2}-(p \mu +\gamma )+\frac{M(t)}{t}, $$

(A.13)

then

$$ \limsup_{t\rightarrow \infty }\frac{\ln I(t)}{t}\leq \beta _{1}- \frac{\sigma ^{2}}{2}-(p\mu +\gamma ),\quad \text{a.s.} $$

(A.14)

That is to say, if \(\sigma ^{2}\leq \beta _{1}\) and \(\beta _{1}<\frac{\sigma ^{2}}{2}+(p\mu +\gamma )\), we have

$$ \limsup_{t\rightarrow \infty }\frac{\ln I(t)}{t}\leq \beta _{1}- \frac{\sigma ^{2}}{2}-(p\mu +\gamma )< 0, \quad \text{a.s.} $$

Hence, we can get

$$ \lim_{t\rightarrow \infty }I(t)=0,\quad \text{a.s.} $$

Furthermore, we have

$$ \lim_{t\rightarrow \infty } \bigl\langle I(t) \bigr\rangle =0, \quad\text{a.s.} $$

(A.15)

For the system (1.3), we have

$$ d(S+I)= \bigl\{ (1-\alpha )\mu -\mu S- \bigl[\mu (1-\alpha q)+\gamma \bigr]I \bigr\} \,dt. $$

(A.16)

Integrating Eq. (A.16) from 0 to *t* and dividing by *t* on both sides, we have

$$ \frac{S(t)-S(0)+I(t)-I(0)}{t}=(1-\alpha )\mu -\mu \bigl\langle S(t) \bigr\rangle - \bigl[\mu (1-\alpha q)+\gamma \bigr] \bigl\langle I(t) \bigr\rangle . $$

(A.17)

Then

$$ \bigl\langle S(t) \bigr\rangle =(1-\alpha )- \frac{[\mu (1-\alpha q)+\gamma ]}{\mu } \bigl\langle I(t) \bigr\rangle +\Psi (t), $$

(A.18)

where

$$ \Psi (t)=\frac{1}{\mu t} \bigl[S(0)+I(0)-S(t)-I(t) \bigr], $$

we obtain

$$ \lim_{t\rightarrow \infty }\Psi (t)=0. $$

(A.19)

Combining (A.15) and (A.19), we have

$$ \lim_{t\rightarrow \infty } \bigl\langle S(t) \bigr\rangle =1- \alpha,\quad \text{a.s.} $$

This completes the proof. □

### 1.4 Proof of Theorem 3.3

### Proof

Making use of Itô’s formula, we obtain

$$\begin{aligned} d\ln I&= \biggl[ \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){S}-(p\mu + \gamma )-\frac{\sigma ^{2}S^{2}}{2} \biggr]\,dt+\sigma S\,dB(t) \\ &\leq \biggl[\beta _{1} S-(p\mu +\gamma )-\frac{\sigma ^{2}S^{2}}{2} \biggr] \,dt+\sigma S\,dB(t). \end{aligned}$$

(A.20)

Integrating Eq. (A.20) from 0 to *t* and dividing by *t* on both sides, we have

$$\begin{aligned} \frac{\ln I(t)-\ln I(0)}{t}&\leq \beta _{1} \bigl\langle S(t) \bigr\rangle -(p \mu +\gamma )-\frac{\sigma ^{2}}{2} \bigl\langle S^{2}(t) \bigr\rangle + \frac{M(t)}{t} \\ &\leq \beta _{1} \bigl\langle S(t) \bigr\rangle -(p\mu +\gamma )- \frac{\sigma ^{2}}{2} \bigl\langle S(t) \bigr\rangle ^{2}+ \frac{M(t)}{t}. \end{aligned}$$

(A.21)

In view of (A.18), we have

$$ -\frac{\sigma ^{2}}{2} \bigl\langle S(t) \bigr\rangle ^{2}=- \frac{\sigma ^{2}}{2} \biggl[(1-\alpha )+\Psi (t)-\frac{\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle \biggr]^{2}. $$

So

$$\begin{aligned} &\frac{\ln I(t)-\ln I(0)}{t} \\ &\quad\leq \beta _{1} \bigl\langle S(t) \bigr\rangle -(p \mu +\gamma )-\frac{\sigma ^{2}}{2} \bigl\langle S(t) \bigr\rangle ^{2}+ \frac{M(t)}{t} \\ &\quad=\beta _{1} \biggl\{ (1-\alpha )-\frac{\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle +\Psi (t) \biggr\} -(p\mu +\gamma ) \\ &\qquad{}-\frac{\sigma ^{2}}{2} \biggl[(1-\alpha )+\Psi (t)- \frac{\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle \biggr]^{2}+ \frac{M(t)}{t} \\ &\quad=\beta _{1}(1-\alpha )-\frac{\beta _{1}\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle +\beta _{1}\Psi (t)-(p\mu +\gamma ) \\ &\qquad{}-\frac{\sigma ^{2}}{2}(1-\alpha )^{2}-\frac{\sigma ^{2}}{2}\Psi ^{2}(t)- \sigma ^{2}(1-\alpha )\Psi (t)+\frac{M(t)}{t} \\ &\qquad{}+ \frac{\sigma ^{2}[(1-\alpha )+\Psi (t)][\mu (1-\alpha q)+\gamma ]}{\mu } \bigl\langle I(t) \bigr\rangle - \frac{\sigma ^{2}[\mu (1-\alpha q)+\gamma ]^{2}}{2\mu ^{2}} \bigl\langle I(t) \bigr\rangle ^{2} \\ &\quad\leq \beta _{1}(1-\alpha )-(p\mu +\gamma )-\frac{\sigma ^{2}}{2}(1- \alpha )^{2}+\beta _{1}\Psi (t)+\frac{M(t)}{t} \\ &\qquad{}+ \frac{\sigma ^{2}[(1-\alpha )+\Psi (t)][\mu (1-\alpha q)+\gamma ]}{\mu } \bigl\langle I(t) \bigr\rangle -\frac{\beta _{1}\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle . \end{aligned}$$

(A.22)

Hence

$$\begin{aligned} \bigl\langle I(t) \bigr\rangle \leq {}&\frac{\mu [\beta _{1}(1-\alpha )-(p\mu +\gamma +\frac{\sigma ^{2}}{2}(1-\alpha )^{2})]}{[\mu (1-\alpha q)+\gamma ] (\beta _{1}-\sigma ^{2}[(1-\alpha )+\Psi (t)])} \\ &{}+\frac{\mu }{[\mu (1-\alpha q)+\gamma ](\beta _{1}-\sigma ^{2}[(1-\alpha )+\Psi (t)])} \biggl[\beta _{1}\Psi (t)+\frac{M(t)}{t} \biggr]. \end{aligned}$$

(A.23)

In view of (A.10) and (A.19), we have

$$ \limsup_{t\rightarrow \infty } \bigl\langle I(t) \bigr\rangle \leq \frac{\mu [\beta _{1}(1-\alpha )-(p\mu +\gamma +\frac{\sigma ^{2}}{2}(1-\alpha )^{2})]}{ [\mu (1-\alpha q)+\gamma ][\beta _{1}-\sigma ^{2}(1-\alpha )]}=I_{1},\quad \text{a.s.} $$

(A.24)

On the other hand,

$$\begin{aligned} d\ln I&= \biggl[ \biggl(\beta _{1}-\frac{\beta _{2}I}{\eta +I} \biggr){S}-(p\mu + \gamma )-\frac{\sigma ^{2}S^{2}}{2} \biggr]\,dt+\sigma S\,dB(t) \\ &\geq \biggl[(\beta _{1}-\beta _{2}) S-(p\mu +\gamma )- \frac{\sigma ^{2}S^{2}}{2} \biggr]\,dt+\sigma S\,dB(t). \end{aligned}$$

(A.25)

So,

$$\begin{aligned} &\frac{\ln I(t)-\ln I(0)}{t} \\ &\quad\geq(\beta _{1}-\beta _{2}) \bigl\langle S(t) \bigr\rangle -\frac{\sigma ^{2}}{2} \bigl\langle S^{2}(t) \bigr\rangle +\frac{M(t)}{t} \\ &\quad\geq (\beta _{1}-\beta _{2}) \biggl[(1-\alpha )- \frac{\mu (1-\alpha q)+\gamma }{\mu } \bigl\langle I(t) \bigr\rangle +\Psi (t) \biggr]- \frac{\sigma ^{2}}{2}+\frac{M(t)}{t}. \end{aligned}$$

(A.26)

Hence,

$$ \bigl\langle I(t) \bigr\rangle \geq \frac{\mu }{(\beta _{1}-\beta _{2})[\mu (1-\alpha q)+\gamma ]} \biggl[ ( \beta _{1}-\beta _{2}) \bigl[(1-\alpha )+\Psi (t) \bigr]- \frac{\sigma ^{2}}{2}+\frac{M(t)}{t} \biggr]. $$

(A.27)

In the light of (A.10) and (A.19), we have

$$ \liminf_{t\rightarrow \infty } \bigl\langle I(t) \bigr\rangle \geq \frac{\mu ((\beta _{1}-\beta _{2})(1-\alpha )-\sigma ^{2})}{2 (\beta _{1}-\beta _{2})[\mu (1-\alpha q)+\gamma ]}=I_{2}\quad \text{a.s.} $$

(A.28)

Thus from (A.24) and (A.28), we have

$$ I_{2}\leq \liminf_{t\rightarrow \infty } \bigl\langle I(t) \bigr\rangle \leq \limsup_{t\rightarrow \infty } \bigl\langle I(t) \bigr\rangle \leq I_{1}\quad \text{a.s.} $$

□