Appendix A: The proof of Lemma 2.1
Since the coefficients of model (5) is local Lipschitz continuous, there is a unique local solution \(y(t,q)\in {\mathbb {R}}^2_+\) on \(t\in [0, \tau _e)\) for any initial value \(\psi (\theta )\), where \(\tau _e\) denotes the explosion time. Next, we only need to show that \(\tau _e=+\infty \ a.s\).
Define a \(C^2\)-function V: \({\mathbb {R}}^2_+\rightarrow {\mathbb {R}}_+\) as follows:
$$\begin{aligned} V_1= & {} y_1(t,q)-1-\ln y_1(t,q)+c[y_2(t,q)-1-\ln y_2(t,q)]\nonumber \\&+\int _t^{t+\tau ^{1-q}_u\tau ^q_l}\frac{c}{4\epsilon _0}b^{1-q}_lb^q_uy^2_1(s-\tau ^{1-q}_u\tau ^q_l,q)ds, \end{aligned}$$
(A.1)
where
$$\begin{aligned} c=a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}/(b^{1-q}_lb^q_u)^2, \ \epsilon _0=0.5a^{1-q}_{2u}a^q_{2l}/b^{1-q}_lb^q_u. \end{aligned}$$
(A.2)
Using Itô’s formula, one has
$$\begin{aligned} dV_1= & {} [(y_1(t,q)-1)(r^{1-q}_{1l}r^q_{1u}-a^{1-q}_{1u}a^q_{1l}y_1(t,q))+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2]dt\\&+c[(y_2(t,q)-1)(r^{1-q}_{2l}r^q_{2u}-a^{1-q}_{2u}a^q_{2l}y_2(t,q)+b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q))\\&+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2]dt+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t)\\&+c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t)\\&+[\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t,q)-\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)]dt\\\le & {} \big \{-a^{1-q}_{1u}a^q_{1l}y^2_1(t,q)+(r^{1-q}_{1l}r^q_{1u}+a^{1-q}_{1u}a^q_{1l})y_1(t,q)+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\\&-ca^{1-q}_{2u}a^q_{2l}y^2_2(t,q)+c(r^{1-q}_{2l}r^q_{2u}+a^{1-q}_{2u}a^q_{2l})y_2(t,q)+\frac{c}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2\\&+cb^{1-q}_lb^q_u[\epsilon _0y^2_2(t,q)+\frac{1}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)]\\&+\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t,q)-\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)\big \}dt\\&+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t)+c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t)\\= & {} LV_1dt+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t) +c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t), \end{aligned}$$
where
$$\begin{aligned} LV_1= & {} (-a^{1-q}_{1u}a^q_{1l}+\frac{cb^{1-q}_lb^q_u}{4\epsilon _0})y^2_1(t,q)+(r^{1-q}_{1l}r^q_{1u}+a^{1-q}_{1u}a^q_{1l})y_1(t,q)+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\\&\quad +(-ca^{1-q}_{2u}a^q_{2l}+cb^{1-q}_lb^q_u\epsilon _0)y^2_2(t,q)+c(r^{1-q}_{2l}r^q_{2u}+a^{1-q}_{2u}a^q_{2l})y_2(t,q)+\frac{c}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2. \end{aligned}$$
Recalling (A.2) we can obtain
$$\begin{aligned}&-a^{1-q}_{1u}a^q_{1l}+\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}=-\frac{a^{1-q}_{1u}a^q_{1l}}{2},\\&-ca^{1-q}_{2u}a^q_{2l}+cb^{1-q}_lb^q_u\epsilon _0=-\frac{a^{1-q}_{1u}a^q_{1l}(a^{1-q}_{2u}a^q_{2l})^2}{2(b^{1-q}_lb^q_u)^2}. \end{aligned}$$
It is easy to prove that \(LV_1\) is bounded. The rest proof is standard and we omit the details.
Now, we are in the position to prove \(\limsup _{t\rightarrow +\infty }{\mathbb {E}}[y^p_i(t,q)]\le \mathcal K(p), i=1,2\). For any \(p>0\), we show, by Itô’s formula, that
$$\begin{aligned} \begin{array}{lll} d(e^ty^p_1(t,q))=L(e^ty^p_1(t,q))dt+pe^ty^p_1(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t), \end{array} \end{aligned}$$
(A.3)
where
$$\begin{aligned} L(e^ty^p_1(t,q))= & {} e^t\Bigg \{[1+pr^{1-q}_{1l}r^q_{1u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2]y^p_1(t,q)\nonumber \\&-\,pa^{1-q}_{1u}a^q_{1l}y^{p+1}_1(t,q)\Bigg \}. \end{aligned}$$
(A.4)
Then there exists a positive constant \({\mathcal {K}}_1(p)\) such that
$$\begin{aligned} \begin{array}{lll} {\mathcal {K}}_1(p)=\max \limits _{y_1\in \mathbb R^2_+}\left[ 1+pr^{1-q}_{1l}r^q_{1u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\right] y^p_1(t,q)-pa^{1-q}_{1u}a^q_{1l}y^{p+1}_1(t,q). \end{array} \end{aligned}$$
Hence, it follows from (A.3) that
$$\begin{aligned} \begin{array}{lll} d(e^ty^p_1(t,q))\le \mathcal K_1(p)e^tdt+pe^ty^p_1(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t). \end{array} \end{aligned}$$
(A.5)
Integrating both sides of (A.5) from 0 to t and then taking the expectations lead to
$$\begin{aligned} {\mathbb {E}}[e^ty^p_1(t,q)-y^p_1(0,q)]\le \mathcal K_1(p)(e^t-1). \end{aligned}$$
(A.6)
As a consequence, one has
$$\begin{aligned} {\mathbb {E}}[y^p_1(t,q)]\le \mathcal K_1(p)+e^{-t}[y^p_1(0,q)-{\mathcal {K}}_1(p)]. \end{aligned}$$
(A.7)
Similar to (A.3), applying Itô’s formula to \(e^ty^p_2(t,q)\) yields
$$\begin{aligned} \begin{array}{lll} d(e^ty^p_2(t,q))=L(e^ty^p_2(t,q))dt+pe^ty^p_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t). \end{array} \end{aligned}$$
For sufficiently large integer \(\lambda >0\), we compute
$$\begin{aligned} L(e^ty^p_2(t,q))= & {} e^t\left\{ \left[ 1+pr^{1-q}_{2l}r^q_{2u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2\right] y^p_2(t,q)\right. \nonumber \\&\left. -pa^{1-q}_{2u}a^q_{2l}y^{p+1}_2(t,q)+pb^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)y^p_2(t,q)\right\} \nonumber \\\le & {} e^t\left\{ [1+pr^{1-q}_{2l}r^q_{2u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2]y^p_2(t,q)\right. \nonumber \\&-pa^{1-q}_{2u}a^q_{2l}y^{p+1}_2(t,q)+p[\frac{p}{(p+1)}\lambda ^{-\frac{p+1}{p}}b^{1-q}_lb^q_uy^{p+1}_2(t,q)\nonumber \\&\left. \quad +\frac{\lambda ^{p+1}}{p+1}b^{1-q}_lb^q_uy^{p+1}_1(t-\tau ^{1-q}_u\tau ^q_l,q)]\right\} . \end{aligned}$$
(A.8)
Define
$$\begin{aligned} V_2(y(t,q))=e^{\tau ^{1-q}_u\tau ^q_l}\frac{p\lambda ^{p+1}}{p+1}b^{1-q}_lb^q_u\int _{t-\tau ^{1-q}_u\tau ^q_l}^te^sy^{p+1}_1(t,q)ds. \end{aligned}$$
(A.9)
Applying Itô’s formula to \(e^ty^p_2(t,q)+V_2(y(t,q))\), together with (A.8) and (A.9), we deduce
$$\begin{aligned} L(e^ty^p_2(t,q)+V_2(y(t,q)))\le & {} e^t\left\{ \left[ 1+pr^{1-q}_{2l}r^q_{2u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2\right] y^p_2(t,q)\right. \nonumber \\&-p[a^{1-q}_{2u}a^q_{2l}-\frac{p}{p+1}\lambda ^{-\frac{p+1}{p}}b^{1-q}_lb^q_u]y^{p+1}_2(t,q)\nonumber \\&\left. +e^{\tau ^{1-q}_u\tau ^q_l}\frac{p\lambda ^{p+1}}{p+1}b^{1-q}_lb^q_uy^{p+1}_1(t,q)\right\} . \end{aligned}$$
(A.10)
Assign \(V_3(y(t,q))\) with the form
$$\begin{aligned} V_3(y(t,q))=\hbar e^ty^p_1(t,q)+(e^ty^p_2(t,q)+V_2(y(t,q))), \hbar =e^{\tau ^{1-q}_u\tau ^q_l}\lambda ^{p+1}b^{1-q}_lb^q_u(a^{1-q}_{1u}a^q_{1l})^{-1}. \end{aligned}$$
By Itô’s formula, together with (A.4) and (A.10), there is a constant \({\mathcal {K}}_2(p)>0\) such that
$$\begin{aligned} LV_3(y(t,q))\le & {} e^t\big \{[1+pr^{1-q}_{2l}r^q_{2u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2]y^p_2(t,q)\nonumber \\&-p[a^{1-q}_{2u}a^q_{2l}-\frac{p}{p+1}\lambda ^{-\frac{p+1}{p}}b^{1-q}_lb^q_u]y^{p+1}_2(t,q)\nonumber \\&+\hbar [1+pr^{1-q}_{1l}r^q_{1u} +\frac{p(p-1)}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2]y^p_1(t,q)\nonumber \\&-e^{\tau ^{1-q}_u\tau ^q_l}\frac{p^2\lambda ^{p+1}}{p+1}b^{1-q}_lb^q_uy^{p+1}_1(t,q)\big \}\nonumber \\\le & {} e^t{\mathcal {K}}_2(p). \end{aligned}$$
(A.11)
By a simple calculation, we can see that
$$\begin{aligned} {\mathbb {E}}[e^ty^p_2(t,q)]\le {\mathbb {E}}[V_3(y(t,q))]\le V_3(y(0,q))+{\mathcal {K}}_2(p)(e^t-1). \end{aligned}$$
(A.12)
Furthermore
$$\begin{aligned} {\mathbb {E}}[y^p_2(y(t,q))]\le \mathcal K_2(p)+e^{-t}[V_3(y(0,q))-{\mathcal {K}}_2(p)]. \end{aligned}$$
(A.13)
Hence, according to (A.7) and (A.13), we have
$$\begin{aligned} \limsup _{t\rightarrow +\infty }{\mathbb {E}}[y^p_1(t,q)]\le \mathcal K_1(p), \ \limsup _{t\rightarrow +\infty }\mathbb E[y^p_2(t,q)]\le {\mathcal {K}}_2(p). \end{aligned}$$
Choosing \({\mathcal {K}}(p)=\max \{{\mathcal {K}}_1(p),\mathcal K_2(p)\}\), one obtains that
$$\begin{aligned} \limsup _{t\rightarrow +\infty }{\mathbb {E}}[y^p_i(t,q)]\le \mathcal K(p). \end{aligned}$$
(A.14)
This proof is complete.
Appendix B: The proof of Theorem 2.1
Using Itô’s formula to model (5), one gets that
$$\begin{aligned}&\begin{array}{lll} d\ln y_1(t,q)=[m_1(q)-a^{1-q}_{1u}a^q_{1l}y_1(t,q)]dt+\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t), \end{array} \end{aligned}$$
(B.1)
$$\begin{aligned}&\begin{array}{lll} d\ln y_2(t,q)=[m_2(q)-a^{1-q}_{2u}a^q_{2l}y_2(t,q)\\ +b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)]dt+\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t). \end{array} \end{aligned}$$
(B.2)
Integrating both sides of equations (B.1) and (B.2) from 0 to t leads to
$$\begin{aligned}&\begin{array}{lll} \ln y_1(t,q)-\ln y_1(0,q)=m_1(q)t-a^{1-q}_{1u}a^q_{1l}\int _0^ty_1(s,q)ds+\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t), \end{array} \end{aligned}$$
(B.3)
$$\begin{aligned}&\begin{array}{lll} \ln y_2(t,q)-\ln y_2(0,q)=m_2(q)t-a^{1-q}_{2u}a^q_{2l}\int _0^ty_2(s,q)ds\\ +b^{1-q}_lb^q_u\int _0^ty_1(s-\tau ^{1-q}_u\tau ^q_l,q)ds+\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(B.4)
By a further calculation, one has
$$\begin{aligned}&\begin{array}{lll} t^{-1}[\ln y_1(t,q)-\ln y_1(0,q)]=m_1(q)-a^{1-q}_{1u}a^q_{1l}\langle y_1(t,q)\rangle +t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t), \end{array} \end{aligned}$$
(B.5)
$$\begin{aligned}&\begin{array}{lll} t^{-1}[\ln y_2(t,q)-\ln y_2(0,q)]=m_2(q)+b^{1-q}_lb^q_u\langle y_1(t-\tau ^{1-q}_u\tau ^q_l,q)\rangle \\ -a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle +t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(B.6)
(1) When \(m_1(q)<0\), by (B.5) we have
$$\begin{aligned} \begin{array}{lll} t^{-1}\ln y_1(t,q)\le t^{-1}\ln y_1(0,q)+m_1(q)+t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t). \end{array} \end{aligned}$$
Note that \(\lim _{t\rightarrow +\infty }t^{-1}B_i(t)=0(i=1,2)\) and \(m_1(q)<0\), we can obtain
$$\begin{aligned} \limsup _{t\rightarrow +\infty }t^{-1}\ln y_1(t,q)\le m_1(q)<0. \end{aligned}$$
Furthermore,
$$\begin{aligned} \lim _{t\rightarrow +\infty }y_1(t,q)=0 \ a.s. \end{aligned}$$
(B.7)
Based on (B.7), (B.6) can be rewritten as
$$\begin{aligned} \begin{array}{lll} t^{-1}\ln y_2(t,q)\le t^{-1}\ln y_2(0,q)+m_2(q)+t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
Similarly, when \(m_2(q)<0\), one derives that \(\lim _{t\rightarrow +\infty }y_2(t,q)=0 \ a.s.\)
(2) When \(m_1(q)<0\), we can get (B.7) and hence for arbitrary \(\epsilon >0\), there is \(T>0\) such that for \(t\ge T \),
$$\begin{aligned} -\epsilon /2\le b^{1-q}_lb^q_u\langle y_1(t-\tau ^{1-q}_u\tau ^q_l,q)\rangle \le \epsilon /2, \ -\epsilon /2\le t^{-1}\ln y_2(0,q)\le \epsilon /2.\nonumber \\ \end{aligned}$$
(B.8)
It follows from (B.4) and (B.8) that
$$\begin{aligned} \begin{array}{lll} \ln y_2(t,q)\le [m_2(q)+\epsilon ]t-a^{1-q}_{2u}a^q_{2l}\int _0^ty_2(s,q)ds +\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t),\\ \ln y_2(t,q)\ge [m_2(q)-\epsilon ]t-a^{1-q}_{2u}a^q_{2l}\int _0^ty_2(s,q)ds +\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(B.9)
Since \(m_2(q)>0\), there is a \(\epsilon >0\) sufficiently small such that \(m_2(q)-\epsilon >0\). An application of Lemma 2 in [21] to (B.9) shows that
$$\begin{aligned}{}[m_2(q)-\epsilon ]/a^{1-q}_{2u}a^q_{2l}\le {\langle y_2(t,q)\rangle }_*\le {\langle y_2(t,q)\rangle }^*\le [m_2(q)+\epsilon ]/a^{1-q}_{2u}a^q_{2l}.\nonumber \\ \end{aligned}$$
(B.10)
It follows from the arbitrariness of \(\epsilon \) that \(\lim _{t\rightarrow +\infty }\langle y_2(t,q)\rangle =m_2(q)/a^{1-q}_{2u}a^q_{2l} \ a.s.\)
(3) Similar to (B.10), when \(m_1(q)>0\) we apply Lemma 2 in [21] to (B.3) and get
$$\begin{aligned} \lim _{t\rightarrow +\infty }\langle y_1(t,q)\rangle =m_1(q)/a^{1-q}_{1u}a^q_{1l} \ a.s. \end{aligned}$$
(B.11)
Substituting (B.11) into (B.5) and using \(\lim _{t\rightarrow +\infty }t^{-1}B_i(t)=0\), we see
$$\begin{aligned} \lim _{t\rightarrow +\infty }t^{-1}\ln y_1(t,q)=0 \ a.s. \end{aligned}$$
(B.12)
Also, we know that
$$\begin{aligned} \begin{array}{lll} \int _0^ty_1(s-\tau ^{1-q}_u\tau ^q_l,q)ds=\int _0^ty_1(s,q)ds-\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds- \int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ], \end{array} \end{aligned}$$
which together with (B.6) yields
$$\begin{aligned} \begin{array}{lll} t^{-1}[\ln y_2(t,q)-\ln y_2(0,q)] +t^{-1}b^{1-q}_lb^q_u\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds-\int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ]\\ \quad =m_2(q)-a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle +b^{1-q}_lb^q_u\langle y_1(t,q)\rangle +t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(B.13)
Then, computing \((\mathrm{B}.5)\times b^{1-q}_lb^q_u+(\mathrm{B}.13)\times a^{1-q}_{1u}a^q_{1l}\) leads to
$$\begin{aligned} \begin{array}{lll} t^{-1}{b^{1-q}_lb^q_u}[\ln y_1(t,q)-\ln y_1(0,q)]+t^{-1}{a^{1-q}_{1u}a^q_{1l}}[\ln y_2(t,q)-\ln y_2(0,q)]\\ +t^{-1}{a^{1-q}_{1u}a^q_{1l}b^{1-q}_lb^q_u}\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds-\int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ]\\ =\triangle _2-{\bar{\triangle }}_2-a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle \\ +t^{-1}{b^{1-q}_lb^q_u}\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t) +t^{-1}{a^{1-q}_{1u}a^q_{1l}}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(B.14)
Thus
$$\begin{aligned} \begin{array}{lll} t^{-1}a^{1-q}_{1u}a^q_{1l}\ln y_2(t,q) \le t^{-1}b^{1-q}_ub^q_l\ln y_1(0,q)+t^{-1}a^{1-q}_{1u}a^q_{1l}\ln y_2(0,q)+\triangle _2-{\bar{\triangle }}_2\\ +t^{-1}b^{1-q}_lb^q_u\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t) +t^{-1}a^{1-q}_{1u}a^q_{1l}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
Similar to (1), when \(\triangle _2-{\bar{\triangle }}_2<0\) we can obtain that \(\lim _{t\rightarrow +\infty }y_2(t,q)=0\) a.s.
(4) When \(m_1(q)>0\), we can get (B.11). Let us consider species \(y_2\). By (B.11) we have
$$\begin{aligned}&\lim _{t\rightarrow +\infty }\frac{1}{t}\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds\nonumber \\&\quad =\lim _{t\rightarrow +\infty }\frac{1}{t}\big (\int _0^ty_1(s,q)ds-\int _0^{t-\tau ^{1-q}_u\tau ^q_l}y_1(s,q)ds\big )=0. \end{aligned}$$
(B.15)
By (B.12) and (B.15), we obtain, for any \(\epsilon >0\), there is \(T>0\) such that for \(t\ge T\),
$$\begin{aligned} \begin{array}{lll} -\epsilon /3\le b^{1-q}_lb^q_ut^{-1}[\ln y_1(t,q)-\ln y_1(0,q)]\le \epsilon /3, \ -\epsilon /3\le a^{1-q}_{1u}a^q_{1l}t^{-1}\ln y_2(0,q)\le \epsilon /3,\\ -\epsilon /3\le a^{1-q}_{1u}a^q_{1l}b^{1-q}_lb^q_ut^{-1}\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds -\int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ]\le \epsilon /3, \end{array} \end{aligned}$$
which together with (B.14) leads to
$$\begin{aligned}&\begin{array}{lll} a^{1-q}_{1u}a^q_{1l}\ln y_2(t,q) \le [\triangle _2-{\bar{\triangle }}_2+\epsilon ]t-a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}\int _0^ty_2(s,q)ds\\ +b^{1-q}_lb^q_u\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t) +a^{1-q}_{1u}a^q_{1l}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t), \end{array}\\&\begin{array}{lll} a^{1-q}_{1u}a^q_{1l}\ln y_2(t,q) \ge [\triangle _2-{\bar{\triangle }}_2-\epsilon ]t-a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}\int _0^ty_2(s,q)ds\\ +b^{1-q}_lb^q_u\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t) +a^{1-q}_{1u}a^q_{1l}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
Since \(\triangle _2>{\bar{\triangle }}_2\), it follows from Lemma 2 in [21] that
$$\begin{aligned} \lim \limits _{t\rightarrow +\infty }\langle y_2(t,q)\rangle =(\triangle _2-{\bar{\triangle }}_2)/(a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}), a.s. \end{aligned}$$
(B.16)
Appendix C: The proof of Theorem 2.2
We shall divide the whole proof of Theorem 2.2 into three steps.
Step 1 We first prove \(\lim _{t\rightarrow +\infty }\mathbb E\big |y_i(t, q; \psi )-y_i(t, q; \phi )\big |=0(i=1,2)\), where \(y_i(t, q; \psi )\) and \(y_i(t, q; \phi )\) are two solutions of model (5). Define
$$\begin{aligned} V_4(t,q)= & {} \sum \limits _{i=1}^2|\ln y_i(t, q; \psi )-\ln y_i(t, q; \phi )|\\&+b^{1-q}_lb^q_u\int _{t-\tau ^{1-q}_u\tau ^q_l}^t|y_1(s, q; \psi )-y_1(s, q; \phi )|ds. \end{aligned}$$
Calculating the right differential of \(V_4(t,q)\), we have
$$\begin{aligned} d^+V_4(t,q)= & {} \mathrm{sgn}(y_1(t, q; \psi )-y_1(t, q; \phi ))\big [-a^{1-q}_{1u}a^q_{1l}(y_1(t, q; \psi )-y_1(t, q; \phi ))\big ]dt\\&+\mathrm{sgn}(y_2(t, q; \psi )-y_2(t, q; \phi ))\big [-a^{1-q}_{2u}a^q_{2l}(y_2(t, q; \psi )-y_2(t, q; \phi ))\\&+b^{1-q}_lb^q_u(y_1(t-\tau ^{1-q}_u\tau ^q_l, q; \psi )-y_1(t-\tau ^{1-q}_u\tau ^q_l, q; \phi ))\big ]dt\\&+b^{1-q}_lb^q_u|y_1(t, q; \psi )-y_1(t, q; \phi )|dt\\&-b^{1-q}_lb^q_u|y_1(t-\tau ^{1-q}_u\tau ^q_l, q; \psi )-y_1(t-\tau ^{1-q}_u\tau ^q_l, q; \phi )|dt\\\le & {} -\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}|y_i(t, q; \psi )-y_i(t, q; \phi )|dt +b^{1-q}_lb^q_u|y_1(t, q; \psi )-y_1(t, q; \phi )|dt\\= & {} -(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)|y_1(t, q; \psi )-y_1(t, q; \phi )|dt\\&-a^{1-q}_{2u}a^q_{2l}|y_2(t, q; \psi )-y_2(t,q; \phi )|dt. \end{aligned}$$
Integrating both sides of the above equation from 0 to t gives
$$\begin{aligned} \begin{array}{lll} {\mathbb {E}}[V_4(t,q)]\le V_4(0,q)-(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)\int _0^t{\mathbb {E}}|y_1(s, q; \psi )-y_1(s, q; \phi )|ds\\ -a^{1-q}_{2u}a^q_{2l}\int _0^t{\mathbb {E}}|y_2(s, q; \psi )-y_2(s, q;\phi )|ds. \end{array} \end{aligned}$$
It therefore follows from \(V_4(t,q)\ge 0\) that
$$\begin{aligned} \begin{array}{lll} (a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)\int _0^t{\mathbb {E}}|y_1(s, q; \psi )-y_1(s, q;\phi )|ds\\ +a^{1-q}_{2u}a^q_{2l}\int _0^t\mathbb E|y_2(s, q; \psi )-y_2(s, q; \phi )|ds\le V_4(0,q)<+\infty , \end{array} \end{aligned}$$
and hence \({\mathbb {E}}\big |y_i(t, q; \psi )-y_i(t, q; \phi )\big |\in L^1(0,+\infty ), \ i=1,2\). By (5), one has
$$\begin{aligned} d\mathbb E[y_1(t,q)]=y_1(0,q)+\int _0^t\big [r^{1-q}_{1l}r^q_{1u}\mathbb E[y_1(s,q)] -a^{1-q}_{1u}a^q_{1l}\mathbb E[y^2_1(s,q)]\big ]ds.\nonumber \\ \end{aligned}$$
(C.1)
In other words, \({\mathbb {E}}[y_1(t, q)]\) is differentiable. Together with (A.14), we have
$$\begin{aligned} \begin{array}{lll} \frac{d{\mathbb {E}}[y_1(t,q)]}{dt}=r^{1-q}_{1l}r^q_{1u}\mathbb E[y_1(t,q)]-a^{1-q}_{1u}a^q_{1l}{\mathbb {E}}[y^2_1(t,q)] \le r^{1-q}_{1l}r^q_{1u}{\mathbb {E}}[y_1(t,q)] \le r^{1-q}_{1l}r^q_{1u}{\mathcal {K}}(p). \end{array} \end{aligned}$$
As a consequence, \({\mathbb {E}}[y_1(t, q)]\) is uniformly continuous. Similarly, we can prove that \({\mathbb {E}}[y_2(t, q)]\) is also uniformly continuous. Thus, it follows from Barbalats lemma [35] that
$$\begin{aligned} \lim _{t\rightarrow +\infty }{\mathbb {E}}\big |y_i(t, q; \psi )-y_i(t, q; \phi )\big |=0, \ i=1,2. \end{aligned}$$
(C.2)
Step 2 We prove that there exists a unique probability measure \(\varpi \) such that for any initial value \(\psi (\theta )\), the transition probability \(u(t,q;\psi (\theta ),\cdot )\) converges weakly to \(\varpi \) as \(t\rightarrow 0\). We define that \(\mathcal U([-\tau ,0],{\mathbb {R}}^2_+)\) is the space of all probability measures on the \(C([-\tau ,0],{\mathbb {R}}^2_+)\). For any \(H_1, H_2\in {\mathcal {U}}([-\tau ,0],{\mathbb {R}}^2_+)\), assign
$$\begin{aligned} \begin{array}{lll} d_{{\mathbb {L}}}(H_1,H_2)=\sup _{v\in {{\mathbb {L}}}}\big |\int _{\mathbb R^2_+}v(y)H_1(dy)-\int _{{\mathbb {R}}^2_+}v(y)H_2(dy)\big |, \end{array} \end{aligned}$$
where \({{\mathbb {L}}}=\big \{v:C([-\tau ,0],{\mathbb {R}}^2_+)\rightarrow {\mathbb {R}}\Big ||v(y)-v({\bar{y}})|\le \Vert y-{\bar{y}}\Vert , |v(\cdot )|\le 1\big \}.\) Let \(u(t,q;\psi (\theta ),dy)\) be the transition probability of the process y(t, q). For any \(v\in {{\mathbb {L}}}\) and \(t, s > 0\), we compute
$$\begin{aligned}&d_{{\mathbb {L}}}(u(t+s,q;\psi (\theta ),\cdot ),u(t,q;\psi (\theta ),\cdot ))\\&\quad =\sup _{v\in {{\mathbb {L}}}}\big |\int _{{\mathbb {R}}^2_+}v(y(t+s,q;\psi (\theta )))u(t+s,q;\psi (\theta ),dy)\\&\qquad -\int _{{\mathbb {R}}^2_+}v(y(t,q;\psi (\theta )))u(t,q;\psi (\theta ),dy)\big |\\&\quad =\sup _{v\in {{\mathbb {L}}}}\big |{\mathbb {E}}[v(y(t+s,q;\psi (\theta )))]-{\mathbb {E}}[v(y(t,q;\psi (\theta )))]\big |\\&\quad =\sup _{v\in {{\mathbb {L}}}}|{\mathbb {E}}[{\mathbb {E}}[v(y(t+s,q;\psi (\theta ))|{\mathcal {F}}_s)]-{\mathbb {E}}[v(y(t,q;\psi (\theta )))]|\\&\quad =\sup _{v\in {{\mathbb {L}}}}\big |\int _{{\mathbb {R}}^2_+}{\mathbb {E}}[v(y(t,q;{\bar{\psi }}(\theta )))u(s,q;\psi (\theta ),d{{\bar{\psi }}(\theta )})]-{\mathbb {E}}[v(y(t,q;\psi (\theta )))]\big |\\&\quad =\sup _{v\in {{\mathbb {L}}}}\big |\int _{{\mathbb {R}}^2_+}{\mathbb {E}}[v(y(t,q;{\bar{\psi }}(\theta )))]u(s,q;\psi (\theta ),d{{\bar{\psi }}(\theta )})\\&\qquad -\int _{{\mathbb {R}}^2_+}{\mathbb {E}}[v(y(t,q;\psi (\theta )))]u(s,q;\psi (\theta ),d{{\bar{\psi }}(\theta )})\big |\\&\quad \le \sup _{v\in {{\mathbb {L}}}}\int _{\mathbb R^2_+}\big |{\mathbb {E}}[v(y(t,q;{\bar{\psi }}(\theta )))]-\mathbb E[v(y(t,q;\psi (\theta )))]\big |u(s,q;\psi (\theta ),d{{\bar{\psi }}(\theta )}). \end{aligned}$$
Applying Chebyshev’s inequality and (A.14) we obtain that the family of \(u(t,q;\psi (\theta ),\cdot )\) is tight. It then follows from (C.2) that there exists a \(T>0\) such that \(t\ge T\),
$$\begin{aligned} \sup _{v\in \mathbf{{\mathbb {L}}}}\big |\mathbb E[v(y(t,q;{\bar{\psi }}(\theta )))]-\mathbb E[v(y(t,q;\psi (\theta )))]\big |\le \epsilon . \end{aligned}$$
Namely, for arbitrary \(t\ge T\) and \(s>0\), \(\big |\mathbb E[v(y(t+s,q;\psi (\theta )))]-\mathbb E[v(y(t,q;\psi (\theta )))]\big |\le \epsilon .\) Therefore, \(\sup _{v\in \mathbf{{\mathbb {L}}}}\big |\mathbb E[v(y(t+s,q;\psi (\theta )))]-\mathbb E[v(y(t,q;\psi (\theta )))]\big |\le \epsilon .\) Consequently,
$$\begin{aligned} d_{\mathbb L}(u(t+s,q;\psi (\theta ),\cdot ),u(t,q;\psi (\theta ),\cdot ))\le \epsilon . \end{aligned}$$
So \(\big \{u(t,q;\psi (\theta ),\cdot ):t\ge 0\big \}\) is Cauchy in \({\mathcal {U}}([-\tau ,0],{\mathbb {R}}^2_+)\). Similarly, \(\big \{u(t,q;\phi (\theta ),\cdot )\big \}\) is Cauchy in \(\mathcal U([-\tau ,0],{\mathbb {R}}^2_+)\), where \(\phi (\theta )=(0.2,0.2)\). Hence, there exists a unique \(\varpi (q;\cdot )\in \mathcal U([-\tau ,0],{\mathbb {R}}^2_+)\) such that \(\lim _{t\rightarrow +\infty }d_{\mathbb L}(u(t,q;\phi (\theta ),\cdot ),\varpi (q;\cdot ))=0.\) As a consequence
$$\begin{aligned} \lim _{t\rightarrow +\infty }d_{\mathbb L}(u(t,q;\psi (\theta ),\cdot ),u(t,q;\phi (\theta ),\cdot ))=0. \end{aligned}$$
Thereby
$$\begin{aligned} \begin{array}{lll} \lim _{t\rightarrow +\infty }d_{{\mathbb {L}}}(u(t,q;\psi (\theta ),\cdot ),\varpi (q;\cdot ))\le \lim _{t\rightarrow +\infty }d_{{\mathbb {L}}}(u(t,q;\psi (\theta ),\cdot ),u(t,q;\phi (\theta ),\cdot ))\\ +\lim _{t\rightarrow +\infty }d_{\mathbb L}(u(t,q;\phi (\theta ),\cdot ),\varpi (q;\cdot ))=0. \end{array}\nonumber \\ \end{aligned}$$
(C.3)
Step 3 We prove \(\varpi \) is ergodic. By Corollary 3.4.3 in [34] and the uniqueness of \(\varpi \), one obtains that \(\varpi (q;\cdot )\) is strong mixing. It then follows from Theorem 3.2.6 in [34] that \(\varpi (q;\cdot )\) is ergodic. An application of (3.3.2) in [34] leads to
$$\begin{aligned} \lim _{t\rightarrow +\infty }\langle y_i(t,q)\rangle =\int _{\mathbb R^2_+}y_i\varpi (dy_1,dy_2,q),i=1,2. \end{aligned}$$
(C.4)
Combining Steps 1–3, one obtains the proof of Theorem 2.2.
Appendix D: The proof of Theorem 2.3
Define
$$\begin{aligned} V_5(y(t,q))=y_1(t,q)+y_2(t,q). \end{aligned}$$
(D.1)
One derives, by Itô’s formula to (D.1), that
$$\begin{aligned} dV_5(y(t,q))= & {} \left[ \sum \limits _{i=1}^2y_i(t,q)(r^{1-q}_{il}r^q_{iu}-a^{1-q}_{iu}a^q_{il}y_i(t,q))\right. \\&\left. +b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)y_2(t,q)\right] dt\\&+\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t)). \end{aligned}$$
Assign
$$\begin{aligned} V_6(y(t,q))=e^t\ln V_5(y(t,q)), \end{aligned}$$
(D.2)
by Itô’s formula
$$\begin{aligned} dV_6(y(t,q))= & {} e^t\Big \{\ln V_5(y(t,q))+\frac{1}{V_5(y(t,q))}\big [\sum \limits _{i=1}^2(r^{1-q}_{il}r^q_{iu}y_i(t,q)-a^{1-q}_{iu}a^q_{il}y^2_i(t,q))\\&+b^{1-q}_lb^q_uy_2(t,q)y_1(t-\tau ^{1-q}_u\tau ^q_l,q)\big ]\\&-\frac{1}{2V^2_5(y(t,q))}\big [\sum \limits _{j=1}^2(y^2_j(t,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&+2y_1(t,q)y_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}dt\\&+\frac{e^t}{V_5(y(t,q))}\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t))\\\le & {} e^t\Big \{\ln V_5(y(t,q)) +\frac{1}{V_5(y(t,q))}\sum \limits _{i=1}^2(r^{1-q}_{il}r^q_{iu}y_i(t,q)-a^{1-q}_{iu}a^q_{il}y^2_i(t,q))\\&+b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q) -\frac{1}{2V^2_5(y(t,q))}\big [\sum \limits _{j=1}^2(y^2_i(t,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&+2y_1(t,q)y_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}dt\\&+\frac{e^t}{V_5(y(t,q))}\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t)). \end{aligned}$$
We construct a non-negative function
$$\begin{aligned} U_1(y(t,q))=b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}\int _{t-\tau ^{1-q}_u\tau ^q_l}^te^sy_1(s,q)ds. \end{aligned}$$
(D.3)
It follows that
$$\begin{aligned} d(V_6(y(t,q))+U_1(y(t,q)))\le & {} e^t\Big \{\ln V_5(y(t,q))+b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(t,q)\\&+\frac{1}{V_5(y(t,q))}\sum \limits _{i=1}^2(r^{1-q}_{il}r^q_{iu}y_i(t,q)-a^{1-q}_{iu}a^q_{il}y^2_i(t,q))\\&-\frac{1}{2V^2_5(y(t,q))}\big [\sum \limits _{j=1}^2(y^2_i(t,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&+2y_1(t,q)y_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}dt\\&+\frac{e^t}{V_5(y(t,q))}\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t)). \end{aligned}$$
Integrating both sides of the above inequality shows that
$$\begin{aligned} \begin{array}{lll} V_6(y(t,q))+U_1(y(t,q))\le V_6(y(0,q))+U_1(y(0,q))+F_1(t,q)\\ +\int _0^te^s\Big \{\ln V_5(y(s,q))+b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(s,q)\\ +\frac{1}{V_5(y(s,q))}\sum \limits _{i=1}^2(r^{1-q}_{il}r^q_{iu}y_i(s,q)-a^{1-q}_{iu}a^q_{il}y^2_i(s,q))\\ -\frac{1}{2V^2_5(y(s,q))}\big [\sum \limits _{j=1}^2(y^2_j(s,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\ +2y_1(s,q)y_2(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}ds, \end{array} \end{aligned}$$
(D.4)
where \(F_1(t,q)=\int _0^t\frac{e^s}{V_5(y(s,q))}\sum \limits _{j=1}^2(y_j(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}dB_i(s)).\) \(F_1(t,q)\) is a local martingale and its quadratic variation is given by
$$\begin{aligned} \langle F_1,F_1\rangle _t= & {} \int _0^t\frac{e^{2s}}{V^2_5(y(s,q))}\big [\sum \limits _{j=1}^2y^2_j(s,q)(\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\nonumber \\&+2y_1(s,q)y_2(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]ds. \end{aligned}$$
(D.5)
By the exponential martingale inequality (see [35]), for any \(\epsilon \in (0, 1), \vartheta> 1, \rho >0\) and \(n>0\), we get
$$\begin{aligned} {\mathbb {P}}\Big \{\sup _{0\le t\le {\rho n}}\Big [F_1(t,q)-\frac{\epsilon }{2e^{\rho n}}\langle F_1,F_1\rangle _t\Big ]\ge \frac{\vartheta e^{\rho n}\ln n}{\epsilon }\Big \}\le \frac{1}{n^{\vartheta }}. \end{aligned}$$
By the Borel-Cantelli lemma (see [35]), there is an \(\Omega _0\subset \Omega \) with \({\mathbb {P}}(\Omega _0)=1\) such that for any \(\omega \in \Omega _0\), there exists an integer \(n_0=n_0(\omega )\) satisfies
$$\begin{aligned} F_1(t,q)\le \frac{\epsilon }{2e^{\rho n}}\langle F_1,F_1\rangle _t+\frac{\vartheta e^{\rho n}\ln n}{\epsilon }, \end{aligned}$$
(D.6)
for \(n\ge n_0(\omega ),0\le t\le n,i=1,2\). Thus (D.4) results in
$$\begin{aligned}&V_6(y(t,q)) \le V_6(y(0,q))+U_1(y(0,q))+\frac{\vartheta e^{\rho n}\ln n}{\epsilon } +\int _0^te^s\Big \{\ln V_5(y(s,q))\\&\qquad +b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(s,q)+\frac{1}{V_5(y(s,q))}\sum \limits _{i=1}^2(r^{1-q}_{il}r^q_{iu}y_i(t,q)-a^{1-q}_{iu}a^q_{il}y^2_i(s,q))\\&\qquad -\frac{1}{2V^2_5(y(s,q))}\big [\sum \limits _{j=1}^2(y^2_j(s,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&\qquad +2y_1(s,q)y_2(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\\&\qquad +\frac{e^{s-\rho n}\epsilon }{2V^2_5(y(s,q))}\big [\sum \limits _{j=1}^2(y^2_j(s,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&\qquad +2y_1(s,q)y_2(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}ds\\&\quad \le V_6(y(0,q))+U_1(y(0,q))+\frac{\vartheta e^{\rho n}\ln n}{\epsilon } +\int _0^te^s\Big \{\ln V_5(y(s,q))-0.5\Theta V_5(y(s,q))\\&\qquad +\Lambda +b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(s,q) +\frac{1}{2V^2_5(y(s,q))}\big [\sum \limits _{j=1}^2(y^2_j(s,q)\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&\qquad +2y_1(s,q)y_2(s,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ](1+\epsilon e^{s-\rho n})\Big \}ds\\&\quad \le V_6(y(0,q))+U_1(y(0,q))+\frac{\vartheta e^{\rho n}\ln n}{\epsilon }+\int _0^te^s\Big \{\ln V_5(y(s,q))+\Lambda \\&\qquad -[0.5\Theta -b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}]V_5(y(s,q)) +\frac{1+\epsilon e^{s-{\rho n}}}{2}\Gamma \big ]\Big \}ds, \end{aligned}$$
where
$$\begin{aligned} \begin{array}{lll} \Lambda =\max \big \{r^{1-q}_{1l}r^q_{1u}, r^{1-q}_{2l}r^q_{2u}\big \}, \ \Theta =\min \big \{a^{1-q}_{1u}a^q_{1l}, a^{1-q}_{2u}a^q_{2l}\big \},\\ \Gamma =\max \big \{\sum \limits _{i=1}^2((\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2+\prod _{j=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}), \sum \limits _{i=1}^2((\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2+\prod _{j=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu})\big \}. \end{array} \end{aligned}$$
Assume that the condition of Theorem 2.3 holds, then there is a constant \(\digamma >0\) independent of n such that
$$\begin{aligned} \begin{array}{lll} \ln V_5(y(s,q))-[0.5\min \limits _{i=1,2}\big \{a^{1-q}_{iu}a^q_{il}\big \}-b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}]V_5(y(s,q)) +\Lambda +0.5(1+\epsilon )\Gamma \le \digamma . \end{array} \end{aligned}$$
Therefore
$$\begin{aligned} e^t\ln V_5(y(t,q))=V_6(y(t,q))\le V_6(y(0,q))+U_1(y(0,q))+\vartheta e^{\rho n}\ln n/\epsilon +\digamma [e^t-1]. \end{aligned}$$
That is to say,
$$\begin{aligned} \ln V_5(y(t,q))\le e^{-t}( V_6(y(0,q))+U_1(y(0,q)))+\vartheta e^{-t}e^{\rho n}\ln n/\epsilon +\digamma [1-e^{-t}]. \end{aligned}$$
Hence, if \(\rho (n-1)\le t\le \rho n\) and \(n\ge n_i(\omega )\), then we can easy show that
$$\begin{aligned} \frac{\ln V_5(y(t,q)}{\ln t}\le \frac{e^{-t}( V_6(y(0,q))+U_1(y(0,q)))}{\ln t} +\frac{\epsilon ^{-1}\vartheta e^{-\rho (n-1)}e^{\rho n}\ln n}{\ln t}+\frac{\digamma [1-e^{-t}]}{\ln t}, \end{aligned}$$
which implies that \(\limsup _{t\rightarrow +\infty }\ln V_5(y(t,q)/\ln t\le \vartheta e^{\rho }/\epsilon \ a.s.\) Let \(\vartheta \rightarrow 1,\epsilon \rightarrow 1\) and \(\rho \rightarrow 0\), one yields that
$$\begin{aligned} \limsup _{t\rightarrow +\infty }\ln |y(t,q)|/\ln t\le \limsup _{t\rightarrow +\infty }\ln V_5(y(t,q)/\ln t\le 1. \end{aligned}$$
(D.7)
Appendix E: The proof of Theorem 2.4
Assign
$$\begin{aligned} V_7(y(t,q))=e^t\ln [y_1(t,q)y_2(t,q)]=e^t[\ln y_1(t,q)+\ln y_2(t,q)]. \end{aligned}$$
(E.1)
Together with (D.3), using Itô’s formula to \(V_7(y(t,q))+U_1(y(t,q))\), we give
$$\begin{aligned}&d(V_7(y(t,q))+U_1(y(t,q)))\nonumber \\&\quad =e^t[\ln y_1(t,q)+\ln y_2(t,q)+m_1(q)-a^{1-q}_{1u}a^q_{1l}y_1(t,q)+m_2(q)-a^{1-q}_{2u}a^q_{2l}y_2(t,q)\nonumber \\&\qquad +b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(t,q)]dt +e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t) +e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t). \end{aligned}$$
(E.2)
Then
$$\begin{aligned}&V_7(y(t,q))+U_1(y(t,q))\nonumber \\&\quad =V_7(y(0,q))+U_1(y(0,q))+\int _0^te^s\big [\ln y_1(s,q)+m_1(q)+m_2(q)-a^{1-q}_{1u}a^q_{1l}y_1(s,q)\nonumber \\&\qquad +\ln y_2(s,q)-a^{1-q}_{2u}a^q_{2l}y_2(s,q)+b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(s,q)\big ]ds+F_2(t,q)+F_3(t,q), \end{aligned}$$
(E.3)
where \(F_2(t,q)=\int _0^te^s\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(s), \ F_3(t,q)=\int _0^te^s\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(s).\)
Similar to the derivation of (D.6), we can get the following conclusions
$$\begin{aligned} F_i(t,q)\le \frac{\epsilon }{2e^{\rho n}}\langle F_i,F_i\rangle _t+\frac{\vartheta e^{\rho n}\ln n}{\epsilon }, \ i=2,3. \end{aligned}$$
Thus, for \(n\ge n_i(\omega )(i=1,2)\) and \(0\le t\le n\) when \(\omega \in \Omega _i\), let \(\Omega _0=\bigcap _{i=1}^2\Omega _i\), which leads to \({\mathbb {P}}(\Omega _0)=1\). Define \(n_0=\max _{1\le i\le 2}n_i(\omega )\), for all \(\omega \in \Omega _0\), it then follows from (E.3) that
$$\begin{aligned} V_7(y(t,q))\le V_7(y(0,q))+U_1(y(0,q))+2\vartheta e^{\rho n}\ln n/\epsilon +\int _0^te^s\sharp (y(t,q)) ds, \end{aligned}$$
where
$$\begin{aligned}&\sharp (y(t,q))=\left\{ \left[ \ln y_1(s,q)+r^{1-q}_{1l}r^q_{1u}-(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l})y_1(s,q)\right] \right. \\&\left. \quad +[ \ln y_2(s,q)+r^{1-q}_{2l}r^q_{2u}-a^{1-q}_{2u}a^q_{2l}y_2(s,q)] -\frac{1-\epsilon e^{s-\rho n}}{2}\sum \limits _{j=1}^2\left( \sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2\right) \right\} . \end{aligned}$$
If \(a^{1-q}_{1u}a^q_{1l}>b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}\) and sufficiently small \(\rho > 0\), then there exists a positive constant \(\mathfrak {R}\) such that \(\sharp (y(t,q))\le \mathfrak {R}\) for all \((y_1,y_2)\in \mathbb R^2_+\). Furthermore
$$\begin{aligned} V_7(y(t,q))\le V_7(y(0,q))+U_1(y(0,q))+2\vartheta e^{\rho n}\ln n/\epsilon +\mathfrak {R}(e^t-1). \end{aligned}$$
If \(\rho (n-1)\le t\le \rho n\) and \(n\ge n_i(\omega )\), then we obtain
$$\begin{aligned} \ln [y_1(t,q)y_2(t,q)]/\ln t\le [Ce^{-t}+2\vartheta e^{\rho }\ln n/\epsilon +\mathfrak {R}+\mathfrak {R}e^{-t}]/\ln t, \end{aligned}$$
where \(C=V_7(y(0,q))+U_1(y(0,q))\). The above inequality implies that
$$\begin{aligned} \limsup _{t\rightarrow +\infty }\ln [y_1(t,q)y_2(t,q)]/\ln t\le [2\vartheta e^{\rho }]/\epsilon \ a.s. \end{aligned}$$
Letting \(\vartheta \rightarrow 1,\epsilon \rightarrow 1\) and \(\rho \rightarrow 0\) yields the desired assertion.
Appendix F: Proof of Theorem 2.5
Let
$$\begin{aligned} V_8(y(t,q))=\ln [y_1(t,q)y_2(t,q)]=\ln y_1(t,q)+\ln y_2(t,q). \end{aligned}$$
(F.1)
Recalling (B.1) and (B.2), one can obtain
$$\begin{aligned} dV_8(y(t,q))= & {} [\sum \limits _{i=1}^2m_i(q)-\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}y_i(s,q)+b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)]ds\\&+\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}dB_i). \end{aligned}$$
Similar to (D.3), a non-negative functional is constructed as follows
$$\begin{aligned} U_2(y(t,q))=b^{1-q}_lb^q_u\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds. \end{aligned}$$
(F.2)
Together with (F.2), we apply Itô’s formula to \(V_8(y(t,q))+U_2(y(t,q))\) and integrate on both sides to get
$$\begin{aligned} V_8(y(t,q))+U_2(y(t,q))= & {} V_8(y(0,q))+U_2(y(0,q))+F_4(t,q)+\int _0^t[\sum \limits _{i=1}^2m_i(q)\nonumber \\&-(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)y_1(s,q)-a^{1-q}_{2u}a^q_{2l}y_2(s,q)]ds, \end{aligned}$$
(F.3)
where \(F_4(t,q)=\int _0^t\sum \nolimits _{j=1}^2(\sum \nolimits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}dB_i)\). Its quadratic form is given by
$$\begin{aligned} \begin{array}{lll} \langle F_4,F_4\rangle _t =\int _0^t\left[ \sum \limits _{i=1}^2(\sum \limits _{j=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2+\prod _{j=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu})\right] ds. \end{array} \end{aligned}$$
Let \(\epsilon \in (0, 1)\) and \(\vartheta >1\), by the exponential martingale inequality [35], we can show that for every integer \(n\ge 1\),
$$\begin{aligned} {\mathbb {P}}\Big \{\sup _{0\le t\le n}\Big [F_4(t,q)-\frac{\epsilon }{2}\langle F_4,F_4\rangle _t\Big ]\ge \frac{\vartheta \ln n}{\epsilon }\Big \}\le \frac{1}{n^{\vartheta }}. \end{aligned}$$
Since the series \(\sum \nolimits _{n=1}^{\infty }\frac{1}{n^{\vartheta }}\) converges, we have, by the Borel-Cantelli lemma [35], that there exists an \(\Omega _0\subset \Omega \) with \(\mathbb P(\Omega _0)=1\). For any \(\omega \in \Omega _0\), \(n\ge n_0(\omega )\) and \(0\le t\le n\), there is an integer \(n_0=n_0(\omega )\) such that
$$\begin{aligned} F_4(t,q)\le \frac{\epsilon }{2}\langle F_4,F_4\rangle _t+\frac{\vartheta \ln n}{\epsilon }. \end{aligned}$$
(F.4)
Substituting (F.4) into (F.3) one can see that
$$\begin{aligned} \begin{array}{lll} V_8(y(t,q))+U_2(y(t,q))\le V_8(y(0,q))+U_2(y(0,q))+\frac{\vartheta \ln n}{\epsilon }+\int _0^t[\sum \limits _{i=1}^2m_i(q)\\ \quad -(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)y_1(s,q)-a^{1-q}_{2u}a^q_{2l}y_2(s,q)+\epsilon \Gamma ]ds. \end{array} \end{aligned}$$
In other words,
$$\begin{aligned} \begin{array}{lll} V_8(y(t,q))\le V_8(y(0,q))+U_2(y(0,q))+\frac{\vartheta \ln n}{\epsilon }+\int _0^t\natural (y(t,q))ds, \end{array} \end{aligned}$$
where \(\natural (y(t,q))=\{\sum \nolimits _{i=1}^2m_i(q)-(a^{1-q}_{1u}a^q_{1l}-b^{1-q}_lb^q_u)y_1(s,q)-a^{1-q}_{2u}a^q_{2l}y_2(s,q)+\epsilon \Gamma \}.\)
If \(a^{1-q}_{1u}a^q_{1l}>b^{1-q}_lb^q_u\), then there is a constant \(\aleph >0\) such that \(\natural (y(t,q))\le \aleph \) for all \((y_1,y_2)\in {\mathbb {R}}^2_+\). Thus, for \(n-1\le t\le n\) and \(n\ge n_0(\omega )\), we have
$$\begin{aligned} \limsup \limits _{t\rightarrow +\infty }\frac{V_8(y(t,q))}{t}\le \limsup \limits _{n\rightarrow +\infty }\left[ \frac{V_8(y(0,q))+U_2(y(0,q))}{n-1}+\frac{\vartheta \ln n}{\epsilon (n-1)}+\aleph \right] =\aleph \ \ a.s. \end{aligned}$$
Appendix G: The proof of Lemma 3.1
Since the proof of Lemma 3.1 is similar to that of Lemma 2.1, we only give the formulas which are different from those in “Appendix A”. Let us consider the same function as (A.1) and obtain
$$\begin{aligned} dV_1= & {} \left[ (y_1(t,q)-1)(r^{1-q}_{1l}r^q_{1u}-a^{1-q}_{1u}a^q_{1l}y_1(t,q))+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\right] dt\\&+c\left[ (y_2(t,q)-1)(-R^{1-q}_{2l}R^q_{2u}-a^{1-q}_{2u}a^q_{2l}y_2(t,q)+b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q))\right. \\&\left. +\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2\right] dt+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t)\\&+c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t) +\left[ \frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t,q)-\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)\right] dt\\\le & {} \big \{-a^{1-q}_{1u}a^q_{1l}y^2_1(t,q)+(r^{1-q}_{1l}r^q_{1u}+a^{1-q}_{1u}a^q_{1l})y_1(t,q)+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\\&-ca^{1-q}_{2u}a^q_{2l}y^2_2(t,q)+ca^{1-q}_{2u}a^q_{2l}y_2(t,q)+cR^{1-q}_{2l}R^q_{2u}+\frac{c}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2\\&+cb^{1-q}_lb^q_u \left[ \epsilon _0y^2_2(t,q) +\frac{1}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)\right] \\&+\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t,q)-\frac{cb^{1-q}_lb^q_u}{4\epsilon _0}y^2_1(t-\tau ^{1-q}_u\tau ^q_l,q)\big \}dt\\&+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t)+c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t)\\= & {} L\bar{V_1}dt+(y_1(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t) +c(y_2(t,q)-1)\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t), \end{aligned}$$
where
$$\begin{aligned} L\bar{V_1}= & {} (-a^{1-q}_{1u}a^q_{1l}+\frac{cb^{1-q}_lb^q_u}{4\epsilon _0})y^2_1(t,q)+(r^{1-q}_{1l}r^q_{1u}+a^{1-q}_{1u}a^q_{1l})y_1(t,q)+\frac{1}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{1il}\sigma ^q_{1iu})^2\\&+(-ca^{1-q}_{2u}a^q_{2l}+cb^{1-q}_lb^q_u\epsilon _0)y^2_2(t,q)+ca^{1-q}_{2u}a^q_{2l}y_2(t,q)+cR^{1-q}_{2l}R^q_{2u}+\frac{c}{2}\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2. \end{aligned}$$
Clearly, \(L\bar{V_1}\) is bounded. The rest details are omitted.
Appendix H: The proof of Theorem 3.1
This proof is similar to that of Theorem 2.1. Here we only show the different parts.
Assign \(m_3(q)=R^{1-q}_{2u}R^q_{2l}+\frac{1}{2}\left[ \sum \nolimits _{j=1}^2(\sigma ^{1-q}_{2jl}\sigma ^q_{2ju})^2\right] .\) Similar to (B.6), we have
$$\begin{aligned} t^{-1}[\ln y_2(t,q)-\ln y_2(0,q)]= & {} -m_3(q)+b^{1-q}_lb^q_ut^{-1}\int _0^ty_1(s-\tau ^{1-q}_u\tau ^q_l,q)ds\nonumber \\&-a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle +t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{aligned}$$
(H.1)
(1) When \(m_1(q)<0\), we can get (B.7). Together with (B.7), (H.1) can be rewritten as
$$\begin{aligned} \begin{array}{lll} t^{-1}\ln y_2(t,q)\le t^{-1}\ln y_2(0,q)-m_3(q)+t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
Similar to (1) in “Appendix B”, we obtain that \(\lim _{t\rightarrow +\infty }y_2(t,q)=0 \ a.s.\)
(2) Similar to (3) in “Appendix B”, we obtain an equation which is parallel to (B.13)
$$\begin{aligned} \begin{array}{lll} t^{-1}[\ln y_2(t,q)-\ln y_2(0,q)] +t^{-1}b^{1-q}_lb^q_u\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds-\int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ]\\ \quad =-m_3(q)-a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle +b^{1-q}_lb^q_u\langle y_1(t,q)\rangle +t^{-1}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(H.2)
Then, computing \(\mathrm{(B.5)}\times b^{1-q}_lb^q_u+\mathrm{(H.2)}\times a^{1-q}_{1u}a^q_{1l}\) yields
$$\begin{aligned} \begin{array}{lll} t^{-1}{b^{1-q}_lb^q_u}[\ln y_1(t,q)-\ln y_1(0,q)]+t^{-1}{a^{1-q}_{1u}a^q_{1l}}[\ln y_2(t,q)-\ln y_2(0,q)]\\ \qquad +t^{-1}{a^{1-q}_{1u}a^q_{1l}b^{1-q}_lb^q_u}\big [\int _{t-\tau ^{1-q}_u\tau ^q_l}^ty_1(s,q)ds-\int _{-\tau ^{1-q}_u\tau ^q_l}^0y_1(s,q)ds\big ]\\ \quad =\triangle _3-{\bar{\triangle }}_2-a^{1-q}_{1u}a^q_{1l}a^{1-q}_{2u}a^q_{2l}\langle y_2(t,q)\rangle \\ \qquad +t^{-1}{b^{1-q}_lb^q_u}\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}B_i(t) +t^{-1}{a^{1-q}_{1u}a^q_{1l}}\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}B_i(t). \end{array} \end{aligned}$$
(H.3)
The rest proof is similar to that of Theorem 2.1 and we omit it.
Appendix I: The proof of Theorem 3.2
The proof of this conclusion is the same as that of Theorem 2.2, and we omit it.
Appendix J: The proof of Theorem 3.3
Recalling (D.1) and (D.2) and applying Itô’s formula to \(V_6(y(t,q))\), one has
$$\begin{aligned}&dV_6(y(t,q))\\&\quad =e^t\left\{ \ln V_5(y(t,q))+\frac{1}{V_5(y(t,q))}\big [r^{1-q}_{1l}r^q_{1u}y_1(t,q)-R^{1-q}_{2u}R^q_{2l}y_2(t,q)-\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}y^2_i(t,q)\right. \\&\qquad +b^{1-q}_lb^q_uy_2(t,q)y_1(t-\tau ^{1-q}_u\tau ^q_l,q))\big ] -\frac{1}{2V^2_5(y(t,q))}\left[ \sum \limits _{j=1}^2y^2_j(t,q)(\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\right. \\&\left. \left. \qquad +2y_1(t,q)y_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\right] \right\} dt\\&\qquad +\frac{e^t}{V_5(y(t,q))}\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t))\\&\quad \le e^t\Big \{\ln V_5(y(t,q))+\frac{1}{V_5(y(t,q))}\big [r^{1-q}_{1l}r^q_{1u}y_1(t,q)-\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}y^2_i(t,q)\\&\qquad +b^{1-q}_lb^q_uy_2(t,q)y_1(t-\tau ^{1-q}_u\tau ^q_l,q))\big ] -\frac{1}{2V^2_5(y(t,q))}\big [\sum \limits _{j=1}^2y^2_j(t,q)(\sum \limits _{i=1}^2(\sigma ^{1-q}_{jil}\sigma ^q_{jiu})^2)\\&\qquad +2y_1(t,q)y_2(t,q)\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}\sigma ^{1-q}_{2il}\sigma ^q_{2iu}\big ]\Big \}dt\\&\qquad +\frac{e^t}{V_5(y(t,q))}\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}y_j(t,q)dB_i(t)). \end{aligned}$$
The remaining proof is similar to (D.3)–(D.7), which is omitted.
Appendix K: The proof of Theorem 3.4
Recalling (E.1) and (D.3) and applying Itô’s formula to \(V_7(y(t,q))+U_1(y(t,q))\),
$$\begin{aligned}&d(V_7(y(t,q))+U_1(y(t,q)))\\&\quad =e^t\left[ \ln y_1(t,q)+\ln y_2(t,q)+m_1(q)-a^{1-q}_{1u}a^q_{1l}y_1(t,q)-m_3(q)-a^{1-q}_{2u}a^q_{2l}y_2(t,q)\right. \\&\qquad \left. +b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(t,q)\right] dt+e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t) +e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t)\\&\quad \le e^t\big [\ln y_1(t,q)+\ln y_2(t,q)+m_1(q)-a^{1-q}_{1u}a^q_{1l}y_1(t,q)\\&\qquad -0.5\sum \limits _{i=1}^2(\sigma ^{1-q}_{2il}\sigma ^q_{2iu})^2-a^{1-q}_{2u}a^q_{2l}y_2(t,q)+b^{1-q}_lb^q_ue^{\tau ^{1-q}_u\tau ^q_l}y_1(t,q)\big ]dt\\&\qquad +e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{1il}\sigma ^q_{1iu}dB_i(t) +e^t\sum \limits _{i=1}^2\sigma ^{1-q}_{2il}\sigma ^q_{2iu}dB_i(t). \end{aligned}$$
The rest proof is similar to that of “Appendix E”, and the desired assertion can be obtained.
Appendix L: The proof of Theorem 3.5
Recalling (F.1) and making use of Itô’s formula to \(V_8(y(t,q))\), one can obtain
$$\begin{aligned}&dV_8(y(t,q))=[m_1(q)-m_3(q)-\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}y_i(s,q)\\&\qquad +b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)]ds+\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}dB_i)\\&\quad \le [m_1(q)-\sum \limits _{i=1}^2a^{1-q}_{iu}a^q_{il}y_i(s,q)\\&\qquad +b^{1-q}_lb^q_uy_1(t-\tau ^{1-q}_u\tau ^q_l,q)]ds+\sum \limits _{j=1}^2(\sum \limits _{i=1}^2\sigma ^{1-q}_{jil}\sigma ^q_{jiu}dB_i).\\ \end{aligned}$$
Similar to “Appendix F”, we can complete the rest proof.
